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Consider a free theory with one real scalar field: $$ \mathcal{L}:=-\frac{1}{2}\partial _\mu \phi \partial ^\mu \phi -\frac{1}{2}m^2\phi ^2. $$ We write this positive coefficient in front of $\phi ^2$ as $\frac{1}{2}m^2$, and then start calling $m$ the mass (of who knows what at this point) and even interpret it as such. But pretend for a moment that you've never seen any sort of field theory before: if someone were to just write this Lagrangian down, it's not immediately apparent why this should be the mass of anything.

So then, first of all, what is it precisely that we mean when we say the word "mass", and how is our constant related to this physical notion in a way that justifies the interpretation of $m$ as mass?

If it helps to clarify, this is how I think about it. There are two notions of mass involved: the mathematical one that is part of our model, and the physical one which we are trying to model. The physical mass needs to be defined by an idealized thought experiment, and then, if our model is to be any good, we should be able to come up with a 'proof' that our mathematical definition agrees with the physical one.

(Of course, none of this at all has anything to do with this particular field theory; it was just the simplest Lagrangian to write down.)

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"But pretend for a moment that you've never seen any sort of field theory before: if someone were to just write this Lagrangian down, it's not immediately apparent why this should be the mass of anything." - It's a basic feature of science that one is actually allowed - and encouraged - to use the brain and learn things that aren't obvious at the beginning. In the QFT above, one analyzes physics and finds that it contains particles whose mass is $m$. What's the problem? Are you serious that you would want all things in science to be obvious from the beginning and without thinking? –  Luboš Motl Mar 14 '13 at 19:35
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@Luboš I think Jonathan is trying to shed some light on the black box represented by "one analyzes physics" in your comment. IMO it's an excellent question. –  David Z Mar 14 '13 at 19:42
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I agree with David that it's an excellent question. @LubošMotl Your question "Are you serious that you would want all things in science to be obvious from the beginning and without thinking?" is an inaccurate, unproductive manipulation of statements made by the OP. –  joshphysics Mar 14 '13 at 19:46
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Related: physics.stackexchange.com/questions/9000/… –  John Rennie Mar 14 '13 at 20:02
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I would probably go with QFT=>Rep of Poincare alg.=>Casimir invariant $m^2$. Single particle interpretation: $P^2=m^2$ on shell, and we know that $m$ here is precisely the invariant mass. I don't know anything deeper. The parameters in QFT are given their specific names because they correspond to the appropriate things in the appropriate limits (classical and non relativistic). I think, though, the particle interpretation is key to calling that Lagrangian parameter "mass". –  twistor59 Mar 15 '13 at 7:48

3 Answers 3

There's two (ultimately related) answers.

For the first answer, just forget about $\hbar$ (but say $c=1$), we are doing a classical relativistic field theory.

The first is that you can consider the field profile around a static, spherically symmetric source of mass $M$ (you need to add a coupling to the action of the form $g \phi J$, where $J$ is an external source, to do this).

So we write the static spherically symmetric source as

$J=J_0 \delta^3(r)$

(if you don't like delta functions you can make it a top hat or a shell and you will find the same conclusions below, this kind of manipulation should be familiar from E/M classes)

You can work out the equations of motion for $\phi$, you will find

$\nabla^2\phi + m^2 \phi = g J_0 \delta^3(r)$

This looks exactly like Poisson's equation, except with this extra $m^2$ term (which I haven't called a mass yet). If you solve it (say using green's functions, or you could just make an explicit ansatz) you will find the solution is

$\phi = g J_0 \frac{e^{-m r}}{4\pi r}$

This is the famous yukawa potential with mass $m$. It describes a force with a range $m^{-1}$, which you may have heard before is the smoking gun for force carrying particle having mass (the weak interactions are so weak at human distances because $m$ is so big)

Now at this point you are still justified in asking why we are calling this quantity $m$, which in the classical analysis above is just an inverse length, a mass.

The second answer addresses this more directly. It comes when we quantize the theory. This is discussed in great detail in many qft textbooks, but the gist is relatively simple. At this point I will say $\hbar=1$, if you really want to keep track of the $\hbar$ dependence you can get it back by dimensional analysis.

If you take the equations of motion for the above system without the source (note we are no longer working with static systems):

$-\partial_t^2\phi + \nabla^2 \phi + m^2 \phi^2 = 0$

and go to fourier space

$\tilde{\phi}(\vec{k},t)\equiv\int\frac{d^3 k}{(2\pi)^3}e^{i\vec{k}\cdot\vec{x}}\phi({\vec{x},t})$

then you will find the equations of motion are

$-\partial_t^2 \tilde{\phi} +\vec{k}^2\tilde{\phi} + m^2\tilde{\phi}=0$

This is just a set of harmonic oscillators, labeled by $\vec{k}$ with frequency

$\omega^2=\vec{k}^2+m^2$

We can think of $\tilde{\phi}(\vec{k},t)$ for a single value of $k$ as being a single quantum variable obeying a harmonic oscillator equation, so there when we quantize these variables we will have one wave function for each value of $\vec{k}$, each of which obeys a schrodinger equation for a harmonic oscillator with the above frequency. Each harmonic oscillator will be quantized and have discrete energy levels. These energy levels are what we call 'particles'. The idea is that for a given momentum, the field $\phi$ can only have certain values of energy, and the interpretation is that these quantized wiggles in $\phi$, or in other words discrete packets of energy, are particles.

Now we remember we are doing quantum mechanics, and we are supposed to identify frequency with energy and momentum with wavelength, using

$E=\omega$, $\vec{p}=\vec{k}$

So we are describing a system with a relationship between energy and momentum given by

$E^2=\vec{p}^2+m^2$

This is einstein's famous formula, and now we see that the parameter $m$ appears in exactly the place the mass of a particle would appear. It is the energy that a particle has when it has $0$ momentum. So we identify the particle like excitations we discovered above with particles of mass $m$.

However I emphasize that there are many, many, many other treatments of this and other ways of looking at it.

A good question would be, what does a particle mean if you don't have a free theory and don't get a harmonic oscillator equation above?

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The concept of mass is related to the 4-momentum, or 4-momentum density. It is not just a “quadratic term” in the Lagrangian.

By varying Lagrangian one can obtain the equations of motion. But the momentum, or momentum density need to be defined independently from Lagrangian, in order to enable one to introduce the concept of mass or mass density.

For instance, the momentum can be defined as the sum of two isotropic 4-vectors constructed from spinor and co-spinor:

$\xi = \begin{vmatrix} \xi^1 \\ \xi^2 \end{vmatrix}$ - spinor

$\eta = \begin{vmatrix} \eta_{\dot1} \\ \eta_{\dot2} \end{vmatrix}$ - co-spinor

$p_\mu = \frac{1}{2} (\xi^{+}\sigma_\mu \xi)$ - covariant vector

$\tilde p^\mu = \frac{1}{2} (\eta^{+}\tilde\sigma^\mu \eta)$ - contravariant vector

The vector constructed from spinor is covariant, while the vector constructed from co-spinor is contravariant. This is due to different transformations properties of spinor and co-spinor. Both $p_\mu$ and $\tilde p^\mu$ are isotropic (i.e. $p_\mu p^\mu =\tilde p_\mu \tilde p^\mu =0$), but their sum is not isotropic:

$P_\mu = p_\mu +g_{\mu\nu}\tilde p^\nu $ - momentum or momentum density - depending on the model.

The mass, or mass density, is therefore defined as the "square" of the momentum (density) vector:

$m^2=P_\mu P^\mu$

In general, this definition of mass is independent from the choice of the Lagrangian.

However, the momentum and/or momentum density cannot be defined arbitrarily. It has to be defined in such a way that

  1. The total momentum is conserved (as a consequence of the equations of motion),
  2. The components of the momentum 4-vector are all real-valued, and
  3. The momentum 4-vector is time-like to preserve causality.

In general, we need two ingredients to say that mass in our model is really a mass:

  1. Lagrangian, or at least equations of motion, and
  2. Appropriate definition of momentum or momentum density (see conditions 2 and 3 above).

The conservation of momentum is required to be the consequence of the equations of motion (except for the Theory of gravity where conservation of momentum is an identity!).

It is also easy to demonstrate that we do not necessarily need the “quadratic” and/or constant mass term in the Lagrangian in order to introduce the concept of mass. Here you can find an example of the model where the “mass term” is variable and complex valued, but still allows for real-valued “mass square” of the momentum density vector.

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We can define mass this way, and I already know how to relate this definition to the term appearing in the Lagrangian. The question is, how do we relate this mathematical definition of mass to a precise, physical, operational definition of mass. Does that make sense? –  Jonathan Gleason Mar 15 '13 at 12:48
    
Please explain what in your opinion is "precise, operational definition of mass". To me the only relativistic physical definition of mass is the square of 4-momentum. –  Murod Abdukhakimov Mar 15 '13 at 13:19
    
Ironically, the word "precise" here is not meant to be precise; it is open to interpretation. The word that really matters here is "operational": to define mass via some sort of (thought) experiment. In section 2.2 of academia.edu/829613/… , I give a classical definition in the spirit of which I am looking, except now, I want to do this in a relativistic setting. I never thought of this until just now, but maybe the idea behind the classical definition could just be modified? –  Jonathan Gleason Mar 15 '13 at 13:27
    
If it helps to clarify, this is how I think about it. There are two notions of mass involved: the mathematical one that is part of our model, and the physical one which we are trying to model. The physical mass needs to be defined by an idealized experiment, and then, if our model is to be any good, we should be able to come up with a 'proof' that our mathematical definition agrees with the physical one. Does that make sense? –  Jonathan Gleason Mar 15 '13 at 13:30
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I think I understand your question now. Think of the total energy of two photons resulting from annihilation of particle and its antiparticle in the rest frame of their center of inertia. Energy of photon can be expressed (and measured) via its wavelength in difraction experiment. The total energy will be $2m$. –  Murod Abdukhakimov Mar 15 '13 at 14:00

I've been thinking about this question on and off since it was posted, and I have some thoughts that I hope will shed some light on this stuff. I think it helps to begin with the following:

An Analogy from Mechanics.

Consider the following expression that you'll often encounter in classical mechanics: $$ L(x, \dot x) = \frac{1}{2}m\dot x^2 $$ and let's say that you've never seen Lagrangian mechanics before. Someone comes and tells you that this Lagrangian "describes" a free particle of mass $m$ moving in one dimension. You could then ask the question "how can we identify the parameter $m$ that appears in this expression with the physical mass of a particle?

How would you answer this question? This isn't just meant to be rhetorical, I think it would help to think about this before reading on!

My thought process in answering this question is as follows. Well, we want to identify the parameter $m$ with the physical definition of mass in classical mechanics. How do we define mass operationally in classical mechanics? Well, we exert a force on the mass, and we see the extent to which it accelerates. In other words, we need to interact with the particle and watch what happens to determine its mass. Next, we return to the Lagrangian above. If it were able to predict how a particle reacts to an interaction, then we could use it to identify the parameter $m$ as the mass of a particle.

Unfortunately, the free particle Lagrangian makes no such dynamical predictions. The corresponding Euler-Lagrange equations lead to the equation of motion $$ \ddot x = 0 $$ which tells us nothing about how a particle will react to an interaction.

Now, you might try to identify $m$ as the mass by defining the energy and momentum in this mathematical model to be the generators of time and space translations respectively and then say "oh look, these expressions correspond to the expressions for energy and momentum of a non-relativistic particle that we're used to, so $m$ must be the mass!" I don't think there is much content or validity to this argument, because making these "definitions* in no way allows one to identify the parameter in the Lagrangian with the physical mass operationally defined in terms of interactions.

In order to identify $m$ as the mass, we need to put terms in the Lagrangian that correspond to interactions (like a potential term), see what the Euler-Lagrange equations of motion predict about the resulting dynamics, and then compare to experiments to identify $m$ as the mass.

Quantum Field Theory.

Using these analogies, I feel dissatisfied by user20797's response where we obtains the relativistic energy-momentum relationship for a particle because as far as I can tell, that procedure amounts to little more than making definitions for energy and momentum that give you the relation you want. It does not tell you how to relate the parameter $m$ to some empirically defined mass, and that, I feel, was your original (quite excellent/subtle in my opinion) question. I think his first response is probably closer to what is necessary.

However, it seems to me that a proper answer to your question requires an analysis of the following (schematic) form whose details (especially empirically), are quite non-trivial.

  1. We note that the mass of particles can be defined by their interactions with one another. In particular, we can use scattering experiments to bring about these interactions.

  2. We add interaction terms to the Lagrangian density you wrote down, and we develop a prescription by which the resulting quantized field theory can predict what will happen in scattering experiments.

  3. We compare the results of our scattering experiments to the predictions of the quantum field theory, and we find that we can identify the parameter $m$ in the Lagrangian density as the physical mass as it was defined through scattering experiments.

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"In order to identify m as the [m]ass, me need to put terms in the Lagrangian the correspond to interactions (like a potential term), see what the Euler-Lagrange equations of motion predict about the resulting dynamics, and then compare to experiments to identify m as the mass." Agreed that this would be a good operational definition of mass. I suggested this in a comment on the question. You should be able to derive $F=ma$ through some straightforward steps in the appropriate limits. –  Michael Brown Mar 26 '13 at 13:10
    
@MichaelBrown Woops thanks :). The more I thought about this, the more I felt dissatisfied with "the standard dreck" in which one shows that one can reproduce the appropriate free particle spectrum and energy-momentum relationship because it doesn't make contact with some such sensible operational definition of mass via dynamics e.g. scattering. –  joshphysics Mar 26 '13 at 13:18

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