Relativistic charged particle in a constant, uniform EM field

In summary, the conversation discusses finding the four momentum of a particle with mass m and charge q in a constant and uniform electric field E parallel to the y-axis and a magnetic field B parallel to the z-axis. The equations are solved by substituting equation (1) into (2) and (3) and then uncoupling the equations using the relation E^2 - B^2 = 0. This allows for the solution of the y-component of the four momentum, which can then be used to find the other components.
  • #1
Giuops
1
1
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.

246552


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

246553


... 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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  • #2
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
 
  • #3
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:
  • #4
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
 

Related to Relativistic charged particle in a constant, uniform EM field

1. What is a relativistic charged particle in a constant, uniform EM field?

A relativistic charged particle in a constant, uniform electromagnetic (EM) field is a particle that has both mass and an electric charge, and is moving in a uniform electromagnetic field that does not change with time.

2. What is the equation for the motion of a relativistic charged particle in a constant, uniform EM field?

The equation for the motion of a relativistic charged particle in a constant, uniform EM field is known as the Lorentz force equation. It can be written as F = q(E + v x B), where F is the force on the particle, q is the particle's charge, E is the electric field, v is the particle's velocity, and B is the magnetic field.

3. How does the motion of a relativistic charged particle differ from a non-relativistic one?

The motion of a relativistic charged particle differs from a non-relativistic one in that it takes into account the effects of special relativity, such as time dilation and length contraction. This means that the particle's mass, velocity, and energy will all be affected by its high speed and the presence of the EM field.

4. What are some real-life applications of studying relativistic charged particles in EM fields?

Studying relativistic charged particles in EM fields has many real-life applications, including in particle accelerators (such as the Large Hadron Collider), where physicists study the behavior of particles at high energies. It is also important in understanding the effects of radiation on spacecraft and astronauts, and in developing new technologies such as particle beams for medical treatments.

5. How does the strength of the EM field affect the motion of a relativistic charged particle?

The strength of the EM field affects the motion of a relativistic charged particle by increasing the force acting on the particle and altering its trajectory. A stronger field will exert a greater force on the particle, causing it to accelerate more quickly and possibly change direction. This can lead to complex and fascinating behaviors, such as spiral motion and synchrotron radiation.

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