Keplerian motion in 4 dimensions

In summary, the equation in the introductory part of the section 2 says that V^2 has to be 1 in order for t', x', y', z' to lie on the sphere, but for that to be the case, E=-m/2 has to be true. However, since E=-1/2, m has to be half the mass, and that is what k=1 is all about.
  • #1
Caneholder123
10
0
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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  • #2
Caneholder123 said:
I have to show that vector (t', x', y', z') lies on the sphere.
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.
 
  • #3
mfb said:
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.

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 , 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.
 
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  • #4
Caneholder123 said:
the problem I have to solve
Please post the full problem statement, otherwise this is becoming wild guesswork.

Caneholder123 said:
but why does this make a general case?
All other energy values give trajectories that are scaled versions of this.

Caneholder123 said:
but why does it have to be half the mass
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.
 
  • #5
mfb said:
Please post the full problem statement, otherwise this is becoming wild guesswork.

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.
 
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  • #6
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.
 
  • #7
mfb said:
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.

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)?
 
  • #8
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.
 
  • #9
mfb said:
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.

Thank you very much for the help!
 

1. What is Keplerian motion in 4 dimensions?

Keplerian motion in 4 dimensions is a mathematical model that describes the motion of objects in a four-dimensional space, taking into account the effects of gravity.

2. How is Keplerian motion in 4 dimensions different from 3-dimensional Keplerian motion?

In 3-dimensional Keplerian motion, objects move in a two-dimensional plane under the influence of gravity. In 4-dimensional Keplerian motion, the objects move in a three-dimensional space under the influence of gravity and also take into account the curvature of space in the fourth dimension.

3. What are the implications of Keplerian motion in 4 dimensions?

Keplerian motion in 4 dimensions has implications in the field of astrophysics, as it allows for a more accurate understanding of the motion of objects in space, such as planets orbiting a star or galaxies moving through the universe. It also has applications in general relativity and the study of the curvature of space-time.

4. How is Keplerian motion in 4 dimensions calculated?

Keplerian motion in 4 dimensions is calculated using mathematical equations that take into account the mass and position of the objects in space, as well as the curvature of space in the fourth dimension. These equations are derived from Newton's laws of motion and Einstein's theory of general relativity.

5. Can Keplerian motion in 4 dimensions be observed in real life?

While we cannot directly observe the motion of objects in 4-dimensional space, the effects of Keplerian motion in 4 dimensions can be observed in the behavior of objects in our 3-dimensional world. For example, the precession of Mercury's orbit around the sun can be explained by the effects of 4-dimensional Keplerian motion.

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