# conservation of momentum II

Discussion in 'Pseudoscience Archive' started by DRZion, Jun 21, 2010.

1. ### AlphaNumericFully ionizedRegistered Senior Member

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Yes, that's the way accelerators detect neutrinos and also how they'll look for supersymmetry. However in systems involving only massive charged particles and photons the photons can carry plenty of momentum and are experimentally observed to do so.

But the only way the momentum of one electron can be transferred into another (for energies below 90GeV) is by it emitting a photon of that amount of momentum.

For instance suppose you have two electrons heading at one another, one with momentum p and the other q. They scatter and come out with momentum p' and q' and by conservation of momentum you'd have q' = q-(p'-p). The main contribution to this process is the tree level exchange of a photon and it'll have momentum p'-p. It carries the momentum they exchange. When you go further and do loop contributions you encounter photons of momentum as high as you wish, its part of the reason you have to renormalise. So even in electron-electron scattering theory and experiment disagree with you.

And then there's hard Compton scattering, an Xray photon slams into a charged particle and gives it a huge amount of momentum. Its like the photo-electric effect but more pronounced.

And this means they carry negligible momentum because.....? Their momentum/energy ratio is independent of how much energy or momentum they can carry.

So why are you claiming you've demonstrated them to be wrong, which is the implication of momentum conservation violation you're talking about. How can you reach conclusions like "Its simply IMPOSSIBLE for momentum to be conserved if photons are the only momentum carriers that are created!! It is also impossible for the hypothetical momentum carriers to be massless. ". You admit you're cutting corners and not using the full models and yet you're proclaiming those models are either wrong or don't do what people think they do, based entirely on your very superficial back of an envelope guessing.

I disagree with your application of them to this problem and I disagree with your conclusions as they are based on simplifications and mashing together different models, some of which aren't terribly appropriate. When experiment disagrees with you you're wrong. QED models bremsstrahlung processes with enormous accuracy, its tested every second of every day there's an accelerator running somewhere and QED says momentum is conserved, which is what we see. We've seen in experiments photons with momentum of the same order of magnitude as massive particles. You are arguing that not only is QED wrong but the universe is too.

3. ### PeteIt's not rocket surgeryRegistered Senior Member

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Well, you can have an interesting and useful exploration of an idea (ie a productive argument) without having any vested belief in any conclusion.

But, that's not what I asked.
Why are you convinced that your claim is correct?
Why not withhold judgement unless and until you work out the details properly?

If you properly work out the kinetic energy and momentum of the massive object and emitted photon, you will find that energy and momentum are conserved without any extra particles involved.

That's why you're geting it wrong. The difference between newtonian mechanics and relativistic mechanics is miniscule for low-speed massive objects, but it is not zero. Just as the momentum-energy exhange between the photon and the object is miniscule but not zero. You can't ignore one and not the other.

Last edited: Jun 23, 2010

5. ### PeteIt's not rocket surgeryRegistered Senior Member

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Newtonian mechanics.
In this scenario, the difference between newtonian mechanics and relativistic mechanics is on the same scale as the effect of the photon momentum and energy.

If the difference between newtonian mechanics and relativistic mechanics is negligible in this situation, then the photon momentum and energy is also negligible.

If you work it out properly using relativistic mechanics, the discrepancy goes away.

Try it. An object of mass m1 has velocity v1:

The object emits a photon of frequency f. The object now has mass m2 and velocity v2:

Here are the equations.
Before emission, the object has momentum p_1, energy E_1:

\begin{align} p_1 & = & \gamma m_1v_1 \\ & = & \frac{m_1v_1}{\sqrt{1 - v_1^2/c^2}} \\ E_1^2 & = & p_1^2c^2 + m_1^2c^4 \\ & = & \frac {m_1^2v_1^2c^2}{1 - \frac{v_1^2}{c^2}} + m_1^2c^4 \end{align}

After emission, the object has momentum p_2, energy E_2, and the photon has momentum p_p, energy E_p::

\begin{align} p_2 & = & \gamma m_2v_2 \\ & = & \frac{m_2v_2}{\sqrt{1 - v_2^2/c^2}} \\ E_2^2 & = & p_2^2c^2 + m_2^2c^4 \\ & = & \frac {m_2^2v_2^2c^2}{1 - \frac{v_2^2}{c^2}} + m_2^2c^4 \\ p_p & = & \frac {hf}{c} \\ E_p & = & hf \\ \end{align}

If energy and momentum are conserved, then:

\begin{align} p_1 & = & p_2 + p_p \\ E_1 & = & E_2 + E_p \end{align}

Now, if you are correct then there are no physically realisable values for m_2 and v_2 that will make the conservations equations work.

We can test this by using the above equations to solve for m_2 and v_2. If you're right, then the solution will be something physically impossible, like an imaginary, complex, or infinite mass or velocity, or a negative mass.

Want to try?

7. ### DRZionTheoretical ExperimentalistValued Senior Member

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I am not very informed on all of this, but I think I read somewhere that electrostatic interactions are mediated by virtual photons?

Are these regular photons? By regular, I mean are these the same kind as free-travelling em radiation? . . . my sources tell me the momentum of a photon is just its energy / c . But what do i know about virtual photons!

Well, momentum can't be created or destroyed, so effectively its just the momentum of the photon. I've heard of cosmic rays with energies in the joules range, so this wouldn't surprise me.

Oh sure, they can carry all the momentum in the world, but thats because there are lots of them. If you have limited energy but want to get maximum momentum, photons are quite literally the worst momentum carriers you will find ( at least regular free-travelling photons ).

You're right, my statement is entirely too broad. I see now that most electrostatic or magnetic interactions are facilitated through photons and that they conserve momentum.

You know, I have forgotten just how complicated real physics is.. and by no means can I argue over the details. But, the scenario I have come up with is extremely clever in that it excludes almost everything but the barest laws, to me it seems. I have reformulated it for you so that it is more like electron-electron scattering. Here it is-

Suppose that instead of clumps you have protons. One bundle is composed of two protons (no electrons) and the other is composed of one proton and one neutron.

You bring these clumps very close together so they have considerable potential energy. You then let them go (optical tweezers or whatever). Their masses are close to identical and so they accelerate equally. However, one loses energy to bremsstrahlung twice as fast as the other. The only source of this bremsstrahlung is kinetic energy. SO, via the most basic logic, the two proton bundle will lose more kinetic energy. It will move more slowly. And SO, according to newton's laws it will have less momentum than the proton-neutron bundle.

Now, there are photons which are emitted (other than the photons which mediate the electrostatic force). These photons also carry momentum. BUT, these photons CANNOT carry as much momentum as that which was lost by the protons. This is because photons carry the minimum amount of momentum/energy. Any amount of kinetic energy lost as bremsstrahlung will express only the minimum (physically possible) momentum.

But, the fact that bremsstrahlung is such a well studied process and the fact that there are entire textbooks written on the subject would mean that it has been explored... throughly. So I guess I'll have to learn for myself to see where the momentum goes.

I agree. But I think that an argument should have a purpose otherwise its just arguing for the point of arguing.. which is useful sometimes i suppose.

I think my scenario is really clever an I don't see any way to resolve it other than a new particle or violation of conservation.

No way. Photons cannot account for the momentum lost. Every joule of em radiation will carry away only the minimum amount of momentum (e/c). The kinetic energy lost this way will play into
p=mv
e=1/2mv^2

I have said it before .. maybe I'm just making a huge error somewhere in my conceptualization. But so far no one has pointed out where. I hope the above scenario will give people something to ponder about.

Its like, since photons don't have mass you will expend most of the kinetic energy to create mass rather than momentum.

Last edited: Jun 23, 2010
8. ### PeteIt's not rocket surgeryRegistered Senior Member

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Done, but not checked.
For simplicity, I set c to 1, and express the result in terms of p_1 and p_p rather than v_1 and f, and I don't expand the m_2 term in the expression for v_2:

$m_2^2 = m_1^2 + 2p_1p_p - 2p_1\sqrt{p_1^2 + m_1^2}$

$v_2 = \frac{p_1 - p_p}{\sqrt{m_2^2 - (p_1 - p_p)^2}}$

This isn't the most elegant way to achieve this, but it's all I can do with the tools I understand. If I had a better grasp of energy-momentum four vectors in Minkowski space, I suspect that this would be much easier.

9. ### PeteIt's not rocket surgeryRegistered Senior Member

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Your error has been pointed out several times - you're using a newtonian approximation for the change in the energy and momentum of the massive object.

Why don't you work it out properly?

10. ### PeteIt's not rocket surgeryRegistered Senior Member

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Of course. The ideal point of arguing is twofold:
To learn something, to find out whether some idea is right or wrong, to check whether all the pieces fit.
To demonstrate your understanding of that process, to explain the details and show how the pieces fit.

If you don't understand the details, then it makes no sense to be convinced of a conclusion. You should be withholding judgement until/unless you do understand.

11. ### DRZionTheoretical ExperimentalistValued Senior Member

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TOO MUCH MATH - to put it bluntly.

I'll really give it my best shot though. It may take a few days.

The object would actually lose momentum due to the energy emitted. This would put it further away from conservation as I stated earlier.

Maybe equal and opposite forces don't apply 100% in the quantum world, that would also make it possible for momentum to be conserved.

12. ### PeteIt's not rocket surgeryRegistered Senior Member

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If you don't put in the effort, you're just waving your arms and wasting your time. If the required investment is too much (do you have a few years to spare?) that's OK - but it also means that you'll have to concede that you just don't know. I've been told by many smart people that "I don't know" is one of the smartest things anyone can say.

My comment before about loss of mass was off the mark. Yes, there is a reduction in rest mass, but that's only part of the story.

Edit:
Actually, it wasn't too far off the mark, in a handwaving kind of way (note - handwaving is no substitute for rigour).
Your argument is that when the object loses momentum to match the photon's momentum gain, the object doesn't lose enough kinetic energy to match the photon's energy gain.
Well, the photon's energy gain doesn't just come from the object's kinetic energy loss. It also comes from the object's mass loss. Part of the object's mass is converted to photon energy.

Last edited: Jun 23, 2010
13. ### DRZionTheoretical ExperimentalistValued Senior Member

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Yes, exactly!

Well, the two are interchangeable, is that what you mean?

Otherwise I have a good argument that bremsstrahlung's source is pure kinetic energy -
http://en.wikipedia.org/wiki/Bremsstrahlung#X-ray_tube

And I am still looking at the math.

Last edited: Jun 23, 2010
14. ### PeteIt's not rocket surgeryRegistered Senior Member

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No. I mean that the rest mass of the object must decrease, contributing to the photon energy.

I'm not touching Bremsstrahlung. That's a much more difficult case than I can handle, due to the acceleration and interaction with the electric field. In that case, I suspect that the extra mass/energy comes from the electron-field system in some way, but like I said - it's too hard for me. I'm not going to pretend I know how it works.

I'm only focussing on the simple case of a moving clump of matter emitting a photon - no other objects, fields, or anything. I'm directly addressing your "underlying argument":

Last edited: Jun 23, 2010
15. ### AlphaNumericFully ionizedRegistered Senior Member

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If you weren't so utterly unwilling to accept there's some results you don't understand applying to this then you'd know that conservation is preserved and Newton's 3rd law still applies.

Experimentally momentum is conserved. The models of this process in QED predict conservation of momentum. There's nothing else to say. You're admitting you don't know how to do the problem properly because you don't know the relevant mathematics and yet you seem unwilling to accept that because of that you don't end up with the right answers. Momentum and energy conservation come from the same source, the invariance of a Lagrangian under space-time translation. If $t \to t + \delta t$ is a symmetry then energy is conserved. If $x_{i} \to x_{i} + \delta x_{i}$ is a symmetry then momentum is conserved. They are symmetries in QED and thus ALL processes in QED preserve them. The fact you don't understand variational principles upon which Noether's theorem is based doesn't mean you can just dismiss them and then only use mathematics you understand, which you pick and choose from other things in an inconsistent way.

Your claim the photons 'cannot' carry away enough momentum is wrong. Every single person whose done QED at university knows that, having done calculations on such things as radiation braking and obtained predictions which conserve energy and momentum and which match experiments. Its one thing for you to deny mathematics you don't understand but you're denying experimental fact.

16. ### DRZionTheoretical ExperimentalistValued Senior Member

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Oh geez, if only one photon is emitted, how on earth can momentum be conserved? This is not a stochastic process - it has to be perfectly deterministic in order to be possible. Not impossible. This stuff is really complicated..

My simplest counter argument is in the form of another example. Suppose you heat up a baseball to several thousand degrees (its made of the same stuff as the space shuttle so it doesn't melt). Its easy enough - you take your space-ball and focus an enormous amount of sunlight on it using a lens (btw can this lens ever be so large as to heat the baseball to hotter than the surface of the sun?). You then toss this ball in outer space. As the ball travels it emits plenty of photons of thermal radiation. This decreases the rest mass of the baseball.. however, i do not expect to the baseball to lose a considerable amount of momentum even as it cools to 2.7 degrees kelvin.

Yes, I admit i know very little. But I take bremsstrahlung to be as basic as force as gravity ... its easy to conceptualize with. Just like you know an apple will drop to the ground, you know that em radiation will be emitted when a charge accelerates. You can't tell me I don't understand this much. (although the underlying physics are FAR from simple)

Here are the possibilities I accept so far-
1. No equal and opposite forces on the quantum scale.
2. Creation of a different momentum carrying particle.
3. Violation of conservation of momentum.
The last on is the least likely.

I simply haven't found a satisfying answer to my problem yet

I know plenty of experimental facts which are wrong, have been cooked, and I know plenty of academics who will satisfy their human needs by throwing around certain words and facts (jargon). I have learned this much in biology (a very dynamic field). Sometimes entire cabinets of papers become obsolete; its something we will have to deal with. I know that there are all forms of corruption at the university just as there is everywhere else ( except inbetween my ribs where my blessed soul rests

), and I know that people make mistakes, people get comfortable in their existence, and people defend their beliefs. I do it too, I am defending my beliefs as a skeptic. I am not saying the everybody who does QED is wrong, and from what I know it is amazingly accurate at predicting and describing natural processes. If there are some exceptions to it's rules, then I'm sure there will be good explanations for them.

And more directly - photons are horrible carriers of momentum. You'd be far more energy efficient at accelerating your spacecraft by flicking pebbles off the back than churning a hand-crank which powers a laser beam. Photons are massless particles and for this reason they are piss poor at carrying momentum. Anything (everything else) with mass is better at storing momentum (but not necessarily at transmitting it as you have pointed out earlier in the example of electron scattering).

17. ### PeteIt's not rocket surgeryRegistered Senior Member

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You're not paying attention?
Momentum conservation is easy: the momentum of the object is reduced by exactly the momentum of the photon. What's not to understand?

You might naively think that this means that energy would not be conserved, but I suggest (though I'm not completely sure) that the extra energy of the photon comes from the mass of the massive body.

The maths is all there. Work it through.

18. ### DRZionTheoretical ExperimentalistValued Senior Member

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Sorry I was partly expressing my exasperation with the complexity of this 'simple' problem. What I was also saying is that the direction of emission is also important in conservation of momentum, and in the case of 1 photon emitted there would only really be 1 possibility ... but I might be getting ahead of myself.

I'm on it.

19. ### DRZionTheoretical ExperimentalistValued Senior Member

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Yeah, there is the whole issue with the vectors. Even if I use the equations you have given, I'm not sure if the vectors wouldn't change the outcomes.

If we work without vectors then we don't know which way the photons are going ... which is okay, because my assumption is that it doesn't even matter which way they go, it still won't be enough momentum.

Hmm - this looks relevant.
http://en.wikipedia.org/wiki/Four-m..._the_presence_of_an_electromagnetic_potential

20. ### DRZionTheoretical ExperimentalistValued Senior Member

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The more I read random physics the more I doubt my scenario is the way it is. This stuff is complex.. But I still think I'm right.

Of all people, you AlphaNumeric should be able to get to the bottom of this. From the above link I learned about canonical momentum, which is apparently a big part of symplectic manifolds, something you know a lot about

apparently momentum gets affected by charge :shrug:

I will go to another example. Its again the heated baseball in vacuum. It emits photons of thermal radiation. Each photon carries momentum. Conservation of momentum states that the center of mass of any system must remain constant. The 'mass' carried by the photon is what causes a recoil of the object. There is conservation of momentum here.

Now, if you have emission of bremsstrahlung you will also have this recoil.. in addition to reduction in kinetic energy. I guess I'm just restating what I said before..

And yes pete, I have pencil and paper here and I am working out these equations.

21. ### PeteIt's not rocket surgeryRegistered Senior Member

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Consider the baseball to be moving.
How is kinetic energy exchange different between blackbody radiation and bremsstrahlung?

22. ### AlphaNumericFully ionizedRegistered Senior Member

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What is there for me to get to the bottom of, the issue is your understanding so nothing I do or read or calculate is going to have any effect on your understanding. Yes, there's a lot of symplectic geometry when you get knee deep in the mathematics of such things as phase space, where the canonical 2-form is $dp \wedge dq$ for p canonical momentum of q. I know such things arise and I know a bit of the detail when it relates to something I've worked on. So what? What should I be getting to the bottom of other than my work? Are you really surprised you had no clue how deep and detailed the work on such things goes? It literally fills libraries, I could read from dawn till dusk about symplectic geometry and I'd never learn it all. Most people don't ever do symplectic geometry, even people doing a maths or physics degree and likewise I haven't even heard of things which friends of mine might be doing every day as part of their research. Think of all the knowledge a 65 year old professor in maths or physics has and then realise that they haven't even heard of 99.9% of the rest of maths and physics. They write the textbooks you might not even have bothered to read in high school and yet you're surprised your grasp might not be as good as you might hope?!

23. ### DRZionTheoretical ExperimentalistValued Senior Member

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Thats an interesting question. Its hard to get a definite answer as to where thermal radiation comes from. Some people say its quantum energy levels spontaneously degenerating, others say its charged oscillators. The two theories aren't necessarily exclusive if I understand it correctly. In some sense it is charged particles accelerating back and forth according to the kinetic theory of heat. But I wonder if its also due to an electron orbiting a nucleus. Probably both, but electrons are not required for thermal radiation so that is only a portion. I have studied thermal radiation quite a bit