Getting from $E^2 - p^2c^2 = m^2c^4$ to $E = \gamma mc^2$ [closed]

Question

What is each mathematical step (in detail) that one would take to get from:

$E^2 - p^2c^2 = m^2c^4$

to

$E = \gamma mc^2$,

where $\gamma$ is the relativistic dilation factor.

This is for an object in motion.

NOTE: in the answer, I would like full explanation. E.g. when explaining how to derive $x$ from $\frac{x+2}{2}=4$, rather than giving an answer of "$\frac{x+2}{2}=4$, $x+2 = 8$, $x = 6$" give one where you describe each step, like "times 2 both sides, -2 both sides" but of course still with the numbers on display. (You'd be surprised at how people would assume not to describe in this detail).

Answer 1

Starting with your given equation, we add $p^2 c^2$ to both sides to get $$ E^2=m^2 c^4 + p^2 c^2$$ now using the definition of relativistic momentum $p=\gamma m v$ we substitute that in above to get $$E^2 = m^2 c^4 +(\gamma m v)^2 c^2=m^2 c^4 +\gamma^2 m^2 v^2 c^2$$ Now, factoring out a common $m^2 c^4$ from both terms on the RHS in anticipation of the answer we get $$E^2=m^2 c^4 (1+\frac{v^2}{c^2}\gamma^2)$$ Now using the definition of $\gamma$ as $$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ and substituting this in for $\gamma$ we get $$E^2=m^2 c^4 \left(1+\frac{\frac{v^2}{c^2}}{1-\frac{v^2}{c^2}}\right)$$ and making a common denominator for the item in parenthesis we get $$E^2=m^2 c^4 \left( \frac{1}{1-\frac{v^2}{c^2}} \right)=m^2 c^4 \gamma^2$$ Taking the square root of both sides gives $$E=\pm \gamma mc^2$$ Hope this helps.

Answer 2

First you set $c=1$.

$$ E^2 - p^2 = m^2$$

Then you think about it, it is saying that the relativistic length of the energy momentum vector is "m". The ratio of p to E is the velocity, since that's what happens to a four-vector under a boost, it gets space-components and time component whose ratio is the velocity. From $|p|=v|E|$, you substitute,

$$ E^2(1-v^2) = m^2 $$

And

$$ E= { m\over\sqrt{1-v^2}}$$

Done.

It is generally a sign of total incompetence to not set c to 1, it just makes ridiculously trivial geometrical formulas, which, as you can see above, are absolutely transparent, look like they are sophisticated or complicated.

Answer 3

Starting with relativistic momentum $$p^2 = \left( \gamma m v \right)^2 = \frac{m^2 v^2}{1 - \frac{v^2}{c^2}}$$ one than gets $$E = \pm \sqrt{ m^2 c^4 + p^2 c^2 } = \pm \sqrt{ m^2 c^4 + \frac{m^2 v^2 c^2}{1 - \frac{v^2}{c^2}} } = \pm mc^2 \sqrt{\frac{1- \frac{v^2}{c^2}}{1- \frac{v^2}{c^2}} + \frac{\frac{v^2}{c^2}}{1- \frac{v^2}{c^2}}} = \pm \gamma mc^2$$