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A Killing vector notation confusion time translation

  1. Feb 12, 2017 #1
    Okay so when there is time-translation symmetry because the metric components do not have any time- dependence, ##\partial_x^0## is a Killing vector.

    I'm just confused what this means explicitly, since a derivative doesn't make sense without acting on anything really?

    But by 'spotting the pattern' for example I know that for Minkowski space it is ##(1,0,0,0)## and for Schwarzschild space-time it is## ((1-\frac{2GM}{r}),0,0,0) ##, i.e the component multiplying ##dt^{2}## when the metric takes diagonal form anyway,

    How is this explicitly?

    Many thanks
  2. jcsd
  3. Feb 12, 2017 #2


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    This is backwards. The correct statement is that when there is a time translation symmetry, it is possible to choose coordinates such that the metric components do not depend on the "time" coordinate (traditionally ##x^0##). But there is nothing that requires you to choose such coordinates, and the presence of the time translation symmetry does not depend on any such choice.

    Your notation does not look correct. The correct expression would be ##\partial / \partial x^0##. Also, as above, this assumes that you have chosen coordinates appropriately.

    ##\partial / \partial x^0## is a vector; it's the zeroth coordinate basis vector. This notation takes advantage of the fact that there is a one-to-one correspondence between vectors at a point and partial derivatives at that point. This correspondence is used extensively in GR, so it's a good idea to get used to it. IIRC Carroll's lecture notes on GR discuss this in one of the early chapters:



    No. The Killing vector in Schwarzschild spacetime (again, assuming an appropriate choice of coordinates) is just ##\partial / \partial x^0##, i.e., ##(1, 0, 0, 0)##.

    What you appear to be thinking of is the 4-velocity vector of a static observer; but your expression is not correct for that either--see below.

    No, for several reasons:

    (1) The expressions appearing in the line element, multiplying ##dt^2## and other coordinate differentials, can be used to derive expressions for covectors, not vectors.

    (2) You get covectors from the square roots of expressions in the line element (strictly speaking, it's only this simple if the metric is diagonal, but that is sufficient for this example). So the unit timelike covector in Schwarzschild spacetime, in Schwarzschild coordinates, is ##(\sqrt{1 - 2M / r}, 0, 0, 0)##.

    (3) The unit timelike vector in Schwarzschild coordinates is the vector that has a unit inner product with the above covector. This will therefore be ##(\frac{1}{\sqrt{1 - 2M / r}}, 0, 0, 0)##. This is the 4-velocity vector of a static observer. And, as above, this is not the same as the timelike Killing vector (although the two vector fields have the same integral curves).
  4. Feb 12, 2017 #3


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    Just a little point: There is good reason for using [itex]\frac{\partial}{\partial x^\mu}[/itex] to mean the basis vector [itex]e_\mu[/itex]. But it's not really relevant for solving the Killing equation. The equation

    [itex]\nabla_\mu V_\nu + \nabla_\nu V_\mu = 0[/itex]

    just assumes that [itex]V[/itex] is some vector field, which can be written in terms of basis vectors as:

    [itex]V = \sum_\mu V^\mu e_\mu[/itex]

    The fact that [itex]e_\mu[/itex] is secretly a directional derivative doesn't come into play.
  5. Feb 12, 2017 #4
    Would this be important when we "apply" the vector ##V## on a function, say ##c(x^{\mu})##, e.g. when ##c(x^{\mu})## map points along a curve?
  6. Feb 13, 2017 #5


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    I'm not sure what you mean by "apply" here. When we identify vectors with directional derivatives, then there is a notion of applying a vector to a scalar field: [itex]V(\phi) \equiv \sum_\mu V^\mu \frac{\partial \phi}{\partial x^\mu}[/itex]. Is that what you mean?
  7. Feb 13, 2017 #6
    Yes. It's exactly what I mean.
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