Miles Mathis' Charge Field
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The Cause of Gravity - the next major chapter

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Vexman
Russ T
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Jared Magneson
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Post by Nevyn Mon Apr 01, 2019 3:25 am

Thanks Jared, good to know it is making sense. It shows how much information an equation can hide from you, that's for sure. I don't fault Vexman for looking at it the way he has been. When you can't see how something is working, you look for bits that you do understand and try to work from there. He is right about the units, too. They are a problem. Why mass times density? Another good question. I'm afraid I don't have any good answers right now. Although, I might try some anyway.

The units come out to kg^2/km^3. I have previously asked if kg^2 makes any sense and no-one seems to have a good answer, myself included. However, instead of just throwing everything out because some units don't look right, I'm going to accept that they are right and see if I can explain them.

So far, we have thought that a density is kg/km^3. A quantity distributed over a volume. So why don't we accept that, because it does make sense, but instead of thinking about it as a kg, we think about it as a quantity. We remove the units for now and focus on the concept of density and what it wants.

Looking at it like that means that we have a quantity which has the units kg^2. We don't have 2 quantities multiplied together, we have a single quantity with those units. Now we have to explain what that means. We still have to explain what a kg^2 is, but we are doing it from a slightly different position. It is a subtle difference, but it is there.

To do that, we go back to our original problem and look at what it is we are trying to find. We want a value that represents the amount of charge distributed over a unit volume. So the equation is telling us that charge can be thought of as kg^2. The reason it can look like that is because we are using one mass to emit another mass. We are looking at the secondary mass through the eyes of the primary mass. It is because we are using the bodies mass as input that we get kg^2 for the charge. The body is primary because that is what the input values represent. The charge is a derived value from that, so we are seeing it from the perspective of the body, not the charge.

In this way, we can probably ignore the kg^2 and just treat it as kg. By doing that, we are just switching perspectives from the body to the charge. Once the charge density is created, we don't care about the body anymore. It can't change that density. Only the drop-off equation can change it by letting some charge escape while keeping the unit volume the same size.

This is getting into philosophical territory. I'm practically inventing math at this point. But that is how math is invented or discovered. Reality presents a problem. We find concepts to explain it. We use or create math to represent the concepts. Heuristic math allows us to get numbers but not reasoning. But if those numbers match reality, then we have to take the math seriously and accept what it is telling us. The math might not be in the right form to tell us what it is really doing. We might not understand the concepts that it is expressing. So we must figure out how to extract reasoning from it.

As long as we stay logical and sensible, we can let the math teach us a few things.

Love that video, Jared. Great work. Can't wait to see it with the proper masses, etc.
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Post by Vexman Mon Apr 01, 2019 9:54 am

There are more than enough words in your posts that should've helped me understand your argument. In a way, I do understand it, math-wise. We disagree about the applied method. As well, I think it is wrong to assume that absolute values are not allowed to be used in C=M*D equation. Only by inserting the absolute values into a mathematical construct, we are able to check the result as correct or wrong. The math has to represent the reality.

In reality, the Earth's magnetopause is at the distance of roughly 11,5x Earth radii from the surface. That means cca 11,5 * 6378km or 73347 km. You will tell me that it confirms your thesis and method being used, but I think you didn't interpret it correctly in terms of relative radius units. You will then try to explain to me that C=M*D is not allowed to be used in the way I understand it, so we would be going in circles about this.

A short explanation of my point, by using your suggested method:

s = distance between both bodies
l = distance from body 2
Therefore:
F1/((s-l)/R1)^4 = F2/(l/R2)^4


What is in my opinion not correct here, is relativizing the radius expressed within 1/R^4 against the radius of two individual bodies; R1 vs R2 accordingly. Why? Because it won't give you the correct ratio of values in question. The ratio of radii of two bodies in focus is set as R1:R2= 695799 km : 6738 km= 103,26. The spherical zone of influence around each body is measured as an absolute value, just like a radius is the absolute measurement of distance. So in above equation, this part is completely ignored if you want to compare in relative terms. What we need is one single definition of radius in 1/R^4.

In my opinion, the correct way to make a relative comparison is by letting  the radius of 1/R^4 represent a fixed value, but in terms of either Sun's or Earth's radius. Of course, only by doing so, you can say that the length of R represents the same absolute value of such sphere in the case of both bodies. In the end, the result has to tell you the same answer, regardless if it's expressed in relative or absolute units. What do I mean, specifically? Well, if you let Sun's zone of influence to inflate to, say, 214 Sun's radii from Sun, and we have a fixed distance between Sun-Eart at roughly 149,5x10^6km, than it means we have only left 600.000km of distance for the Earth's zone of influence. It doesn't matter if you say 600.000 km is 94,07 Earth radii or 0,8623 Sun radii, in any interpretation we are looking at the absolute value of 600.000 km. Just like we are looking at either 214 Sun radii or 148900986 km or 22099 Earth radii. When comparing in relative terms, you need to arrive at the same value. But comparing in relative terms means you need to choose one relative value, and express all distances in that chosen term until you arrive at the result. Only that would be a correct relative comparison.

In order to check the theory, we need to test it against the reality. I want to believe that it is possible to do so only by inserting the absolute values into a mathematical construction of reality. If that is not possible or the values don't match, then either the concept is wrong or the method is not correctly applied. In my opinion your method doesn't treat distances (and radii) on both sides of equation in the same way.

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Post by LongtimeAirman Mon Apr 01, 2019 3:28 pm

.
Jared, Congratulations on creating your own charged particle system!  king

The discussion is doing fine. The charge density function of radius and distance is well neigh defined.

Given the opportunity I needed to update the charge field diagram to correct the earth’s magnetosphere radius.

The Cause of Gravity - the next major chapter - Page 6 Sem12o13
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Post by Nevyn Mon Apr 01, 2019 10:03 pm

Vexman wrote:As well, I think it is wrong to assume that absolute values are not allowed to be used in C=M*D equation. Only by inserting the absolute values into a mathematical construct, we are able to check the result as correct or wrong.

That is not what I said and certainly not what I meant. There is no question that we are using absolute values as inputs. I have never disallowed that. I said that C is not an absolute value, not M or D. I said, or at least implied, that your requirement that if you put in absolutes, then you get out absolutes, is wrong. Just take any ratio and it disproves your point. Absolute values in, unit-less value out. But that is moot now, because my last post was an attempt at making C absolute. I won't blame you for not accepting that, it is a bit of conjecture. It is just an idea that may be valid or it may not be. No-one has a better idea, though.

Vexman wrote:The math has to represent the reality.

Yes, it does, and yours doesn't. You say that we need to use one radius and one only, but you have no justification for that. Only your desire for it to work that way. I have ripped this equation apart, put it back together and explained every step of the way. I have explained exactly what needs to happen and why it needs to happen. I have shown the tricky points and why they work the way they do. The only part I struggle with is explaining the C=MD, but we can actually ignore that because that is not where your problem is. Your problem is in the drop-off equation which I deconstructed in full detail. If you read my last 5-10 posts about this, you will see my understanding growing. You will see me trying to explain something at the start and then slowly gaining more traction on the problem and seeing how it all fits together. Each post gets a little bit deeper and gains more focus on the problem.

Vexman wrote:In reality, the Earth's magnetopause is at the distance of roughly 11,5x Earth radii from the surface. That means cca 11,5 * 6378km or 73347 km. You will tell me that it confirms your thesis and method being used, but I think you didn't interpret it correctly in terms of relative radius units.

So your position is that I have the right answer, but I have the wrong method? Even though the method works for multiple bodies? You can fluke one body, but you can't fluke two. Since the radius, mass and density are all different for each planet, to get the right answer for multiple bodies with the wrong method is impossible.

Vexman wrote:You will then try to explain to me that C=M*D is not allowed to be used in the way I understand it, so we would be going in circles about this.

Your problem isn't even with C=MD. So I'm not going to tell you anything about that. Your problem is with how to treat the density drop-off and the distances used in it. For some reason, you think that the earth has a say in the sun's charge density. You haven't explained how it does or why it does. You just keep asserting it.

You say we must use one radius. You don't say which one, so I assume we could use either. Does that make any sense? Not to me. If you choose the earth's radius, then you are using the earth to compute numbers about the sun. Explain how the earth can effect the sun. But then when we calculate the earth's charge density, we don't need the sun to explain that. Explain why one body needs the other, but the other doesn't. When I've done one thing to one side but not the other, I have explained exactly why that is needed and how it relates physically.

Vexman wrote:A short explanation of my point, by using your suggested method:

s = distance between both bodies
l = distance from body 2
Therefore:
F1/((s-l)/R1)^4 = F2/(l/R2)^4


What is in my opinion not correct here, is relativizing the radius expressed within 1/R^4 against the radius of two individual bodies; R1 vs R2 accordingly. Why? Because it won't give you the correct ratio of values in question.

You think I'm doing that just to get the right answer? Did you actually read any of my posts? Did you stop and think about them? I explicitly explained why I was doing that. I explained the physics and the math. I don't think you are understanding the difference between a relationship and an equation. A relationship is inherently relative. It doesn't express any absolute information, it takes out all of the specific variables and just gives you the major players that are common. When we say that gravity has an inverse square relationship, we are ignoring all of the mass, we are ignoring G, and just showing the drop-off relationship. The masses don't have anything to do with that drop-off, but they do tell you how much is dropping off. The law applies to any mass, so the masses are not part of the law. The same applies to the density drop-off. Any density will drop-off, but how quickly it drops off is totally reliant on the size of that body. It relies on the curvature of its surface. So please tell me how the earth changes the sun's curvature, because that is what you are doing in your math.

A unit sphere will drop-off at a rate set by 1/d^2 (let's ignore the Relativity part for now). The sun is not a unit sphere. The earth is not a unit sphere. The sun/earth ratio is not a unit sphere. So we have to make them a unit sphere in order to use the relationship. The relationship wants the unit sphere, not me, not Miles. The math demands it because it was a unit sphere that was used to derive that relationship. But that relationship is not being applied to the sun-earth combination. It is being applied to each body individually. We are not calculating a single result using both bodies. We are calculating each body individually, and then comparing those values. There is no connection between the 2 sides, except for the distance between them and the equality point that must be somewhere on that distance. I have handled that equality point differently on each side of the equation and I have explained why that is required. Each charge density at a distance calculation is operating from the perspective of the body that is emitting the charge. We don't calculate the sun's charge density from the earth's perspective. We don't calculate it from some obscure combination of them. We calculate it from that perspective because that is what the equations require. The equations are written from that perspective and we have to respect that to use them correctly.

This is what I mean when I say that you are treating all of this mathematically and ignoring the physics. You don't understand the equations you are using and what they require. I'm not trying to insult you. I am a bit incredulous at the moment, and this post is probably coming across a bit too strongly. I apologize for that, but I find it difficult to understand your position and how you have failed to understand my last few posts. I thought that I had it laid out pretty well yesterday.

Vexman wrote:The ratio of radii of two bodies in focus is set as R1:R2= 695799 km : 6738 km= 103,26. The spherical zone of influence around each body is measured as an absolute value, just like a radius is the absolute measurement of distance. So in above equation, this part is completely ignored if you want to compare in relative terms. What we need is one single definition of radius in 1/R^4.


I don't want to compare in relative terms. Given my assumption about kg^2 being an emission from the perspective of that which is emitting, I have gotten back to absolute values. Those numbers represent the actual charge density of each body at the same position between them. I admit that the density equation may need a bit more added to it, but whatever is added must be added to both sides and be equal. So the ratio of the sun's charge density to the earth's charge density remains the same, but the actual numbers change. For example, suppose we wanted to include the 19:1 charge to matter emission ratio into this equation. We would do that by changing C=MD into C=19MD. But 19A/19B = A/B, so they have not changed relatively, only in absolute values. There is no room for any other kind of change. We can't add in another variable that is different for each body, because that will change the ratio.

Vexman wrote:In my opinion, the correct way to make a relative comparison is by letting  the radius of 1/R^4 represent a fixed value, but in terms of either Sun's or Earth's radius. Of course, only by doing so, you can say that the length of R represents the same absolute value of such sphere in the case of both bodies. In the end, the result has to tell you the same answer, regardless if it's expressed in relative or absolute units. What do I mean, specifically? Well, if you let Sun's zone of influence to inflate to, say, 214 Sun's radii from Sun, and we have a fixed distance between Sun-Eart at roughly 149,5x10^6km, than it means we have only left 600.000km of distance for the Earth's zone of influence. It doesn't matter if you say 600.000 km is 94,07 Earth radii or 0,8623 Sun radii, in any interpretation we are looking at the absolute value of 600.000 km. Just like we are looking at either 214 Sun radii or 148900986 km or 22099 Earth radii. When comparing in relative terms, you need to arrive at the same value. But comparing in relative terms means you need to choose one relative value, and express all distances in that chosen term until you arrive at the result. Only that would be a correct relative comparison.

I repeat, I am not performing a relative comparison. So I have no requirement to use a common radius. C=MD/(l/R)^4 is being used on each body, completely separately from the other. Only the result of that is used in the comparison. One side refers to the sun and the sun alone. The other side refers to the earth and the earth alone. Anything else would be assuming that the sun effects the earth and the earth effects the sun. If that were the case, then we would need to bring in all of the other planets too. They would also be effecting the sun and the earth, so how could we leave them out?

Vexman wrote:In order to check the theory, we need to test it against the reality. I want to believe that it is possible to do so only by inserting the absolute values into a mathematical construction of reality. If that is not possible or the values don't match, then either the concept is wrong or the method is not correctly applied. In my opinion your method doesn't treat distances (and radii) on both sides of equation in the same way.

It does treat them in the same way, it just doesn't treat them with some idea of common ground. That is because the equations being used can't work that way. It treats them with respect to the body creating the density that it is using. C=MD/(l/R)^4 refers to one body and one body only. That single body supplies the M, D and R. We don't see any values from the other body in there because it has nothing to do with the other body. The l variable is shared between both sides because it represents the point of equality, or more specifically, it represents the point between the bodies that we are testing in the current iteration. The equation absolutely needs a distance that is relative to the body that is emitting. It can't use the same value on both sides because the 1/d^4 relationship wants a distance from the body and the point is not the same distance from each body. The relationship is written from the perspective of the body. It represents the charge as it moves away from the body. You have to make sure that you understand the variables that you are using, or you will mess it up. You can't just look at an equation and know what the variables mean. You can't see a distance variable and assume that you can just throw in any distance. You have to know what the equations want, because they will give you the wrong answers if you don't.

So I am going to tell you that the equations can't work the way you want them to, because the equations can't work the way you want them to and I've explained exactly why that is. Understand the tools you are using before you make demands of them. Understand the problem before you judge the math. Understand how the math fits the problem.

At the end of the day, I have an algorithm that works. It does calculate the correct answer, as tested against reality. You have nothing. You have some idea that you think should apply, but you have no equations to implement it. You have no explanation for it. You have no answer. You imply that it is wrong to accept my algorithm, in spite of the fact that it gets the right answer. You keep saying that we need to test against reality, but when we do, you ignore it. Then you accuse me of doing things just to get the right answer. Firstly, I have explained why I am doing everything that I am doing. Secondly, why would we want to follow a path to the wrong answer? Thirdly, the one place where I do that (C=MD), I have plainly stated that that is what I am doing and supplied the reasons for using M and D. I might think that C=MD looks a bit weird or too simple, but on the other hand, it works. Why would I run away from that?

I apologize again for the heat in my words. To be honest, I can't remove them right now. I am flabbergasted by most of your reply, and annoyed about some of it. That doesn't mean that I don't want to discuss this anymore. Although I feel that I have explained it enough and it is up to you to understand what I have said. Study the problem and study the equations. Read what I have said about it, again and again, until you can see why I have said it. If you still disagree, then explain what you disagree with and why that is so. Maybe I haven't explained it in a way that makes sense for you. I don't know. I can't know that. All I can do is explain it in my own way and hope that it conveys the information I want it to. If you want more detail about a certain part, then I will see if I can supply it.

I feel that I have attempted to understand your arguments about various things. I even made attempts to follow your path to see if it helped me understand your position better. I even thought that I had found a fault in Miles work, and published that. I didn't hide it. It didn't ignore it. I came back here and stated it plainly for all to see. I have been open and honest. I have supplied detail after detail. In the early posts, I got some things wrong, or didn't explain them enough, but by the last ones, I think it is stated quite clearly. You're talking generically. I'm talking specifically. You say that a relative comparison must have a relative base (a general statement). I say that it isn't a relative comparison and has no requirement for a relative base and show you what it does want and why it wants that (a specific statement about a specific problem). Can you see how you are talking about a general problem while I am talking about a very specific one? Have you entertained the idea that you might be wrong? I accept that I can be wrong, and my posts show that. I doubt myself all the time. I think that is healthy in this type of work. It keeps me humble. In this case, though, I think I have shown that my understanding of this problem is deep. I can, and have, explained every single step. I can, and have, explained why those steps are needed. I'm not throwing around high level concepts and hoping they stick to the wall. I'm giving you logical and precise explanations that fit the problem and give the correct answer.
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Post by Nevyn Mon Apr 01, 2019 11:47 pm

Vexman wrote:In my opinion, the correct way to make a relative comparison is by letting  the radius of 1/R^4 represent a fixed value, but in terms of either Sun's or Earth's radius. Of course, only by doing so, you can say that the length of R represents the same absolute value of such sphere in the case of both bodies.

We don't want a fixed value (I'm not even sure I understand what you mean by that in this context). We don't want to use the same value on both sides. We can't use the same value on both sides. The distance in use here (R in your statement) represents the distance from the body to the point of measurement. How can that possibly remain the same distance from both bodies? The point of measurement is what we have to let slide between the bodies. It is the movable entity in this problem because we are using it to find the point of equality. We want to find the point between the bodies where both sides are equal.

Basically, what we are doing in this problem is placing a ball half way between the sun and earth. We then watch which way it goes. The stronger field will push it away from the source of that field. It will move towards the earth. So we pick up our ball and place it 1/4 distance from the earth. We watch which way it goes. It will move towards the earth. We pick up our ball and place it 1/8 distance from the earth, etc, etc. When the ball doesn't move, we have found the point of equality. We may jump too far towards the earth, so it will move back towards the sun, but it will find the point of equilibrium.

When we are at 1/8th the distance between the earth and sun, from the earth, we can't also be 1/8th the distance from the sun. We must be 7/8ths from the sun. Any distance you take away from one side, you have to put onto the other. It is a zero-sum game.

I think your problem is more about using the radius to scale the distance, but I thought this might be worth mentioning. I've already explained why we scale by the radius, which is just using it as a scaling term so that the 1/d^4 relationship can be used. In any case, it might help you to see what I am not understanding about your statements.
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Post by Vexman Tue Apr 02, 2019 11:35 am

I did not explain it properly in my previous attempt. I feel responsible for your frustration because of that. Let me try again, I think I have much clearer thoughts about it this time.

"That is because the equations being used can't work that way. It treats them with respect to the body creating the density that it is using. C=MD/(l/R)^4 refers to one body and one body only. That single body supplies the M, D and R. We don't see any values from the other body in there because it has nothing to do with the other body."


The Cause of Gravity - the next major chapter - Page 6 C

In the diagram, we have "r" representing body's radius, and we have "R" representing body's zone of influence radius. Both radii have one end at the center of their sphere, which is in this case body's center.

That is the crucial part where we have different opinion.

The single body supplies M and D, while R is the radius of body's zone of influence we're expanding M*D into.

If you express this R with one body's own radius (r), you have set the metering of R in terms of that body's radius. And by doing so, you have to use the same metering on the other side of the equation (when comparing two bodies and their zones of influence). That's what I meant by saying we need to choose one body's radius and stick to it on both sides of your equation. Maybe you don't see it, but using two diferent radius values is defining two different meters of / for distance, which actually represents the radius of body's zone of influence.

Looking at C/R^4 , the "R" in 1/R^4 is not about the curvature of emitting body, it's about the volume of the spherical zone of influence that we're expanding body's charge density into and its curvature. But we're not interested in the curvature at this point, as much as we're not interested in surface areas. We need to find out how much of body's charge density do we have left within a larger sphere, its zone of influence. This zone of influence has its own radius, being spherical  in shape since it is created by the sphere emitting charge. Since charge field, when emitted from the spherical surface, is diminished by the rule 1/R^4, we're now focused at the new, inflated sphere (of influence) where we want to know how much of initial charge density is present within new volume unit. That is why this radius has no relation to the emitting sphere, except in terms that we are applying M and D of that particular body as the input data.

And I think that's where Miles had made a mistake. He chose Sun's radius to express R, but then interpreted it as Earth radius when talking about the results he made on the other side of the equation. And after your first iteration, which matches my results acquired by inserting absolute numbers into equations, you changed the method to match Miles' results again.

I do understand "l-s" and "l" used in a fraction, it's a ingenious algorithmic addition to help calculate the balance point by exhaustion. It has nothing to do with what we're discussing, but thanks for explaining that again for me.

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Post by Vexman Tue Apr 02, 2019 3:59 pm

I should say something on a personal note as well and I believe I should have said it earlier. I was steaming each time when reading your above post, but you can't tell from my reply that was actually so. I can take your criticism without any issues as long as you refrain from direct insults, Nevyn. I don't blame you for using heated words as I'm familiar with a feeling when you can't get your point through, even though you think you have made it crystal clear. So maybe we'll simply have to agree that we disagree here or at some other future issues. It may benefit you and the ongoing discussion here, if you took that into consideration. There are too many points in your previous post that I wanted to reply to. Some of them make sense, while others don't. But I don't want to dissect your every sentence only for the sake of the argument. I don't need that in my life. We have different opinions that are coming out of different perspective or viewpoints, as simple as that. What is essential is that I'm aware of your capabilities of comprehension, so you won't hear me saying how you don't understand problems at hand. I may or will only say that we have differing opinions, like possibly in the case of 1/R^4 and its proper use and/or interpretation.  

As much as you were flabbergasted with my post, so was I with your reply. What actually bothers me is how your are coming across, as if you were a superior instance in both physics and math. Maybe you really are, but nobody is infallible. Maybe you don't even care what I think of you. But still, I want you to know that I didn't come here for a battle of egos, my friend. Taking part in this discussion was inspiring so far with an exception of one of your posts, filled with mostly frustration. I already apologized for causing it, even though I shouldn't really have. In reality, I regret English isn't the language of my thoughts so I could express myself as good as you.

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Post by Jared Magneson Tue Apr 02, 2019 7:51 pm

Thank you both for taking the time to learn how to communicate with each other. As I've said before, I value both of you highly and for similar reasons, if on different topics, so it's really cool to see you join us here, Vexman, and meet the team and dive into these problems. I know you pretty well and know your intentions are only for the best here, and hopefully the others will come to know this about you as well.

Personally, I consider Nevyn and Miles the only "authorities" on physics I would almost always defer to, but not because they are authorities at all - they simply have the best track record for the correct (or most correct) answers and solutions, thus far in my experience. That doesn't mean you can't also provide answers and solutions too, Vexman, so please know I'm not taking sides here but trying to follow all the logic AND the math, the second part not being my strongest suit.

But that doesn't mean any of us are infallible, and I think Nevyn made it pretty clear he doesn't feel that he is. Nor do I think Miles feels like HE is. This topic is heavy in every way so I expected some heat, indeed, expected Miles to exile me outright at worst over this one but he has not, and I just heard from him today about my 3-body charge video so all is well. Hell, LongtimeAirman and I just had our first "fight" here a few pages back as well, but we moved on in good spirits and I hope you two can also.

Let's keep going and see what else we can glean from these discussions. Let's try to keep it cool too, but of course how we deal with things AFTER we lose our cool can be just as important as before. We're all looking for the same thing here - answers!

And given all our discussions elsewhere on this forum, I think we're a pretty good team to find them. Smile

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Post by Nevyn Tue Apr 02, 2019 9:51 pm

I didn't like the way that post was coming across either. I didn't like it as I was writing it, but couldn't express it in any better way at the time. I let my frustration get the better of me. I usually control that much better, but this time I was on a bit of a high from the previous day. I was quite proud of what I had written. How I had deconstructed the whole equation and explained it in detail. I thought I had nailed it and you would see what I was talking about. But you didn't, and I came crashing down. Pride before the fall, as they say. That's all on me. You stood up and were the better man. I thank you for that. The last thing I want to come across as is superior. However, when I have confidence in my own understanding, and I can see every little detail, it is hard to not come across that way. I can usually re-word it to sound better, but this time I just couldn't find the words and couldn't bring myself to delete them. Once I calmed down a bit, I got worried that I might push you away and I certainly don't want that. You haven't been here long, but I already see you as a valuable member of the team.

So, on to the problem...

The first thing I want to point out is that C=MD/(l/R)^4 is an equation to calculate the charge density of a body at a distance. The key points are of a body and at a distance. That equation does not calculate the charge density with respect to any other body. We can use that in isolation. We can imagine that the sun is the only thing in existence and we can still use that equation. If we make the changes that you want, then we can't do that. Where does the other radius come from? Why would we use some other radius in this situation? What could we make it relative to?

The distance, l, is specified in absolute units. It will be km, for example. We are not feeding it a relative distance (except that it is positionally relative, but that is not what you are talking about). Inside of the equation, it is then making it relative to the radius of the body in order to satisfy the 1/d^4 relationship. It is making the sun a unit sphere since that is what that relationship requires. 1/d^4 only works on a unit sphere. It doesn't work for all spheres of any radius. Look into how that relationship is derived to prove that.

Let's look at what it is trying to represent. The numerator is creating the charge density. The denominator is reducing that density based on the distance from the center of the body. The density can only be reduced when it is outside of the body. At the surface of the body, it is at its strongest. Not inside of it, but only outside. That actually limits the values that we can use for the l variable: l must be >= R. The equation breaks down if you try to use an l that is less than R. It just doesn't work inside of the body, it actually starts making the density greater, the farther in you go. That doesn't make any sense, so we can't use it inside. Which fits our problem perfectly. We are trying to represent the charge density from the surface and moving outwards. If we change R though, we don't get that. It no longer represents the surface of the body when we set l to the radius of the body.

Now let's look at the units. The drop-off part of the equation needs to reduce the quantity of charge, without reducing the volume. The numerator has a density already (ignoring issues with kg^2), so it can't change the units of that density because we still want a density. We absolutely require a density at any location from the surface, we just want to reduce the amount of it. My version does that perfectly. Since I am dividing l by R, the units of those values are lost. However, if we use a relative distance, which is unit-less, then we have a left over unit. We now don't have a density per unit volume, kg/km^3, we have kg/km^3/km to give kg/km^2. You have converted the volume into an area. I don't think that was your intention, but that is what it is doing regardless.

In my version, we set l=R and still get the same charge density as we calculated for the surface to begin with. At R, there is no drop-off, so that is working as we need it to. The drop-off part equates to 1 at that distance, so it doesn't change the density. That won't happen if we change R though. So by changing R, you are creating some space between the surface of the body and where it starts to drop-off. It also depends on how you change R. If we use the earth to make it relative, then we get a different location than if we used jupiter, for example. That doesn't make any sense. The point where the sun's charge starts to drop-off is not dependent on any other body. It is only dependent on the radius of the sun.

I think the problem arises when we try to put that equation into our comparison. Think about the equation in isolation, because that is how it is defined and how it works. We are then using it to calculate values, but we are still using it in isolation. We are never trying to express one field in terms of the other. We are just giving it different values for l and seeing what it gives us. You can build a table from that data with a column for the sun and another for the earth, and every other planet if you want to. Each column is unrelated to the others. They only relate to the body used to calculate them. Each column is an expression of that bodies charge field at the distances that we calculate for. Remembering that those distances are actually relative to the position of that body.

That table then becomes the data for our comparison. So when we get to comparing values, we have left the equation behind. We don't need it anymore. It has performed its job and given us what we want from it. Now, all we have to do is compare the correct values with each other. We pick some point between the sun and earth, say. We find the distance of that point from the sun and retrieve the value for the sun at that relative-to-the-sun distance. We then find the distance from the earth and retrieve its value at that relative-to-the-earth distance. Now we have 2 separate charge densities at the same point between those bodies, so we can compare them to determine which one is the larger field. Note that I am still talking about relative position, not relative size.

It is the measurement point that needs to be common to both sides. We always want that to be with respect to the body we are calculating a charge density for in the equation, but the only requirement that the comparison has is that it is the same point in space. We do need common ground, which is what I think you are trying to get at, but it is the location of the measurement that needs to be common, not anything to do with the calculation of the charge density.

So you are right to want some common ground, but you are looking in the wrong place to implement it. It all gets confusing when we put it all together into one big problem, as we are doing here. You have to keep an eye on what the equations are doing, but also when they are doing it, and what they are doing it to. In this case, we are never calculating a charge density for 2 bodies at the same time. We are never using the charge density equation on 2 bodies at the same time. We only ever use it for one body, so the other body has no say in that calculation. Which makes sense. There is no way that the earth can effect the sun's charge density at a distance. The sun emits its charge with no care for anything else. That charge will move away from the sun with no care for anything else. As long as there is space for it to move into, that is. Of course it can collide with bodies, etc, but in this case, we assume a clear space between the 2 bodies.

Some may also be thinking that when the fields meet at the point of equality, that they annihilate each other. They don't. Both fields keep going on past that point. One side of that point is now defined as the sun's charge field, and the other side is defined as the earth's charge field because each is the strongest field at those locations. I should say that the ambient field is defined by that. On the earth side, the ambient field is defined as the earth's field, but on the sun side, the ambient field is the sun's field.

Which brings us full circle. My original argument was that the earth does not define the ambient field at the moon. However, I am not so sure that that matters anymore. I mentioned a while ago that it looks like Miles is talking about objects on the surface, or close to it, in the new gravity papers, and orbits are more about net force at the location of the orbiter. Hopefully he writes a new paper about that soon.
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Post by Nevyn Tue Apr 02, 2019 10:58 pm

Your image above implies that we are looking at the full output of the body, and R is expressing a radius. But R is not really a radius, it is a distance. You can use it as a radius, in the way that you have, but that is not how this algorithm is using it. If you added field lines to that image, just like Airman has been, then this algorithm is only using the charge between 2 of those field lines. The distance between those field lines is what bounds our unit volume for the density. Since those field lines are radial, they will diverge as they move away from the surface. However, we don't want our volume to expand, we want to keep the same volume and let the density drop. So we are really creating 2 new lines that start at the same point where the radial lines touch the surface, but they remain parallel to each other as they move away from the body. That keeps the volume the same, but lets the density drop. We do the same for both bodies, and use the same distance between those parallel lines for both of them. That gives us a comparable volume and lets us compare the densities alone. R is not expressing a zone of influence, it is representing the distance that the charge has traveled since leaving its source. Which is then used to determine how much charge has escaped from our unit volume.

I am referring to your R in the diagram here, not my R in the charge density equation.


Last edited by Nevyn on Wed Apr 03, 2019 7:54 am; edited 1 time in total (Reason for editing : Added a note about what R refers to)
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Post by Nevyn Wed Apr 03, 2019 1:28 am

This is very rough. Certainly not up to the standard set around here recently. It should help show what I am talking about though.

The Cause of Gravity - the next major chapter - Page 6 Cd-0110

We have a body in yellow. The red lines represent the field lines. They are where the charge density created at the surface will move into. They diverge from each other because of the radial emission. Each red line should be at a right angle to the tangent at the point where it crosses the surface. If we imagine a volume starting at the surface and ending a little bit away from it, and bounded by those red lines, that is our unit volume of charge.

So far we have only used the MD part of our equation to create the charge density at the surface.

If we let that charge move away from the body, it will remain within those red lines. Those red lines represent the same charge photons all along them. Obviously, that means that the volume must be growing. There is more distance between the red lines, the farther we move out, so the volume is increasing. Before anyone asks, no, you can't change it in the other dimension.

The green lines contain the same volume as the red lines at the surface. If you imagine another line half way between the green lines that goes all the way back to the center of the body, that line is orthogonal to the tangent at the surface. The green lines are parallel with that line. Since the distance between the green lines does not change the farther we move away from the body, its volume is not changing either. This is how we keep the unit volume steady. However, we can see that charge is escaping. The space between the top red line and the top green line represents the lost charge. So does the bottom red and green lines. That area is what our drop-off equation is removing from our numbers.

Now we do the same for the other body, making sure to use the same distance between the green lines as we did with the first body. That gives us equal volumes to compare. That is another bit of common ground in this problem.
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Post by Jared Magneson Wed Apr 03, 2019 2:28 am

Might this help distinguish the variables?

The Cause of Gravity - the next major chapter - Page 6 I5zCxZT

I don't know if it's important in this situation, but SOME measure of charge outside the selected surface emission area would be crossing or joining the direct, perpendicular charge. There's definitely some spherical dropoff involved, but I don't think it's negligible at all. For the same reasons we see these bodies as discs and not just points or points with halos. But I don't mean to digress from your efforts, Nevyn. You're making very solid sense to me and by all means, keep going.

If it would help I could plug in all the proper known numbers in a big diagram, but zoom it in at the the Earth/Moon level and use similar gradients to help us express it. Sort of a merging of LongtimeAirman's gradient lines and your theory? Then we'd have a pretty clean visual representation of the concepts, at least.

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Post by Nevyn Wed Apr 03, 2019 7:50 am

Yeah, that's much better. It even shows the density drop-off. Our unit volume is a small rectangle super-imposed over the green area. As we increase the distance l, that rectangle moves to the right. The amount of charge in that rectangle is what we are calculating.
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Post by Vexman Wed Apr 03, 2019 1:46 pm

Speaking of the the equation C/R^4 strictly in light of its units:

C=kg^2 / km^3
R= km => R^4 = km^4

C/R^4= (kg^2 / km^3) / km^4= kg^2 / km^-1= kg^2 * km

Well,  the unit of kg^2*km doesn't make much sense either. What is its dimension? How do we interpret that? Maybe we can avoid expressing units if we let all values to be expressed in relative terms. But we're able to avoid it only up to a point when we have to interpret the result. Let assume we have calculated C/R^4 and we're looking at some value which is unit-less since we cancelled them out during the procedure. What is that particular value other than a ratio between charge density (C) and R? Let's ignore real values for a moment and  focus only on the result itself, for instance, Cs/Rs^4= 3. What is the value of 3 representing? The strength of charge field at distance R ? Charge density drop-off at the distance R?

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Post by Nevyn Wed Apr 03, 2019 4:05 pm

Oops, I missed the exponent. Still, the point is the same. The units in the denominator change what is in the numerator and we can't have that. The point of the drop-off part is only to reduce the density, not convert it into something else.

I don't know what a kg^2km is, or even a kgkm. So Cs/Rs^4 doesn't work. If it was Cs/(l/Rs)^4, then l/Rs loses its units and keeps Cs as whatever it was to begin with. Which is supposed to be a density. So if Cs/(l/Rs)^4 = 3, then 3 is the amount of charge, at distance l from the center of the body, in a unit volume. Exactly what we want it to be.

C is not just a ratio. It is an expression of a quantity over a volume. It is a ratio that tells us we will have X amount of something in Y volume of space. It is a ratio, but it has units, so it is a bit more than just a ratio. It is a ratio with meaning. It is a ratio that expresses a particular relationship.

Yes, you can try to make everything relative, but then we are losing the units that we need as well as the ones we don't. We also have to explain why we are using other things to make it relative. Not just mathematically, but physically. What does it represent in the problem? Why are we doing it? Doing that also means that we require 2 bodies to calculate the charge density of 1. How does that make sense? Why do we want to jump through all these hoops when we have an equation that makes sense, fits the problem, and gets the right answer?
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Post by LongtimeAirman Thu Apr 04, 2019 10:41 am

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The Cause of Gravity - the next major chapter - Page 6 Channe10

Nevyn wrote. I mentioned a while ago that it looks like Miles is talking about objects on the surface, or close to it, in the new gravity papers, and orbits are more about net force at the location of the orbiter. Hopefully he writes a new paper about that soon.

Does charge binding apply to orbits? It appears you’re waiting for Miles' next gravity paper. May I suggest we assume charge binding works for orbital bodies or show why not?

I've mostly been considering your main objection to orbital charge binding; the magnetosphere somehow interferes with or prevents orbital charge binding.

With respect to the discussion, I believe we’ve made progress. We are all more cognizant of each body’s charge field and charge density and a little of the number theory behind it. Has the discussion changed your thinking at all? For example, unless you disagree, it appears the moon’s overall charge field is stronger than the earth’s.

Do you still deny the apparent "fact" that charge channels exist at all times between orbiting bodies? Do you honestly believe the so-called magnetosphere ion ‘boundary’ prevents two way charge channeling between orbiting bodies?

Charge binding - if it exists for orbital bodies, must involve the charge channel between the orbital bodies.
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Post by Jared Magneson Thu Apr 04, 2019 12:37 pm

LongtimeAirman wrote:Does charge binding apply to orbits?

It doesn't appear to, which is one of the most critical flaws in the new charge-binding "gravity" theory.

LongtimeAirman wrote:For example, unless you disagree, it appears the moon’s overall charge field is stronger than the earth’s.

Where do you get that from? The Earth's field is much stronger than the moon's, simply due to its mass and rotation.

LongtimeAirman wrote:Do you still deny the apparent "fact" that charge channels exist at all times between orbiting bodies?

I don't see how that's remotely a "fact". Charge emission isn't a channel, it's an emission. What is channeling the photons? What is telling them to move back and forth, as though in circuit? It may just be semantics, but for example when the moon is eclipsed by the Earth (from the sun), no charge "channels" from the sun to the moon exist at all. It's in the Earth's shadow, so the Earth blocks insolation to the moon some amount of time. An eclipse also causes a slight wobble as the moon is only feeling the Earth's charge "pressure" and not the sun's, so it moves in towards both slightly more than normal due to gravity.

Same with all of Jupiter or Saturn's moons. While those orbits are HUGE compared to the Earth/moon orbit, they still get eclipsed occasionally. They even eclipse each other, from their host planet, on occasion.

Here's a quick sim of Saturn with all its major moons, for example. Our previous, working model of gravity vs charge (expansion/universal spin, whatever) can explain and calculate this. So far, charge-binding gravity cannot account for it*.
The Cause of Gravity - the next major chapter - Page 6 SAlWuwH
(right-click "View image" to zoom in)

*as I understand it, currently

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Post by Nevyn Thu Apr 04, 2019 6:20 pm

LongtimeAirman wrote:
Nevyn wrote. I mentioned a while ago that it looks like Miles is talking about objects on the surface, or close to it, in the new gravity papers, and orbits are more about net force at the location of the orbiter. Hopefully he writes a new paper about that soon.

Does charge binding apply to orbits? It appears you’re waiting for Miles' next gravity paper. May I suggest we assume charge binding works for orbital bodies or show why not?

I think I have shown why it can't work at the moon. I was initially wrong about some of it (requiring the downward charge to be channeled) but it does still require the upward charge to be channeled in order to avoid all (or most) of the upward pressure. Remember that the up vector is now 9.81 and the down vector is 0.009545. We have shown that the earth's field is not the primary field at the moon, it may not even be the secondary field, although to say that I am counting the moon itself, and I can't do that when talking about the effects on the moon. The sun's field is primary, so that is what the moon will align to, if it does.

However, maybe we don't want the moon to avoid all of that upwards pressure. Maybe orbits want the push. You can easily look at the moon and imagine it rolling along the earth's charge field. That push was always a part of Miles orbital theory. So I am actually hoping that Miles is only talking about objects close to the surface and not orbits. I think that will actually help binding gravity.

LongtimeAirman wrote:I've mostly been considering your main objection to orbital charge binding; the magnetosphere somehow interferes with or prevents orbital charge binding.

With respect to the discussion, I believe we’ve made progress. We are all more cognizant of each body’s charge field and charge density and a little of the number theory behind it. Has the discussion changed your thinking at all? For example, unless you disagree, it appears the moon’s overall charge field is stronger than the earth’s.

I don't think it has changed it yet. But there is more to think about.

Yes, I believe the moon's field is stronger than the earth's, at least close to the moon. I'm not sure how far it reaches from the surface. The Moon Gives Up a Secret paper mentions the moon's field being larger than the earth's. Well, I should say that it is stronger, it is not necessarily larger, and I can't remember if that is only when measured at the moon or if it means at the same distance from each body. That is, if we measured at 10,000km from the earth and 10,000km from the moon. I would think that it only refers to a single position close to the moon, because the smaller size of the moon has a larger curvature, so it will dissipate much quicker.

LongtimeAirman wrote:Do you still deny the apparent "fact" that charge channels exist at all times between orbiting bodies? Do you honestly believe the so-called magnetosphere ion ‘boundary’ prevents two way charge channeling between orbiting bodies?

Charge binding - if it exists for orbital bodies, must involve the charge channel between the orbital bodies.
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It depends on what that charge channel is. Do I believe that charge from the earth reaches the moon? Yes. Do I believe that charge from the moon reaches the earth? Yes. Those the only things I see as facts. But that is just photons traveling from one to the other. I don't see how that creates some sort of force, other than the push away from the source of the charge, and magnetic effects. I don't see how it creates an alignment of nuclei when the sun is also there and creating a stronger field. Miles mentions something about them being a straight alignment of fields and that does something, but all bodies have a straight alignment with the earth and moon, so what makes that one so special?
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Post by LongtimeAirman Thu Apr 04, 2019 6:22 pm

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Jared wrote. Charge emission isn't a channel, it's an emission. What is channeling the photons? What is telling them to move back and forth, as though in circuit? It may just be semantics, but for example when the moon is eclipsed by the Earth (from the sun), no charge "channels" from the sun to the moon exist at all. It's in the Earth's shadow, so the Earth blocks insolation to the moon some amount of time. An eclipse also causes a slight wobble as the moon is only feeling the Earth's charge "pressure" and not the sun's, so it moves in towards both slightly more than normal due to gravity.

As I've mentioned before, I’ve adopted the term charge channel a bit loosely; nevertheless, I believe it’s appropriate. The charge channel results from the addition of not one but two emission fields. Unless there is a third body present to physically block the path between them, there is always a direct two-way charge connection between two bodies in an orbital system, the direct angular width and direction to the other body where the emission traffic from either of the two bodies will meet head to head.

Within my so-called charge channel between the earth and moon there are many, many relative simultaneous charge coherent links between the proton matter of the two bodies’ charge sources. Both overlapped emission fields have the ability to deflect, or redirect the photons from the other charge source near that direct head to head radial.

Note. Eclipses are NOT a problem. True, they occur. During eclipses the orbiting body doesn’t lose gravity, the eclipsed body has lost the charge source that was creating one of the velocities that made up its total orbital motion. While in the eclipse, the body still moves with the forward velocity of body's orbital inertia. The occupants on an eclipsed body do not feel a loss of gravity, they are charge cohered directly to the celestial body beneath them. The orbital body’s normal acceleration is resumed when the source is no longer blocked.
 
Jared wrote . Where do you get that from? Earth's field is much stronger than the moon's, simply due to its mass and rotation.

Miles shows that EM = 110.12 EE,   EE = .009545 m/s2 and EM = 1.051 m/s2
in The Moon Gives up a Secret. http://milesmathis.com/moon.html  Numbers from the Moon are analyzed to show that "gravity" is a compound field. 4pp. Thanks for that source when I needed it Nevyn.

The Sun/Earth/Moon relative field density diagram I put together clearly indicates the Moon’s charge field emission is stronger than the Earth’s. In order to match the sun’s field strength’s at the magnetosphere and earth, the diagram suggests the moon’s relative charge field is 180/40 = 4.5 times stronger than the earth’s. Of course the diagram is a two dimensional cross section of spherical charge fields; note that the cube root of 110 is 4.79 – close to 4.5. So I feel some justification in believing the relative field density diagram is a good model/diagram. That would make the sun’s charge field density (360/.00482)^3 = 416*10^12 (!?) roughly stronger than the earth’s.

Nice image. Saturn has an amazing system, rings and all. I don’t see how the image itself supports or contradicts the charge binding interpretation of gravity nor of the existence of direct two-way emission ‘channels’ between all Saturn’s moons and Saturn itself. We might all agree; the space about Saturn, besides being filled with satellites, is primarily filled with Saturn’s charge. I'm doing my best to determine whether we can show charge binding exists for orbital bodies - or not.  
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Post by Jared Magneson Fri Apr 05, 2019 5:15 am

Getting back to the theory:

Miles wrote:Now we get to the hard part.  Spin mechanics.  We take the photon to be a spinning sphere.  It can bespinning in any direction, but once it enters a pre-existing field it will be made coherent to that field.How?  By collisions.  Photons are colliding all the time, and due to their small size they tend to hit edge-to-edge. Edge hits cause spin changes rather than speed changes.  Hits can either cause spin-upsor spin-downs.   (emphasis Jared's) Photons can collide side-to-side, moving in the same direction; or head-on.
http://milesmathis.com/grav4.pdf

I think this is not accurate. Edge-hits do not always cause spin-ups or spin downs, but primarily vector changes. Bounces. Only certain edge-hits at certain angles can cause a spin-up or spin-down, as I have attempted to show in my videos detailing why the vector is so important. It has to be orthogonal at the point of contact to the current/existing spins of either photon particle. And it has to be pretty close to the edge to work, simply because the closer you get to the center, the more it's not an "edge-hit" at all but a direct, stifling collision.

In such a dense charge field as the Earth, we might see a great many spin-ups and spin-downs, a vast and amazing many. But that would still be as a matter of probability far, far fewer than simple vector changes (bounces). This isn't detrimental to the theory but I feel like it should be heavily clarified: in my study, MOST collisions will not and cannot cause spin-ups or spin-downs, but rather simply bounces. Redirection of the existing energy.

This isn't just semantics though - consider the angle of the Earth's tilt. For insolation to travel through the core it would HAVE to be bounced at least once. Why? Because photons coming straight out from the sun (either the disc sun or the point sun, which doesn't exist but for some reason keeps getting used in this thread) cannot enter the Earth's poles at all, because they are moving in the wrong vector. There can BE no through-charge without simple redirecting bounces. For charge to even enter the poles from the sun, it must be bounced charge. Otherwise it's too oblique.

A boy doesn't ATTRACT charge to its poles or bend charge in, it can only REPEL any such charge by emitting its own. A solar photon moving out at intersection with the Earth doesn't simply get closer enough and turn 90° down into one pole or other, it has to be bounced into that pole. Up or down. It's going sideways, and we need it going up or down as it nears or reaches our body-of-study.

What I'm getting at is that I think this bounce-charge has been severely underestimated in the maths and concepts. Without it, even the proton cannot work per the theory. For any charge to enter ANY pole of any body, it must be moving parallel to the poles to some extent or other at the time it enters, which is perpendicular to (say) solar emission. In general orbits, we have charge moving perpendicular to the poles almost entirely.

We can use Miles' own theory of Uranus's tilt to make this even more clear:
The Cause of Gravity - the next major chapter - Page 6 E7hvbjb

Now I'm not saying that incoming charge has to be at 90°, but definitely some angle that would cause a throughput. And sure, a volume warming at some incident angle would produce some heavy effects, but we can calculate these things and simulate them so I'll probably work on this next. Just to illustrate the point.

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Post by Nevyn Fri Apr 05, 2019 6:24 am

I plugged the numbers for the moon into my charge-pause calculator. The moon values are in the sun fields.

The Cause of Gravity - the next major chapter - Page 6 Charge19

So the moon's charge-pause should be near 16 moon radii from its center. That's about 7% of the distance to the earth.
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Post by Nevyn Fri Apr 05, 2019 6:41 am

Every planet is below 1%, except Jupiter and Saturn which are below 2%, of the distance to the Sun. But the moon is over 7% from the Earth. I'm not sure what that means, but it looks interesting.
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Post by Jared Magneson Fri Apr 05, 2019 1:47 pm

Nevyn wrote:Every planet is below 1%, except Jupiter and Saturn which are below 2%, of the distance to the Sun. But the moon is over 7% from the Earth. I'm not sure what that means, but it looks interesting.

What about Ganymede to Jupiter, or Titan to Saturn? I'm curious what this means as well.

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Post by Nevyn Fri Apr 05, 2019 5:00 pm

PlanetMoonCharge-pause
JupiterIO0.2%
JupiterEuropa0.18%
JupiterGanymede0.37%
SaturnTitan0.66%
Mathematically, it just represents the relationship of mass and density between the bodies, but I find it interesting that everything I have calculated so far is under 2%, with most of that under 1%. Our moon is quite special. So far.
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Post by Nevyn Fri Apr 05, 2019 7:00 pm

I thought it might mean that the moon is not in equilibrium, so it wants to move farther out, which we know it is actually doing. I tried to increase the distance between the Earth and Moon, but it doesn't get any where near 1%, or even 2%, without the equations breaking. The results come out as half way between the bodies, which I think happens when the fields are not strong enough to reach that far. Since 0=0, it thinks it has found an answer, but it has just crapped its pants. I will look further into that. It could also be the mathematical precision of the numbers causing the problems. Computers are usually limited by the size of the numbers they can use, but there are ways around it.

However, let's assume that it is the equation breaking down. Maybe that implies that the Moon is actually too big to be a moon of the Earth at its proper distance. If it gets far enough away, we may lose it and the Moon becomes another planet. Unfortunately, we will all be long gone before that happens, so we may never know if my prediction is correct.

We may also be able to use this data to find the force exerted on the Moon, by the Earth's charge field. We know that the Moon is moving away from the Earth at a rate of 4cm per year, but slowing down. We can calculate the Earth's charge field at the surface of the Moon. Bring in the Moon's mass and we should be able to calculate the actual force. I'll look into that too.
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Post by Nevyn Fri Apr 05, 2019 7:02 pm

The larger percentage for Jupiter and Saturn also show Miles point about the solar system being out of order. Those big planets need to move outwards, and they will move out to where their charge-pauses are within 1%.
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Post by LongtimeAirman Fri Apr 05, 2019 9:06 pm

.
Just a quick update. Mostly confused.

Nevyn wrote. Yes, I believe the moon's field is stronger than the earth's, at least close to the moon. I'm not sure how far it reaches from the surface. The Moon Gives Up a Secret paper mentions the moon's field being larger than the earth's. Well, I should say that it is stronger, it is not necessarily larger, and I can't remember if that is only when measured at the moon or if it means at the same distance from each body. That is, if we measured at 10,000km from the earth and 10,000km from the moon. I would think that it only refers to a single position close to the moon, because the smaller size of the moon has a larger curvature, so it will dissipate much quicker.

Nevyn wrote. So the moon's charge-pause should be near 16 moon radii from its center. That's about 7% of the distance to the earth.

Airman. I am mentally wrapped around the radius of the newly identified, moon’s earth-pause, and the moon’s extraordinary strength in that regard, thank you very much. I suppose the relative charge field diagram absolutely needs a modification of some sort - something that indicates the field's fall-off rate. An idea does occur to me. I can replace each current individual infinitely long radial with a series or train of radials, from longer to shorter, ending at the object's Kuiper limit.  

Jared wrote. What I'm getting at is that I think this bounce-charge has been severely underestimated in the maths and concepts. Without it, even the proton cannot work per the theory. For any charge to enter ANY pole of any body, it must be moving parallel to the poles to some extent or other at the time it enters, which is perpendicular to (say) solar emission. In general orbits, we have charge moving perpendicular to the poles almost entirely.

The Cause of Gravity - the next major chapter - Page 6 Satura10
Airman. I agree, bounce-charge is present. It increases the charge density and effective charge receiving width about the planet. Looking at the Saturn/Uranus/Neptune inclinations. I believe Miles considers it a circuit diagram. From our previous discussion, the direct radial charge emission contribution we’ve been considering is the primary component of the Sun/planet circuit. The planet is also recycling charge. The charge field includes larger charged particles. These particles are massed over the planetary poles. They get their primary charge from the planet’s poles or the sun. Emissions from the larger charge helps extend and direct more photons into the planetary poles.  

I believe the diagram shows the equilibrium tilts that result from adding both the direct charge and recycling secondary charge. A proton aligns it's pole to the planet, but the charge mix the planet receives makes that direct (sun/planet pole) alignment impossible - although Uranus gets pretty close.  
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Post by Jared Magneson Sat Apr 06, 2019 2:32 am

Nevyn wrote: Maybe that implies that the Moon is actually too big to be a moon of the Earth at its proper distance. If it gets far enough away, we may lose it and the Moon becomes another planet. Unfortunately, we will all be long gone before that happens, so we may never know if my prediction is correct.

But wouldn't it just slowly "suck" back in, due to gravity, at some eventuality? You know, the ol' ping-pong ball held underwater analogy?

I guess that's a bit naive considering it's NOT a 3-body problem, but rather an every-body problem. Even so, since Luna and the Earth are both orbiting the sun, the moon can hardly propel itself into a higher or lower orbit. It would require some force that would trump the Earth's gravity field, it seems like, to pull it faster or slower and thus change orbits, but it would have to change orbits enough to escape the Earth's gravity, which seems pretty farfetched to me.

That said, it needn't change orbits outright, but perhaps would "tail" the Earth like the C-orbit horseshoe asteroids do?

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Post by Nevyn Sat Apr 06, 2019 3:44 pm

I've found that the reason my app does't find the charge-pause at extreme distances is because of the tolerance value. At those distances, the charge fields of the bodies are so weak that they are already below the tolerance. Setting the tolerance to 0 makes it work. I might remove the tolerance. I thought it would be needed to avoid looping forever.
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Post by Nevyn Sat Apr 06, 2019 4:05 pm

I can't get Jupiter's charge-pause below 1.77%, even at 7786e+78km between it and the Sun. At 7786e+79km they are too far apart. So maybe these percentages don't mean as much as I first thought.
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Post by Russ T Mon Apr 08, 2019 8:31 am

Blimey! You guy's have been busy. I have some catching up to do.

I notice someone mentioned this dual property of the charge field i.e. gravity and charge pressure (?).
I haven't read the last 11 blog pages so need some time to absorb...but...in the meantime, is this duality/dual field a bit like the dual field in electrical circuits? The E & B fields, electric and magnetic? The charge photons being charge anyway. Moving charge/moving photons creating a magnetic field? Aren't they all properties of the same field i.e. electric charge, magnetism, gravity, heat, all describing different properties of the charge field?
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Post by Jared Magneson Mon Apr 08, 2019 1:06 pm

Russ T wrote: Aren't they all properties of the same field i.e. electric charge, magnetism, gravity, heat, all describing different properties of the charge field?

Yes indeed, they are. The problem we're having with charge-binding gravity (the one I'm having so far) is that it's hard to get that huge vector DOWN out of it. But it's a work in progress and we'll see where it leads. Miles has emailed me about it and he's trying to flesh it out as the muses allow, but there's a lot of great questions and issues here that the theory can confront down the road too. I'm not expecting him to drop every damn thing and just cater to us here. This is just the best place for such discussions, with the best team around on the task at hand. Read up. Maybe you'll see something we missed?

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Post by LongtimeAirman Mon Apr 08, 2019 2:35 pm

.
Russ T wrote. I notice someone mentioned this dual property of the charge field i.e. gravity and charge pressure (?).
Airman. Hi Russ T. There is only the charge field. Before this thread, Miles described the charge field as a real exclusionary emission field in opposition to gravity; now, gravity is an aspect of charge. Most everyone - I know I am - is having difficulties fitting charge binding in with the rest of their own charge field understanding.

Russ T  wrote. ... is this duality/dual field a bit like the dual field in electrical circuits? The E & B fields, electric and magnetic? The charge photons being charge anyway. Moving charge/moving photons creating a magnetic field?
Airman. According to my understanding, both B and E fields are the effects of the charge field observed at the electron scale or larger. All charge photons spin and travel at light speed. The E field is the result of charge photons pushing electrons or ions in the forward direction, at a resulting velocity indicating momentum or energy. The B field is the amount of spin coherency imparted on the electrons and ions by the spinning of those same charge field photons.

Russ T  wrote. Aren't they all properties of the same field i.e. electric charge, magnetism, gravity, heat, all describing different properties of the charge field?
Airman. Absolutely, yes Sir, I agree, although I must admit, the charge gravity explanation of electromagnetism is a bit beyond me at present.
.

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Post by Jared Magneson Mon Apr 08, 2019 4:10 pm

LongtimeAirman wrote:Airman. Absolutely, yes Sir, I agree, although I must admit, the charge gravity explanation of electromagnetism is a bit beyond me at present.

It's actually mildly comforting that I'm not the only one struggling here. That said, I do have a bit of faith that Miles will either make sense of it all or jettison it. I don't see him chasing a theory too hard if it turns out to have more problems than it fixes, just because it was his. He's corrected himself before but at the same time I'd love for him to be RIGHT too, so we could source gravity once and for all! Or once and for now, anyway!

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Post by Chromium6 Sun Jul 16, 2023 4:52 am

Notes on "Gravity" affecting photons and a mostly classical explanation for it. Miles has rewritten some of this below in his UFT-Gravity papers. The site link has full equations:
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https://profoundphysics.com/how-are-photons-affected-by-gravity-if-they-have-no-mass/

How Gravity Affects Photons: The Physics Explained

We know that photons have no mass  Question and we like to think that gravity only affects things that have a mass. However, photons still get deflected by the Sun and can even orbit around a black hole. How exactly are photons being affected by gravity then?

Photons have no mass, but they are nonetheless affected by gravity due to the bending of spacetime itself. In the presence of gravity, photons travel along geodesics. Geodesics depend on the geometry of spacetime and photons moving along a curved geodesic will appear to be affected by gravity.

In this article, we’ll look at how Newton didn’t quite get it right with gravity, what gravity really is, how general relativity describes gravity and how this all relates to tell us how massless photons are affected by gravity.

If you’re wondering why exactly photons do not have mass in the first place, I have a full article covering that here. I also cover why photons still have momentum, even though they have zero mass in this article.


Does Newtonian Gravity Affect Photons?

The first and longest standing theory of gravity was Newton’s theory. This comes nicely within the framework of Newton’s laws,

Law 1: A body stays at rest, or travels in a straight line at constant speed, unless acted on by a force.
Law 2: Force equals mass times acceleration F = ma.
Law 3: Every action has an equal and opposite reaction.
Newton’s theory of gravity fits on the left-hand side of the equation in his second law. This is the formula telling us the force of gravity due to a body of mass M on a second body of mass m that are separated by some distance r:

F=-G\frac{Mm}{r^2}
Here G is the gravitational constant – a number fundamental to the universe that tells us how strong gravity is.

The minus sign is a bit of mathematical convention that tells us that the force is attractive!
Let’s put this force to action with a body such as a star with mass M and a photon with mass m. Starting with what we know, photons have mass m=0, so let’s plug that in here:

F=-G\frac{M\cdot0}{r^2}=0
Here we’ve arrived at Newton’s interpretation of how light is affected by gravity – it isn’t! No massive object will affect a photon according to Newtonian gravity!

We can apply all of Newton’s laws here – his third law tells us that the photon also doesn’t exert a force on the star and his first law tells us that since there is no force of gravity acting on the photon, the photon travels in a straight line!


These laws seemed infallible for a long time – they described everything we saw on Earth well for a long time. An issue came up with the orbit of Mercury.

The technical term for this is orbital precession but we’ll get on to what that exactly means in a moment. First, we just need to know one fact: orbits in Newtonian gravity are ellipses – they look like squashed circles.

These ellipses slowly rotate over time as the planet orbits – this is orbital precession and is predicted in Newtonian gravity. However, the amount Mercury should precess according to Newtonian gravity versus how much astronomers saw did not agree.

This was one of the first hints that Newtonian gravity were not complete!

However, Newton’s laws stood the test of time until Einstein came along with a new perspective on gravity known as the theory of general relativity.

Why Gravity Affects Photons In General Relativity
Next, we’ll look at how Einstein’s theory of general relativity, a more accurate description of gravity, explains why photons indeed are affected by gravity like all other forms of matter.

Specifically, we’ll discover that:

In general relativity, photons always travel along geodesics.

Geodesics can be thought of as the shortest paths between two points.

In general relativity, the shortest path between two points might not always be a straight line.
Sometimes the shortest distance a photon can take between two points may actually be along a curved path, in which case it would appear to us that gravity has an effect on the photon’s path.

It may be funny to think that the shortest distance between two points could possibly be anything other than a straight line, but in general relativity, it indeed can. This is one of the many peculiarities of the theory.
Now, what determines the shortest distance between two points? In general relativity, it is the curvature of spacetime. The geodesics of photons appear as different paths depending on how spacetime is curved.

Einstein formulated the concept of “spacetime”, meaning that we look at space and time equally as one greater concept, rather than as time being some universal ticking clock.

Einstein’s theory of relativity does away with the idea that gravity is a force and replaces it with the idea that gravity is the bending of spacetime due to matter! This can be summarized as:

“Spacetime tells matter how to move, matter tells spacetime how to bend.”

Einstein’s theory has to be a geometric theory in order to talk about the shape of spacetime so it is all described in a new mathematical language.

To truly understand how and why gravity affects photons, we need to dive into the mathematics of general relativity a little bit and see how exactly Einstein predicted the bending of light and the trajectories of photons.

If general relativity is something you'd be interested to learn more about, I have a full introductory article called General Relativity For Dummies: An Intuitive Introduction. The article gives you a full overview of what general relativity is about and teaches you the most important concepts in an intuitive sense.

I also have a full guide on learning general relativity on your own, if that's something you're interested in doing.

A Brief Introduction To The Mathematics of General Relativity

In Newtonian physics, we can talk about how things change with time, we can talk about, say, the position of a car at time t = 0, t = 1, t = 10 etc. We would write this position x(t) as x(0), x(1), x(10). Time is special and can “parameterize” the path that the car takes.

In general relativity, time is no longer special and is put on an equal footing with spatial coordinates. We describe everything using the concept of spacetime, which means that paths need to also be parameterized in a different way than by using time.

Paths in spacetime are called worldlines and we write them as xµ(λ). The upper µ labels four coordinates
µ = 0,1,2,3 and can be represented as a vector (called a four-vector):

x^{\mu}\left(\lambda\right)=\begin{pmatrix}
x^0\left(\lambda\right)\\
x^1\left(\lambda\right)\\
x^2\left(\lambda\right)\\
x^3\left(\lambda\right)
\end{pmatrix}

We’ve now seen the notation xµ(λ) or x(t) but what does this mean? We treat the coordinates like functions. In these two cases they are either functions of λ or of t, time!

This means that if you plug in some number for time or the “parameter variable”, the function will give you what xµ or x is at that time or parameter value.

Using a general path parameter λ in general relativity is just a mathematical tool to allow us to parameterize paths of particles in a similar way as in Newtonian physics, where time does this for us.

Now, most of us have heard of Pythagoras’ theorem; this relates two sides of a right angled triangle to its hypotenuse, most famously as a2+b2=c2.

What this is secretly telling us is that the diagonal line between any two points is the shortest distance between those two points (mathematically, this follows from the fact that for two positive numbers a and b, √(a2+b2)
The maths of general relativity uses ideas exactly like this. If we want to talk about the surface of a shape, we think about how the shape changes on very small length scales – this is the same intuition as what derivatives are!

Since we’ve identified c2 as the squared size of the hypotenuse, this is our “line element“, which we’ll call ds2.

We can then treat a and b as being some small distances on an (x,y)-plane. Small changes are often written with the letter d in from of them (representing differentials), like dx and dy.

With this in mind, Pythagoras’ theorem would read:

ds^2=dx^2+dy^2

This is what we call the line element in Euclidean geometry. This extends (like Pythagoras’ theorem) to three dimensions as ds2=dx2+dy2+dz2.

This line element describes a small distance in space. However, in general relativity, we model everything by describing not only space, but spacetime.

In space, we can walk forwards, backwards, turn to the side, and jump – we have motion that we can control in our three spatial dimensions. However, time acts differently. To model this, we write time slightly different in our line element as:

ds^2=-dt^2+dx^2+dy^2+dz^2

This is what we call the Minkowski spacetime line element. This describes a small distance in flat spacetime. The Minkowski line element resembles a “straight line” in spacetime, meaning that there is no curvature.

If you’d like to read an intuitive introduction to special relativity, you’ll find one here. The article covers everything discussed here, but in much more detail.

The important thing for us is that the line element encodes all the information about gravity in general relativity. Minkowski spacetime is the special case where there is no gravity in our spacetime!

Much like how we economically write xµ for coordinates (worldlines), we often write the line element in a slightly different way as:

ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}
The nice thing about this form of the line element is that it is completely general; we can write all line elements (even in curved spacetime) in this form.

This is also an example of index notation and Einstein’s summation convention. The rules are as follows:

Greek letters represent the numbers 0, 1, 2, 3.
Roman letters represent the numbers 1, 2, 3.
If an index appears the same in an upper and lower position, we sum over all values of the index (according to rules 1 and 2).
Before we see an example of this, let’s talk about this gµν that we introduced. This is called the metric tensor. For most purposes, this is a 4×4 symmetric matrix.

The metric tensor tells us the coefficients of our small distances dxµ. The metric also describes how distances are measured in spacetime; if the spacetime is curved, the shortest distance between two points may not be a “straight” line anymore and this is all encoded in the metric!

In the Minkowski line element above, we use the special letter gµν = ηµν with:

\eta_{\mu\nu}=\begin{pmatrix}
\eta_{00}&\eta_{01}&\eta_{02}&\eta_{03}\\
\eta_{10}&\eta_{11}&\eta_{12}&\eta_{13}\\
\eta_{20}&\eta_{21}&\eta_{22}&\eta_{23}\\
\eta_{30}&\eta_{31}&\eta_{32}&\eta_{33}
\end{pmatrix}=\begin{pmatrix}
-1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&1
\end{pmatrix}
In the vocabulary of linear algebra, we say that this is a diagonal matrix and many metrics that are interesting to study are diagonal. The indices µ and ν label each entry in this matrix. Since it is diagonal, the non-zero entries are η00=-1, η11=1, η22=1 and η33=1.
Example: Flat Spacetime Minkowski Line Element
Whilst having no gravity in a spacetime is the simplest case, that is not the question at hand; We want to see what happens to light in the presence of gravity!

Perhaps the most widely known spacetime with gravity (which also describes the bending of light near a star, for example) is called Schwarzschild spacetime. The Schwarzschild solution to general relativity describes how spacetime reacts to a massive, spherical object such as a star!

Before we see that, let’s recap spherical coordinates as these will be used throughout this article (and everywhere else in physics). This is the last thing we need to go over before looking at photons specifically.

Quick tip: Spherical coordinates are one of the many important things used in physics that I cover in my Advanced Math For Physics: A Complete Self-Study Course (link to the course page). In fact, vector calculus is one of most important topics you should learn for understanding relativity, electromagnetism or even just mechanics. This course will teach you that, along with giving you all the tools you need for applying everything in practice.

We can describe the position of something in space by three coordinates (x,y,z) which are great in general but become difficult if we want to think about things that are symmetric under rotations, such as a sphere (which has a radius of √(x2+y2+z2) which is often difficult to work with!).

Instead, we use spherical coordinates (r, θ, φ) which describes a radius r and two angles of rotation:


Essentially, we describe a point in space by specifying two angles and a radial coordinate (distance from the center).

In this way, spherical coordinates cover all the same space as (x, y, z) do and are equivalent but sometimes much easier to work with!

The three dimensional line element in spherical coordinates is written as:

ds^2=dr^2+r^2d\theta^2+r^2\sin^2\theta d\varphi^2
In fact, this line element is equal to ds2=dx2+dy2+dz2. This signifies the fact that all distances (line elements) are the same regardless of which coordinates we describe them in; physics doesn’t care about your coordinate system!

This looks a little more complicated but despite its appearance, is much easier to work with in the case of a spherical star and in many other gravitational spacetimes as well!

Now, the line element in a Schwarzschild spacetime (which describes all distances near a gravitating spherical star) looks somewhat similar to this, but is written as:

ds^2=-\left(1-\frac{2M}{r}\right)dt^2+\frac{1}{1-\frac{2M}{r}}dr^2+r^2d\theta^2+r^2\sin^2\theta d\varphi^2

M here is the mass of our star and all physical stars will have r > 2M since we look outside the star. This is relevant because 1−2Mr is zero at r = 2M. If we were modelling a non-rotating, uncharged black hole, this would correspond to the event horizon of a black hole.
The last term terms are exactly the same as the Minkowski (flat or non-gravitational spacetime) line element – this tells us that in terms of gravity, it doesn’t matter how we rotate the star, only the distance from it does.

There’s now a prefactor in front of the time portion of the line element – this is telling us that time acts differently due to gravity.

This gives us amazing features such as gravitational time dilation; we can see that the coefficient in front of dt2 gets smaller the closer we get to r = 2M, we interpret this as time slowing down!

For an interesting example on why exactly time slows down near a black hole, I have an entire article on that, which you’ll find here.

In the same manner, gravity also affects distances in the Schwarzschild spacetime. It turns out that the shortest paths for photons in Schwarzschild spacetime are actually curved trajectories, leading to the deflection of light around a star.

With line elements, metrics, and worldlines safely under our belts, we can tackle the question at hand: How does gravity affect matter? And most importantly for us, how does gravity affect photons if they have no mass?

How Does Gravity Affect The Path of a Photon?
We’ll start with one important fact: The universe is lazy. Everything – planets, photons, everything – travels on the shortest path it can.

If we consider only gravity, we want to consider the paths or worldlines that all matter follows without any external forces (since gravity is no longer a force in general relativity).

These paths have special names – geodesics! They can be assigned three different types: timelike, null and spacelike.

Timelike geodesics are for matter that has mass and travels slower than the speed of light.
Null geodesics are for matter without mass (such as photons) which travels at the speed of light.
Spacelike geodesics are for matter which travels faster than the speed of light (hypothetical particles known as tachyons).

To tell us about these paths, we define the line element in a specific way since this is telling us intimate details about the geometry of our spacetime. By convention, we say:


The important thing is that light travels on a null geodesic – We can now understand how gravity affects a photon by looking at these null geodesics in any given spacetime with gravity.

Let’s think about this physically for a moment: we said before that ds2 is like a distance in spacetime, so a null geodesic means that light travels on paths that have zero spacetime distance.

This sounds funny but in general relativity, this is indeed possible; a photon can still move through space without moving in spacetime (this is because of the minus sign we saw earlier in front of the dt-part of the line element).

So essentially, photons travel along the shortest paths through spacetime and at the same time, these paths always have zero spacetime length. In this sense, it doesn’t make sense to talk about a “shortest distance” in spacetime for a photon, since the spacetime distance is always zero.

In any case, photons move along null geodesics in spacetime. The shape and form of these geodesics depends on the spacetime we’re in.


Now, how do we actually find the geodesics of photons? The simplest and most brute force approach to get the trajectory is via the geodesic equation:

\frac{d^2x^{\alpha}\left(\lambda\right)}{d\lambda^2}+\Gamma_{\mu\nu}^{\alpha}\frac{dx^{\mu}\left(\lambda\right)}{d\lambda}\frac{dx^{\nu}\left(\lambda\right)}{d\lambda}=0
We can see the parameterized worldlines xα(λ) in this equation that we discussed earlier appearing in three places here; the purpose of the geodesic equation is to solve for these to get the spacetime trajectories.

The first and second derivatives of the wordlines are taken in the above – these derivatives describe how the wordlines xα(λ) change as we vary the path parameter λ.

Finally, we have the Christoffel symbols, denoted by Γ. In short, these encode any changes in coordinates if we look at our system from different perspectives – just like when we changed from Cartesian (x,y,z) coordinates to spherical coordinates earlier!

For those who are interested, the Christoffel symbols are mathematically given by:

\Gamma_{\mu\nu}^{\alpha}=\frac{1}{2}g^{\alpha\beta}\left(\frac{\partial g_{\nu\beta}}{\partial x^{\mu}}+\frac{\partial g_{\mu\beta}}{\partial x^{\nu}}-\frac{\partial g_{\mu\nu}}{\partial x^{\beta}}\right)
The main part we see is the metric tensor that describes our spacetime as well as its derivatives.

The Christoffel symbols give rise to phenomena such as artificial or fictitious forces like the centrifugal force when rotating something – this arises effectively from changing a coordinate system to another.

I actually have a full guide on Christoffel symbols, which you’ll find here. It covers everything from the physical and geometric meanings of the Christoffel symbols all the way up to how to actually calculate and use them in practice.

In case you’re interested to see where the geodesic equation really comes from, you’ll find its full derivation below. This uses some advanced concepts, which are presented as intuitively as possible.

Where Does The Geodesic Equation Come From? (Full Derivation)

Now, in flat or Minkowski spacetime, the geodesics of photons are straight lines, just like Newton’s laws would predict. You’ll see how this comes about down below.

However, things change greatly when we consider other, more complicated spacetimes and metrics, which correspond to spacetimes in which gravity is present.

In these cases, a photon may not travel in a straight line anymore, differing from the predictions of Newtonian gravity.

The most extreme case of this may be for a photon orbiting around a black hole. I have a full article explaining how this happens, in case you’re interested.

Example: Geodesics In Flat Spacetime

As we’ve seen, the geodesic equation in specific circumstances can tell us all about Newtonian physics but it can do much more.

For any spacetime, if you can write down its metric, you can plug it into the geodesic equation and find the equations of motion for any particle! However, that doesn’t mean the equation is always necessarily solvable, but if it is, then you can find the trajectories of a photon (or any other particle) under gravity.

Mathematically, a more elegant approach is to take the metric, look at its geodesic Lagrangian (explained earlier), calculate its Euler-Lagrange equations, and combine this with the fact that we are looking at null geodesics for light!

We can easily get the geodesic Lagrangian by taking the line element, replacing any variable (such as dt, dr, dx etc.) by the same variable with a dot over it (ṫ, ṙ, ẋ etc.), representing a derivative with respect to λ and putting a half in front of the whole thing!

For example, in the Schwarzschild spacetime we briefly looked at earlier, we have (see the similarity between the line element and the geodesic Lagrangian?):

ds^2=-\left(1-\frac{2M}{r}\right)dt^2+\frac{1}{1-\frac{2M}{r}}dr^2+r^2d\theta^2+r^2\sin^2\theta d\varphi^2
L=-\frac{1}{2}\left(1-\frac{2M}{r}\right)\dot{t}^2+\frac{1}{2}\frac{1}{1-\frac{2M}{r}}\dot{r}^2+\frac{1}{2}r^2\dot{\theta}^2+\frac{1}{2}r^2\sin^2\theta\dot{\varphi}^2
In fact, this can also be used as an efficient method for calculating Christoffel symbols. I cover this “trick” in this article.

Now, to answer the main question: if photons are massless, how are they affected by gravity – under the influence of gravity, photons travel on null geodesics (ds2 = 0) and geodesics are described by the Euler-Lagrange equations of their geodesic Lagrangian (or equivalently by the geodesic equation; both describe the same thing).

The equations we get are determined by the metric gµν and in general relativity, gravity is the curving of spacetime rather than a force so all the effects of gravity are wrapped up in the metric.

Photons, like all matter, want to follow a geodesic because of the laziness of the universe and the easiest path to take is to follow how matter has bent and curved spacetime, causing gravity.

The point is that it doesn’t matter whether the photons are massless or not; they still travel along geodesics and IF the metric describes a curved spacetime (in which gravity is present), then the photons will inevitably move along curved paths as well. This is how gravity affects photons!

The only place where the fact that photons are massless actually matters is that the geodesics of photons are null (ds2 = 0), which is different in the case for massive particles (with ds2 = -1 instead).

This doesn’t change the fact that photons are still affected by gravity, it simply causes the paths of photons and massive particles to look slightly different.

For example, light can orbit a black hole at only one possible distance, while a massive particle could have two different orbits. You can read more about orbits of light around a black hole in this article.

With all this theory, let’s see it altogether fully in an example spacetime!

How Gravity Affects Photons Near a Star

We have seen already that in the presence of no external forces and without gravity, all matter travels in straight lines. If we introduce gravity, that is no longer true – think about planetary orbits!

Let’s look at what happens to a photon (light) when it passes a perfectly spherical star. In Newtonian gravity, we would expect for the photons to keep moving in a straight line, as gravity does not affect them.

In general relativity, this is not true – the key result is that a ray of light passing a star gets deflected by an angle:

\delta=\frac{4GM}{c^2D}
G is the gravitational constant, c is the speed of light, M is the mass of the star, and D is the smallest distance radially that the light gets to the star.
Essentially, this deflection angle describes how much a light ray would get bent as it passes near a star. In other words, how much the path of the photon differs from being a straight line.

This can be observed by looking at light rays (photons) coming from a distant star – since the light rays get deflected as they pass the Sun, for us, the distant star would appear to be in a different position in the sky compared to where we would expect it to be.


For some context, if we consider light just grazing the sun, this gives a measurement of 1.75 arcseconds – Arthur Eddington verified this empirically in 1919 and it was a key result in verifying general relativity experimentally!

The deflection angle δ is typically very small. For scale, an arcsecond is 1/3600th of a degree – so the result is very very small as expected – but crucially it is not zero as we would expect in Newtonian gravity!

This shows directly how a massless photon is affected by gravity – it must follow the natural bending of spacetime due to matter!

Now, where does this result come from? Let’s get down to the details – the key ingredient is geometry!

Essentially, we will discover that the geodesic of a photon as it passes by a star, is described by the following equation:

r\left(\varphi\right)=\frac{D^2}{M}\frac{1}{1+C\cos\varphi+\frac{D}{M}\sin\varphi+\cos^2\varphi}
Note; this is in units where G=c=1. In case you’re familiar with standard orbital mechanics, this may look somewhat similar to Kepler’s orbit equation describing, for example, the elliptical orbits of planets. In a sense, this is a more complicated “orbit equation” that describes the orbit of a photon.
This describes the distance r of the photon to the star as a function of the angle φ in polar coordinates (see picture below).


From this, we can derive the deflection angle δ=4GM/c2D. You’ll see the full derivation of this below.

Derivation of Photon Geodesics & Deflection Angle Near a Star
More importantly than the actual result of deflection, this is an example that directly shows that photons are indeed affected by gravity – how they are affected by gravity will depend on the particular spacetime we look at.

In Minkowski spacetime, we saw that photons travel in straight lines. This corresponds to the case with no gravity and is consistent with what we expect in Newtonian physics!

However, in Schwarzschild spacetime (under the gravity of a spherical mass), a photon will travel in a curved path and get deflected. In this case, the photon will be affected by gravity.

In other spacetimes, photons will also generally be affected by gravity but in different ways – near a rotating black hole (described by the so-called Kerr spacetime), for example, a photon’s trajectory may look incredibly complicated.


Ville Hirvonen

I'm the founder of Profound Physics, a website I created to help especially those trying to self-study physics as that is what I'm passionate about doing myself. I like to explain what I've learned in an understandable and laid-back way and I'll keep doing so as I learn more about the wonders of physics.

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Post by Jared Magneson Sun Jul 16, 2023 1:55 pm

Newton’s theory of gravity fits on the left-hand side of the equation in his second law.

False. Gravity is not the F, gravity is the a (acceleration).

Also space-time was falsified absolutely by Miles in a series of papers.

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Post by LongtimeAirman Mon Jul 17, 2023 12:24 pm

.
Cr6, thanks for bringing up this old thread, maybe this site’s most liveliest. I don’t think we resolved Miles’ "gravity as charge idea", but I certainly enjoyed the discussion at the time, as well as in re-reading some of it yesterday.

Hi Jared! Happy to hear from you, was thinking about you just yesterday. You made some great animations, do you still make them?

How Gravity Affects Photons: The Physics Explained
https://profoundphysics.com/how-are-photons-affected-by-gravity-if-they-have-no-mass/
Of course I disagree with the article’s main assertions: 1. Photons have no mass; and 2. Mass curves spacetime. Profoundphysics does a good job providing and explaining the mainstream math involved and how to use it.

Back to the discussion.
LongtimeAirman wrote:...nuclear matter will align its main n/s axis with the local dominant charge source. Objects in the dominant emission (gravitational) field will acquire energy and acceleration in the downward direction.
Jared wrote. I'm fine with the first statement at the atomic level anyway, but the second statement contradicts it and itself.

"dominant emission" in this case would be the Earth, correct? So if it's emitting UP, how does this cause an acceleration DOWN? ...
Airman. Yes, the “dominant emission" in this case would be the Earth. The proton aligns its n/s spin axis to the Earth directly below.

The proton will feel a net downward force not from the photons it receives and mainly channels as through-charge upward from the earth. The proton is accelerated downward by photons it receives from space, and incoming charge spiraling downward entering the topside of the proton.
.

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Post by LongtimeAirman Wed Jul 19, 2023 12:33 pm

.
26 February 2019 at 3:00 pm. Jared wrote. I'm actually stuck on this one, myself. I keep thinking back to what Nevyn said in his first response here, about the orbits of planets.

While I can kinda see how charge might "create" or cause gravity at, say, a planet's surface as the binding energy for matter, I don't yet see how this could apply to the planetary orbits around the sun. They aren't at the surface. They are very, very far away from it - so it seems like any binding energy should be long lost by the time it could reach a planet, even Mercury. What would cause Mercury then to move towards the sun?
Airman. There is no binding energy, gravity is not a “force of attraction”. Objects within the Sun’s gravitational field are pushed toward the Sun by charge approaching the Sun from the rest of the galaxy. The Sun provides some charge resistance to any large proton matter objects simply falling within its gravitational field, emitted solar charge will not stop that object from striking the Sun’s surface. Gravity keeps us on the Earth's surface. So far this sounds a lot like the old Push gravity idea.

The charge field has something the old Push gravity ideas lacked - lift.* As we know, objects traveling with a velocity orthogonal to the Sun’s center, constantly break the charge channels between the object and the Sun, re-directing the solar charge received outward creating lift.  
Mercury’s orbit is in balance between the Sun’s emission field and galactic charge headed toward the Sun, based on Mercury’s size and distance from the Sun, etc.

*
264a. Lift on a Wing. http://milesmathis.com/lift.pdf Plus extended comments on buoyancy and on the raindrop problem. 14pp.
264b. The Magnus Effect, Lift, and Charge. http://milesmathis.com/magnus.pdf Where we learn more about bending soccer balls as well as about lift on a wing. 3pp.
http://milesmathis.com/index.html
.

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Post by LongtimeAirman Wed Jul 19, 2023 9:01 pm

.
Jared, Please don’t think I’m trying to pick a fight with you, I'm not. You started this thread. If I’m trying to argue with anyone, it would be Miles.

I’ve re-read Miles’ paper, “The Cause of Gravity, the Next Chapter”. I didn’t understand it then and I don’t understand it now. I didn’t get the impression anyone else here understood "magnetic resonance" either.

Miles wrote.  I suppose it bears repeating that this theory is not a variant of push gravity. About the only thing Le Sage got right was proposing a field of corpuscles (photons). He was completely right there. Well, to be honest, he also got some other things right, such as that gravity is not a pull and that the mainstream was wrong. So in general he was on the right track. But the theory of blocking was far too naïve to answer data, which is why his theory never made much headway. However, the current mainstream theory of gravity (Newton's) is also far too naïve to answer data, and they have known that for centuries. As a field theory, it is nearly as oversimplified as Le Sage's.

Dismissing Le Sage’s push theory on the basis of it being too naïve is a bit rash. In my opinion the addition of charge field's  ‘Lift’ makes push gravity work perfectly. A clear explanation of Gravity as a function of charge. Much easier to understand than “magnetic resonance”.

My apologies everyone. I'll keep trying. I'll re-read More on Gravity again too.

NEW PAPER, added 2/20/19, http://milesmathis.com/grav3.pdf The Cause of Gravity, the Next Chapter. Possibly the most important paper I have published in several years.

NEW PAPER, added 3/3/19, http://milesmathis.com/grav4.pdf More on Gravity. This may clarify for some readers.
.

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Post by Jared Magneson Wed Jul 19, 2023 9:53 pm

LongtimeAirman wrote:
Jared, Please don’t think I’m trying to pick a fight with you, I'm not. You started this thread. If I’m trying to argue with anyone, it would be Miles.


I feel no animosity here at all, only genuine curiosity and desire to understand!

It just occurred to me as well that if we have charge "pushing" down and charge pushing up (from the Earth, for example), there would be some level of stasis or balance between the two net forces.

But if charge-gravity pushing down was caused by photons, why would it be so consistent? How could it stay at a near-constant 9.8m/s², all across the land and see, where different charge emissions FROM the Earth would affect that stasis/balance ratio?

And what about when the Earth is receiving more insolation, such as when we're closest to the sun (in January), or less insolation, in June? Wouldn't that affect gravity as well?

As far as I know, objects don't weigh more or less in January or June. I imagine such a variance would be detectable. But all that said, one might argue that if the Earth were receiving more insolation, it would also be receiving more through-charge as well, and offset that difference.

I just don't know. Kinda spit-balling, here.

And to answer your other question, I haven't done any new animations in quite some time. But I still CAN! It was the photon stacked-spin stuff that kinda... Broke me. It's so complex, and I wasn't able to script any of it really. Not in my program, Maya.

The last things I did were generate an actual gravity/charge balanced "orbit" and then illustrate the solution to the 3-Body Problem (charge, obviously). Neither of which are using charge-gravity, but here they are:

Orbit:
https://vimeo.com/538340408

3-Body Problem:
https://vimeo.com/327650447

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Post by LongtimeAirman Sun Jul 23, 2023 6:27 pm

.
Well I certainly needed and sure enough enjoyed, the review.  

After re-reading “The Cause of Gravity, the Next Chapter” and, “More on Gravity” I also re-read “Why Gravity is not a Function of Charge” *. Then I re-read them again. I’m over my confusion and embarrassment and feeling much better. Don’t know why my mind latched onto push gravity.  

During charge recycling, the Earth’s emission field has been split, compressed and reversed compared to downward, incoming solar charge. Gravity is a measure of binding force due to matter’s spin augmentation occurring where the emission and incoming charge fields meet, which falls off above the Earth surface according to gravity’s inverse square gradient. Or something like that.

Glad the only charge field change Lift needs is in explaining lift as a loss of charge binding with the Earth below, which sounds like breaking vertical charge channels to me.

Jared, Thanks for kindly stepping in.
I enjoyed your latest animations as well as some of your older ones including:
PhotonCharge_X2-spin Infrared_paths
Planetary Spin - Charge Dynamics
Alpha Particle - Charge Channels (Helium) v5
I've been thinking a lot about Helium and proton stacks too lately, wondering how binding energy, or gravity as charge, might effect them.

182b. Why Gravity is not a Function of Charge. http://milesmathis.com/gravmag.pdf  A short paper, explaining why I have unified gravity and charge, but not resolved them into one field. 3pp.
http://milesmathis.com/index.html
.

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Post by Jared Magneson Sun Jul 30, 2023 9:39 pm

Thank you so much for your patience and persistence, Airman. I too re-read all those papers and it does make sense to me now. I'm not sure I completely agree (not that it matters), but I can see where Miles is coming from a LOT better after some time and distance and a fresh look at it all.

I can't quite put my finger on my primary rebuttal, though. He definitely answered many of them, but I need to go back and re-read all of Nevyn's replies here to feel it out, again. It's good stuff, all around!

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Post by Chromium6 Sat Aug 12, 2023 2:43 am

Found this that might be kind of interesting to look at from 2013. Keep in mind Miles' Pyramid papers. They use LIDAR with oil/gas discovery which can give hints of what gravity is doing on the surface:

LIDAR for geospatial discoveries:
LIDAR
https://www.degruyter.com/document/doi/10.1515/geo-2020-0257/html?lang=en
Performance comparison of the wavenumber and spatial domain techniques for mapping basement reliefs from gravity data
https://www.degruyter.com/document/doi/10.1515/geo-2020-0321/html
https://www.bluefalconaerial.com/from-exploration-to-refining-lidars-impact-on-oil-and-gas-operations/

Salt Dome Gravity for changes...they use above and subsurface to find O/G:

MULTIDISCIPLINARY INVESTIGATION OF SURFACE
DEFORMATION ABOVE SALT DOMES IN HOUSTON, TEXAS

https://uh-ir.tdl.org/bitstream/handle/10657/636/Zheng%20Huang%20MS%20Thesis%20Geology.pdf?sequence=1
Abstract
Surface deformation has been an ongoing problem in the Houston Metropolitan
area because of the city’s location in a passive margin where faulting and subsidence are
common. According to previous studies the causes of the surface deformation are
typically attributed to anthropogenic activities, mainly the subsurface withdrawals of oil,
gas, and groundwater. However, the majority of the studies done have not accounted for
the vast amount of salt underneath the Houston area and its role in the surface
deformation. The objective of this study was to identify areas of surface deformation in
the greater Houston area and their possible relationship with subsurface salt movements.
To accomplish this, I integrated three kinds of data:
1) GPS,
2) LiDAR (Airborne and TLS), and
3) Gravity. GPS data revealed subsidence and uplift in Harris County. DEMs
generated from airborne LiDAR revealed changes between salt domes and their
surrounding areas. TLS data collected over the Pierce Junction site, chosen for
accessibility and depth, revealed vertical changes over the surface above the salt dome.
Gravity data acquired over Pierce Junction salt dome also revealed changes in the
subsurface. Groundwater withdrawal may be a large influence in the surface deformation
of the Houston area, but salt-related surface deformation should be more closely studied
to quantify its influence.

https://agupubs.onlinelibrary.wiley.com/doi/full/10.1002/grl.50838
Geophysical Research Letters

New ultrahigh-resolution picture of Earth's gravity field


Christian Hirt, Sten Claessens, Thomas Fecher, Michael Kuhn, Roland Pail, Moritz Rexer
First published: 12 August 2013 https://doi.org/10.1002/grl.50838Citations: 146
2 Data and Methods

[4] Our ultrahigh resolution picture of Earth's gravity field is a combined solution based on the three key constituents GOCE/GRACE satellite gravity (providing the spatial scales of ~10,000 down to ~100 km), EGM2008 (~100 to ~10 km), and topographic gravity, i.e., the gravitational effect implied by a high-pass filtered terrain model (scales of ~10 km to ~250 m).

[5] Regarding the satellite component, we use the latest satellite-measured gravity data (release GOCE-TIM4) from the European Space Agency's GOCE satellite [Drinkwater et al., 2003; Pail et al., 2011], parameterized as coefficients of a spherical harmonic series expansion, that currently provides the highest-resolution picture of Earth's gravity ever obtained from a space gravity sensor. Resolving gravity field features at spatial scales as short as 80–100 km, GOCE confers new gravity field knowledge, most notably over poorly surveyed regions of Africa, South America, and Asia [Pail et al., 2011].

[6] Compared to pure GOCE models, complementary GRACE satellite gravity [Mayer-Gürr et al., 2010] is superior in the spectral range up to degrees 70–80 [Pail et al., 2010]. Therefore, first, a combined satellite-only solution based on full normal equations of GRACE (up to degree 180) and GOCE (up to degree 250) is computed [see, e.g., Pail et al., 2010]. The GRACE/GOCE combination is then merged with EGM2008 [Pavlis et al., 2012] using the EGM2008 coefficients as pseudo-observations. Since for EGM2008 only the error variances are available, the corresponding normal equations have diagonal structure. In our combination, GRACE/GOCE data have dominant influence in the spectral band of harmonic degrees 0 to 180 with EGM2008 information taking over in the spectral range 200 to 2190, leaving the main spectral range of transition from GRACE/GOCE to EGM2008 in spectral band of degrees 181 to 200. The relative contributions of EGM2008 and GRACE/GOCE satellite gravity are shown in Figure 1.

Details are in the caption following the image
Figure 1
The Cause of Gravity - the next major chapter - Page 6 Grl50838-fig-0001-m

Relative contribution of GOCE/GRACE data per spherical harmonic coefficient in the combination with EGM2008 data (in percent) for the degrees 0 to 250.
[7] The spherical harmonic coefficients of the combined GRACE/GOCE/EGM2008 (GGE) gravity model were used in the spectral band of degrees 2 to 2190 to synthesize a range of frequently used gravity field functionals at the Earth's surface. For accurate spherical harmonic synthesis at the Earth's surface, as represented through the Shuttle Radar Topography Mission (SRTM) topography, the gradient approach to fifth order [Hirt, 2012] was applied. This numerically efficient evaluation technique takes into account the effect of gravity attenuation with height. Applying the gradient approach as described in Hirt [2012] yielded numerical estimates for radial derivatives (gravity disturbances) and horizontal derivatives (deflections of the vertical) of the disturbing potential and quasi-geoid heights from the GGE data set at 7.2 arc sec resolution (about 3 billion surface points) within the SRTM data coverage.

[8] For the Mount Everest region, Figure 2 exemplifies the associated resolution of GOCE/GRACE satellite gravity (a) and their combination with EGM2008 gravity (b). The spatial resolution of the GGE gravity field functionals is limited to about ~10 km (or harmonic degree of 2190) which leaves the problem of modeling the field structures at short scales, down to few 100 m resolution at any of the surface points.

Details are in the caption following the image
Figure 2
Open in figure viewer
PowerPoint
Gravity field at different levels of resolution over Mount Everest area. (a) Satellite-only (free-air) gravity from GOCE and GRACE satellites, (b) GGE gravity (satellite gravity combined with EGM2008 gravity), and (c) GGMplus as composite of satellite gravity, EGM2008, and topographic gravity. Shown is the radial component of the gravity field over a ~400 × 400 km area covering parts of the Southern Himalayas including the Mount Everest summit area (marked), units in 10−5 m s−2. The spatial resolution of the gravity modeling increases from ~100 km, ~10 km to ultra-fine ~200 m spatial scales.

[9] Because ground gravity measurements at a spatial density commensurate with our model resolution do not exist over most parts of Earth [e.g., Sansò and Sideris, 2013]—and will not become available in the foreseeable future—alternative solutions are required to estimate the gravity field signals at scales shorter than 10 km. High-resolution topography data is widely considered the key to ultrahigh-resolution gravity modeling and used successfully as effective means to estimate short-scale gravity effects [Sansò and Sideris, 2013; Tziavos and Sideris, 2013; Pavlis et al., 2012; Forsberg and Tscherning, 1981]. This is because the short-scale gravity field is dominated by the constituents generated by the visible topographic masses [Forsberg and Tscherning, 1981]. However, forward estimation of the short-scale gravity field constituents from elevation models near-globally at ultrahigh (few 100 meters) resolution is computationally demanding. Yet we have accomplished this challenge for the first time through advanced computational resources.

[10] Massive parallelization and the use of Western Australia's iVEC/Epic supercomputing facility allowed us to convert topography from the Shuttle Radar Topography Mission (SRTM) [cf. Jarvis et al., 2008]—along with bathymetric information along coastlines [Becker et al., 2009]—to topographic gravity at 7.2 arc sec resolution everywhere on Earth between ±60° latitude with SRTM data available. Based on nonparallelized standard computation techniques, the calculation of topographic gravity effects would have taken an estimated 20 years, which is why previous efforts were restricted to regional areas [Kuhn et al., 2009; Hirt, 2012].

[11] The conversion of topography to topographic gravity is based on the residual terrain modeling technique [Forsberg, 1984], with the topography high-pass filtered through subtraction of a spherical harmonic reference surface (of degree and order 2160) prior to the forward modeling. We treated the ocean water masses and those of the major inland water bodies (Great Lakes, Baikal, Caspian Sea) using a combination of residual terrain modeling with the concept of rock-equivalent topography [Hirt, 2013], whereby the water masses were “compressed” to layers equivalent to topographic rock. These procedures yield short-scale topographic gravity that is suitable for augmentation of degree 2190 spherical harmonic gravity models beyond their associated 10 km resolution [cf. Hirt, 2010, 2013]. The topographic gravity is based on a mass-density assumption of 2670 kg m−3 and provides the spatial scales of ~10 to ~250 m, which is complementary to the GGE gravity (spatial scales from ~10,000 km to ~10 km).

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