# Relativistic Electron

1. Sep 15, 2007

### strangequark

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

a) What potntial difference will accelerate an electron to the speed of light according to classical physics?
b) With this potential difference, what speed will the electron acheive relativistically?
c) What is the relativistic mass at that speed?
d) What is the relativistic kinetic energy?

2. Relevant equations

see solution
3. The attempt at a solution

a) $$q_{e} \Delta V = \Delta K = \frac{1}{2}m_{e}c^{2}$$

$$\Delta V =\frac{m_{e}c^{2}}{2q_{e}}=255,499 Volts$$

b) here i'm assuming that i can say (is this assumption valid??):

$$q_{e} \DeltaV = \gamma m_{e}c^{2}-m_{e}c^{2}$$

then solve for $$\gamma$$,

$$\gamma=\frac{q_{e}\DeltaV}{m_{e}c^{2}}+1$$

the velocity follows immediately

c) $$m = \gamma m_{e}$$

d) here's where I get confused, if i calculate kinetic energy using:

$$K=\gamma m_{e}c^{2}-m_{e}c^{2}$$

I get the same kinetic energy ($$q_{e} \Delta V$$) I assumed in part b (obviously)... now this makes some sense to me, in the respect that the classical kinetic energy at the speed the electron would obtain relativistically is lower than this, but it is also equal to $$\frac{1}{2}m_{e}c^{2}$$... i don't think this is a contradiction, but it seems strange... is my underying assumption in part b wrong?

2. Sep 16, 2007

### dynamicsolo

This is fine. (If you have learned about masses of particles expressed in electron-volts, you'll see that this value makes sense. Otherwise, don't worry about that for now...)

For some reason, your delta-V isn't displaying; but your intended expression (as seen in the TeX code) is correct, I believe, so you get gamma = 3/2 .

The rest of this seems correct to me as well. You *will* get the same KE as you had classically (your instructor is being cute here). There is no contradiction: the Lorentz factor alters the velocity corresponding to various values of KE. As you note, the speed at which KE is one-half the rest mass-energy of the electron is lower than the classical result. The function now permits KE to grow without limit as speed approaches c ; the illustration of this is, I think, the main point of this exercise.

3. Sep 16, 2007

### strangequark

Great, thanks... I looked at it quite a few times, and couldn't figure out how it could be wrong, but it seemed too convenient... I have another quick question actually if you don't mind...

A different problem is looking at a particle decay in the center of momentum frame (S) and the laboratory frame (S')...

It gives me the total energy of the parent particle,$$E_{total}= 498 MeV$$, and the individual rest energies of the daughter particles, 135 MeV each. The particle decays at some point, and the two daughter particles leave w/ energy and momentum $$p_{1},E_{1}$$ and $$p_{2}, E_{2}$$. In frame S, the parent particle is at rest. In frame S' it moves with a velocity of $$v = \frac{3c}{5}$$, and the daughter particles have energies and momenta $$E_{1}',p_{1}'$$ and $$E_{2}',p_{2}'$$.

For the S frame, (I've assumed that they energies/momenta were equal) and said that:

$$E_{total}=E_{1}+E_{2}=2E_{daughter}$$

Then obtained the momentum for each from:

$$E^{2}=(pc)^{2}+(m_{0}c^{2})^{2}$$

For the S' frame, I'm thinking that if I find the velocities of each particle with respect to S' using the Lorentz velocity transform, I can use that velocity to find the momentum via:

$$p=\gamma m_{0} u_{S'}$$

where u will be velocity relative to S'... Now, if I am looking at the situation the right way, I will have a different $$\gamma$$ for each of the particles when figuring their momentum in S'. So for the momentum of particle 1, I'll have:

$$p_{1}=\frac{m_{0}u_{S'}}{\sqrt{1-(\frac{u_{S'}}{c})^{2}}}$$

Would that be correct?

Last edited: Sep 16, 2007