Relativistic charged particle in a constant, uniform EM field

AI Thread Summary
The discussion focuses on finding the four-momentum pμ(τ) of a charged particle in a uniform electromagnetic field, where the electric field E is parallel to the y-axis and the magnetic field B is parallel to the z-axis, with E equal to B. The user has derived four differential equations and solved the z-axis equation, but struggles with the x and y components. They have attempted to uncouple the equations using the electromagnetic field tensor Fμβ, but suspect there may be typos in their calculations. The conversation highlights the method of uncoupling equations as effective, while also acknowledging potential errors in the matrices presented. The user seeks clarification on their approach and the correct formulation of the equations.
Giuops
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Homework Statement
Find the four-momentum of a charged particle in an external, constant electric and magnetic field.
Relevant Equations
dpμ/dτ=q*Fμν*uν
pμ=muμ
I have to find pμ(τ) of a particle of mass m and charge q with v(0) = (vx(0), vy(0), vz(0)) in a electric field E parallel to the y-axis and a magnetic field B parallel to z axis, both constant and uniform, with E = B.

Here follows what I have done (see pictures below):

I wrote 4 differential equations (using that E=B) and called qE/m = a. The last equation (z axis) is immediately solved.

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To solve the others, I substituted equation (1) in (2) and (3), obtaining this:

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... but I don't know how to solve these. How do I find the x and y components of the four momentum?

Thanks for the read.
 
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ful disclosure i have seen this method online and not something i came up with myself

##
\frac{d P^{\mu}}{d \tau} = q F^{\mu \beta}u_{\beta}\\
##
i think ##F^{\mu \beta}## looks like this
##
\begin{pmatrix}
0&0&E&0\\
0&0&B&0\\
-E&-B&0&0\\
0&0&0&0
\end{pmatrix}
##
##
m\frac{d^2 u^{\mu}}{d \tau^2} = q F^{\mu \beta} \frac{d u_{\beta}}{d \tau}\\
m\frac{d^2 u^{\mu}}{d \tau^2} = \frac{q}{m} F^{\mu \beta} \frac{d P_{\beta}}{d \tau}\\
m\frac{d^2 u^{\mu}}{d \tau^2} = \frac{q^2}{m} F^{\mu \beta} F_{\beta \alpha}u^{\alpha}\\
F^{\mu \beta} F_{\beta \alpha} =
\begin{pmatrix}
E^2&-EB&E&0\\
-EB&-B^2&B&0\\
0&0&E^2-B^2&0\\
0&0&0&0
\end{pmatrix}
##
this has uncoupled the ##u^y## equation and since##E^2 - B^2 = 0## it is quite easy to solve for it and then you plug ##u^y## into earlier equations to get other components
someone tell me if it is wrong
 
timetraveller123 said:
##
F^{\mu \beta} F_{\beta \alpha} =
\begin{pmatrix}
E^2&-EB&E&0\\
-EB&-B^2&B&0\\
0&0&E^2-B^2&0\\
0&0&0&0
\end{pmatrix}
##
Looks like there are some "typos" here. We can let @Giuops find them if he/she wants to. :oldsmile:

This way of uncoupling the equations is pretty nifty.
 
Last edited:
TSny said:
Looks like there are some "typos" here.
oh i think i see them i think forgot to change some values after copying the first matrix
 
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