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Killing Vectors conserved quantity along geodesic proof

  1. Jan 24, 2015 #1
    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.
     
  2. jcsd
  3. Jan 24, 2015 #2

    martinbn

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    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: Jan 24, 2015
  4. Jan 24, 2015 #3

    Matterwave

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    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.
     
  5. Jan 24, 2015 #4
    Thanks very much.
     
  6. Jan 25, 2015 #5
    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.
     
  7. Jan 25, 2015 #6

    Matterwave

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    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.
     
  8. Jan 25, 2015 #7

    DrGreg

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    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.
     
  9. Jan 25, 2015 #8

    Matterwave

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    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.
     
  10. Jan 26, 2015 #9

    martinbn

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    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.
     
  11. Jan 26, 2015 #10

    Matterwave

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    I think you mean ##U^v\nabla_v (V_u U^u)##... otherwise that overloaded index is still there.
     
  12. Jan 27, 2015 #11

    martinbn

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