# Homework Help: Frequency of a proton moving in a circular orbit

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1. Dec 26, 2017

### Pushoam

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Assuming that the relation $v = \omega r$ is valid in relativistic calculation.

As the speed is greater for higher kinetic energy, $\omega_2$ is greater.

This is shown by only (d). So, I think (d) should be the correct answer.

$K.E._2 = (\gamma_2 -1) m_p c^2 = m_p c^2$

$v_2 = \frac { \sqrt{ 3}} 2 c$

Similarly, $v_1 = 0.6 c$

$\frac { \omega_1}{\omega_2} = \frac { v_1}{v_2} = \frac { 1.2}{\sqrt{ 3}}$

Is this correct?

Last edited by a moderator: Dec 26, 2017
2. Dec 26, 2017

### kuruman

It is not correct. Look up "cyclotron frequency" and take the appropriate ratio.

3. Dec 26, 2017

### Pushoam

qvB = $\frac { mv^2} r$

$\frac { v} r = \frac { qB} m$

So, the frequency is constant (independent of kinetic energy of the proton). Hence, the answer is option (a).

But then, v is different. Does it mean that r increases as v increases, that r is not constant?

Is this correct?

4. Dec 26, 2017

### kuruman

Only if the proton is non-relativistic and $m$ is the rest mass. Is this the case here?

5. Dec 27, 2017

### Pushoam

The proton's kinetic energy reaches to its rest mass energy. So, the calculation is relativistic.

Classically, $\frac { mv^2} r = qvB$ .....(1)

Relativistically,

$\vec F = \frac { d(\gamma m \vec v )}{dt} = \frac { d(\gamma)}{dt} m \vec v + \gamma \frac { d(m \vec v )}{dt}$

$= \gamma ^3 \frac { v \dot v}{c^2} mv ~\hat \phi + \gamma \frac { mv^2} r (~ - \hat r)$ .....(2)

Now, the $~\hat \phi$ component increase the speed, but it does not have any role in calculating frequency.

$\gamma \frac { mv^2} r = qvB$ .....(3)

Assuming $v = \omega r$ .....(4)

$\omega = \frac { q B}{\gamma m}$ .....(5)

$\frac { \omega_1}{\omega_2} = \frac { \gamma _2}{\gamma _1} = \frac {0.5 }{0.8} = \frac 5 8$ .....(6)

But if $v = \omega r$ is valid, then why $\frac { \omega_1}{\omega_2} \neq \frac { v_1}{v_2}$ ?

Is $r_1 \neq r_2$ ?

Is this correct?

Last edited: Dec 27, 2017
6. Dec 27, 2017

### kuruman

Yes. The proton spirals out as it gains energy.
Also, I don't know where you got the $\gamma$ values in equation (6), but I know that $\gamma$ is always greater than or equal to 1. To find $\gamma$, I would use the relativistic kinetic energy expression, $K=(\gamma-1)m_0c^2$.

7. Dec 27, 2017

### Pushoam

The following is wrong. While doing calculation, I took $\gamma = \sqrt { 1 - \frac {v^2}{c^2}}$ instead of $\frac 1 {\sqrt { 1 - \frac {v^2}{c^2}}}$.
The correcct one is $\frac { \omega_1}{\omega_2} = \frac { \gamma _2}{\gamma _1} = \frac {\frac 1 {0.5 }} {\frac 1 {0.8}} = \frac 8 5$

So, the frequency decreases as kinetic energy increases.

8. Dec 27, 2017

### Staff: Mentor

Right. In classical mechanics speed and radius change at the same rate, but relativistic the radius increases more (and the velocity is limited of course), so the frequency goes down with increasing energy.