Charged particle in a B field, tensor notation

In summary, the particle experiences an electric field with a constant value and a magnetic field with a changing value. The equations of motion are solved to find the particle's position and momentum in a plane.
  • #1
fluidistic
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Homework Statement


A charged particle of charge q with arbitrary velocity ##\vec v_0## enters a region with a constant ##\vec B_0## field.
1)Write down the covariant equations of motion for the particle, without considering the radiation of the particle.
2)Find ##x^\mu (\tau)##
3)Find the Lorentz transformation such that the motion of the particle is restricted to a plane.
4)Calculate the dipole term of the fields and the average radiated EM power by the particle.

Homework Equations


##\tau=t\gamma##

The Attempt at a Solution


1)##\frac{dp^\alpha}{d\tau}=qU_\beta F^{\alpha \beta}## would be my reply.
Where ##p^\alpha## is the four-momentum, ##U_\beta## is the four-velocity and ##F^{\alpha \beta}## is the electromagnetic tensor.
2)I solved the equation found in 1) by writing the tensors under matrix form. I got that ##x^\alpha=(ct,x^1(t),x^2(t),0)## where ##x^1(t)=\frac{mv_{0x}}{qB_0}\sin \left ( \frac{qB_0 t}{m} \right ) + x^1_0## and ##x^2(t)=\frac{mv_{0x}}{qB_0}\cos \left ( \frac{qB_0 t}{m} \right ) + x^2_0##. Now I just realize that they asked in terms of ##\tau## and not t. Well I get ##x^1(\tau)=\frac{mv_{0x}}{qB_0\sqrt{\gamma}}\sin \left ( \frac{qB_0\sqrt{\gamma}\tau}{m} \right )##.
3)I'm out of ideas on this one. I know I can express the tensor of the Lorentz transformation as a matrix (I guess this way: https://en.wikipedia.org/wiki/Lorentz_transformation#Matrix_forms) but I'm not sure what they're asking for when they ask for the motion to be in a plane. Isn't the motion already in the x-y plane in my case? I picked ##\vec B_0## as ##B_0\hat z## originally.
 
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  • #2
fluidistic said:
but I'm not sure what they're asking for when they ask for the motion to be in a plane. Isn't the motion already in the x-y plane in my case?

Yea, your motion is in a plane, but it is not the most general solution to the equations of motion. In trying to solve 2, you solved the special case 3.
 
  • #3
Orodruin said:
Yea, your motion is in a plane, but it is not the most general solution to the equations of motion. In trying to solve 2, you solved the special case 3.
I see. Is it because I chose the magnetic field to be oriented in the z direction? I would have thought that I'd need a boost in a particular direction to solve 3), not picking a convenient orientation.
 
  • #4
fluidistic said:
I see. Is it because I chose the magnetic field to be oriented in the z direction?
It is the combination of selecting the B-field in the z-direction and assuming that there is no velocity component in the z-direction.
 
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  • #5
Orodruin said:
It is the combination of selecting the B-field in the z-direction and assuming that there is no velocity component in the z-direction.
Well then I miserably failed, even at solving part 3 since there's no way I can pick the velocity to vanish in the z direction (unless I apply a boost maybe?). I'll try to solve the problem in a correct way. Which means, I believe, that I can pick the orientation for the B field to be in the z direction but I can't assume that there's no speed in the z direction.
 
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  • #6
Ok I've redone part 2. I had made some mistakes previously I think.
I get [tex]x^\mu (\tau) = (c\gamma \tau , \gamma ^2 v_{0x}\frac{B_0q}{m}\sin \left ( \frac{m\tau }{q\gamma B_0} \right ) +x_0 , -\frac{mv_{0x}}{qB_0}\cos \left ( \frac{m \tau}{q\gamma B_0} \right ) +y_0, z_0 )[/tex].
The fact that I get different amplitudes in front of the sine and cosine really bothers me. This would make an elliptical trajectory in the x-y plane projection instead of circular path. Am I doing things wrong?
 
  • #7
The easiest way to check is to insert your result into the original differential equation ...
 

1. What is a charged particle in a B field?

A charged particle in a B field refers to a particle that is moving in a magnetic field. The particle experiences a force due to the interaction between its charge and the magnetic field, causing it to move in a curved path.

2. What is the significance of tensor notation in this concept?

Tensor notation is used to describe the direction and magnitude of the magnetic field. It helps in understanding the complex relationship between the charged particle's motion and the magnetic field.

3. How does the direction of the magnetic field affect the motion of the charged particle?

The direction of the magnetic field affects the direction of the force experienced by the charged particle. The force is always perpendicular to the direction of the magnetic field and the velocity of the particle, causing it to move in a circular or helical path.

4. What is the difference between a positively charged particle and a negatively charged particle in a B field?

The direction of the force experienced by a positively charged particle is opposite to that of a negatively charged particle in a B field. This is due to the opposite direction of the magnetic force on particles with opposite charges.

5. How does the strength of the magnetic field affect the motion of the charged particle?

The strength of the magnetic field affects the magnitude of the force experienced by the charged particle. A stronger magnetic field will result in a larger force, causing the particle to move in a tighter curved path.

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