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Boosting a current loop

  1. Aug 26, 2012 #1


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    Suppose we have a neutral uncharged square current loop, carrying a constant current I. If we boost it in the x direction, [itex] x' = \gamma \left( x + \beta t \right) [/itex], so that x' increases as t increases (hence the plus sign), what happens to the current?

    I get that the current density in the y direction should be unchanged by the boost, but the cross sectional area decreases, so the total current gets divided by gamma. The current density in the x direction increases by gamma, and the area is unchanged, so the total current increases by gamma. In addition, the loop is not neutral everywhere, though the total net charge remains zero. This gives the following picture (the x axis is from left to right, y runs up the page and + and - represent the charge).


    This result is not very intuitive, but I don't see any error in it. I wouldn't mind a second opinion, though the problem seems simple enough. The + and - signs show the charge distribution in the boosted frame.

    I was particularly worried about the continuity equation, but if we micro-model the current density in the lower left hand corner in the rest frame as

    [tex]J_x = \frac{kX}{L} \hspace{30 mm} J_y = k \frac{y-L}{L} \hspace{30 mm} 0 < x,y < L [/tex]

    it seems to satisfy the continuity equations [itex]\nabla_a J^a = 0[/itex] in the rest frame and in the boosted frame.

    So it appears to me that Kirchoff's current law takes an apparent hit.

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    Last edited: Aug 26, 2012
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  3. Aug 27, 2012 #2


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    When I teach E&M, I like to do electrostatics, then a tiny bit of relativity, and then present a simplified form of the classic argument from Purcell that relativity requires magnetism. Basically you boost an infinite, straight, current-carrying, neutral wire. I think Purcell was the first to do this at the undergrad level. It's not at all difficult to present even at the "physics for poets" level: http://www.lightandmatter.com/html_books/7cp/ch06/ch06.html#Section6.2 [Broken]

    A common source of confusion with this approach is that students feel uneasy about the issue of where the extra charge came from. Basically the answer is that it's an example of the relativity of simutaneity -- different observers have different opinions about which charges are where at which instant in time. Your treatment would have the advantage that it avoids any such whiff of paradox. However, I find it a bit mind-bending to try to visualize the process by which the excesses of charge are replenished and depleted at the corners. Kirchoff's laws seem like a non-issue to me, since the boosted loop is basically a capacitor, and we don't expect the junction rule to hold when charge is being stored or depleted in a circuit.

    Your analysis using the current four-vector seems right to me. It might be nice to check and understand it through other approaches as well.

    Just speaking in terms of length contraction and time dilation, it makes sense that the upright wires have their current decreased by gamma; this is just time dilation. You probably can't successfully analyze the currents along the horizontal wires in this way, since the off-diagonal elements of the Lorentz transform play a role, and those elements aren't described by length contraction and time dilation.

    Another cross-check would be to transform the fields near the wires.

    About 10 years ago, I think there was a discussion of boosting a loop like this in The Physics Teacher. I no longer subscribe, so I don't have access to the back issues. I think they used circular loops. I remember getting into some intense discussions about it with a couple of my colleagues.
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  4. Aug 27, 2012 #3


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    Me too - but I think it's starting to make some intuitive sense. Because there's more current flowing out of the lower left hand corner than into it, it must be discharging. Which is indeed the case. Similarly, there's more current flowing into the lower right corner than is flowing out of it, charging it. Other than at the corners, there's as much current flowing in as out, so the charge stays constant.

    Overall, there's a current of I / gamma flowing in the loop, which as you point out would be expected from time dilation.
  5. Oct 20, 2012 #4
    This post is a consequence of following the link provided here
    The unphysical implications depicted in currentloopf.png of #1 - continual charge pile-up in the upper left/lower right corners, and similarly ever growing depletion in the other two corners, cannot be correct. The boost does uncontroversially reduce current in the vertical legs by factor 1/γ as shown, owing to reduced clock-rate hence flow speed, sure. But unless current in horizontal arms is equally reduced, continuity fails and SR is 'disproved'. First clue here is to note that in the loop rest frame (denoted here as S), results for a number-counter recording charges/second passing a given point is independent of being placed in a vertical or horizontal leg or a corner. In lab frame S' the observed 'flash-rate' of said recorder must likewise be a position invariant factor 1/γ reduced wrt proper rate in S. This logically guarantees current in lab frame S' is independent of orientation/position in loop. No absurd differences in current magnitude allowed - not imo anyway.

    Choose the horizontal leg where boost velocity v = βc and charge proper flow velocity u adds. Then in S', conduction charge velocity u' is given by SR addition formula as
    u' = (u+v)/(1+uv/c2) - (1).
    Velocity relative to the lattice is what defines the net current flow velocity in S' and is
    u'0 = u'-v = (u+v(1-(1+uv/c2)))/(1+uv/c2) = (u(1-v2/c2)/(1+uv/c2) = (u2)/(1+uv/c2) - (2)
    Proper linear charge number density of ρ in S becomes, for u<<v, in S'
    ρ' ~ γρ -(3)
    Multiplying u'0 in (2) by ρ' in (3) then gives
    I' ~ u'0ρ'q = (I/γ)/(1+uv/c2) - (4)
    This approximate result only disagrees with being precisely the same in magnitude as for vertical leg value I/γ because in (3) a negligible speed u<<v was assumed and was conveniently set to zero. In fact by requiring, as necessary, uniformity of |I'| = I/γ, it is the value of ρ' in (3) that is then derived to be
    ρ' = ργ(1+uv/c2) - (5)
    Above is necessary consequence of continuity relation ∂ρ/∂t + ∇.j = 0 holding in any frame. There can be no continual charge pile-up or depletion anywhere in one frame if none occur in any other. Or is it I that am missing the obvious here? (Gotten used to being shunned here so not particularly expecting a response, but one never knows...)
    [Better add that above relates to the loop conduction current - the one that concerns anything supposedly 'funny' happening in corners. Even for a loop having just an excess uniform static charge distribution in S, there is a non-zero convection current in horizontal legs in S'. I have not checked that the net convective (includes lattice charges) + conduction currents reduce to above result, but see little point in doing so, as the convective part clearly has nothing to do with continuity 'paradoxes' - or lack thereof.]
    Last edited: Oct 20, 2012
  6. Oct 20, 2012 #5


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    The apparent pile-up of charge in the lower left hand corner doesn't violate the continuity equations - but first I'll provide the derivation of the above result, then I'll explain intuitively how it satisfies the continuity equations in intuitive terms, then review the continuity equations I originally presented in post #1.

    If I've made an error, I don't see it. I'll go through the derivation (which uses nothing more than basic 4 vectors) in more detail.
    [add] I"ll only do the bottom of the wire, since we agree on the sides

    The above result may be surprising, but it certainly doesn't "disprove relativity". I would prefer we not see such wild claims on PF.

    One can derive the above picture by boosting the charge-current 4-vector


    In the rest frame, we can write for the bottom of the loop of cross section A , using geometric units where c=1

    (p, j_x, j_y, j_z) = [itex]\left(\rho, j_x, j_y, j_z\right)[/itex] = 0,I/A,0,0

    The wire is uncharged in the rest frame (hence p=0) and carrying a current I. For convenience we'll set A, the cross-sectional area of the loop, to unity. Then we can write

    [itex]\rho = 0 \hspace{1cm} j_x = I[/itex]

    Thee transformation for a 4-vector (t,x,y,z) is

    x' = \gamma \left(x + \beta t\right) \hspace{1cm} t' = \gamma \left(t + \beta x\right)

    Thus the transformation for the charge-current 4-vector is the same as it is for any other 4-vector, in particular for the above

    \rho' = \gamma \left(\rho + \beta j_x\right) \hspace{1cm} j'_x = \gamma \left(j_x + \beta \rho\right) \hspace{1cm} j'_y = j_y \hspace{1cm} j'_z = j_z

    Substituting p=0 and j_x = I gives

    [tex]\rho' = \beta \, \gamma \, I[/tex]
    [tex] j'_x = \gamma j_x[/tex]

    Which is just what is drawn.

    Intuitively, what is happening is that we have a bar of positive charge (on the bottom wire) moving to the right. To make the charge move through space, something needs to neutralize the charge in the lower left hand corner so that the charge remaining when the wire "moves on" is zero, and move it to the lower right hand corner. This is accomplished by the "extra" current [itex]\left( \gamma - 1/\gamma\right) I[/itex] flowing to the right.

    The continuity equations, as mentioned previously, can be written compactly as
    [itex]\nabla_a j^a = 0[/itex] or in the wiki-style http://en.wikipedia.org/w/index.php?title=Four-current&oldid=517543860 as

    [tex]\frac{\partial \rho}{t} + \frac{\partial j_x}{\partial x} + \frac{\partial j_y}{\partial y} + \frac{\partial j_z}{\partial z} = 0 [/tex]

    The equations posted in #1 for the current densities in the corner satisfy these continuity equations.

    If you want to use traditional SI units rather than geometric ones, you need to include the extra factors of 'c' in the Lorentz transform.

    c \rho' = \gamma \left( c \rho + \beta j_x \right)
    j'_x = \gamma \left( j_x + \beta c \rho\right)
    Last edited: Oct 20, 2012
  7. Oct 20, 2012 #6
    what if we consider relativistic effects as physically affecting the resistance of the wire, thus changing the current accordingly
  8. Oct 20, 2012 #7
    Honestly now. Was it not clear just from context, even without the inverted commas around 'disproves', that no such 'wild claim' was implied - quite the reverse in fact. But thanks for at least responding.

    The rest of your derivation looks to be formally correct but something is imo logically out here. I introduced the concept of number counter early on in #4 as a reality check against which differing conclusions need to be referenced. Don't have time now to finger exactly what's going on, but suspect it has something to do with convection/conduction current differences. I'm not inclined to buy the argument about one corner 'eating' the excess charge in leg since that charge is co-moving with the leg. I have no problem with that in the setup phase, when either boost speed v or current I is changing, there is a transitory current imbalance in lab frame which finally settles down to yield the steady values of excess/depleted charge in horizontal legs. What I cannot buy is the inferred ever growing charge excess/depletion that a fixed difference in conduction currents in vertical vs horizontal legs implies. Must go. :zzz:
  9. Oct 20, 2012 #8


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    Note that one can break down the charge- current 4 vector as a charge density multiplied by a 4-velocity.

    If we consider stationary + charges and moving - charges in the frame where the loop is at rest, we get:

    rho+ for the charge density of positve charges and (1,0,0,0) for the 4-vector

    rho- for the charge density of negative charges and
    (1/sqrt(1-B^2), B/sqrt(1-B^2), 0, 0) for the 4-vector

    where B = v/c is the average velocity of the negative charges, assumed to be the charge carriers.

    rho+ can't equal rho-, because we demand rho+ * 1 + rho- /sqrt(1-B^2) = 0, so that the wire be uncharged in the lab frame.

    The current in the lab frame is just the sum of the current due to the + charges (0) and the current due to the - charges , B

    We can then boost the 4-velocities, and re-add, to get the currents in the frame where the wire is moving.
  10. Oct 20, 2012 #9
    Sorry if I misunderstand you.
    I don't understand why your definition of I/γ ( not Iγ ) satisfies ∂ρ/∂t + ∇.j = 0.
    And your equation (3) needs to be added the current term of j, which leads to (5).
    This current j term is not combatible with I/γ , I think.

    And I don't see why Eq.(4) is equal to I/γ.
  11. Oct 20, 2012 #10
    Perhaps the issue is that a truly spatially directed current is unphysical, and we should instead model the current density as (a) positive current directed in the time direction and (b) negative current directed along some other four-velocity.

    This would lead to a small amount of built-up charge density even in the "rest" frame, increasingly more as the velocity increases. Of course, we can demand that timelike component of the net four-current be zero by playing with the magnitude of the negative current, but only in one frame.
  12. Oct 21, 2012 #11
    It all gets down to what is being defined as 'current'. As per bracketed entry in #4, distinguish between conduction current Icond - that which flows around the loop in either frame, and convection current Iconv - that owing to gross motion of all charge from boost velocity v only. For sure the discrepancy is that the total current It = Icond+Iconv is what OP has produced, whereas for me the meaningful entity here is just Icond. And the latter, which is referenced wrt rest frame of loop S ('the lattice') must be a position invariant quantity in lab frame S'. Total current I't, referenced wrt a given fixed point in S', can, as per #4, be non-zero even when zero conduction current flows in loop - if static charge excess is present in S. Take the case in rest frame S of a spinning spoked wheel with charges evenly placed around periphery. Call the resulting peripheral time-averaged current Icond. Transform that into a boosted situation such that in lab frame S' wheel moves at some translational speed v as before for loop. [STRIKE]Confident it will then be found that for positions where rim velocity and boost velocity are parallel/anti-parallel, total time-averaged current I't = I'cond+I'conv = γIcond,[/STRIKE] whereas for the time-averaged conduction current I'cond = Icond/γ at all positions around the 'loop'.
    [Strikethrough part cannot be true in general - if convective component dominates then sign but not magnitude of total current will be the same in parallel and anti-parallel locations, whereas in #1 scenario sign reverses but magnitude is constant. So somehow it needs a neutral loop in S for things to all add-up consistently.]
    Last edited: Oct 21, 2012
  13. Oct 21, 2012 #12
    Perhaps you can elaborate on this. Cannot see how any charge accumulation/depletion can occur in a rest frame steady loop-current. What I can see is that working from a Lienard-Wiechert retarded field expression there is increasingly a detachment between 'virtual' or 'apparent' position of moving charges in respect of retarded vs present values, with increasing separation distance from circuit. Apparent motions/positions can in that perspective be wildly different to actual motions in the circuit. But we are not working from any retarded field perspective here.
  14. Oct 21, 2012 #13
    Q-reeus, you are right. I may have misunderstand it

    Now I'm calculating separating minus charges and plus charges.
    Also in the case that the currect i and boost v are in the same direction, the currect i is increased.
    ( Only puls charges are supposed to be moving. )
    [tex]\rho_+ = \rho_- \qquad i_x = \rho_+ v[/tex]
    From S' frame ( which moves at v in the +x direction with respect to S ), the plus charge density becomes
    [tex]\rho'_+ = \frac{\rho_+ - v^2/c^2 \rho_+ }{\sqrt{1-\beta^2}}[/tex]
    and the minus charge density becomes
    [tex] \rho'_- = \frac{\rho_-}{\sqrt{1-\beta^2}}[/tex]
    and the total current i becomes
    [tex]i' = \frac{\rho v}{\sqrt{1-\beta^2}} [/tex]
    As shown above, the minus charge density is increased and the plus charge density is decreased.
    And both these charges are moving in the minus x direction with the electric current from S' frame, so we have to eliminate this effect. So the "real" current i'' is
    [tex]\frac{\rho v}{\sqrt{1-\beta^2}} - \frac{\rho v (v^2/c^2)}{\sqrt{1-\beta^2}} = \rho \sqrt{1-\beta^2}[/tex]
    Surely, the total current is not accumulating. You are right. ( I misunderstood it. )
    This current accumulating case is good for the relativity.
    But the basic problems remain.
    This current stability completely relies on the fact that in each upper or Lower current wire, minus or plus charges are not conserved, though the electron (or proton) charge e is the constant value.
    ( One is increased , and another is decreased in one wire in the upper case. )
    If both charge densities in one wire are the same, the current is accumulating somewhere.

    This causes serious paradox. How do you think about it ?
  15. Oct 21, 2012 #14
    Let there be two four-currents, ##I_+ = |I_+| e_0## represents the positive charge-current. ##I_- = -|I_-| W (e_0 + v e_1)## represents the negative charge current, where ##W = (1-v^2)^{-1/2}## is the Lorentz factor. (I've specifically chosen not to call it ##\gamma## and not to call the velocity ##\beta## to not get confused with a boost.) This is in ##c=1## units.

    It should be clear that even if ##|I_-| = |I_+|##, the net current is ##I = I_+ + I_- = |I_0|[(1-W) e_0 - Wv e_1]##. The time component of this four-vector is nonzero, leading us to conclude there is a slight residual charge density.

    While it is not difficult to play with the magnitudes of the two currents such that the time component cancels, I think having equal magnitudes may be more similar to a physical case. One naturally expects that, in the case that there is no three-current, both magnitudes must be the same so that the whole circuit is electrically neutral.
    Last edited: Oct 21, 2012
  16. Oct 21, 2012 #15
    Thanks ytuab - I agree entirely with the substance of your derivation. Only quibble is that while you initially stated we want conduction and convective velocities adding, in S' the +q current is opposed to convective motion at v, but that does not effect the results.
    It's when the neutral loop in S is treated in S' as superposition of a 'fixed' +ve charged lattice and circulating -ve charged conduction current (my convention) that it gets interesting. In one view the lattice charges simply form a uniformly moving structure and there is certainly no relative motion of lattice charges in or out of corners etc. Also the horizontal leg currents so formed act equally in the same direction. On the other hand, it seems this 'accumulation/depletion' viewpoint arises by picking a fixed reference point X in S', and noting that in the horizontal legs the lattice charges constitute a current vρ'q+ that 'begins' when one end of leg hits at X, and 'ends' when the other end of leg passes X. One then can interpret this as one corner 'feeding' charge/current into leg, the other corner 'consuming' the same. Similarly for conduction current, which although having a position/orientation invariant magnitude and thus not accumulating anywhere in S', when superposed on top of lattice current, yields this 'funny' result of a boosted net current in horizontal legs. Seems more or less what commentary in #3 is on about.

    This could be an interesting playground if extended to say cases where the wire x-section varies along a given leg, which may be e.g. at an oblique angle. Anyway I believe there is no final paradox here, just a matter of understanding what constitutes a current in this scenario, and what perspective gives rise to notion of 'accumulation/depletion' at corners.
  17. Oct 21, 2012 #16
    OK so basically this is the situation where the moving -ve charges are trying to maintain their rest value density = +ve lattice charge density - in their locally boosted frame? If that's so then it gets down to that in a circuit number density is constrained to be invariant in the circuit rest frame. It's then in the proper frame of a -ve charge that things look 'strange' - spacing between neighboring -ve charges has increased, while spacing between lattice charges is compressed. Would you agree with this?
  18. Oct 21, 2012 #17
    In my scenario...

    Do the negative charges have the same rest charge density as the positive charges have as their rest charge density? Yes, I think that's correct.

    It does seem the number densities of the positive vs. negative charges would no longer be equal, though, and overall this construction leads to the conclusion that the loop overall picks up negative charge (in the rest frame of the loop), which seems peculiar. Hm, I think I will just have to mull on this further.
  19. Oct 21, 2012 #18
    Right. And one little pointer here might be the matter of charge invariance! :wink:
  20. Oct 21, 2012 #19


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    The 3-current has to be spatial, it lacks a time component to make it otherwise. The 4-current, http://en.wikipedia.org/w/index.php?title=Four-current&oldid=517543860 is already timelike in all frames. So I don't see your point. Your proposal seems to me to be both imprecise ('directed along some-other four velocity - which one?") and rather speculative. Do you have any sort of reference that suggests four currents should be handled in this manner?

    Four currents are very standard - while the wiki article has at least one minor error, it's not so badly fouled up as to require a fundamental redaction.

    The proof that the 4-current is timeike simply requires that the charged particles be moving at less than "c". Then their contribution to the 4-curent will be p u, where p is the charge density and u is the 4 velocity. This will be a timelike vector, because u is timeike. If you have groups of charges moving at differently velocity, conceptually you break the group up into parts, such that each part consists of charges moving at the same 4-velocity, use the above equation to find the 4-current contribution for that group of charges, and add all the groups together. This will yield the weighted sum of timelike vectors, which will be timelike.

    Furthermore, the fact that the wires build up charges is EXPECTED from relativity. You can explain t his in various ways, one way looks at the moving B-field and says that you expect a vertical electric field, the other way looks at the charge densities and says you expect an uneven charge density in the top and bottom wires due to Lorentz contraction.
  21. Oct 21, 2012 #20
    A reference, no, but I just did a calculation about a current of positive charges and a current of negative charges, and the net four-current is spacelike. Seemed to me that everyone was starting from a neutral current loop with nonzero 3-current. That is by definition a spacelike four-current, is it not?

    There are positive and negative charges, and if the charge densities of two four-currents almost cancel, the overall direction of the four-current could be spacelike.

    Ultimately, I was trying to probe whether the initial setup is nonphysical and there should "really" be built up charge even for original loop, which would make the "appearance" of built up charge in boosted frames not so different after all. By no means am I standing by my original assertion that the setup of the problem is nonphysical. I'm still mulling over that one.

    I am, however, reasonably certain that net four-currents can be spacelike. The theory of steady spacelike four-currents is what gives rise magnetostatics, so to say net four-currents can't be spacelike doesn't make sense to me.
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