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Struggling with the relationship between the Lagrangian L and the interval ds?

  1. May 16, 2010 #1
    I am going through my notes trying to get things straight in my head and something is confusing me. I also have the feeling it will turn out not to be very complicated, which is why I am a bit frustrated with this.
    I know that there is something used in general relativity called the Lagrangian, which is given by [tex] L = g_{\nu\mu}\frac{dx^\nu}{d\tau}\frac{dx^\mu}{d\tau}[/tex], where the dx[tex]\mu[/tex] and dx[tex]\nu[/tex] refer to the coordinates of the chosen coordinate system, and g is the metric, where the subscript refer to elements of the matrix it can be written as.

    I also know there is something called the spacetime interval ds. As I understand things, in flat space, we use Cartesian coordinates and we have a metric, which (I think) is called the Minkowski metric:
    [-1 0 0 0]
    [ 0 1 0 0]
    [ 0 0 1 0]
    [ 0 0 0 1] and the metric elements and the interval are related by:
    [tex] ds^2 = g_\nu_\mu dx^\nu dx^\mu [/tex], where [tex] \mu,\nu = 0,1,2,3 [/tex].
    As I understand things, if the spacetime involved was not flat, then these diagonal elements would be different, right?

    Then we say that [tex] dx^0 = cdt, dx^1 = dx, dx^2 = dy, & dx^3 = dz [/tex], and noting that if [tex] \nu \neq \mu [/tex] then the corresponding metric coefficients are zero.

    this gives us, using Einstein's summation convention, that:
    [tex] ds^2 = g_0_0(cdt)(cdt) + g_1_1(dx)(dx) + g_2_2(dy)(dy) + g_3_3(dz)(dz) [/tex]
    [tex] = -c^2dt^2 + dx^2 + dy^2 + dz^2 [/tex].
    I understand this as being the separation between 2 events separated by position or time or both.

    Then looking back at the Lagrangian equation above, it gives this:
    [tex] L = -c^2(\frac{dt}{d\tau})^2 + (\frac{dx}{d\tau})^2 + (\frac{dy}{d\tau})^2 + (\frac{dz}{d\tau})^2 [/tex]

    My first question is, what is the interpretation of the Lagrangian exactly? All I have seen in my notes is that it gets stuck into the Euler-Lagrange equation, and sometimes conserved quantities or equations of motion pop out. I remember using the Lagrangian L=T-V in classical mechanics, and I do not see the relationship between the two (nor did I ever really understand what the significance of the difference between kinetic and potential energy was).
    Does it really have a physical meaning or is it just some weird mathematical trick I have to blindly use to make equations of motion appear?

    The other thing that was making me confused was a part in my notes where the interval/Schwarzschild metric for a particle on a radial trajectory towards a spherical mass was given, using spherical polar coordinates, as:

    [tex] ds^2 = -c^2dt^2(1 - \frac{2GM}{rc^2}) + \frac{dr^2}{1-\frac{2GM}{rc^2}} [/tex].

    Apparently I am supposed to be able to find a Lagrangian for the particle from this interval, but I don't get what I am supposed to do with it to find this. I am not really even comfortable with the way "interval" and "metric" seem to be interchangeable.

    Why then, if the interval is given by [tex] ds^2 = g_\nu_\mu dx^\nu dx^\mu [/tex], is the Lagrangian given by [tex] L = g_{\nu\mu}\frac{dx^\nu}{d\tau}\frac{dx^\mu}{d\tau}[/tex]? These two things look similar, i mean, there is only a derivative with respect to tau to tell between them, but its clearly not as simple a thing as, say, [tex] L = \frac{ds}{d\tau} [/tex].

    Anyone able to set me straight here?
    Last edited: May 16, 2010
  2. jcsd
  3. May 16, 2010 #2


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    A lagrangian is always a weird trick to make the correct equations appear;) After all, the correct equations are themselves a weird trick - the only reason they are correct, is that they match experiment. So you can postulate them directly, or indirectly via the Lagrangian. Often the Lagrangian is easier because if we want the equations to have certain symmetries, it is easier to enforce their presence via the Lagrangian.

    Anyway, in this case, this is is the "Lagrangian" (I'm not sure this is the right term) for the geodesic equation which describes the motion of a free falling particle. A geodesic extremizes the proper time between two points (roughly speaking, there can be more than one geodesic connecting two points, in which case, both cannot be the "most extreme").
  4. May 16, 2010 #3
    The rationale why this works is explained very well in Rindler (Relativity, Special, General and Cosmological) in chapter 102 (Geodesics), pages 204-6.
    Once you get the Lagrangian, you can use the standard approach of variational mechanics in order to get the equations of motion. Therein lies the power of the Hamilton-Lagrange approach.

    The Lagrangian is simply:

    [tex]L=-c^2 \frac{dt^2}{ds^2}(1 - \frac{2GM}{rc^2}) + \frac{dr^2}{ds^2}\frac{1}{1-\frac{2GM}{rc^2}} [/tex]

    From the above you can get the Euler-Lagrange equations without any difficulty.
    Last edited: May 16, 2010
  5. May 16, 2010 #4


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    The idea behind the Lagrangian is that, in this example, it is the function

    [tex]L(t,r,u,v) = -c^2u^2(1 - \frac{2GM}{rc^2}) + \frac{v^2}{1-\frac{2GM}{rc^2}}[/tex]​

    When you take a specific worldline and set [itex]u=dt/d\tau, v=dr/d\tau[/itex], you will get

    [tex]L(t,r,\frac{dt}{d\tau},\frac{dr}{d\tau}) = -c^2[/tex]​

    but the Lagrangian equations involve calculating [itex]\partial L/\partial t, \partial L/\partial r, \partial L/\partial u, \partial L/\partial v[/itex] before you make that substitution.

    Books on the subject don't always use separate symbols such as u and v, like I did, but instead just use the "dotted" symbols

    [tex]\dot{t}, \dot{r}[/tex]​
    Last edited: May 16, 2010
  6. May 16, 2010 #5


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    They are the same thing essentially. The line element dss represents the metric, which is a tensor that eats two vectors and spits out a number. You can think of each dx in the metric as things that each eat one vector and spit out a number. You can think of the line element as the metric BEFORE it has eaten any vectors.

    In the Lagrangian, you are integrating a number along a worldline. At each point on the worldline, there is a vector tangent to that worldline. You can let the metric eat two of these tangent vectors at every point on the worldline, and it will give you a number at every point on the worldline. You have a number at every point in the path, so now you can integrate along the path. So you can think of the Lagrangian as the what the metric spat out AFTER it ate the tangent vectors on the worldline.

    More formally: in the line element, dx represents a "one-form" or a "covariant vector"; in the Lagrangian, dx/dT represents a tangent vector or a "contravariant vector". For more details, see http://homepages.physik.uni-muenchen.de/~Winitzki/T7/GR_course.pdf [Broken]; the right hand column of p28 just before section 1.5.2 gives the relationship between the metric and the Lagrangian, and section 1.5.3 explains the usual sloppy physics notation, which is good for calculating, but can be confusing mathematically.
    Last edited by a moderator: May 4, 2017
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