# 4-velocity of a timelike curve in special relativity

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## Main Question or Discussion Point

In special relativity we can view spacetime as $\mathbb{R}^4$ with its standard smooth structure, and a metric $\eta_{ab} = \sum\limits_{\mu, \nu = 0}^3 \eta_{\mu, \nu} (dx^\mu)_a (dx^\nu)_b$ where $\nu_{\mu \nu} = \mathrm{diag}(-1, 1, 1, )$. Given a curve $\gamma: I \rightarrow \mathbb{R}^4$ (where $I$ is an interval), let $T$ be the tangent vector field on $\gamma$. Then we define the proper time as $\tau = \int (-\eta_{ab}T^aT^b)^{1/2} dt$, where $t$ parametrizes $\gamma$. We define timelike curves as curves whose proper time is negative. Now, Wald says that "we may parametrize timelike curves by the proper time $\tau$". How are we parametrizing the curve using $\tau$? Are we parametrizing it in the sense that, for each given $t$, some portion of the curve is traced out, and its proper time (its length, in a sense) is given by $\tau$? I believe this is what he must mean, but please correct me if I am wrong.

But onto the main reason for my post: In the same notation as above, given a timelike curve parametrized by $\tau$, the tangent vector $u^a$ is defined as the $4-$velocity of the curve. Wald says that it follows directly from the definition of $\tau$ that the $4-$velocity has unit length: $\eta(u, u) = -1$. I don't see how this is true?

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PAllen
2019 Award

As to your second question, the key is the chain rule. For typing ease, I'll use X to mean all for coordinates parametrized by t. Then:

dX/dτ = dX/dt * dt/dτ = (dX/dt) / (dτ /dt)

Compute this using your definition of tau, and the definition of the tangent vector. It will fall right out (take the norm of your result).

• JonnyG
Thanks! It was a straight forward calculation.

pervect
Staff Emeritus
Her'es another way of getting the same result. Suppose you have a stationary observer. We compute the 4-velocity u. $u^0 = dx^0/d\tau=1$, and $u^i = dx^i/d\tau=0$ for i = 1,2,3.
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