# Possible Charged Particle Field

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## Re: Possible Charged Particle Field

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Jared, I don't know about you, but I'm not script smart. I'd be more than happy to work with you to learn this app to the extent that you might need it. Helping you would help me learn it. I can't say I can teach it since there's a great deal about it I haven't begun to figure out and it's not finished yet. As Nevyn says, programming is a tool. It's more important to understand how to work with it.

I’ve got a tack for you Nevyn, but fair’s fair, you deserve credit. By expansion theory, we know charge particles are expanding. We also know, given an interval of time, a charged particles’ emissions will expand outward from the particle, as on an expanding spherical layer. Have you told us something we don’t know? I don’t think so. On the other hand, I believe this development of yours improves our understanding of relativity. Now we can say – of course! We can clearly see the both the expanding charge particle and its expanding emission field, easily adjusting for micro or macro.

I'm surprised you feel implementing the R function is a problem. I believe we already know the radius function R. All objects in the universe are expanding – accelerating - at the same rate. For Earth, the radius doubles every 19 minutes or so. Why there’s a post or two on the subject here at this site. As requested, just poking.

All radii and distances are time dependent variables – the relativistic perspective. We know that the particle radius should be time dependent, equal to a distance, r0 at some position p0 at time t0, r(t)= p(0) + r(0)*expansion(t). Normally one can overlook such details, but if you’re trying to model the universe, relativity must be included. Previously I could always claim I didn't understand relativity, but that excuse is worn pretty thin now, although you or Miles have always surprised me with things I hadn’t considered.

So the CPIM app already includes spin and a latitude function – great, it may go all the way. With all due respect, when I said you should use spin to tie all your formulas together, app aside, the physics comes first. If you’ve already built CPIM with these ideas in mind, not just a charge field but a relativistic charge field I’d say you’re doing a fine job.

Otherwise, as we're all aware, it's not quite done. It needs gravity - or this great new expansion alternative you're putting together.

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Jared, I don't know about you, but I'm not script smart. I'd be more than happy to work with you to learn this app to the extent that you might need it. Helping you would help me learn it. I can't say I can teach it since there's a great deal about it I haven't begun to figure out and it's not finished yet. As Nevyn says, programming is a tool. It's more important to understand how to work with it.

I’ve got a tack for you Nevyn, but fair’s fair, you deserve credit. By expansion theory, we know charge particles are expanding. We also know, given an interval of time, a charged particles’ emissions will expand outward from the particle, as on an expanding spherical layer. Have you told us something we don’t know? I don’t think so. On the other hand, I believe this development of yours improves our understanding of relativity. Now we can say – of course! We can clearly see the both the expanding charge particle and its expanding emission field, easily adjusting for micro or macro.

I'm surprised you feel implementing the R function is a problem. I believe we already know the radius function R. All objects in the universe are expanding – accelerating - at the same rate. For Earth, the radius doubles every 19 minutes or so. Why there’s a post or two on the subject here at this site. As requested, just poking.

All radii and distances are time dependent variables – the relativistic perspective. We know that the particle radius should be time dependent, equal to a distance, r0 at some position p0 at time t0, r(t)= p(0) + r(0)*expansion(t). Normally one can overlook such details, but if you’re trying to model the universe, relativity must be included. Previously I could always claim I didn't understand relativity, but that excuse is worn pretty thin now, although you or Miles have always surprised me with things I hadn’t considered.

So the CPIM app already includes spin and a latitude function – great, it may go all the way. With all due respect, when I said you should use spin to tie all your formulas together, app aside, the physics comes first. If you’ve already built CPIM with these ideas in mind, not just a charge field but a relativistic charge field I’d say you’re doing a fine job.

Otherwise, as we're all aware, it's not quite done. It needs gravity - or this great new expansion alternative you're putting together.

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## Re: Possible Charged Particle Field

No, I haven't shown anything we didn't already know, but I have used it in a more logical, consistent, and precise manner.

I don't know where that doubling every 19 minutes comes from, so I don't know if I can trust it. Doubling is also a relative term, so it doesn't really tell us anything about the actual radius. You also have to know what that 19 minutes applies to. You see, from a non-expanding perspective, both distance

Finding this change in radius will actually allow me to implement gravity at the same time. Since that change in radius

I don't know where that doubling every 19 minutes comes from, so I don't know if I can trust it. Doubling is also a relative term, so it doesn't really tell us anything about the actual radius. You also have to know what that 19 minutes applies to. You see, from a non-expanding perspective, both distance

**and time**expand, since time is based on distance. If the meter expands, then so does the second. This keeps the ratio the same in spite of the expansion, from an expanding perspective. I am trying to find math that fills the gap between what we see from an expanding perspective and what really happens from a non-expanding perspective. I am not exactly in either camp, but somewhere between them. I need to step through it with great care to keep everything in its proper place.Finding this change in radius will actually allow me to implement gravity at the same time. Since that change in radius

*is*gravity. I just need to account for time discrepancies. I'm looking over Miles Third Wave papers for equations I can use and have found a few, but I'm still trying to figure out exactly how he calculated an initial radius for the proton. Miles states that you can't just use the equations of motions to find an older radius, or you can but you can only use it as a proportionality. I found another equation this morning, but haven't had any time to play with it.**Nevyn**- Admin
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## Re: Possible Charged Particle Field

I found the pin!

While the above analysis is still applicable, I have started from the wrong position. It is not volume that I should be analyzing, but surface area. My method is still the same, but we start from SA = 4Pi * r^2.

Which gives us a 1/r^2 relationship.

Therefore, to find the charge at a given distance, we use: D(d) = C(1)/d^2

The relativistic version is: D(d) = C( R( -d/c ) )/d^2

Miles then went on to include gravity, which he said brings in another 1/r^2 relationship, so we end up with 1/r^4. However, I stand by my analysis and in the case of charge density, gravity must be included in the initial determination of charge density at the boundary of the emitter.

While the above analysis is still applicable, I have started from the wrong position. It is not volume that I should be analyzing, but surface area. My method is still the same, but we start from SA = 4Pi * r^2.

Radius | Area | A(r) / A(1) | A(1) / A(r) |

1 | 12.5663706144 | 1 | 1 |

2 | 50.2654824574 | 4 | 0.25 |

3 | 113.0973355292 | 9 | 0.11111111 |

4 | 201.0619298297 | 16 | 0.0625 |

5 | 314.159265359 | 25 | 0.04 |

6 | 452.3893421169 | 36 | 0.0277777778 |

7 | 615.7521601036 | 49 | 0.0204081633 |

8 | 804.247719319 | 64 | 0.015625 |

9 | 1017.8760197631 | 81 | 0.012345679 |

10 | 1256.6370614359 | 100 | 0.01 |

Which gives us a 1/r^2 relationship.

Therefore, to find the charge at a given distance, we use: D(d) = C(1)/d^2

The relativistic version is: D(d) = C( R( -d/c ) )/d^2

Miles then went on to include gravity, which he said brings in another 1/r^2 relationship, so we end up with 1/r^4. However, I stand by my analysis and in the case of charge density, gravity must be included in the initial determination of charge density at the boundary of the emitter.

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## Re: Possible Charged Particle Field

To be clear, charge density is a function of the volume of the emitter at the time of emission. It is the emission

*of*a volume,*over*a surface. However, the drop-off of that emission is only dependent on surface area. We are taking the emission of 1 dt, and watching it as it moves out into space. We don't care about the space between that emission and its source, which using volume would include, so we use the surface area equation.**Nevyn**- Admin
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## Re: Possible Charged Particle Field

That makes perfect sense to me, Nevyn. The math even makes sense, which is a good thing since I'm not great with these kinds of maths and would have never seen these relationships the way you do.

I apologize to just be a sort of cheerleader here, but it's still interesting and I'm trying to follow along as best I can. Wish I could actually help on this one though.

I apologize to just be a sort of cheerleader here, but it's still interesting and I'm trying to follow along as best I can. Wish I could actually help on this one though.

Last edited by Jared Magneson on Mon Aug 13, 2018 9:59 pm; edited 1 time in total

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## Re: Possible Charged Particle Field

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That makes sense to me. You may have noticed I've been thinking of your finding strictly in terms of surface area. In my last post I interpreted your 'emission volume' as an expanding dt surface - "We also know, given an interval of time, a charged particles’ emissions will expand outward from the particle, as on an expanding spherical layer". The entire emission volume would involve all subsequent time differentials which we don't need.

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That makes sense to me. You may have noticed I've been thinking of your finding strictly in terms of surface area. In my last post I interpreted your 'emission volume' as an expanding dt surface - "We also know, given an interval of time, a charged particles’ emissions will expand outward from the particle, as on an expanding spherical layer". The entire emission volume would involve all subsequent time differentials which we don't need.

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## Re: Possible Charged Particle Field

Yeah, I was thinking the same way, only really seeing it as the top layer of the volume, but still didn't put it together that I should be using the surface area instead. Miles had to remind me as I looked over one of his papers. But my method is sound and it was only the initial equation that needed to be fixed, so the general outcome is the same.

It is good to know that you can understand what I have written. That I am making sense. So you are doing a bit more than just cheer-leading. I want to show how I find these things, and why the math is the way it is, so others can see what

It is good to know that you can understand what I have written. That I am making sense. So you are doing a bit more than just cheer-leading. I want to show how I find these things, and why the math is the way it is, so others can see what

*being mechanical*means. So they can run with it themselves and maybe find things that everyone else has missed.**Nevyn**- Admin
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## Re: Possible Charged Particle Field

Let's have a look at the Radius function.

The radius function,

We use the equation of motion s = (at^2)/2 to find the displacement of the radius given the time it occurs over, and the acceleration of the radius. We already know the time, d/c, but we need to know the value of gravity. I have found differing values for the gravity of the proton and need to decipher which one is applicable and why. For now, we will let it stand as the variable

Let:

r = current radius of emitter

a = gravity of emitter

t = amount of time the motion occurs over

Then:

If t < 0

R(t) = r - at^2 / 2

Else

R(t) = r + at^2 / 2

Unfortunately, passing in a negative time does not work as well in the math as I wanted it to. The time is squared which means the negative sign is lost so it needs to be extracted from the time or used in a different way to avoid this. I am going to redefine the R function so that it doesn't need to handle that part.

Let

That leaves the subtraction from the current radius out of the

So

R(t) = (at^2)/2

Putting that into

D(d) = C( r - R( d/c ) )/d^2

= C( r - (a/2)(d/c)^2 )/d^2

Now we can see a bit more clearly why the relativistic equation reduces to the quantum version. I said earlier that as d becomes very small, the ratio of d:c approaches 0. In the above equation, that causes the change in radius to be so small that it can be thought of as just

The values for

D( r, a, d ) = C( r - (a/2)(d/c)^2 )/d^2

to reflect that

The radius function,

**R**, must return the radius of a particle given a time differential, t, where t is in unit-time, for example, the number of seconds. When t=0, the current radius is returned. When t>0, a future radius is returned. When t<0, a past radius is returned.We use the equation of motion s = (at^2)/2 to find the displacement of the radius given the time it occurs over, and the acceleration of the radius. We already know the time, d/c, but we need to know the value of gravity. I have found differing values for the gravity of the proton and need to decipher which one is applicable and why. For now, we will let it stand as the variable

**a**.Let:

r = current radius of emitter

a = gravity of emitter

t = amount of time the motion occurs over

Then:

If t < 0

R(t) = r - at^2 / 2

Else

R(t) = r + at^2 / 2

Unfortunately, passing in a negative time does not work as well in the math as I wanted it to. The time is squared which means the negative sign is lost so it needs to be extracted from the time or used in a different way to avoid this. I am going to redefine the R function so that it doesn't need to handle that part.

Let

**R**be a function that returns the change in radius given an amount of time, t, where t>=0.That leaves the subtraction from the current radius out of the

**R**function and we can deal with the minus sign because we know that we are moving backwards in time in this equation.So

**R**becomes:R(t) = (at^2)/2

Putting that into

**D**gives us:D(d) = C( r - R( d/c ) )/d^2

= C( r - (a/2)(d/c)^2 )/d^2

Now we can see a bit more clearly why the relativistic equation reduces to the quantum version. I said earlier that as d becomes very small, the ratio of d:c approaches 0. In the above equation, that causes the change in radius to be so small that it can be thought of as just

*being*the radius. i.e. r - (a/2)0^2 = r.The values for

**r**and**a**are treated as constants here, but they really represent the emitter. It might be a proton or it might be a star. For a given problem they may be constants but in others they will need to be specified. So I would rewrite the equation like this:D( r, a, d ) = C( r - (a/2)(d/c)^2 )/d^2

to reflect that

**r**,**a**, and**d**are all parameters. I'm not sure if that is how mathematicians specify such things but I know programmers will understand.**Nevyn**- Admin
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## Re: Possible Charged Particle Field

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After rereading a couple of Miles’ third wave and part of his long Mercury precession papers earlier today, I’d say you’re using the motion equation s = (at^2)/2, to find the displacement of the radius exactly as Miles has done.

Critically speaking, that’s all I’ve got.

You’re making charge field history. What next?

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I’m sure mathematicians and programmers would agree,So I would rewrite the equation like this:D( r, a, d ) = C( r - (a/2)(d/c)^2 )/d^2

to reflect that r, a, and d are all parameters. I'm not sure if that is how mathematicians specify such things but I know programmers will understand.

**D( r, a, d )**appears radical. The charge density received from a charged particle depends on the particle’s radius, gravity and distance. Standard math doesn’t list all the parameters used on the right side of an identity equation as an index, or whatever that is, on the left. I would say**a**- the gravity of the emitter - is the only emitter “constant”. It’s difficult to tell the difference between distances and an acceleration without accompanying units - meters, seconds or sec^2.After rereading a couple of Miles’ third wave and part of his long Mercury precession papers earlier today, I’d say you’re using the motion equation s = (at^2)/2, to find the displacement of the radius exactly as Miles has done.

Critically speaking, that’s all I’ve got.

You’re making charge field history. What next?

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## Re: Possible Charged Particle Field

Yeah, I thought it would be non-conformant, but I wanted to express that they

What's next?

Seriously though, I am working on the Charge Density function,

Question: What do you get if you take the volume of a sphere and divide it by the surface area of that sphere?

Conceptually, you should get the volume per unit-area.

A Dimensional Analysis tells you that the value is a meter (assuming you are using meters as your length units). It is 1 dimensional, not 3, so how does it represent a volume? Because it is a radius!

If you take the equation for the volume of a sphere (4/3 * PI * r^3) and divide it by the equation for the surface area of a sphere (4 * PI * r^2), you get r/3. Therefore, if my interpretation is correct, the volume per unit-area is equal to the volume of a sphere with a radius that is 1/3 of the original sphere's radius.

Does that sound reasonable?

Also, gravity. With the equation s = at^2/2, we can find the change in radius given an amount of time (which will be the time between frames but also include any time discrepancy as a result of distance separation) to find the amount of space that the other particles have taken up as a result of their expansion. We convert that into a vector that pulls our particle in that direction, by that amount. We do that for every other particle and sum them all together to get a resultant gravity vector. I'm not sure this is correct, since the other particles will expand regardless of what our current particle is doing. But in this math, 2 particles on either side of our particle will result in no motion. It doesn't feel quite right, but I can't see another way to do it, yet.

*may*be variables in some circumstances while they*may*be constants in others. This is especially true when dealing with stars, planets and moons. Every entity will have its own radius and gravity value, where-as if we are looking at protons, then they all share the same values for those. But electrons and neutrons will be different, so it isn't just a micro vs macro type of thing. Even for our purposes in this app, we will need to handle different values once we bring electrons into the mix. We could probably use the same values for protons and neutrons though.What's next?

Brain, in every episode of Pinkie and the Brain, wrote:The same thing we do every night, Pinkie, plan to take over the world!

Seriously though, I am working on the Charge Density function,

**C**, to figure out if charge density can be calculated in the way that I want it to be. I want an expression that gives us the charge density, but only for a unit-area of the emitter, that can then be expressed on the receiver.Question: What do you get if you take the volume of a sphere and divide it by the surface area of that sphere?

Conceptually, you should get the volume per unit-area.

A Dimensional Analysis tells you that the value is a meter (assuming you are using meters as your length units). It is 1 dimensional, not 3, so how does it represent a volume? Because it is a radius!

If you take the equation for the volume of a sphere (4/3 * PI * r^3) and divide it by the equation for the surface area of a sphere (4 * PI * r^2), you get r/3. Therefore, if my interpretation is correct, the volume per unit-area is equal to the volume of a sphere with a radius that is 1/3 of the original sphere's radius.

Does that sound reasonable?

Also, gravity. With the equation s = at^2/2, we can find the change in radius given an amount of time (which will be the time between frames but also include any time discrepancy as a result of distance separation) to find the amount of space that the other particles have taken up as a result of their expansion. We convert that into a vector that pulls our particle in that direction, by that amount. We do that for every other particle and sum them all together to get a resultant gravity vector. I'm not sure this is correct, since the other particles will expand regardless of what our current particle is doing. But in this math, 2 particles on either side of our particle will result in no motion. It doesn't feel quite right, but I can't see another way to do it, yet.

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## Re: Possible Charged Particle Field

Brain, in every episode of Pinkie and the Brain, wrote:

The same thing we do every night, Pinkie, plan to take over the world!

Lol!!!

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## Re: Possible Charged Particle Field

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Pinky and the Brain are about to implement gravity.

Curious about the outcome? See:

https://gifer.com/en/oaf

////\\\\////\\\\////\\\\////\\\\////

I thought we were going to implement gravity by redrawing expanding particles. The screen dimensions would need to be recalculated along with the acceleration of the expanding particles.

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Pinky and the Brain are about to implement gravity.

Curious about the outcome? See:

**animaniacs pinky and the brain**https://gifer.com/en/oaf

////\\\\////\\\\////\\\\////\\\\////

I thought we were going to implement gravity by redrawing expanding particles. The screen dimensions would need to be recalculated along with the acceleration of the expanding particles.

.

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## Re: Possible Charged Particle Field

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Sorry Nevyn, I don't understand your resultant gravity vector idea. If particles are outside each other's charge range will gravity still be calculated? Real expanding particles seems easier.

Concerning the reasonableness of the notion of a volume of a sphere divided by the sphere’s surface area. You’ve shown how they – the surface area and volume - are related. A “volume of particle emission” becomes an expanding spherical layer, a surface area with a differential time thickness dt. The further an object is, the smaller the amount of the expanding surface area the receiver will intercept. Sounds perfectly reasonable to me.

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Sorry Nevyn, I don't understand your resultant gravity vector idea. If particles are outside each other's charge range will gravity still be calculated? Real expanding particles seems easier.

Concerning the reasonableness of the notion of a volume of a sphere divided by the sphere’s surface area. You’ve shown how they – the surface area and volume - are related. A “volume of particle emission” becomes an expanding spherical layer, a surface area with a differential time thickness dt. The further an object is, the smaller the amount of the expanding surface area the receiver will intercept. Sounds perfectly reasonable to me.

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## Re: Possible Charged Particle Field

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http://physicsinsights.org/sphere-volume-1.html

I don't see a problem. Check your thinking against this site.Therefore, if my interpretation is correct, the volume per unit-area is equal to the volume of a sphere with a radius that is 1/3 of the original sphere's radius.

**Area and Volume of a Circle and Sphere Using Simple Geometry**http://physicsinsights.org/sphere-volume-1.html

.(14) Volume(sphere) = (1/3)*r*A

where A now represents the surface area of the sphere. ...

(15)The Surface(sphere) = 4Pi*r^2

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## Re: Possible Charged Particle Field

LongtimeAirman wrote:

I thought we were going to implement gravity by redrawing expanding particles. The screen dimensions would need to be recalculated along with the acceleration of the expanding particles.

No, we can't do that for at least 2 reasons. The first one mathematical and the second physical.

1) The numbers explode!

We are working within a certain precision which is set by the computer that the application is running on and the language used to implement it. In our case, we have 64bit floating point numbers in Javascript, but on a GPU, such as in a shader, we may only have 32bit floating point values. On a phone, we might only have 16bit numbers. No matter how large that is, it is still finite. We could be working with 1024bit numbers and it still would not be enough. If you just keep increasing the radii, then you will eventually, actually rather quickly, reach the upper limit of what your numbers can store.

There are ways around this limit. You can work with arbitrarily sized numbers that do not have an upper limit, except for the amount of memory in your system. However, they introduce a speed limit. The larger your numbers are, the slower they are to use in any calculation. Eventually, again rather quickly, the app grinds to a halt just trying to add 2 numbers together.

2) It doesn't represent the actual physics.

A 3D system is a Newtonian system, rather than a Relativistic one. The 3D engine will render all geometry

*as it is now*, no matter how far it is from the camera. It does take distance into account to make things smaller the further they are from the camera, but it does not take the time discrepancy that that distance represents into account. We can make it do that, but only with considerable effort, and it is not as easy as it first looks.

You see, it isn't just the size that we see in the past, it is also the position (and velocity, spin, etc). When we look at the Sun from the Earth, we are seeing it at a size that it was 8 minutes ago, but we are also seeing it in the position that it was in 8 minutes ago. This is also a problem for charge calculations. My equation above takes care of the time discrepancy by using a smaller radius to calculate the initial emission, but it does not take the position difference into account. In its current state, it can't do that. In fact, it might not be feasible to do that because

*while*that charge was travelling from the source to the receiver, the emitter could have had a collision and changed trajectory. We have no way of knowing that.

So we are left to find another way. It isn't going to be perfect, but we may be able to get pretty close and hopefully minimize the differences.

My idea is to treat everything as if it is always at the same size, but use adjusted values from all other particles that the current particle is interacting with. You always need a reference point, which is the current particle for us, so that you can determine how things look from that point. They may look completely different from a different point, but that is irrelevant to our current particle, for it is not over there, it is here.

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## Re: Possible Charged Particle Field

LongtimeAirman wrote:Sorry Nevyn, I don't understand your resultant gravity vector idea. If particles are outside each other's charge range will gravity still be calculated? Real expanding particles seems easier.

I'm not sure about that. There is some distance where the change becomes so small as to be irrelevant. We can choose to find that distance and ignore everything outside of it just like we are with charge. Or we might just add it in anyway. It will probably be just as much computational effort to determine if it is too far away as it is to just calculate and add it in, which is not the case with charge as that is expensive to calculate.

LongtimeAirman wrote:Concerning the reasonableness of the notion of a volume of a sphere divided by the sphere’s surface area. You’ve shown how they – the surface area and volume - are related. A “volume of particle emission” becomes an expanding spherical layer, a surface area with a differential time thickness dt. The further an object is, the smaller the amount of the expanding surface area the receiver will intercept. Sounds perfectly reasonable to me.

Not quite correct. Or it is in some ways, but maybe a bit misleading. The further an emitter is from the receiver, the

*larger*the surface area is at the location of the receiver. That is, if we calculate a unit-area on the emitter at the time of emission, then that unit-area expands as that emission travels away from the emitter. That is what the outer 1/d^2 is doing in my equation. It is finding the amount of emission over the

*same*unit-area on the receiver and the d^2 reduces the charge density to reflect the same unit-area at the receiver. The unit-area stays the same and we let the charge density change but we could just as easily find the new area that the emission covers at the receiver.

In fact, I think we should do that. We should let the emission travel out uninhibited until it reaches the receiver. We then calculate the size of the original unit-area at that distance. Then we divide that, let's call it the received-area, by the area that it interacts with on the receiver. This creates an equation that calculates the interaction of one area, on the emitter, and another area, on the receiver. Therefore, it can handle the difference in area that the receiver has with respect to the emitter. I might investigate that at some point. It may just lead to the same 1/d^2 relationship, but it is worth finding out.

LongtimeAirman wrote:I don't see a problem. Check your thinking against this site.

Area and Volume of a Circle and Sphere Using Simple Geometry

http://physicsinsights.org/sphere-volume-1.html

I knew the math was correct, not because I searched for it, but because I plugged some values into it and saw the relationship directly. What I am unsure of is my interpretation of it being the radius of a sphere to represent a volume. That site does not explicitly state it, but I was using the same concepts as they are (as in a pyramid-like shape from the surface to the center).

What I am trying to get at with that analysis, is that it is the whole volume that is emitting, not just the surface, even though it only emits

*at the surface*. How can that be? Easy, because not all of that emission was created (for want of a better term) at the same time, but it is all emitted from the surface at the same time.

This is easier to see with stars and planets, than protons or small particles. I am actually wondering if there is a different equation for particles as compared to composite entities like stars. Anyway, with a star or planet, we have a body made up of smaller particles and it is each of those particles that is actually emitting. The sum of that emission becomes the emission of the body. However, the emission that came from the very center of that body is older than the emission that came from the surface particles. But age doesn't really matter. It is the sum of emission at the surface that we care about because that is the amount of emission that will travel away from the emitter and eventually collide with a receiver.

That gets us the total charge emission from the volume (with an assumption that the whole sphere is emitting, which we know is incorrect) at the surface. I then divided that by the surface area to get to a unit-volume or the amount of volume that is behind a unit-area. That should represent the amount of emission that is moving towards the receiver, with some margin of error. What we really should do, is determine the surface area of reception, translate that into an area of emission and use that to calculate the amount of emission. I may investigate

*that*as well.

**Nevyn**- Admin
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