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Arman777

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Arman777

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Ibix

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But you must know the symmetries of FLRW spacetime. Can you write down an ansatz for at least some of the Killing fields?

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Arman777

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Then it seems the problem is hard to solve as you have said...I am not sure that I can

But you must know the symmetries of FLRW spacetime. Can you write down an ansatz for at least some of the Killing fields?

- #4

Ibix

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Well, can you name some symmetries? I can think of six in two sets of three. Hint: what is the defining property of FLRW spacetimes?Then it seems the problem is hard to solve as you have said...I am not sure that I can

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Arman777

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Homoegenous and Isotropic spaceHint: what is the defining property of FLRW spacetimes?

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Ibix

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I suggest starting with the case of a spatially flat RW universe. The symmetries should be familiar.

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PeterDonis

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NoIs there a simple/general equation that gives the killing vectors for a given metric?

No. What the comments from @Ibix and @Orodruin are telling you is that, instead of looking for a mechanical process to crank out Killing vectors from the metric, you should try to understandOr do I have to solve ten independent killing equations simultaneously to find the killing vectors?

As John Wheeler said, you should never do a calculation in physics unless you already know the answer.

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Yeah, I can't imagine solving the Killing equation in any other way than guessing a coordinate system where the metric is independent of some coordinate.I don't think so, beyond the usual "if the metric coefficients are independent of a coordinate" trick. At least, Carroll comments that it's hard work.

- #10

PeterDonis

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That method won't always find all of the Killing vectors, though. For example, for spherical symmetry there is a 3-parameter group of Killing vectors, but looking for an adapted coordinate chart will only find one of them.I can't imagine solving the Killing equation in any other way than guessing a coordinate system where the metric is independent of some coordinate.

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Well. There’s nothing stating that you cannot use other coordinate systems to find the remaining ones.That method won't always find all of the Killing vectors, though. For example, for spherical symmetry there is a 3-parameter group of Killing vectors, but looking for an adapted coordinate chart will only find one of them.

In fact, given a Killing field, it should always be possible to find a coordinate system where that field is a coordinate tangent field.

There are also other ways to construct additional Killing fields however. For example, the commutator of two Killing fields is also a Killing field.

- #12

PeterDonis

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Yes, that's true, although it does exchange the problem of finding all the Killing vectors for the problem of finding all the coordinate systems.There’s nothing stating that you cannot use other coordinate systems to find the remaining ones.

Again, true, but I'm not sure how helpful it is if you don't already know what the Killing field is.given a Killing field, it should always be possible to find a coordinate system where that field is a coordinate tangent field.

Yes, good point; I believe this is one way to get all three of the spherical symmetry Killing fields, for example.There are also other ways to construct additional Killing fields however. For example, the commutator of two Killing fields is also a Killing field.

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Sure, you are still guessing. But for some seeing coordinate systems where symmetry is apparent may be easier than seein the Killing fields.Yes, that's true, although it does exchange the problem of finding all the Killing vectors for the problem of finding all the coordinate systems.

If you have two of them, then yes. Taking their connutator will give the third.Yes, good point; I believe this is one way to get all three of the spherical symmetry Killing fields, for example.

It will not always generate a new Killing field though.

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