# Interesting energy problem

1. Jul 1, 2011

### guss

Mechanical energy against gravity? (Interesting energy problem)

Why doesn't the above picture work?
(I think it has to do with relativity, not 100% sure)

Last edited: Jul 1, 2011
2. Jul 1, 2011

### danR

To start, light doesn't accelerate. c is fixed, although that may not be the fundamental problem.

3. Jul 1, 2011

### guss

Oops, yeah, just meant the photons are gaining energy.

4. Jul 1, 2011

### danR

You can simplify the assembly of gears by replacing them with a train of levers (which is what gears of this sort are over the short haul), and then reducing the assembly to a single lever. You push the lever right at the bottom, and it goes left at the top and can do work. Or just one big wheel.

That done there is the relation of the lever action moving through a gravity gradient from bottom to top. Whether that matters or not to the energy-transfer would be for the experts to determine.

5. Jul 1, 2011

### Staff: Mentor

Gravitational time dilation means that the bottom gear would turn more slowly than expected for a given frequency. This exactly offsets the increased energy.

6. Jul 1, 2011

### bcrowell

Staff Emeritus
That was the kind of thing that occurred to me when I looked at it, but I wasn't able to figure out a detailed analysis. Do you think you understand this in detail, or is this just a guess as to the outlines of the solution?

It seems to me that there may be a variation on the Ehrenfest paradox here: http://en.wikipedia.org/wiki/Ehrenfest_paradox . The Ehrenfest paradox is SR, not GR, but I think there may be similar issues in that it may not be correct to assume that perfect rigid-body rotation of the gears is possible in a gravitational field.

To avoid making an impossible assumption about rigidity, maybe it would be interesting to recast the machine as one in which the energy is transmitted upward mechanically through a vibration. If I tap the bottom end of a vertical rod, the frequency of the vibrations should decrease as they propagate up the rod, due to gravitational time dilation. If their amplitude stays the same (which I think they have to if the rod is going to end up in equilibrium at the end), then they should lose kinetic energy, although it seems like the energy loss would go like the square of the time-dilation factor, which wouldn't be the right factor to patch up conservation of energy in the perpetual motion machine as originally proposed.

7. Jul 1, 2011

### Staff: Mentor

Definitely just a guess as to the outlines. But you can imagine a synthesizer locally generating a signal at the same frequency as the bottom gear and then transmitting that signal up. It would be redshifted on the higher side meaning that according to the top, the bottom is turning slower than it thinks it is.

8. Jul 1, 2011

### I like Serena

That does not seem right for 2 reasons.

First, if you set the machine to work horizontally (that is, you eliminate gravity), the machine will work, since there is no loss due to gravitational energy.
The Ehrenfest paradox would not be applicable for this reason, since the Lorentz-contraction of the circumference would be present in both cases.
In other words, it matters whether there is gravity or not, meaning this is GR and not SR.

Second, I thought GR and SR do not depend on nature being perfect or not. It's about the math. The math must still be right with perfect bodies. I believe this is essential in relativity theory: it's all about thought experiments that must mathematically work out.

9. Jul 1, 2011

### bcrowell

Staff Emeritus
I'm not suggesting that it's the same as the Ehrenfest paradox, just that there might be some ideas in common.

The resolution of the Ehrenfest paradox is that the math doesn't allow perfect bodies. That is, it's not kinematically possible to impart an angular acceleration to a disk while keeping the disk perfectly rigid.

10. Jul 1, 2011

### BruceW

As you went further from the source of gravity, you would start to see that the chain of gears carried less energy.
So what is the property of the chain of gears which indicates its energy? If it is simply the rotational speed of the gears, then that must mean as you went further away from the source of gravity, you would eventually notice that the gears are moving less quickly.
This seems like a paradox, since surely the gears wouldn't fit together if they were moving at different speeds. But there is no paradox, since local spacetime is flat, so the gears would connect properly.

11. Jul 1, 2011

### bcrowell

Staff Emeritus
This makes some sense to me, but:

(1) Let the difference in gravitational potential between the top and bottom of the apparatus be ΔΦ, so that time dilation gives t'=kt, where k=1+ΔΦ (with c=1). It seems to me that the kinetic energy of the gears would scale by k2, whereas the energy of the photons only scales by k.

(2) I don't think it really works to invoke locality. If you just use a single big gear, clearly "local" isn't big enough to include both the top and the bottom of the gear. This is why I'm guessing that it's not possible to have the gears be rigid to the approximation required for transmission of energy that is lossless compared to the energy gained by the photons on the way down.

Here are a couple of references on the generalization of rigidity from SR to GR:

F.A.E. Pirani and Gareth Williams, "Rigid motion in a gravitational field," Séminaire Janet. Mécanique analytique et mécanique céleste, tome 5, (1961-1962), exp. no 8-9, p. 1-16, available for free at permanent url http://www.numdam.org/item?id=SJ_1961-1962__5__A8_0

Boyer, Rigid Frames in General Relativity, Proc. R. Soc. Lond. A 19 January 1965 vol. 283 no. 1394 343-355

12. Jul 1, 2011

### guss

These responses do make sense to me, but I don't quite understand where the energy goes. Or maybe I do. If 1 J = kg*m^2/s^2, and we move that energy perpendicular to the gravitational field, is the energy in each reference frame going to differ by a factor of something like the inverse of the Lorentz factor squared?

If that's correct, then the change in energy of the beam should be the same. If it's not correct, maybe someone else can do it out properly (sorry).

13. Jul 1, 2011

### danR

If you use one big gear, and cut away all but the top and bottom, you have a lever and can solve for the instantaneous case. What happens to an arbitrarily small amount of work at the bottom in terms of the work done at the top, factoring in all GR effects? I wouldn't even know how to set it up.

[For simplicicy, you could have a mirror at the bottom to reflect the photon into a perfect absorber (pretending such a thing could theoretically exist) that would convert all the photon's energy into lateral momentum.]

14. Jul 1, 2011

### danR

You'll make it simpler for me if you simplify the gear assembly to a single lever, as per my comment just above. Really that's all we need if we go one photon at a time. I'm not the sharpest candle in the drawer. How does the energy 'travel up' the lever? Is it subject to GR distortion in its corresponding movement at the top, so that it does not move as far above as below, and cannot do as much work as expected?

15. Jul 1, 2011

### PAllen

I'll throw in a few abstract observations, without getting into the mechanical details of impossible devices.

Time and energy are intimately connected in SR and GR. (They also are in quantum mechanics; I suspect some argument can be made for this from Noether's theorem). Therefore, whatever the details, what looks like energy x in one place, will look like less than x where time flows faster. This contraption ultimately is equivalent to 'imagine a magic fiber optic that carries blueshifted light up the gravity will while preserving its frequency'. Then use it with a mirror in a device analogous to the OP and you can rapidly transform light to gamma rays.

Again, without looking at details, noting that perfect rigidity is mathematically excluded by both SR and GR, any energy carried mechanically is going to encompassed in displacement and/or compression waves. These should clearly be affected by time dilation the same way as light.

16. Jul 1, 2011

### bcrowell

Staff Emeritus
I don't know either. But the lever has the same issue as the gear. I don't see how it can maintain rigidity while rotating in a gravitational field. Time dilation says that the top has to rotate more slowly than the bottom.

Here's a really strange way of thinking about it. Suppose that the bottom of the lever oscillates back and forth through n cycles in 1 second as measured by a clock at the bottom. You could be silly and imagine that due to time dilation, the top of the lever oscillates n/k cycles in 1 second as measured by a clock at the top. Then the mechanical work done at the top is reduced by a factor of k, which is exactly the factor needed in order to preserve conservation of energy. I suspect that this is essentially the solution to the problem. A rigid lever can't exist, but if you replace it with something that can exist, time dilation reduces the mechanical work by a factor of k.

17. Jul 1, 2011

### guss

A lever does make more sense, and I'd edit the diagram, but I don't have access to my computer for a few days.

18. Jul 1, 2011

### bcrowell

Staff Emeritus
My knowledge of condensed matter physics is pretty weak, but you could probably argue that the waves are phonons with frequency f and energy E=hf, so they are affected the same way as photons.

So I don't know if we're converging on a solution, but anyway, I want to thank guss for posing a very fun brain-teaser!

19. Jul 2, 2011

### I like Serena

I've been trying to wrap my head around the notion that perfectly rigid bodies are not allowed and that this is part of the Ehrenfest resolution.
I thought you were saying that we need to accept nature's imperfections, which simply did not make sense to me.

Now I think I have it. I think what you're saying is that Euclidean rigid bodies are not allowed, which does make sense, since in GR we have to let go of Euclidean geometry!

I think we need to add adjectives to the use of the word rigidity.
It's important whether it actually deforms like it would in reality, or whether it is Euclidean or Riemannian rigid.
It seems to me that it still has to be mathematically rigid somehow.

I think you're right with your concept of an oscillating lever, where the moving part gets behind more and more, which is the effect of centripetal acceleration.
That means btw, that it only "looks" that way to an observer!
And with the added effect of gravity that is stronger at the bottom than it is at the top, we have our loss of energy.

That leaves the question how exactly the lever (or a wheel) should still be mathematically rigid.
I find that the wiki article on the Ehrenfest paradox, and also the related articles, are not quite clear on this.
It is suggested that the Langevin-Landau-Lifschitz metric approximates Riemannian rigidity "in the small".
But that just doesn't make mathematical sense to me.
In mathematics we don't do approximations. That's typically an engineering solution, meaning we don't quite understand what's going on yet.

Last edited: Jul 2, 2011
20. Jul 2, 2011

### I like Serena

I'd suggest to make it 2 opposing oscillating levers, to maintain symmetry.

This would decouple the effect of centripetal acceleration from the effect of gravity.