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Killing vectors and Geodesic equations for the Schwarschild metric.

  1. Jan 4, 2013 #1
    Hello Everybody,

    Instead of solving the geodesic equations for the Schwarzschild metric, in many books (nearly in all books that I consulted), conserved quantities are looked at instead.

    So take for eg. Carroll, he looks at the killing equation and extracts the equation

    [tex] K_\mu \frac{dx^\mu}{d \lambda}= constant, [/tex]

    and he then writes:"In addition we have another constant of the motion for geodesics", and he writes the normalization condition:

    [tex] \epsilon = -g_{\mu \nu} \frac{dx^\mu}{d \lambda} \frac{dx^\mu}{d \lambda}. [/tex]

    Now I don't understand why this set of equations is equivalent to the geodesic equations. And I do not understand why we are allowed to use these equations to extract information about the geodesics.
    Maybe the questions are the same, but I hope you get my point.

    Any help would be greatly appreciated!!
  2. jcsd
  3. Jan 4, 2013 #2


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    For most any mechanics problem, the equations of motion are second-order, and you're always welcome to solve them directly. But also in most mechanics problems there are conserved quantities such as energy and angular momentum, which mathematically are first integrals. You can derive them as a consequence of the EOM each time, or you can take a shortcut and write them down immedately, simplifying the problem.

    The geodesic equations for the Schwarzschild metric have three first integrals: energy conservation corresponding to the dt Killing vector, angular momentum conservation corresponding to the dφ Killing vector, plus the norm of the velocity vector.
  4. Jan 4, 2013 #3
    Thank you very much!

    I definetely got the point. I was already thinking that these conservation laws are some form of integrated EOMs.

    But I did never encounter this formulation, or let me say, this way of looking at the EOMs, by integrating them and then solving.

    Did I miss something in my years at uuniversity as a physics student??
  5. Jan 4, 2013 #4


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    Consider the conserved energy, (1-2m/r) (dt/dtau)

    Write this as a function of tau

    E(tau) = (1-2m/r(tau) ) ( dt(tau) / dtau)

    Then take the derivatrive with respect to tau, using the chain rule. You should find that dE/dtau = 0 is identical to one of the geodesic equations done via the Christoffel symbols.

    [tex]E = \left( 1-2\,{\frac {m}{r \left( \tau \right) }} \right) {\frac {d}{d
    \tau}}t \left( \tau \right)

    \frac{dE}{d\tau} = 2\,{\frac {m \left( {\frac {d}{d\tau}}r \left( \tau \right) \right) {
    \frac {d}{d\tau}}t \left( \tau \right) }{ \left( r \left( \tau
    \right) \right) ^{2}}}+ \left( 1-2\,{\frac {m}{r \left( \tau
    \right) }} \right) {\frac {d^{2}}{d{\tau}^{2}}}t \left( \tau \right)

    So, you can verify easily enough that E is a conserved quantity given the geodesic equations.

    The normalization condition is a bit sneaky - it turns out you need to assume it to derive the geodesic equations in their standard form. If you don't have the normalization condtion, it's very messy.
  6. Jan 4, 2013 #5
    Yep, checked it and you're right.

    In Hartle there is a very nice quote in page 176 also:" Conservation laws, such as those for energy and angular momentum, lead to tractable problems in Newtonian mechanics. Conservation laws give first integrals of the equations of motion that can reduce the order and number of the equations that have to be solved". Exactly as mentioned by Bill K.

    And basically doing the calculation as guessed by myself and proposed by pervect indeed yields to the geodesic equations in their "standard form".
  7. Jan 4, 2013 #6


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    I don't know if what I'm going to say is redundant with what other people have said, but I want to go through it for my own benefit, if for nobody else's.

    For massive particles, the proper time [itex]\tau[/itex] is defined by:

    [itex] \sqrt{- g_{\sigma \mu} dx^\sigma dx^\mu} = d\tau[/itex]​

    (Depending on the convention, the minus sign on the left-hand side may be a plus sign instead.) This implies the following identity:

    1. [itex] g_{\sigma \mu} \dfrac{dx^\sigma}{d\tau} \dfrac{dx^\mu}{d\tau} = -1[/itex]​

    In terms of [itex]\tau[/itex] the geodesic equation can be written (in a coordinate basis):

    [itex]\dfrac{d}{d\tau}(g_{\sigma \mu} \dfrac{dx^\mu}{d\tau}) - (\dfrac{1}{2} \partial_\sigma g_{\eta \mu}) \dfrac{dx^\eta}{d\tau} \dfrac{dx^\mu}{d\tau} = 0[/itex]​

    In the special case in which the metric is independent of the time coordinate, then the time component of the geodesic equation becomes:

    [itex]\dfrac{d}{d\tau}(g_{0 \mu} \dfrac{dx^\mu}{d\tau}) = 0[/itex]​

    which implies:

    2. [itex]g_{0 \mu} \dfrac{dx^\mu}{d\tau} = E[/itex]​

    for some constant E. So the geodesic equations agree with the Killing vector equations if we identify [itex]K_\mu = g_{0 \mu}[/itex] and the parameter [itex]\lambda[/itex] is chosen to be proportional to [itex]\tau[/itex]:

    [itex]\lambda = \dfrac{1}{\sqrt{\epsilon}}\tau[/itex]​

    In terms of [itex]\lambda[/itex] and [itex]K_\mu[/itex], equations 1. and 2. become:

    1'. [itex] g_{\sigma \mu} \dfrac{dx^\sigma}{d\lambda} \dfrac{dx^\mu}{d\lambda} = -\epsilon[/itex]​

    2'. [itex]K_{\mu} \dfrac{dx^\mu}{d\lambda} = \sqrt{\epsilon}E[/itex]​

    So the Killing vector approach is completely consistent with the geodesic equation, but is more general, since it doesn't require you to find a coordinate system in which the metric components are independent of time. Also, I think that the Killing vector equations are valid even for massless particles, while the derivation in terms of proper time only works for massive particles.
  8. Jan 4, 2013 #7


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    I'd like to share my approach - though just the outline of it.

    Let's consider only a 2d problem, with one space parameter x, and one time parameter t. And thus we can omit the usual tensor notation. And we'll do timelike geodesics, as they're the easiest.

    We assume we have some path parameterized by lambda, i.e. we have [itex]t(\lambda)[/itex] and [itex]x(\lambda)[/itex]

    We'll let [itex]\dot{x} = dx / d \lambda[/itex] and [itex]\dot{t} = dt / d \lambda[/itex]

    The path that extremizes the proper time will extremize the Lagrangian (note that the metric coefficients are functions of x and t):

    L(\lambda, t, x, \dot{t}, \dot{x}) = \sqrt{(g_{00}(t,x)\dot{t}^2 + 2 g_{01} (t,x) \dot{t} \dot{x} + g_{11}(t,x) \dot{x}^2}

    This is becuase s = g_00 dt^2 + 2*g_01 dtdx + g_11 dx^2, and dt = (dt/d lambda) d lambda, dx = (dx / dlambda) d lambda

    Then the geodesic equations are just

    \frac{\partial L}{\partial x} - \frac{d}{d \lambda} \left( \frac{\partial L}{\partial \dot{x}} \right)

    If you look at the terms for [itex]\partial L / \partial \dot{t}[/itex], you'll see that it's

    [tex]\frac{g_{00} \dot{t} + g_{01} \dot{x}} {\sqrt{L}}[/tex]

    Now we need to take [itex]d / d\lambda[/itex] of the above. If we don't impose any restrictions on our curve parameteriztion, this becomes very difficult. But if we choose to parameterize our courve by proper time, the very troublesome sqrt(L) goes away.

    It's quite a chore to work it all out, but at the end you get the geodesic equations, and an understanding of why you needed to add the constraint on your curve parameterization to make it managable.

    You can find a more detailed, fleshed-out example of this approach in MTW's "Gravitation".
    Last edited: Jan 4, 2013
  9. Jan 5, 2013 #8


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    As someone pointed out, the results are the same as if you started with the quadratic Lagrangian

    [itex]L = \dfrac{1}{2}g_{\mu \nu} \dfrac{dx^\mu}{d\lambda} \dfrac{dx^\nu}{d\lambda}[/itex]

    (with no square-root)

    The thing that's weird is that the geodesic equation works for both massive and massless particles, although in the latter case, it can't be understood as the extremization of an effective Lagrangian (because L in that case is zero for a lightlike path, so any variation of the path that is still lightlike will have the same "action", namely zero).
  10. Jan 5, 2013 #9


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    If you use the squared action, but don't use an affine parameterization, you'll get wrong results. The affine parameterization is still required.

    There is a mathematical trick called the "Einbein" which justifies in more detail when and why you can minimize the squared Lagrangian. It turns out you're really introducing another variable.

    There's an old (short) thread in PF, where I first learned about them:



    [tex]L = -m \sqrt{1 - \dot{x}^2}[/tex]

    can be minimized by minimizing

    [tex]L = -\mu (1 - \dot{x}^2) / 2 - m^2 / 2 \mu[/tex][/QUOTE]

    u being the "Einbein field".

    As far as the distinction between lightlike, timelike, and spacelike geodesics, I'd have to agree that it's more obvious that the "parallel transport" technique works that the Lagrangian one. But light still follows an effective Lagrangian. See for instance http://en.wikipedia.org/w/index.php?title=Hamiltonian_optics&oldid=517305783

    Keywords are "Fermat's principle", "Lagrangian optics", or "optical path length'.

    Rather than extremizing proper time, light extremizes the optical path length.
  11. Jan 6, 2013 #10
    I'll go through the comments this evening or tomorrow and write my results.
  12. Jan 6, 2013 #11


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