# Schwarzschild Four Velocity

1. Mar 19, 2008

### Kreizhn

1. The problem statement, all variables and given/known data
Calculate the four velocity $V^i$, the four acceleration $A^i$ and the scalar $A^i A_i$ for an observer at $r=r_0, \theta = \theta_0, \phi = \phi_0$ in the Schwarzschild spacetime with r>2m.

2. Relevant equations

The Schwarzschild Metric

$$ds^2 = -\displaystyle \left(1-\frac{2m}{r} \right) dt^2 + \left( 1- \frac{2m}{r} \right)^{-1} dr^2 + r^2 \left( d\theta ^2 + sin^2(\theta) d\phi ^2 \right)$$

3. The attempt at a solution
I'm not too sure how to even start this problem in that I don't see how the four velocity will depend on the metric, and so the only place I can see it being needed is in calculating the scalar $A^i A_i$. Is the spacial part of the four velocity given by the spherical velocity $( \dot{r}, r \dot{\theta}, r \dot{\theta} sin(\theta) )$ and the time element would just be 1? (we've scaled c= 1).

2. Mar 19, 2008

### kdv

V^i is given by $$\frac{dx^i}{d \tau}$$ Al the x^i are constant except for $$x^0 = t$$ so you will have to take the derivative of this with respect to the proper time and this is where the metric will be needed.

3. Mar 19, 2008

### Kreizhn

The metric should become
$$ds^2 = -\displaystyle \left(1-\frac{2m}{r} \right) dt^2$$

I figured that the space like components would be zero, but as for the time coordinate, if I want to examine $\frac{dt}{d\tau}$ then wouldn't I also need to consider $\frac{ds}{d\tau}$ or should I make some assumptions based on the fact that we assume it's a massive particle and must follow a time like geodesic?

But if it follows a time like geodesic, it seems to me that the time will trivially be the proper time, when I have a strange inkling that it should be something with
$$\displaystyle \left( 1- \frac{2m}{r} \right)$$.

4. Mar 20, 2008

### kdv

It's not following a geodesic since it is staying at constant r, theta and phi!

5. Mar 20, 2008

### Kreizhn

It's not following a spatial geodesic, but wouldn't it still be following a geodesic in time through space time? And in that case then ds=0 and time is just proper time.

6. Mar 20, 2008

### George Jones

Staff Emeritus
No.

No.

In special relativity, what does it mean for the interval ds to be zero?

In special relativity, when is the time coordinate also proper time?

7. Mar 20, 2008

### Kreizhn

Okay, so ds = 0 would correspond to a null geodesic. Since the object is at rest, it's following a time-like geodesic, and so I would solve

$$ds^2 = -1 = -\displaystyle \left(1-\frac{2m}{r} \right) \left( \frac{dt}{d \tau}\right)^2$$

Does this seem right?

8. Mar 21, 2008

### George Jones

Staff Emeritus
No, an observer that is freely falling follows a timelike geodesic. A geodesic worldline is one on which an https://www.physicsforums.com/showpost.php?p=981470&postcount=10" would read zero. An observer at rest in Schwarzschild coordinates follows a timelike worldline, but, as kdv said, this worldline is not a geodesic. As you are supposed to show, the 4-acceleration of such an observer is non-zero.

Not quite. On any observer's timelile worldline, $ds^2 = -d\tau^2$.

Last edited by a moderator: Apr 23, 2017
9. Mar 21, 2008

### Kreizhn

Okay yes, I realize I wrote my above equation to be solved in a rather bad manner, but setting $ds^2 = -d\tau^2$ will give me the same result so I think that's right.

10. Mar 21, 2008

### George Jones

Staff Emeritus
Yes, if you get rid of everything to the left of the -1, the equation is right.

11. Mar 21, 2008

### Kreizhn

So we find that $V^i = \displaystyle \left( \frac{1}{\sqrt{1-\frac{2m}{r}}} , 0, 0, 0 \right)$. Now in order to find $A^i$ I must find $\frac{D V^i}{d\tau} = \frac{\partial V^i}{\partial d\tau} + \Gamma^i_{bc} V^b V^c$.

The only non-obvious partial is $\frac{\partial V^t}{\partial d\tau}$. So,
$$\frac{d^2 t}{d\tau^2} = \displaystyle \frac{d}{d\tau} \left( \frac{dt}{d\tau} \right)$$

But since there's no explicit dependence on $\tau$I tried using the chain rule. That is

$$\displaystyle \frac{d}{d\tau} \left( \frac{dt}{d\tau} \right) = \frac{dr}{d\tau} \frac{d}{dr} \left( \frac{dt}{d\tau} \right)$$
$$= -\left(\frac{dr}{d\tau}\right) \left(\frac{m}{r^2 \left( 1- \frac{2m}{r} \right) ^{\frac{3}{2}} }\right) = 0$$

We conclude that $\frac{\partial V^i}{\partial d\tau} =0$

Furthermore, for non-zero $\Gamma^i_{bc} V^b V^c$ we are required to use $\Gamma^i _{tt}$

Now the only way $\Gamma^i _{tt}$ is non-zero is if we have an r derivative and so we consider $\Gamma^r _{tt} =\frac{1}{2}g^{rr}(-g_{tt,r})$

But I'm not too sure what $g^{rr}$ is. Our reduced metric seems to only be 1x1 and if I increase it to 2x2 then it's singular and doesn't have an inverse since dr²=0. What do I do?

Edit: Is it really just $1-\frac{2M}{r}$ ?

Last edited: Mar 21, 2008
12. Mar 21, 2008

### Kreizhn

It makes sense that $g^{rr} = (1-\frac{2m}{r} )$ since then the four acceleration would be

$$A^i = (0, \frac{M}{r^2}, 0, 0)$$

which makes sense since we expect to exert a force to keep the observer at rest, and $\frac{M}{r^2}$ happens to correspond beautifully with the Newtonian force of gravity where G=1.

13. Mar 22, 2008

### George Jones

Staff Emeritus
Yes, this is $g^{rr}$

and these are components of the 4-acceleration with respect to the coordinate basis.

To find the actual physical acceleration of a stationary observer, the 4-acceleration should be expressed with respect to an orthonormal frame, not with respect to a coordinate basis. Since the magnitude of the 4-acceleration is independent of the basis used, it suffices to calculate $A^\mu A_\mu$ in any basis, even a coordinate basis, to find the acceleration experienced by the observer. This expression is quite interesting.