Frequency of a proton moving in a circular orbit

In summary, the conversation discusses the relationship between velocity, kinetic energy, and frequency in a relativistic calculation. It is determined that the frequency is constant and independent of the proton's kinetic energy, but the velocity and radius do change. It is also noted that the gamma values used in the calculation were incorrect and the correct ratio of frequencies is 8:5.
  • #1
Pushoam
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Homework Statement



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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?
 

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  • #2
It is not correct. Look up "cyclotron frequency" and take the appropriate ratio.
 
  • #3
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
Pushoam said:
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?
 
  • #5
kuruman said:
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?
 
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  • #6
Pushoam said:
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##.
 
  • #7
kuruman said:
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}}} ##.
Pushoam said:
## \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.
 
  • #8
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.
 
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Related to Frequency of a proton moving in a circular orbit

1. What is the frequency of a proton moving in a circular orbit?

The frequency of a proton moving in a circular orbit is determined by the speed and radius of the orbit. It can be calculated using the formula f = v / 2πr, where v is the speed of the proton and r is the radius of the orbit.

2. What factors affect the frequency of a proton in a circular orbit?

The frequency of a proton in a circular orbit is affected by the speed and radius of the orbit. As the speed increases, the frequency also increases. Similarly, as the radius decreases, the frequency increases. Additionally, the strength of the magnetic field and the charge of the proton also affect the frequency.

3. How does the frequency of a proton in a circular orbit relate to its energy?

The frequency and energy of a proton in a circular orbit are directly proportional. As the frequency increases, so does the energy. This is because the frequency is determined by the speed of the proton, which is related to its kinetic energy.

4. Can the frequency of a proton in a circular orbit be changed?

Yes, the frequency of a proton in a circular orbit can be changed by altering the speed or radius of the orbit. It can also be changed by changing the strength of the magnetic field or the charge of the proton.

5. How is the frequency of a proton in a circular orbit measured?

The frequency of a proton in a circular orbit can be measured using various techniques such as spectroscopy or particle accelerators. These methods involve analyzing the energy and velocity of the proton to calculate the frequency.

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