# Derive a formula for momentum in terms of kinetic energy

1. Oct 14, 2010

### martinhiggs

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

Using:
particle velocity, beta
particle momentum, p
total energy, E
Lorentz factor, gamma
kinetic energy, KE

Derive an equation for momentum as a function of kinetic energy. The functions have to depend either on the variable in the bracket, p(KE), or on a constant.

3. The attempt at a solution

This is what I've done so far, and I am now stuck, and unsure if the way I am doing it is correct or if there is a different approach.

$$E^{2} = p^{2}c^{2} + m^{2}c^{4}$$

$$KE = E - m_{0}c^{2}$$

$$KE = \sqrt{p^{2}c^{2} + m^{2}c^{4}} - m_{0}c^{2}$$

$$p^{2} = \frac{KE^{2}}{c^{2}} - m^{2}c^{2} - m_{0}^{2}c^{4}$$

The only thing I could think of doing next is:

$$KE = \frac{p^{2}}{2m_{0}} , m_{0} = \frac{p^{2}}{2KE}$$

$$p^{2} = \frac{KE}{c^{2}} - m^{2}c^{2} - \frac{p^{4}}{4KE^{2}}c^{2}$$

$$p^{2} + \frac{p^{4}}{4KE^{2}}c^{2} = \frac{KE}{c^{2}} - m^{2}c^{2}$$

$$p^{2}(1 + \frac{p^{2}}{4KE^{2}}c^{2}) = \frac{KE}{c^{2}} - m^{2}c^{2}$$

I'm not sure if this is the best or easiest way to do this, as it seems to be pretty messy, and I also have one more m in the equation that I need to get rid of but am not sure of the best way of doing so.

Any help will be greatly appreciated :)

2. Oct 15, 2010

### martinhiggs

Ok, so I've been working on this problem for about 24 hours and I think I'm finally getting somewhere with it. In class we were given a sheet of useful formulae, and this included:

$$p = \gamma \beta m_{0} c = \frac{m_{0} \beta c}{\sqrt{1 - \beta^{2}}}$$

$$= \frac{\sqrt{E_{tot}^{2} - m_{0}^{2}c^{4}}}{c}$$

From this final equation, I noticed that

$$KE = \sqrt{E_{tot}^{2} - m_{0}^{2}c^{4}}$$

So this means that I have the relation:

$$p = \frac{KE}{c}$$

Which is momentum which is only dependent on KE or a constant!

The only problem I have now is working out where that equation from p comes from, can anybody help?

3. Oct 15, 2010

### kreil

4. Oct 15, 2010

### martinhiggs

Great, finally I've figured it all out! Thank you for your help! :)