I Energy of a Rotating Binary system

1. Dec 26, 2015

greswd

- We have two identical point masses, they are moving in uniform circular motion and in the same plane.

- Both masses are 180 degrees out of phase, they are always on opposite ends of their circular path.

-The "orbit" radii for both masses are the same.

We consider only SR effects, no GR.

Now we transform to another inertial frame that is moving parallel to the plane of the masses' orbits.

Is the total energy (γm0c2) of both masses in the second frame constant over time?

2. Dec 26, 2015

greswd

Or does it fluctuate periodically?

3. Dec 26, 2015

Staff: Mentor

So then we are not considering a graviational system. I guess that you are thinking of a system composed of two oppositely charged classical particles? That is probably easier to analyze anyway. I don't think that there is a closed form solution for the two-body problem in GR.

The total energy of both masses will not be constant in the first frame as the field radiates away energy. Also, point charges have an infinite energy, so you may need to consider extended charges, particularly as the orbit decays to collision.

4. Dec 26, 2015

greswd

haha, my question is actually way simpler.

just ignore all EM effects as well.

two masses moving in uniform circular motion as viewed from one frame. In another frame the velocities are transformed.

In the second frame, is the quantity γm0c2 constant? Just simple Lorentz transformations, nothing else.

Or just γ since the values of the rest masses don't change.

5. Dec 26, 2015

Staff: Mentor

Or maybe the even easier problem of a gravitational system in which the masses and orbital speeds are small enough that GR effects can be ignored? Or even better, two masses joined by a string? If so, we can ignore radiative effects and the answer to OP's question is that the total kinetic energy is conserved in both frames - it does not fluctuate.
[Edit: This post crossed the one just above by greswd]

Last edited: Dec 26, 2015
6. Dec 26, 2015

greswd

thanks. how do we go about proving that?

7. Dec 26, 2015

Staff: Mentor

There's an easy way and a hard way.

The hard way is to write down the position and velocity of each mass in the first frame. Then transform these values to the moving frame and grind through the algebra to find that even though $v$, $\gamma$, and hence the kinetic energy of each mass is varying with time the sum of the kinetic energies of the two masses is not - the changes always cancel. Only a total complete masochist would try doing it this way.

The easy way is to remember that the Lorentz transformations of special relativity are derived from, among other things, the assumption that the laws of physics are the same in all inertial frames. Conservation of energy is one of those laws, so you won't violate it by applying Lorentz transformations, and changing frames is just applying those transformations.

8. Dec 26, 2015

Staff: Mentor

Oh, I see what you are asking. "Total energy" meaning rest mass energy plus KE, not rest mass energy plus KE plus PE. You are just interested in the kinematics.

9. Dec 26, 2015

pervect

Staff Emeritus
Let's suppose there is a string holding the two masses in their circular orbit. The string is under tension. Then the system of the two masses and the string is a closed system, and it's energy is constant in every frame.

If you omit the string from your calculation, I believe you will get a non-constant energy. But it would be worth calculating it to be sure.

To include the string in your calculation, some familiarity with the stress-energy tensor is very helpful. This is technically SR, but it's usually taught in GR classes. A string with more tension than it's rest energy is known to be a form of exotic matter, it can and does have a negative energy density in some reference frames even though it has a positive or zero energy density in its rest frame.

10. Dec 27, 2015

greswd

ahh, yes. that term had escaped my mind.

Let's omit the string.

11. Dec 27, 2015

greswd

I'll do the hard way.

We have to take into account the RoS. I'm assuming that in the 2nd frame, when one mass is at the top of the "wheel", the other mass is simultaneously at the bottom.

Then we consider the left and rightmost parts of the wheel, the midpoints. One mass will reach the left before the other reaches the right in the 2nd frame. Travelling from the midpoint to the top is slower than from the midpoint to the bottom.

I assume that there is symmetry of the motion, that the magnitude of the y-component of velocity only depends on the distance from the midpoint and not on which direction the mass is moving.

12. Dec 27, 2015

Staff: Mentor

If you omit the string, and you also assume that the gravity of the point masses is negligible, you don't have a self-consistent solution. Something has to cause the masses to move in circles instead of straight lines.

13. Dec 27, 2015

pervect

Staff Emeritus
I basically agree, though I'd call it an incomplete system rather than inconsistent system. Something unspecified that's not accounting for is making the masses move in circles, that "something" is exchanging momentum with the masses to make them move in circles. It's also important to note that the exchange process takes time, it doesn't happen instantly in SR, the concept doesn't even have a well defined meaning in SR (unlike in Newtonian physics). I think it's simpler to give the unspecified something a name and a conceptual basis and call it a string. It's a bit more specific, which on one hand makes it easier to understand, though its specificity may draw some attention from the general abstract nature of what is happening.

Anyway, since momentum and energy are related by the Lorentz transform, (to use the jargon, they're both part of the energy-momentum 4 vector), a system that is exchanging momentum but not exchanging energy in the frame of reference S is exchanging BOTH energy and momentum in another, moving, frame of reference S'. Since one is omitting something that is exchanging energy in frame S', and because the exchange process takes time, one shouldn't be terribly surprised or shocked to find that energy isn't conserved in S' without the missing, unspecified, piece.

To make the books balance, one need to include the unspecified something, at which point it becomes convenient to specify it. Making it a string seems the obvious choice, and there is a way to account for the energy and momentum stored in the string. When one does so (which requires the stress-energy tensor), one finds that the total energy of the whole system is conserved as one expects from the Newtonian case.

The details may vary if we specify the "something" differently, the string is probably the simplest and most convenient though. Abstractly, if one wants to know the distribution of energy in S', one does so by finding the energy-momentum density in system S (including the energy-momentum stored in any fields), and transforming the energy-momentum density into system S'. The one selects out the part of the energy-momentum density that is energy in S', and integrates over the appropriate volume element in S'. The density of energy-momentum turns out to be mathematically described by the rank-2 stress-energy tensor, so a really detailed solution can't really avoid discussing the issue.

14. Dec 27, 2015

Staff: Mentor

I think that there is also an intermediate way where you write down the Lagrangian and note that it is independent of time. Therefore there is a conserved quantity, the energy.

15. Dec 27, 2015

Staff: Mentor

More precisely, it's independent of time in appropriately chosen coordinates. The coordinate-free way to say it, at least the one that I'm familiar with, would be to say that there is a timelike Killing vector field. That helps to avoid the possible misconception that the conserved "energy" is coordinate-dependent; it isn't.

16. Dec 29, 2015

greswd

Yes, it is not a realistic situation.

So is it confirmed that if we just transform the kinematics without any definite force that accelerates the masses, the energy in the 2nd frame will fluctuate periodically?

17. Dec 29, 2015

pervect

Staff Emeritus
I'd stop short of saying it's "confirmed" until I'd actually calculated it and/or found a reference that calculated it. But I would expect it to fluctuate.

18. Dec 29, 2015

Staff: Mentor

There is an obvious argument for this, which is that, since the two point masses are always 180 degrees opposite each other, their velocities relative to the moving frame will vary in the same way (i.e., they will both increase and decrease in sync), so their energies will as well. However, this argument, as it stands, has a flaw: it ignores relativity of simultaneity (the simultaneity assumption is hidden in the phrase "always 180 degrees opposite", which is false in the moving frame, as we will see). Without trying to do a full-blown calculation, let me try to put that back in to see if it changes the intuitive conclusion.

Suppose that, at the instant $t = 0$ in the rest frame, the two point masses are aligned exactly along the direction of relative motion of the moving frame. Then, in the moving frame, one mass will be aligned along that direction--call it, for concreteness, the $x$ direction--before the other. So at the instant $t' = 0$ in the moving frame, one mass (the one that was aligned along the $x$ direction earlier) will be further along in its orbit by some angle $\alpha$, while the other (the one that was aligned along the $x$ direction later) will be not as far along by some angle $\beta$.

Furthermore, $\alpha$ and $\beta$ will not, in general, be the same. How can we tell? Because, at the instant when the two point masses, in the rest frame, are aligned along the $y$ direction, they must also be aligned along the $y'$ direction in the moving frame, because the $y$ axis is not changed by the Lorentz transformation. So from the same time $t'$ in the moving frame, somewhat before $t' = 0$, one mass (the one that ends up further along past the $x'$ axis at $t' = 0$ by angle $\alpha$) must have moved further than the other (the one that ends up short of the $x'$ axis at $t' = 0$, by angle $\beta$). In other words, each point mass travels slower along one half of the orbit than along the other. So their velocities at the same time in the moving frame will not always be the same, and the simple argument above breaks down.

Furthermore, I'm not sure we can tell, without a full-blown calculation, whether there is still a version of the argument (that the total energy should fluctuate) that works. Consider the instant $t' = 0$ described above. At that instant, both point masses are in the "slower" semicircle of the orbit; the one that is further along by $\alpha$ is slowing down, and the one that is not as far along by $\beta$ is speeding up. Since the point masses line up along the $y'$ axis, where one is at its slowest and the other is at its fastest, it will always be true that one is slowing down while the other is speeding up; so it is still possible that their total energy could be constant. (The fact that, at the instant $t' = 0$, they are both in the "slower" semicircle does not rule this out, since their velocities will not vary linearly with orbital angle in the moving frame.)

Further comment on how putting the string in affects all this in a follow-up post.

19. Dec 29, 2015

Staff: Mentor

To follow up my previous post, what effect does the string have?

In the rest frame (i.e., the frame in which the center of mass of the entire system is at rest), the string can be modeled as having some fixed rest mass per unit length $\sigma$ and a tension which is zero at the center and increases out towards each end. By symmetry, the tension as a function of radius $r$ must be the same on both sides and must not vary with time, and the total energy of the string is then just the sum of its rest energy (constant) plus its kinetic energy (constant because each infinitesimal piece of the string stays at the same radius and therefore has constant velocity) plus the energy stored in its tension (which can be taken to be the integral of the tension from one end of the string to the other, and is therefore constant).

In the moving frame, the kinetic energy and tension of each infinitesimal piece of the string will vary; furthermore, the string will actually be curved in this frame, because its center (which is the center of mass of the whole system) must describe a straight line in this frame. (Intuitively, as the point masses speed up and slow down, parts of the string will become slack and then go taut again as their tension varies, so it won't always describe a straight line as it does in the center of mass frame, where the tension of each piece is constant.) The total energy of the string will still be its rest energy (still constant) plus its kinetic energy plus its tension, where both of the latter will now have to be integrated over the length of the string at an instant of time in the moving frame. It's not clear to me, without a detailed calculation, whether this integral will be constant either; if it isn't, then the kinetic energies of the point masses themselves will also have to change with time, since the total energy of the system as a whole must be conserved in every frame.

20. Jan 1, 2016

pervect

Staff Emeritus
I took a stab at calculating it - I can't solve the equations analytically, but if I didn't make an error, the numerical results support the energy not being constant.

A somewhat quick description of how I approached it:

First we write a parametric equation for the position of one of the rotating masses as a function of proper time in the frame S, where the object is at rest. We will later transform this to a moving frame. We get
$$x(\tau) = a \cos w \tau \quad y(\tau) = a \sin w \tau$$

Where a is some constant, and w can be regarded as the proper frequency (because $\tau$ is proper time, not coordinate time). When we get around to finding the total energy, we will sum the energy of two particles, one at "a" and one at "-a".

We also need t(\tau). Using the relationship $d\tau^2 = dt^2 - dx^2 - dy^2$ we can solve for $dt / d\tau$

$$\frac{dt}{d\tau} = \frac{1}{ \sqrt{ 1 - (dx/dt)^2 - (dy/dt)^2}}$$

which gives as one solution (setting the additive constant to zero)

$$t(\tau) = \frac{\tau}{\sqrt{1-w^2 a^2}}$$

I'll point out what is hopefully obvious: $\tau$ and t are both time coordinates, but the proper time $\tau$ is independent of the observer, while the coordinate time t depends on the specific choice of frame. I am presuming some familiarity of the reader with the concept of proper time, which may be problematical in some instances :(, but I don't think I have much choice with the amount of time I'm willing to spend writing this up.

Furthermore there is a 1:1 correspondence between t and $\tau$ for each value of a, so that if you know $\tau$ you can find t, and because its a 1:1 correspondence you can invert the relationship to find $\tau$ if you know t.

Now it is time to consider the moving frame $S^\prime$. The coordinates in this moving frame will be given by a Lorentz transform
$$t^\prime = \gamma (t - v \, x) \quad x^\prime=\gamma(x - v \, t) \quad y^\prime = y \quad \gamma = \frac{1}{\sqrt{1-v^2}}$$

The 4-velocity will be the derivatives of $(t^\prime, x^\prime, y^\prime)$ respect to $\tau$, so the energy in frame $S^\prime$ will be $m dt^\prime / d\tau$. We'll set m=1 at this point to get it out of the way. We have $t^\prime$ as a function of $\tau$ from our above results. We will call this function f, so we have

$$t^\prime = f(a, \tau) = \gamma \left( \frac{\tau}{\sqrt{1-w^2 a^2}} - v a \cos w \tau \right)$$

Again, $\tau$ is the observer independent proper time, while $t^\prime$ is the observer-dependent coordinate time. Just as was the case in frame S, for any given value of a there is a 1:1 correspondence between $\tau$ and $t^\prime$, but in frame $S^\prime$ the relationship is much more complex.

What we need to proceed further is an expression for the proper time $\tau$ as a function of the coordinate time $t^\prime$, $\tau = f^{-1} (a, t^\prime)$, the inverse of the function f. This is the part I can only do numerically. The energy in frame $S^\prime$ will be given by the expression

$$E(t^\prime) = \frac{df}{d\tau} | _ {a=d, \tau = f^{-1}(t^\prime)}+ \frac{df}{d\tau} | _ {a=-d, \tau=f^{-1}(t^\prime)}$$

which is just to say the total energy for the two particles (sans string) is the energy of the particle at a=d plus the energy of the particle at a=-d, evaluated at a constant value of the time coordinate $t^\prime$. Because of the relativity of simultaneity, a constant value of $t^\prime$ is not a constant value of $\tau$, this is the tricky part of the problem, both conceptually and in solving for it analytically.

To summarize the conclusion, though, using Maple I threw in some numbers for w and a, and graphed the results of the energy of each particle and the total energy versus coordinate time $t^\prime$, and my results were that it wasn't constant (as I expected). But I didn't spend a lot of time checking my work, though it's probably right.

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