# Normalization of 4-velocity in general relativity

1. Apr 10, 2010

### jmlaniel

I know that in Minkowsky space, the 4-velocity is normalize according to the following relation:

$$\eta_{\mu\nu} U^{\mu} U^{\nu} = -1$$

Can someone explain to me ho this can be generalized to a normalization in a curved space with the following relation :

$$g_{\mu\nu} U^{\mu} U^{\nu} = -1$$

The reference that I am using just use the previous expression without any justification and this bothers me.

Thanks!

2. Apr 10, 2010

### tiny-tim

Welcome to PF!

Hi jmlaniel! Welcome to PF!

The norm is the inner product U.U, or UμUμ.

And Uμ is defined as gμνUν.

So U.U = gμνUνUμ.

In Minkowski space, gμν = ημν.

3. Apr 10, 2010

### bcrowell

Staff Emeritus
One way of stating the correspondence principle is that GR is locally the same as SR. Therefore the generalization of this relationship to GR has to be the way you're describing it.

4. Apr 10, 2010

### jmlaniel

Thanks for the reply!

I also thought that one of the justification was simply the correspondence principle which states that one just need to replace $$\eta_{\mu\nu}$$ by $$g_{\mu\nu}$$ to get the equivalent results in general relativity (from special relativity).

However, let's just take for example the Schawrzschild metric:

$$ds^2 = -\left( 1-R_s/r \right) dt^2 + \left( 1-R_s/r \right)^{-1} dr^2 + r^2 d\Omega^2$$,

(here c = 1 and $$R_s$$ is the Schwarzschild radius).

If I position an observer in a rest frame, therefore (according to the normalization relation for the 4-velocity) its 4-velocity vector needs to be :

$$U^\mu$$ = $$\left( \frac{1}{\sqrt{1-R_s/r}},0,0,0\right)$$

so that $$g_{\mu\nu} U^\nu U^\mu = -1$$.

Is this right? If so, it means that the evolution of time is dependent on where you are with respect to the infinity where there is no curvature of space-time (which I think is normal according to general relativity).

It confuses me a little since in special relativity, when an observer is at rest, its 4-velocity vector is just the intuitive $$U^\mu$$ = (1,0,0,0) (and $$\eta_{\mu\nu} U^\nu U^\mu = -1$$ is clearly satisfied).

5. Apr 12, 2010

### yuiop

Yes. See post #9 by Dr Greg in this thread -> https://www.physicsforums.com/showthread.php?t=383884
Yes again, (but with caveats). The 4 velocity you have given above, is the 4 velocity of a stationary particle at r, according to an observer at infinity in Schwarzschild coordinates. The 4 velocity of the same particle according to a local observer at r is simply (1,0,0,0) just as in Special Relativity, because locally the spacetime is Minkowski.

If it helps, the Schwarzschild metric can be stated in a more general way as:

$$ds^2 = -\frac{\left( 1-R_s/r \right)}{(1-R_s/R)} dt^2 + \frac{(1-R_s/R)}{\left( 1-R_s/r \right)} dr^2 + r^2 d\Omega^2$$,

where r is the radial location of the particle and dt, dr and $d\Omega$ are measured by an observer at R. When R=r the Minkowski metric (in spherical coordinates) is obtained.

6. Apr 27, 2010

### jmlaniel

Thanks kev for this other way to express the Schwarzschild metric. It answers all my question.

However, do you have any reference on how to obtain this expression? Or how to obtain it from the standard derivation for the Schwarzschild metric?

Thanks!

7. Apr 27, 2010

### jmlaniel

In order to obtain the "restated" metric, I tried the following change of variable:

$$t \rightarrow t (1-R_s/R)$$

$$r \rightarrow \frac{r}{(1-R_s/R)}$$

and I got the following metric :

$$ds^2 = -\frac{(1-R_s/r)}{(1-R_s/R)} dt^2 + \frac{(1-R_s/R)}{(1-R_s/r)}dr^2 + r^2 (1-R_s/R)^2 d\Omega^2$$

This results is quite similar to the "restated" metric, except for the last term which contains a correction term.

Is my approach correct? Is there another way to obtain the "restated" metric?

8. Apr 28, 2010

### yuiop

This might work better:

$$dt \rightarrow \frac{dt}{\sqrt{(1-R_s/R)}}$$

$$dr \rightarrow dr \sqrt{(1-R_s/R)}$$

$$d\Omega \rightarrow d\Omega$$

To be honest I have no idea how to formally or rigorously get the result I obtained, other than it seems to work. I'm not a mathematician.

9. Apr 28, 2010

### jmlaniel

I have a problem with your change of variable. You cannot make a change of variable only with differential element since it implies the derivative of the variable. Furthermore, since there a change of variable on r, it will influence the last term of the metric since it carries the factor which is clearly r-dependant.

I also realized that I made a mistake on my last post. So here is the corrected version with new variable (which is less confusing than keeping the same r and t.

If :

$$\bar{t} = t \sqrt{1-R_s/R}$$

$$\bar{r} = \frac{r}{\sqrt{1-R_s/R}}$$

then we get the following metric :

$$ds^2 = -\frac{(1-R_s/\bar{r})}{(1-R_s/R)} d\bar{t}^2 + \frac{(1-R_s/R)}{(1-R_s/\bar{r})}d\bar{r}^2 + \bar{r}^2 (1-R_s/R) d\Omega^2$$

where $$d\Omega^2 = sin^2\theta d\phi^2 + d\theta^2$$

I suspect that there must a change of variable for $$\phi$$ and $$\theta$$ in order to absord the factor of $$(1-R_s/R)$$. Since it is nonlinear, I am not quite sure how to do the change of variable. I will have to think about it...

Finally, kev, I am sorry to say that I am not entirely satisfied with your statement that "it seems to work"... And I don't think that we need a mathematician to solve this!!!

Last edited: Apr 29, 2010
10. Apr 28, 2010

### starthaus

Is this what you are looking for?

11. Apr 28, 2010

### jmlaniel

Thank you starhaus for your suggestions. Unfortunately, it is not what I am looking for. The Eddington-Finkelstein is just a change of coordinate that remove the "false" singularity located at the event horizon.

What I am looking for is just a change of variables, or an explanation, that gives the metric given by kev as shown above.

12. Apr 28, 2010

### starthaus

Ok, I understand now what you are looking for, this is impossible, no matter what transformation one attempts to apply to $$\phi$$ and $$\theta$$. The trigonometric function kills any such attempt.

13. Apr 28, 2010

### jmlaniel

I completely agree with you. Which means that the metric in my post #9 is the closest that I can get to the metric given by kev in post #5.

14. Apr 28, 2010

### starthaus

Your metric is the correct one. What do you use it for?

15. Apr 28, 2010

### George Jones

Staff Emeritus
What is $R$?

16. Apr 28, 2010

### jmlaniel

My sudy of that metric is actually motivated by the exploration of a stationnary particle located within the Schwarzschild metric at a certain distance r.

If you look at post #4, I was exploring the 4-velocity of a stationnary particle and asking why the time component was so strange (after normalizing the 4-vector). I understood that the time was dilated due to the fact that the rest frame was establish with respect to an observer located at infinity (where the spacetime is flat).

It was kev who proposed another way of expressing the metric so that the observer is now located anywhere (actually, at r = R) which allow a non-dilated time component. I then tried to understand how kev obtained that metric which led me to the latest posts.

17. Apr 28, 2010

### jmlaniel

R is the place where the rest particle is located.

18. Apr 28, 2010

### George Jones

Staff Emeritus
What does this mean?

Do you mean that the particle is at a fixed $r$-coordinate with $r=R$, where $R$ is a constant? I don't think that this is what you mean, but if you do not mean this, I have no idea what you do mean.

19. Apr 28, 2010

### jmlaniel

Sorry about my answer, I realize that it is not really clear. R is the location of the observer.

In the standard Schwarzschild metric, the observer is located at infinity where spacetime is flat. With the change of variable I used in post #9, the observer can now be located at R. I see it as some kind of change of referential (although I am not convince that it is really the case).

Basically, it simplifies the 4-velocity of a particle at rest located at r = R since the metric becomes:

$$ds^2 = -dt^2 + dr^2 + R^2 (1-R_s/R) d \Omega^2$$

which gives the following 4-velocity of a particle at rest :

$$U_\mu = (1,0,0,0)$$

instead of the one I showed in post #4. There is no longer any time dilation since the observer is at the same place than the particle at rest. And infinity is still flat as expected.

20. Apr 28, 2010

### starthaus

$$ds^2 = -dt^2 + dr^2 + R^2 d \Omega^2$$ (look at post #9)