# Is Spacetime Flat in Uniform Gravitational Fields and Acceleration?

• Dale
In summary, the usual "windowless room" thought experiment involves two equivalent situations - one with uniform acceleration and one with a uniform gravitational field. The spacetime in the uniform acceleration case is flat, so the equivalent spacetime in the uniform gravitational field must also be flat. This means that there are no tidal gradients in the field and the spacetime can be considered "uniform". The metric for a uniform gravitational field is identical to that of a uniformly accelerating frame of reference, and the Einstein tensor vanishes for such a spacetime, making it empty.
Dale
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.

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.

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.

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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
$$x'=(x+a^{-1}c^2)\cosh(c^{-1}at)-a^{-1}c^{2}, t'=c^{-1}(x+a^{-1}c^2)\sinh(c^{-1}at), y'=y,z'=z. [/itex] (You might want to check the details - I'm not guaranteeing there's no mistakes.) Last edited: gel, I find no non-zero Christoffels for your metric. Does this not mean zero acceleration ? btw, I think this is what's called Born rigid motion. It's flat because the curvature tensor vanishes, and can be transformed to Minkowski space. You can also work out a metric for a uniform gravitational field, but it isn't flat and the Einstein tensor is non-zero, so it wouldn't be empty space. Mentz114 said: gel, I find no non-zero Christoffels for your metric. Does this not mean zero acceleration ? I got [tex] \Gamma^x_{tt}=a(1+c^{-2}ax),\ \Gamma^t_{tx}=a(c^2+ax)^{-1}.$$

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

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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.

Correction to my calculation above, Christoffel symbs are

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

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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.

gel said:
btw, I think this is what's called Born rigid motion. It's flat because the curvature tensor vanishes, and can be transformed to Minkowski space.

You can also work out a metric for a uniform gravitational field, but it isn't flat and the Einstein tensor is non-zero, so it wouldn't be empty space.
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

DaleSpam said:
In the gravitational/accelerated non-inertial frame what is the form of the gravitational field (as a function of position)?
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.
If the field is a non-constant function of position then how is it considered either uniform or flat?
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

pmb_phy said:
The metric for a uniform gravitational field is identical to that of a uniformly accelerating frame of reference. ... Such a spacetime must be empty too.

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

Regards,

Bill

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

Regards,

Bill
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

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pmb_phy said:
There is more than one way to generate a uniform gravitational field.

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.

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.

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

Regards,

Bill

pmb_phy said:
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

Pete

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?

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.

Phrak 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.
you found this via electrostatics, right?
Well I did have this as an exam question in an EM class but I was reminded of this in a GR text.

Pete

Antenna Guy said:
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.
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

So a closed room affected by a uniform gravitational field cannot exist.
No.

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

Regards,

Bill

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

Regards,

Bill

Let me start from this point. I had stated that
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.
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.
Hmm. So a closed room affected by a uniform gravitational field cannot exist.
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

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..."

pmb_phy said:
No.
Pete

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

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pmb_phy said:
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.

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

gel said:
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 $\frac{a}{1+c^{-2}ax}$.
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.

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.

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DaleSpam said:
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.

I think the problem is that a stationary particle would not be accelerating.

Shouldn't there be a dx/dt in u for the accelerating particle?

Regards,

Bill

DaleSpam,
possibly your SR calculations don't take into account the variable speed of light in the space-time.

In Minkowski (1+1) space for a null worldline

$$0 = c^2dt^2 - dx^2 \rightarrow c^2 = \frac{dx^2}{dt^2}$$

but if you bend space, the null line element is

$$0 = c^2dt^2 -f(x)dx^2 \rightarrow c^2 = f(x)\frac{dx^2}{dt^2}$$

which seems to say that light-speed in the x-direction depends on x .

M

Warning: I am not a GR expert, I know more about SR than GR.

DaleSpam said:
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.

The equivalence principle is usually considered in a small room so that tidal effects can be ignored. In a large room, what does “uniform acceleration” mean? Well, an observer who is static (relative to the room) should experience proper acceleration that is constant over time. Beyond that:
(a) Do you want all static objects to experience the same proper acceleration as each other?
(b) Do you want the room to remain a constant size (according to its occupants)?
Conditions (a) and (b) are incompatible. (a) is the acceleration in paradox[/url]. (b) is Born rigid acceleration.

I believe I am correct is saying that in relativity “uniform acceleration” is defined to be (b), even though that may seem a counterintuitive name. This also raises the question of how does an observer in the room measure the room’s size. Radar won’t give a valid answer because the observer is accelerating. What the observer must do is jump in the air and make radar measurements while airborne. Whilst airborne, the observer is an inertial observer (ignoring air resistance). The measurement deemed to be simultaneous with the top of the jump, when the observer is stationary relative to the room, is deemed to be the valid measurement for the accelerating frame at that instant. The jumping observer defines a “co-moving inertial frame” at the top of the jump. Simultaneity, distance and time for the accelerating frame are all defined via the appropriate co-moving inertial frame, i.e. a jumping observer. The proper acceleration is also the acceleration of the room relative to the co-moving inertial frame.

If you have a number of jumping observers (reaching their peaks at different times) they can use the Lorentz transform to convert between each other’s points of view.

gel said:
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
$$x'=(x+a^{-1}c^2)\cosh(c^{-1}at)-a^{-1}c^{2}, t'=c^{-1}(x+a^{-1}c^2)\sinh(c^{-1}at), y'=y,z'=z.$$
This is correct. $(t, x, y, z)$ are Rindler coordinates, relative to the room. $(t', x', y', z')$ are inertial coordinates. In fact, they are the co-moving inertial frame at all the events where $t = 0$.

More generally at any event E where $t = t_0$, it can be proved that the co-moving inertial frame is given by

$$x'_E = \left( x + \frac{c^2}{a} \right) \cosh \frac {a (t - t_0)} {c} - \frac{c^2}{a}$$
$$t'_E = \frac {1}{c} \left( x + \frac{c^2}{a} \right) \sinh \frac {a (t - t_0)} {c}$$​

$t$ is the proper time of the observer at $x = 0$. For any other object stationary in the room, the proper time is

$$\tau = t \left(1 + \frac {ax}{c^2} \right)$$​

This illustrates gravitational time dilation. Two events with the same $t$ coordinate are simultaneous in the room’s frame (i.e. in the co-moving inertial frame).

gel said:
btw, I think this is what's called Born rigid motion. It's flat because the curvature tensor vanishes, and can be transformed to Minkowski space.
This is also true, Any equation for the metric that can be transformed by change of variable into the standard Minkowski equation must be flat, by definition.

How to calculate the proper acceleration.

Use the equations I gave above for the co-moving inertial frame. You can simplify the maths a bit by moving the spatial origin (in both frames) to get

$$x'_E = x \cosh \frac {a (t - t_0)} {c}$$
$$t'_E = \frac {x}{c} \sinh \frac {a (t - t_0)} {c}$$​

For a constant value of $x$, calculate $dx' / dt$ and $dt' / dt$. Divide these to get the velocity $dx' / dt'$ measured in the co-moving frame. Differentiate this with respect to $t$ and divide by $dt' / dt$ to get the acceleration $d^2x' / dt'^2$ measured in the co-moving frame. Finally put $t = t_0$ to get the event when the inertial frame is co-moving; this, by definition, is the proper acceleration, which should come out as $c^2 / x$, which, after the change of origin, is equivalent to the value given by gel in post #4.

On a final note, uniform acceleration gives rise to an event horizon. Any photons in the region

$$x' + \frac{c^2}{a} \leq c t'$$​

(in gel’s original coordinates) will never reach the accelerating room.

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## 1. What is a flat uniform and equivalent surface?

A flat uniform and equivalent surface is one that has the same properties and characteristics throughout its entire surface. This means that no matter where you measure or observe on the surface, you will find the same physical properties, such as height, texture, and composition.

## 2. How is a flat uniform and equivalent surface different from a non-flat surface?

A non-flat surface can have variations in its physical properties, such as bumps, valleys, or changes in texture, throughout its surface. This makes it non-uniform and non-equivalent, as different areas of the surface may have different properties.

## 3. What are some examples of flat uniform and equivalent surfaces?

Some examples of flat uniform and equivalent surfaces include a smooth sheet of paper, a polished mirror, and a flat table or desk surface. These surfaces have consistent properties throughout and can be measured or observed as such.

## 4. Why is it important for a surface to be flat uniform and equivalent?

Having a flat uniform and equivalent surface is important in many scientific and practical applications. It allows for accurate measurements and observations, as well as consistent results in experiments and processes. It also provides a stable and reliable foundation for building structures and devices.

## 5. How can a surface be made flat uniform and equivalent?

A surface can be made flat uniform and equivalent through various methods, such as polishing, grinding, and using specialized tools and equipment. It is also important to carefully control the materials and processes used in creating the surface to ensure uniformity and equivalence.

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