I Calc Precession of Mercury Orbit: Stress-Energy Tensor=0?

[pixel]

In the calculation of the precession of Mercury's orbit, why is the stress-energy tensor equal to 0? There is energy and momentum at the location of the planet.

 

[Orodruin]

To a good approximation, the stress energy tensor is zero and you can consider Mercury a test planet moving on a geodesic world line in the static Schwarzschild background.

 

[pixel]

I assumed it was that the stress-energy tensor is small and can be neglected, which you have confirmed. It seemed to me that if you are calculating an orbit (i.e. position, momentum etc.) of an object and the stress-energy tensor has components made up of the object's momentum, etc., then those components would be comparable to the things you are trying to calculate and hence not negligible. But I guess you are saying that they are negligible compared to the influence of the much larger, nearby sun.

 

[bcrowell]

This isn't specifically a GR issue. An object can't make a nonzero net force on itself, because of conservation of momentum. Therefore its motion isn't affected by its own field. This is true for a test mass in Newtonian gravity and also for a test charge in electricity and magnetism.

It seemed to me that if you are calculating an orbit (i.e. position, momentum etc.) of an object and the stress-energy tensor has components made up of the object's momentum, etc., then those components would be comparable to the things you are trying to calculate and hence not negligible..

The part of the stress-energy tensor due to Mercury itself, at the position of Mercury, isn't just comparable to the part due to the sun. It's infinitely large compared to the part due to the sun, since the sun's stress-energy tensor there is zero.

In terms of Newtonian gravitational fields, or the Riemann tensor in GR, Mercury's contribution in its own neighborhood isn't just comparable to the sun's, it's orders of magnitude greater.

 

[pixel]

Still a little confused. One answer is that it's a good approximation to set the stress-energy tensor to zero. The other answer is that it's a fundamental fact that an object is not affected by its own field.

 

[bcrowell]

Still a little confused. One answer is that it's a good approximation to set the stress-energy tensor to zero. The other answer is that it's a fundamental fact that an object is not affected by its own field.

I don't know what Orodruin intended when he said, "To a good approximation, the stress energy tensor is zero[...]" Doesn't make sense to me. Maybe he could clarify.

The second half of his sentence is this: "[...] and you can consider Mercury a test planet moving on a geodesic world line in the static Schwarzschild background." This makes sense, and this is an approximation.

So the way it reads to me is that Orodruin wrote the sentence in haste, and the first half of the sentence is not actually the correct justification for the validity of the approximation used in the second half of the sentence.

The test-particle approximation is valid because Mercury's mass is too small to have a big effect on the sun. For instance, Mercury must raise some tiny tidal bulge on the sun, but that bulge is much too small to measure. That bulge in turn would have a gravitational effect on Mercury's orbit, but that is even more negligible.