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Is the asymmetry mentioned in 1905 SR paper fully removed?

  1. Jan 12, 2013 #1
    In his 1905 paper introducing special relativity, Einstein calls attention to the asymmetry in the classical treatment of the relative motion of a magnet and a conductor:
    In a previous post, it was explained to me that, when the conductor is moving, the electromotive force has "in itself ... no corresponding energy" because the integral of work along the path of the moving conductor cannot always be resolved to a scalar quantity. The integral of the work done along the path of the moving magnet, on the other hand, does resolve to a scalar in all cases. Hence the asymmetry. (I don't understand the math well enough to understand why the asymmetry arises. If in my ignorance I've butchered the explanation somewhat, my apologies to BCrowell.)

    Einstein claims to have eliminated this asymmetry by transforming the event to the frame of the moving conductor. In that frame, the force on the charge is due to the electric field, which, as noted, has "a certain definite energy."

    It seems to me that the asymmetry has been only partially removed. I explain in the following what I mean by "partially removed."

    Here is Einstein on the removal of the asymmetry:
    In essence, the asymmetry is removed by always considering the charge to be at rest. This way, the force is always due to the electric field, and there is consequently always "a certain definite energy" associated with the force.

    First, the success of the theory in establishing that the force can be legitimately considered as due to an electric field must be recognized. Prior to Einstein, it was not possible to consider the moving electric charge to be at rest, because it was the movement relative to the ether which generated all electromagnetic effects. Movement relative to the ether is an absolute fact; it cannot be transformed away. With the elimination of the ether, only the relative motion of magnet and charge come into play, and it is legitimate to consider the conductor to be at rest.

    It is the rest of the claim that I am having trouble accepting. It is not at all clear to me (as Einstein says it should be) that the asymmetry has been removed. Rather, it seems to me that the theory establishes the rest frame of the conductor as the preferred frame.

    According to the principle of relativity, as declared by Einstein himself, the laws of nature should be of the same form for all observers. Yet, as I understand the text above, the event as seen by the charge at rest is of a different form than the event as seen by the magnet at rest. The need for the concept of electromotive force is driven by this difference in form.

    Further, Einstein says that the view from the rest frame of the charge is the true view, while the view from the rest frame of the magnet is affected by the motion of that frame relative to the frame of the charge.

  2. jcsd
  3. Jan 12, 2013 #2


    Staff: Mentor

    No. The asymmetry is removed by the derived transformation equation for the fields. You don't have to consider the charge to be at rest, although you certainly can use the frame where it is at rest if you wish.

    That seems like a rather untenable interpretation. What would you do if you had two identical conductors moving relative to each other? Which would you designate as preferred?

    The laws of nature are the differential equations governing the behavior of a physical system. Here we are specifically talking about Maxwell's equations. In what way do you think that the form of Maxwell's equations is different in the frame of the charge or in the frame of the magnet?

    Nonsense. He said no such thing.
  4. Jan 12, 2013 #3
    That's not what Einstein says. He says that the force is ascertained by transforming to the frame in which the charge is at rest. "...the force acting upon it is equal to the electric force which is present at the locality of the charge, and which we ascertain by transformation of the field to a system of co-ordinates at rest relatively to the electrical charge."

    To follow the instructions from Einstein, one would consider each conductor separately, transforming the field to the rest frame of each conductor.

    In the general form, Maxwell's equations include a term which indicates the electromotive force on a moving charge. When that term is non-zero, there exists the asymmetry which troubled Einstein. The transformation to the frame of the charge makes that term go to zero. The upshot of Einstein's argument is that the electromotive force term is not quite valid. The valid form of the equation is without the electromotive force term.

    Not in those words, no. If I were speaking to him directly, I would not have been so free with my words. But I would suggest to him that what he says amounts to the same thing as saying that the rest frame is the true frame, insofar as ascertaining the electrical force on the charge is concerned. His meaning is clear: if one wants to know the electrical force on the charge, one must transform the field to the rest frame of the charge. In any other frame, one must introduce the notion of electromotive force, which puts one in precisely the situation which Einstein complained about in the introduction of the paper.
  5. Jan 12, 2013 #4


    Staff: Mentor

    If you do the same thing for both frames then neither frame is prefered.

    None of that represents a change in the laws of physics between frames.

    This is certainly not the upshot of his argument. It is not even a remote logical consequence of his argument. In fact, the implication of his argument is almost exactly opposed to your assertion here. The implication of his argment is that the standard form of Maxwell's equations, including the EMF term, is valid in every frame, provided the frames are related by the derived transforms.

    Then don't pretend that he did. Appeal to authority is not a valid form of argument anyway, and when you try to use it by completely misquoting the authority it destroys any vestige of credibility that you may have. Not only are you presenting an illogical argument, you are not being factual in the presentation.
    Last edited: Jan 12, 2013
  6. Jan 13, 2013 #5
    I see that I misread your question. I took it to mean that you were asking about two conductors moving in the vicinity of the magnet. My response was directed to that case.

    To respond to your actual question, I am not asserting that Einstein asserts that one or the other of the conductors is truly at rest, and the other moving. I am asserting that Einstein asserts that for each conductor, and in general for any charge, there is only one frame in which the force on the charge can be determined without appeal to a force "to which in itself there is no corresponding energy". That one frame is the rest frame of the charge. That is why Einstein says that the force on the charge is to be determined by transforming the field to the rest frame of the charge. In that limited sense, I (not Einstein) assert that the rest frame of a charge is the preferred frame for determining the force on the charge.

    Yes, it does, in the limited sense of considering whether the force on a charge is one "to which in itself there is no corresponding energy". In one frame only, the equation for the force on the charge has just one term, the term for the electric field. In all other frames, there is another term, for the electromotive force.

    Of course Maxwell's equations are valid in all frames; that is not merely implied by Einstein, it is explicitly stated. In all frames, the equations correctly predict the current which rises in the conductor. But in all frames except one, those equations include a term for a force to which in itself there is no corresponding energy. For that reason, the force is not quite valid. If it were without fault, Einstein would not have objected to it in the introduction, nor would he call it an auxiliary concept in the body of the paper.

    I am sorry to have let myself be undisciplined, and put words in someone else's mouth.

    Perhaps you believe that I am claiming that the rest frame of the charge is preferred in the sense of being absolutely at rest. In that case, my argument would be illogical and completely at odds with the text of Einstein's paper. I am not making that case. I am asserting that, by declaring that the force on a charge is to be determined by transforming the field to the rest frame of the charge, and by further declaring that in all other frames there is present in Maxwell's equations a term for the auxiliary concept of electromotive force, Einstein is in effect establishing the rest frame of the charge as the preferred frame, for the limited purpose of establishing the force on the charge. I believe that this is a logical argument, soundly based on the text.
  7. Jan 13, 2013 #6


    Staff: Mentor

    Actually, the point I was making is a little more basic than that. For a frame to be "preferred" in the physical sense means that the laws of physics must be different in that frame than in other frames. Furthermore, those laws of physics must be able to describe all objects in that one prefered frame. So if the prefered frame is the one where one conductor is at rest then in the prefered frame the other conductor is moving. So you still need your laws of physics to be able to deal with moving conductors because you cannot have a prefered frame where all possible conductors are at rest.

    This is correct.

    This does not follow, unless by "preferred" you simply mean "I personally prefer it". It is not "preferred" in any physical sense. There is one force law, and that same force law can be used in all frames to correctly determine the force on the charge.

    The term is there in all frames.

    Consider an example with a magnet and two charges, A and B, all moving relative to each other. There is one law which governs the force on both charges in all frames, the Lorentz force law. In the magnet's frame they are both moving, so you need the v term in the magnet's frame. In A's frame B is moving, so you still need the v term in A's frame. In B's frame A is moving, so you still need the v term in B's frame.

    There is no frame in which the correct force law lacks a v term.

    Nonsense. It is still a perfectly valid force. There are lots of valid forces without a corresponding energy. Friction, for example.

    The term is there in all frames, including the frame where a given charge is at rest. See the example above.
  8. Jan 13, 2013 #7
    Okay. I see that "preferred frame" is a technical term, and that I have misused it.

    The force on the charge can be calculated in all frames using the Maxwell/Lorentz equations. The practical effect on the charge indicated by the calculation will be the same in all frames. But the force is not of the same character in all frames. In every frame except the rest frame of the charge, the force is one "to which there is no corresponding energy". Only in the rest frame of the charge does the force have "a certain definite energy". Einstein considers this distinction to be of importance. And, to use your term of expression, he prefers the force in the rest frame.

    It would not be there if the equation were developed from observations made by an observer on the charge. That observer would rightly object to its introduction into the equation.

    True. But according to Einstein, "the force acting upon it [the charge] is equal to the electric force which is present at the locality of the charge, and which we ascertain by transformation of the field to a system of co-ordinates at rest relatively to the electrical charge." To correctly (that is, according to the instructions given by Einstein) calculate the force on each charge in the system, one must transform the field to the rest frame of the charge.

    See above.

    Granted that whether Einstein would use the word "valid" or not is a matter of speculation. The clear sense of the text is that he considers the force which has "a certain definite energy" to be somehow better than the force "to which there is no corresponding energy". He instructs us to calculate the force based on the field in the rest frame. He calls the force in every other frame an auxiliary concept.
  9. Jan 13, 2013 #8


    Staff: Mentor

    OK, no problem. You have a personal preference which coincides with Einstein's personal preference. Neither personal preference represents a prefered frame in the standard physics sense.

    No, it would still be there for such an observer. They would require that term to explain the forces they observe acting on other charges which are moving wrt them.

    No. One "may" transform the field to the rest frame of the charge. The word "must" is incorrect. That is essentially the point of the first postulate of relativity. You are free to use any frame you like.

    Sure, but again, this is not "prefered" in the physics sense, nor is it invalid.
    Last edited: Jan 13, 2013
  10. Jan 13, 2013 #9


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    The total symmetry is, I think, manifest if one writes the equations using tensor notation.

    I'll give a brief exposition and encourage the OP to look into the tensor formulation (background permitting) as it should answer his question. For a fuller explanation, any E&M textbook that covers tensors should suffice, I'd suggest Griffiths.

    I apologize in advance if the material is too advanced, but I thought I should at least mention it.

    Proceeding, one can, then write the usual non-relativistic force law

    f = dp/dt = q(e+v x B)

    using tensors as

    [itex]f\prime^a = dp/d\tau = F^{a}{}_{b} u^b[/itex]

    here [itex]f\prime[/itex] is the f-fource, the 4-vector equivalent of the 3-fource f. u is the 4-velocity, the 4-vector equivalent of the 3-velocity v. [itex]F^{a}{}_{b}[/itex] is the Faraday tensor, which combines both the electric and magnetic fields into a single unified geometric entity.

    F can be thought of more or less as a 4x4 matrix, though in tensor terms it would be called a rank 2 tensor.

    There's a brief description of F in the wiki at http://en.wikipedia.org/w/index.php?title=Electromagnetic_tensor&oldid=520946298

    As this is a tensor equation, summation over the repeated index is implied, i.e. the above equation is actually 4 equations, one for each of the 4 values of a, giving the 4 components of f', and each equation is the sum of four terms , the sum of b=0,1,2,3.

    All of the entites, f', u, and F, transform as tensors should, the equation is symmetrical, and neither E nor B is singled out for special treatment, being combined into the single gemoetric entity, the Faraday tensor.
  11. Jan 13, 2013 #10


    Staff: Mentor

    Greg, one quick comment before I call it a night:

    I think you are getting too caught up in small details of the wording of one seminal paper. Much more important than that is the clear and unambiguous math. It is mathematically clear that Maxwells equations are not invariant under the Galilean transform. It is also mathematically clear that they are invariant under the Lorentz transform. Those two clear mathematical facts should resolve any confusion or ambiguity introduced by the English.
  12. Jan 14, 2013 #11
    I do see that the Maxwell equations are invariant under the Lorentz translation. I have sort-of-known that for while; meaning that I have read about it, and what I read made sense to me. Reading the Maxwell paper and stumbling for a bit as I tried to work out the meaning of the equations, until it dawned on me that the equations as he wrote them are all with reference to the ether, solidified what I had sort-of-known.

    I agree that the issue of whether the emf on a moving charge is an "asymmetry" or an "auxiliary concept" is a small detail in the large scheme of special relativity. The reason I have pursued it is that there have been many times, in my journey toward understanding special relativity, that I have not followed up on a "small detail", only to find out later that it in skipping over the detail I failed to uncover a major misconception on my part. Relativity is not at all intuitive; I can't trust my normal reasoning processes to get me to the correct answer.

    Then, too, discussing these issues with people who know what they are doing gives me a chance to learn some things that are not going to be covered in a popular treatment of relativity. And it gives me a chance to learn (sometimes the hard way) to be more disciplined in my thinking.

    My biggest problem re relativity at this point is that I do not have the mathematical tools to process the technically oriented books. Unfortunately, I have to prioritize my effort, and learning the math is a couple of levels down in priority. Understanding relativity is part of a larger project, which is mostly metaphysical. Where the project touches on physics, I must have my facts straight. But, much as I think I would enjoy the exercise, mastering the math is not top priority. (I figure it would take me, including GR, two years as a full-time student, and perhaps triple that in my spare time. I bought some books 20 months ago to start; then I broke my hip in a cycling accident and lost eight months.)
  13. Jan 14, 2013 #12
    Well, it is advanced for me because I do not know tensors. I do know linear algebra and differential equations, so I have some sense of what is going on. And I have read that tensors are independent of coordinate system, in the sense that the value of the tensor is the same in all coordinate systems. So I think I get the gist.

    I find it curious, then, that Einstein makes a distinction in the 1905 paper between the force on the charge in its rest frame and the force in other frames. The one is due to E only; the other is some combination of E and v × B. That (in more precise terms than I originally framed my thoughts) is why I decided to ask the question in the original post. Given that the Faraday tensor makes no geometric distinction between the E and B fields, why does Einstein say that the emf due to velocity in the B field is an auxiliary concept, appearing only because (paraphrasing) the electric and magnetic fields are not independent of the movement of the frames?
  14. Jan 14, 2013 #13


    Staff: Mentor

    Tensors are a bit challenging if you really need GR. But if you are only interested in SR then you can learn the math very easily. I would estimate less than a week if you already know algebra and matrices and vectors.

    Personally, I never understood SR until I learned the mathematical framework of four-vectors. It is very simple, and once I learned it everything "clicked" almost immediately.
  15. Jan 15, 2013 #14
    I think the question raised in this thread is important, however I don't share the views expressed so far. What does the "asymmetry" noticed by Einstein consist in? It relates to the fact that according to the former electric and magnetic theories, a change of the reference frame affects the description of the EM field whereas it leaves invariant the relative speed between the magnet and the conductor. Here lies the asymmetry. In other words, the description of the EM field provided by the former theories is dependent on something else than the relative speed between both items, and in this context the only candidates are the velocities of both items in respect to the reference frames. This infringes the principle of relativity of motion: the cause of the observed current should depend only on the relative motion of both items in respect to each other.
    The only way to remove or resolve this asymmetry consists in proposing a new EM theory in which the change of reference frame leaves the expression of the EM field invariant, because it leaves invariant the relative motion between the magnet and the conductor... But this is not what Einstein proposes.
    Indeed in his 1905 SR paper Einstein invokes his new SR theory to state that the EM field (E,B) corresponding to the reference frame where the conductor is at rest can be transformed, using the Lorentz transformation derived in the SR context, into (E',B') corresponding to the same EM field as described in the reference frame where the magnet is at rest. Both descriptions lead to the same prediction concerning the observed current. So Einstein's strategy consists in demonstrating that the descriptions of the EM field provided by the former theories are equivalent in spite of their apparent incompatibility.... But the "asymmetry" is still there.
  16. Jan 15, 2013 #15


    Staff: Mentor

    No, it isn't. What do you think still depends on something other than relative speed?
    Last edited: Jan 15, 2013
  17. Jan 15, 2013 #16
    A bargain at twice the time. Can you recommend a book or on-line tutorial?

    [later] I went back to a 2010 post and saw that pervect recommended Taylor and Wheeler's Spacetime Physics because it has a four-vector approach. I bought that book then, and was partially through it when I broke my hip. I didn't remember it as particularly math-intensive; didn't think of it for four-vectors. I'll get back into it now.
    Last edited: Jan 15, 2013
  18. Jan 16, 2013 #17


    Staff: Mentor

    I don't have Taylor and Wheeler, but I have heard good things about it. I stumbled on 4-vectors at hyperphysics, and I still consider it a good intro:


    Then, of course, there are a bunch of good Wikipedia pages:

    You also need to know about spacetime intervals:

    And spacetime diagrams will help too:

    There are also many good university pages, but many of them seem to have a tendency to jump into tensor notation which can be a little intimidating at first. So I personally tended to stick with less formal sources in the beginning.
  19. Jan 16, 2013 #18


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    Taylor has the first chapter of the older 1965 editio of "Spacetime Physics" available for dowload at his website:


    Probably not enough to get the complete story, but enough to see if you like the book.
  20. Jan 16, 2013 #19
    Thank you both.
  21. Jan 18, 2013 #20
    Your question pushed me into second thoughts and I must withdraw my previous statement. Indeed a change of reference frame is a purely mental act by a theoretician who describes the same experiment in two different ways: it cannot affect the physical quantities involved in the experimental setup. However, according to Einstein's paper, E and B should no longer be considered as genuine physical quantities, but somehow "components" of a physical entity at an higher level, the EM field. Thus, and contrary to my previous statement, it is no longer a contradiction that their value changes through the swap between two reference frames, provided the overall EM field itself remains invariant. Then it is correct to state that the "asymmetry" has been removed by establishing the equivalence of (E,B) and (E',B') .
    But still there is one aspect which puzzles me: according to SR the impact of applying the Lorentz transformation to (E,B) is negligible for small values of v (in respect to c) envisaged here, so that (E',B') should be nearly identical to (E,B). How can one expect that the Lorentz transformation will justify the equivalence of both descriptions at stake? I hope you can explain this otherwise Einstein's strategy seems to fall apart.
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