Relativistic Coulomb Force

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

  1. Q-reeus Valued Senior Member

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    2,605
    From that it seems there is the intent to keep saying "something funny is going on". But no desire to put in the effort of testing that hunch by analytically evaluating a simple model, satisfying your 'anisotropic' stipulation, such as laid out in #112. Hmm....checking back, found an earlier thread where you were suggesting SR itself was flawed:
    http://www.sciforums.com/threads/re...ion-of-fizeau-experiment-fresnel-drag.155634/
    Also now recall visiting your website at http://www.physicsmyths.org.uk/
    It soon becomes obvious there you have an argument with virtually all of established physics. Seems to me it's a futile exercise then to keep presenting arguments relating to EM theory that can be derived from and assume the absolute correctness of SR. Have a nice day.
     
    danshawen likes this.
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  3. tsmid Registered Senior Member

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    The formula I gave in my initial post is completely appropriate to make my case here: Relativity predicts that the electric field has a velocity dependent correction factor

    f=(1-v^2/c^2) / ( 1-v^2*sin^2(θ)/c^2 )^1.5

    multiplied to the classical Coulomb field. Retardation does not enter into this factor at all. The only thing that would be retarded is the classical Coulomb term q/r^2 itself, but the retardation for a negative and positive charge at the same point will obviously be the same, and even for different positions in the box, the difference in the retardation will become arbitrarily small for large enough distances of the observer. So any moving charges will produce a field different to that of a charge at rest according to the above factor. And unless the velocity distribution function is 3D-isotropic, this difference will not average out

    You are still missing the point: the charge density inside the box may be zero everywhere (on average), but that doesn't mean the net electric field outside the box is (on average) zero as well. Unless you have the same velocity distribution functions for the two charge constituents, they should produce a different field strength at a given point by virtue of the different relativistic factors involved.

    Electric field flucuations of about 10^-14 sec enable you for instance to see, so you should care about it (even though not in this context).
     
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  5. przyk squishy Valued Senior Member

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    You're not applying it appropriately. I keep telling you this and you keep ignoring it and carrying on as if nothing were wrong.


    Excuse me? I am not assuming the electric field will be zero if the average charge distribution is zero. I've told you multiple times you can calculate that from, e.g., the Liénard-Wiechert potentials or the Jefimenko equations. Put \(\rho(\boldsymbol{r}, t) = 0\) and \(\boldsymbol{j}(\boldsymbol{r}, t) = 0\) into these equations and you immediately get out \(\boldsymbol{E}(\boldsymbol{r}, t) = 0\) and \(\boldsymbol{B}(\boldsymbol{r}, t) = 0\).

    Yet you insist the field is different from this based on the equation for the electric field around a constantly-moving charge from your first post. There aren't many possible explanations for that discrepancy:
    • The formulae we're using aren't equivalent and make contradictory predictions about the electromagnetic field.
    • You're getting a wrong result due to applying your equation improperly or in the wrong context.
    And you yourself have already said that all these different formulae for the electric field are equivalent.
     
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  7. tsmid Registered Senior Member

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    I have not been ignoring anything. I have replied to all your objections and shown that they are inappropriate. You tried to make the point initially that the acceleration term would cancel the difference in the velocity terms and I have shown that this is not possible a) due to the short duration of the acceleration field and b) due to the fact that the acceleration field is a time dependent transverse field rather than a static radial field (I note that you completely ignored the latter point repeatedly).

    I would be happy to consider any valid arguments, but so far the only valid argument was that of Q-reeus who mentioned that with my initial assumption of a 3d-isotropic velocity distribution function the relativistic deformation of the electric field would be cancelled out everywhere when intergrated over all charges. That's why I am now considering an anisotropic distribution where it will not cancel out.


    If you write

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


    then this clearly contradicts the formulae we have been discussing here, because you can have \(\rho = \rho_{+} + \rho_{-} = 0\) but the resulting field would in general not be zero if the velocities of the two charge components are different and the electric field of a charge depends on the velocity.
     
    Last edited: Jun 8, 2017
  8. przyk squishy Valued Senior Member

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    3,171
    No. I pointed out that the equation in your first post gives the electromagnetic field for a charge assuming a) it wasn't accelerating at the retarded time and b) it continued to travel with constant velocity in a straight line since the retarded time. If a charge behaves differently than this (e.g., because it bounced off the edge of your box) between the retarded time and "now" then that projected position will be different from where the charge actually is.

    If you naively apply the formula from your first post ignoring this then you are making an error corresponding to how far the charges can move between the retarded time and "now". This is not constant. It depends on both the velocity of the charges and where and how far away the test charge is compared with your box.

    You haven't addressed this at all.


    So let's be clear on this: you are claiming that at least two well-known formulae for the electromagnetic field contradict each other. Is that correct?
     
  9. tsmid Registered Senior Member

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    368
    And I pointed out repeatedly that you won't see it accelerating at the retarded time. The chance for that to happen is much smaller than winning the lottery.

    Well, you emphasized yourself that for a homogeneous distribution of an equal number of positive and negative charges in the box the net charge density is zero everywhere and therefore also the potential outside the box. But this is directly contradicted by Lienard-Wiechert potential (Eq(8) in http://users.wfu.edu/natalie/s13phy712/lecturenote/lecture27/lecture27latexslides.pdf )

    Φ = q/(R-vR/c) (leaving out the constants i.e. using cgs units)

    If you assume the particles move only along the x-axis in the box (which is centered on the origin) and the observer is located on the y-axis i.e. R=sqrt(y^2+x^2), we have

    Φ = q/sqrt(y^2+x^2) / (1-v/c*cos(θ)) = q/sqrt(y^2+x^2) / (1-v/c*x/sqrt(y^2+x^2))

    Taylor expansion of the denominator gives

    Φ = q/sqrt(y^2+x^2) * ( 1 + v/c*x/sqrt(y^2+x^2) + v^2/c^2*x^2/(y^2+x^2) + ....)

    Now if you average over particle positions on the positive and negative side of the x-axis, the linear term disappears, but not the quadratic term, which means that particles with opposite charge q but different velocities will not average to a zero net potential, not even if there is no net current i.e. an equal number of charges moving with v and -v.
     
  10. exchemist Valued Senior Member

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    6,633
    Tsmid, I've following this with interest up to now but I am beginning to wonder what you are trying to do.

    It is plain that charge is, as a matter of observed fact, conserved, so the exercise can only consist in finding the flaw in your reasoning. Several people have tried to point this out. Why don't make more of an effort yourself to look for the flaw?
     
  11. przyk squishy Valued Senior Member

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    3,171
    And I've pointed out since literally my first post in this thread what I've just highlighted in red for you. You ignored it again.

    There is nothing wrong with the equation from your first post itself. But if you put the wrong numbers into it then you're going to get the wrong numbers out of it.


    That might sound superficially convincing if it didn't blatantly contradict what I've already told you: you can solve the same problem by substituting \(\rho(\boldsymbol{r}, t) = 0\) and \(\boldsymbol{j}(\boldsymbol{r}, t) = 0\) into Eqs. (4) and (5), in which case you immediately get \(\boldsymbol{\Phi}(\boldsymbol{r}, t) = 0\) and \(\boldsymbol{A}(\boldsymbol{r}, t) = 0\), and then for Eq. (10) you get \(\boldsymbol{E}(\boldsymbol{r}, t) = - \nabla \Phi - \frac{\partial \boldsymbol{A}}{\partial t} = 0\).

    Your attitude here is bizarre and missing the big picture. You are pointing out a situation where you can solve the same problem in 2 or 3 different ways that should be equivalent and yet apparently getting different results. That is a mathematical contradiction and a certain sign that there's a mistake somewhere. Pointing out yet another way to calculate the electric field isn't going to change this. The sensible thing to do in a situation like this is to study the different formulae and how they relate to each other carefully and try to pinpoint where and how the discrepancy arises. You don't just pick the one you like and ignore the others as if everything were normal.

    Do you understand this? There is no point discussing again what specific technical error you are making if you can't see such an obvious contradiction in what you are saying.
     
    Last edited: Jun 11, 2017
  12. tsmid Registered Senior Member

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    368
    Just hang around a little bit longer. The flaw should soon become evident.
     
  13. tsmid Registered Senior Member

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    368
    a) the effect of retardation on the state of motion of the particle can be made arbitrarily small by choosing the dimension of the problem accordingly.
    b) if you have a spatially (quasi)-homogeneous distribution of particles, retardation doesn't matter anyway; particles will just exchange their positions with each other, and since they are indistinguishable, they might just as well not be moving at all.

    I am actually not quite sure anymore what your position is here now. Assume that at time t=0 you have a positive charge at rest at the origin of the coordinate system, and at the same time t=0 you have a negative charge moving with velocity v also at the origin of the coordinate system (or sufficiently close to it, ignoring the interaction between the two). What in your opinion is the potential Φ at distance R and time R/c in this case?
     
    Last edited: Jun 12, 2017
  14. przyk squishy Valued Senior Member

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    3,171
    This is an empty assertion. You haven't done any calculation or analysis to justify it.

    I never said anything about radiation being an issue.

    The result of adding the corresponding potentials:
    \(\displaystyle \frac{q}{4 \pi \epsilon_{0}} \biggl[ - \frac{1}{R - \frac{\boldsymbol{v} \cdot \boldsymbol{R}}{c}} + \frac{1}{R} \biggr] \,.\)​
    Obviously this is nonzero. But this is for one negative moving charge and one stationary positive charge that are in the same place at just one moment in time. The situation with the box you described is different. In that case the negative charges were moving in both directions and the charges were distributed throughout the box and not all in the same place.

    I honestly don't know what yours is, and I'm trying to get you to think about the implications of what you say before you post here.

    Forget your equations for a moment. If you have a region of space in which there is locally no net charge or current (like the one-dimensional box you described, if you ignore what's happening at the microscopic level) then electromagnetism predicts no overall electric or magnetic field. I justified this in terms of Eqs. (4) and (5) from the presentation slides Q-reeus originally linked to, but this doesn't really matter: this is among the most basic situations you can possibly have in electromagnetism and anyone who has learned the subject at a basic level (understood Maxwell's equations, etc.) will tell you that zero net local charge and current density means -- according to Maxwell's equations -- that there is no (non-background) electromagnetic field.

    Now, you're saying that it looks like the equation from your first post and Eq. (8) from the presentation slides predicts something different. I can sympathise that it might look that way, but what I'm frustrated with here is that you don't seem to show any interest in actually thinking about what you're saying and why you're getting an apparent prediction that obviously contradicts basic electromagnetism. So I'm trying to press you to explain this. What, in your own words, do you think is happening here? Do you think I (and pretty much everyone) misunderstand basic electromagnetism? Do you think the equation from your first post and Eq. (8) from the presentation slides are wrong? Those are extremely unlikely but would at least represent some kind of consistent stance.
     
  15. tsmid Registered Senior Member

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    368
    It is not an empty assertion. I gave numerical estimates for this already. Unless you make the size of the box suitably small, a particle travels only a very small fraction of the width of the box during the light travel time to the observer, and in principle you can make it arbitrarily small. So your argument that the particle would change its velocity during that time interval does not hold here. You are at best again clutching at straws here.

    It seems you misread something: I said

    If you have a spatially (quasi)-homogeneous distribution of particles, retardation doesn't matter anyway; particles will just exchange their positions with each other, and since they are indistinguishable, they might just as well not be moving at all.

    But previously you insisted that the the potential must be zero if the charge density is zero everywhere as per

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

    and clearly the net charge density is zero everywhere if you have an equal positive and negative point charge at the origin (r'=0)

    How do you reconcile these two contradictory statements?
     
    Last edited: Jun 13, 2017
  16. exchemist Valued Senior Member

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    6,633
    I think you may be close to exhausting everybody's patience. You admit there has to be a flaw in your reasoning. Why don't you start looking for it, using the remarks that have been made as pointers?
     
  17. przyk squishy Valued Senior Member

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    3,171
    But I did an exact calculation that is trivially simple, and you haven't been able to point to anything wrong with it. Your own estimates are nowhere near a full calculation for the problem, and you keep denying that the retardation is important to account for despite the fact 1) you consistently refuse to actually analyse it properly (so there is simply no reason for anyone to believe you) and 2) you get an obviously wrong result due to not taking account of it.

    Under the circumstances I would say it is you who is grasping at straws.


    I told you that the potential is zero if the change density is \(\rho(\boldsymbol{r}, t) = 0\). If you have a single stationary positive point charge (at the origin, let's say) and and a negative point charge moving with velocity \(\boldsymbol{v}\) (crossing the origin at time \(t = 0\), let's say) then the charge density is
    \(\rho(\boldsymbol{r}, t) = q \, \delta(\boldsymbol{r}) - q \, \delta(\boldsymbol{r} - \boldsymbol{v} t)\)​
    where \(\delta(\,\cdot\,)\) is the Dirac distribution. This is nonzero except at time \(t = 0\), and the equation you quoted above correctly gives the resulting nonzero potential in that case.

    That I need to explain this makes it plain you haven't understood the equations you keep referring to here.
     
    Last edited: Jun 13, 2017
  18. tsmid Registered Senior Member

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    368
    Yes, \(\rho(\boldsymbol{r=0}, t=0) = q \, \delta(\boldsymbol{0}) - q \, \delta(\boldsymbol{0} - \boldsymbol{v}* 0) = 0 \)

    Now insert this into the equation for the potential

    \(\begin{eqnarray} \Phi (\boldsymbol{R}, R/c) &=& \frac{1}{4 \pi \epsilon_{0}} \int \mathrm{d}^{3}r' \, \frac{q \, \delta(\boldsymbol{0}) - q \, \delta(\boldsymbol{0} )}{\boldsymbol{R}} \end{eqnarray} =0 \)


    At no location other than the origin could the moving particle contribute to the field at distance R and time R/c (unless you violate the principle of the constancy of c)
     
    Last edited: Jun 14, 2017
  19. exchemist Valued Senior Member

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    6,633
    And your point is??.....
     
  20. przyk squishy Valued Senior Member

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    3,171
    No, that's not how you take an integral, especially one with a Dirac delta in it.

    The presentation slides linked to by Q-reeus explicitly say that Eq. (8) is derived from Eq. (4) for the special case of a moving point charge. Didn't you notice this? Did you even try to understand how that works?

    Here's an explanation of how to derive Eq. (8) for the charge density \(\rho(\boldsymbol{r}, t) = q \, \delta(\boldsymbol{r} - \boldsymbol{v} t)\). (After that, taking the difference of two potentials with different velocities is trivial.) First, Eq. (4) from the presentation slides is the same thing as the equation I wrote for the potential, except I'd already taken out the integration over \(t'\). It's more convenient to leave it in for now, so we start with

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

    where \(t_{\mathrm{R}} = t - \frac{1}{c} \lVert \boldsymbol{r} - \boldsymbol{r}' \rVert\). Substituting in \(\rho(\boldsymbol{r}', t') = q \, \delta(\boldsymbol{r}' - \boldsymbol{v} t')\),

    \(\begin{eqnarray} \Phi(\boldsymbol{r}, t) &=& \frac{q}{4 \pi \epsilon_{0}} \int \mathrm{d}^{3}r' \int \mathrm{d}t' \, \frac{\delta(\boldsymbol{r}' - \boldsymbol{v} t')}{\lVert \boldsymbol{r} - \boldsymbol{r}' \rVert} \, \delta(t' - t_{\mathrm{R}}) \\ &=& \frac{q}{4 \pi \epsilon_{0}} \int \mathrm{d}t' \, \frac{1}{\lVert \boldsymbol{r} - \boldsymbol{v} t' \rVert} \, \delta(t' - t_{\mathrm{R}}) \,, \end{eqnarray}\)​

    where now \(t_{\mathrm{R}} = t - \frac{1}{c} \lVert \boldsymbol{r} - \boldsymbol{v} t' \rVert\). Now, when we integrate over \(t'\) to take out the Dirac delta, we need to divide by the derivative of the argument \(t' - t_{\mathrm{R}}\). (This is the reason the difference of the potentials for the two charges ends up being nonzero even at \(t = r / c\), and what you claim above is wrong. See https://en.wikipedia.org/wiki/Dirac_delta_function#Composition_with_a_function or a similar explanation if you need a reminder.) So we get

    \(\begin{eqnarray} \Phi(\boldsymbol{r}, t) &=& \frac{q}{4 \pi \epsilon_{0}} \, \frac{1}{\lVert \boldsymbol{r} - \boldsymbol{v} t_{\mathrm{R}} \rVert} \, \biggl[ \left. \frac{\mathrm{d}}{\mathrm{d}t'} (t' - t_{\mathrm{R}}) \right\rvert_{t' = t_{\mathrm{R}}} \biggr]^{-1} \\ &=& \frac{q}{4 \pi \epsilon_{0}} \, \frac{1}{R} \, \biggl[ \left. \frac{\mathrm{d}}{\mathrm{d}t'} (t' - t_{\mathrm{R}}) \right\rvert_{t' = t_{\mathrm{R}}} \biggr]^{-1} \,, \end{eqnarray}\)​

    where I introduced \(R = \lVert \boldsymbol{R} \rVert\) with \(\boldsymbol{R} = \boldsymbol{r} - \boldsymbol{v} t_{\mathrm{R}}\) in the second line. Using that here \(t_{\mathrm{R}} = t - \frac{1}{c} \lVert \boldsymbol{r} - \boldsymbol{v} t' \rVert = t - \frac{1}{c} \sqrt{(\boldsymbol{r} - \boldsymbol{v} t') \cdot (\boldsymbol{r} - \boldsymbol{v} t')}\),

    \(\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}t'} (t' - t_{\mathrm{R}}) &=& 1 + \frac{1}{c} \, \frac{\mathrm{d}}{\mathrm{d}t'} \lVert \boldsymbol{r} - \boldsymbol{v} t' \rVert \\ &=& 1 + \frac{1}{c} \, \frac{1}{2} \, \frac{1}{\lVert \boldsymbol{r} - \boldsymbol{v} t' \rVert} \, 2 \, (\boldsymbol{r} - \boldsymbol{v} t') \cdot (-\boldsymbol{v}) \\ &=& 1 - \frac{\boldsymbol{\beta} \cdot (\boldsymbol{r} - \boldsymbol{v} t')}{\lVert \boldsymbol{r} - \boldsymbol{v} t' \rVert} \,, \end{eqnarray}\)​

    with \(\boldsymbol{\beta} = \boldsymbol{v} / c\). Substituting in \(t' = t_{\mathrm{R}}\) and using the definitions of \(R\) and \(\boldsymbol{R}\) above,

    \(\begin{eqnarray} \left. \frac{\mathrm{d}}{\mathrm{d}t'} (t' - t_{\mathrm{R}}) \right\rvert_{t' = t_{\mathrm{R}}} &=& 1 - \frac{\boldsymbol{\beta} \cdot \boldsymbol{R}}{R} \\ &=& \frac{R - \boldsymbol{\beta} \cdot \boldsymbol{R}}{R} \,. \end{eqnarray}\)​

    So the potential works out to

    \(\begin{eqnarray} \Phi(\boldsymbol{r}, t) &=& \frac{q}{4 \pi \epsilon_{0}} \, \frac{1}{R} \, \frac{R}{R - \boldsymbol{\beta} \cdot \boldsymbol{R}} \\ &=& \frac{q}{4 \pi \epsilon_{0}} \, \frac{1}{R - \boldsymbol{\beta} \cdot \boldsymbol{R}} \,. \end{eqnarray}\)​
     
    Last edited: Jun 15, 2017
  21. tsmid Registered Senior Member

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    368
    Yes, I noticed it, but it contradicts you previous statement

    I told you that the potential is zero if the charge density is \(\rho(\boldsymbol{r}, t) = 0\). If you have a single stationary positive point charge (at the origin, let's say) and and a negative point charge moving with velocity \(\boldsymbol{v}\) (crossing the origin at time \(t = 0\), let's say) then the charge density is

    \(\rho(\boldsymbol{r}, t) = q \, \delta(\boldsymbol{r}) - q \, \delta(\boldsymbol{r} - \boldsymbol{v} t)\)


    where \(\delta(\,\cdot\,)\) is the Dirac distribution. This is nonzero except at time \(t = 0\), and the equation you quoted above correctly gives the resulting nonzero potential in that case.



    So, according to what you said above, if \(\rho(\boldsymbol{r}, t=0) = 0\), this must be reflected in a zero potential at some instant in time at a given point (e.g. t=R/c at distance R). Or are you qualifying your statement now to hold only for stationary charge configurations?


    Yes, I know that (even more detailed derivations can be found at http://www.hep.man.ac.uk/u/rmj/PHYS30441/Radiation and Lienard Wiechert Potentials Part 2.pdf or http://www.physicspages.com/2014/11/24/lienard-wiechert-potentials-for-a-moving-point-charge/ ). If only you could make it consistent now with your statement that the potential is zero if the charge density is zero.
     
    Last edited: Jun 15, 2017
  22. przyk squishy Valued Senior Member

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    3,171
    Nothing I said above implies that, and the derivation I just did even disproves it.

    I don't see what is so complicated about this. \(\rho(\boldsymbol{r}, t)\) is a function (or distribution) of four variables. If it is the function \(\rho(\boldsymbol{r}, t) = 0\) then the potential is zero, as should be evident from the general equation for the potential that I used. If it is a different function or distribution, as it is with your two point charges passing by each other, then the result will be something different, and you can determine what it is by computing the integral.

    I specifically posted the equation for the potential at one point so that you could see for yourself what I was basing the claim on. So if you disagree that \(\Phi(\boldsymbol{r}, t) = 0\) if \(\rho(\boldsymbol{r}, t) = 0\) then, please, go ahead and tell me what you think the correct potential is for that case and how you managed to derive something different than zero from the equation for a general charge distribution that I posted.


    If you already knew how to derive the potential for the special case \(\rho(\boldsymbol{r}, t) = q \, \delta(\boldsymbol{r} - \boldsymbol{v} t)\) then you would have known that this is wrong:
    So why did you write this? Why would you post a derivation that you already knew was completely wrong?


    If you have a point to make here, please make it. This thread is on its 7th page and the only thing I'm getting at this point is that you're determined to prove I've done ... something ... wrong, but you can't seem to take a clear stance on what that is.
     
    Last edited: Jun 16, 2017
  23. tsmid Registered Senior Member

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    368
    You seem to overlook that there is no explicit velocity dependence in the equation for the potential

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

    The potential is only given by the superposition of the momentary (retarded ) locations of the charges. Their speed does not go directly into it. So instead of a moving charge, you might just as well have an array of charges at rest on the 'trajectory' that somehow are briefly activated at the appropriate time. So if in this sense you assume a charge at the origin that pops briefly into existence at the origin of the coordinate system at t=0, you can represent the charge density by

    ρ(r,t) = q*δ(r)*δ(t)

    and thus (ignoring the usual SI-constants)

    Φ(r,t) = q/r *δ(t-r/c)

    so an observer at distance r from the origin will see a potential q/r at time t=r/c originating from the charge event at the origin at time t=0. This is the only time the observer at r will see anything at all from this event. If we have now another event at time Δt at a distance Δr from the origin, we would instead have

    ρ(r,t) = q *δ(r-Δr)*δ(t-Δt)

    and thus

    Φ(r,t) = q/(r-Δr) *δ(t-Δt-r/c)

    so again it is the classical potential exptected from the distance and time of the event.

    OK, I figured out now is that there is an error on the derivation that you quoted, as the delta function is not properly normalized. If you write the δ-function in terms of e.g. the generating exponential function (as given in the animation at https://en.wikipedia.org/wiki/Dirac_delta_function ), the properly normalized function should read (I am using ε instead of 'a' here for the limiting value, and apply this to our special case t=r/c (i.e. emission at r=0 at t=0))

    (1) dt'* δ(t'*(1-v/c)) = dt'* (1-v/c)/sqrt(π)/ε * exp(-(t'*(1-v/c))^2/ε^2)

    We can change the integration variable to

    (2) τ = t'*(1-v/c)

    so

    (3) dt' = dτ/(1-v/c)

    and thus in your new variable the delta function would be

    (4) dτ* δ(τ) = dτ* 1/sqrt(π)/ε * exp(-τ^2/ε^2)

    which is again properly normalized, but the factor (1-v/c) has dropped out.

    So the potential can now be written as (substituting t' in the denominator by (2))

    (5) Φ(r,r/c) = q* Int [ dτ * δ(τ) /(r-v*τ/(1-v/c)) ] = q/r .
     
    Last edited: Jun 17, 2017

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