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My Background:

In high school, I completed AP Physics C Mechanics and Electricity and Magnetism. In my first year of undergrad, I completed a course on Newtonian Mechanics and a course on Special Relativity and Electromagnetism which both approximately followed the sections on those topics in the Feynman Lectures on Physics.

The Question:

In my free time, I am starting to learn tensor analysis and general relativity. I wanted to explain what my current understanding of GR is and was wondering if what I understand of it so far could be verified and if it is not correct the problems with it could be explained.

My Current Understanding:

  1. Objects follow geodesic in spacetime which extremize the total spacetime distance (proper time) along that geodesic. These geodesics can be found through the geodesic equation if you know the metric tensor.
  2. The energy-momentum tensor measures how much energy density/flow, momentum density/flow is in a certain region of spacetime.
  3. The energy-momentum tensor determines what the metric tensor is through the Einstein field equation.
  4. If the energy-momentum tensor is known, the Einstein field equation can be used to solve for the metric tensor (i.e. the Schwarzschild metric is the solution for the metric tensor if the energy-momentum tensor is that of a spherical star or black hole). Then the geodesic equation can be used to calculate the trajectory of any object in spacetime.

To summarize, the energy/mass existing at a point in spacetime causes the spacetime around it to curve and this curvature influences the movement of objects that travel through the "shortest path" through spacetime.

Additional Questions:

  1. Does the energy-momentum tensor vary with the spacetime coordinates just like the metric tensor does and is it determined by the distribution of energy and momentum throughout spacetime (i.e. if a massive body exists somewhere)?
  2. If that is so, does the value of the energy-momentum tensor at a point in spacetime influence the curvature of spacetime only at that specific point or does it influence the curvature of surrounding points in spacetime as well (i.e. does the Sun cause spacetime to curve in a large region around it or just at the points in spacetime where the Sun exists)?
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Most of what you said is right.

If the energy-momentum tensor is known, the Einstein field equations can be used to solve for the metric tensor

This is wrong. For example, suppose the energy-momentum tensor is zero. There are still many possible metrics, including Minkowski space, versions of Minkowski space with nonstandard topologies, spacetimes containing gravitational waves, and black-hole spacetimes.

the Schwarzschild metric is the solution for the metric tensor if the energy-momentum tensor is that of a spherical star or black hole

The energy-momentum tensor of the Schwarzschild metric is zero everywhere. The mass of the black hole is hard to pin down. You can think of it as being at the singularity, but the singularity is a spacelike surface in the future and is not part of the spacetime manifold. Or you can think of the mass as being in the spacetime but not being localized, but then it's not measured by the energy-momentum tensor.

Does the energy-momentum tensor vary with the spacetime coordinates just like the metric tensor does and is it determined by the distribution of energy and momentum throughout spacetime (i.e. if a massive body exists somewhere)?

The energy-momentum tensor does vary from point to point. Its value at a point only describes the energy and momentum density at that point, not far away.

If that is so, does the value of the energy-momentum tensor at a point in spacetime influence the curvature of spacetime only at that specific point or does it influence the curvature of surrounding points in spacetime as well (i.e. does the Sun cause spacetime to curve in a large region around it or just at the points in spacetime where the Sun exists)?

This depends on what you mean by "influence" and "curvature." There is curvature that is not measured by the Einstein tensor, such as the curvature of a gravitational wave. The direct influence of the stress-energy is only on the part of the local curvature measured by the Einstein tensor.

This is actually pretty similar to electromagnetism. The divergence of the electric field is determined locally by the charge density, but electric fields propagate.

It's great that you're formulating questions of this kind. These are all good, natural questions to be asking as a beginner at GR. Good luck!

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  • $\begingroup$ Some follow up questions: So many possible metrics can solve the Einstein field equations for a given stress-energy tensor? Is the Schwarzchild metric only for a black hole or is it also the metric used to solve for the trajectory of a planet orbiting the Sun? I think the problem is I am thinking of the pop-science demo of GR where a ball is placed on a trampoline (spacetime) which gets curved and causes objects to move differently around it. I know this is probably a bad way to think of it but does the mass at a point affect spacetime similar to that? $\endgroup$ – mihirb Jul 18 at 0:22
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    $\begingroup$ @mihirb the Schwarzchild metric is for spherically symmetric objects. So, yes it can be used to solve the Sun's effect on spacetime. A black hole is just a special case which arises when a parameter in the metric goes to 0. It was actually discovered as a mistake, but now the Schwarzchild metric is known for the black hole :) $\endgroup$ – PNS Jul 18 at 6:31
  • $\begingroup$ @PNS Thanks! Do you know of a better way of thinking of the Sun's effect on spacetime than the trampoline analogy used which is clearly very false and misleading? I guess I'm wondering how far away from the Sun does the curvature of spacetime get affected due to the Sun's mass/energy since it clearly must affect the spacetime near the Earth to cause the Earth to change its motion. Does this mean the energy-momentum tensor affects spacetime curvature at an area not local to it through some mechanism such as waves through spacetime? $\endgroup$ – mihirb Jul 18 at 16:48
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    $\begingroup$ @mihirb I think you'll find this answer by John Rennie & its links useful. John talks briefly about the river model here, which is a superior analogy to the trampoline analogy. $\endgroup$ – PM 2Ring Jul 21 at 9:03
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If the energy-momentum tensor is known, the Einstein field equations can be used to solve for the metric tensor (i.e. the Schwarzschild metric is the solution for the metric tensor if the energy-momentum tensor is that of a spherical star or black hole).

The metric tensor depends also on the symmetries. For example: if the energy-momentum tensor is zero in a region outside a spherical mass, and this mass is not rotating, we can say that there is a spherical symmetry, and the field is only a function of $R$. After calculating all components of the Ricci tensor, we come to differential equations that leads to the Schwartzschild metric.

In this approach we don't use any information about mass or energy values or densities. It is forcing the equation to match Newtonian gravity for weak fields that brings the product $GM$ to the metric.

But if this mass is rotating, the spherical symmetry is no more valid and the metric is different.

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  • $\begingroup$ What about if we are solving something completely new without knowing the Newtonian limit? So given the energy-momentum tensor in space and the symmetries, the Einstein field equations can be uniquely solved for the metric tensor? And then once the metric tensor is known, the geodesic equation can be used to find the trajectory of an object in spacetime? I'm just trying to figure out the exact steps of how GR would be used to solve a general problem. Such as going from a mass/energy distribution in an area of spacetime to solving for the trajectory of an object in that spacetime. $\endgroup$ – mihirb Jul 18 at 16:40
  • $\begingroup$ In the case of a null energy-momentum tensor and without the symmetry considerations mentioned, there is the solution of gravitational waves. That is completely new, not present in the Newtonian theory. $\endgroup$ – Claudio Saspinski Jul 18 at 16:56
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First, you need boundary conditions as well as the energy momentum tensor to determine a solution to Einstein's equation for gravity.

In answer to Q1, tensors, including the energy-momentum tensor and the metric tensor, are coordinate independent. In practice, calculation requires a choice of coordinates. It is the representation of the tensor in given coordinates which varies, not the tensor itself.

In answer to Q2, Einstein's equation

$$ G^{ab} = 8\pi G T^{ab} + \Lambda g^{ab}$$

states that Einstein curvature $G^{ab}$ is specified at a point by the energy momentum tensor (and cosmological constant). It does not specify the Riemann curvature tensor $R^a_{bcd}$. The Riemann curvature tensor can be found from the solution of Einstein's equation given $T^{ab}$ together with boundary conditions. IOW the Sun does indeed cause spacetime to curve in a large region around it.

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  • $\begingroup$ I have a doubt: is that the contravariant form of the EFEs? I did not know that was possible! $\endgroup$ – PNS Jul 19 at 11:01
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    $\begingroup$ @PNS, indices can be raised and lowered at will (so long as it is done consistently). $\endgroup$ – Charles Francis Jul 19 at 11:38
  • $\begingroup$ So, what would these equations physically mean? I mean, is there any noticeable difference between these and the covariant version? $\endgroup$ – PNS Jul 19 at 11:46
  • $\begingroup$ @PNS, it is the same equation with the indices raised and means the same thing. $\endgroup$ – Charles Francis Jul 19 at 11:58
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    $\begingroup$ That is pretty much right, but the Riemann curvature tensor is only really used to define Ricci. It is actually the metric which tells you what spacetime "looks like" (the Riemann curvature tensor is determined by the metric, but it is probably too complicated to tell you anything much). What you actually do is map spacetime (usually on a flat map), then the metric tells you the scaling distortions at each part of the map (c.f. a map of the surface of the Earth). As with mapping the Earth, many such maps are possible, but they all describe the same unique solution. $\endgroup$ – Charles Francis Jul 20 at 19:58

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