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I Electromagnetic field acting on a charged particle

  1. Nov 18, 2016 #1
    Can anyone help me find any mistake in this expansion ? (I've asked it also in other places but I got no answer))

    Pα= e Fαβ Uβ

    c = speed of light
    m = "rest" mass
    e = charge
    a = sqr(1 - v2/c2)
    v2 = vx2 + vy2 + vz2
    dτ = dt a (proper time)
    momentum 4 vector : Pα = [mc/a , mvx/a , mvy/a , mvz/a ]

    velocity 4 vector : Uβ = [c/a , vx/a , vy/a , vx/a ]

    electromagnetic tensor matrix Fαβ =
    | 0 -Ex/c -Ey/c -Ez/c |
    | Ex/c 0 -Bz By |
    | Ey/c Bz 0 -Bx |
    | Ez/c -By Bx 0 |

    Expending Pα= e Fαβ Uβ we get

    - for P0 :
    d (m c/a) / dt = - e/(c a) (Ex vx+ Ey vy + Ez vz)
    m c / a' = - e/(c a) (Ex vx+ Ey vy + Ez vz) + m c / a

    - for P1 :
    d (m vx)/(a dt) = e/a (Ex- Bz vy + By vz)
    m v'x/a' = (e/a) (Ex- Bz vy + By vz) dt + m vx/a

    - for P2 :
    d (m vy)/(a dt) = (e/a) (Ey+ Bz vx - Bx vz)
    m v'y/a' = (e/a) (Ey + Bz vx - Bx vz) dt + m vy/a

    - for P3 :
    d (m vz)/(a dt) = (e/a) (Ez- By vx + Bx vy)
    m v'z/a' = (e/a) (Ez - By vx + Bx vy) dt + m vz/a


    All up:
    (0) m c/a' = - e/(c a) (Ex vx+ Ey vy + Ez vz) + m c / a = D
    (1) m v'x/a' = (e/a) (Ex- Bz vy + By vz) dt + m vx/a = A
    (2) m v'y/a' = (e/a) (Ey + Bz vx - Bx vz) dt + m vy/a = B
    (3) m v'z/a' = (e/a) (Ez - By vx + Bx vy) dt + m vz/a = C

    So:
    A/v'x = B/v'y = C/v'z = D/c = m/a'

    Are there any mistakes here ?
    v'x = A c / D
    v'y = B c / D
    v'z = C c / D
    a' = m c / D

    where the new U'β = [c/a' , v'x/a' , v'y/a' , v'z/a'] and a' = sqr(1 - v'2/c2)

    Thanks.
     
  2. jcsd
  3. Nov 18, 2016 #2

    PeterDonis

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    2016 Award

    Staff: Mentor

    The formula you are starting with is not correct as you write it there (although you appear to be including a derivative on the LHS later in your post--but that's also not quite right as you do it, see below). The correct formula is

    $$
    \frac{dP_{\alpha}}{d\tau} = e F_{\alpha \beta} U^{\beta}
    $$

    where ##\tau## is the particle's proper time. Note that ##\tau## is not the same as ##t##, the coordinate time. See below.

    This isn't quite right because you have written ##P_\alpha## with a lower index and ##U^\beta## with an upper index, and that makes a difference. The components you give for ##U^\beta## are correct; but the "0" or "t" component of ##P_\alpha## should have a minus sign because of the lower index. (This is assuming we are using the ##-+++## sign convention for the metric signature.)

    I see three issues here. First, you should be evaluating ##d(mc/a) / d\tau##, not ##d(mc/a) / dt##. Second, you don't appear to be taking the derivative on the LHS correctly: ##d(1/a) / d\tau## should evaluate to ##-1/a^2 \left( da / d\tau \right)## (and even then you probably want to use the chain rule treating ##a## as a function of ##v## in order to get an expression in terms of ##dv / d\tau##). Third, I don't understand where the ##mc/a## term on the RHS in the second line comes from.

    I haven't looked at the rest of your derivation because it looks to me like you need to start again from scratch in the light of the issues above.

    One other suggestion: use standard notation. What you are calling ##1/a## is usually called ##\gamma##. It is also easier to use units where ##c = 1##.
     
  4. Nov 18, 2016 #3
    Hi and thanks for the reply!

    - I wrote Pα but i really actually used Pα = " Pα/dτ " . Sorry.

    - You are correct - I got the sign wrong for P0. It's -mc/a

    - d(mc/a)/dτ, d(mvx/a)/dτ, etc, which is what I ment to write (my original notes are correct), sorry my transcription got a little wrong.

    - regarding differentials :
    d(mc/a) = mc/a' - mc/a = new - old

    So
    (new - old)/dτ = - e/(c a) (Ex vx+ Ey vy + Ez vz)

    new - old = - e/(c a) (Ex vx+ Ey vy + Ez vz) dτ

    new = - e/(c a) (Ex vx+ Ey vy + Ez vz) dτ + old

    (-mc/a') = - e/(c a) (Ex vx+ Ey vy + Ez vz) dτ + (-mc/a)

    -mc γ' = - (e/c) γ (Ex vx+ Ey vy + Ez vz) dτ - mcγ

    mc γ' = (e/c) γ (Ex vx+ Ey vy + Ez vz) dτ + mcγ

    Correct ?
    So i don't need to use derivative since I'm dismanteling the differential and putting in RHS what is related to old and on LHS what is related to new. Right ?

    Thanks anyway !
     
  5. Nov 18, 2016 #4

    PeterDonis

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    2016 Award

    Staff: Mentor

    I assume you mean ##dP_\alpha / d\tau##.

    They aren't differentials, they're derivatives.

    I don't see the point of approximating this way since you can easily evaluate the derivative of ##mc / a## (or ##m \gamma## if you use my suggested notation and units) with respect to ##\tau## exactly, without having to approximate.
     
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