[General Relativity] Find the acceleration of an object

[eoghan]

Homework Statement

Given the metric
$$ds^2=-e^{2\phi}dt^2+\frac{1}{1-\frac{b(r)}{r}}dr^2$$
find the time component of the 4-acceleration of an object moving with velocity v in the r direction.

The Attempt at a Solution

The four-velocity of the object is $$u^a=(t', v)$$
where the prime stands for the derivative with respect to the proper time of the object.
I know that the four velocity satisfies $$g_{ab}u^au^b=-1$$ so I find
$$t'=\left(1+\frac{v}{1-\frac{b}{r}}\right)\exp(-2\phi)$$
Now I use the formula
$$a^0=u^a\nabla_au^0=u^0(\partial_0\dot t+\Gamma_{0c}^0u^c) +u^1(\partial_1\dot t+\Gamma_{1c}^0u^c)$$
and I find
$$a^0=t'\dot\phi v+v\left[\dot\phi t'+\frac{v}{2\exp(2\phi)} \frac{b/r^2}{\left(1-b/r\right)^2}\right]$$
where the dot stands for the derivation with respect to r and I've supposed that $\partial_0 t'=\partial_1 t=0$.
Could this be right?

 

[George Jones]

moving with velocity v in the r direction

Velocity v with respect to whom?

 

[gabbagabbahey]

I know that the four velocity satisfies $$g_{ab}u^au^b=-1$$ so I find
$$t'=\left(1+\frac{v}{1-\frac{b}{r}}\right)\exp(-2\phi)$$

I get something a little different... perhaps you should show your calculation

 

[eoghan]

Velocity v with respect to whom?

With respect to an observer at rest who sees the metric $ds^2$

I get something a little different... perhaps you should show your calculation

I'm a little tired...I did some stupid mistakes
The metric is
$$ds^2=-e^{2\phi}dt^2+\frac{1}{1-\frac{b(r)}{r}}dr^2$$
and I get
$$t'=\pm\sqrt{1+\frac{v^2}{1-\frac{b}{r}}}\:\:e^{\phi}$$
Now I'm going to correct all the others formula (I've also found an error in a Christoffel symbol )...

 

[George Jones]

With respect to an observer at rest who sees the metric $ds^2$

Sorry, but I don't know what this means. Do you mean "At rest with respect to an observer who has contant $r$"?

 

[George Jones]

Do you mean "At rest with respect to an observer who has contant $r$"?

I think that this must be the case. Another qestion. Is $v$ a coordinate velocity, or is it an actual physical velocity?

Here is a (slick) way to calculate relative physical (not coordinate) speed between two observers who are coincident at an event. Suppose the two 4-velocities are $u$ and $u'$. Then, $-\gamma = -\left( 1 - v^2 \right)^{-1/2} = g \left( u , u' \right) = g_{\alpha \beta}u^\alpha u'^\beta.$ This is an invariant quantity, and, consequently, can be calculated using any coordinate system/basis.

This works in all coordinate systems in both special and general relativity.

In order to use this method, you first need to (easily) find the 4-velocity of an observer who sits at constant $r$.

 

[eoghan]

I think that this must be the case

Yes, I mean that. And v is the physical velocity.

Suppose the two 4-velocities are $u$ and $u'$. Then, $-\gamma = -\left( 1 - v^2 \right)^{-1/2} = g \left( u , u' \right) = g_{\alpha \beta}u^\alpha u'^\beta.$

I don't understand this. If I am at constant r, then my four velocity is
$u^a=(e^{-\phi}, 0)$ while the 4-velocity of the moving object is
$u'^b=(k, v)$ with k chosen so that $u'^bu'_b=-1$. Where am I wrong?

 

[George Jones]

$u'^b=(k, v)$

This isn't correct. Is v the spatial component of 4-velocity in special relativity?

 

[George Jones]

There two ways that you can do this:

1) in a coordinate basis;

2) in an orthonormal basis (like in the other thread).

1) In a coordinate basis, the 4-velocity of the stationary observer has components $\left( e^{-\phi} , 0\right)$ and call the components of the other 4-velocity $\left( u'^0 , u'^{1}\right)$. Use the method I give above to find $u'^0$.

2) In an orthonormal basis, what are the component of the 4-velocity of the stationary observer? The components of the other 4-velocity are the same as the components of 4-velocity in special relativity. What are these?

Notes: you don't have to do 2) to do 1); don't confuse the two methods.

 

[eoghan]

I got it! I used the coordinate basis method ad I've found the right answer
$$u'=(\gamma e^{-\phi}, \pm v\gamma\sqrt{1-\frac{b}{r}})$$.
But I don't understand why
$$-\gamma = -\left( 1 - v^2 \right)^{-1/2} = g \left( u , u' \right) = g_{\alpha \beta}u^\alpha u'^\beta$$
In flat spacetime where $u=(\gamma, \gamma v)$ this is obvious, but how can you generalize it to a generic curved spacetime?

 

[George Jones]

Great!

In the orthonormal basis, the components of the two 4-velocities are $\left( 1 , 0 \right)$ and $\left( \gamma , \gamma v \right)$, just like in special relativity. Consequently, $u^\mu u'_\mu = - \gamma$ in the the orthonormal basis. But, $u^\mu u'_\mu$ is a scalar, and thus its value is independent of basis. Consequently, $u^\mu u'_\mu = - \gamma$ in every basis, including a coordinate basis.

 

[eoghan]

First of all, thank you very much for your help!
As you can see my knowledge of general relativity is very limited...
Why do you say that the components of the 4-velocities in an orthonormal basis is just like in special relativity? I'm studying GR on "Wald" and it doesn't explain these things