# B Energy relativity derivation

1. Jan 25, 2017

### Glenn G

Hi community,
There is a step on the wiki page about a derivation.

I don't seem to get the same thing when I try and solve for (u/c)^2 and then plug into Lorentz factor to get the new Lorentz factor so you can get the full energy momentum relation

My gamma term looks more like 1/ ((mo^2u^2/p^2))^-1/2 when i've subbed for (u/c)^2??

Can anyone help or any other resources to show easy as poss how to derive the full equation?
Thanks
G.

2. Jan 25, 2017

### Orodruin

Staff Emeritus
If you have a u left in your expression you have not solved for (u/c)^2 ...

3. Jan 25, 2017

### Glenn G

I rearranged the top expression and made (u/c)^2 the subject but then you still have the u^2 that was in the numerator of the momentum expression? Confused.

4. Jan 25, 2017

### Orodruin

Staff Emeritus
Yes, so you have not solved for u^2/c^2. You need to actually solve for u^2/c^2.

Edit: Hint: Solve for u^2. The speed of light is just a constant.

5. Jan 25, 2017

### Glenn G

I got this far and became a bit convoluted. Will try for u^2 then.
Thanks.
G.

6. Jan 25, 2017

### Orodruin

Staff Emeritus
Hint:
$$1 = \frac{m^2 c^2 + p^2}{m^2 c^2 + p^2}$$
...

7. Jan 25, 2017

### Glenn G

Ooh, that's cheeky. I'll have another play this evening, thanks Orodruin.

8. Jan 25, 2017

### Glenn G

Got it finally, quite satisfying ...