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Qmechanic ♦
Qmechanic ♦
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In this question, see Why is general relativity in (2+1) dimensions different from cylindrical systems in (3+1) dimensional GR?Why is general relativity in (2+1) dimensions different from cylindrical systems in (3+1) dimensional GR?, it is mentioned "The gravitational potential Φ of an infinite rod in Newtonian gravity is Φ∼ln(r)"

"The gravitational potential Φ of an infinite rod in Newtonian gravity is $Φ\sim\ln(r)$"

and further that the same is true for Einstein's gravity.

So it seems assuming a certain symmetry implies the gravitational potential inherent to that. Is that correct?

If we e.g. assume an infinite disc (instead of the rod) and thus cylindrical symmetry what would we expect regarding the gravitational potential (logarithmic or not?) comparing Newton vs. Einstein's GR? Or does in the GR case the answer require to derive e.g. the vacuum solution of said disc?

In this question, see Why is general relativity in (2+1) dimensions different from cylindrical systems in (3+1) dimensional GR?, it is mentioned "The gravitational potential Φ of an infinite rod in Newtonian gravity is Φ∼ln(r)" and further that the same is true for Einstein's gravity.

So it seems assuming a certain symmetry implies the gravitational potential inherent to that. Is that correct?

If we e.g. assume an infinite disc (instead of the rod) and thus cylindrical symmetry what would we expect regarding the gravitational potential (logarithmic or not?) comparing Newton vs. Einstein's GR? Or does in the GR case the answer require to derive e.g. the vacuum solution of said disc?

In this question, see Why is general relativity in (2+1) dimensions different from cylindrical systems in (3+1) dimensional GR?, it is mentioned

"The gravitational potential Φ of an infinite rod in Newtonian gravity is $Φ\sim\ln(r)$"

and further that the same is true for Einstein's gravity.

So it seems assuming a certain symmetry implies the gravitational potential inherent to that. Is that correct?

If we e.g. assume an infinite disc (instead of the rod) and thus cylindrical symmetry what would we expect regarding the gravitational potential (logarithmic or not?) comparing Newton vs. Einstein's GR? Or does in the GR case the answer require to derive e.g. the vacuum solution of said disc?

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timm
timm
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Is there a deeper relationship between symmetry and gravitational potential comparing Newton's and Einstein's gravity?

In this question, see Why is general relativity in (2+1) dimensions different from cylindrical systems in (3+1) dimensional GR?, it is mentioned "The gravitational potential Φ of an infinite rod in Newtonian gravity is Φ∼ln(r)" and further that the same is true for Einstein's gravity.

So it seems assuming a certain symmetry implies the gravitational potential inherent to that. Is that correct?

If we e.g. assume an infinite disc (instead of the rod) and thus cylindrical symmetry what would we expect regarding the gravitational potential (logarithmic or not?) comparing Newton vs. Einstein's GR? Or does in the GR case the answer require to derive e.g. the vacuum solution of said disc?