Frequency of a proton moving in a circular orbit

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

[/B]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?

 

[kuruman]

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

 

[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?

 

[kuruman]

So, the frequency is constant (independent of kinetic energy of the proton).

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

 

[Pushoam]

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

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?

 

[kuruman]

Is $r_1 \neq r_2$ ?

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, $K=(\gamma-1)m_0c^2$.

 

[Pushoam]

I don't know where you got the γγ\gamma values in equation (6), but I know that γγ\gamma is always greater than or equal to 1.

The following is wrong. While doing calculation, I took $\gamma = \sqrt { 1 - \frac {v^2}{c^2}}$ instead of $\frac 1 {\sqrt { 1 - \frac {v^2}{c^2}}}$.

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

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.

 

[mfb]

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.