Relativistic particle in uniform magnetic field (solution check)

[Adgorn]

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?

 

[PeroK]

$$T=\frac {2\pi} {\omega}=2\pi \sqrt {\frac {m^2c^2} {q^2B_0^2}+\frac {R^2} {c^2}}.$$ 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 $v << c$.

 

[Orodruin]

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

 

[Orodruin]

The answer must reduce to the non-relativistic result in the case of $v << c$.

LaTeX pet peeve … $v \ll c$

 

[TSny]

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: $$T = 2 \pi \frac{\gamma mc}{qB_0} = 2 \pi \frac{ mc}{qB_0\sqrt{1-v^2/c^2}} $$

 

[TSny]

Are you sure? Did you try dimensional analysis on that?

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

Maybe using Gaussian or Heaviside-Lorentz units.

 

[Orodruin]

Maybe using Gaussian or Heaviside-Lorentz units.

The thing is that the examiner does not seem to…

 

[TSny]

The thing is that the examiner does not seem to…

How can you tell? [Edited to correct a typo]

 

[Orodruin]

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.

 

[TSny]

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 $qB/mc$.

 

[Orodruin]

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.

 

[TSny]

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.

 

[Delta2]

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 $$v=\omega R,\omega=\frac{B_0cq}{\sqrt{B_0^2q^2R^2+m^2c^4}}$$

 

[Orodruin]

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.

 

[Adgorn]

OK. Maybe the OP will get back to us about the 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.

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!