# Flat uniform and equivalent

1. Jun 8, 2008

### Staff: Mentor

In the usual "windowless room" thought experiment it is stated that there is no experiment which can be performed which will determine if the room is undergoing "uniform accleration" far from any massive body or is in a "uniform gravitational field". The two situations are therefore said to be "equivalent". Also, since the spacetime in the uniform acceleration case is obviously flat the equivalent spacetime in the uniform gravitation must also be flat.

So, my question is a pretty basic one. How do we construct the right coordinate systems for doing the "equivalent" windowless room thought experiments? Specifically:

In the gravitational/accelerated non-inertial frame what is the form of the gravitational field (as a function of position)?

If the field is a non-constant function of position then how is it considered either uniform or flat?

How are simultaneity, distance, and time experimentally defined in the uniform field?

What is the correct transformation between the accelerated and an inertial frame?

Can you simply use the Lorentz transform to boost between observers with uniform (non-inertial) motion in the gravitational field? (i.e. since spactime is flat do four-vectors still transform like usual, especially the four-momentum)

Any other hints would be appreciated.

2. Jun 9, 2008

### Mentz114

My guess is that g_00 and g_11 ( x^0 = t, x^1=r say) must depend on r linearly, so that d/dr(g_00) = constant, d/dr(g_11) = constant. This means the Christoffel symbols will exist and will not change over space, but the Riemann tensor is zero.

The problem with this setup is that it diverges at infinity.

It's a tough question, let's see what the GR experts say.

3. Jun 9, 2008

### Mentz114

Uniform red-shift

I've done some work and find that this non-physical vacuum solution gives time dilation in the x-direction proportional to x. Otherwise the space is flat, no Christoffels or Riemann components.

$$ds^2 = -B^{-1}x^{-2}dt^2 + Ax^2dx^2 + dy^2 + dz^2$$

This must be the space outside an infinite slab of matter, and the red-shift depends not at all on any properties the matter may have - probably because there's an infinite mass.

Will this do for the acceleration comparison ?

 No it has no acceleration as required.

Last edited: Jun 9, 2008
4. Jun 9, 2008

### gel

For a gravitation acceleration of a in the -ve x direction at the origin, I get the following metric
$$d\tau^2 = \left(1+c^{-2}ax\right)^2dt^2-c^{-2}(dx^2+dy^2+dz^2)$$
where $\tau$ is the proper time.
It is non-uniform, the gravitational acceleration at coordinate x is $\frac{a}{1+c^{-2}ax}$. Note, it blows up at x=-c2/a.

And, you can transform to inertial coords by

8. Jun 9, 2008

### gel

Google search came up with http://www.mathpages.com/home/kmath422/kmath422.htm" [Broken]. I haven't read it all, but the diagram is what I had in mind.

Last edited by a moderator: May 3, 2017
9. Jun 9, 2008

### Mentz114

gel,
$$\Gamma^x_{tt}=a(1+c^{-2}ax),\ \Gamma^t_{tx}=a(c^2+ax)^{-1}.$$

I agree. I entered 'r' instead of 'x' into my calculator and zapped all the derivatives.

Phew - thought I had a bad problem.

Correction to my calculation above, Christoffel symbs are

x-xx: x^-1
x-tt: -A^-1*B^-1*x^-5
t-tx: -x^-1

Last edited: Jun 9, 2008
10. Jun 9, 2008

### gel

To correct what I said above. For a uniform gravitational field, but non-flat space, you have the metric $d\tau^2=\exp(2c^{-2}ax)dt^2-c^{-2}(dx^2+dy^2+dz^2)$. This isn't flat, so not what the OP asked for, but the Einstein tensor vanishes, so it is empty space.
In fact, the Einstein tensor must vanish for similar reasons that it does in 2d manifolds.

11. Jun 9, 2008

### pmb_phy

The metric for a uniform gravitational field is identical to that of a uniformly accelerating frame of reference. This follows as an immediate consequence of the equivalence principle. Also, the spacetime associated with a uniform gravitational field is flat. The Einstein tensor is vanishes for such a spacetime. Such a spacetime must be empty too.

Pete

12. Jun 9, 2008

### pmb_phy

There are two quantities normally associated with a field. One is the potential and the other is given in terms of derivatives of the potential. In the case of gravity the potential is a tensor known as the metric tensor. The 10 independant components of the metric are gravitational potentials. The field quantity in terms of the derivatives are the components of the affine connection or the Christoffel symbols. The Christoffel symbol for a uniform g-field is given above I believe.
A gravitational field is said to be "uniform" when there are no tidal gradients in the field. By definition this means that the spacetime is flat.

Pete

13. Jun 9, 2008

### Antenna Guy

Aside from the "infinite slab of matter" that Mentz mentioned - otherwise I'd assume that the gravitational field were uniformly zero.

Regards,

Bill

14. Jun 9, 2008

### pmb_phy

There is more than one way to generate a uniform gravitational field. An infinite slab of matter is just one way which occurs in Newtonian mechanics. I don't believe that such a distribution of matter will generate a uniform g-field in the strong field limit in GR.

Another distribution of matter which generates a uniform gravitational field is a spherical cavity cut out of a sphere of matter whose center is off center of the original sphere. This is easier to draw than describe. A diagram is shown here

http://www.geocities.com/physics_world/gr/image_gif/gr10-img-01.gif

Inside such a cavity the field is uniform, at least in the Newtonian approximation. Whether or not this represents a uniform g-field in general relativity is another matter. I know it will hold in the weak field limit though. And yes. I was speaking about outside the distribution of matter which generates such a field. Inside matter the spacetime will always be curved, i.e. where there is matter there is spacetime curvature.

Pete

Last edited: Jun 9, 2008
15. Jun 9, 2008

### Antenna Guy

Yes. The more common version being that the gravitational field about a spherical body of mass of uniform density is uniform (in magnitude) over all theta and phi associated with a constant radius from the center of gravity.

Hmm. So a closed room affected by a uniform gravitational field cannot exist.

Regards,

Bill

16. Jun 9, 2008

### Phrak

too cool, pmb. If you're sitting on an attractive ball of r1 and (an intersecting) repulsive ball of r2. the field is unform on the surface of r2...

you found this via electrostatics, right?

17. Jun 9, 2008

### Staff: Mentor

Thanks for all of the great information everyone. I will post follow-up questions soon. At present, I am not interested in whether or not a uniform field can actually be produced by some specific arrangement of matter, I just want to get a better understanding of the mathematical details of the equivalence principle.

18. Jun 9, 2008

### pmb_phy

No.
Well I did have this as an exam question in an EM class but I was reminded of this in a GR text.

Pete

19. Jun 9, 2008

### pmb_phy

If I understand what you're referring to then I don't see how.
[quote
So a closed room affected by a uniform gravitational field cannot exist.
[/QUOTE]No.

Pete

20. Jun 9, 2008

### Antenna Guy

Do you feel that the statement is true or false? I cannot tell from your response.

Regards,

Bill

Last edited: Jun 9, 2008
21. Jun 10, 2008

### pmb_phy

Let me start from this point. I had stated that
For some reason, which I'm not quite sure of, you took this to mean something besides what I meant it to mean since you responded with the following comment.
I believe this statement to be incorrect. So yes, it is false.

Suppose we're speaking about the gravitational field inside that cavity I mentioned above. Then field inside the cavity is uniform. If you're at rest inside that cavity and inside an enclosed a room then there is no experiment that you could execute that would tell you whether you're in a a uniform gravitational field or a uniformly accelerating frame of reference. This doesn't mean that a uniform gravitational field cannot exist, by far. I've already given an example of such a field. I hope that helps.

Pete

22. Jun 10, 2008

### Phrak

I said "too cool, pmb. If you're sitting on an attractive ball of r1 and (an intersecting) repulsive ball of r2. the field is unform on the surface of r2..."

No??

From your drawing r1 and r2 are radii inside their respective spheres. Fill-in the cavity with attractive matter, as well as (hypothetical) repulsive matter of the same density. The idea being that they cancel. The problem is now simplified into two spherical 'masses' as long as we remember that the smaller sphere has to live inside the larger one.

From Newton's Principia, the shell of matter of radius greater than r1 has no unbalanced force on an object at radius r1. The same with the hypothetical repulsive matter; the shell of material at radius greater than r2 would not effect a mass at r2. So my statement's directly derivable given your claim of a uniform field. I added the ellipses at the end 'cause you were speaking of more than just the surface described by a constant r2.

Where r1 and r2 are allowed to range it should be trivial to verify a uniform field

Last edited: Jun 10, 2008
23. Jun 10, 2008

### Antenna Guy

The issue wasn't whether or not a uniform gravitational field could be devised, but whether you could introduce a hollow cube of matter in which to conduct experiments (without perturbing the otherwise uniform field).

Regards,

Bill

24. Jun 10, 2008

### Staff: Mentor

How did you get this? I was trying to derive the proper acceleration for a stationary particle from your metric and kept on getting zero as follows:

For a stationary particle the metric reduces to:
$$d\tau^2=\left(\frac{a x}{c^2}+1\right)^2 dt^2$$
or
$$\frac{d\tau}{dt}=\frac{a x}{c^2}+1$$

So the stationary particle has a worldline
$$f = \left( c t,x,y,z \right)$$
$$u=\frac{df}{d\tau}=\frac{df}{dt}}\frac{dt}{d\tau}=\frac{(c,0,0,0)}{\frac{a x}{c^2}+1}$$
$$a=\frac{du}{d\tau}=\frac{du}{dt}}\frac{dt}{d\tau}=\frac{(0,0,0,0)}{\frac{a x}{c^2}+1}$$

This kind of thing works fine for inertial frames in SR, but it isn't working right for me here.

25. Jun 10, 2008

### Mentz114

By solving the geodesic equation for gel's metric I get

$$\frac{d^2x}{d\tau^2} = \gamma^2a(1 + ax)$$

where $$\gamma = \frac{d\tau}{dt}$$

As always, I could be wrong.

Last edited: Jun 10, 2008