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I am a bit confused as to how the Equivalence Principle implies a curved spacetime. Or if it doesn’t imply a curved spacetime, then what exactly makes it necessary to have a curved space time?

I could very well have local inertial frames in flat spacetime in arbitrary coordinates. Particles on which only gravity acts could follow curved paths in flat spacetime and their trajectories would be straight locally.

So what exactly forces the manifold to be curved? How exactly is the result that our spacetime should be curved reached?

Why not just study gravity as a background force field on a flat space time?

Edit after the answer:

Why can we just suppose that the world lines of freely falling particles are curved and not straight but in a flat spacetime?

Then the divergence of the geodesics which is explained by the Riemann Tensor would not need that explanation. Two falling bodies could very well come near each other because their worldliness would be curved in the flat spacetime.

Why don’t we model gravity as an external force field on a flat spacetime just like other force fields?

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First, consider spacetime without gravity. You notice that inertial objects have worldlines which are straight lines in spacetime and accelerometers measure how much a worldline bends in spacetime, with an accelerometer reading of zero corresponding to an inertial object’s straight worldline. Importantly, inertial objects at rest with respect to each other have parallel worldlines and they never intersect.

Now, an inertial frame consists of a coordinate grid of straight lines on the spacetime, and an accelerating frame consists of a coordinate grid of lines curving in the direction of acceleration. Describing the straight worldline of an inertial object in the accelerating coordinates produces a fictitious force (Christoffel symbols). By the equivalence principle that fictitious force is locally equivalent to gravity. Since fictitious forces do not change the fact that an inertial worldline (accelerometer reading 0) is straight, so also gravity cannot change a worldline from being straight. Otherwise, gravity would not be locally equivalent to a fictitious force.

Then, you extend those ideas to a global spacetime with gravity. Objects in free fall have accelerometers which read zero, so their worldlines are straight. But two objects in free fall initially at rest with respect to each other can eventually intersect. So we have straight lines which are initially parallel to each other but eventually intersect. This is impossible in a flat spacetime, but easily happens in a curved spacetime.

The equivalence principle only applies locally in small regions where gravity is approximately uniform and spacetime is approximately flat. Over large regions of spacetime gravity is non-uniform, and it is that non-uniform gravity (tidal gravity) that is spacetime curvature.

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  • $\begingroup$ Okay I understand that. But I am saying that why can’t we assume that worldliness of freely falling particles are curved and not straight in flat space time. Then two freely falling bodies could very well come near each other without the requirement of a Riemann tensor? Why don’t just model gravity as a an external force field on a flat space time like other force fields? Please see the edit for the details. $\endgroup$ – Shashaank Feb 9 at 13:38
  • $\begingroup$ @Shashaank precisely because of the equivalence principle. The equivalence principle says that locally gravity is equivalent to a fictitious force. Fictitious forces leave the worldline straight and do not curve it. I guess I didn’t convey that clearly. I will add some explanation in the answer $\endgroup$ – Dale Feb 9 at 13:55
  • $\begingroup$ Let me put it like this-"why does the fact that locally gravity behave like a fictitious force forces our manifold to be curved or why does this fact not allow gravity to be studied as an external force field acting in th background on flat space time. $\endgroup$ – Shashaank Feb 9 at 14:53
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    $\begingroup$ I think I get it. Curved space time would not have been needed if gravity behaved as a fictitious force globally and not just locally. Like an accelerating frame. The fictitious force is the same globally and hence the Riemann tensor for the metric vanishes in that case. But just because gravity can be eliminated locally, means any metric for which the Riemann tensor vanishes would not be to explain such a situation. A constant gravitational field will just act as a fictitious force and thus could be studied in Minkowski space time, like in the Newtonian approximation we consider $\h$ $\endgroup$ – Shashaank Feb 9 at 16:39
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    $\begingroup$ (cont).. as a background perturbation. We could consider the perturbation on the Minkowski metric because it was constant throughout. Just because real gravitational sources don’t produce constant gravitational fields, it becomes necessary to model them in a curved space time. I think I get it. Is that all right? Or would you like to point out any correction? $\endgroup$ – Shashaank Feb 9 at 16:42
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An observer in an uniform accelerated frame, that is the typical example for equivalence principle, has a flat spacetime.

But it is difficult to imagine a bunch of matter that can generate a field like that. Matter tends to concentrate in an approximately spherical form, generating a non uniform field, where the Riemann tensor is not zero.

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The best simple answer I've found for this is in The Feynman Lectures.

By the principle of equivalence, clocks at different heights measure time at different rates.

[In a rocket accelerating at g, (Rate at Receiver at height H)=(Rate of Emission)($1+gH/c^2$)]

But that doesn't give you curvature. To get to curvature, Feynman asks you to consider two paths in spacetime that would meet if there was no gravity. The first path is along the time axis 100 seconds to B. You don't move in space; only in time. Then you go up in space H feet to point D. The second path is up in space H feet to point C. Then you wait 100 seconds. You will find that you are not at point D in spacetime. Because time went faster at height H. That's spacetime curvature.

The Feynman Lectures. Volume 2. Chapter 42.

https://www.feynmanlectures.caltech.edu/II_42.html

So, the principle of equivalence alone doesn't give you spacetime curvature. To get curvature you have to also compare different regions in spacetime.

Why don’t we model gravity as an external force field on a flat space spacetime just like other force fields?

The difficulty Einstein had was that he needed a relativistic theory. And he also wanted it to explain the effect of gravity on light. It's not easy to come up with such a field theory of gravity. They're restrictive assumptions. And then his theory went on to be confirmed by experiments.

There also are models of small areas of spacetime as Newtonian. The tidal "forces" of GR are modeled as regular forces. Newton's theory is an approximation of GR too. But there are a lot of GR things that you won't get out of Newtonian gravity, obviously. You're never going to get clocks slowing down from the equations of Newtonian gravity.

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  • $\begingroup$ I liked this answer. Do you know of any attempts to study gravity as an external force field. What exactly were the issues In doing so? Was it not possible to have a finitely propagating influence as a background on the Minkowski space time? When the field strength is less then we can indeed study it as a background field on Minkowski space tii me. Was the finite propagation of gravity in the strong field regime a problem to study it as an external field on Minkowski space time $\endgroup$ – Shashaank Feb 10 at 14:42
  • $\begingroup$ Given that Newton's law is similar to the electrostatic force, I'm sure someone must have tried to model gravity on electrodynamics. But light has no mass or charge, so that theory wouldn't be able to explain bending of light rays and gravitational lensing. I would reverse it and ask why, given that GR is true, you can't describe the same stuff without GR. The main thing is that GR uses vectors and tensors that mix space and time and a non-relativistic theory won't do that. But I'm not an expert. And I do believe that there is much more to say about this subject. $\endgroup$ – user288251 Feb 11 at 2:10
  • $\begingroup$ (E&M is relativistic, of course.) $\endgroup$ – user288251 Feb 11 at 3:16

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