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

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


$E = \gamma mc^2$,

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).


closed as off-topic by tpg2114, akhmeteli, Emilio Pisanty, user10851, Qmechanic Nov 10 '13 at 16:08

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – tpg2114, akhmeteli, Emilio Pisanty, Community, Qmechanic
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ I've changed $m$ to $m_0$ to make it clear it's the rest mass. If you didn't mean this please revert my edit. $\endgroup$ – John Rennie May 18 '12 at 15:46
  • $\begingroup$ Oh wait, I think this is for when it's moving, because this equation $E=\gamma mc^2$ goes to show how if an object reached the speed of light it would have used infinite energy (although it's not strictly true) and as far as I know $E^2 - p^2c^2 = m^2c^4$ is for objects that are moving, so surely $E=\gamma mc^2$ would be for a moving object? $\endgroup$ – ODP May 18 '12 at 15:53
  • $\begingroup$ @JohnRennie I have reverted your edit, but please do correct me if I'm wrong $\endgroup$ – ODP May 18 '12 at 15:58
  • $\begingroup$ There's nothing wrong with using $m$ for rest mass, as that is the modern convention. $\endgroup$ – David Z May 20 '12 at 2:45
  • 2
    $\begingroup$ By the way, while you're allowed to ask for it, the level of mathematical detail you seem to want is more than an answers on this site are expected to give. In other words, it's really not our job to explain basic algebra. $\endgroup$ – David Z May 20 '12 at 2:52

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.


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,

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


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


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.

  • $\begingroup$ Ya, having every formula peppered with multiplications and divisions by c when you could of just done it in the beginning, results in great deal of confusion and is just plain bad for readability (I would not want such in software code, let alone in text that is written for humans). I think though its bit like hard to read code written by clever programmers - not a sign of incompetence, just a nasty habit - it is easy to read if you have done that stuff a lot. $\endgroup$ – Dmytry May 19 '12 at 20:34
  • $\begingroup$ @Dmytry: That's a good analogy, it's like (slightly) obfuscated C--- every experienced person ignores the c's in the head, and sees the derivation as I wrote it above, no matter how it is written on the page. But if you then transcribe your thoughts adding in the c's, it obscures your reasoning in a layer of algebra that a student can't see through, and makes you look like a wizard to a student. This type of thing is what kills physics pedagogy, and is the reason to remove all extraneous constants from all formulas on a first pass. $\endgroup$ – Ron Maimon May 20 '12 at 17:05

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 $$E = \pm \sqrt{ m^2 c^4 + p^2 c^2 } = \pm \sqrt{ m^2 c^4 + \frac{m^2 v^2 c^2}{1 - \frac{v^2}{c^2}} } = \pm mc^2 \sqrt{\frac{1- \frac{v^2}{c^2}}{1- \frac{v^2}{c^2}} + \frac{\frac{v^2}{c^2}}{1- \frac{v^2}{c^2}}} = \pm \gamma mc^2$$

  • $\begingroup$ The sort of thing I was hoping for would be an explanation as to how "one then gets" the next part. $\endgroup$ – ODP May 18 '12 at 16:19
  • $\begingroup$ @Olly the explanation is right there in the equations in this answer. $\endgroup$ – David Z May 20 '12 at 2:47

Not the answer you're looking for? Browse other questions tagged or ask your own question.