# 4-velocity of a timelike curve in special relativity

• I
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?

PAllen

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
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