Angular momentum conservation.

[andrewr]

I have been studying spin, angular momentum, etc. And became curious about how relativity would affect a classical problem: eg: that of a mass rotating around a center of mass;

In the classical case; two point masses of the same value, are separated by a distance 2r (mass-less attachement rod). This system has an angular frequency 'ω' which is in radians/second. The center of mass would be the midpoint r, and it can be considered at rest in our lab frame.

Now, classically, this is the kind of approach that can be taken with a figure skater, etc: If the momentum is conserved (energy too?), but the radius cut in half -- the velocity will exactly double to maintain conservation of momentum. (The angular frequency will quadruple.).

I intuitively know that increased velocity acts as increased mass under relativity. The rest mass is transformed/increased by a factor of motion. So, talking about a massive particle, I wanted to get an idea of how much the same problem would be affected in relativity. ( No crack ups about the figure skater ... it's hypothetical.)

I expect it isn't going to be a mass invariant problem; and so, I am thinking to define energy in a way that makes the actual mass irrelevant. EG: If I define energy to be in units of rest mass energy: eg: Tr=T/Em, so that when Tr=1 it means that for the amount of rest mass present, the kinetic energy is equal to the rest mass energy; so the total energy is 2 times the rest mass energy.

If I do this, and momentum is still conserved in the same problem -- but velocity corrected for relativity, I am thinking intuitively that the angular frequency will almost (but not quite) quadruple if the radius is cut in half; more mass=less velocity required for the same total energy.

Is this correct, and could someone estimate what halving the radius while keeping the total momentum constant would do to the angular frequency or velocity (or some other factor say 1/3, if that's more convenient than 1/2), I am not quite sure how to do it using the Lorentz transform, and when I looked around the web I came across mildly conflicting answers.

 

[Meir Achuz]

I didn't read your whole tract, but there is no useful concept of "center of mass" in relativity. What is useful is the "barycentric system", which means the total momentum is zero. In NR physics that corresponds to the center of mass system. One confusion is that almost all physicist call the barycentric system the "center of mass system", even though we know that is wrong. As with most colloquialisms, we know what we mean, but others may get confused.

 

[Bill_K]

It's actually pretty easy. Don't try to make it harder than it is by thinking in terms of rotating frames and so forth. Angular momentum in special relativity has the same value as it has in Newtonian mechanics, L = r x p. These are now 4-d quantities, but the value is the same and it's still conserved.

So the answer your question is, if r is halved and the angular momentum stays the same, then p must be doubled. What does this mean in terms of v?

p = γmv, solve for v and get
v = p/sqrt(m² + p²/c²)

From this relationship, for a particular value of m you can work out what effect doubling p will have on v. The angular frequency? ω = v/r still holds.

 

[andrewr]

It's actually pretty easy. Don't try to make it harder than it is by thinking in terms of rotating frames and so forth. Angular momentum in special relativity has the same value as it has in Newtonian mechanics, L = r x p. These are now 4-d quantities, but the value is the same and it's still conserved.

So the answer your question is, if r is halved and the angular momentum stays the same, then p must be doubled. What does this mean in terms of v?

p = γmv, solve for v and get
v = p/sqrt(m² + p²/c²)

From this relationship, for a particular value of m you can work out what effect doubling p will have on v. The angular frequency? ω = v/r still holds.

OK. Bill.
If I take your approach and say the linear momentum is doubled, then I come up with:<br /> 2mv_{0}\gamma_{0}=\gamma mv=\frac{mv}{\sqrt{1-(\frac{v}{c})^{2}}}<br />

which solves to a v, of:
nv_{0}\frac{\sqrt{1-(\frac{v}{c})^{2}}}{\sqrt{1-(\frac{v_{0}}{c})^{2}}}=v\Longrightarrow n^{2}v_{0}^{2}\frac{1-\frac{v^{2}}{c^{2}}}{1-\frac{v_{0}^{2}}{c^{2}}}=v^{2}
n^{2}v_{0}^{2}(c^{2}-v^{2})=v^{2}(c^{2}-v_{0}^{2})

v=\left\{ \left(nv_{0}\right)^{-2}+\left(1-\frac{1}{n^{2}}\right)c^{-2}\right\} ^{-\frac{1}{2}}

n=2, means 1/2 radius.

When I plug in small numbers it is almost exactly double, but not quite. So my intuition was correct.
This relationship happens to be true regardless of the value of mass.

That looks decent, did I miss anything?