# I Electromagnetic field acting on a charged particle

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1. Nov 18, 2016

### Mihai_B

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. Nov 18, 2016

### 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 $c = 1$.

3. Nov 18, 2016

### Mihai_B

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 !

4. Nov 18, 2016

### 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.