In my preceding post I have pointed out an interesting mathematicalquestion about the exact solutions of the scalar field theory that I use in this paper

given by

that holds for

If you compute the Hamiltonian the energy does not appear to be finite, differently from what the relation dispersion is saying. This is very similar to what happens to plane waves for the wave equation. The way out is to take a finite volume and normalize properly the plane waves. One does this to get the integral of the Hamiltonian finite and all amounts to a proper normalization. In our case where must this normalization enter? The striking answer is: The coupling. This is an arbitrary parameter of the theory and we can properly rescale it to get the right normalization in the Hamiltonian. The final result is a running coupling exactly in the same way as I and others have obtained for the quantum theory. You can see the coupling entering in the right way both in the solution and in the computation of the Hamiltonian.

If you are curious about these computations you can read the revised version of my paper to appear soon on arxiv.

Marco Frasca (2010). Mass generation and supersymmetry arxiv arXiv: 1007.5275v1

I do not understand how coupling can be arbitrary parameter. Does that mean it has no influence?

Also, I think that mass is a phenomenological parameter that is simply written as such by definition. Why one should “generate” mass? What if the “generated mass” does not correspond to any experiment?

No, not really. In a classical theory there is no constraint on the coupling. Mathematicians use to put it to 1 and this avoid to take it further in the computations. In order to make a coupling somewhat a different object one needs quantum field theory but otherwise it is seen to run and to be modified in different ways. Anyhow, one should remember that making a finite volume is a kind of artifact to make also computations finite in a situation where the solutions of the equations extend all over the specetime.

Mass is something that needs to be undersood being one of the fundamental objects of physics. And when Einstein reduced to energy we become aware that it was not a fundamental concept but something that could be reduced to more fundamental ones. But it belongs to our work of physicists to try to get in understanding as deeper as we can and whatever concept used in physics is exposed to this.

Finally, so far the generated mass depends on arbitrary parameters and fitting is not a real problem. But for QCD this would imply a failure of the theory. Since now, the agreement is excellent as obtained from lattice.

Hi Marco

I do not understand how coupling can be arbitrary parameter. Does that mean it has no influence?

Also, I think that mass is a phenomenological parameter that is simply written as such by definition. Why one should “generate” mass? What if the “generated mass” does not correspond to any experiment?

Hi Vladimir,

No, not really. In a classical theory there is no constraint on the coupling. Mathematicians use to put it to 1 and this avoid to take it further in the computations. In order to make a coupling somewhat a different object one needs quantum field theory but otherwise it is seen to run and to be modified in different ways. Anyhow, one should remember that making a finite volume is a kind of artifact to make also computations finite in a situation where the solutions of the equations extend all over the specetime.

Mass is something that needs to be undersood being one of the fundamental objects of physics. And when Einstein reduced to energy we become aware that it was not a fundamental concept but something that could be reduced to more fundamental ones. But it belongs to our work of physicists to try to get in understanding as deeper as we can and whatever concept used in physics is exposed to this.

Finally, so far the generated mass depends on arbitrary parameters and fitting is not a real problem. But for QCD this would imply a failure of the theory. Since now, the agreement is excellent as obtained from lattice.

Cheers,

Marco