Background Gravity – Time Dilation in a Flat Field

I got into an argument with a physics buddy not that long ago (a year, maybe), about gravity. We have an intermittent arrangement where we go drink beer and talk about physics at two or three pubs in San Luis Obispo. Usually, the physics becomes a little less coherent as the evening wears on.

One of the discussions centered on whether there is a “background field” of gravity or not, or whether it’s even sensible to discuss such a thing since, in an infinite field of equally distributed mass (or gas, or 1 atom per cubic light year, whatever), all the forces around you seem to cancel out. The mass of the universe to your left is equal in size to the mass of the universe on your right; you feel a net acceleration of zero. I argued that even though the field was “flat”, there was still a field there. He argued that a field implied a gradient; there is always a force.

We did not come to a satisfactory conclusion. It might merely have been the fact that we were defining the same terms in different ways in our heads. I’m not sure. I thought my argument was rock-solid.

So, here is my side of it.

Some of you are probably familiar with Newton’s Shell Theorem. It’s in his Principia Mathematica, and if I remember right, he solved it without using calculus. Basically, what it says is this; if you are inside a spherically symmetric shell of mass, then you feel no gravity pulling you any direction. It’s a bit non-intuitive. Let’s say the Earth is hollow, and the entire planet’s mass has been compressed into a thin spherical shell a few centimeters (or meters, it doesn’t matter) thick. If you are floating around in your Nike Space Suit inside this shell, you will not be pulled toward the center, or the inner surface of the shell, or anywhere else inside the shell. Wherever you are put, you will remain.

Personally, I think this is one of the coolest theorems ever.

It’s also true that if you are outside the surface of a spherically symmetric planet, then it doesn’t matter how dense it is, at a given radius you will feel the pull of a certain amount of gravity. If you are in orbit above the Earth, and the Earth suddenly becomes a black hole of the exact same mass, you will remain in orbit, totally unaffected by that change. That’s pretty cool, too. Given that the gravitational force is based on F=GMm/r2, this should be kind of obvious. Neither your mass (m) or the mass of the Earth (M) has changed, your orbital radius is the same, and G is the gravitational constant. Ergo, the density of the object you are orbiting at a radius “r” from the center of the object is irrelevant.

So, that was me drifting from the actual subject. Shell Theorem—let’s get back to that.

As you might know, the clock of anyone in a gravity field runs slower than that of a clock outside of that gravity field. This is called gravitational time dilation (and is equal to ∆t’ = ∆t √(1-2MG/rc2) for a non-rotating sphere). A person on Earth actually ages slower than a person in deep space, according to relativity. This was verified with clocks flying around the Earth in the Hafele-Keating experiment. Before you ask, yes, they took into account Earth’s rotational speed, the speed of the airplane (in both directions relative to Earth’s rotation) with regard to Special Relativity’s time dilation due to velocity. It was a nice experiment.

Let’s say we’re using a hollow Earth from Newton’s shell theorem. As you get closer to the Earth, you are in a deeper gravity well, and the outside observer sees your clock slow down. There’s a small hole in the planet, and you pass into the planet, where everything is pulling you in opposite directions equally, so you seem to feel no force. And yet, your time dilation effect does not suffer a discontinuity, jumping suddenly to that of the outside observer. You are in a denser gravity field, but a flat gravity field. [to the physics majors out there, for god’s sake, if my terminology sucks, please correct me]. Your time dilation will be just the same as if you were standing on the surface of the planet.

So now you have a flat gravitational field (no “force” pulling you in one direction, that is, all forces pulling you equally in all directions). And yet, even in this apparent lack of gravity, where you can’t actually tell that you’re in a gravity well, your time runs slower than the time of someone far from the planet.

I extended this argument to the rest of the universe. If mass was distributed equally around you, even though you felt no force one way or the other, there would still be a background gravitational field. Gravitational time dilation implies a gradient; for time dilation to be relevant, you need someone in a weaker gravitational field measuring your time. However, both the measured and the measurer can be in locally flat gravitational fields.

Does a flat gravitational field curve spacetime by itself? Or is it only the gradient between two different gravitational fields that curves spacetime? My general opinion is that you don’t need the gradient for the curvature of spacetime. If you have an infinite universe with equally distributed mass, then from some arbitrary center, it will appear to curve spacetime until it closes the dimensions of universe into closed loops, like the inside surface of an event horizon (though other arbitrary centers will have different, yet overlapping event horizons – a subject I will touch upon another day). Likewise, if you have a large, thin flat sheet of soapy water in the air, fluctuations are going to cause it to form bubbles, closing up the edges. In spacetime, there may be a similar tension (gravity?) that closes the edges together into a 4D hypersphere.

How would you test the curvature of space inside a shell? I’m not entirely sure. I think the universe we have is a good test case, however.

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