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Magnetic field, frames of reference

  1. Aug 29, 2007 #1
    In the link:
    that I found here 5 minutes ago ( thanks for the link ) there are two drawings: the first and the second.

    In the first the link explains that there is a magnetic field, because the - charges are moving.

    In the second the link explains that there is not a magnetic field, although the + charges are moving.

    What I see is that the two drawings are completely symmetrical -> if there is a B field in the first there must be a B field in the second.

    To make the question simpler, forget the + charge that is outside the wire. Focus in the + and - charges in the wire and the B field.
    Do you agree with me?
  2. jcsd
  3. Aug 29, 2007 #2
    Read on in the text... the whole point of the page is to explain the paradox!
  4. Aug 29, 2007 #3
    It says that there is no magnetic force. There is non-zero magnetic field, but the test charge is in rest, so there's no magnetic force.
  5. Aug 30, 2007 #4
    I did read the text again, and I was wrong, there is a magnetic field in the two drawings.

    But, now including the + charge that is outside the wire, the two drawings are the same:
    - in the first drawing the test charge sees the + charges of the wire moving from right to left and the - charges at rest.
    - in the second drawing the test charge sees the + charges of the wire moving from right to left and the - charges at rest.

    There is one thing that is implicit ( not clearly stated ) in the link that I think its not true: the origin of the magnetic field is the movement of the - charges.

    As the test charge is at rest respect to the - charges there could not be any magnetic interaction between them. The magnetic field is due to the movement ( relative ) between the test charge and the + charges of the wire.

    Do you agree now ?
    Thanks by your patience.
  6. Aug 30, 2007 #5
    They are not insisting that - charges would be the source of magnetic field in both frames.

    The Biot-Savart law says that the magnetic field is proportional to a term [tex]v_{\textrm{source}}\times r / |r|^3[/tex], and the Lorentz force says that the magnetic force is proportional to a term [tex]v_{\textrm{test}}\times B[/tex]. These are absolute velocities in some chosen frame, not the relative velocities of the source and test charge! And notice that [tex]v_{\textrm{test}}=0[/tex] is sufficent to make magnetic force zero, even though the magnetic field was non-zero.

    Feel free to be confused, this is difficult stuff. In fact I can remember being in a debate, where one guy started insisting that the magnetic forces should be calculated using relative velocities. I wonder what he meant. Something like this? [tex]v_{\textrm{source}}\mapsto v_{\textrm{source}}-v_{\textrm{test}}[/tex] and [tex]v_{\textrm{test}}\mapsto v_{\textrm{test}}-v_{\textrm{source}}[/tex]. :surprised

    The famous claim that "laws of physics are the same in all inertial frames" is not an easy piece you can just understand through accepting it. You'll have to fight with these paradoxes.

    Edit: It could be I understood this
    wrong. If my response doesn't have anything to do with your question, I appologize. But it doesn't make sense to say magnetic field to be caused by relative velocities. Explain more clearly.
    Last edited: Aug 30, 2007
  7. Aug 30, 2007 #6
    From jostpuur:
    Yes, that is what I think ( so far )
    Ill try. In the first drawing of the link:
    1- What would happen if you remove the - charges of the wire ? My answer: there will be a magnetic force on the test charge.
    2- What would happen if you remove the + charges of the wire ? My answer: there will not be any magnetic force.

    Of course you can explain the way you want, but, please answer 1 and 2 short and clear.
  8. Aug 30, 2007 #7

    There will not be a magnetic force. When the (-) charges are removed, there is no magnetic field anymore. (+) charges don't move in this frame, and don't produce magnetic field.

    There will be a magnetic force. The (-) charges are moving, so they create a magnetic field, and then the test charge moves in this field and experiences the magnetic force.

    I might guess that your confusion stems from an assumption, that the forces are absolute. If in one frame a magnetic force exists, it could be that at the same instant in other frame the magnetic force does not exist.

    More about claims I have heard: Some folks claim, that if two electrons fly parallel to each others, then there will not be a magnetic force, because the electrons are in rest relative to each others. That is incorrect. It is true that there is no magnetic force in the frame where electrons are in rest, but in a frame where both electrons move, they do feel magnetic force. I emphasize: Forces are not absolute. They are relative, and depend on the chosen frame. Ultimatelty that is so because the time derivative of the three momentum, [tex]dp/dt[/itex], is not a Lorentz invariant quantity. This can be counter intuitive, because forces remain invariant in Galilean transformations.
    Last edited: Aug 30, 2007
  9. Aug 30, 2007 #8
    If I understand, in the case of two electrons flying parallel to each others:
    1 - if you are at rest and they are moving, you will see them moving away because there will be a magnetic force.
    2 - if you are running at the same velocity than them, they will not move away !!

    Im shure that there must be an experimental answer. Do you know any?
    Last edited: Aug 30, 2007
  10. Aug 30, 2007 #9
    If you check the equations more closely, you will see that the electric force will move the electrons away from each others, but the magnetic force slightly slows down that acceleration. The magnetic force is in fact very small compared to the electric force, and if the charges are not moving close to the speed of light, the electric force dominates this, and you don't see much difference in different frames.

    However, if the electrons move with very great speeds, close to the speed of light, then the magnetic force becomes stronger and the electrons accelerate away from each others a lot slower. This effect can be explained alternatively, without magnetism, merely by the time dilation! In the electrons rest frame they accelerate away with greater speed, but when you observe them from a different frame, the acceleration is slower. That is time dilation, or magnetism, which ever you want it to be :wink:
  11. Aug 31, 2007 #10
    Two charged spheres moving the same way the electrons you said.
    The spheres are joint ( physically connected ) with a spring. The spring acts, behaves, as a Spring scale, so you can see what is the force between the two spheres.

    1- You are at rest. You see the spheres when they pass in front of you and you can measure the force between the spheres.
    2 - You now move at the same velocity than the spheres. You measure the force.

    The spring is 90º respect the movement -> no Lorentz contraction.
    There is no aceleration -> no time dilation.

    Is there any difference between 1 and 2 measures ?
  12. Sep 1, 2007 #11
    I haven't thought about this spring scale thing before (as I had thought the previous stuff a lot), so I'm not fully sure. The correct answer most surely is that the spring scale measures the force in the electrons' (or perhaps they should be macroscopic charged balls now) rest frame. Only because you observe the spring scale from different frame won't change its reading. But why precisely does the spring scale not measure force correctly when it is moving? I think it's related to the way how the scale works in the first place. I'll try to not guess anything more right now.
  13. Sep 1, 2007 #12
    A tension is not absolute either. It is part of the energy-momentum tensor, and transforms somehow in boosts. You should take this into account when dealing with relativistic springs.
  14. Sep 1, 2007 #13
    Thanks for the answer. May be someone else could help, specifically if:

    is true.
  15. Sep 2, 2007 #14
    Assuming your reasoning,the only way the two situations are equivalent is suppose that the spring becomes "weaker" when seen from a rest frame.

    Ive come to this thinking about the following:

    A clock made up of a mass M and a spring whit k constant has a period T proportional to
    According to SR this period must be slowed down, when seen by a frame at rest, by gamma.
    The mass is increased by a factor of gamma.
    So the k must be divided by gamma when seen from a rest frame.

    Do you agree ?

    k of the spring as defined in "http://en.wikipedia.org/wiki/Hooke%27s_law"
    The clock consist in a mass attached by a spring at a fixed point. The mass moves up and down (no gravity ).
    gamma = 1 / ( 1- v^2/c^2)^0,5
  16. Sep 15, 2007 #15
    I disagree. I haven't seen an analysis of a relativistic oscillator, but I'm sure it's not that easy. You cannot just take a non-relativistic result, and substitute relativistic mass.
  17. Sep 15, 2007 #16


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    It's possible to reasonably assume that Hooke's law works in the lab frame. One must first ask if Hooke's law is a correct description of the dynamics in a boosted frame, before one can ask "what is the value of k". If Hooke's law no longer works, there is no sense in asking what the value of 'k' is.

    I believe we've been over much of this in another thread. If you orient the spring perpendicular to the boost, I haven't run across any problems so far in assuming that Hooke's law still holds. (This doesn't mean I haven't missed something).

    If you orient the spring parallel to the boost, there are some definite issues, though they may or may not show up depending on the value of the boost.

    Consider, rather than a spring-mass system, a sound wave travelling through a steel bar. Let [itex]v_{\mathrm{sound}}[/itex] be the speed of sound in the bar.

    We know that the correct solution in a boosted frame will be given by the relativistic velocity addition formula, i.e. we will have sound waves in the two directions with speeds v+ and v- equal to

    [tex]v_{+} = \frac{v+v_{\mathrm{sound}}}{1+v \, v_{\mathrm{sound}}/c^2}[/tex]

    [tex]v_{-} = \frac{v-v_{\mathrm{sound}}}{1- v \, v_{\mathrm{sound}}/c^2}[/tex]

    When v is small enough so that [itex]v_{+} - v = v - v_{-}[/itex] there may not be any issue with using a linearized version of the wave equation, but I don't see any way to make a linearized wave equation give the correct answer if this condition does not hold. For moderate boosts, this condtion may hold, but eventually an extreme enough boost will cause the condition to fail.

    Under extreme boosts one the problem is that sound waves moving in one direction (the + direction) must havae a different speed than sound waves in the other direction (the - direction). I do not believe this is compatible with Hooke's law, which implies that the sound wave must travel at some velocity that is independent of direction. One can probably avoid this issue by making the speed of sound "slow enough", but this is possible only if one restricts the problems being considered, it won't always work.

    So I believe that Hooke's law cannot always work - thus one must spend some time in any particular problem considering whether it works or not before even attempting to apply it.

    And the closest thing to a correct approach that I know of for relativistic dynamics I've discussed in https://www.physicsforums.com/showthread.php?t=183382

    The full version of the actual PDE's are quite intractable as one will see. Only by linearizing them can one work with them. Unfortunately, non-linear effects can be important in some scenarios, like a steel bar boosted to a high enough velocity.

    The simple approach, which I would strongly recommend, is this:

    work the problem in the lab frame, then use the Lorentz transform to describe what the results "look like" in any other frame. I.e. do the dynamics in the lab frame, and use kinematical principles to describe how the results work in different frames. The problem is that doing the dynamics requires some approximations to be tractable even with the simplest assumptions, and these approximations which work in the lab frame may not work in the boosted frame.

    Assuming that the Newtonian dynamics one used to use are compatible with relativity is not guaranteed to give one a correct answer. One may get lucky and get the right result, one may not be so lucky. This is not the right way to do physics IMO, - the right way involves getting the correct answer without relying on luck.
    Last edited: Sep 15, 2007
  18. Sep 15, 2007 #17
    From "https://www.physicsforums.com/showthread.php?t=182838"
    Do you ( both, or anybody who master relativity ) think the answer is true ?

    You didnt answer: ( me asking pervect )
  19. Sep 15, 2007 #18


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    I'm not quite sure of the point of this question, since I believe I've already answered it. As I said in my last post, the slow down of an oscillator can be predicted purely from kinematical considerations. Since I suspect I'm not being understood, I'll belabor the point a bit - please forgive me if you already know some of what I write below, because I'm not sure what it is that's missing from the communication, it's just that I get a sense that I'm not "getting through".

    The point I am making is that relativistic dynamics "done right" is difficult, and that you get non-linear PDE's (partial differential equations). But one knows what the result must be in advance because of the kinematics. The kinetmatics (see above) is easy. Therfore, the sensible way to solve the problem is to do the dynamics in the lab frame

    and then to use kinematics to express the results in an arbitrary frame.

    This is much easier than attempting to do relativistic dynamics, in part because it's difficult to see what approximations to make - as I pointed out at length in my last post, one can't always linearize the non-linear PDE's in an arbitrary frame, even though one knows one can linearize them in the lab frame.
    Last edited: Sep 15, 2007
  20. Sep 16, 2007 #19
    To pervect:
    I dont understand what you are talking about.
    I just want to know if the approach I used was correct.
    Im thinking that it can be applied to many problems:
    - A LC oscillator ( the L and C can be parallel or 90º to the boost )
    - radiation by black bodies
    - ...

    I dont mind if this is the "sensible way to solve the problem" but if the results I obtain are corrects.
  21. Sep 16, 2007 #20
    alvaros: yes, it is correct. However, you must be extremely careful about accounting for all the relativistic corrections. In practice, this is too hard to be doing all the time, so people just transform to a frame where things are easy/easier.
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