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SR explanation of Magnetism

  1. Apr 15, 2015 #1
    Hi all,

    I have been struggling mightily through Purcell's book on Electricity and Magnetism, but chapter 5, which deals with the SR explanation of magnetism has me somewhat confused.

    Suppose we take the following scenario:


    In the Lab frame, F, electron 2 (e2) is moving towards the right with velocity v, while electron 1 (e1) is stationary. In Purcell's section "Force on a moving charge", he shows that

    Force12 = γ.Force12'

    where Force12 is the force on e1 due to e2. This makes sense to me since the electric field due to e2 will be larger in the Lab frame because e2 is moving. He then goes on to state a simple rule that, "The transverse component of the force on a particle is larger in the frame of the particle than in any other frame". This statement obviously applies in the when the source of the E field is moving relative to the particle.

    Anyway, in the following example in the same page, he assumes both e1 and e2 are at rest in the Lab frame, and then calculates

    Force12' = γ.Force12

    To me, this makes sense also, because in the moving frame, e2 would then appear to be moving, and so its E field is increased by γ. However, Purcell argues that this cannot be true since the force on a particle is larger in the frame of the particle than in any other frame. I don't understand the basis for this arguement. I thought the statement was just a general rule of thumb which he introduced assuming the particle was moving relative to the source of the E field acting on it.

    1) So my first question is what have I understood wrongly?

    In his next section he talks about interaction between a moving charge and other moving charges. In his setup, e2 is replaced by a current carrying wire. He first explains the assuming the electrons are moving to the right with speed v, and the protons are fixed, the linear density of electrons will be equal to that of the protons, and therefore e1 will experience no electric force.

    2) My second question is how can that be true? Shouldn't the density of the electrons be increased by γ thereby introducing an E field perpendicular to v?

    I appreciate any help here.

  2. jcsd
  3. Apr 15, 2015 #2


    Staff: Mentor

    I am not certain, but it sounds a little like you may be attributing a velocity to the E field itself rather than just the objects that create the E field.

    The comment itself seems to be more a statement of how force transforms, rather than a statement specifically about E-fields.
  4. Apr 16, 2015 #3
    Hi DaleSpam,

    thanks for the reply.

    In what way do you think I am attributing velocity to the E-field? My understanding is quite simple: if a charged object is considered to be moving in the frame of reference, then the E-field which it creates will be larger than if it were at rest. This E-field would therefore exert a greater force on a test charge irrespective of the motion of the test charge.

    My contention is that Purcell seems to spinning the argument around when he says that "the force on a particle is larger in the frame of the particle than in any other frame". This statement IMO is usually true because (going back to my example), in the rest frame of the test charge, the E-field acting on it is generated by a moving object. However, the statement is not true when the source of the E-field (the object) is also at rest in the rest frame of the test charge.

    Regarding your second comment, yes I agree. Purcell actually describes the whole section in terms of forces rather than E-fields. However, I find it much easier to understand in terms of E-fields. One of the reasons I posted this thread is because I'm not sure if I am correct in using E-fields rather than force. Purcell spends some time deriving the transformation laws for force (rate of change of momentum), which to me just complicates the issue when we have perfectly good E-fields to work with. Please let me know if I'm wrong here.


    btw, I realise that my second question wasn't right as Purcell included a disclaimer that the situation was an idealised one...
    Last edited: Apr 16, 2015
  5. Apr 16, 2015 #4
    "The transverse component of the force on a particle is larger in the frame of the particle than in any other frame".

    In other words moving forces are weaker. At least I like to think it that way.

    Are moving E-fields weaker or stronger? Let's consider a small charged accelerometer inside a charged plate capacitor, I mean between the plates. The accelerometer measures always the same acceleration caused by the E-field, velocity of the accelerometer does not matter. But according to the accelerometer the charges on the plates move closer as accelerometer gains velocity transverse to the E-field. So transversely moving E-fields are weaker, according to the accelerometer.

    "When the plates Lorentz-contracted the charges moved closer to me, but E-fields around the charges became weaker, that's why the E-field here where I am is unchanged" That's what the accelerometer says.

    (How do I know that the reading on the accelerometer stays constant? Well, a moving accelerometer is 1: more sensitive by factor of gamma, because all forces inside the accelerometer are weaker. 2: more massive by factor of gamma, because of relativistic mass increase.)
    Last edited: Apr 16, 2015
  6. Apr 16, 2015 #5
    Hi Jartsa,

    I'm sorry, but your explanation is very difficult to follow...

    If the charges move closer, then the charge density is higher. Shouldn't this result in a larger E-field? Why weaker?

    I don't understand why the charges move closer to the accelerometer in its frame. I thought it is moving transversely to the E-field. So there is no Lorentz contraction in the axis connecting the two plates.

    Actually in Purcell's book, he doesn't talk about relativistic mass increase. Perhaps this is something I overlooked.

  7. Apr 16, 2015 #6

    No problem I'll clarify.

    If the plates are quite large, some charges must be quite far from our test charge. If the plates happened to become less large, because of Lorentz-contraction, those charges would move closer to the test charge.
    Charge density increases = There are more charges near the test charge = Charges move closer to the test charge

    There was an accelerometer attached to the test charge. Its job is to measure how hard the E-field pulls the test charge in the frame of the test charge. Now I claim that the reading on the accelerometer is always the same. From that we can conclude that the E-field is always the same in the frame of the test charge.

    Let's temporarily forget the part about weaker E-fields around the charges.

    It's a little bit challenging to understandably explain the constant reading on the accelerometer ... maybe experts here can confirm that it's constant?

    Here's one more thought: The plates of a charged capacitor pull each other with some force. That force is largest in the rest frame of the plates. So in other frames it's smaller. Why would we not conclude that E-field is smaller in those frames where the force is smaller?
    Last edited: Apr 16, 2015
  8. Apr 17, 2015 #7
    I see. Thanks for the tips. I was missing the fact that the transverse momentum in the particles rest frame is minimum. I can see from the equations that it is true, but intuitively, not yet...

  9. Apr 17, 2015 #8

    Sorry Experts, don't bother. The reading is not constant.

    Instead it's like this: An accelerometer that is shaken measures larger accelerations in the direction transverse to the shaking motion, compared to an unshaken accelerometer.
  10. Apr 17, 2015 #9


    Staff: Mentor

    No worries. It was just an impression that I got from reading your post, but I think it may have had more to do with previous conversations with other people and I just got the wrong impression that you were asking the same thing.

    This is not always true. If you look at the Lienard Wiechert potentials you can see that the magnitude of the E-field is greater in the direction of travel, but less in directions perpendicular to the travel (unless I made a mistake in my math).

    I don't think that is correct since the E-field is not necessarily the only force. There is also the B-field. The force that transforms is not just the E-field force, but rather the total force.

    I believe that the point of Purcell's line of reasoning is to use the known way that forces transform in order to show that the E-field alone does not account for the force and therefore infer that there must also be another field, the B-field. I don't think that you can just jump to transforming the E-field, unless you are already assuming the existence of the B-field, which I believe is the conclusion he wishes to reach rather than an assumption he wants to start with.

    That isn't to say that Purcell's approach is the only valid approach, just that if you want to follow it then I think that you have to stick with the forces for a bit longer and not jump right to the fields.
  11. Apr 23, 2015 #10
    Yes, I managed to reach the same conclusion.

    This is how I understood Purcell's derivation of the magnetic force for any who are interested. Assume you have a wire with electrons moving at v0 and protons at rest, and a test charge moving parallel to the wire moving at v1. His method to calculate the magnetic force was as follows.

    1) First assume that the electron density in the wire is equal to the proton density in the lab frame. In the lab frame, this assumption results in the electric force being zero so that any force we calculate in the rest frame must be the magnetic force. (There is nothing wrong with this assumption, and I have derived the same final answer without this assumption).

    2) Calculate the E field due to the moving electrons/protons in the test charge's rest frame. The test charge's rest frame is used to calculate the E field because the B field is known not to exist in this frame. This allows us to directly use the classic formula F = qE.

    3) Using the E field in the test charge's rest frame (E'), the force acting on the test charge is qE'.

    4) Now transform the force in the test charge's rest frame to the lab's rest frame. The resulting force is the magnetic force since the experiment was set up so that the electric force in the lab frame would be zero.

    If we simply calculte the force in the lab's frame from the beginning, we only see the classic electrostatic answer. It is only when we calculate the force in the electron's frame and the transform it to the lab frame that we see the magnetic force.

    Last edited: Apr 23, 2015
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