Photon collides with stationary electron

[BenjyPhysics]

Hey, this question did not come with a mark scheme and I want to make sure that I am going about it in a correct manner, please could someone check what I've done (and answer a couple of questions I have raised)? Thank you in advance! :D

Homework Statement

Give expressions for the total energy E and the momentum p of a relativistic particle with mass m and speed u. Use these to show that E2 = p2c2 + m2c4. State the form this takes for photons, which have zero mass, and show that this is consistent with the de Broglie relations E = h f and p = h/λ. [4]

An electron (mass m) is at rest in a laboratory, when a photon with energy E0, equal to the rest energy mc2 of the electron, collides with it. Find the initial momenta of the photon and the electron in a frame moving with speed v in the same direction as the photon. Hence show that the speed of the zero momentum (ZM) frame is c/2 . [4]

In the ZM frame, after the collision, the photon is observed to leave at 90◦ to its original direction. Find the speed and direction of the electron in the ZM frame. By transforming both velocity components of the electron back to the lab frame, find the speed and direction of the electron in the lab frame. [5]

Find also the final energy of the photon in the lab frame

Homework Equations

$${u'_x} = \frac{{{u_x} - v}}{{1 - \frac{{{u_x}v}}{{{c^2}}}}}$$

$${u'_y} = \frac{{{u_y}}}{{\gamma (1 - \frac{{{u_x}v}}{{{c^2}}})}}$$

$$f' = \sqrt {\frac{{1 - \frac{v}{c}}}{{1 + \frac{v}{c}}}} f$$

The Attempt at a Solution

I have $$E = \gamma m{c^2}$$ for the total energy of the particle, and $$p = \gamma mu$$ for its momentum, and put these into the E-p invariant to show the result given holds true.


I then stated the momentum of a photon as $$E = pc$$ and used the two relationships to demonstrate their consistency with this.

Main part of question

Now, in the laboratory frame the photon has energy $${E_0} = mc{}^2$$

When the frame is moving at a speed v, the velocity of the electron is given by $${{u'}_x} = \frac{{0 - v}}{{1 - 0}} = - v$$ so that it has a momentum of $${\gamma _v}mv$$

The photon has an initial frequency of $$\frac{{{E_0}}}{h}$$, and in the new frame this frequency becomes $$f' = \sqrt {\frac{{1 - \frac{v}{c}}}{{1 + \frac{v}{c}}}} \frac{{{E_0}}}{h} = \sqrt {\frac{{1 - \frac{v}{c}}}{{1 + \frac{v}{c}}}} \frac{{m{c^2}}}{h}$$ which leads me to a momentum for the photon of $$p = \sqrt {\frac{{1 - \frac{v}{c}}}{{1 + \frac{v}{c}}}} mc$$

For the ZM frame, these two momenta are equal and opposite... so I solved $$\sqrt {\frac{{1 - \frac{v}{c}}}{{1 + \frac{v}{c}}}} mc = \frac{{mv}}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }}$$ for $$v = \frac{c}{2}$$

ZM Frame

Photon goes off at 90 degree angle to initial direction of travel, this implies the electron moved off at a 180 degree angle to the photon (in the opposite direction), as this is a must for the ZM frame. The electron moves off (in the ZM frame), at a speed $$v = \frac{c}{2}$$

Transforming back to lab frame

Electron velocity components...

$${u_x} = \frac{{{{u'}_x} + v}}{{1 + \frac{{{{u'}_x}v}}{{{c^2}}}}}$$ with $$v = \frac{c}{2}$$ and $${{u'}_x} = 0$$ yields $${u_x} = \frac{c}{2}$$

$${u_y} = \frac{{{{u'}_y}}}{{\gamma (1 + \frac{{{{u'}_x}v}}{{{c^2}}})}}$$ with $${{u'}_y} = - \frac{c}{2}$$ yields $${{u'}_y} = \frac{{ - \frac{c}{2}}}{\gamma }$$

Is gamma 1 in this case? If so, could someone explain why this is true? Is it because the frame has no velocity component in the y-direction?

Using the transforms I calculated that in the lab frame the electron goes in a south-east direction (hence 45 degree angle to initial trajectory of photon) at a speed of $$\frac{{\sqrt 2 }}{2}c$$

Final photon energy
I'm not quite sure how to find the energy of the photon afterwards, but since we are back in the lab frame surely the observed frequency is back to normal and the energy is once again $${E_0} = m{c^2}$$ ?

 

[BvU]

Hello again, Benjy. Making progress and now at B8, eh? Good you typed it out. You also read it while typing ?
What is a mark scheme ? Not the [numbers] then ?

I have $$E = \gamma m{c^2}$$ for the total energy of the particle, and $$p = \gamma mu$$ for its momentum, and put these into the E-p invariant to show the result given holds true.

On the surface this looks great. How did you do that? I can't find a $\gamma$ for a photon... Unless you want to allow $\infty$ and then claim that $\infty \cdot 0 = 1$ with dry eyes.

 

[BenjyPhysics]

Hello again, Benjy. Making progress and now at B8, eh? Good you typed it out. You also read it while typing ?
What is a mark scheme ? Not the [numbers] then ?

On the surface this looks great. How did you do that? I can't find a $\gamma$ for a photon... Unless you want to allow $\infty$ and then claim that $\infty \cdot 0 = 1$ with dry eyes.

An interesting response no doubt but I was referring to a particle of mass 'm' and speed 'u', this was the first part of the question - before a photon got involved!

 

[vela]

$${u_y} = \frac{{{{u'}_y}}}{{\gamma (1 + \frac{{{{u'}_x}v}}{{{c^2}}})}}$$ with $${{u'}_y} = - \frac{c}{2}$$ yields $${{u'}_y} = \frac{{ - \frac{c}{2}}}{\gamma }$$

Is gamma 1 in this case? If so, could someone explain why this is true? Is it because the frame has no velocity component in the y-direction?

No, $\gamma$ corresponds to $v$, the velocity of the lab frame relative to the ZM frame.

Using the transforms I calculated that in the lab frame the electron goes in a south-east direction (hence 45 degree angle to initial trajectory of photon) at a speed of $$\frac{{\sqrt 2 }}{2}c$$

Final photon energy
I'm not quite sure how to find the energy of the photon afterwards, but since we are back in the lab frame surely the observed frequency is back to normal and the energy is once again $${E_0} = m{c^2}$$ ?

No, the energy of the photon won't turn out to be the original energy. Some of the energy of the initial photon was given to the electron so it could recoil.

 

[BenjyPhysics]

No, $\gamma$ corresponds to $v$, the velocity of the lab frame relative to the ZM frame.No, the energy of the photon won't turn out to be the original energy. Some of the energy of the initial photon was given to the electron so it could recoil.

Cool, so $${{u'}_y} = \frac{{ - \frac{c}{2}}}{{{\gamma _{\frac{c}{2}}}}} = \frac{{ - \frac{c}{2}}}{{\frac{{2\sqrt 3 }}{3}}} = - \frac{{\sqrt 3 }}{4}c$$ and the electron in the lab frame has a velocity of $$\frac{{\sqrt 7 }}{4}c$$ in a direction 49.1 degrees to the initial photon trajectory.

So the electron has kinetic energy $$({\gamma _{\frac{{\sqrt 7 }}{4}}} - 1)m{c^2} = \frac{1}{3}m{c^2}$$ which is provided by the photon, so the final photon energy is $${E_0} - \frac{1}{3}m{c^2} = m{c^2} - \frac{1}{3}m{c^2} = \frac{2}{3}m{c^2}$$

Is this correct?

 

[BvU]

Check the direction of the electron. I get 90 - 49.1

 

[BenjyPhysics]

Check the direction of the electron. I get 90 - 49.1

Yep you're right, thanks :D