I Convert Metric Tensor to Gravity in GR

max_zhou
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I am still learning general relativity (GR). I know we can find the path of a test particle by solving geodesic equations. I am wondering if it is possible to derive/convert metric tensor to gravity, under weak approximation, and vice versa.

Thanks!
 
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max_zhou said:
I am wondering if it is possible to derive/convert metric tensor to gravity,
What do you mean “convert” to gravity? It already is gravity.
 
max_zhou said:
I am wondering if it is possible to derive/convert metric tensor to gravity
This is much like asking how you convert gravitational potential to gravity in Newton's gravity. It is already a part of the description of how gravity works.
 
Another possibility is that the OP is interested in how to calculate the proper acceleration of a worldline, so that he'd know how much a test mass would weigh if an observer on said worldline weighed it with a scale.
 
pervect said:
Another possibility is that the OP is interested in how to calculate the proper acceleration of a worldline, so that he'd know how much a test mass would weigh if an observer on said worldline weighed it with a scale.
We can read and try to squeeze our own knowledge about what you can do in GR into the OP. However, all of that is going to be conjecture unless the OP can clarify what he is asking about.
 
Thank everyone for the discussion so far. I should have clarified my question. My bad. So my question is, let us say we have known the metric tensor distribution, and I want to calculate the corresponding gravity force distribution (magnitude and direction of gravity at every point). The direction here is a direction defined on every local coordinate system. Not sure if this is clear enough? Maybe this is not a valid question. Maybe with metric tensor we can only solve the geodesic equation to get the path of a test particle. But I imagine at least under weak limitation/approximation, i.e. weakly curved spacetime, there would be a way to still use gravity to describe the world, just as what we do with Newtonian gravity. In this case, we could have a conversion between metric tensor and the gravity.
 
max_zhou said:
gravity force distribution (magnitude and direction of gravity at every point)

This still isn't enough because we don't know what you mean by "magnitude and direction of gravity". Gravity in GR is not a force. It's spacetime curvature.

I think you need to give a concrete example instead of just talking about abstractions. For example, say you have a spaceship with a rocket engine and any measuring instruments you want aboard. How would you use the ship to measure the "magnitude and direction of gravity" in the space around a gravitating body like the Earth?
 
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max_zhou said:
Maybe with metric tensor we can only solve the geodesic equation to get the path of a test particle. But I imagine at least under weak limitation/approximation, i.e. weakly curved spacetime, there would be a way to still use gravity to describe the world, just as what we do with Newtonian gravity. In this case, we could have a conversion between metric tensor and the gravity.

Newtonian gravity is just an approximation to GR when speeds are low compared to the speed of light and spacetime curvature is small. So anything that Newtonian gravity can solve, GR can solve as well. (And GR will give more accurate predictions that Newtonian gravity since the latter is only an approximation to the former.)
 
PeterDonis said:
Newtonian gravity is just an approximation to GR when speeds are low compared to the speed of light and spacetime curvature is small. So anything that Newtonian gravity can solve, GR can solve as well. (And GR will give more accurate predictions that Newtonian gravity since the latter is only an approximation to the former.)
Thanks for the reply!
Let us say we have a spaceship that is stationary at a location, and we do the free falling experiment in the spaceship. Let us say the falling distance is h, and the falling time is dt. Because spaceship is very small in a slightly curved spacetime, like the earth, I think we will be able to approximate the relationship between h and dt with
h = 1/2 * a * dt * dt
Then we define the coefficent a as the gravity acceleration. Notice that *** h *** and *** a *** here will be a vector (having direction), not just a scalar.
Assume we know the metric tensor at the spaceship location g, is there a formula that relates the metric tensor g to the acceleration a? And will this formula apply to every location?
Again, the whole scenario is set up in a weakly curved spacetime.
 
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  • #10
max_zhou said:
Assume we know the metric tensor at the spaceship location g, is there a formula that relates the metric tensor g to the acceleration a?

Yes. Modulo a few technical assumptions that I'll leave out for now, the acceleration ##a## will be given by the geodesic equation, which of course we know given the metric tensor.

max_zhou said:
will this formula apply to every location?

Yes.
 
  • #11
PeterDonis said:
Yes. Modulo a few technical assumptions that I'll leave out for now, the acceleration ##a## will be given by the geodesic equation, which of course we know given the metric tensor.

So we can only first solve the geodesic equation first, then to calculate the `gravity acceleration`? I take this as implicit relationship.
Is it possible to give an explicit form of the relationship between the gravity acceleration and the metric tensor? By explicit, I mean some formula that can be written out directly, like ##a## = ##covariant## ##derivative## ##of## ##metric## ##tensor##, or like ##a## = ##integration## ##of## ##metric## ##tensor##, etc, just to give some instances to illustrate my point.
 
  • #12
max_zhou said:
So we can only first solve the geodesic equation first, then to calculate the `gravity acceleration`?

The geodesic equation is the equation for what you are calling "gravity acceleration". Of course you have to solve the equation for something to figure out what the formula for it is.

max_zhou said:
I take this as implicit relationship.

No, it isn't. The geodesic equation gives you an equation with what you are calling "gravity acceleration" on one side, and other stuff on the other. That's as explicit as it gets.
 
  • #13
PeterDonis said:
The geodesic equation is the equation for what you are calling "gravity acceleration". Of course you have to solve the equation for something to figure out what the formula for it is.
No, it isn't. The geodesic equation gives you an equation with what you are calling "gravity acceleration" on one side, and other stuff on the other. That's as explicit as it gets.
Ah, I see. Thank you!
 
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