Special relativity - collision of a photon and an electron

In summary, the conversation discusses the collision of a photon and an electron in its rest system. The relevant conservation equations are discussed and the general method for solving these types of problems is explained. The final solution is that the photon's energy prior to the collision is equal to the electron's rest energy. The question of finding the photon's energy in the center of mass frame is also addressed.
  • #1
shwouchk
5
0
Hi!
Just wanted to say that this is my first post here, and I thank in advance anyone who spends time on this. Hopefully I will be able to return the favor someday.
Anyway...,

I thought I knew how to do this, but apparently I was wrong... I need to analyze the collision of a photon at an electron.

Homework Statement



A photon hits an electron in it's rest system. After the collision, the photon's direction is perpendicular to it's prior direction, having half the starting energy (first it was going x+, now it is going y+).
After the collision, the electron is traveling at some speed v at an angle -a to the x-axis (i.e. x+ y-)

1. write the relevant preservation equations.
2. Find the hitting photon's energy
... (if I'll understand these I'll do fine with the rest)


The "knowns" are C - the speed of light and E0 - the electron's rest energy.

Homework Equations


The Attempt at a Solution



I have to say that I'm new to using the Lorentz transformation and it's consequences so I likely made a few mistakes, but..:

I had the conservation of energy as:
(E1 is the photon's starting energy)
E_0 + E_1 = E_0 * \omega + (E1 / 2);

I wrote the conservation of momentum as follows:
(E_1 / C) = (E_0 / C) * \beta * cos a (X axis)
0 = (E_1 / 2C) - (E_0 / C) * \beta * sin a (Y axis)

Then (after hours of thought and tackling this) I realized that E=pc is only true for particles traveling very close to (or at) the speed of light, and revised it to:
P_1 = P_e * cos a (X axis)
0 = P_2 - P_e sin a (Y axis)
and used:
(E_1 / 2) = P_2 * C
P_2 = ([P_e]^2 + [m_e]^2 * C^2)^(1/2)

However, it got somewhat messy and then I got an impossible answer, so I suspect something is wrong here as well...

I would really appreciate if someone explains the general way of solving this kind of problems (or rather how to treat the momentum conservation, I guess) and/or what is wrong about my way.

Thanks in advance,
Shwouchk.
 
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  • #2
Let p1 and p2 denote the magnitude of the momenta of the photon before and after, and g=gamma(v), m0 the rest mass of the electron. So, m=m0g and E0=m0c^2.

From energy consvn,

p1c + m0c^2 = p2c + mgc^2.

From momentum consvn toward the initial direction pf photon, that is, the x-axis,

p1 = mv(cos a) = m0gv(cos a).

From momentum consvn toward the final direction pf photon, that is, the y-axis,

p2 - m0gv(sin a) = 0.

Solve for p1 and E1=p1*c.
 
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  • #3
Thanks for the reply!

These are my exact equations (from the second time around), so I know I did something right, but on the other hand, I still don't know what I did wrong because I didn't get a solution.

I will double check all my calculations though, since your post confirms that the initial equations were correct...

Thanks again - Hopefully I will reply in a short time, saying I've done it...
 
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  • #4
Alright, I have reached a solution, and it seems somewhat too simple, esp. after all the time I put into this. Can someone please confirm this is correct?

Here is what I did:
P_1 = 2P_2 = The photon's momentum before the collision.
P_2 = The Photon's momentum after the collision.
E_1 = P_1*C = The energy of the photon prior to the collision.
m_0 = The electron's rest mass.
E_0 = m_0*C^2 = The electrons rest energy.
v = the electron's velocity after the collision.
g = (1-[v^2/C^2])^(-1/2)
a = The angle between the X axis and the velocity of the electron after the collision.

Conservation of momentum:
Along X:
P_1 = 2P_2 = m_0*g*v*cos a (i)

Along Y:
0 = P_2 - m_0*g*v*sin a =>
P_2 = m_0*g*g*sin a (ii)

Conservation of Energy:
P_1*C + m_0*C^2 = P_2*C + m_0*g*C^2 =>
2P_2*C + m_0*C^2 = P_2*C + m_0*g*C^2 =>
2P_2 + m_0*C = P_2 + m_0*g*C =>
m_0*C = m_0*g*C - P_2 (iii)

Calculations:
(i)^2+(ii)^2:
4(P_2)^2 + (P_2)^2 = (m_0*g*v)^2 =>
P_2 = [5^(-1/2)]*m_0*g*v (iv)

(iv) -> (iii):
m_0*C = m_0*g*C - [5^(-1/2)]*m_0*g*v =>
C*5^(1/2) = g(C*5^(1/2) - v) =>
5C^2 = [5C^2 - 2*(5^{1/2})*C*v + v^2]/[1 - (v^2)/(C^2)] =>
5(C^2 - v^2) = 5C^2 - 2*[5^(1/2)]*c*v + v^2 =>
6v^2 = 2*[5^(1/2)]*C*v =>
v = sqrt(5/9)*C; (v)
g = 1/sqrt[1-5/9]=1/(2/3) = 1.5

(v) -> (iv):
P_2 = sqrt(1/9)*m_0*1.5*C =>
E_1 = 2P_2*C = 2*1.5*(1/3)*m_0*C^2 = m_0*C^2 = E_0;

So the final answer is that the photon's energy prior to the collision is equal to the electron's rest energy.

Is that correct?
 
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  • #5
Alright, now for the really hard part - how do I find the photon's energy (prior to the collision) in the center of mass system?
 
  • #6
shwouchk said:
So the final answer is that the photon's energy prior to the collision is equal to the electron's rest energy.

Is that correct?

That is correct. If you had done Compton scattering, that can be derived in a very short time. Good work.

shwouchk said:
Alright, now for the really hard part - how do I find the photon's energy (prior to the collision) in the center of mass system?

In relativity, a “centre of momentum frame” or CM frame is better defined than the centre of mass. It’s an IFR in which the total momentum is zero. The “centre of mass” is not a unique point but frame dependant.

If the speed of the CM frame is ‘u’, wrt the lab frame, then u = [(total momentum in lab frame/total energy in lab frame)]c^2 = c^2*p1/(E1 + m0c^2). From there, you can find the energy of the photon in a frame traveling with speed u to the right.
 
  • #7
Thanks, got it :)
 

Related to Special relativity - collision of a photon and an electron

1. What is the theory of special relativity?

The theory of special relativity, proposed by Albert Einstein in 1905, describes the relationship between space and time and how they are affected by the motion of objects at high speeds.

2. How does special relativity explain the collision of a photon and an electron?

In special relativity, the collision of a photon and an electron is described as an elastic collision, where the total energy and momentum remain conserved. The energy of the photon is transferred to the electron, causing it to gain energy and move at a higher speed.

3. Can the speed of light be exceeded in special relativity?

No, according to special relativity, the speed of light is the maximum speed at which any object can travel. This concept is known as the speed of light postulate and is a fundamental principle of the theory.

4. How does special relativity affect the concept of time dilation?

Special relativity predicts that time is not absolute and can be affected by the relative motion of objects. The faster an object moves, the slower time passes for that object. This phenomenon is known as time dilation.

5. What are the implications of special relativity for space travel?

Special relativity has significant implications for space travel, as it explains the effects of time dilation and length contraction at high speeds. This means that astronauts traveling at high speeds will experience time differently than those on Earth and will also perceive distances differently.

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