# About bare and physical mass, Juan R

1. Sep 13, 2005

### EL

Since my discussion with Juan R in the thread "photon's mass is zero?" under Special and General relativity went off topic I will try to continue it here:

As I wrote before (#88) books often start with assuming that m in the Lagrangian is the ordinary mass (i.e. the one you can find in tables) just to later find out that this leads to infinities when calculating higher order processes.
Then this problem can be solved by noticing that if we in the Lagrangian substitute m with the bare mass m0 instead, the amplitudes turn out to be finite when we express them in terms of the physical mass. (Of course we also have to do a charge renormalization, but let's just stick to the mass for simplicity.)

Hence the correct Lagrangian density should include the bare mass (as well as the bare charge), and not the physical (since that leads to infinities). However, all results will of course be expressed in terms of the physical mass (i.e. the one we find in tables).

Please could any mentor or advisor verify or crank down on what I am saying, so we can get an end to this...

2. Sep 13, 2005

### Physics Monkey

EL,

I haven't been following the discussion in the other thread, but I think I can answer your question. The Lagrangian is written in terms of the bare mass, bare field, etc. (i.e. the nice pretty looking Lagrangians in your book always refers to the bare quantities). Let us focus on the behavior of the bare mass. The physical mass of a particle is usually defined in terms of the pole of the propagator, and at tree level the pole is located at the bare mass. Therefore, at tree level the bare mass is the physical mass. However, if the propagator is evaluated beyond tree level then one finds that the pole shifts. In other words, the pole of the propagator is not given by the simple parameter m^2 that appears in the Lagrangian. Mass renormalization is then the procedure whereby we correct the pole structure of the propagator so that the propagator maintains its pole at the physical mass.

The really interesting thing to me is that renormalization is not inherently associated with removing infinities. Even in a theory where all momentum integrals converged, mass renormalization would still be necessary because you still have to correct the pole.

If I may, let me recommend Weinberg's magnificent text on quantum field theory to the interested people out there. Everything is wonderfully clear. In particular, an excellent discussion of this very subject (including a reference to the importance of renormalization apart from infinities) can be found around p. 438 in volume 1.

3. Sep 14, 2005

### EL

Juan R, have a look at this:

I will try Weinberg as soon as I get some time over.

4. Sep 14, 2005

### EL

So in that case the relation beween bare and physical mass won't include any "infinities"?

5. Sep 14, 2005

### Juan R.

Thanks by continue this interesting discussion!

My point is as follow. The mass of an electron is m, its rest mass, which appears in handbooks or tables of universal constants.

The Lagrangian contains that m. I already cited to you three textbooks. In Weinberg, the lagrangian of QED appears in volume 1 equation 8.6.1. It contain m.

Then, there are problems with infinites in the computation of interactions and that mass may be changed via renormalization procedure. However "physical" mass after procedure and measured exp is not the mass of the electron "nude". It is really the mass of the electron nude more cloud of virtual particles surrounded the electron. The same about the physical charge.

I also cited a book on quantum physics where this problem of changing of mass -mass before renormalization is different from mass after it- is claimed to be one of main flaws of current theoretical physics. The infinites are artificial.

6. Sep 14, 2005

### EL

Yes but that is the physical mass.

Yes but that is the bare mass.

However, as Physics Monkey pointed out, at tree level they are the same, and that's why it's possible to calulate first order processes without encountering infinities. But for the theory to be consistent at higher order corrections you need to start from a Lagrangian where m is the bare mass. Hence the correct Lagrangian is written in terms of the bare mass (and charge).

You can cite as many books as you want, the problem is that it seems you don't understand them.

7. Sep 14, 2005

### Physics Monkey

Juan R,

Unfortunately, Weinberg is being a bit careless here. He does not specify whether this is the physical or bare mass. As I said, it is true that the the paramters of the Lagrangian correspond to the physical particle parameters at tree level. Notice that he does not calculate anything beyond tree level in Ch. 8, so his statement is harmless at that point: physical and bare mass are the same. However, please note eq. 11.1.1 in Weinberg where he explictly indicates that the mass, etc. in the Lagrangian you are talking about is the bare mass, etc. He proceeds to break the Lagrangian into a term which looks like the free Lagrangian (but now written with physical paramters), a renormalized interaction term, and the renormalization counterterms.

EL,

You right about the corrections being finite if the momentum integrals are all finite. All this means is that in any interacting field theory, the bare charge and mass get 'dressed' by the interaction.

8. Sep 14, 2005

### EL

Great. Could you give an example of a theory where this happens? I would guess this could occur when using quantum field theory in solid state physics?

9. Sep 15, 2005

### vanesch

Staff Emeritus
Just a guess: any finite field theory would be good enough, no ? Like phi^4 in less than 4 spacetime dimensions ?

10. Sep 15, 2005

### Juan R.

No, you are not fixing my point of the change of mass!

I already cited several books on the topic saying the same i said, including the "Fisica quantica" where clearly states that re-definition of mass and charge is one of flaws of QFT that nobody has solved.

I began again now from Weinberg

Equation (8.6.2) for Lagrangian (i ignore the field "FF" contribution by comodity)

$$\mathcal{L} = - \overline{\Psi} (\gamma^{\mu}[\partial_{\mu} + ieA_{\mu}] + m) \Psi$$

$$m$$ is the rest mass that appears in special relativity, in Maxwell electromagnetism, and in the Dirac equation.

Now, this does not work well for the computation of higher orders in scattering, and then one may ad hoc change it a posteriori.

Then the (11.1.1) is (again I omit "FF" terms)

$$\mathcal{L} = - \overline{\Psi_{B}} (\gamma_{\mu}[\partial^{\mu} + ieA^{\mu}] + m_{B}) \Psi_{B}$$

where the new variables are related to previous one via (11.1.2), (11.1.3), etc. For example the mass is (11.1.5)

$$m \equiv m_{B} +\delta m$$

But i maintain again my initial points: the bare mass $$m_{B}$$ is not identical to initial $$m$$ mass that appears in the lagrangian. Note also that $$m$$ does not disappear of Lagrangian after of the use of bare constants even if equation (11.1.1) suggests that at the first look. In the (11.1.7) the free Lagrangian is defined in terms of $$m$$. instead of $$m_{B}$$.

Also the E.g. equation 12.1.1 of Michel le Bellac. Quantum and statistical field theory. Oxford university Press, 1991. i cited begin with $$m$$ and change after to $$m_{B}$$.

Also the J. Sanchez Guillen and M. A. Braun. Física cuántica. Alianza Editorial S.A., 1993. does and add in its pag 362.

It is saying that [my translation]

In fact, that Weinberg is doing in chapter 11 that Physics Monkey cited is adding ad hoc counter terms via $$\mathcal{L}_{2}$$ related to new renormalized constants Z, etc to the initial lagrangian (8.6.2).

It was my point in the other thread, that one may begin with physical rest mass and after changes it ad hoc if one want compute higher orders, if one want compute lower orders, one may use the Lagrangian (8.6.2). In fact, if one want obtain the Dirac equation one may use the (8.6.2) with $$m$$, and i do not see because i would am wrong. Please any comment!

Last edited: Sep 15, 2005
11. Sep 15, 2005

### EL

Yes. But why not start from the same Lagrangian in both lowest order and higher order calculations? Since both the Lagrangian with the physical mass and the Lagrangian with the bare mass gives the same result for tree diagrams, which one would you say is the "correct" one? The one which is correct to all orders, or the one which only works in a special case?

12. Sep 15, 2005

### EL

That's fine to me.

13. Sep 15, 2005

### Physics Monkey

EL,

Most field theories in condensed matter still have divergent integrals, but the cutoff is a very physical thing. The lattice spacing is the natural limit to the high energy behavior of the system. As vanesch said, you usually need to look at field theories in low space-time dimensions to see examples of finite theories.

Juan R,

You are correct that bare mass and physical mass are not the same. What is also true is that the Lagrangian,
$$\mathcal{L} = - \overline{\Psi} (\gamma^{\mu}[\partial_{\mu} + ieA_{\mu}] + m) \Psi - \frac{1}{4} F^{\mu \nu} F_{\mu \nu}$$ does not describe particles of mass m. One finds that at tree level approximation, the theory does contain a particle of mass m. However, as the interaction energy is included in higher order loop calculations, the mass of the particle predicted by the theory is altered. Now we want the mass of the particle predicted by theory to be the mass we observe, so we renormalize. We adjust the parameters in the Lagrangian in such a way that the pole of the propagator is always located at the physical mass no matter what order of perturbation theory we are doing. The confusion stems in part from the fact that the electron we observe is not really described by a free Lagrangian but rather is described by the interacting Lagrangian "summed to all orders". The electron we see is already dressed by electromagnetic interactions.

14. Sep 15, 2005

### Juan R.

I'm sorry but I do not understand to you.

If you want compute self-reaction effects and radiative corrections correctly you may work with

$$\mathcal{L} = - \overline{\Psi_{B}} (\gamma_{\mu}[\partial^{\mu} + ieA^{\mu}] + m_{B}) \Psi_{B}$$

but if want compute energy spectra of H atom to the Dirac level of precision (without self-reaction), the Dirac-like equation for the electron field $$\Psi(x)$$ is

$$(\gamma_{\mu}[\partial^{\mu} + ieA^{\mu}] + m) \Psi(x) = 0$$ (1)

not

$$(\gamma_{\mu}[\partial^{\mu} + ieA^{\mu}] + m_{B}) \Psi(x) = 0$$ (2)

or similar

unless you assumed a priori that $$\delta m = 0$$ and $$Z_{2} = 1$$ and then counterterms from (11.1.1) are cancelled. But then you are assuming that $$m_{B}$$ and $$m$$ are the same and by using (2) you are really using (1) which is the correct.

Renormalization may be done order by order. There is not general prescription. Moreover the correction of mass cannot be computed, only obtained via experiments. I always thought that correct mass was $$m$$ and introduction of other mass was a "mathematical trick" for accounting effects like interaction of electron with itself and polarization of vacuum. I think that is the reason that all textbooks I know begin with Lagrangian defined in terms of rest mass of the electron $$m$$ and only in precision computation change the mass, charge used.

Last edited: Sep 15, 2005
15. Sep 15, 2005

### Juan R.

Yes i agree. This was precisely my point in the other thread. The mass and charges of electron nude are altered due polarization of vacuum. E.g. the induced virtual cloud around the electron modifies his initial mass. Electron when moves may also move the cloud surrounding it and his inertial properties vary.

However, standard QFT states that only physical electron is the observed electron, which is the "dressed electron". I claim that

dressed electron = nude electron + virtual cloud

In fact, standard QFT claims that $$\delta m$$ has no physical sense, since virtual cloud form part of that QFT calls the "observed electron". I think that above decomposition between electron and polarization of vacuum would offer full physical value to $$\delta m$$

Last edited: Sep 15, 2005
16. Sep 15, 2005

### vanesch

Staff Emeritus
I agree that this there is a temptation to do so, especially in theories which have small coupling constants (such as QED) and where successive orders seem to have physical meaning (so that the lines in a Feynman diagram seem to have some physical meaning). However, I think it is fundamentally misleading. After all, what do we have ?

We have a theory that is supposed to crank out quantum amplitudes for different measurements (usually scattering experiments) as a function of a few parameters, here mB and eB. It turns out that certain behaviours of those quantum amplitudes are very similar to those of a free field theory, or have in other ways behaviours which make us think of classical particle theories. So we identify certain approximate properties of these quantum amplitudes (in certain limiting conditions) as defining something we call "the physical mass" mP or "the physical charge" eP of the particle. This comes down in setting up an experiment (satisfying the said limiting conditions) to extract these quantities from the experimental results (which are predicted by the quantum amplitudes). Let us for the moment assume that our theory is finite.

This means that the theory gives us a function f1(mB,eB) which gives us mP and f2(mB,eB) which gives us eP:

mP = f1(mB,eB)
eP = f2(mB,eB)

We could then use our experimental knowledge of mP and eP to fix the parameters mB and eB.

However, in the perturbative approach, we introduce an extra perturbation parameter lambda in our theory, and do a series devellopment wrt lambda. So we've now introduced a new function f1(mB,eB,lambda) such that:

f1(mB,eB) = f1(mB,eB,lambda=1), and we write the second term out in a series in lambda:

f1(mB,eB,lambda) = f1_0(mB,eB) + lambda f1_1(mB,eB) + lambda^2 f1_2(mB,eB) + ...

It now turns out that f1_0(mB,eB) = mB

So we have that to zeroth order, mP is equal to mB.

In the same way, we can inverse the relation, and use mP as an input. mB will now be a function of mP and eP:

mB = g1(mP,eP) (the inverse of f1 and f2)
eB = g2(mP,eP)

In a similar series devellopment, we now have:

mB = g1_0(mP,eP) + lambda g1_1(mP,eP) + ....

and it turns out that g1_0 = mP.

All other quantum amplitudes, for other experiments, are of course just a function of eB and mB: A(mB,eB) = A( g1(mP,eP), g2(mP,eP) ) = A{mP,eP}

It is a priori not clear to me why the mathematical trick of writing f1 as a series in lambda, should give physical meanings to the different orders in lambda.

In the case of infinite (but renormalizable) theories, the function f1(mB,eB) is ill-defined, so we introduce an extra parameter C (cutoff). With C, the theory becomes finite, and we are again in the same situation as above, only now with f1(mB,eB,C).

This means that we can have a g1(mP,eP,C) ...

It also means that OTHER quantum amplitudes, A(mB,eB) now become a function of C too: A(mB,eB,C), and the trick of renormalization is that:

A( g1(mP,eP,C), g2(mP,eP,C), C) is, in the limit of large C, asymptotically not dependent anymore on C. If that's true for all A, we say that the theory is renormalizable, because it means that in the end, A is only a function of mP and eP.

The point I'm trying to make is that it is somehow a coincidence that the 0-th order terms have mP and mB coincide. It is not really a coincidence, because these parameters where choosen so that they corresponded to the corresponding parameters in free field theories, which themselves were of course designed to describe free particles with mass mB for instance, so we shouldn't be surprised in fact that this comes out of it again when we look at zeroth order. But there's nothing in fact wrong or magical that the approximate property we understand as "physical mass" is a complicated function of the parameters of our theory.

17. Sep 15, 2005

### EL

Ok, now I'm lost again. Question to you who know this subject well:
Who is right, me or Juan R? Or are we partly right both of us? Or maybe none of us?

18. Sep 15, 2005

### vanesch

Staff Emeritus
My opinion is that you are right, but Juan is also right of course if you stick to lowest-order interactions because both are the same. On top of that, there's an extra confusion if you ask different people, because of the technique of counter terms in the Lagrangian. In that case, you keep the physical mass and charge in the lagrangian, but you add correction terms to the lagrangian. So people doing that would say that you have the physical mass in the lagrangian, not the bare one.

19. Sep 15, 2005

### EL

Nice to here that! (And of course Juan is right to lowest order, I never doubted that.)

Yes sure it's possible to express the Lagrangian in terms of the physical mass if we add correction terms instead, but I don't think that is the case we have been discussing, right Juan R?

20. Sep 16, 2005

### Juan R.

Are you claiming that, in rigor, there are not particles?