# conservation of momentum II

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

1. ### DRZionTheoretical ExperimentalistValued Senior Member

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Here is a scenario I have cooked up that seems to violate the law of conservation of momentum.

Take two clumps of matter of equal weight. Give one a positive charge x, and the other a negative charge 2 x.

On a stationary spaceship ( p = 0 ) release the two charged clumps of matter. They accelerate towards one another according to coulomb's law. The acceleration on both is equal since forces are equal and opposite.

Both emit braking radiation (bremsstrahlung). But the negative charge is twice as large as the positive charge and so it releases twice as much energy due to bremsstrahlung. As the two clumps come into contact the positively charged clump has more momentum since it has not lost any kinetic energy to bremsstrahlung. Net momentum has been created.

Surely there is some crazy physics involved, because according to basic physics there is creation of momentum in this extremely simple scenario.

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5. ### rpennerFully WiredValued Senior Member

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Please submit mathematics demonstrating this point with a concrete example.
For example, you neglect that momentum is conserved in bremsstrahlung.
If momentum is conserved at every interaction, why do you feel momentum is not conserved in the aggregate of interactions?

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*snap*

8. ### DRZionTheoretical ExperimentalistValued Senior Member

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Yes, and it is p = e/c
e energy
p momentum
c speed of light

I can't really illustrate this mathematically, I would appreciate if somebody could.

It seems that both you and pete are getting at the same thing, that radiation carries away the momentum. The example is concrete. Momentum is not conserved in bremsstrahlung. Even if all the kinetic energy of the clump was emitted unidirectionally (which seems unlikely to me) it still could not carry away enough momentum.

The ratio of kinetic energy to momentum has no lower bound, but at most [kinetic energy : momentum] is [c : 1] . We know that a photon's [kinetic energy : momentum] is [c:1]. This is the maximum and it is only possible for massless objects. IE a photon has the lowest amount of momentum per unit energy as is allowed in physics.

If you apply conservation of energy to this, you will see that even if all the kinetic energy lost by the clump is emitted unidirectionally as photons it will still not offset as much momentum as was lost by the charged particles of the clump. It isn't possible according to conservation of energy.

Pretend the clumps each weigh 1 gram. Now take any amount of energy which is to be emitted as bremsstrahlung. If you take that energy and convert it to photons, calculate the momentum of the photons. Then calculate the momentum lost by the clump if it loses an equal amount of energy. They are not equal.

9. ### rpennerFully WiredValued Senior Member

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Various physics texts say you are wrong.

See Eberhard Haug, Werner Nakel. The elementary process of Bremsstrahlung (2004) p. 15

10. ### DRZionTheoretical ExperimentalistValued Senior Member

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Well I'm not sure if its all of bremssrahlung which violates conservation of momentum. Instead what I'm trying to say is that if one transforms kinetic energy to radiation momentum cannot be conserved.

If you take the above scenario and make the charges equal, momentum is conserved whether or not bremsstrahlung is emitted.

But if one is twice as large, it will lose twice the kinetic energy as radiation. This difference in energy emitted cannot be offset by the momentum of the departing radiation since photons have the MINIMAL momentum : kinetic energy ratio. Anything that has any mass will have a higher momentum : kinetic energy ratio than electromagnetic radiation, meaning it would have to lose more kinetic energy to radiation in order for that radiation to carry enough momentum, but this in turn would lead to more loss of momentum.

Last edited: Jun 21, 2010
11. ### AlphaNumericFully ionizedRegistered Senior Member

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Its easy to prove that the QFT model of bremssrahlung radiation does conserve momentum. By 'easy' I mean you can argue it without having to do any actual calculations, only know various results which pertain to QFT and conversation laws.

As per Noether's theorem a conserved quantity is associated to one and only one continuous symmetry in a model. The well known ones are conservation of linear momentum being related to spacial translation, angular momentum being related to rotations and energy being related to time translations. Thus if you construct a model which has an invariance under translations, $\mathbf{x} \to \mathbf{x} + \mathbf{x}_{0}$ then you are certain to have momentum conservation, if you do a calculation which doesn't then you've done your calculation incorrectly.

The QED Lagrangian is Poincare invariant, making it invariant under rotations, boosts and translations, thus has conserved momentum, angular momentum and energy. As a result all equations of motion derived from the QED Lagrangian are also Poincare invariant and have the same conserved quantities. Bremssrahlung radiation processes in QED thus conserve momentum. Getting bogged down in the actual calculations is not needed, that's the beauty and power of Noether's theorem.

12. ### DRZionTheoretical ExperimentalistValued Senior Member

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This isn't a very convincing proof. You are basically saying that my novel approach is wrong because it is novel. I would prefer it if someone addressed my scenario since that is what makes me believe momentum is not conserved.

In case momentum is conserved, there will have to be some momentum carrying particle which is created other than a photon. This would be more plausible and would still be enough for some kind of new propulsion technology which isn't necessarily recoil-less.

Going back to the scenario. Lets say the clumps start out with 100 joules of electric potential each. One loses 10 milli joules and the other loses 20 milli joules as radiation. This means that at contact one will have 99.990 joules and the other 99.980 joules of kinetic energy. .003 joules have been emitted as photons. This is hypothetical of course, as i can't really do the calculations for bremsstrahlung. The amount of radiation would likely be much smaller than this since bremsstrahlung scales to the ^4 or something like that.

Ke=1/2mv^2
P= mv
Pr = e/c

e/c
.003 j / c =
1e-8
Pcl (momentum that the clump lost) =
Ke/2m = v^2
vm = p
.003/2 = v^2
v = .0387 m/s
p = .0387

Pr = 1e-8
Pcl = .0387

Theres a difference.... no matter how much radiation is emitted, it cannot carry the momentum lost by the clumps of matter because photons carry the least amount of momentum per unit energy of all matter. 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.

13. ### AlphaNumericFully ionizedRegistered Senior Member

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I said nothing of the sort. I said that conservation of momentum is unavoidably built into quantum field theory, so no matter how you rephrase it you'll never be able to get a violation of that using quantum field theory without doing something like breaking Lorentz invariance. Experimentally momentum is conversed in bremssrahlung processes and its also conserved in the model we have of it. Where's the problem?

My argument applies to ALL quantum processes, as modelled by quantum field theory. By restricting yourself to a single example you're actually making life harder for yourself because you can't use powerful arguments like Lorentz invariance in a general way. By putting in numbers you've removed the ability to see underlying structure.

Yes, that's how the neutrino was originally predicted, a 'gap' in the momentum but that's not needed for bremssrahlung radiation in quantum field theory, it preserves momentum.

They are not Lorentz invariant expressions, they are Newtonian approximations. Yes, momentum is conserved in Newtonian mechanics for exactly the same reasons, translation invariance, but if you're dealing with energies which are enough to get an electron to close to the speed of light in the centre of mass frame then you're not able to use Newtonian mechanics. Furthermore Newtonian mechanics is not compatible with expressions like $E=mc^{2}$ or $E = hf$. And even in relativity $E=mc^{2}$ doesn't apply to the photon. You're applying a mishmash of equations from different models and confusing yourself.

If you make the mistake of thinking you can use $E=mc^{2}$ for the photon you're going to get nonsense answers. The full equation is $E^{2} = (mc^{2})^{2} + |\mathbf{p}|^{2}c^{2}$ where $\mathbf{p}$ is the objects usual 3-momentum. This is Lorentz invariant because $-(mc^{2})^{2} = p_{\mu}p^{\mu} = -E^{2} + \mathbf{p}\cdot \mathbf{p} c^{2}$. $E=mc^{2}$ is a special case. A photon has m=0 and thus its energy is $E = |\mathbf{p}|c$ which becomes $E=hf$ if you use the DeBroglie result $|p| = \frac{h}{\lambda}$. $E=mc^{2}$ is for a particle at rest, as you get it by setting $\mathbf{p} = 0$ in the full expression. A photon can never be at rest.

You're not doing something novel which is freaking people out, you're failing to realise that your very scattered knowledge of very simple expressions is insufficient to grasp what the models of bremssrahlung radiation actually involve.

14. ### AlphaNumericFully ionizedRegistered Senior Member

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By the way, bremssrahlung radiation is something which has been examined enormously because the emission of radiation by an accelerating charge is the biggest obstacle in accelerator design. Synchrotron radiation bleeds energy out of particle beams in circular accelerators, making it harder and harder to speed particles up even over and above the issue of relativistic speeds. Its the reason the LHC collides protons, not electrons. Colliding electrons and positrons is very 'clean' compared to colliding protons because protons are not fundamental and they spew out loads of mesons and baryons which waste collision energy and make it difficult to extract data. However, the proton is 2000 times the mass of the electron and thus emits much much less synchrotron radiation when carrying the same kinetic energy. The next generation of colliders will be linear, removing the vast majority of radiation loss. If we couldn't model bremssrahlung radiation very very accurately machines like the LHC and LEP would be impossible to build, we'd never how much force we need to apply (via electromagnets) to keep the beams inside the central tube. The sorts of energies a beam can carry at the LHC goes into the megajoules range.

I suggest you spend more time reading up on experiments and less time coming up with 'novel approaches' to problems which don't exist.

15. ### DRZionTheoretical ExperimentalistValued Senior Member

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For this reason I am not going to use quantum field theory but a combination of different theories , all in their extreme simplest forms.

Maybe you are right and this is why I have tried to simplify the process as much as possible. I will try to work it out until someone is convinced of something

The underlying argument is that a massive particle with finite kinetic energy and momentum cannot lose all (or any) of its kinetic energy solely as electromagnetic radiation without breaking either conservation of energy or conservation of momentum... a bold claim, I know.

Right, there is also thermal bremsstrahlung which is found in plasmas, whose ionized components move around at very fast speeds and bounce off of each other creating large accelerations and heavy bremsstrahlung. The thing is that in a cyclotron one of the charges is permanently FIXED in place by being anchored to the earth. Any discrepancy is completely unobservable. We can agree that an electron has different momentum after deflection, and that the momentum of the earth changes as well. But we also agree that momentum is conserved. But how are you going to measure the momentum of the earth? Its simply considered negligible (afaik...).

Furthermore, the scenario I am talking about is far from relativistic (no need to get into relativity) and different than a cyclotron because both charges are mobile and are acted upon by exactly the same forces except for the bremsstrahlung which should be very close to twice as large for one of the clumps.

It does not! A photon carries next to no momentum! It carries far less momentum than a bowling ball does per joule of energy which it carries. If the bowling ball loses energy as photons, then those photons carry less momentum than the bowling ball did! It doesn't matter whether its electromagnetic or kinetic energy, they are interchangeable as long as they are conserved.

I don't see a problem, I see an opportunity

For instance a propellant-less thruster of some kind.

Last edited: Jun 22, 2010
16. ### DRZionTheoretical ExperimentalistValued Senior Member

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you can further simplify the scenario by assuming you have an enormous charge-less massive body and a small ball on its surface which has a charge. by flinging the ball into the air it will emit bremsstrahlung when it is accelerated by gravity. the chargeless massive body will emit no bremsstrahlung and so it will hit the ball harder than the ball hits the massive body and so momentum is not conserved. You can't argue with that*. One will emit radiation while the other will not. One will lose energy while the other will not. If energy and momentum are at all related then you have a violation of something here.

*That is, if gravitational acceleration is enough to create bremsstrahlung; this i cannot be certain of, which is why i used the above scenario to illustrate my point.

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

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Hi DrZion,
I see what you are saying, but you are forgetting that the departing radiation must also carry away some mass/energy.
The clump of matter that emits the radiation loses mass as well as momentum, so you can't apply simple kinetic energy equations to get the energy difference.

18. ### DRZionTheoretical ExperimentalistValued Senior Member

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Hi Pete!

The amount of mass carried away as energy can be considered negligible. Its kind of like considering the mass of a bullet before and after it comes to rest... however, the amount of energy radiated due to bremsstrahlung of this kind will indeed be very very small and the energies may be comparable. In a particle accelerator velocities approach the relativistic limit and acceleration tend to be dramatic, hence large amounts of radiation is emitted. In the case of two electric charges attracting one another, it is quite different and the radiation is quite tiny. I don't know if it is larger or smaller or comparable to the mass difference due the radiated energy. I am not sure if e=mc^2 would be applicable in this case, maybe AN could help out.

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

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Then so can the momentum.

If you want to consider the momentum lost due to the radiation, then you can't ignore the mass lost.
It's hard to work the maths, but it appears to exactly balance out if you do it properly - which means not neglecting anything.

20. ### DRZionTheoretical ExperimentalistValued Senior Member

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Yes, this is true. in order to make the definitive proof one will have to know all the details or do the experiment. I guess I will try to convince people that the experiment is worth doing by showing that the evidence for my claim far outweighs the evidence against it. This is the only course of action I can take because i sure as hell don't know all the details!

What you are actually saying above is that the momentum disparity is greater than what I believed it to be before! By losing mass, the clump also loses momentum. This is an addition to the momentum lost due to reduction of velocity (kinetic energy). Meaning that the em radiation has to carry that much more momentum. But this means more mass is lost and more energy is lost, and more momentum is lost! I am not saying this is an endless cycle, but it does point to the fact that there is something weird going on here.

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

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Why are you convinced that your claim is correct, when you don't know all the details?

What I am actually saying is that if you don't work it out properly, you're just waving your arms and wasting your time.

22. ### AlphaNumericFully ionizedRegistered Senior Member

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Mashing together different models and making simplifications is almost certainly going to give you a non-exact model. Newtonian mechanics is much simpler than relativity but if you try to apply it to objects moving at 0.9c then you'll get incorrect answers. You're doing that, misapplying models and yet complain its not your fault. QED, which is the defacto model of the radiation in question, doesn't violate momentum conservation. All other complaints are mute.

And its a false claim. Experiment and theory both disagree with you. If you have a model which says otherwise then it contradicts experiment and thus is wrong. Simply because you can make a prediction doesn't mean its right.

How is this difficult to grasp, if you have an incorrect theory then you shouldn't trust its predictions!!

No, the force the accelerator applies to the beam is precisely measurable and thus the momentum changes are measurable. You're making claims about experiments you simply know nothing about.

And you think no one has done experiments with these? The simplest examples of QFT calculations done as homework involve bremsstrahlung processes. And before you complain about mobile charges, it arises in electron-electron scattering processes inside colliders or as positron-electron processes, when they don't collide but deflect one another. That's something LEP saw in the billions. You're not talking about some unexplored phenomenon, you're talking about the bread and butter of colliders.

Wrong. Flat out fucking wrong. Experiments say you're wrong and working models say you're wrong. Electrons only interact with one another via photons. An electron bouncing off another electron doesn't mean they physically hit one another, they do it by exchanging photons and so ALL of the momentum change an electron will experience is carried by photons!!! ALL of it (well, until you get into the 90GeV range and you get Z bosons too).

Then there's the photo-electric effect, the original experiment which demonstrated the ballistic nature of photons. A single photon has enough energy and momentum to knock an electron out of the metal and move through an electric potential. Even Maxwell's classical electromagnetism has the photon carrying plenty of momentum at high enough frequency. You can see it carries plenty of momentum by considering ionizing radiation. If gamma rays didn't have enough momentum to knock electrons out of bonds in our body then we'd not need to worry about radiation like that! Instead the fact the photon has enough energy is the reason why its deadly.

Learn some fucking science before shooting your mouth off.

You aren't going to get a grasp of relativistic quantum mechanics by thinking about purely Newtonian classical mechanics. Your grasp of this is very very bad. That in itself wouldn't be a problem, we all were like that once, but you seem unable to accept any correction from anyone who has done this kind of physics. You're denying experimental reality.

Your problem is that you don't want to consider you might be incorrect on something. There's a reason quantum field theory isn't taught till the 4th year (at the earliest!) of physics or maths degrees, its difficult. If someone whose spent 3 years doing the basic requirements struggles with it you have no chance. Is it too much to expect you to behave mildly rationally and think "Hmmm, I don't know any of the basics of this and people who do are telling me I'm wrong. Perhaps I should do some more reading and learning before making claims about this thing I know nothing about....." ?

Last edited: Jun 22, 2010
23. ### DRZionTheoretical ExperimentalistValued Senior Member

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Thread closed??! This is fascism! I'm mining for the truth here!

It wouldn't be much of an argument if I didn't even believe in what I was saying right?

And I am leaving room for discussion - it could be the emission of another momentum carrying particle and not violation of conservation.

I'm hardly using any models. I'm just using simple newtonian mechanics for objects which will move much more slowly than .9 c. And the momentum of a photon I take for granted as a fact, something I read on the internet. How else would I treat the momentum of a photon? Its simply energy divided by the speed of light.

Is not!

This is slightly different. In the electron-electron scattering processes virtual photons are intermediates, they are not the final momentum carriers. You basically have transfer of momentum between identical bodies.

I'm not saying they have zero momentum, im just saying they have the lowest momentum/energy ratio. Of all matter. They are incredibly poor momentum carriers! Thats why radiation thrusters are so terribly inefficient!!

I'm not trying to go into any of that, it is FAR beyond my capabilities. I am really only dealing with the following-

newtonian mechanics (this is easy enough for me to understand)
photon momentum (something which is inarguable; it is in fact energy/c)
bremsstrahlung (also something which is inarguable; accelerating charges release energy in the form of radiation)

Can you disagree with any of the above? I think that my assumption about bremsstrahlung may be incorrect. Is it possible no bremsstrahlung will be emitted when a charge is accelerated in the wrong circumstances? The way the wiki states it it is only dependent on charge and acceleration.

Last edited: Jun 22, 2010