I'm reading General Relativity by Robert Wald and his notation is confusing me.

On page $61$ of the book, Wald claims that by setting $c=1$ we can derive $E=mc^2$ from the following information.

(1) $u^au_a=-1$ where $u^a$ is a tangent vector to a timelike curve.

(2) $p^a=mu^a$ where $p^a$ is an energy-momentum $4$-vector of the rest mass $m$.

(3) The energy of a particle as measured by an observer, present at the site of the particle, whose $4$-velocity is $v^a$ is written as $E=-p_av^a$.

(4) The particle is at rest with respect to the observer, so $u^a=v^a$ is assumed.

As far as I am aware, it is impossible to algebraically derive $E=mc^2$ from this as written unless I am completely misunderstanding Wald's notation. However, if we set $E=-p^av_a$ then we have $E=m$, but this is extremely silly since we can also have $E=mc^{100}$ or something if $c=1$. Is there something I am missing here? Is there a typo, or is Wald saying something different?

A similar issue arises with more notation introduced on the next page. He writes $\partial^a$, which was never used prior to page 62 (unless Wald means $\partial^a=\partial_a^{-1}$...?). What is he talking about here?

Up until this point, the book has been a pretty nice read and I've understood most of the content since the first three chapters are primarily math review and understanding Wald's new notational conventions. I just need help with this notational discrepancy in Chapter 4. Thanks for the help.

  • 1
    $\begingroup$ If you understand the difference between $u^a$ and $u_a$, the difference between $\partial^a$ and $\partial_a$ is similar. $\endgroup$
    – Ghoster
    Commented Oct 23, 2023 at 3:46
  • 6
    $\begingroup$ $E=mc^{100}$ wouldn’t be dimensionally correct when using SI units. $\endgroup$
    – Ghoster
    Commented Oct 23, 2023 at 3:48
  • 2
    $\begingroup$ $E=m$ is not “extremely silly”. Is this the first time that you have encountered the use of natural units such as $c=1$? Natural units are generally first encountered when learning SR, not GR. Likewise with upper vs. lower spacetime indices. $\endgroup$
    – Ghoster
    Commented Oct 23, 2023 at 3:55
  • $\begingroup$ See en.wikipedia.org/wiki/… and en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation $\endgroup$
    – The Tiler
    Commented Oct 23, 2023 at 6:13
  • 3
    $\begingroup$ I don't really see how setting $c=1$ simplifies the argument here. Maybe clearer to leave $c$ as $c$ and then get an explicit factor of $c^2$ at the end. $\endgroup$
    – gandalf61
    Commented Oct 23, 2023 at 9:03

1 Answer 1


First, note that

$$p^a = mu^a \rightarrow \eta_{ab}p^a = m\eta_{ab}u^a \rightarrow p_b = mu_b$$

Then, relabel the dummy indice to get $p_a = mu_a$. We can use this to make use of your (4) relations. We start with energy (3)


Then, insert (2)

$$E=-mu_av^a$$ Then, insert (4)


Finally, use (1)

$$E = m.$$

As Ghoster said, this is not ridiculous at all when using natural units. Wald does use the notation that $\partial^a = \partial_a^{-1}$. This is common notation in GR.

Finally, to emphasize Ghoster's correct point, $E = mc^{100}$ wouldn’t be dimensionally correct when using SI units. It is, indeed, a power of 2 that is the correct input of $c$ despite $c$ being $1$ in natural units.

  • $\begingroup$ "it being $1$" referring of course to $c$, not its exponent, which in natural units is taken as $0$. $\endgroup$
    – J.G.
    Commented Oct 23, 2023 at 6:47
  • $\begingroup$ yes, let me rephrase @J.G. $\endgroup$ Commented Oct 23, 2023 at 6:54

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