# Thread: verify that gamma = 1 + p^2/[(m^2)(c^2)]

1. ## verify that gamma = 1 + p^2/[(m^2)(c^2)]

Verify that 1/sqrt(1-(v^2)/(c^2)) = 1 + p^2/[(m^2)(c^2)]

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OK, so on the right side I changed p^2 into (m^2)(v^2) and cancelled out the m^2 in the numerator and denominator. This leaves me with:

1 + (v^2)/(c^2) = 1/sqrt(1-(v^2)/(c^2))

This seems tantalizingly close, but I have no clue what to do next.

Any guidance would be appreciated.

Yes, this is a homework problem, but homework isn't turned in. I'd still like to know how to do it though.

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By the way, how do you guys type math? I'd like to make it more readable.

2. Are you sure you're not supposed to be using the relativistic definition of momentum here: $p = \gamma mv$ . After all, you are dealing with the lorentz factor here.

Also, it looks like your right hand expression turns out to be $\gamma ^2$ . Are you sure you typed in the question correctly?

3. Thanks for the prompt response.

I thought of using the relativistic definition, but it seems the nonrelativistic one got me closer. When using p = gamma mv, I get 1+gamma^2-gamma^4 for the right hand side. Can't factor that . . . or can I?

Yes, I typed it as it appears in the book. I'll look for a list of errata, though.
I'm taking gamma to be one over the square root of (1-(v^2)/(c^2))

4. Although your two expressions look similar, they are not equal. Just plug in some values of $v$ and you'll find out for yourself.

\begin{aligned} 1 + \frac{p^2}{m^2c^2} &= 1 + \frac{{\color{red}\gamma ^2} {m^2} v^2}{m^2 c^2} \\ & = 1 + {\color{red}\frac{1}{1 - \frac{v^2}{c^2}}} \cdot \frac{v^2}{c^2} \\ & = 1 + \frac{v^2}{c^2 - v^2}\end{aligned}

Keep going with this. Combine the fractions and simplify. You should end up with $\gamma ^2$.

5. Yes, I do end up getting gamma squared. After combining the fractions, I multiplied top and bottom by 1/(c^2).

So I guess it was a bad problem then. International editions do have a lot of errors (probably intentional).