2
$\begingroup$

I know that inner product between 4-velocity is invariant under Lorentz transformation and I know that inner product between any 2 vectors under general coordinate transformation is invariant. Therefore, the inner product between two 4-velocities $u^{i}=\left(d x^{i} / d \tau\right)$ should also be invariant under arbitrary coordinate transformation. In my book it is written that $u_{i} u^{i}=1$ and here is my try to get this result : $$d s^{2}=g_{a b} d x^{a} d x^{b}=g_{a^{\prime} b^{\prime}} d x^{a^{\prime}} d x^{b^{\prime}}=d s^{\prime 2}$$ $$d s'^{2}=g_{0^{\prime} 0^{\prime}} d t'^{2}=g_{0^{\prime} 0^{\prime}} d \tau^{2}$$ $$\begin{aligned} u^{a} u_{a} &=\frac{d x^{a} d x_{a}}{(d \tau)^{2}} \\ &=g_{0^{\prime} 0^{\prime}}\frac{d x^{a} d x_{a}}{d s^{2}} \\ &=g_{0^{\prime} 0^{\prime}} \end{aligned}$$ Here I have assumed that clock is moving in some gravitational field in the frame O' and used $ dt'=d \tau$. Clearly my result is wrong as $g_{0^{\prime} 0^{\prime}}$ is not invariant but I am not getting where I have done something silly.

$\endgroup$
2
$\begingroup$

The definition of proper time you used is wrong. To find the proper time, you can't just substitute it into the "regular" time. In special relativity, you don't notice this problem as $g_{00}$ has unitary norm, but in general relativity you have to remember that

$ d\tau^2=ds'^2 = g_{\mu\nu}dx^\mu dx^\nu \Rightarrow u^{a} u_{a} =\frac{d x^{a} d x_{a}}{(d \tau)^{2}} =\frac{d x^{a} d x_{a}}{d s^{2}} =1$

wich is invariant. See the example in this discussion or the definition on Wikipedia.

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

Working in the (+ - - -) convention. For simplicity, assume the metric is diagonal (this proof still holds if it's not, it's just longer):

\begin{equation} d \tau^2 = g_{00} dt^2 + g_{xx} dx^2 + g_{yy} dy^2 + g_{zz} dz^2 \\ 1 = g_{00} \left(\frac{dt}{d\tau} \right)^2 + g_{xx} \left( \frac{dx}{d\tau}\right)^2 + g_{yy} \left(\frac{dy}{d\tau}\right)^2 + g_{zz} \left( \frac{dz}{d\tau} \right)^2 \\ 1 = g_{00} \left(\frac{dx^0}{d\tau} \right)^2 + g_{xx} \left( \frac{dx^1} {d\tau}\right)^2 + g_{yy} \left(\frac{dx^1}{d\tau}\right)^2 +g_{zz} \left( \frac{dx^2}{d\tau} \right)^2 \\ 1 = g_{00} \left(u^0 \right)^2 + g_{xx} \left( u^1\right)^2 + g_{yy} \left(u^2\right)^2 +g_{zz} \left( u^3 \right)^2\\ 1 = g_{\mu \nu} u^{\mu} u^{\nu} \implies u^{\mu} u_{\mu} = 1. \end{equation}

I think your mistake is on line 2, but I don't quite understand what you are doing there so I'm not sure I can help.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for your response. I didn't get how did you write your first equation. $\endgroup$ – RandomXYZ Jun 9 at 11:06
  • 1
    $\begingroup$ It's the definition of the proper time. See either the wiki page or any standard reference (e.g Misner, Thorne, Wheeler). $\endgroup$ – Stratiev Jun 9 at 12:13
  • $\begingroup$ Thanks. I got it. $\endgroup$ – RandomXYZ Jun 9 at 13:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.