# Schwarzschild Four Velocity

## Homework Statement

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.

## Homework Equations

The Schwarzschild Metric

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

## 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).

Related Advanced Physics Homework Help News on Phys.org
kdv

## Homework Statement

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.

## Homework Equations

The Schwarzschild Metric

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

## 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).
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.

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
$$\displaystyle \left( 1- \frac{2m}{r} \right)$$.

kdv
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
$$\displaystyle \left( 1- \frac{2m}{r} \right)$$.
It's not following a geodesic since it is staying at constant r, theta and phi!

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.

George Jones
Staff Emeritus
Gold Member
It's not following a spatial geodesic, but wouldn't it still be following a geodesic in time through space time?
No.

And in that case then ds=0 and time is just proper time.
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?

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?

George Jones
Staff Emeritus
Gold Member
Okay, so ds = 0 would correspond to a null geodesic. Since the object is at rest, it's following a time-like geodesic,
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.

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

Last edited by a moderator:
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.

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

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:
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.

George Jones
Staff Emeritus
Gold Member
It makes sense that $g^{rr} = (1-\frac{2m}{r} )$
Yes, this is $g^{rr}$

since then the four acceleration would be

$$A^i = (0, \frac{M}{r^2}, 0, 0)$$
and these are components of the 4-acceleration with respect to the coordinate basis.

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.
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.