Question

How does the geometry of a circular orbit change as velocity approaches the speed of light? @physics, @astronomy

Answer 1

I'm not sure exactly what you mean, could you possibly rephrase the question?

Answer 2

OK say you are circling a point (the sun say), and you have a fixed radius, you are orbiting the sun.
The is also a second orbiter with a smaller radius than you,
As you both approach the speed of light how does your relative position in radius effected

Answer 3

because the faster you are going the more probability density you have in the orbital...

Answer 4

Whoa now. Are we talking about probability density and atomic orbitals or spaceships around the sun?

Answer 5

i am trying to merge the concepts

Answer 6

ok so im confusing myself also.

perhaps the question im tring to ask is does the distance around the circumference have to do with the velocity of orbiting.

Answer 7

Yes, it does. Equating the gravitational force and the centripetal force required to keep one in orbit,
\[G\frac{Mm}{R^2} = \frac{mv^2}{R} \rightarrow R = G\frac{M}{v^2}\]
So the radius of the orbit is actually uniquely determined by the velocity. Is that what you mean?

Answer 8

That is really close to what i mean, but does the circumference change at a fixed radius as you approach the speed of light/

Answer 9

Okay, so what you're really talking about is relativistic circular motion then, nothing about planets or gravity. Because if you are talking about gravitational orbits, then by changing the velocity you can't help but change the radius, assuming gravity is the only force involved.

Answer 10

ok that is true i guess but i was imagining a (powered) rocket zooming around the sun

Answer 11

Right, but if you're saying that the rockets are used in such a way to keep the radius fixed then we have gravitational attraction as well as rocket impulses to deal with, so it's not purely gravitational attraction anymore.

Answer 12

yeah im assuming that the rocket is set to remain at fixed radius by accelerating along the right path

Answer 13

tell me all about relativistic circular motion

Answer 14

well, the mass of the rocket will increase, using the standard Lorentzian dilation formula
\[ m = m_0 / \sqrt{1-v^2/c^2} \]
But as the classical formulae above show, the radius of the rocket is independent of its mass. So there is no reason to first approximation why the radius of the orbit should change.

However ...

Answer 15

However, if the rocket begins to move very close to the speed of light indeed, a time will come when m/M is not a trivial number any more and the sun (or whatever the M body is) will begin to rotate significantly as well.

Answer 16

Which is to say, when we write down the classical formulae above for circular motion, we are using the fact that m/M is a tiny number and that M doesn't move appreciably.

Answer 17

Where does the value of π fit into this?

Answer 18

pi isn't important here, per se. If m/M is non-trivial, then I'm sure pi will appear somewhere in the solution of the system. But it's not a pivotal constant in as much as its precise value swings the system one way or another.

Even in the spherically symmetric solutions of the equations of General Relativity, it's not that important.

Answer 19

is π~3 for all conceivable geometries ? does spinning do anything?

Answer 20

Pi is just a number, so yes.

Answer 21

depends by what you're associating with pi. If you're associating with 4.arctan(1) or something like that, it is what it is.

But if you're associating it with the sum of the angles of a triangle, then "pi" in that context can be quite a bit smaller or larger.

Answer 22

yeah thats what i meant, can you explain that a little more please

Answer 23

For example, a triangle on the sphere has sides that are arcs of great circles. Great circles are the straight lines in this geometry--the technical term is geodesics--and they look like the equator or lines of longitude that circle around the sphere, every point on a great circle having a corresponding point opposite. Lines of latitude except for the equator are not great circles.

That being the case, the sum of the angles of a triangle on a sphere is always greater than pi. See, for example, here:
http://www.wolframalpha.com/entities/mathworld/spherical_triangle/lz/0b/ki/

In other sorts of geometries (hyperbolic to be precise), the sum of the angles is less than pi.

If you want a very, very accessible book on this topic (one in fact I read in High School and stayed with me for a few years), check out of your university library this:
http://www.amazon.com/Geometry-Relativity-Fourth-Dimension-Rudolf/dp/0486234002/ref=sr_1_1?s=books&ie=UTF8&qid=1324524076&sr=1-1

Answer 24

i find this stuff fascinating

Answer 25

It is fascinating!

Answer 26

http://www.gvsu.edu/math/about/gazette/2010/tessellations.html

Answer 27

yes, for example.

Answer 28

what's not clear from that diagram is why those are triangles. Why are those the straight lines?

Answer 29

This is something the little Rucker book I gave you will help you develop an intuition for what's going on here.

These topics are all dealt with formally in modern mathematics courses by Differential Geometry, which is usually offered in 3rd or 4th year.