Relativistic Coulomb Force

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

  1. przyk squishy Valued Senior Member

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    No, the equation I posted relates the electric field at a location P to the "retarded" location of the charge producing it, i.e., the position of the charge at the earlier time when it crossed the past light cone of P. You recover the equation in your OP by rewriting it in terms of where the charge is now assuming it continued to follow a rectilinear trajectory since that earlier time, not by removing some "contribution due to acceleration" term.

    You're still assuming, for no reason, that the difference made by acceleration is only due to the amount of time spent accelerating. If a charge is not accelerating 99% of the time that does not mean the equation in your OP applies 99% of the time. It doesn't work that way. The charges could be changing direction instantaneously and accelerating 0% of the time as far as I'm concerned. It would still matter and it would still need to be accounted for.
     
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  3. Q-reeus Valued Senior Member

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    The fully relativistic field eqns for a point charge in arbitrary motion is given by eqns (16), (17) here:
    http://users.wfu.edu/natalie/s13phy712/lecturenote/lecture27/lecture27latexslides.pdf
    Inspection of (16) should make it obvious that for modest speeds v << c, magnitude of acceleration field component is directly proportional to acceleration only. However what you are really looking for is the time averaged field impulse. I'll leave as an exercise for you to find that, to a first approximation at least, that impulse is independent of the peak value of charge acceleration - i.e. the same for a hard or soft bounce at the box walls.
     
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  5. tsmid Registered Senior Member

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    A charge can only be at one location at a certain time. The first term in that equation gives the electric field associated with the location itself at that time (the classical Coulomb law), the second term gives the electric field associated with the velocity of the charge at that same location and time, and the third term gives the electric field associated with the acceleration at that same location and time. All three terms relate to the same location and time of the particle. Retardation doesn't change anything about this, it only leads to an identical time offset for all three terms (as the information goes from that one location to the location of the test charge).
    So the test particle would see the acceleration term only for the very short times when the charge was in the process of colliding with the wall i.e. when it was being accelerated (and even that could not possible cancel out the other terms as the electric field due to the acceleration term is transverse whereas the others are longitudinal ; http://www.tapir.caltech.edu/~teviet/Waves/empulse.html ). At all other times (i.e. 99.9999999% of the time in this case) the test charge would not see any acceleration field at all.
     
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  7. przyk squishy Valued Senior Member

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    I think you're conflating reasoning for two different equations. If you use the equation I copied from Feynman then you may be able to say that an "acceleration term" only applies a small fraction of the time (assuming it also didn't depend on the magnitude of the acceleration, which, as Q-reeus has pointed out, it does). And you could start with the equation from Feynman and try to work out what kind of field is produced by your box of bouncing charges and find out if it is zero or not. But you didn't do that. Your reasoning was instead based on the equation you posted in your opening post, which doesn't relate the electric field to the charge's retarded position but instead to its projected position: where the charge would be now if it continued to travel with the same velocity it had at the retarded time.

    For example: suppose you're one light year away from your box, and the electrons are moving at one tenth of the speed of light while the positive charges are stationary. Then, one year later, the electric field you feel is the same as if the negative charges had continued to travel so that they are now a tenth of a light year outside the box while the positive charges are still inside it. So in that case if you apply your OP equation and you pretend the acceleration didn't happen (because it is nearly instantaneous) and just use the particles' current positions, then you make an error corresponding to a distance of 1/10 light years for the positions of the negative charges.


    But this doesn't settle anything. Ultimately the problem here is that you're making claims about what a mathematical result will be that contradict the mathematics that has already been done, which apparently you haven't felt motivated to actually look at. I was considering looking for some derivations to show the relations between different formulae here or just working it out myself to settle this, but the slides linked to by Q-reeus saved me the trouble:

    http://users.wfu.edu/natalie/s13phy712/lecturenote/lecture27/lecture27latexslides.pdf

    This makes everything easy. There are two key things to notice there (and verify their derivations if you don't believe them) that should settle this:

    1) You can easily derive the electric and magnetic fields around your box from Eqs. (4), (5), and (8), (9). If the charges are moving in such a way that the charge density is constant over time (\(\rho(\boldsymbol{r}, t) = \rho(\boldsymbol{r})\)) and there is no net flow of charges anywhere (\(\boldsymbol{j}(\boldsymbol{r}, t) = 0\)) then the electric field given in Eq. (8) will just reduce to integrating the Coulomb field. If furthermore the charge density is also zero (every negative charge is compensated by a positive charge so that \(\rho(\boldsymbol{r}) = 0\)) then you just get \(\boldsymbol{E} = 0\) for the field around your box. (This is for smooth distributions, or the limit \(N \to \infty\) particles, to keep the problem simple.)

    These starting equations are linear, so if you add individual solutions of Eqs. (4) and (5) and the charges and currents add up to zero everywhere then the corresponding electric fields must add up to zero everywhere.

    2) The electric field for a single moving charge is derived (as Eq. (16) and repeated as Eq. (19)) as a special case of Eqs. (4) and (5). If you drop the acceleration term it is not difficult to see that it is the same as the equation you posted at the beginning of this thread.
     
    Last edited: May 28, 2017
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  8. tsmid Registered Senior Member

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    The test charge at its fixed location and at the local time t should detect an electric field

    E(t) = E1(t)+E2(t)+E3(t)

    Note that all three terms have to be evaluated at the same time t. It would not make any sense to add the fields for different times. So if these are the fields produced by a single charged particle in the box, they must have been produced at the same point and thus the same retarded time t_ret because otherwise they could not have arrived at the test charge simultaneously (assuming that the speed of light is constant).
    The point is that if E3(t) relates to the acceleration of the charge in the box, it will practically always be zero at all times t, apart from these extremely short blips when the charge collides with the wall. In contrast, the terms E1(t) and E2(t) are always non-zero. What's more, they are longitudinal fields, so E2 could not possibly be cancelled out by the transverse field E3 even when the latter is non zero.
    And as that applies to each individual charge in the box separately, it also applies to the total field of all charges.

    The retarded location of the electron is the location where the electric field was produced that was detected at time t by the test charge. Since the electron never leaves the box, I don't know how you come to this conclusion.

    It is basically the same equation we have been discussing, only in a slightly different form.
     
  9. przyk squishy Valued Senior Member

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    Like Q-reeus pointed out, the contribution from the acceleration is proportional to the acceleration, so e.g. if you take the period of acceleration to be half as long then the magnitude of the acceleration doubles, so the average over time of (fraction of time spent accelerating) * (magnitude of acceleration) is the same.


    I was referring to an error you make if you apply the equation you started with at the beginning of this thread, which is what you based your reasoning on and (so far) the only equation I've seen you do any calculation with. That equation, the one you were using, is not expressed in terms of the retarded position. It is expressed in terms of where the charge is now assuming it was travelling in a straight line with constant velocity since (at least) the retarded time. And that projected position can very well be a position outside the box where the charge is actually confined to.

    To be more explicit: suppose you want to calculate the electric field \(\boldsymbol{E}(\boldsymbol{x}, t)\) at position \(\boldsymbol{x}\) at time \(t\) produced by some charge in your box, and you want to use the formula from your first post to do it. Suppose the charge was at the retarded position \(\boldsymbol{x}_{\mathrm{ret}}\) at the retarded time \(t_{\mathrm{ret}}\), and suppose it was moving with velocity \(\boldsymbol{v}_{\mathrm{ret}}\) and it happened to not be accelerating at that particular time. Then, in that case, it is OK to use the equation in your first post, but only if you use the projected position \(\boldsymbol{x}_{\mathrm{ret}} + (t - t_{\mathrm{ret}}) \boldsymbol{v}_{\mathrm{ret}}\) for the position of the charge (equivalently \(\boldsymbol{r} = \boldsymbol{x} - \boldsymbol{x}_{\mathrm{ret}} - (t - t_{\mathrm{ret}}) \boldsymbol{v}_{\mathrm{ret}}\) for the vector \(\boldsymbol{r}\) appearing in the equation), regardless of where the charge really is at time \(t\).

    So, loosely speaking (and ignoring the acceleration), you can think of the equation from your first post as applying, but for a sort of fictitious distribution for the locations of the charges that extends outside the box for the negative charges (which are moving) but is confined inside the box for the positive charges (which aren't moving).


    But this would be calculating the electric field in the most complicated way for no good reason. Like I said, you can calculate the electric field easily for the distribution of charges in your box (zero charge and zero net current in the limit of an infinite number of infinitesimally charged particles, which I'm assuming is the limit you're most interested in, so zero field) as a special case of Eqs. (4), (5), and (8) from the slides Q-reeus linked to, and the slides summarise how to derive the electric field around a moving charge also as a special case of the same Eqs. (4), (5), and (8). Given that, at this point I don't see what point you're trying to argue.
     
    Last edited: May 29, 2017
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  10. tsmid Registered Senior Member

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    Only that the period of acceleration is completely negligible here. You can make it in fact arbitrarily small in relation to the period of non-acceleration by making the box arbitrarily large.

    The reference I quoted in my opening post assumes the retarded position to be the origin and the retarded time to be t=0 by definition (see below Eq.5.20 in http://www.damtp.cam.ac.uk/user/tong/em/el4.pdf ). The field that emanates from this is then detected at time t=r'/c where r' is the distance from the origin. Classically the field would only depend on r' but relativistically it also depends on the polar angle θ of the radius vector to the test charge.

    And this is exactly the same as Eq.(22) in the reference you and Q-reeus gave above
    http://users.wfu.edu/natalie/s13phy712/lecturenote/lecture27/lecture27latexslides.pdf only that here the test charge is located on the y-axis (so θ=90 deg) and the charge is allowed to be offset from the origin on the x-axis (along which it is moving). The latter produces obviously an x-component of the electric field (which is missing in the other derivation) but we are only interested here in the y-component (the x-component will cancel out anyway for a symmetrical charge distribution). The crucial point is that the y-component (the radial component) is completely unchanged by changing the sign of v (or the sign of θ in the other equation), so all particles moving with constant speed inside the box along parallels to the x-axis (that is all particles in the box if they are all forced to move that way) will contribute the same to the total electric field at any time. The acceleration term in the more general Eq.(19) obviously drops out for a=0 and even during the short periods of velocity reversal at the walls it would not contribute anything to the y-component of the field (i.e. R in Eq.(19)) as it would be perpendicular to R (along the direction of a)
     
  11. przyk squishy Valued Senior Member

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    Only that the magnitude of acceleration is inversely proportional to the period of acceleration, and this is not the only thing you haven't accounted for.

    As I keep telling you, the easiest way to calculate the electric field from your box, particularly since it seems you've defined everything in such a way that the net charges and currents will be zero everywhere, is starting from Eqs. (4) and (5) for the scalar and vector potentials from the reference Q-reeus gave:
    \(\begin{eqnarray} \Phi (\boldsymbol{r}, t) &=& \frac{1}{4 \pi \epsilon_{0}} \int \mathrm{d}^{3}r' \, \frac{\rho \Bigl(\boldsymbol{r}',\, t - \frac{\lVert \boldsymbol{r} - \boldsymbol{r}' \rVert}{c} \Bigr)}{\lVert \boldsymbol{r} - \boldsymbol{r}' \rVert} \,, \\ \boldsymbol{A} (\boldsymbol{r}, t) &=& \frac{1}{4 \pi \epsilon_{0} c^{2}} \int \mathrm{d}^{3}r' \, \frac{\boldsymbol{j} \Bigl(\boldsymbol{r}',\, t - \frac{\lVert \boldsymbol{r} - \boldsymbol{r}' \rVert}{c} \Bigr)}{\lVert \boldsymbol{r} - \boldsymbol{r}' \rVert} \,. \end{eqnarray}\)​
    Clearly, if the net charge density \(\rho = 0\) and the net current density \(\boldsymbol{j} = 0\) everywhere then you're just left with \(\Phi (\boldsymbol{r}, t) = 0\) and \(\boldsymbol{A} (\boldsymbol{r}, t) = 0\), in which case you get \(\boldsymbol{E} = 0\) and \(\boldsymbol{B} = 0\) for the electric and magnetic fields from Eqs. (10) and (11).

    That is the prediction made by electromagnetism for any source (like your box) assuming all the positive and negative charges and the movement of the charges perfectly balance out so that the net charge and current densities are zero everywhere. The electric field will be zero. You cannot derive a different result using a different approach for this situation without making a mistake somewhere.


    No. Eq. (5.20) gives the electric field at time \(t\) (ignoring the primes left over from the Lorentz boost) for a charge with trajectory \(x = - vt\). The equation at the beginning of the next page (the equation from your first post) is just Eq. (5.20) evaluated at \(t = 0\). So it gives the electric field at \(t = 0\) for a charge moving with constant velocity and passing through the origin at time \(t = 0\).

    Now, as you point out, the origin is the retarded position of the charge in relation to the position \(\boldsymbol{r} = (x, y, z)\) at the later time \(t = r / c\), so you can get the electric field in relation to the retarded position just by substituting \(t = r / c = \sqrt{x^{2} + y^{2} + z^{2}} / c\) into Eq. (5.20). If you do that and do some simplification you'll get a formula similar to Eqs. (16) and (19) in Q-reeus's reference, except for the sign in front of the velocity and the presence of the acceleration term.


    Huh? Eq. (22) is the same thing as Eq. (5.20) from the lecture notes you linked to except evaluated for \(x = z = 0\) and with the opposite sign in front of the velocity. In particular, it is, like Eq. (5.20) and like the equation in your first post, expressed in relation to the charge's current location assuming it didn't change velocity since the retarded time, and the complications I've told you relating to that are the same.

    You could in principle calculate the net field for your box using formulae expressed in relation to the retarded positions of the charges (like Eqs. (16) and (19) of Q-reeus's reference, or the formula I copied out from Feynman earlier) and adding the fields, but there are complications to deal with in that case too. For example, the charges may be evenly distributed inside your box at a given time, but this is generally not true of the retarded positions.

    But what's the point of this? You can calculate the resulting field just starting from the general formulae for the scalar and vector potentials (or the similar Jefimenko equations, or just the integral forms of Maxwell's equations for that matter) like I've pointed out multiple times now. The result is that there is no electric field. The derivation in the rest of Q-reeus's reference shows how to derive the electric and magnetic fields for a single moving charge starting from the same formulae for the scalar and vector potentials, so adding Eq. (16) or (19) for all the individual charges in your box cannot produce a different result if you do that calculation correctly. The only thing you accomplish doing the calculation in this more complicated way is to increase the chance of making a mistake.


    An aside, but this doesn't really make sense. In classical (i.e., non quantum) electromagnetic theory the prediction for the electric field around a moving charge (or for any situation) is ultimately just whatever Maxwell's equations say it is. There aren't different relativistic and nonrelativistic versions of the prediction that you can choose from.
     
    Last edited: May 31, 2017
  12. tsmid Registered Senior Member

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    OK, then let's put in some more figures here: the ratio of the (transverse) acceleration field to the (longitudinal) is in the non-relativistic approximation a*r/c^2 where a is the acceleration, r the distance from the charge and c the speed of light (see the formulae at http://www.tapir.caltech.edu/~teviet/Waves/empulse.html ) Assuming an electron speed of about 10^8 cm/sec (a few eV) the time to transverse a distance the size of an atom (10^-8 cm) is about 10^-16 sec, that is the acceleration a=10^8/10^-16 = 10^24 cm/sec^2.
    So a test charge at a distance of 100 cm from the box would see the acceleration field 10^24*100/9*10^20 = 10^5 times stronger than the static field, but that field lasts only for 10^-16 sec, whereas the static field is permanently on. Even considering only one 'box-cycle', it takes 10^-7 sec to traverse a box length of 10 cm for an electron with a speed of 10^8 cm/sec, which is 10^9 times as long as the acceleration period. So even though the acceleration field is 10^5 times stronger, it will affect the test charge overall only a fraction 10^-4 compared to the static charge over one 'box-cycle'. What's more, when the electron gets to the other wall, the acceleration is in the opposite direction and exactly reverses again the previous effect. The only constantly acting electric field is thus the static Coulomb field.

    Even in classical physics, the net charge of a ensemble of point charges can never strictly be zero, even with the same number of positive and negative charges in a given volume. An electric dipole has an electric field ~ 1/r^3 for large distances, a quadrupole ~ 1/r^4, an octupole ~ 1/r^5 and so on. For a large number of particles, the multipole field decreases of course very quickly but is still not exactly zero, even not if the charges are moving. Only no charges at all would give a net charge density zero.

    But the crucial point here is that in Relativity there is actually not something like a 'charge' anymore as the electric field is not spherically symmetric and thus the charge effectively anisotropic dependent on its velocity. Conservation of charge only holds if you integrate over the whole sphere (as shown earlier in this thread), but not with regard to a certain direction as defined by the location of the test charge.


    I have not tried to derive anything. I have just applied the results that can be found in many resources, and that in fact are based on exactly the equations you quoted above. The various versions of the results are exactly equivalent, only depending on from which reference frame you are considering it (see https://mysite.du.edu/~jcalvert/phys/relem.htm ). And since we are interested in what the test charge sees, not the charge in motion, the formula I gave initially is actually the correct one here.
     
  13. przyk squishy Valued Senior Member

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    You still haven't accounted for the effects of retardation properly.


    This is true but I don't see the relevance. Yes, if you have a finite number of charges moving around then the electric field will not be exactly zero, and matter in real life does, for instance, emit thermal EM radiation as a result of not being exactly homogeneous or static at the microscopic level.

    But in the situation you described, you were interested in the kind of field you will have if you have charges moving around, but moving in such a way that the charges and movement of the charges locally balances out on average, and I don't think what you're interested in is the field you'll have due to there being \(N\) charges instead of \(\infty\) charges. So the easiest situation to analyse in order to get an exact result is the situation where the charge distributions and currents are all smooth (the same thing as the limit where the number of particles becomes infinite and the charge of the individual particles becomes infinitesimal). You can model this as having three charge/current distributions superposed along the length of your box:
    • A charge density \(-\rho\) and current density \(-\boldsymbol{j}\) for the negative charges moving in one direction.
    • A charge density \(-\rho\) and current density \(+\boldsymbol{j}\) for the negative charges moving in the other direction.
    • A charge density \(2 \rho\) and current density \(\boldsymbol{j} = 0\) for the stationary positive charges.
    Since everything in electromagnetism is linear, the only thing that matters for finding the resulting electromagnetic field is that the net charge density \(-\rho - \rho + 2\rho = 0\) and \(- \boldsymbol{j} + \boldsymbol{j} + 0 = 0\), and the result for this is \(\boldsymbol{E} = 0\) and \(\boldsymbol{B} = 0\) everywhere.

    Do you expect it to make an important difference if there are \(\sim 10^{9}\) discrete charges in your box instead of smooth charge distributions?


    I don't see what this is supposed to mean. A charge and the electromagnetic field it produces are not the same thing. In electromagnetism, the electric field around a moving point charge is just generally not the spherically-symmetric Coulomb field.


    I have no idea what this is supposed to mean either. Mathematically, conservation of charge in the context of classical electromagnetism means that
    \(\displaystyle \frac{\mathrm{d}}{\mathrm{d}t} \int_{V} \mathrm{d}^{3}r \, \rho = - \int_{\partial V} \mathrm{d}\boldsymbol{s} \cdot \boldsymbol{j} \,,\)​
    i.e., the amount of charge in any volume \(V\) changes over time at a rate equal to the net flow of charge into \(V\) through its boundary \(\partial V\). This is required by electromagnetic theory. Maxwell's equations actually become self-contradictory if you try to solve them for a situation where the conservation equation above isn't satisfied.

    An idealised moving point charge can be modelled with a charge and current density of the form \(\rho(\boldsymbol{r}, t) = q \, \delta \bigl( \boldsymbol{r} - \boldsymbol{r}_{0}(t) \bigr)\) and \(\boldsymbol{j}(\boldsymbol{r}, t) = q \, \dot{\boldsymbol{r}}_{0}(t) \, \delta \bigl(\boldsymbol{r} - \boldsymbol{r}_{0}(t) \bigr)\), where \(q\) is the charge and \(\boldsymbol{r}_{0}(t)\) is the charge's position at time \(t\).


    I am at a loss as to what you are even trying to argue here. All of the formulae you, I, and Q-reeus have linked and referred to apply in all reference frames. The formula in your first post, for example, gives the electric field around a charge moving with constant velocity in any reference frame in terms of the charge's velocity in that frame. Eqs. (4) and (5) from Q-reeus's reference gives the scalar and vector potentials and (through Eqs. (10) and (11)) the electric and magnetic fields in any reference frame in terms of the charge and current densities \(\rho\) and \(\boldsymbol{j}\) in that frame.

    The only reference frame anyone has been interested in as far as I know has been the one in which the box you described is at rest. I have no idea why you think there is suddenly now an issue with the choice of reference frame.
     
    Last edited: Jun 2, 2017
  14. tsmid Registered Senior Member

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    As I said already, retardation produces noting more than a constant time offset for all charges here: with the test charge 100 cm from the box with a width of 10 cm, the path difference for a charge at the center of the box to one at the edge is only about 0.1 cm i.e. the variance of the time for the field to reach the test particle is only about 0.1% i.e. 3*10^-12 sec, during which a particle with a speed of 10^8 cm/sec moves only a fraction 3*10^-5 of the width of the box. And for this geometrical configuration, the variation of the angle θ is merely 90±3 deg, which is insignificant for the relativistic correction factor

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

    First of all, I mentioned the multipole field here only as a side note. The point is that Gauss's law relates to the integrated electric flux over a closed surface around the charges. Only that circumstance enables you to derive the net charge within the volume. But if you measure the electric flux at a specific point only, you won't be able to strictly derive the net charge from this as this measured flux depends on the unknown positions of the charges inside the volume. And in the relativistic case it depends additionally on the velocities, and that is much more important for a large number of charges if those move collectively.

    If on the other hand you assume a continuous charge distribution and the net charge =0 everywhere, you would have no charges at all (as you can't have the same positive and negative charge density in one point).

    A charge manifests only through its electromagnetic field. What else would contribute to the concept of charge?

    Yes, sorry, I misread the article at https://mysite.du.edu/~jcalvert/phys/relem.htm (I just skimmed over it first time around). The equations here and elsewhere relate apparently both to the lab frame (the frame of the test charge), but they are exactly identical. The proof is given in the article in one line (although part of the first terms have been left trailing in the previous line). The denominator for the electric field can thus be transformed as follows

    b^2 + γ^2*v^2*t^2 = b^2 + v^2*t^2 + (γ^2 - 1)*v^2*t^2 = r^2 + (γ^2 -1)*v^2*t^2 = r^2*[1 + γ^2*β^2*cos^2(θ)] = r^2*[1 + γ^2*β^2*(1 - sin^2(θ))] = r^2*γ^2*(1 - β^2*sin^2(θ))

    using γ^2-1 = β^2*γ^2 and v^2*t^2 = r^2*cos^2(θ) (I inserted my previous variable θ here instead of ψ used in the reference)

    The important point is that the radial field does not depend on the sign of θ i.e. the sign of the velocity vector of the charge, that is all negative charges moving back and forth along the x-dimension in the box contribute the same radial electric field at the test charge (which should be stronger than the field of the non-moving positive charges).
     
  15. Q-reeus Valued Senior Member

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    Continuing on with 'something different' might be there overall, based on piecemeal qualitative arguments will just get nowhere useful. As evidently something of an obsession that just must be resolved, the way to do it is to thoroughly work through quantitatively a simplest possible case scenario. Over 30 years ago I did just that.

    Consider then a very slender, straight length of perfectly conducting coaxial line, shorted at each end, and carrying a steady current I circulating within. Assuming a current normalized drift velocity β = v/c << 1, after an expansion of the field eqns, only terms up to order β² need be retained. Because the line is very slender, countercurrents within the line can be treated as spatially coincident.
    Pick a field point p at some distance r, measured from the centre point of the coax line, and polar angle θ measured wrt the line axis. Knowing the retarded field equations for a point charge, it becomes an almost straightforward matter to then perform the necessary integrations over the whole line. Except at each end, there are only velocity fields to consider, while at the ends, only acceleration fields need be considered, since they are inherently β² dependent quantities.

    You might like to start with the easier case of finding the net potentials φ, A, at field point p. If not zero, have a rethink and start again. At that point, conventional theory dictates the fields must be zero, purely on the basis of how the fields E & B are defined in terms of the potentials. But I suggest persevering with a direct calculation of the fields themselves, using the retarded field eqns. Won't say here what results I got, except to say it was a rewarding experience! If you are prepared to do the actual work, present the final expressions to be integrated here, and I will tell you if you have them right. Very likely not.
     
    Last edited: Jun 5, 2017
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  16. danshawen Valued Senior Member

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    An electron within atomic structure absorbing or emitting a photon CHANGES THE SHAPE of the entire electron cloud in a substantial way.

    Does your model of Relativistic Coulomb Force take this into account? Even if it does, different energies of photons will have different effects on how the electron cloud changes shape, and the probability as well as energy densities change, and this says nothing at all about observational quantum effects which are well known. There are forces holding together the fundamental particles of atomic structure which do not appear at all in the math associated with this dynamic, so why not just come clean and admit, a simplified model like this one is of limited value in terms of any real application?

    That atomic structure is electrially neutral is basically all you are telling us. We knew that already. Simple instruments tell us that much. Does charge distribution change when an atom absorbs a photon? Tell us something with your model we don't know anything about.

    I'm pleasantly surprised if your model does, although I will admit, having never gone very far past a bachelor's in physics, I wasn't exposed to very much of the math you seem to be able to bring to bear on the problem. In some respects, I'm glad I wssn't; it looks like computational abuse of a human being to me.

    If your model does not take any of this into account, I'm not surprised. All of the quantum physics math I have seen treats those relativistic electric charges as points that do not expand or contract depending on the energy of the photons they absorb or emit. And the mathematicians who constructed this model couldn't possibly care less, even if they did.

    There is such a thing as a benefit to being able to look at a problem in physics and understand that even the math that can be applied to it with vastly simplified assumptions is not going to be applicable to every problem in a way that will be productive. Possibly this will be one of the first problems to fall to analysis with quantum computational power at our command, but I hold out little hope that this is likely to happen anytime soon, and even if it does, it won't be very meitculously based on this simpler model. The initial conditions will require a lot more attention.

    Exchemist can tell you that not every atom of every element on the periodic table selectively absorbs or emits photons of a given wavelength or at a given temperature. This is a by product of quantum mechanics being about discrete packets of energy from first principles. Observational reality based principles, not purely mathematical ones.

    The analysis that is the subject of this thread has been on a different page from mine from the beginning. I'm pleased however at how far it managed to get with Q-reeus's assistance. I hope the person who originated the OP got everything they bargained for out of it.
     
    Last edited: Jun 5, 2017
  17. exchemist Valued Senior Member

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    Dan thanks for the mention but this is all supremely irrelevant to the thread topic, which is to do with how charge conservation is preserved, when applying relativity to charges in relative motion. That's all. The problem as presented has nothing to do with quantum theory.
     
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  18. danshawen Valued Senior Member

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    3,942
    I went back and read the reference in the OP, and now I understand (finally).

    But now I also understand why Maxwell's equations appear only superficially to be compatible with relativity. There really is more than just two layers here (interchangeable electric and magnetic fields, depending on relative motion). These darned things (fields) are actually recursive, and no indication of any of this in any of Maxwell's equations. Even relativity itself has relativity, if that makes any sense (I know, it doesn't).

    You have no idea how long this problem has been bedeviling me. Worth the effort, it looks like. I have some serious work to do now. Thanks to you and to Q-reuus!
     
  19. przyk squishy Valued Senior Member

    Messages:
    3,171
    No. It is more complicated than that. In particular, the charges might be equally spaced out at some fixed time but this is not generally true of the retarded positions. The precise way this is affected depends on the location of the test charge relative to the direction the charges are moving in and is different for the charges moving in different directions.

    I don't see what you are even trying to argue here. Your stance seems self contradictory and largely based on a refusal to analyse things properly to me. I've already pointed out, using an appropriate formula, that you very clearly get exactly zero field (at least in the limit where the charge distributions are taken to be smooth), and you yourself have accepted that all these formulae for the electric field are ultimately equivalent. Yet you insist that the result will be different based on a different formula you found and handwaving away its limitations instead of doing a proper calculation. Why are you doing that?


    It is perfectly fine as an idealisation, which is what your situation with the charges moving in the box is anyway. Real point charges wouldn't behave like that left to their own accord.

    When you described your box in post #87 you yourself said to ignore collisions. In your first post you happily took an average of \(\sin(\theta)^{2}\) as \(1/2\), as if you could have an infinite number of charges all passing through the same point at the same time in all directions. So why are you objecting to exactly this kind of idealisation now?

    Concerning taking continuous charge distributions, I asked you up front: do you really expect it to drastically change the electric field if all the negative charge is divided among \(\sim 10^{9}\) particles, or \(\sim 10^{100}\) particles, or the limit of \(\infty\) particles?

    Or look at it a different way: the situation described in your box is cyclic over time. If the charges are \(10^{-8}\) cm apart and moving at \(10^{6}\) cm/s, then the box returns to the same state every \(10^{-14}\) seconds, and the average distribution of the negative charges over time is the same as a continuous charge distribution. Do you really care about fluctuations in the electric field over timescales of \(10^{-14}\) seconds?
     
  20. exchemist Valued Senior Member

    Messages:
    6,637
    Thank Q-reeus, then. All I have done is prevent you introducing extraneous ideas into the discussion!

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  21. tsmid Registered Senior Member

    Messages:
    368
    The quantitative equation for the field of a uniformly moving charge (following strictly from conventional theory) has been discussed now already at length here from the first post of this theory. And that results in a net electric field if two particles with an opposite charge at the same (or close enough) location have different velocities. As you yourself pointed out earlier, this net electric field will average to zero in case of many particles with an isotropic (in 3D) velocity distribution function (as then effectively Gauss's law applies) but for an anisotropic velocity distribution the net electric field will in general not be zero.
     
  22. danshawen Valued Senior Member

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    3,942
    Not at all, exchemist. It is now very clear to m that Maxwell defined the speed of propagation of light starting with no assumption that this was proportional to time and that nothing was faster. He could not make this assumption because the fields that elicit its propagation already change (the partial derivative expressions) faster than the speed of the wave. Those derivatives are with respect to time. WHICH frame?

    I had long forgotten this problem. No wonder no one who did not bother to think deeper believed that Minkowski had walked on water. Trust me, it's more complicated and also nuanced than his math based on spacetime distortions of the fields and charge densities of slowly drifting electric current suggests. Only time and energy exist. Space is an artifact of light travel time. In the case of an electron, we.need to take these ideas to a deeper level. Relativity was never complete.
     
    Last edited: Jun 5, 2017
  23. danshawen Valued Senior Member

    Messages:
    3,942
    This is like clearing up an itch I couldn't quite reach. Thanks again for the assist.
     

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