# Product of tangent vectors \pm1 affine parameter

1. Apr 17, 2015

### binbagsss

If $\sigma$ is an affine paramter, then the only freedom of choice we have to specify another affine parameter is $a\sigma+b$, a,b constants. [1]

For the tangent vector, $\xi^{a}=dx^{a}/du$, along some curve parameterized by $u$

My book says that ' if $\xi^{a}\xi_{a}\neq 0$, then by suitable affine parameterization we can arrange such that $\xi^{a}\xi_{a}=\pm1$,

Question:

What does it mean by some suitable affine parameterization? so say if $\sigma$ is a affine parameter and we do not have $\xi^{a}\xi_{a}=\pm1$, is it saying that we can use [1] and carefully choose $a$ and $b$ such that this is the case?

I've often seen proper time used such that $\xi^{a}\xi_{a}=\pm1$ is the case.
Why is this?

Or Is this part of the definition of proper time, are there any other 'known' parameters for which $\xi^{a}\xi_{a}=\pm1$ or is the affine parameter for which this holds unique?

2. Apr 18, 2015

### Orodruin

Staff Emeritus
Yes, the point is that if $\xi^a \xi_b \neq 0$, then you can pick $a$ such that $\xi^a \xi_a = \pm 1$ ($b$ does not really enter into it, it is just a translation along the curve). The point is that this parametrises the curve using the curve length as parameter. In Minkowski space, for time-like curves, this means parametrising the curve with the proper time.

3. Apr 18, 2015

### binbagsss

Ok. So if i use an affine parameter $\tau+b$ I Still achieve $\xi^a \xi_a =\pm 1$?

For a null geodesic we can't use $\tau$ as it is always zero.
However am I correct in thinking that the paramter $\tau+b$ would be plausible?

Thanks.

4. Apr 18, 2015

### Orodruin

Staff Emeritus
Yes, if you have $\tau' = \tau + b$, you would get $d\tau'/d\tau = 1$ and thus $dx^\mu/d\tau = dx^\mu/d\tau'$. It is just a change of what you call "proper time equals zero".

No, it would not be. Just by the same argumentation as above. You would still be trying to parameterise with the proper time, just with a different definition of proper time equal to zero. It does not work for light-like curves. You can still find an affine parameter, but it cannot be based on curve length (i.e., proper time).

5. Apr 22, 2015

### binbagsss

Is a re-scale of prper time ok, so $\lambda=a\tau$?