# Two formulas for relativistic energy - what is the difference?

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

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

or

$$E=\sqrt{m^2c^4+p^2c^2}\tag{2}$$

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

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

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.
• They are in fact the same equation, as discussed here en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation Jan 30 '21 at 12:47
• @Frobenius Yes, it's simple algebra. So why are you writing it in a complicated way? Jan 30 '21 at 14:00
• @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. Jan 30 '21 at 14:52
• @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. Jan 30 '21 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.

• @Frobenius Thank you for that. But I addressed only the conceptual aspect. Jan 30 '21 at 17:46

First note that

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

so

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

• @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. Jan 30 '21 at 13:46
• @Frobenius I stand corrected. However, my point is that the question was not tagged as h-and-e when I answered. Jan 30 '21 at 14:02
• I'm not sure that the OP realised that establishing the equivalence IS simply a matter of algebra. Jan 30 '21 at 14:23
• 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. Jan 30 '21 at 16:24