Getting Used to Killing Vector Fields

  • #1
Wledig
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I'm struggling to get the hang of killing vectors. I ran across a statement that said energy in special relativity with respect to a time translation Killing field ##\xi^{a}## is: $$E = -P_a\xi^{a}$$ What exactly does that mean? Can someone clarify to me?
 

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  • #2
Dale
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What is ##P##? I would guess it is the four momentum, but I would rather not guess.
 
  • #4
haushofer
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I'd also suggest you give your references; here at PhysicsForums we have great scientific powers, not psychic ones.

I can only place your statement in the context of general relativity. Usually, conserved quantities follow from symmetries of spacetime. Flat spacetime has a lot of symmetries, which are (partly) broken if spacetime becomes curved. Only in the directions where symmetries are preserved there will be conserved quantities. That's what your equation (I guess) states: the projection of the 4-momentum on the direction of symmetry is conserved. If the Killing vector is timelike, we call this conserved quantity the "energy (of the particle)".
 
  • #5
Wledig
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Where? Please give a specific reference.
Hi, sorry for not providing the reference. I read it on Wald (page 287).
the projection of the 4-momentum on the direction of symmetry is conserved. If the Killing vector is timelike, we call this conserved quantity the "energy (of the particle)
Can we be more precise about this statement? If I take the derivative with respect to time of ##E = -P_a \xi^a## it should yield zero, right? But how exactly do I calculate this derivative?
 
  • #6
Orodruin
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Hi, sorry for not providing the reference. I read it on Wald (page 287).

Can we be more precise about this statement? If I take the derivative with respect to time of ##E = -P_a \xi^a## it should yield zero, right? But how exactly do I calculate this derivative?
"With respect to time" is ambiguous. "With respect to proper time" is more precise. You compute the derivative as you usually would with ##P## being the 4-momentum of the observer and the Killing field ##\xi## is evaluated and differentiated along the world-line of the observer.
 
  • #7
Wledig
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Sorry, I'm still confused. If I do something like: $$\frac{d}{dt}(-P_a \xi^a) = \frac{1}{c}\frac{d}{d\tau}(-P_a\xi^a)=-\frac{1}{c}\left(\frac{dP_a}{d\tau}\xi^a + P_a\frac{d\xi^a}{d\tau}\right)$$
It vanishes because the derivative of the killing vector is zero by definition and we are imposing momentum conservation? Am I thinking this through correctly?
 
  • #8
Orodruin
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Sorry, I'm still confused. If I do something like: $$\frac{d}{dt}(-P_a \xi^a) = \frac{1}{c}\frac{d}{d\tau}(-P_a\xi^a)=-\frac{1}{c}\left(\frac{dP_a}{d\tau}\xi^a + P_a\frac{d\xi^a}{d\tau}\right)$$
It vanishes because the derivative of the killing vector is zero by definition and we are imposing momentum conservation? Am I thinking this through correctly?
No. There is nothing being imposed in terms of momentum conservation (in fact, momentum conservation is a result from this very type of argumentation for the cases where it applies). You need to put some more ideas into what ##dV^a/d\tau## actually means for a vector field ##V## on a manifold.
 

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