# When emitting a photon, does the body exert an opposite force?

1. Sep 13, 2010

### Barwick

I'm wondering something, when a body emits a photon, that photon has a specific energy (if it emits it at UV frequencies of 10^15 Hz, it's got a rather "low" energy. If it emits it at gamma frequencies of 10^23 Hz, it's got significantly more energy, but still rather "low" when we're talking about "massive" bodies emitting them (like anything the size of a bowling ball wouldn't even "feel" an "equal and opposite reaction").

The question is, is there an opposite force "felt" by the object when it emits a photon? I'm thinking no, because it's not a transfer of kinetic energy, its emitting already present energy.

Matter of fact, if it does make an equal and opposite reaction, then I think perpetual motion would exist from it, but I may be wrong there, I dunno...

Now I'm wondering because, if in theory something emitted a photon at 10^40 Hz, there is 16 million joules of energy in that photon (that's a lot, over 4 megawatt hours). Now, if that happened, if there was in fact an "equal and opposite" reaction being felt by that object, that object would *move*... fast...

2. Sep 13, 2010

### ZapperZ

Staff Emeritus
Photons have momentum. So yes, the body does recoil upon such emission.

Zz.

3. Sep 13, 2010

### Barwick

Now I'm wondering though, lets say that energy was stored in an electron at an absurdly high energy state in an atom. All that energy is already present in the atom, which is moving at a slow velocity of X. That electron drops down to a lower energy state, and the atom emits a photon at whatever frequency (say 10^28 Hz) in one direction (call it 0 degrees on the Y axis)...

I would think the "action" would be the electron changing state, and the "equal and opposite" reaction would be the emitted photon. If the atom changed velocity, there's extra energy in there somewhere.

4. Sep 13, 2010

### maimonides

Hello Barwick,
Yes, conservation of momentum is valid for photon emission by atoms and nuclei.
Yes, there is recoil energy. Energy conservation is valid as well, and the recoil energy of the atom is taken from the photon.
No, there is no "free" energy and you can´t get perpetual motion from this.
The effect has been known for some time. It is involved in the Mössbauer effect and you can read about it here.

maimonides

5. Sep 13, 2010

### Barwick

I was looking at Mossbauer effect, and it seems like there's no way to calculate the recoil energy. Basically, the following must be true:

Recoil Energy + Photon Energy = Energy of Electron State Change

From what I know of fluorescence, the Photon Energy is always the same wavelength for a given element or molecule. I presume that means the recoil energy is also always the same? Is there a formula that tells what the recoil energy would be?

6. Sep 13, 2010

### maimonides

You´ll find the formulae and a model calculation in the link.

7. Sep 13, 2010

### Barwick

I keep looking over that link (I had seen it previously before it was mentioned here), and can't see where they know what the energy of recoil is going to be for any given photon (gamma ray in this case) emission.

I can see where momentum of the recoiling body has to equal momentum of the gamma ray, but then why did he observe a "fraction of the events occurring virtually without recoil"? If it emitted a gamma ray, it emitted a gamma ray, which has a known momentum, and the recoil should equal the momentum of that gamma ray, correct?

It's just not making sense to me, it seems like this recoil is somewhat random, but if the recoil changes, than the energy of the gamma ray emitted changes.

8. Sep 13, 2010

### pallidin

To my knowledge this can not be true.

9. Sep 13, 2010

### pallidin

Notice the term "virtually".
This is considerably different than the statement: "fraction of the events occurring without recoil"

10. Sep 13, 2010

### Barwick

Right, but it implies that the recoil induced by gamma ray emission isn't a constant. That's what I'm wondering about...

11. Sep 13, 2010

### pallidin

Are you under the impression that all gamma rays are of a single, specific frequency?
They are not. Hope that helps.

12. Sep 14, 2010

### maimonides

I hope I see where the problem is.
Recoil energy (for a given photon) is dependent on the recoiling mass. For a free atom the mass is simply the atomic mass, and we can calculate the recoil energy.
Now we put our atom into a diatomic molecule (near absolute zero). It´s extremely improbable that the momentum and energy are exactly right for exciting the next vibrational level, so the molecule will recoil as a whole. That means doubled mass, and half recoil energy (which we can calculate). At high enough temperatures you will get one-atom-recoils as well, because the molecule can be in different vibration states and the state density is higher. Using appropriate statistics, you can calculate the ratio of both kind of events.
Now we put the emitter into the crystal lattice of a solid.
quoted from http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/mossfe.html
"sufficiently low temperatures" means liquid helium and below.
"lattice as a whole" means many atom masses and therefore nearly recoil-free. (but calculable)
The quantized states of lattice vibrations are called phonons. They have defined energy and momentum. And there is a temperature dependent distribution function. Since we do not know exactly which mode the absorbing/emitting atom is in, and which phonon states are occupied and not available for the transition, there is a bit of statistic "randomness". Some events are (nearly) recoil-free, and some not, and the ratio (and the recoil energies) can be calculated. The randomness would disappear at absolute zero, but we can´t get there.

Maybe this helps
maimonides

13. Sep 14, 2010

### pallidin

Recoil free?
With momentum, I thought this was not possible.

14. Sep 15, 2010

### maimonides

By making the recoiling mass very big, you can make the "energy loss" by recoil very small and even negligible for practical purposes. That´s what is meant by (nearly/virtually) recoil free. Of course, momentum is conserved and there is recoil.

15. Sep 15, 2010

### pallidin

To this day, I still can not wrap my head around the idea that a feather thrown against a bowling ball makes the bowling ball move.
I know it's just me, but for some reason I have a problem with it.

16. Sep 15, 2010

### Barwick

No, in my first post I mentioned a conceptual situation with three different frequencies for photons being emitted.

So, I resorted to MS Paint... Which is the case, A or B (or neither?) I'm guessing B, or something else. In case you're running the numbers in your head, I was using a mass of about 5*10^-24 kg for the molecule, assuming something emitting a stupidly high energy photon would be a yet-undiscovered element over 120, if it were even possible.

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17. Sep 16, 2010

### maimonides

The Wikipedia article uses the classic energy-momentum formula. With the ratio of energy to rest mass in your example, you have to go relativistic. (There are two variants: the high energy approximation for v~c and the full formalism).
As you have correctly noticed in your first post ("*move*... fast...") v > c is impossible. Using relativistic mechanics will take care of that.

18. Sep 16, 2010

### Barwick

HA! I had an atom going faster than the speed of light. That's hilarious. You can tell I'm a newb at this stuff when I don't even check for that.

Ok, so this would accelerate that atom to roughly 7.88 * 10^-22 short of the speed of light, if I did the formulas correct. That's one fast atom, ok.

Now, would it be 7.88 * 10^-22 shy of the speed of light, or would it be the equivalent of half that energy (I'm not about to re-do the calculations)...

Basically, does that atom have 16000 Joules of energy in recoil, or 8000 Joules of energy in recoil?

19. Sep 16, 2010

### lucasizit

doesn't it just follow the conservation of momentum ? so the body would travel in the opposite direction to the emitted photon at a speed which is dependant on its mass relative to the speed and mass of the emitted photon

20. Sep 17, 2010

### maimonides

Above .99c, most people don´t care for speed anymore.
Think energy, not speed. You know that your atom is moving highly relativistic, so use the approximation. It will give you the energy in terms of momentum without having to calculate $$\beta$$ .(There are two high energy E-p approximations. You´ve already used one of them. Hint: the photon is a relativistic particle).
One (the last I hope) complication: since the atom carries away a considerable part of the energy, you´ll have to account for that in the calculation of the photon momentum from the start. Using the approximation, the math is not too bad. (think potential energy split into two parts: one for the atom, one for the photon)
And, btw: high energy physicists think eV, not J