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Keplerian motion in 4 dimensions

  1. Apr 29, 2015 #1
    This is the link to the relevant paper. I have to show that vector (t', x', y', z') lies on the sphere. But for that to be, V^2 has to be 1 according to the equation in the introductory part of the section 2.

    That, by definition, means that E=-m/2. What does that mean, and why is this solution not just a special case?
     
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  3. Apr 29, 2015 #2

    mfb

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    What does "sphere" mean (the elements have a different dimension) and why do you have to show that?

    V=1 does not make sense in that context.
     
  4. Apr 29, 2015 #3
    t' is defined in the problem I have to solve as r, and that makes the units in that context OK, but V^2 still remains after calculation is done. John Baez writes about this in his http://blog [Broken], and he defines m=1, E=-1/2 and k=1, and that solves the problem. I am not familiar with this procedure, though. He says that he is working with a single fixed energy, but why does this make a general case? I understand that E has to be negative to have the ellipse and that m and k are just scaling factors, but why does it have to be half the mass, and what is the meaning of that altogether?

    The point of the problem is to derive from this symmetry the conservation of Runge-Lenz vector.
     
    Last edited by a moderator: May 7, 2017
  5. Apr 29, 2015 #4

    mfb

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    Please post the full problem statement, otherwise this is becoming wild guesswork.

    All other energy values give trajectories that are scaled versions of this.

    I'm quite sure it does not, but without problem statement it is hard to tell. It is convenient to consider this energy for sure because the factors of V go away.
     
  6. Apr 29, 2015 #5
    Starting from the equations of motion, show that the "vector of speed" of particle [tex]\vec V = (t', \vec r')[/tex] (where the symbol [itex]'[/itex] is used for differentiation in respect to [itex]s[/itex]) lies on the 4D sphere. Determine the center and radius of that sphere. [itex]s[/itex] is defined with the following relation [tex]\frac {ds}{dt} = \frac 1r[/tex].

    EDIT: I am translating this, and there are no different words for speed and velocity in my mother tongue, so the syntagma "vector of speed" may sound a little bit weird.
     
    Last edited: Apr 29, 2015
  7. Apr 29, 2015 #6

    mfb

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    An odd sphere, but you can always scale the system to make it spherical. Yes, then V=1 and the energy/mass relation you posted is satisfied. The general solutions are then scaled version of this special case.
     
  8. Apr 29, 2015 #7
    This kind of thinking is new to me. I am not sure that I completely understand this, probably because I am stuck with kgs and Joules. Does this mean that we fix mass to some value, let's say 10 of something and then we define energy to be -20*m*(some unit speed squared)?
     
  9. Apr 29, 2015 #8

    mfb

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    It is sufficient to fix the ratio, but that is the idea.

    There is no fundamental difference between orbits like the one of Mercury and orbits similar to Neptune, for example. They have the same possible shapes. One is just larger.
     
  10. Apr 29, 2015 #9
    Thank you very much for the help!
     
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