PDA

View Full Version : Thought experiment on conservation laws



Astrowannabe
2006-Oct-01, 12:32 AM
Hello everyone, I've got a question about a thought experiment I, well thought up recently. However, it seems like it would violate the conservation of energy law, and so I'm wondering what I'm missing. I'm certainly not naive enough to believe that I'm the first person to think this up, so I'm sure I'm just missing something here:

So I was thinking about doppler shift, and how if a light source is coming towards you then the frequency of the photon goes up. But then I starting thinking that that means the energy of the photon would be higher, since the energy of a photon is given by E=hf.

So my the setup for this experiment is this (since this is a hypothetical, we are assuming that all machines work perfectly and friction can be ignored):

Take a laser beam and put it on the edge of a centrifuge arm, and then spin the centurfuge up so that the laserbeam is moving at 99% the speed of light. Also make sure that the laser beam is pointing in the direction of motion. We are also going to line the walls of the room with detectors so that they can measure the energy of the photons being emitted. While we're at it, lets also say that these detectors can absorb the photon and turn the photon's energy into usable energy (to be used by a machine or a lightbulb, whatever).

The formula for relativistic dopplershift is:

f=F*(sqrt[1-(v^2/c^2)]/(1-v/c))

f=observed frequency
F=emitted frequency
v=speed of source

The laser will emit photons with a frequency of 3*10^6 Hz, however the detectors will absorb photons of frequency 4.2*10^7 Hz. Because of this the laser will only have to use up 1.98*10^-27 J of energy, however the detector will absorb and be able to use 2.7*10^-26 J of energy, over 10 times more then what was used to emit the photon.

Now I know that the act of emitting the photon from the laser will give a small "kick", slightly slow down the spin of the laserbeam. However the momentum loss is only 6.6*10^-36. We can even use some of the energy that was absorbed from the detector to keep the laser spinning and there will still be extra energy leftover.

So my question is: Where did this extra energy come from?

Thanks to anyone who can respond!!!

hhEb09'1
2006-Oct-01, 01:51 AM
Now I know that the act of emitting the photon from the laser will give a small "kick", slightly slow down the spin of the laserbeam. However the momentum loss is only 6.6*10^-36. We can even use some of the energy that was absorbed from the detector to keep the laser spinning and there will still be extra energy leftover.

So my question is: Where did this extra energy come from?The leftover energy? How do you know there is extra?

Thanks to anyone who can respond!!!You're welcome :)

Astrowannabe
2006-Oct-01, 04:11 AM
Because the momentum the spinning laser lost is exactly equal to the momentum of the emitted photon. However, when that same photon strikes the detector it's momentum will be higher by slightly more then a factor of 13 due to the doppler shift.

hhEb09'1
2006-Oct-01, 04:34 AM
Because the momentum the spinning laser lost is exactly equal to the momentum of the emitted photon. However, when that same photon strikes the detector it's momentum will be higher by slightly more then a factor of 13 due to the doppler shift.I thought the idea was to use some of that energy to keep the mechanism spinning, and then there would still be some left over?

Ken G
2006-Oct-01, 09:25 AM
So my question is: Where did this extra energy come from?


The recoil effect is crucial for energy conservation, as you suspected but you didn't get the numbers right. It might simplify this to use a nonrelativistic situation-- a gun and bullet works for posing the same question. Imagine you have a gun attached to the Earth, and it fires a bullet with kinetic energy E in the frame of the Earth. The bullet has momentum P, and the Earth acquires negative momentum -P, of course. But the Earth will acquire essentially no kinetic energy, owing to its huge mass. Now let's look at the same problem from the Sun's point of view. The Earth is now moving at speed v (which is about 30 km/s), and if the bullet is fired in the direction of motion, that means the energy is E + Pv, where the extra Pv is due to the fact that the Earth is moving at v in this frame. But in this reference frame, the energy change in the Earth does matter, because of the speed v, and indeed it works out that the loss of kinetic energy by the Earth is -Pv. So that's where the energy goes-- into the lost motion of the planet (or in your case, the lost motion of the laser). So the moral is, when you change to a reference frame where the laser (or the Earth) is moving, you get an additional energy gain by the photon (bullet), and an equal energy loss by the laser (Earth). This is inescapable, it's just the way energy transforms when you change reference frames. You were pretty much on target, you just didn't do the numbers correctly, perhaps because you mixed up momentum and energy. Using the Doppler shift and a relativistic situation doesn't change the basic fact that energy will be conserved when you change reference frames, it's just that the actual amounts of energy are different in the two frames.

Astrowannabe
2006-Oct-01, 07:02 PM
It might simplify this to use a nonrelativistic situation-- a gun and bullet works for posing the same question.

I don't think a gun and a bullet is an analogy to the situation I'm proposing, and here's why:

The analogy is correct as far as if you spin a gun on a centerfuge and fire a bullet, the energy of the bullet impacting a target will be higher then the energy used to fire the bullet. In this case however there is no overall energy gain because of the energy expended to bring the centerfuge up to speed. Once you fire the bullet the gun now has less mass, so even if you could get 100% energy return when you slowed down the centerfuge you won't get all of that kinetic energy back. That's because some of that kinetic energy went into the bullet that was fired which is why it hit the target with greater force.

In the situation with the laser, however, that doesn't happen. The photons are being created in the laser, and the mass of the laser sitting on the end of the centerfuge doesn't change. Which means that you can (in theory) regain all of the energy you expended when you brought the centerfuge up to speed.

My issue is that the energy required to create the photon is less (much less actually) then the energy the photon had when it struck the detector, and I don't see where in the system that energy came from.

hhEb09'1
2006-Oct-01, 09:03 PM
In the situation with the laser, however, that doesn't happen. The photons are being created in the laser, and the mass of the laser sitting on the end of the centerfuge doesn't change. Why would it not? You have to produce the photons from something, don't you?

If not, there's your free source of energy! :)

Astrowannabe
2006-Oct-02, 12:21 AM
A-HA!!!

That's it! At least, I think it is. This whole time I was assuming that the mass loss would be trivial, and so I never considered it. However, after doing the calculations on the mass loss and how much energy it would take out of the system, it almost makes up the supposed energy gain. Almost. It actually covers about half of it. It could just be that I'm forgetting something, so here is what I did (remember that I was figuring that we'd gain 2.5e-26 J per photon):

The formula for relativistic kinetic energy is:

KE=MrC^2-MoC^2
Mr=relativistic mass
Mo=rest mass

And to calculate the relativistic mass:

Mr=M0/(sqrt[1-v^2/c^2])

So I get 1.56e-43 for the relativistic mass. After I plug that into the KE forumula, I get 1.2e-26 J lost per photon emitted, or half of the supposed energy gain.

Still, great catch heb!

Any ideas on what else I'm missing?

Ken G
2006-Oct-02, 02:00 AM
That's it! At least, I think it is. This whole time I was assuming that the mass loss would be trivial, and so I never considered it.
No, that isn't it. The energy you are missing does indeed come from the motion of the centrifuge, just as in the bullet example. You are forgetting that photons carry momentum, just like bullets. And what happens when the laser makes a photon? It loses that momentum, pure and simple. If you account for the motion of the centrifuge, you will always find that the energy gained by the photon due to that motion (which is not the energy to make the photon in the laser frame, that's a whole different issue) is lost by the centrifuge when it gives over that momentum. This is no coincidence, physics is built that way, or we would not have the two key principles of conservation of momentum and conservation of energy. If you find that one or the other of these is conserved in the laser frame but not in the lab frame, then you have simply made a mistake, and if you are not including the recoil of the laser when it creates the photon momentum, you will never get the right answer.