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Classical Charged Particle backward in time/Feynman Stueckelberg interpretation

  1. Aug 1, 2012 #1
    Hello Friends,

    I was trying to illustrate intuitively (rather than rigorously) the equivalence of negative energy solutions moving backward in time to be equal to a particle of opposite sign a la Feynman. I begin with a quote from the maestro

    "A backwards-moving electron when viewed with time moving
    forwards appears the same as an ordinary electron, except
    it’s attracted to normal electrons - we say it has positive
    charge. For this reason it’s called a ‘positron’. The
    positron is a sister to the electron, and it is an example
    of an ‘anti-particle’. This phenomenon is quite general.
    Every particle in Nature has an amplitude to move backwards
    in time, and therefore has an anti-particle. (Feynman,1985):98 "

    I picked a super simple 1 dimensional gedanken example of of placing a test charge (particle 1) a fixed distance (x) away from an origin with another oppositely charged particle (A) fixed at the origin. Thus if the test charge began from rest, it would gain speed whilst approaching the origin. If the y axis were time therefore and the x axis the distance I would expect a curve coming into the origin (so as time gets less-> backward propagation, the particle gains momentum whilst approaching A) and the charged particle would ultimately arrive at the origin with some final momentum p oriented into the origin. Now if I were to view this (as Feynman says) with time moving forward, I would see particle 1 with p opposite in orientation, at A, moving away until it got to the position where particle 1 was at rest, tracing the trajectory this particle would also be at rest at that point. However according to Feynman's statement, I should see it continue to accelerate (if indeed it is being repelled!) So what am I doing wrong ?
    Many thanks for any input ! Simple arguments please !
  2. jcsd
  3. Aug 6, 2012 #2
    Hi azfar, and welcome to PF :smile:
    It's a little difficult to follow your post, because I can't picture the graph you're trying to describe with words. In particular, I can't figure out how you're thinking of time as a y-coordinate, starting out from a point (x,0) and drawing a curve that flows into the origin (0,0) at increasing (or decreasing) time!

    However, it sounds like you're starting off with two oppositely charged particles AND time flowing backwards, which would be a mistake; an electron going backwards in time looks like a positively charged particle going forwards in time.

    To be honest, I'm not convinced that your approach makes sense. I don't think you can really try to picture the trajectory of one particle relative to another as a process evolving in time at the same time as you think about one particle moving forwards in time and another moving backwards in time. I think you're trying to think about this by saying that "If two like charges repel, then they get further away from each other with time; so if I run time backwards, then the two get closer together, so it's like you've changed the sign of the charge on one of them." This is a mistake, because if you look at Newton's law
    [tex]m\frac{d^2 x}{dt^2}=k\frac{q_1 q_2}{x^2}[/tex]
    changing t->-t doesn't alter the left hand side at all, whereas changing the sign on one of the charges does alter the right hand side.

    I hope that helps, but like I said your post is a little hard to follow, so I may have misunderstood you.
  4. Aug 8, 2012 #3

    million thanks for your answer. You are right, what I am trying to say is not very clear, probably because I overthought it. Let me do this a different way and I would appreciate your input at each enumerated step below. Please remember I am trying to give the simplest example possible to my students, your critique of my first attempt is very much appreciated-and valid.

    This time I will assume a constant electric field in the negative x direction everywhere in space. I now place a test particle of charge +q and mass m at point x_f, in the far future at point t=t_f, with 0 initial velocity.

    1) I will now propagate this backward in time, thus with each succsessive t_j < t_f, I will see the charge move in the -x direction as t_j gets smaller. At some time t_i I make a final measurement of its veclocity. Of course v_f was 0, the magnitude of v_i is given by qE (t_f-t_i)/m, with v_i directed toward -x.

    2) If I now observe the trajectory with my sense of time moving forward, I will see at the earliest time (t_i) a velocity oriented toward +x with magnitude qE(t_f-t_i)/m propagating in the +x direction until it slows to a stop at time t_f with t_f>0. With the field oriented toward -x, I conclude that this particle carries charge +q.

    I am getting the same charge as in 1) or have I thoroughly confused myself ?

    thanks for any input,

  5. Aug 8, 2012 #4
    You know, in addition to anything you have to say about the above it would be very interesting to hear how you would explain this in the simplest terms possible.

  6. Aug 8, 2012 #5
    Thanks for clearing that up. This is a lot easier to understand now, and it looks as if I identified the source of your confusion correctly.

    To repeat my earlier post, what you've done is to think about the motion of a particle moving in an electric field, with an equation of motion given by Newton's law, and found that reversing the flow of time doesn't induce a change in the sign of the charge. This much is clear from the form of the equation that I wrote out above; all that's different is you've changed the electric field.

    The concept at work is a fundamentally more subtle one; and the best I can do is try and chop the maths out of a talk on this subject from Feynman himself. I don't know what your background is, but if you have a basic undergraduate background (specifically, if you know what a spacelike interval in special relativity is, and about perturbation theory in quantum mechanics) you might like to have a look at the original:
    http://books.google.co.uk/books?id=...ir_esc=y#v=onepage&q=Feynman weinberg&f=false

    You might find it helpful to follow that link and look at page 11 when having a look at what follows.
    Imagine some electron happily travelling forwards through time, starting out from some time t_0 at which you start observing it.. Now suppose that at some time t_2, at a position x_2, for whatever reason it randomly "decides" to travel backwards to some time t_1 with t_0<t_1 < t_2. At t_1, the electron has enough of messing about and gets back on with travelling forwards in time. You, as an ordinary observer, keep moving forwards in time. What do you see? Between t_0 and t_1, you see an electron travelling along. Between t_1 and t_2, you see "that" electron on its initial path to t_2, and on its path "back" to t_1 (think of this as being to some point x_1 in space, not on the initial trajectory), and then again on yet another trajectory forwards in time. In other words, you see *three* particles flying through space in the interval (t_1, t_2). At the point x_1, at time t_1, you see two particles being created. One of these is just the ordinary electron, flying off into the future. The second particle flies to x_2 at a time t_2, where it vanishes, along with the first electron.

    Electric charge is a conserved quantity. The only way this process of creation and destruction can be compatible with the conservation of charge is if the second particle, travelling from x_1 to x_2, carries positive charge. Then a positive and negative charge are created together at t_1, and a positive charge and negative charge are destroyed together at t_2.

    I hope that helps a bit.
  7. Aug 9, 2012 #6
    thanks very much ! First of all-silly me- I overthought this of course the force equation is second order I was going to get nowhere with that.

    Secondly thanks for the pointer, In fact, Feynman's bit on "The Reason For Anti-Particles" is what I used to explain this and I got caught up in a tizzy when a student of mine asked me if I could show the same for classical charged particles (my students do know special relativity, and perturbation theory) as a matter of curiosity.

    After reading your explanation, I went off and derived the equation of motion of a charged particle in covariant form and obtained:

    m \frac{d^2 x^\alpha}{d \tau^2} = \frac{q}{c} \frac{d x_\beta}{d \tau} F^{\alpha \beta}

    where F^{\alpha \beta} is the E-M tensor and I am using Gaussian CGS units, and \tau is the proper time. Now if I reverse the sign of the proper time, I have the same effect as reversing the charge on the right. So if I integrate this equation bearing in mind that I am an observer whose sense of time is forward this \int^{\tau_2}_{\tau_1} with \tau_2 greater \tau_1 with respect to d\tau I will get an impulse of opposite sign, thus the particle will have acquired opposite the impulse in the same E-M field corresponding to an opposite charge.

    Is this a valid way of showing this for a classical particle ?

    many thanks
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