This isn't a very convincing proof. You are basically saying that my novel approach is wrong because it is novel.

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?

. I would prefer it if someone addressed my scenario since that is what makes me believe momentum is not conserved.

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.

In case momentum is conserved, there will have to be some momentum carrying particle which is created other than a photon.

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.

It is also impossible for the hypothetical momentum carriers to be massless.

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.