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Deriving the Metric from the EnergyMomentum Tensor 
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#1
Dec1312, 04:28 PM

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Say we were given an expression for the energymomentum tensor (also assuming a perfect fluid), without getting into an expression with multiple derivatives of the metric, are there any cases where it would be possible to deduce the form of the metric?



#2
Dec1312, 04:39 PM

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PF Gold
P: 5,059

I don't see how. Even the case of stressenergy tensor = 0 => vacuum, produces nontrivial system of second order partial differential equations for the metric. The only case, even for vacuum, that is simple to derive, is if you add spherical symmetry. Axial symmetry is already much harder, even for vacuum. 


#4
Dec1312, 05:41 PM

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Deriving the Metric from the EnergyMomentum Tensor



#5
Dec1312, 05:46 PM

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PF Gold
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#6
Dec1312, 05:56 PM

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PF Gold
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#7
Dec1312, 06:21 PM

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#8
Dec1412, 02:36 AM

P: 1,020

outside a given source,i.e. if energy momentum tensor is zero for a region with some specified symmetry(say spherical).one can guess the form of metric but with given energy momentum tensor one has to go with what others have said.



#9
Dec1412, 01:02 PM

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Thanks
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#10
Dec1412, 01:14 PM

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PF Gold
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#11
Dec1412, 02:21 PM

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The Weyl family of exact solutions were among the earliest found. Just as for Schwarzschild, in the derivation of the Weyl solutions there is nothing assumed but symmetry, namely the two Killing vectors that generate axial rotations and time translations.
Even for the stationary case there are a great many exact solutions known. Kerr turns out to be just the simplest member of an infinite family of solutions. 


#12
Dec1412, 02:38 PM

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PF Gold
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#13
Dec1412, 06:18 PM

Mentor
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#14
Dec1412, 06:33 PM

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PF Gold
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#15
Dec1512, 11:02 AM

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#16
Dec1512, 11:12 AM

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#17
Dec1512, 08:03 PM

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PF Gold
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#18
Dec1612, 11:29 AM

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Doesn't the energymomentum tensor contain the metric?



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