Killing Vectors conserved quantity along geodesic proof

In summary: OP he has the expression ##U^u\nabla_v(U^u V_u)## which I assumed was a typo meaning ##U^v \nabla_v (U^u V_u)##, or in better notation ##U^\nu\nabla_\nu(U^\mu V_\mu)##. Otherwise there is an overloaded index. It was this expression in the OP that I was addressing, because I think it was this expression that confused him. The fact that ##\nabla_\nu (U^\mu V_\mu)## happens to be the components of a co-vector is irrelevant I think. The main thing here is that it is a directional derivative and does not have the properties of
  • #1
binbagsss
1,254
11
I am trying to follow a proof that given a Kiling vector ##V^{u}##, the quantity ##V_{u}U^{u} ## is conserved along a geodesic.

I am given the Killiing Equation: ## \bigtriangledown_{(v}U_{u)}=0 ## [1]

Below ## U^{u} ## is tangent vector ## U^{u} = \frac{dx^{u}}{d\lambda} ##

The proof considers ## U^{u}\bigtriangledown_{v}(V_{u}U^{u}) =U^{v}U^{u}\bigtriangledown_{v}V_{u}+V_{u}U^{v}\bigtriangledown_{v}U^{u} ##

It then says that the first term is zero by the Killing equation. I can't see this. I see from [1] that ## \bigtriangledown_{v}U_{u} + \bigtriangledown_{u}U_{v} =0 ##. So unless this somehow implies each term must indivually be zero in here, I don't follow.

It also says the second term is ##0## as ##x^{u}(\lambda) ## is a geodesic. I'm unsure here too, so the covariant derivative of a tangent vector along a geodesic is zero? Could someone point me to some notes on this?

Finally, I see the expression has considered ## U^{u}\bigtriangledown_{v}(V_{u}U^{u}) ## as a pose to ## \bigtriangledown_{v}(V_{u}U^{u}) ## is this just because we know things about the former but the latter would be harder to proof?

And if we proove that the former is zero we are done as ## U^{u} ## is non-zero?
/
Thanks very much in advance.
 
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  • #2
For the first question the expression $$U^{v}U^{u}\bigtriangledown_{v}V_{u}$$ is the same if you change $$u$$ and $$v$$, no matter what the vector fields are. At the same time if you use that $$V$$ is a Killing field the expression will change sign if you change the indices. So it has to be zero.

For the second, yes, that is the definition of a geodesic, its tangent vector is parallel transported along the curve i.e. that derivative is zero.

For the last one you need the derivative in the direction of the tangent vector, because the quantity is conserved only along the geodesic.
 
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  • #3
As a side note, it is generally not advised to use ##V## and ##U## to mean vectors and then use the lower case ##v## and ##u## to be indices on those vectors. That is just horribly confusing notation.
 
  • #4
martinbn said:
For the first question the expression $$U^{v}U^{u}\bigtriangledown_{v}V_{u}$$ is the same if you change $$u$$ and $$v$$, no matter what the vector fields are. At the same time if you use that $$V$$ is a Killing field the expression will change sign if you change the indices. So it has to be zero.

For the second, yes, that is the definition of a geodesic, its tangent vector is parallel transported along the curve i.e. that derivative is zero.

For the last one you need the derivative in the direction of the tangent vector, because the quantity is conserved only along the geodesic.

Thanks very much.
 
  • #5
martinbn said:
For the last one you need the derivative in the direction of the tangent vector, because the quantity is conserved only along the geodesic.

Sorry just realized I don't understnad this, to make some tensor expression in the direction of some tensor, I thought it would only make sense if they are of the same type - isn't ## V_{u}U^{u} ## a scalar, so once you do ## \bigtriangledown_{v}V_{u}U^{u} ## you have a covector, whereas the tangent is a vector?

Thanks.
 
  • #6
binbagsss said:
Sorry just realized I don't understnad this, to make some tensor expression in the direction of some tensor, I thought it would only make sense if they are of the same type - isn't ## V_{u}U^{u} ## a scalar, so once you do ## \bigtriangledown_{v}V_{u}U^{u} ## you have a covector, whereas the tangent is a vector?

Thanks.

Your notation is horrible. But since you won't take my advice and use a less confusing notation, I'll just answer the question using better notation and you can match yours to the answer if you can. ##U^\nu\nabla_\nu(V_\mu U^\mu)=\nabla_u (V_a U^a)## Is neither a covector nor a vector. This expression represents the directional derivative, in the ##U## direction of the scalar function ##V_a U^a##. In other words, in some coordinate system, it merely represents ##U^\nu \frac{\partial}{\partial x^\nu} f(x)## where ##f(x)=V_a(x) U^a (x)## and ##x=x^\mu(P)## are the coordinates.
 
  • #7
Matterwave said:
##U^\nu\nabla_\nu(V_\mu U^\mu)=\nabla_u (V_a U^a)## Is neither a covector nor a vector. This expression represents the directional derivative, in the ##U## direction of the scalar function ##V_a U^a##.
I think you have have been misled by the poor choice of letters. If ##\textbf{U}## is a 4-vector then the symbol ##\nabla_\textbf{U}## can be used to denote the directional derivative ##U^b\nabla_b##. But in this case the ##u## is an index, not a vector, so ##\nabla_b(V^a U_a)## is indeed a covector.

binbagsss, please note the correct LaTeX for the covariant derivative is \nabla, not \bigtriangledown. There is also something wrong in post #1 because in some of the expressions you have 3 ##u## indexes and only one ##v## index, which doesn't make sense.
 
  • #8
DrGreg said:
I think you have have been misled by the poor choice of letters. If ##\textbf{U}## is a 4-vector then the symbol ##\nabla_\textbf{U}## can be used to denote the directional derivative ##U^b\nabla_b##. But in this case the ##u## is an index, not a vector, so ##\nabla_b(V^a U_a)## is indeed a covector.

binbagsss, please note the correct LaTeX for the covariant derivative is \nabla, not \bigtriangledown. There is also something wrong in post #1 because in some of the expressions you have 3 ##u## indexes and only one ##v## index, which doesn't make sense.

In his OP he has the expression ##U^u\nabla_v(U^u V_u)## which I assumed was a typo meaning ##U^v \nabla_v (U^u V_u)##, or in better notation ##U^\nu\nabla_\nu(U^\mu V_\mu)##. Otherwise there is an overloaded index. It was this expression in the OP that I was addressing, because I think it was this expression that confused him. The fact that ##\nabla_\nu (U^\mu V_\mu)## happens to be the components of a co-vector is irrelevant I think.
 
  • #9
binbagsss said:
Sorry just realized I don't understnad this, to make some tensor expression in the direction of some tensor, I thought it would only make sense if they are of the same type - isn't ## V_{u}U^{u} ## a scalar, so once you do ## \bigtriangledown_{v}V_{u}U^{u} ## you have a covector, whereas the tangent is a vector?

Thanks.

Yes, that is a covector but it need not be zero. The scalor ## U^{u}\nabla_{v}V_{u}U^{u} ## is zero, it is called the directional derivative of ## V_{u}U^{u} ## in the direction of ## U^{u} ##. That's why in the original question it says conserved along the geodesic.
 
  • #10
martinbn said:
Yes, that is a covector but it need not be zero. The scalor ## U^{u}\nabla_{v}V_{u}U^{u} ## is zero, it is called the directional derivative of ## V_{u}U^{u} ## in the direction of ## U^{u} ##. That's why in the original question it says conserved along the geodesic.

I think you mean ##U^v\nabla_v (V_u U^u)##... otherwise that overloaded index is still there.
 
  • #11
Matterwave said:
I think you mean ##U^v\nabla_v (V_u U^u)##... otherwise that overloaded index is still there.

Yes, of course. Just when I thought that the bad choice of notation doesn't bother me.
 

1. What is a Killing Vector?

A Killing Vector is a vector field on a manifold that preserves the metric and leaves the equations of motion invariant. This means that if a vector field is a Killing Vector, it will not change the distance or time measurements on the manifold and will not affect the path that a particle takes along a geodesic.

2. What is a conserved quantity along a geodesic?

A conserved quantity along a geodesic is a physical quantity that remains constant as a particle moves along a geodesic, which is the shortest path between two points on a curved manifold. This is due to the fact that geodesics are the paths of least resistance, and therefore a particle will follow this path without any external forces acting on it.

3. How is a Killing Vector related to a conserved quantity along a geodesic?

A Killing Vector is related to a conserved quantity along a geodesic because it is the mathematical representation of the symmetry of a manifold. This symmetry leads to the conservation of physical quantities along geodesics, as the Killing Vector preserves the metric and equations of motion.

4. What is the proof of Killing Vectors being a conserved quantity along geodesics?

The proof of Killing Vectors being a conserved quantity along geodesics involves showing that the Lie derivative of the metric along a Killing Vector is equal to zero. This implies that the metric is invariant under the transformation of the Killing Vector, and therefore the equations of motion and conservation of physical quantities along geodesics are also preserved.

5. Why is the conservation of quantities along geodesics important?

The conservation of quantities along geodesics is important because it is a fundamental principle in physics that arises from the symmetries of the underlying space-time. It allows for the prediction and understanding of the behavior of particles moving along geodesics, and is a key concept in theories of gravity such as General Relativity.

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