To the best of my knowledge, relativistic energy $E$ of a body with rest mass $m$ moving at velocity $v$ can be expressed as either

$$E=\gamma mc^2 \tag{1}$$



where $c$ is the speed of light and $\gamma$ is the Lorentz factor


Since relativistic momentum is $p=\gamma mv$, expression $(2)$ could also be written as

$$E=\sqrt{m^2c^4+(\gamma mv)^2c^2}$$

As far as I could tell, one cannot easily simplify expression $(2)$ to expression $(1)$1. This lead me to the following question:

  • Is there any difference2 between these formulas or can they be used equally?
  • Is there any situation where it would be favourable or "better" to use either one of those3?

1It is however obvious how expression $(2)$ reduces to the "rest energy" formula $E=mc^2$ for $v=0$.
2At least, there is no difference in the final result - I did a test with $v=2\cdot10^8m/s$ and $m=50kg$ and the result was the same (Formula 1, Formula 2).
3The only thing I would think of is that equation $(1)$ seems easier to type, but maybe there is some other aspects besides practical use.

Edit: It has been adressed in some comments and the (by now) two answers that $(1)=(2)$. However, I still wouldn't consider this a homework-like question for two reasons:

  • It isn't homework but just a question that came up when learning SR (of course, I have no proof that it isn't an assignment - you will have to believe me here)
  • While I somewhat agree that the first part of my question could be considered homework-like, I don't think that the second part is.
  • 1
    $\begingroup$ They are in fact the same equation, as discussed here en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation $\endgroup$
    – G.Lang
    Jan 30 at 12:47
  • $\begingroup$ @Frobenius Yes, it's simple algebra. So why are you writing it in a complicated way? $\endgroup$
    – PM 2Ring
    Jan 30 at 14:00
  • 5
    $\begingroup$ @Frobenius You wrote an equation with another equation over the top of a giant "equals" sign. That's not exactly common notation. It took me a minute to realise what you were trying to say. $\endgroup$
    – PM 2Ring
    Jan 30 at 14:52
  • 1
    $\begingroup$ @Frobenius BTW, this OP isn't in the habit of writing bad homework questions. They've been a member for almost a year, but they've written almost as many answers as questions, and IMHO all of their posts are good quality. $\endgroup$
    – PM 2Ring
    Jan 30 at 14:56

(a) As others have said, it is a matter of algebra that the equations are equivalent, if we also throw in $$\mathbf p = m \gamma \mathbf u\ \ \ \ \text {leading to}\ \ \ \ \ p^2= m^2 \gamma^2 u^2$$ and $$\gamma =(1-v^2/c^2)^{-1/2}$$

(b) The second equation that you have quoted can be written as $$E^2 - c^2 p^2 = c^4 m^2$$ This is hugely important conceptually. Regard $E$ as the time component of a 4-vector and $c^2p^2$ as the sum of the squares of the magnitudes of the three spatial components of that vector. Combined using the minus sign we get the magnitude squared of the 4-vector, and this is the frame invariant quantity $c^4m^2$, as $m$ itself is frame invariant. Note that the factors of $c^2$ and $c^4$ are conceptually relatively unimportant.

  • 3
    $\begingroup$ @Frobenius Thank you for that. But I addressed only the conceptual aspect. $\endgroup$ Jan 30 at 17:46

First note that

$$1 + \gamma^2 \frac {v^2}{c^2} = 1 + \frac {v^2}{c^2-v^2} = \frac {c^2}{c^2-v^2} = \gamma^2$$


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

  • 4
    $\begingroup$ @Frobenius Downvoting is your prerogative, but for the record Qmechanic had not yet tagged the question as homework-and-exercises when I answered it. It is still not obvious to me that it is h-and-e and not just a genuine query. $\endgroup$
    – gandalf61
    Jan 30 at 13:46
  • 1
    $\begingroup$ @Frobenius I stand corrected. However, my point is that the question was not tagged as h-and-e when I answered. $\endgroup$
    – gandalf61
    Jan 30 at 14:02
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    $\begingroup$ I'm not sure that the OP realised that establishing the equivalence IS simply a matter of algebra. $\endgroup$ Jan 30 at 14:23
  • 2
    $\begingroup$ I did indeed not notice that one can show the equivalence of the two formulas like this (now I do - thanks!). And FWIW, my question isn't from a homework (I don't have physics in school), but one could of course argue that it is homework-like. I will edit my post to adress this. $\endgroup$
    – Jonas
    Jan 30 at 16:24

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