Killing Vectors conserved quantity along geodesic proof

  • Thread starter binbagsss
  • Start date
  • #1
1,208
9

Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
martinbn
Science Advisor
1,845
558
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.
 
Last edited by a moderator:
  • #3
Matterwave
Science Advisor
Gold Member
3,965
326
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
1,208
9
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
1,208
9
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 realised 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
Matterwave
Science Advisor
Gold Member
3,965
326
Sorry just realised 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
DrGreg
Science Advisor
Gold Member
2,290
895
##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
Matterwave
Science Advisor
Gold Member
3,965
326
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
martinbn
Science Advisor
1,845
558
Sorry just realised 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
Matterwave
Science Advisor
Gold Member
3,965
326
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
martinbn
Science Advisor
1,845
558
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.
 

Related Threads on Killing Vectors conserved quantity along geodesic proof

Replies
6
Views
2K
  • Last Post
Replies
2
Views
918
Replies
1
Views
3K
  • Last Post
Replies
4
Views
6K
  • Last Post
Replies
1
Views
280
  • Last Post
Replies
19
Views
3K
  • Last Post
Replies
1
Views
635
Top