Getting Used to Killing Vector Fields: Explained

In summary, the equation given states that the energy associated with a particle in a special relativistic Killing field is negative.
  • #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
What is ##P##? I would guess it is the four momentum, but I would rather not guess.
 
  • #3
Wledig said:
I ran across a statement

Where? Please give a specific reference.
 
  • #4
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)".
 
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  • #5
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
Wledig said:
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
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
Wledig said:
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.
 

FAQ: Getting Used to Killing Vector Fields: Explained

What is a Killing vector field?

A Killing vector field is a mathematical concept used in differential geometry and physics. It is a type of vector field that preserves the metric of a manifold, meaning that it does not change the distances between points on the manifold. In simpler terms, it is a vector field that represents a symmetry of the manifold.

How do Killing vector fields relate to killing vectors?

Killing vector fields are closely related to killing vectors, which are the infinitesimal generators of isometries (symmetries that preserve the metric) on a manifold. In other words, a killing vector field is a vector field that generates killing vectors. However, not all killing vectors can be expressed as killing vector fields.

Why is it important to understand Killing vector fields?

Killing vector fields have many important applications in physics, particularly in the study of spacetime and general relativity. They are also used in other areas of mathematics, such as differential equations and Lie groups. Understanding killing vector fields can provide insight into the symmetries and properties of a manifold.

How are Killing vector fields calculated?

There are various methods for calculating killing vector fields, depending on the specific manifold and metric being studied. In general, the process involves solving differential equations or using geometric methods to find vector fields that satisfy certain conditions, such as preserving the metric or being parallel to the geodesic curves on the manifold.

Can Killing vector fields be visualized?

Yes, it is possible to visualize Killing vector fields using vector field plots or other graphical representations. This can help in understanding the symmetries and properties of a manifold and how the vector field acts on it. However, since killing vector fields can have a high number of dimensions, it may be difficult to visualize them in certain cases.

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