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Effective potential and geodesics in G.R.

  1. Oct 20, 2009 #1
    The geodesics around a spherical mass (Schwarzschild solution) in G.R. can be described by

    [tex]\frac{1}{2}\left(\frac{dr}{d\lambda}\right)^2 + V(r) = \mathcal{E}[/tex]

    where V(r) is the effective potential

    [tex]\frac{1}{2}\epsilon - \epsilon\frac{GM}{r} + \frac{L^2}{2r^2} - \frac{GML^2}{r^3}[/tex]

    and

    [tex]\mathcal{E} = \frac{1}{2}E^2[/tex]

    For photons [tex]\epsilon = 0[/tex] and for massive particles [tex]\epsilon = 1[/tex].

    To illustrate I have tried to draw the potential for different values of L. Lets start with the null geodesics:

    http://www.dianajuncher.dk/geodesic_light.png [Broken]

    The red line indicates the Newtonian potential. This is easy enough: for larger L it just becomes steeper and lies closer to the y-axis meaning that photons with large L (e.i. photons that point more away from the source of gravity) don't get too close, while photons with smaller L (pointing more directly at the source) get closer before moving away again.

    For the G.R. lines it gets a bit more tricky. Now the effective potential goes to minus infinity which means, that you can acutally 'hit' the center. I just don't understand the whole energy-barrier-thing.

    L is again just the direction of the photon, right? The closer it's path points at the source, the smaller the L?
    And then there is the energy. Apparently, if the energy is larger than the barrier, the light can reach the center. But what exactly is this energy? It can hardly be the frequency of the photon or something like that. I find it strange, that the barrier is high for high L - with both a high energy and high L I would think, that the photon just escaped. I thought high energy prevented things from being 'sucked in'?

    And then there is the massive particles:

    http://www.dianajuncher.dk/geodesic_particle.png [Broken]

    Here the L depends on both the direction of the path and the velocity of the particle, right?

    Again I don't get the whole energy-barrier-thing. My instinct tells me, that the higher energy you have, the better you can escape. But the higher your energy is, the better you can get over the barrier?

    Bonus question: if you actually do manage to cross the barrier, you don't just go straight to the center, right? There must be a spiralling motion for both photons and massive particles.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Oct 20, 2009 #2

    George Jones

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    Right. [itex]E[/itex] is, for a particular trajectory, a constant of the motion that acts somewhat like Newtonian orbital energy.; hence, the label.

    Let's build up some intuition by considering examples. Consider two scenarios that both have [itex]\mathcal{E} < V_{max}[/itex].

    Scenario 1: [itex]r[/itex] is very large and [itex]dr/d\lambda > 0[/itex].

    Scenario 1: [itex]r[/itex] is very large and [itex]dr/d\lambda < 0[/itex].

    Use your Newtonian intuition to predict what happens in each case, i.e., think about the Newtonian force to which the potential gives rise.
    What if L = 0?
     
  4. Oct 20, 2009 #3
    Take a look at this webpage http://www.fourmilab.ch/gravitation/orbits/

    If I take the first equation given on that webpage:

    [tex]\~{V}^2(\~{L},r) = \left(1-\frac{2GM}{r}\right)\left(1+\frac{\~{L}^2}{r^2}\right)[/tex]

    (I have replaced G which they have set equal to 1) and multiply the brackets out and divide both sides by two I get:

    [tex]\frac{1}{2}\~{V}^2(\~{L},r) = \frac{1}{2}-\frac{GM}{r}+\frac{\~{L}^2}{2r^2}--\frac{GM\~{L}^2}{r^3}[/tex]

    which is pretty much the expression you have for V(r) except they have assumed a value of 1 for [itex]\epsilon[/itex] so:

    [tex]V(r) = \frac{1}{2}\~{V}^2(\~{L},r) [/tex]

    The next equation they give is:

    [tex]\left(\frac{dr}{d\tau}\right)^2+\~{V}^2(\~L.r) = \~{E}^2[/tex]

    Dividing both sides by 2

    [tex]\frac{1}{2}\left(\frac{dr}{d\tau}\right)^2+\frac{1}{2}\~{V}^2(\~L.r) = \frac{1}{2}\~{E}^2[/tex]

    and substituting V(r) for [itex]\frac{1}{2}\~{V}^2(\~{L},r) [/itex] I get:

    [tex]\frac{1}{2}\left(\frac{dr}{d\tau}\right)^2 + V(r) = \frac{1}{2}\~{E}^2[/tex]

    It seems that the expression they get for [itex]\frac{1}{2}\~{E}^2 [/itex] is the same as the expression you gave for [tex]\mathcal{E}[/tex] if I assume that the symbol that you have used ([itex]\lambda[/itex]) and the symbol that they have used ([itex]\tau[/itex]) both stand for proper time. Do you know if that is the case?
    If so, it seems that the webpage and your post are talking about the same thing and that L is angular momentum according to that webpage.

    As I understand it, the energy barrier is not what prevents you escaping, but is what stops you falling in. The effective potential is unique to your angular momentum and if you have sufficient angular momentum then there is a barrier to falling in and although you might spiral inwards you simply spiral out again, but if you do not have sufficient angular momentum there is no barrier to falling in.

    effpot.png

    In the diagram above the red particle can not fall into the black hole to the left of it because it is prevented from doing so by the peak in the effective potential between it and the black hole. If the particle had insufficient angular momentum, then the peak would not be there because the effective potential is a function of its angular momentum.

    I think you can now figure out what happens if the angular momentum is zero, but if not play around with the settings of the applet on that webpage ;)
     
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