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Confused by variational principle

  1. Apr 22, 2010 #1
    My notes give the variational principle for a geodesic in GR:

    [tex] c\tau_{AB} = c\int_A^B d\tau = c\int_A^B \frac{d\tau}{dp}dp = \int_A^B Ldp [/tex]

    and then apply the Euler-Lagrange equations. By choosing p to be an "affine parameter" where [tex] \frac{d^2 p}{d\tau^2} [/tex] the Euler-Lagrange equations are then expressed as:

    [tex] \frac{dL^2}{dx^\mu} + \frac{d}{dp}\left( \frac{dL^2}{d\dot{x}^\mu} \right) = 0 [/tex]

    where the dot is wrt p. Apparently the affine parameter is usually chosen to be tau, but then L is just 1 and the equation holds trivially! In fact each term in the equation is zero individually. What am I missing here?
     
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  3. Apr 22, 2010 #2

    Mentz114

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    In my notes, L is

    [tex]
    \sqrt{g_{\alpha\beta}{x'}^{\alpha}{x'}^{\beta}
    [/tex]
    where the primes are differentiation wrt the affine parameter. Solving the E-L equations for L gives the geodesic equation in terms of the affine param. After that's done, we can substitute [itex]\tau[/itex] for the affine parameter.
     
  4. Apr 22, 2010 #3
    Yes that's the same thing, the space-time interval is:

    [tex] ds^2=c^2d\tau^2=g_{\mu\nu}dx^\mu dx^\nu [/tex]

    which relates our equations. I don't like substituting for tau at the end, since if we had tau at the beginning then L would be constant and each term in the differential equation would be zero individually. It seems like dividing by x then afterwards setting x=0.
     
  5. Apr 22, 2010 #4

    Mentz114

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    I don't see the problem. It doesn't matter what the affine parameter is, the Lagrangian does not disappear.
     
  6. Apr 22, 2010 #5
    The problem is that L is defined as:

    [tex] L=c\frac{d\tau}{dp} [/tex]

    and if p = tau then L=c, a constant. And the each term in the Euler-Lagrange equation is zero individually, which it shouldn't be. In this case any path would satisfy the equations.
     
  7. Apr 22, 2010 #6

    Mentz114

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    So

    [tex]L=c\frac{d\tau}{dp} = \sqrt{g_{\alpha\beta}{x'}^{\alpha}{x'}^{\beta}[/tex] ?

    Sorry you've lost me. I can't see what you're talking about.
     
  8. Apr 22, 2010 #7
    [tex] ds^2=c^2d\tau^2=g_{\mu\nu}dx^\mu dx^\nu [/tex]

    divide both sides by (dp)^2 (not rigourous but that's what my notes say):

    [tex] c^2\left(\frac{d\tau}{dp}\right)^2=g_{\mu\nu}\frac{dx^\mu}{dp}\frac{dx^\nu}{dp} [/tex]

    denote differentiation wrt p by a dot and take the square root:

    [tex] c\frac{d\tau}{dp}=\sqrt{g_{\mu\nu}\dot{x}^\mu \dot{x}^\nu} [/tex]
     
  9. Apr 23, 2010 #8

    George Jones

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    Are you sure that total derivatives are used?
    No.

    Consider

    [tex]f\left( x, y \right) = x^2 + y^2 .[/tex]

    Set

    [tex]1 = f\left( x, y \right).[/tex]

    Does this mean that

    [tex]\frac{\partial f}{\partial x} = 2x = 0?[/tex]
     
  10. Apr 23, 2010 #9
    No sorry the total derivatives were a typo. You gave a good example, but now I feel like I don't even understand partial differentiation. Setting f(x,y)=1 means x and y are no longer independent, so partial differentiation wrt x and not y doesn't make sense. Is that right? So is partial differentiation always assumed to be applied to the unconstrained function?
     
  11. Apr 23, 2010 #10

    DrGreg

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    Yes. In this context, [itex]\partial f / \partial x[/itex] means "differentiate with respect to x while holding y constant". You can't vary x and hold y constant if you are keeping f(x,y) = 1.

    The chain rule

    [tex]\frac{df}{dp} = \frac{\partial f}{\partial x}\frac{dx}{dp} + \frac{\partial f}{\partial y}\frac{dy}{dp}[/tex]​

    relates three derivatives of f, the first along the curve (f=1 in this case), the second if you were to hold y constant, the third if you were to hold x constant. Of course in the case where f is constantly 1 along the curve in question, df/dp would be zero.
     
  12. Apr 23, 2010 #11

    George Jones

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    Yes, as DrGreg explained.

    Have you taken Hamiltonians and Hamilton's equations in a mechanics course? If so, what is the Hamiltonian, and what are Hamilton's equation for a harmonic oscillator? This is a similar application of partial derivatives.
     
  13. Apr 24, 2010 #12
    Yes I took a whole course on Hamiltonian dynamics. I assume you're pointing out the constant Hamiltonian whose derivatives wrt p and q are non-zero. I think my problem is that I was very lazy for the first two years of university and only really applied myself when the grades started to count towards my degree. I still have some misunderstandings left over from basic stuff that should be very obvious to me by now.
     
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