# Schwarzschild metric

1. Jul 18, 2012

### eoghan

Hi!
Given the schwarzschild metric
$$ds^2=-e^{2\phi}dt^2+\frac{dr^2}{1-\frac{b}{r}}$$
I can make this coordinate transformation
$$\hat e_t'=e^{-\phi}\hat e_t \\ \hat e_r'=(1-b/r)^{1/2}\hat e_r$$
and I will get a flat metric. Is this correct?
Another thing I'm a lot confused about: if I am at constant r, than my 4-velocity is
$$u^a=(e^{-\phi}, 0)$$
Then I experience an acceleration
$$a^a=u^b\nabla_bu^a$$
and this acceleration has a non-null r-component. So, does this mean I'm accelerating in the r-direction? But I supposed I'm at constant r!

2. Jul 18, 2012

### Matterwave

1) You don't get a "flat metric". You are simply using an orthonormal basis. You can't change the curvature of the manifold by a change of coordinates or change of basis.

2) You can't "assume constant r" without applying a force of some sort to prevent the object from falling naturally as it would in a gravitational field. All you did at the beginning is assume there was no initial velocity in the r-direction.

3. Jul 18, 2012

### George Jones

Staff Emeritus
These components are with respect to the coordinate basis. What are the components with respect to the orthonormal basis?

4. Jul 18, 2012

### eoghan

But with these new coordinates the metric becomes
$$ds^2=-dt^2+dr^2$$
which is flat

5. Jul 18, 2012

### George Jones

Staff Emeritus
No.
These are 4-vectors (tangent vectors), not coordinates. This is a non-coordinate basis, i.e., there is no coordinate system that has these vector fields are tangent vector fields.

6. Jul 18, 2012

### haushofer

You give here the zweibeins which, at every point in spacetime, allow you to transform the metric to the flat one. However, this transformation (zweibein) changes from point to point; it depends on {t,r}. So you won't be able to use one and the same transformation to make the metric flat everywhere, as would be the case for a flat manifold.

What you express here is the equivalence principle: locally, you can always transform the metric to be flat. Globally however you cannot (in general).

A flat manifold is characterized by a vanishing Riemann tensor. Being a tensor, if it vanishes in one coordinate system, it will vanish in all coordinate systems.

7. Jul 18, 2012

### Staff: Mentor

Even if you are at constant r, that doesn't necessarily mean that you are not accelerating in the r direction. Consider the case of objects on the wall of a rotating centrifuge. They are experiencing a radially inward force exerted by the wall, and are accelerating radially inward. In the case of the centrifuge, the two directions involved are the circumferential direction and the radial direction. In the particular case you are considering, the two directions involved are the time direction and the radial direction. The orientation of the 4 velocity is in the time direction, and this orientation is changing for an observer at rest within the particular frame of reference under consideration.

8. Jul 18, 2012

### Matterwave

Since you are using a non-coordinate basis, you need to use somewhat modified formulas for calculating Christoffel symbols and the Riemann tensor. In this case, even though your metric is diag(-1,1,1,1) everywhere, the Riemann tensor will not vanish.

9. Jul 18, 2012

### grav-universe

A constant proper acceleration would need to be applied when originally stationary to the gravitating body at r in order to remain stationary, so that's what that would be. But that's not why I posted. I am curious, for another thread, anybody, what is that local acceleration along r that eoghan posted as it would be expressed algebraicly in terms of m and r?

Last edited: Jul 18, 2012
10. Jul 18, 2012

### Mentz114

In the coordinate basis it is
$$\frac{m}{r^2(1-2m/r)}$$
and in the local frame basis of the stationary observer it is
$$\frac{m}{r^2\sqrt{1-2m/r}}$$

11. Jul 18, 2012

### grav-universe

Ah, thanks. :) Is the first the coordinate acceleration of a free falling body at r according to a distant observer and the second the coordinate acceleration of a free falling body as measured locally by a hovering observer at r (or proper acceleration applied by the hovering observer)? Also, wouldn't the speed of the freefalling body be considered? For instance with a photon travelling radially, the coordinate acceleration measured by a distant observer would be negative as the photon loses speed. I assume what you have would be for a body falling from rest at r. I had thought that the acceleration was supposed to be just a = m / r^2 locally. Then with some original local radial speed v, and if SR applies locally in this regard, I figured the acceleration would apply relativistically, applying the usual relativistic formulas for acceleration, working out to a = (m / r^2) (1 - (v/c)^2)^(3/2) locally, with zero acceleration measured locally for a photon travelling radially. So then, from what you have, should that be a = (m / r^2) (1 - (v/c)^2)^(3/2) / sqrt(1 - 2 m / r) locally?

Last edited: Jul 18, 2012
12. Jul 18, 2012

### Mentz114

These are proper accelerations of the hovering observer, ∇βUαUβ. The first one would be measured by 'the observer at infinity' and the second is that measured by the hoverer.

This is a digression from the hovering observer. The FF observer experiences no proper acceleration so the proper derivative ∇βUαUβ is zero.

Sounds logical.

There is a relativistic factor as you can see, for the local proper acceleration.
Don't know. Null geodesics are another thing altogether. The speed of light at distant points doesn't have much physical significance, being coordinate dependent. I think the same thing applies to coordinate acceleration in general.

Last edited: Jul 18, 2012
13. Jul 18, 2012

### grav-universe

Okay, thanks again Mentz114. :)

14. Jul 18, 2012

### ApplePion

"In this case, even though your metric is diag(-1,1,1,1) everywhere, the Riemann tensor will not vanish."

This is not correct. If you really did get that metric everywhere things really would be flat everywhere--you can see that by plugging into the formula for the Riemann Tensor.

The thing that no one seems to be noticing is that while the coordinate transformation in the original post makes the g_tt and the g_rr metric components become Lorentzian, there are two other metric components, the g theta theta component and the g phi phi component. The coordinate transformation that puts g_tt and g_rr in a simple flat space form puts those other two components in a way not compatible for flat space.

15. Jul 18, 2012

### George Jones

Staff Emeritus
The expressions for the components of the Riemann tensor include terms that depend on the commutators of the basis fields. In a coordinate basis, these commutators are all zero. In a non-coordinate basis, these commutators are not all zero, and not all the components of the Riemann tensor vanish globally.

Did you include the commutators of the non-coordinate basis?

Last edited: Jul 18, 2012
16. Jul 19, 2012

### Matterwave

As I mentioned earlier in my post, the formula for Christoffel symbols and therefore the Riemann tensor will be slightly modified. As George Jones says, these formulas will include Lie bracket operations (what he refers to as commutators) on the basis vectors. If you include these Lie brackets, then neither the Christoffel symbols nor the Riemann tensors would necessarily be 0.