# Question about the physical meaning of normalization of 4-velocity

I'm struggling with a concept of normalization of 4-velocity in Special Relativity (and General Relativity eventually).

Well, after a bunch of introduction to vector/tensor calculus we came up with the mathematical fact that the inner product and the metric tensor have a quite close relation in order, at least, to perform calculations. In order to compute the inner product of four velocity of massive particles (Faraoni; Special Relativity, Springer; page 139), for example, we get:

$$g_{\mu \nu}u^{\mu}u^{\nu} = u^{\mu}u_{\mu} = -c^2$$

Then here arise the condition called Normalization:

$$u^{\mu}u_{\mu} = -c^2$$ or in Geometrized Units $$u^{\mu}u_{\mu} = -1$$

An example on how to use this relation is (Introduction to General Relativity and the physics of compact stars, Livraria da Física, Brazil):

In order to determine the Ergosphere, consider an observer at rest in infinity, which the 4-velocity is $$u^{\mu} = (u^{0},0,0,0)$$. From the Normalization condition we have: $$u^{\mu}u_{\mu} = -1 = g_{00}(u^{0})^{2}$$ and then we get: $$(u^{0})^{2} = \frac{1}{1-\frac{2Mr}{\Sigma ^{2}}}$$ Which is singular at $$\Sigma ^{2} = 2Mr$$

My thing is: I know that Normalization is just a result of an inner product action of a specific two 4-vector and in the example above is just a result from a specific metric with an specific 4-velocity.

But what is the very fundamental physical meaning of using this technique? Why is it so useful (and basic) in order to calculate things like the above example and in SR and GR?

## 2 Answers

The phrase "the metric tensor and the inner product have a close relation at least to perform calculations" makes me a bit uncomfortable since the inner product and the metric are essentially the same thing.

But anyway, suppose you have a timelike curve in spacetime $x^\mu (\lambda)$ parametrized by some parameter $\lambda$. The proper time between two infinitesimally separated points on the curve is

$$d \tau = \sqrt{- g_{\mu\nu} dx^\mu dx^\nu} = d\lambda\, \sqrt{- g_{\mu\nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda}}$$

so if we let $u^\mu = dx^\mu/d\lambda$, the requirement that $u_\mu u^\mu = -1$ is exactly the same as asking that $\lambda = \tau$, i.e., the curve is parametrized by proper time.

Exactly the same thing happens in Euclidean geometry, by the way: a curve with unit speed is one whose parameter is arclength along the curve.

• Thank you for your answer. The reason that I wrote the phrase about the metric tensor and the inner product is because metric tensor is not necessarily positive definite and the inner product map is. – M.N.Raia Jun 7 '18 at 17:58
• @JackClerk The metric tensor is the inner product map, so that is not correct. – J. Murray Jun 7 '18 at 21:57
• One usually requires it to be a nondegenerate symmetric bilinear form along with the signature convention. Not entirely sure if there are more details like how it is allowed to change between points and so on. – Emil Jun 7 '18 at 22:27

Normalization help provide a “unit vector” so that determining components of tensors is made easier.

UPDATE: Another interpretation of the normalized 4-velocity is that it represents one-tick of that observer's clock. A normalized spacelike vector orthogonal to that 4-velocity would represent a unit of length (say one light-tick) for that observer. With the signature having plus or minus signs, a sign might be included for convenience in each of the above.

On the other hand, lightlike vectors can't be normalized. So, there is no lorentz-invariant sense of a unit vector for a lightlike direction.

• I'm sorry, but your answer isn't satisfatory to me. – M.N.Raia Jun 7 '18 at 17:39