I Constants of motion symmetric spacetimes

cianfa72
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About the constants of motion associated with spacetime symmetries (KVF)
As discussed in this thread, for a symmetric spacetime (i.e. with a KVF) there are conserved quantities as constants of motion.

For instance in Schwarzschild spacetime there is a timelike KVF, hence for example the contraction of a geodesic tangent vector (4-velocity) and the timelike KVF doesn't change along a geodesic.

As said in that post the existence of conserved quantities is related to Noether's Theorem.

For Schwarzschild spacetime what does mean that Energy at infinity is conserved as constant of motion associated to the timelike KVF ?

Thanks.
 
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Let a freefalling test particle have four-momentum ##p^a## and the timelike Killing field be ##\xi^a##. ##p^a\xi_a## is a constant of freefall motion. Since in the limit as ##r\rightarrow\infty## we get flat spacetime with ##\xi^a## as the four velocity of an observer at rest with respect to the source of gravity, this constant is the energy measured by such an observer (or minus that, depending on metric signature convention) - hence it's called "the energy at infinity".

You can replace ##p^a## with four-velocity ##u^a## and you get energy per unit mass instead, but it can still be called the energy at infinity, so watch your units.

In the case ##|u_a\xi^a|<1## the test particle does not reach infinity, but the name is still used.
 
Ibix said:
Let a freefalling test particle have four-momentum ##p^a## and the timelike Killing field be ##\xi^a##. ##p^a\xi_a## is a constant of freefall motion. Since in the limit as ##r\rightarrow\infty## we get flat spacetime with ##\xi^a## as the four velocity of an observer at rest with respect to the source of gravity, this constant is the energy measured by such an observer (or minus that, depending on metric signature convention) - hence it's called "the energy at infinity".
Is your statement "observer at rest w.r.t. the source of gravity" an invariant one (i.e. coordinate independent) or isn't ?

Ibix said:
You can replace ##p^a## with four-velocity ##u^a## and you get energy per unit mass instead, but it can still be called the energy at infinity, so watch your units.
MTW defines the four-momentum ##p## as a covector (1-form). Are the two formulation actually equivalent?
 
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cianfa72 said:
Is your statement "observer at rest w.r.t. the source of gravity" an invariant one (i.e. coordinate independent) or is not ?
The observer is following the integral curves of the timelike Killing field. That is invariant. Such an observer at infinity is called "at rest at infinity". Whether you want to take the name literally or not is up to you.
cianfa72 said:
MTW defines the four-momentum as a covector (1-form). Are the two formulation actually equivalent?
In any vector space equipped with a metric there's a unique invertible mapping between vectors and covectors via the metric. You must know that, surely.
 
Ibix said:
The observer is following the integral curves of the timelike Killing field. That is invariant. Such an observer at infinity is called "at rest at infinity". Whether you want to take the name literally or not is up to you.
Ah ok, got it.

Ibix said:
In any vector space equipped with a metric there's a unique invertible mapping between vectors and covectors via the metric. You must know that, surely.
Ok, we're assuming spacetime with a given (Lorentz) metric (that can be used to lower or raise tensor indices).
 
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Since Schwarzschild spacetime is spherically symmetric there are also 3 spacelike KVFs representing the symmetries of 2-sphere.

What about the constants of motion given by the (constant) scalar products of geodesics's 4-velocity and those 3 spacelike KVFs along them ?
 
cianfa72 said:
For Schwarzschild spacetime what does mean that Energy at infinity is conserved as constant of motion associated to the timelike KVF ?
It means exactly what you said in your OP. I don't understand the question.
 
cianfa72 said:
What about the constants of motion given by the (constant) scalar products of geodesics's 4-velocity and those 3 spacelike KVFs along them ?
Angular momentum.
 
cianfa72 said:
Since Schwarzschild spacetime is spherically symmetric there are also 3 spacelike KVFs representing the symmetries of 2-sphere.

What about the constants of motion given by the (constant) scalar products of geodesics's 4-velocity and those 3 spacelike KVFs along them ?
They represent angular momentum.
 
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cianfa72 said:
Ok, we're assuming spacetime with a given (Lorentz) metric (that can be used to lower or raise tensor indices).
In relativity, always.
 
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