# Relativistic particle in uniform magnetic field (solution check)

In summary, a relativistic particle in a uniform magnetic field follows a helical trajectory due to the Lorentz force, with the radius of the helix increasing as the particle's energy and magnetic field strength increase. The particle's energy is quantized in discrete levels, with the spacing between levels increasing as the magnetic field strength increases. This effect, known as the Landau quantization, is a result of the particle's spin interacting with the magnetic field. Additionally, the presence of a magnetic field can also lead to the production of electron-positron pairs, known as the Schwinger effect, due to the strong electric field induced by the particle's acceleration in the magnetic field. These phenomena have been extensively studied and confirmed through experiments
Homework Statement
A relativistic particle with mass ##m## and positive charge ##q## moves in the ##[yz]## plane in a circle of radius ##R## under the influence of a uniform magnetic field ##B=B_0 \hat x##. Find the period of the motion.
Relevant Equations
##\frac {d\overrightarrow p} {dt}=q \frac {\overrightarrow v} {c}\times \overrightarrow B_0##
My solution was as follows:

$$\frac {d\overrightarrow p} {dt}=q \frac {\overrightarrow v} {c}\times \overrightarrow B_0$$
The movement is in the ##[yz]## plane so ##|\overrightarrow v\times \overrightarrow B_0|=vB_0##, therefore: $$\biggr |\frac {dp} {dt}\biggr |= \frac {qvB_0} {c}.$$ On the other hand, $$p=\gamma m v=\gamma m R \omega$$ and since the force is perpendicular to the velocity, hence to the momentum: $$\biggr|\frac {dp} {dt}\biggr|=p \omega=\gamma mR \omega^2.$$ Comparing the expressions, I got $$\frac {qR \omega B_0} {c}=\frac {mR \omega^2} {\sqrt {1-\frac {R^2 \omega^2} {c^2}}}$$
which after some algebra yielded $$T=\frac {2\pi} {\omega}=2\pi \sqrt {\frac {m^2c^2} {q^2B_0^2}+\frac {R^2} {c^2}}.$$ My examiner marked this as false without explaining why, but I can't find any fault with this solution, so is it correct?

Delta2
$$T=\frac {2\pi} {\omega}=2\pi \sqrt {\frac {m^2c^2} {q^2B_0^2}+\frac {R^2} {c^2}}.$$ My examiner marked this as false without explaining why, but I can't find any fault with this solution, so is it correct?
The answer must reduce to the non-relativistic result in the case of ##v << c##.

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My solution was as follows:

$$\frac {d\overrightarrow p} {dt}=q \frac {\overrightarrow v} {c}\times \overrightarrow B_0$$
Are you sure? Did you try dimensional analysis on that?

Also, see https://en.wikipedia.org/wiki/Lorentz_force

PeroK said:
The answer must reduce to the non-relativistic result in the case of ##v << c##.
LaTeX pet peeve … ##v \ll c##

I got ...$$T=\frac {2\pi} {\omega}=2\pi \sqrt {\frac {m^2c^2} {q^2B_0^2}+\frac {R^2} {c^2}}.$$
This looks correct if you want to express ##T## in terms of the radius ##R##. You will often see ##T## expressed in terms of the speed ##v## of the particle: $$T = 2 \pi \frac{\gamma mc}{qB_0} = 2 \pi \frac{ mc}{qB_0\sqrt{1-v^2/c^2}}$$

PeroK and Delta2
TSny said:
Maybe using Gaussian or Heaviside-Lorentz units.
The thing is that the examiner does not seem to…

Orodruin said:
The thing is that the examiner does not seem to…
How can you tell? [Edited to correct a typo]

TSny said:
I can you tell?
In SI units the low velocity angular frequency would be ##qB/m##, not ##qB/mc## because that would have the wrong dimensions.

Orodruin said:
In SI units the low velocity angular frequency would be ##qB/m##, not ##qB/mc## because that would have the wrong dimensions.
But how do we know if the class is using SI units? In Gaussian units I believe the low velocity limit would be ##qB/mc##.

TSny said:
But how do we know if the class is using SI units?
I inferred that they are not using Gaussian or Heaviside-Lorentz units. If they were the answer would not be marked wrong as you already mentioned.

Orodruin said:
I inferred that they are not using Gaussian or Heaviside-Lorentz units. If they were the answer would not be marked wrong as you already mentioned.
OK. Maybe the OP will get back to us about the units.

Also what might fooled your teacher is that the two expressions (yours and @TSny at post #5) seem not to be equivalent, but they are (up to a factor of ##c##, check other posts about unit system ) if you do the algebra and replace $$v=\omega R,\omega=\frac{B_0cq}{\sqrt{B_0^2q^2R^2+m^2c^4}}$$

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Delta2 said:
Also what might fooled your teacher is that the two expressions (yours and @TSny at post #5) seem not to be equivalent, but they are if you do the algebra and replace $$v=\omega R,\omega=\frac{B_0cq}{\sqrt{B_0^2q^2R^2+m^2c^4}}$$
A possibility but ##v## was not given so answering with a ##\gamma## in there seems a bit weird to say the least. At the very least the examiner should be aware that there may be more than one equivalent way of writing the correct result and particularly in SR it tends to happen at a fairly high rate. This was one of the first things I learned when I started teaching SR: Many times looking at a final result that looks different from the one you computed at first glance, they are actually the same.

Of course, we don’t know if the examiner knows this or if the exam was corrected by a TA marking down everything that is not a carbon copy of the model solution.

Delta2
TSny said:
OK. Maybe the OP will get back to us about the units.
Orodruin said:
I inferred that they are not using Gaussian or Heaviside-Lorentz units. If they were the answer would not be marked wrong as you already mentioned.

The class is using Gaussian units, sorry for not being clear on that.

It looks like the examiner was wrong, I just wanted to be sure. The official answer was indeed ##\gamma T## where ##T## is the non-relativistic answer, which I could not match with my answer because I mistook ##v = \frac R T## for ##v = \frac {2 \pi R} T##. Thanks for the help everyone!

malawi_glenn and Delta2

## 1. What is a relativistic particle in a uniform magnetic field?

A relativistic particle in a uniform magnetic field is a theoretical concept in physics that describes the behavior of a charged particle moving at high speeds in a constant magnetic field. This phenomenon is governed by the principles of special relativity and electromagnetism.

## 2. How is the motion of a relativistic particle in a uniform magnetic field described?

The motion of a relativistic particle in a uniform magnetic field is described by the Lorentz force equation, which takes into account the effects of both the magnetic field and the particle's velocity. This equation is a fundamental equation in electromagnetism and is used to calculate the trajectory of the particle.

## 3. What is the significance of the Lorentz force equation in describing the motion of a relativistic particle in a uniform magnetic field?

The Lorentz force equation not only describes the motion of a relativistic particle in a uniform magnetic field, but it also provides insight into the fundamental principles of special relativity and electromagnetism. It is a key equation in understanding the behavior of charged particles in magnetic fields and has many practical applications in fields such as particle accelerators and plasma physics.

## 4. What are the key factors that affect the motion of a relativistic particle in a uniform magnetic field?

The motion of a relativistic particle in a uniform magnetic field is primarily affected by the strength of the magnetic field, the velocity of the particle, and the charge of the particle. The direction of the magnetic field and the initial position of the particle also play a role in determining its trajectory.

## 5. How is the motion of a relativistic particle in a uniform magnetic field different from that of a non-relativistic particle?

The motion of a relativistic particle in a uniform magnetic field is significantly different from that of a non-relativistic particle due to the effects of special relativity. Relativistic particles experience time dilation and length contraction, which can affect their trajectory and velocity. Non-relativistic particles do not exhibit these effects and are described by simpler equations such as the classical Lorentz force equation.

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