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Logarithmic Galactic Potential

  1. Jul 2, 2009 #1
    Hey guys,

    I have a potentially (no pun intended...ok maybe a little) stupid question. People often use log potentials to model galaxies with flat rotation curves and things and i'm currently using it to model a galactic potential for some simulation work. However none of the sources i have looked at show how one can extract the effective mass of a log potential, ie for a given log potential what is M or vice versa. I need to know the mass so i can ensure its conservation when i redistribute it and run simulations in potentials of different 'lumpiness'.

    Cheers
    -G
     
  2. jcsd
  3. Jul 2, 2009 #2

    Wallace

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    It might help if you define exactly what you mean by 'log potential'. I can think of two possibilities, simply the log of the potential, and a potential defined by a logarithmic derivative with respect to the radius. If you can write down exactly how the potential is defined then using log laws you should be able to convert back to the 'normal' definition of potential and extract the mass.
     
  4. Jul 2, 2009 #3
    Hmm good point but i dont think log laws will help with this one: as it currently stands i have:

    ln(x^2+y^2+1) (the one is there so things don't 'splode)
     
  5. Jul 2, 2009 #4

    Wallace

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    But what quantity does that equation represent, that is the question? You've only given us half of the equation. Maybe you don't know the answer, in which case what you need to do is find that out (ask your advisor if you are doing this for a research project?). Once you know that the answer to your question might be trivial.
     
  6. Jul 2, 2009 #5
    I'm simply using phi = ln(what i wrote) as the central potential for a galaxy in my simulations. Basically i want to know how to factor the mass that that potential represents into the equation so i can vary it. I mean i can scale the potential up and down and change depth etc but i need a way to relate that to changes in the mass. Sorry if that doesn't answer your query.
     
  7. Jul 2, 2009 #6

    Wallace

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    Okay, lets go back to basics. Assuming spherical symmetry (not true in your case, but I'll get to that) we have that

    [tex] \phi = \frac{-G M}{r} [/tex]

    Now, that is the purest defintion of 'the potential'. It may for some reason be convenient to work in logs on both sides, e.g.

    [tex] ln (\phi) = ln(G M(r)) - ln(r) + ln(-1) [/tex]

    Note now that the -1 is a problem, if we add one to the original potential (as you did and we are free to do) this goes away and we get

    [tex] ln (\phi) = ln(G M(r)) - ln(r) [/tex]

    Now, lets say M(r) is proportional to r^3 (the case of a uniform density), then we end up with

    [tex] ln (\phi) = ln(K r^3) - ln(r) [/tex]

    [tex] ln (\phi) = ln(K) + ln (r^2) [/tex]

    where K is some constant. This now starts to look like your equation, except that we have the log of the potential on the left hand side, rather than the potential. In the case of axial symmetry we would have x and y as you do.

    So, the answer to your question depends on two things you either haven't told us or need to find out, that is what you are assuming for the density profile and what your left hand side is. You have said it is potential, but are you sure it is not the log of the potential? Again, you either haven't given or don't know enough information. I think if you think this through and find that out the answer will probably be obvious, you just need to find out the definitions of what you are actually dealing with.

    The one constant here is the defintion, from the physics, of what the potential actually is, which is the derivative of the gravitational force. If you can relate your equations back to that you will have solved your problem.
     
  8. Jul 2, 2009 #7
    Ok i see where we're getting mixed messages =) I don't want to take the log of a point mass potential, what the log potential i wrote represents is a particular distribution of point masses such that the sum of their potentials gives a logarithmic function. I found a way to get the mass out i just worked through the Poisson equation and integrated the density. Thanks for your help anyway!
     
  9. Jul 3, 2009 #8

    Chronos

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    Logs are merely a mathematical convenience. Nothing magical there.
     
  10. Jul 3, 2009 #9
    I never said there was anything magical?
     
  11. Jul 3, 2009 #10

    Wallace

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    Ah I see, you mean you want the potential to be a logarithmic function of the radius. Yes, looks like we were having 2 parrallel discussions! Glad you found a solution in the end.
     
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