This question arose in the Space Exploration forum, by non-physicists, so it likely sounds naïve. The question involved the plausibility of using a very strong gravitational field to accelerate an astronaut at ludicrously high $G$s. A comment was made that the field would need to be uniform to prevent lethal tidal forces.

My intuition says that a uniform gravitational field has zero acceleration... that it is the curvature which defines the direction and magnitude of the acceleration. "Mass tell space how to bend, space tells mass how to move." No tide, no bend, no acceleration.

Can gravitational acceleration exist as a "uniform field"?

  • $\begingroup$ Acceleration due to gravity $\endgroup$ Oct 30 '21 at 0:57

We can't create such a thing. But if you could create one, we would expect a uniform field to subject objects in freefall to constant acceleration (proportional to the strength of the field)

The earth's gravitational field magnitude varies only slightly as long as you are near the surface. Close enough that we can consider the field to be "constant" within a building or laboratory.

Tidal effects arise from the non-uniformity of the fields we see. The moon and the sun create gravitational fields that are stronger on the side of the earth than on the other. These differences result in forces that deform the earth and slosh the oceans around.

A person falling through a uniform gravitational field would accelerate and feel no effects (regardless of the strength).

If a hypothetical uniform ("flat") gravitational field could be created, it would be symmetrical. How would it know which direction to accelerate masses?

Ignoring the GR/curvature portions of your question, a uniform field is not necessarily symmetric. The field still has direction. We can create a nearly uniform electric field between parallel plates. But the orientation of the field is unambiguous. Likewise we can imagine a uniform gravitational field. All points in that field would have the same magnitude and direction for perceived gravitational forces.

  • $\begingroup$ If a hypothetical uniform ("flat") gravitational field could be created, it would be symmetrical. How would it know which direction to accelerate masses? Isn't it the curvature of space which defines the direction and magnitude of the acceleration? Maybe I'm confusing "flat" spatial geometry with a flat (uniform) gravitational field. $\endgroup$
    – Woody
    Oct 30 '21 at 2:06
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    $\begingroup$ @BruceWoodburn The root of the confusion might be that the word "acceleration" is overloaded. When people say that an object in free-fall is accelerating, they mean relative acceleration. Even in flat spacetime, an object in free-fall can be accelerating relative to an object that's not in free-fall. When I jump off of a ladder, the earth and I are both accelerating relative to each other, but only the surface of the earth is accelerating in the absolute sense. I am not: before I hit the ground, I am weightless. When on the ground, the surface and I are both accelerating absolutely. $\endgroup$ Oct 30 '21 at 2:18
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    $\begingroup$ @BruceWoodburn ... Both kinds of acceleration, relative and absolute, can occur in flat spacetime (if non-gravitational forces are present), so neither one requires curvature. But relative acceleration between two objects that are both in free-fall does require curvature. That's what tidal effects are. $\endgroup$ Oct 30 '21 at 2:28
  • $\begingroup$ @BruceWoodburn uniform here means homogenous but not isotropic. $\endgroup$
    – Señor O
    Oct 30 '21 at 4:24

Yes, but theoretically you would need an infinite sheet of mass to create a uniform gravitational field (just like you need an infinite sheet of mass to create a uniform electric field).

For any finite acceleration, you could conceive of a very large infinite sheet (or very large planet/sphere) where the tidal forces are small enough across the body of an astronaut to not kill them.


In a strict use of terms, there are two different senses in which a gravitational field may be uniform, and they are not the same. (I'll describe them in a moment). But the main qualitative answer to your question is that to create the kind of field you want you would need a large mass shaped as a very wide flat plane or as a long straight cylinder (see below for the reason for two answers). The field would then cause acceleration of other objects towards that plane or cylinder. So yes, it does cause acceleration. But as usual, two objects in freefall next to one another would have small or zero acceleration relative to each other.

Now for the more technical part.

We describe gravitation by setting up a system of coordinates in some region of spacetime and then providing an equation which says how to find the spacetime interval between neighbouring events. The equation is called the line element or the metric (more strictly, the metric is the set of coefficients that appear in the equation). The line element for the uniform gravitational field has the form $$ ds^2 = -\alpha^2(x) dt^2 + dx^2 + dy^2 + dz^2 $$ where $\alpha(x)$ is a function of $x$, and we should consider two possibilities: $$ \alpha(x) = a x \\ \alpha(x) = c e^{k x} $$ where $a$ and $k$ are constants and $c$ is the speed of light.

The first possibility gives $$ ds^2 = -a^2 x^2 dt^2 + dx^2 + dy^2 + dz^2. $$ This is called the Rindler metric. It describes an example of flat spacetime, so in one sense one might say there is no gravity here. But it describes what goes on if your system of coordinates has constant proper acceleration, like coordinates set up inside a rocket whose engine is constantly providing the same proper acceleration to the rocket. Inside such a rocket you see apples fall to the floor etc. People sometimes call this a "uniform gravitational field". But it is uniform in one sense and not in another. There are no tidal stretching or squeezing effects, which indicates a kind of uniformity, but on the other hand the acceleration due to gravity depends on $x$, so it is not uniform in that sense. For the Rindler metric one finds that the acceleration, relative to the coordinates, of an object released from rest is proportional to $1/x$. On the other hand, if you are sitting high up in the field and dangling a heavy object on a light rope, then the force you have to apply to your end of the rope does not depend on where the object is! How can these two statements not be mutually contradictory? (Exercise for the (expert) reader).

Now for the second case. $$ ds^2 = -c^2 e^{2kx} dt^2 + dx^2 + dy^2 + dz^2. $$ In this case, let's see what happens if we shift the origin of spatial coordinates by some constant $x_0$ and multiply the temporal coordinate by a constant factor $e^{k x_0}$. Define $$ X = x - x_0, \;\;\;\;\;\; T = e^{k x_0} t. $$ One finds $$ ds^2 = -c^2 e^{2kX} dT^2 + dX^2 + dy^2 + dz^2. $$ In other words we have the same metric as before, just with the new coordinate labels. It follows immediately that all gravitational effects of this metric will be the same no matter where you are in spacetime. So this earns the right to be called a "uniform gravitational field". But this uniformity includes that everywhere it produces a tidal stretching/squeezing effect on any object, because the spacetime curvature is now non-zero.

But to answer your question, this second example of a "uniform gravitational field" is impossible, because no configuration of matter can give rise to it (the matter would have to have pressure but no energy density, which is not possible).

Finally, let me return to the opening paragraph. You have near Earth's surface an example of a gravitational field which is close to uniform on distance scales small compared to the radius of the Earth. That is enough to tell you that approximately uniform gravitational fields are possible, and it gives you a hint of how they may be produced. But I get the impression from the question that what you really want to know is whether you can have a large and steady gravitational acceleration without any tidal stretching effects which might make the voyaging astronaut uncomfortable. For that you want the Rindler metric, and the configuration of matter which gives this metric to reasonable approximation (in a plane at least) is one where the gravitational acceleration falls inversely with distance. A large mass shaped as a long cylinder would do that.

But using gravity to accelerate spaceships is never the whole solution, because the spaceship will fall down towards the gravitating body. It will either crash on the body, or pass through a hole if there is one, or it will have to be steered aside. If it gets very fast then it will be difficult to steer aside, and in any case it will then slow down again as it leaves the gravitational potential well.

In view of all this, I suspect that your best bet may be to 'hitch a ride' by approaching a star which is already moving fast relative to your rocket, and then use a sling-shot effect, or just be content to orbit the star for a while if it is moving towards your destination.


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