Deriving the norm of four-velocity from the definition of a time-like interval

Question

I am going through Sean Carroll's Spacetime and Geometry, trying to learn GR. On page 31, he introduces the concept of the four-velocity $U^\mu$ as $$U^\mu = \frac{dx^\mu}{d\tau}$$ where $\tau$ is the proper time along a timelike path, with a timelike interval given by $$d{\tau}^2 = -\eta_{\mu\upsilon}dx^{\mu}dx^{\upsilon}$$

He further mentions that, because of the formula above for the timelike interval, the four-velocity is automatically normalized: $$\eta_{\mu\upsilon}U^{\mu}U^{\upsilon} = -1$$

In trying to derive this fact for myself, I tried to take the derivative of the timelike interval w.r.t the proper time twice. For the first:$$\frac{d}{d\tau}(d\tau^2) = \frac{d}{d\tau}(-\eta_{\mu\upsilon}dx^{\mu}dx^{\upsilon})$$ $$2d\tau = -\eta_{\mu\upsilon} . \frac{d}{d\tau}(dx^{\mu}dx^{\upsilon})$$ $$2d\tau = -\eta_{\mu\upsilon}\left(dx^\mu\frac{dx^\upsilon}{d\tau} + dx^\upsilon\frac{dx^\mu}{d\tau}\right)$$ And for the second derivative: $$\frac{d}{d\tau}(2d\tau) = -\eta_{\mu\upsilon}\frac{d}{d\tau}\left(dx^\mu\frac{dx^\upsilon}{d\tau} + dx^\upsilon\frac{dx^\mu}{d\tau}\right)$$ which results in $$2 = -\eta_{\mu\upsilon}\left(2\frac{dx^\mu}{d\tau}\frac{dx^\upsilon}{d\tau} + dx^\mu\frac{dx^\upsilon}{d\tau^2} + dx^\upsilon\frac{dx^\mu}{d\tau^2}\right)$$ I realize that the normalization value for the four-velocity would hold if only the first term of the last equation were present, and I've seen a lot of second-order terms dropped from equations before (mainly in engineering, though). However, I don't think that I can make that assumption in this case, since these derivatives are more symbolic in nature.

My question is this: Have I made a mistake in either the calculations, or made any assumptions that don't really hold? If not, why doesn't this approach work?

Answer 1

In $-\eta_{\mu\nu}\mathrm{d}x^\mu\mathrm dx^\nu$ there is an implicit tensor product, namely $-\eta_{\mu\nu}\mathrm{d}x^\mu\otimes\mathrm dx^\nu$. This means to contract this one-form with a derivative one must use \begin{align} \frac{\mathrm d}{\mathrm d\tau}\otimes\frac{\mathrm d}{\mathrm d\tau}(-\eta_{\mu\nu}\mathrm{d}x^\mu\otimes\mathrm dx^\nu)=-\eta_{\mu\nu}\frac{\mathrm d x^\mu}{\mathrm d\tau}\frac{\mathrm d x^\nu}{\mathrm d\tau} \end{align} Similarly, $$\frac{\mathrm d}{\mathrm d\tau}\otimes\frac{\mathrm d}{\mathrm d\tau}(\mathrm{d}\tau\otimes\mathrm d\tau)=\frac{\mathrm d\tau}{\mathrm d\tau}\frac{\mathrm d\tau}{\mathrm d\tau}=1$$

Answer 2

It may be not the most rigorous explanation, but essentially you are already working with differentials, so instead of deriving it you can just divide as if they were fractions, so just divide your second equation by $d\tau$ twice and you get the result.

Actually the correct derivative is like: \begin{equation} \frac{d}{d\tau}\tau=1 \end{equation} and not \begin{equation} \frac{d}{d\tau}d\tau=1 \end{equation} (with $d\tau$ instead of $\tau$)

Answer 3

We have: $$ \mathrm{d}\tau^2 = -\eta_{\mu\nu} \mathrm{d} x^\mu \otimes\mathrm{d}x^\nu = -\eta_{\mu\nu} \left(\frac{\mathrm{d}x^\mu}{\mathrm{d\tau}} \mathrm{d}\tau\right) \otimes \left(\frac{\mathrm{d}x^\nu}{\mathrm{d\tau}} \mathrm{d}\tau\right) = -\eta_{\mu\nu} \frac{\mathrm{d}x^\mu}{\mathrm{d\tau}} \otimes \frac{\mathrm{d}x^\nu}{\mathrm{d\tau}} \mathrm{d}\tau^2 $$ for every $\mathrm{d}\tau^2$. For the chain rule, $x^\mu = x^\mu(\tau)$ is implied.