# Relativistic Coulomb Force

Discussion in 'Physics & Math' started by tsmid, May 16, 2017.

1. ### przyksquishyValued Senior Member

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Well if you actually need the clarification, then when I've said "integral equation" (for the potential) I've always meant the equation

$\displaystyle \Phi (\boldsymbol{r}, t) = \frac{1}{4 \pi \epsilon_{0}} \int \mathrm{d}^{3}r' \, \frac{\rho \Bigl(\boldsymbol{r}',\, t - \frac{\lVert \boldsymbol{r} - \boldsymbol{r}' \rVert}{c} \Bigr)}{\lVert \boldsymbol{r} - \boldsymbol{r}' \rVert}$​

or a different but equivalent form of it. This is essentially the only general equation I've been referring to that involves the scalar potential and has an integral in it, so I thought "integral equation" should be unambiguous as well as much easier than copying the entire equation every time I want to refer to it (which has turned out to be quite a lot).

But it's not the name I care about. You can suggest a better name if you'd like.

You seem to be missing the point that you are resorting to handwaving and assumptions which I have already personally disproved in this thread using mathematics. The effect of the retarded time appearing in the integral above is significant enough that it gives a potential proportional to $q / (R - \boldsymbol{\beta} \cdot \boldsymbol{R})$ instead of $q / R$ for a point charge. I have justified that personally and more than once in this thread by direct calculation.

You're just wasting time at this point. If you think there is an error in the calculations I've posted then point out specifically at which step there is an error and what that error is in, for example, post #171. If you can't do that then you are in no position to dispute the result.

That is hypocritical. I have been able to justify mathematical relations I have depended on in this thread. You have not and have generally resorted to handwaving and taking your own assumptions for granted rather than do proper calculations. (Including assuming my calculations must be wrong instead of being open to the possibility that you could understand and learn something from them.) The previous quote from your post is a perfect example.

One thing in particular: despite your "nothing more and nothing less" comment above, you are not taking the integral above for what it is but instead seem to be assuming it behaves like a sum

$\displaystyle \int \mathrm{d}^{3}r' \, \frac{\rho(\boldsymbol{r}',\, t_{\mathrm{R}})}{R} \approx \sum_{i} \frac{q_{i}}{R_{i}}$​

over charges $q_{i}$ with associated retarded distances $R_{i}$. (You've never said this explicitly, but your posts make little sense unless you are assuming something like this.) Treating the integral like this is mathematically unjustified and wrong because (e.g., in the 1 dimensional version) $\rho(x', t_{\mathrm{R}}) \, \Delta x'$ is generally not a good approximation for the charge between the coordinates $x'$ and $x' + \Delta x'$ at the varying retarded time (for instance, along the diagonal dashed line in the figure in post #172).

Which concepts need clarifying?

Unlike you I've tended to define the quantities I use. For example, in post #171 where I calculate the potential for a moving line of charge, I stated right at the beginning what the charge density function was that I was calculating the integral for (something you didn't bother to do in post #168). So if you objected, you could have quoted what you thought was wrong and what you thought the function should be instead. But instead you chose to make vague claims about some issue with charge density and changing lengths that makes no direct reference to the derivation I provided.

For more basic things, like what the fields $\Phi$ and $\rho$ are, I am just using standard concepts and definitions from electromagnetism. (To be blunt, the sort of thing you should already know before you start arguments about electromagnetism on the internet.) For instance, the charge density function $\rho$ represents the amount of charge per unit volume at any given position and time. It is defined so that the volume integral $\int_{V} \mathrm{d}x \, \mathrm{d}y \, \mathrm{d}z \, \rho(x, y, z, t)$ is equal to the total charge contained in the volume $V$ (the integration domain) at the time $t$, for any volume $V$. (Importantly: this integral is taken at constant time, not for varying retarded time like in the integral for the potential above.) This is the same function $\rho$ that appears in the differential form of Maxwell's equations and elsewhere in electromagnetism.

Last edited: Aug 5, 2017

3. ### tsmidRegistered Senior Member

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368
You should call it just 'integral'. Calling it 'integral equation' hurts anybody who has actually worked with integral equations (as I happen to have extensively in the past, see for instance http://www.plasmaphysics.org.uk/radiative_transfer.htm ), and for those who have not come into contact with it, it is just potentially misleading. I am actually surprised you are not aware of this given your emphasis on mathematical formalities in all your posts.

It is not an assumption. Charges are discrete particles and you get their total field by summing up the contributions of the individual particles. For a large number of particles you may use the approximation of a continuous density function (provided you normalize it correctly), but this hardly makes sense for just 1 or 2 particles. In that case you just consider a charge q at the corresponding location and work out the potential from this from Coulomb's law. This avoids having to deal with any normalization issues as the charge q must be invariant by default.

The total charge must also be preserved for the retarded density distribution. Since for each particle there is exactly one retarded location (after all, a particle can not be in two places at the same time), the retarded distribution is just a spatial redistribution of the original (instantaneous) distribution. And all that is relevant here is the spatial locations. For the determination of the potential, you can completely ignore the time dependence of ρ if it describes the correct locations.

So you are maintaining then that the potential Φ(x,t) due to a charge q at the retarded position x'=L/2/(1-v/c) at the retarded time t-|x-x'|/c is different from that of another charge q at the same location but at time t and assuming instantaneous interaction?

5. ### przyksquishyValued Senior Member

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3,150
You're only confirming what I said: you are making assumptions which are unjustified and discredited. The formula for the potential is $\frac{1}{4 \pi \epsilon_{0}} \int \mathrm{d}^{3}r' \, \frac{\rho(\boldsymbol{r}',\, t_{\mathrm{R}})}{R}$. It is not $\frac{1}{4 \pi \epsilon_{0}} \sum_{i} \frac{q_{i}}{R_{i}}$. You are not a priori justified to interpret the former as if it "really meant" the latter and I have already proved it doesn't work that way.

If you put aside your preconception and treat the formula like an integral, seeing as that is what it actually is, you find that it implies $\frac{1}{4 \pi \epsilon_{0}} \, \frac{q}{R - \boldsymbol{\beta} \cdot \boldsymbol{R}}$ for a moving point charge and, by extension, $\frac{1}{4 \pi \epsilon_{0}} \sum_{i} \frac{q_{i}}{R_{i} - \boldsymbol{\beta}_{i} \cdot \boldsymbol{R}_{i}}$ for a collection of moving point charges.

Your objection about matter being composed of particles is irrelevant and unjustified. Classical electromagnetism is not nowadays considered to be a completely accurate model of reality anyway, and while matter does happen to be composed of particles (of a certain kind), the theory of electromagnetism itself does not assume or require this and certainly does not give point charges any special status. As far as I know the history, the atomic nature of matter wasn't even considered a confirmed fact at the time electromagnetism was formulated (mid-to-late 19th century) like it is today, let alone that whatever particles matter might be made of must necessarily be true points of zero size. (Even today that is only "known" according to theory and for a certain meaning of "point particle" used in modern quantum field theories.)

At the mathematical level, point charges can be modelled with Dirac delta distributions (or, equivalently, as small regions of charge in the limit with the size taken to zero), so the integral handles point charges just fine as a special case.

This is irrelevant nonsense. In terms of $\rho$, charge conservation means that the total charge $Q = \int \mathrm{d}^{3}r \, \rho(\boldsymbol{r}, t)$ must be independent of the time $t$. $\rho$ needs to be specified in advance in order to apply the integral for the potential. Calculating that integral does not change $\rho$ and does not change whether it respects charge conservation or not.

Total charge is also "preserved" in the sense that it is the same in different reference frames. Since we are only considering one reference frame that meaning of "preserved" is also not relevant even to begin with here.

If you apply the integral to $N$ point charges (e.g. by making $\rho$ the sum of $N$ Dirac delta distributions) then the integral reduces to the sum of $N$ terms, one corresponding to each point charge. It's just that each charge is multiplied by a factor $(R_{i} - \boldsymbol{\beta}_{i} \cdot \boldsymbol{R}_{i})^{-1}$ instead of $R_{i}^{-1}$ in the sum. This has nothing to do with preservation of charge.

I'm not sure that is even a meaningful question. In classical electromagnetism (the theory based on Maxwell's equations that everyone learns in university), interactions propagate at a finite speed limited by $c$. In order to make a comparison you would first need a different theory of electromagnetism that predicts instantaneous interactions to compare with. (You could maybe consider trying to construct such a theory by taking electromagnetism and taking the limit $c \rightarrow \infty$, but I'm not sure that is all that meaningful, since the numerical value of $c$ depends on the units and you can always choose the units such that $c = 1$.)

If you just mean something like using $t$ instead of $t_{\mathrm{R}}$ in the integral for the potential then yes, that does make an important difference. I don't know why you need to ask, since I calculated that in earlier posts, as I keep telling you. You can refer to those posts for the details.

Last edited: Aug 5, 2017

7. ### tsmidRegistered Senior Member

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368
I really don't know on what basis you are saying this. If you have discrete particles, then the only exact representation is given by the discrete sum of the individual potentials ( see for instance http://farside.ph.utexas.edu/teaching/em/lectures/node28.html ). The integral representation would only be exact if you actually had a continuous charge distribution, which is not the case. So in reality the integral form is only an approximation useful if you consider contributions from volumes containing many particles (and that is assuming that you normalize the function correctly to reflect the correct total number of charges). It does not make any sense for just a few particles or if you are interested in the small scale field between the particles.

Well, then you probably have the reason here for the wrong and misleading concepts that have crept into electrodynamics and that you are repeating here.

The Dirac delta distribution is also based on continuous functions. It just mimics a discrete function as a limiting process.

It is not nonsense. Even in everyday life, if you see two persons at different distances approaching you, you see them actually at different times due to the finite propagation time of the light signal. And with whatever speed the persons are moving, you will at any time always see 2 persons, not 1, not 3, and neither 2.1 or 1.9 persons. So the retardation does not change anything about the total number of persons you see. Likewise the retardation does not change anything about the total number of charges you see here. The integral of the 'retarded density' distribution must be the same as that over the non-retarded contribution.

8. ### przyksquishyValued Senior Member

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3,150
On the basis that the equation is the integral I wrote and not the sum you seem to have mistaken it for. This is what I used in earlier posts and it is how you'll find the equation expressed in any good reference on electromagnetism. It is never written as a sum.

Mathematically, $\frac{1}{4 \pi \epsilon_{0}} \int \mathrm{d}^{3}r' \, \frac{\rho(\boldsymbol{r}',\, t_{\mathrm{R}})}{R}$ is not the same thing as $\frac{1}{4 \pi \epsilon_{0}} \sum_{i} \frac{q_{i}}{R_{i}}$. Not even approximately, as I have shown in previous posts. It is a mathematical error to write one where you meant the other.

As to why it is an integral, that is justified by context and the way it is derived.

In my earlier posts in this thread, I was mainly interested in the integral mathematically as a generalisation of the Liénard-Weichert potential, in the sense that $\frac{q}{R - \boldsymbol{\beta} \cdot \boldsymbol{R}}$ can equivalently be written $\int \mathrm{d}^{3}r' \, \frac{\rho(\boldsymbol{r}',\, t_{\mathrm{R}})}{R}$ for the special case $\rho(\boldsymbol{r}, t) = q \, \delta(\boldsymbol{r} - \boldsymbol{v} t)$. So my own justification for introducing the integral is that it is mathematically equivalent to another equation we happened to be discussing. I justified that mathematical equivalence using different methods in posts #137 and #147 and in the one-dimensional case in post #171. (Why do I need to keep reminding you of this?) I refer you to those posts for the details.

In textbooks and courses on electromagnetism, it and similar integrals are usually derived from Maxwell's equations. Maxwell's equation for the scalar potential in the Lorentz gauge is

$\displaystyle \Box \Phi = \frac{\rho}{\epsilon_{0}}$​

where $\Box = \frac{1}{c^{2}} \frac{\partial^{2}}{\partial t^{2}} - \frac{\partial^{2}}{\partial x^{2}} - \frac{\partial^{2}}{\partial y^{2}} - \frac{\partial^{2}}{\partial z^{2}}$ is the d'Alembertian operator. This happens to be a wave equation with source term. The integral for the potential is a general solution in the case where the function $\rho$ is known and you want to solve for $\Phi$. The sum $\frac{1}{4 \pi \epsilon_{0}} \sum_{i} \frac{q_{i}}{R_{i}}$ you have mistaken the integral for is not a solution to this wave equation except in the static case.

This is false, since you can represent discrete particles with Dirac delta distributions, as I just told you.

There are two problems with your webpage reference.

First, it is only concerned with electrostatics. This is the only situation in electromagnetism where the Coulomb force law is considered generally valid. This is also the only situation where retardation happens to make no difference.

Second, writing the Coulomb force as a sum for discrete particles does not imply it is the only way to accurately represent that situation or that everything in electromagnetism has to be done that way. Your webpage reference claims no such thing and thus does not support you on this.

Classical electromagnetism is not a theory about what matter is really like at the microscopic level and it is not founded based on the idea of particles. It is a classical field theory founded on Maxwell's equations. These are field equations involving the charge and current distribution functions alongside other fields.

I don't particularly want to start doing quantum electrodynamics calculations. Do you? If not then I would propose not worrying too much what matter is really like at the atomic level. I didn't take that to be what this thread is really about anyway.

It seems to me you were under the impression that electromagnetism is fundamentally based on the idea of charged particles and are reacting poorly to finding out you were wrong about this.

Which concepts would those be that are not addressed in more recent and more accurate theories?

What has "crept" into electromagnetism? It is a theory from the nineteenth century, formulated in essentially its modern form by Maxwell.

Regardless, whatever you think of the physical correctness of electromagnetism, I'll remind you it was you who wanted to start a discussion about this theory and not some other theory. It makes no sense for you to start a thread asking questions about electromagnetism and then insist I should use some sum equation in my answers to you that is not considered generally valid in electromagnetism.

Like I just told you, if you apply the integral to a situation with discrete particles then the integral reduces to a sum of terms with one term per particle. It's the same calculation I did in my earlier posts, except with $N$ point particles instead of just one.

(A detail: this is only necessarily true assuming the particles always move slower than the speed of light, otherwise mathematically you can have situations where a particle gets counted more than once, though it is usually taken for granted that things aren't moving faster than light, like in everyday life.)

Assumption. You have not justified anything of the sort.

Last edited: Aug 7, 2017
9. ### tsmidRegistered Senior Member

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368
As far as the static Coulomb potential is concerned (which we are discussing here), it is (at least in some respects) approximately the same if you take the distances in the discrete sum as the retarded distances (which is how you obviously would apply this), assuming of course you have succeeded in fitting a suitable (and normalized) continuous density distribution ρ to the actual distribution of discrete particles.

The fact that your whole argument crucially depends on representing the charge distribution as a continuous one should give you already a hint that there is an error in your derivation related to this assumption.

As I said already, the Dirac delta distribution does not exactly represent a discrete particle. It only mimics one as a limiting process of some suitable continuous function.

It doesn't have to be done that way (not in this case here anyway), but if you do it differently, you have to satisfy some constraints that are consistent with the discrete particle model.
As for the reference I gave ( http://farside.ph.utexas.edu/teaching/em/lectures/node28.html ), they introduce the continuous distribution ρ with the sentence "Suppose that, instead of having discrete charges, we have a continuous distribution of charge represented by a charge density ρ". This clearly introduces the continuous distribution as a completely hypothetical mathematical construct without any connection to the physical reality.

We don't have to go to the atomic (let alone sub-atomic) level. In extragalactic space for instance, electric charges are spaced about 1m apart. I wonder how you want to model the charge distribution there as a continuous function on this scale.

Density distributions in general are certainly useful concepts in macroscopic physics, that is if the physical parameters of interest are determined statistically by a large number of particles which would be too difficult or impossible to consider on an individual particle basis. But if the considered scale becomes small enough so that only a few particles contribute, this concept not only doesn't make sense anymore but can lead to serious errors in physical models if used regardless.

I have not asked you to use the discrete sum method, only to adjust your integral method such as to arrive at the same result and conclusion (or at least consider why you arrive at a different result).

You still did not use discrete sums involving the individual charges q but the density distribution ρ, and this in a way that does not preserve the charge for the retarded distribution.

It is not an assumption. As the retardation only changes the apparent location of each particle, it follows logically that the total number of charges for the retarded distribution is the same as for the unretarded one.

Last edited: Aug 7, 2017
10. ### przyksquishyValued Senior Member

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3,150
I don't know why you need to ask or guess since this is easy. If you have a number of discrete charges $q_{i}$ at rest at positions $\boldsymbol{r}_{i}$ then the charge density is

$\displaystyle \rho(\boldsymbol{r}, t) = \sum_{i} q_{i} \, \delta(\boldsymbol{r} - \boldsymbol{r}_{i}) \,.$​

The total charge is $\sum_{i} q_{i}$.

In units where $4 \pi \epsilon_{0} = 1$, the potential works out to

$\begin{eqnarray} \Phi(\boldsymbol{r}, t) &=& \int \mathrm{d}^{3}r' \, \frac{\rho(\boldsymbol{r}',\, t_{\mathrm{R}})}{\lVert \boldsymbol{r} - \boldsymbol{r}' \rVert} \\ &=& \sum_{i} q_{i} \int \mathrm{d}^{3}r' \, \frac{\delta(\boldsymbol{r}' - \boldsymbol{r}_{i})}{\lVert \boldsymbol{r} - \boldsymbol{r}' \rVert} \\ &=& \sum_{i} \frac{q_{i}}{\lVert \boldsymbol{r} - \boldsymbol{r}_{i} \rVert} \,. \end{eqnarray}$​

This is easy because the retarded time immediately drops out of the calculation in this case, because the distribution $\rho(\boldsymbol{r}, t)$ here doesn't depend on $t$.

You can do the same for any number of discrete moving charges. The derivation is the same as what I already posted before, just done for $N$ charges instead of one.

Why? Charge distributions appear everywhere in electromagnetism, including Maxwell's equations which are part of the definition of the theory. Your objection to using them is arbitrary.

I don't know why you keep talking about distributions being "continuous". Nobody ever said $\rho$ necessarily has to be continuous. The charge distribution function for the line of charge I considered in post #171 is not a continuous function for example. The Dirac delta also isn't continuous. Technically it is also not a function in the usual sense (it is a distribution in the sense defined in distribution theory).

The derivative does not exactly represent the gradient of a function. It only mimics it as a limit of suitable approximations $\frac{f(x + h) - f(x)}{h}$.

No it doesn't. And just how, exactly, would you or whoever wrote the website know what the "physical reality" is in the first place? You have a bizarre obsession with particles. When, in real life, have you ever known for a fact that a particle is exactly a point particle of zero size?

Like I said, classical electromagnetism is not intended to be an exact model at the level of fundamental particles. It doesn't even attempt to model the photon, for example. When we use things like the Coulomb force law or Dirac delta distributions for particles in electromagnetism, it is not because we know for a fact that particles are perfectly pointlike or because we believe this is a perfectly accurate representation of the situation. The reason we do this is because if we're calling something a "particle" the implication is that it is small enough that we don't care about and perhaps don't even know its exact size, if it even has a size. In such a case, taking the particle to have zero size is usually the simplest thing to do and usually as good an approximation as any.

That would make no sense in the context of electromagnetism or of this thread. We were initially discussing the Liénard-Weichert potential and related equations and their implications. I cited the retarded potential as one of the well-known, general relations in electromagnetism that implies the Liénard-Weichert potential mathematically as a special case. Your imagined sum $\sum_{i} \frac{q_{i}}{R_{i}}$ has nothing to do with any of this. It is neither a postulate nor a derived result in electromagnetism and it is not relevant to the discussion we were having before.

I cannot "adjust my integration method" in order to make the integral the same as a sum $\sum_{i} \frac{q_{i}}{R_{i}}$. The only ways to get that result are to compute the integral incorrectly or to start with a different integral.

I've already given three complete derivations explaining why the integral gives the value it does for a moving point charge. Why do I have to keep reminding you of this?

It is worthless to say "it follows logically" when you are unable to give a logical derivation.

You are claiming that the integral $\int \mathrm{d}^{3}r' \, \frac{\rho(\boldsymbol{r}',\, t_{\mathrm{R}})}{R}$ behaves like a sum $\sum_{i} \frac{q_{i}}{R_{i}}$. That is a mathematical claim. Logically it can be confirmed or disproved on mathematical grounds. The result is not just whatever you wish it to be. I have already done the relevant mathematical analysis for this multiple times, which I have already posted here, and my conclusion on the matter is based on this. You have not done this.

Last edited: Aug 8, 2017
11. ### tsmidRegistered Senior Member

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368
Only that your integral does not give the same result as the discrete sum anymore if the charges are moving.

Your argument is a contradiction in terms. Any distribution function describes the distribution of something. If you have 10 charges with charge q distributed over a distance of 1 cm, you can say that the linear charge density is 10 q/cm.But that doesn't mean the charge of 10 q is distributed as some kind of continuous medium over the 1 cm. Maxwell certainly did not suggest this. He was well aware for instance that his Maxwell distribution function is only a statistical description of the velocities of molecules due to mutual collisions. That's indeed how he derived it. You could not have collisions or any kind of particle kinematics with a kind of continuous medium that you seem to be assuming here.

The error you are making is to mistake the distribution function for the actual distribution itself, when in fact it characterizes only certain statistical averages of the latter.

Again, you are contradicting yourself here. Your whole argument so far relied on having a continuous charge distribution ρ(x') over the range considered, because with this you could bypass the requirement of charge invariance that would automatically results from the assumption of discrete particles distributed over the range.

The situation is not as simple as that here. You may suitably approximate the location of a discrete particle by means of a delta distribution, but if applied to the retarded location, you get the charge of the particle wrong here (as I have pointed out now already repeatedly).

a) it is your claim, not mine. I don't require the integral over the charge distribution ρ in order to calculate the potential of a number of point charges.

b) logic does not necessarily require formal mathematics, rather on the contrary. If you have N particles in a volume and merely re-arrange their locations in this volume, you still have N particles in the volume. This is a logical conclusion that holds by design (so effectively it is a tautology). If your maths gives a different number after re-arranging then it must be erroneous.

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13. ### przyksquishyValued Senior Member

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3,150
Nor is there any reason it should.

"Atom" derives from a word that means "indivisible" in Greek. What's your point? You're committing the etymological fallacy. There is no lower limit to the resolution with which you can describe the way charges are distributed, formally to the level of point particles using Dirac deltas if you like.

No, I did mathematics, showing that one equation is mathematically a special case of another equation. You are reading meaning into this that just doesn't exist.

You are effectively claiming that the value of an integral is something different than the result of computing the integral. I don't see how you can possibly consider that logical.

My "maths" computes the scalar potential. Not the number of particles and not the charge. These are just whatever they are, tautologically, like you say, and they need to be specified in advance before you can even calculate the integral.

I saved this for last, since I think it could be the best way to settle this:
So why don't you do it? You claim the integral approximates $\sum_{i} \frac{q_{k}}{R_{k}}$. I claim it approximates $\sum_{k} \frac{q_{k}}{R_{k} - \boldsymbol{\beta}_{k} \cdot \boldsymbol{R}}$. It would be quite easy to compute either of these for a few hundred (or a few hundred million) particles on a computer.

I've already given an example in the one-dimensional case not using Dirac deltas that you can test against. In post #171 I considered a line of charge of length $L$ with total charge $q$ moving with velocity $v$. This is described by a charge density function which is $\rho(x, t) = q/L$ if $x$ is between $vt - L/2$ and $vt + L/2$ (the end points of the line at time $t$) and $\rho(x, t) = 0$ everywhere else. In that case, for $t = x/c$ and $x > \frac{L/2}{1 - v/c}$ (to ensure $x$ is outside the line) I showed that the integral for the potential gives the exact result

$\displaystyle \Phi(x, x/c) = \int_{-\infty}^{\infty} \mathrm{d}x' \, \frac{\rho \bigr(x',\, \tfrac{x}{c} - \tfrac{1}{c} \lvert x - x' \rvert \bigr)}{\lvert x - x' \rvert} = \frac{q}{L} \Biggl[ - \log \biggl(1 - \frac{L/2}{(1 - v/c) x} \biggr) + \log \biggl( 1 + \frac{L/2}{(1 - v/c) x} \biggr) \Biggr]$​

where $\log$ is the natural logarithm. Do you accept this is the result of simply applying the integral in this case? (If not, can you find an error in post #171 and/or explain how you derived a different result, and preferably both?)

You could also consider a moving line of charge of length $L$ made of $N$ discrete particles each with charge $q/N$, with particle $k$ following the trajectory (say)

$\displaystyle x_{k}(t) = v t + (2 k - N - 1) \frac{L}{2N}$​

for $k \in \{1,\, 2,\, \dotsc,\, N\}$ (this divides the line into $N$ segments of length $L/N$, with a charge in the middle of each segment).

According to you, the potential (with the same condition on $x$) should be

$\displaystyle \Phi(x, x/c) = \frac{q}{N} \sum_{k} \frac{1}{x - x_{k}(t_{k})}$​

where $x_{k}(t_{k})$ is the retarded position for particle $k$ when it crosses the line $x = c t$, i.e., the position at the time $t_{k}$ such that $x_{k}(t_{k}) = c t_{k}$.

Do you stand by this? Are you claiming that, for large $N$, you would get approximately the same result from computing this sum as I did in post #171 from the integral?

This is a purely mathematical/computational question. There should be no distractions here about distributions being "statistical" or integrals counting weird numbers of particles or not respecting "preservation of charge" or the like. I am simply asking whether you claim a certain integral I already happen to have worked out is approximately equal to the discrete sum you seem to keep insisting it is.

Last edited: Aug 9, 2017
14. ### tsmidRegistered Senior Member

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Then prove it without using any integral over some hypothetical mathematical functions that you claim represent the ensemble of discrete charges.

It is a question of doing maths correctly. Mathematical symbols and equations, if they should make sense, must represent some elements and relationships in reality. And if the constraint here is that the total number of charges in a volume can't change by just re-arranging them spatially (be it through the retardation effect or any other mechanism), then this constraint must be reflected in the maths as well, i.e. your changed distribution function must be normalized again such as to preserve total charge. Otherwise your maths does not relate to the problem anymore.

15. ### przyksquishyValued Senior Member

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3,150
There is nothing wrong with the mathematics I've used. Charge distribution functions are routinely used and a perfectly valid way to describe how charge is distributed (which can include "not distributed") in electromagnetism, especially at the macroscopic level which is the domain of applicability of the theory. You're just giving spurious excuses to avoid accepting the result that have nothing to do with any derivation I posted.

Notice how, like with your previous spurious complaints about integration bounds, you complain about some "error" I am making but are unable to point out where this supposed error actually occurs. You can't actually point to where I did anything like "change the number of particles" or "violate preservation of charge" in any derivation I gave, for the simple reason that I never did any such thing. Particularly since I mostly concentrated on the case of there being just one charge.

I did, however, point out a subtlety in that the charge per unit distance/volume at fixed time is not the same as the charge per unit "retarded distance"/"retarded volume" along the varying retarded time, and I did clarify that it is the former I was calling "charge density" and using in the integral, in case this was the point you were not understanding.

I've spent the better part of this thread proving and disproving things that you've been unwilling to investigate yourself. (I'm also a bit worried that I asked you direct questions, along the lines "do you stand by this" and you didn't actually answer them.) But so be it.

Let's take the line of charge of length $L$ moving with velocity $v$, in the one-dimensional case, as described in my previous post. The scalar potential for $t = x/c$ and $x > \frac{L/2}{1 - v/c}$ according to the result I gave in post #171 is

$\displaystyle \Phi_{\text{int}}(x, x/c) = \frac{q}{L} \Biggl[ - \log \biggl(1 - \frac{L/2}{(1 - v/c) x} \biggr) + \log \biggl( 1 + \frac{L/2}{(1 - v/c) x} \biggr) \Biggr] \,.$​

I'll assume you accept this is the result of correctly computing the integral. You've had a month to say so if you believe otherwise.

According to you, for large $N$ this should be approximately the same thing as the discrete sum

$\displaystyle \Phi_{\text{discr}}(x, x/c) = \frac{q}{N} \sum_{k = 1}^{N} \frac{1}{x - x_{k}(t_{k})}$​

for the retarded positions $x_{k}(t_{k}) = \frac{(2k - N - 1) L}{2 N (1 - v/c)}$ associated with the family of $N$ trajectories I proposed in my previous post as a discrete version of the moving line.

Well these are not the same. Choosing $x = 10$, $q = 1$, $\beta = v/c = 0.5$, and $L = 1$, the result for the integral is about

$\displaystyle \phi_{\text{int}}(x, x/c) \approx 0.20067070 \,.$​

For comparison, for the same parameters and for $N = 1000$ particles, your sum version only gives

$\displaystyle \phi_{\text{discr}}(x, x/c) \approx 0.10033535 \,.$​

So, here, if you used your sum as an approximation of the integral, or vice versa, you would make an error of a factor of about 1/2, which is what $1 - v/c$ is in this case.

You can try this out for yourself, for the same or different parameters, if you like. I computed the integral result using this simple GNU Octave function:
Code:
function [ phi ] = phi_int(x, q, beta, L)
y = L / (2 * (1 - beta) * x);
phi = (q / L) * log((1 + y) / (1 - y));
end
I computed the sum above with this function:
Code:
function [ phi ] = phi_discr(x, q, beta, L, N)
k = 1:N;
xr = (2*k - N - 1) * L / (2*N*(1 - beta));
phi = (q / N) * sum(1 ./ (x - xr));
end
I picked GNU Octave for this since it is free and open source software and easy enough to use. It can be obtained here: https://www.gnu.org/software/octave/.

Last edited: Aug 10, 2017
16. ### tsmidRegistered Senior Member

Messages:
368
I asked you to derive it without using any charged density function ρ at all but only the the charge q of each particle.

So you are claiming that a simple spatial re-arrangement of the charges would lead to a different total charge when integrated over space, that is the integral over ρ(r) (using scalar notation here) would be different from that over ρ'(r'(r)), where ρ' is the new distribution function resulting from some spatial transformation of the particle at r to r' ?

And I pointed out that it is the latter you are using in the integral for the potential.

Last edited: Aug 10, 2017
17. ### przyksquishyValued Senior Member

Messages:
3,150
And you are in no position to dictate how I am allowed to do things. You seemed to believe that the integral approximates a sum $\sum_{i} \frac{q_{i}}{R_{i}}$. I tested that simply by computing both for some test parameters. Obviously I need to have computed the integral for some charge distribution, since that is what is in the integral, in order to make that comparison.

Now please address my last post properly, including the large part you omitted in your quote. Otherwise I think we're done here.

Last edited: Aug 10, 2017
18. ### tsmidRegistered Senior Member

Messages:
368
It is no surprise that you are getting a different result from a numerical evaluation of your integral because you integral is incorrect. I have pointed out now at least have a dozen times in this thread that you failed to normalize your retarded density distribution (that appears under the integral for the potential). I have illustrated this in the figure below.

The top shows the unretarded positions x' of 10 particles between -L/2 and +L/2, the bottom the retarded positions x'/(1-v/c) between -L/2/(1-v/c) and +L/2/(1-v/c) for v/c=0.5 in this case. And it is readily evident that for the retarded distribution the charge density (defined as the number of charges per unit length) is only half that of the original distribution, therefore exactly cancelling out the increased length of the distribution. Your integral wrongly assumes the charge density shown at the top would apply all the way between -L/2/(1-v/c) and +L/2/(1-v/c) at the bottom, resulting thus in 20 instead of 10 charges and therefore in a potential a factor 2 too high for this particular example.

So in general, you have to apply a further factor 1-v/c to your integral, which exactly eliminates the factor 1/(1-v/c) in the Lienard-Wiechert potential.

Last edited: Aug 12, 2017
19. ### przyksquishyValued Senior Member

Messages:
3,150
And I've already told you that that is the wrong charge density to use in the formula. The charge density is defined as the density of charges at a given fixed time. Here it's simple: the total charge is $q$, the length at any given time is $L$, so the charge density is $q / L$ within the line. Or locally in terms of the particles: each charge is $q / N$ and the distance to the next charge at any given time is $L / N$, so the local charge density is $q / L$.

Now yes, the charges are spread out over a range $L / (1 - v/c)$ of retarded positions. So you can say that the "charge density" is $q (1 - v/c) / L$ in this sense. But this is not the meaning of charge density conventionally used in electromagnetism or in the integral for the potential in particular, nor is it the meaning of charge density I was referring to before in the thread (in post #110 for instance). Like I told you before, the integral for the potential is itself derived in electromagnetism from Maxwell's equations, for the same function $\rho(\boldsymbol{r},\, t)$ used in Maxwell's equations. You cannot interpret $\rho$ in just whatever way you like.

From one of my recent posts (emphasis added):
The first definition normalises $\rho$ so that $\int \mathrm{d}^{3}r \, \rho(\boldsymbol{r},\, t)$ is equal to the total charge with the integral taken with $t$ fixed. The second definition would instead normalise $\rho$ so that $\int \mathrm{d}^{3}r' \, \rho(\boldsymbol{r}',\, t_{\mathrm{R}})$ is equal to the total charge. I told you before that these are different and $\rho$ is consistently normalised so that $\int \mathrm{d}^{3}r \, \rho(\boldsymbol{r},\, t) = q$ or $\int \mathrm{d}x \, \rho(x,\, t) = q$ in the derivations I gave in posts #137, #147, and #171.

It seems to me this matter is resolved. It's obvious you simply misunderstood what meaning of "charge density" is used in the integral.

Last edited: Aug 12, 2017
20. ### tsmidRegistered Senior Member

Messages:
368
There is no integral over time in the definition for the potential. The integral is only over spatial positions. Writing the time argument for ρ only serves as a formal aid to fix these positions. Once we have found these, the time argument is completely redundant; we only integrate of the corresponding spatial distribution; that's why I have left out the time argument on ρ altogether in my diagram

And if you obtain a potential that is a factor 2 too high (1/(1-v/c) in general) from the retarded positions at the bottom, each particle would have a charge 2q instead of q, which would violate charge invariance. In order to have charge invariance you have to normalize ρ, whether it is for the unretarded or retarded distribution, since either distribution consists of the same particles, only at different locations.

Last edited: Aug 13, 2017
21. ### przyksquishyValued Senior Member

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I just told you the way you're interpreting the charge density in the integral is not consistent with the way the integral is derived from Maxwell's equations.

Last edited: Aug 13, 2017
22. ### tsmidRegistered Senior Member

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368
I am puzzled as to how you get to this conclusion, considering that the 'retarded' charge density (=q*(1-v/c)/L in this case) is exactly what appears under the integral defining the potential (q meaning the total charge here, not that of one particle as in my previous post)

$\displaystyle \Phi (\boldsymbol{r}, t) = \frac{1}{4 \pi \epsilon_{0}} \int \mathrm{d}^{3}r' \, \frac{\rho \Bigl(\boldsymbol{r}',\, t - \frac{\lVert \boldsymbol{r} - \boldsymbol{r}' \rVert}{c} \Bigr)}{\lVert \boldsymbol{r} - \boldsymbol{r}' \rVert}$

That's the whole point for it appearing there and there is no other meaning to this variable (I thought I had made it clear that the 'retarded' charge density distribution is a different function to the 'unretarded' density distribution, so it is actually misleading to use the same variable ρ for both functions; that's why I called the retarded distribution ρ_R in my diagram above, but that's exactly what ρ under the integral is referring to).

Last edited: Aug 14, 2017 at 9:41 PM
23. ### przyksquishyValued Senior Member

Messages:
3,150
I thought I had made it clear that it is specifically the unretarded charge density that is used in the integral.