# "Unexpected" Symmetry in Elliptical Orbit

Hello everyone :)

Not too long ago, I was thinking about planetary motion around a sun, both with circular orbits and elliptic orbits. However, when thinking a little longer about these two cases in a broader sense, I spotted a big difference which I found quite odd (assume purely classical mechanics - no relativity):

In the case of a perfectly circular orbit, there is circular symmetry both in the shape of the orbit AND the gravitational field, and the centres of symmetry align at exactly the same point. In other words, the symmetry of the orbit trajectory "aligns" perfectly with the symmetry of the field that is governing it. For this particular example, let us consider only linear symmetry in two perpendicular axes, x and y i.e. In the circular orbit, the x and y lines of symmetry for the orbit are the same as the x and y lines of symmetry for the gravitational field. This is OK so far.

In the case of the elliptic orbit, there IS x and y linear symmetry BOTH in the orbit AND and field, but the lines of symmetry no longer align (one line of symmetry, say x, aligns, but the other, y, is translated by a distance d. d = distance from centre of gravity to centre of elliptic orbit).
To frame this observation in another way: Let us consider the centre of the ellipse as our point of interest. The orbit path is symmetric in BOTH x and y around our point of interest, but the field that is governing it is ONLY symmetric in one of the axes around our point of interest. (picture attached)

I find this quite strange. Why is it the case that a less symmetrical cause (the gravitational field) can result in a more symmetrical effect (orbit path) around our point of interest?.I can derive the trajectory using classical mechanics and the laws of gravitation, but this does not give me any obvious insight into the symmetry mismatch. I am aware that if we also consider the speed of the planet, the problem does becomes a little less strange, as the speed is not symmetrical in one of the axes.

Sorry for the slightly hand-wavey explanation XD Any response, however broad, would be appreciated!!

Nat :D

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ZapperZ
Staff Emeritus
I'm not exactly sure what your issue is here.

In the circular path, you have A LOT of symmetry. You have reflection symmetry along both axis and you have rotational symmetry about the z-axis.

In the elliptical path, you still have reflection symmetry along the x and y-axes (assuming that they are along the major and minor axis of the ellipse), but you lose the rotational symmetry along the z-axis except for 180-degree rotation.

So they are both not identical, and they shouldn't be.

Why one can get such a path requires one to understand conic section, because such elliptical path is only ONE possible solution to the central force problem.

Zz.

• NatanijelVasic
I'm not exactly sure what your issue is here.

In the circular path, you have A LOT of symmetry. You have reflection symmetry along both axis and you have rotational symmetry about the z-axis.

In the elliptical path, you still have reflection symmetry along the x and y-axes (assuming that they are along the major and minor axis of the ellipse), but you lose the rotational symmetry along the z-axis except for 180-degree rotation.

So they are both not identical, and they shouldn't be.

Why one can get such a path requires one to understand conic section, because such elliptical path is only ONE possible solution to the central force problem.

Zz.
Thanks for the response! :) I do agree with what you have said. So my issue with the elliptic orbit is this:

From the point of view of the centre of the sun, the gravitational field has a lot of symmetry (infinity degrees of symmetry), and results in a plnetary orbit with only one degree of symmetry (again, looking only from the sun point of view).

Looking from the centre of the orbit, the orbit has a higher degree of symmetry (2 degrees of symmetry) than the field (1 degree of symmetry)..

So depending on your point of view, either the cause has a higher symmetry than the effect OR the effect has a higher symmetry than the cause. I am used to seeing the former in nature, which is why i find the elliptic case strange.

Cthugha
I am not sure I get your point and I do not really understand your definition of "higher" and "lower" symmetry. The conserved quantities for a two-body system with elliptic orbits may not be obvious, but you have "hidden" constants of motion such as the Laplace-Runge-Lenz vector (which ironically was discovered first by Jakob Hermann and Bernoulli).

• NatanijelVasic
A.T.
So depending on your point of view, either the cause has a higher symmetry than the effect OR the effect has a higher symmetry than the cause.
Which indicates that this "point of view" concept is not very useful for analyzing symmetries?

I am used to seeing the former in nature...
Examples?

• NatanijelVasic
ZapperZ
Staff Emeritus
Thanks for the response! :) I do agree with what you have said. So my issue with the elliptic orbit is this:

From the point of view of the centre of the sun, the gravitational field has a lot of symmetry (infinity degrees of symmetry), and results in a plnetary orbit with only one degree of symmetry (again, looking only from the sun point of view).

Looking from the centre of the orbit, the orbit has a higher degree of symmetry (2 degrees of symmetry) than the field (1 degree of symmetry)..

So depending on your point of view, either the cause has a higher symmetry than the effect OR the effect has a higher symmetry than the cause. I am used to seeing the former in nature, which is why i find the elliptic case strange.
I don't understand this.

Why should the symmetry of the field matches the symmetry of the path/trajectory? After all, you didn't have a problem with the "symmetry" (or lack of it) of a typical projectile motion in a gravitational field.

Zz.

• NatanijelVasic
Dale
Mentor
Why is it the case that a less symmetrical cause (the gravitational field) can result in a more symmetrical effect (orbit path) around our point of interest?
The reason why is simply because you chose that specific point of interest. Why did you choose that point as your point of interest?

• NatanijelVasic
sophiecentaur
Gold Member
Why one can get such a path requires one to understand conic section, because such elliptical path is only ONE possible solution to the central force problem.
It's worth pointing out that this only applies to central attractive fields that follow the inverse square law.

• NatanijelVasic
Which indicates that this "point of view" concept is not very useful for analyzing symmetries?

Examples?
Hmm yes, I guess whatever example I do give, it won't really be that valid because as you say I could just change the point of view. But my thought process went something like this: since many systems in nature are non-linear, a symmetric input will usually result in a non-symmetric, or less symmetric, output, for example, a Karman vortex street, extensions of plastic materials when being subject to an oscillatory force, electron trajectory through a uniform magnetic field..

I don't understand this.

Why should the symmetry of the field matches the symmetry of the path/trajectory? After all, you didn't have a problem with the "symmetry" (or lack of it) of a typical projectile motion in a gravitational field.

Zz.
I don't think that the symmetry of cause and effect SOULD match up in space/time, and I can vaguely understand why by looking at the maths, but I just wanted to understand this observation at some other level, which may not even exist.

For the projectile example in a uniform field, there is less of a problem because the symmetry line of the trajectory coincides with one of the infinite number of symmetry lines in the uniform gravitational field. In the elliptic orbit, this is also the case for one line of symmetry, but there is an extra line that does not match.

Orodruin
Staff Emeritus
Homework Helper
Gold Member
I feel that you are doing yourself a disservice by focusing on spatial symmetries rather than space-time symmetries, which is where the physics lie. A time-independent central potential has a lot of space-time symmetries, including time-translational invariance and full rotational invariance. These symmetry transformations will take one possible solution into another and the continuous symmetries are related to conserved quantities through Noether's theorem. That the solutions themselves do not display many of these symmetries is unavoidable, but again the point is that the symmetry transformations result in new solutions with the same constants of motion.

• NatanijelVasic
I feel that you are doing yourself a disservice by focusing on spatial symmetries rather than space-time symmetries, which is where the physics lie. A time-independent central potential has a lot of space-time symmetries, including time-translational invariance and full rotational invariance. These symmetry transformations will take one possible solution into another and the continuous symmetries are related to conserved quantities through Noether's theorem. That the solutions themselves do not display many of these symmetries is unavoidable, but again the point is that the symmetry transformations result in new solutions with the same constants of motion.
I think this is the solution I was looking for :) Thanks so much! I haven't heard of Noether's Theorem before, but now I wish I came across it earlier.

Nat :D