Firstly I'll say that I know the current standard terminology is to just refer to "mass", but I wanted to be clear about what I was talking about.

I've heard that in standard model of physics (which is compatible with special relativity), it is necessary to treat mass as an emergent property of a particle. I've done some quantum mechanics and some special relativity, but no QFT yet. So where does rest mass come from? What is it fundamentally? We've all seen that $E=mc^2$ (or $E^2 = m^2c^4 + p^2c^2$) , but most intuitions I have for energy come from the Newtonian domain. Is energy (or momentum-energy) the new first class citizen in relativity from which all else is derived, or is something more subtle going on here? What are the grounding base concepts?

I have a decent understanding of 4-vectors and the geometry of spacetime, it's just once we start putting things into it that the confusion starts.

  • 1
    $\begingroup$ have a look at the Higgs-mechanism $\endgroup$
    – Densch
    Commented Apr 9, 2018 at 7:25
  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/64232/2451 , physics.stackexchange.com/q/73225/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Apr 9, 2018 at 8:18
  • $\begingroup$ @Densch The Higgs mechanism is not responsible for mass in general. It only explains mass of gauge bosons. We have no explanation for why particles have mass to begin with. $\endgroup$
    – Apoorv
    Commented Apr 10, 2018 at 12:04
  • 2
    $\begingroup$ @ApoorvKhurasia In the Standard Model, the Higgs gives mass to the quarks, electrons, etc too. I agree that hadron masses, i.e. proton, pion, etc, aren’t from the Higgs mechanism. $\endgroup$ Commented Apr 10, 2018 at 13:33
  • $\begingroup$ @BobJacobsen Thanks. Yes, I should have said that instead. $\endgroup$
    – Apoorv
    Commented Apr 10, 2018 at 14:24

3 Answers 3


The beauty of special relativity is that seemingly disparate concepts from Newtonian physics (such as space and time) are seen to be deeply linked and constrained.

In Newtonian physics mass is just an axiomatic property of particles. Energy and momentum are introduced at an elementary level as separate conserved quantities in closed systems.

A deeper understanding of energy and momentum, even at the Newtonian level, is that they are conserved quantities associated with the symmetries of your system under time translation and spatial translations respectively. Loosely: if it does not matter whether you do the experiment now or later, then there exists an abstract quantity called "energy" which is conserved in the system; and if it does not matter if you do it here or there, then a quantity called "linear momentum" is conserved.

But in special relativity time and space are linked into a spacetime, whose geometry is characterized by the Lorentz invariant spacetime interval. Similarly, energy and momentum get linked into a four vector whose magnitude is Lorentz invariant, namely $E^2 -p^2 =m^2$ in $c=1$ units.

So "mass" in special relativity is just a quantity that characterizes the length of the energy-momentum four vector. It is a Lorentz invariant quantity and so a good quantity to characterize a particle with (other than its intrinsic spin).

Physically, for a particle at rest, $E=mc^2$, so mass is just a form of condensed energy. You can release some of it, eg in fission, or create new particles of mass from pure energy, as in colliders.

The key concept you need to absorb from special relativity is "Lorentz invariant quantities". They play a special role, everything else is relative.

Quantum physics does not explain what "mass" is. It only provides processes for transforming mass to other forms of energy and vice versa.

  • 1
    $\begingroup$ I've always found it useful to explain the term "symmetry" in this context, because it is key to field theories, but often used in ways that imply it is something else. To whit: momentum does not have to be conserved, but it is. That is because of the translational symmetry of our universe, which is essentially spherical. If the universe was not symmetric, let's say it was egg shaped, momentum would not be conserved - everything would roll to the "heavy end". So, conservation implies physical symmetry of something. $\endgroup$ Commented Apr 10, 2018 at 13:30
  • $\begingroup$ How do we ensure that only one such quantity exists? $\endgroup$ Commented Apr 13, 2018 at 15:15
  • $\begingroup$ @user6873235 Can you clarify your question. Do you mean, "is the mass unique for each particle"? or do you mean something else? $\endgroup$
    – rparwani
    Commented Apr 14, 2018 at 2:27
  • $\begingroup$ How do we know only a single conserved quantity exists which would match the properties expected for energy/momentum energy? What if there simply happened to be another quantity that was conserved in the same way, but separate? Or would the existence of such a quantity imply the original symmetry, and thus render all such quantities effectively identical? $\endgroup$ Commented Apr 14, 2018 at 16:37
  • $\begingroup$ The energy-momentum 4-vector is the conserved quantity for invariance of the Lagrangian under spacetime translations. This may sound abstract at first but the definition agrees with what you know in the Newtonian limit. Derivations are via Noether's theorem (see any classical field theory textbook). So the explicit expression you get for the 4-vector depends on the Lagrangian you start with (ie what matter it represents). But the length of the 4 vector is invariant and is by definition $m^2$. $\endgroup$
    – rparwani
    Commented Apr 15, 2018 at 12:30

Consider an electro-magnetic wave. The frequency-wavenumber relationship is:

$$ \omega = kc $$

In the quantum interpretation we consider a single quanta of the wave:

$$ \hbar\omega = \hbar kc $$

as a massless photon traveling at the speed-of-light, and the relationship is:

$$ E = pc $$

Now consider an electromagnetic wave in a waveguide--there is a cutoff frequency:

$$ \omega = \sqrt{\omega_0^2 + (kc)^2} $$

All this means it that there is a minimum frequency at which the wavelength goes to infinity--you simply cannot have a lower frequency in a propagating wave. The existence of the waveguide rules out certain modes.

Apply $\hbar$ and this becomes:

$$ E = \sqrt{(mc^2)^2 + (pc)^2 } $$

so that the cutoff frequency acts just like an effective mass:

$$ m_0 = \frac{\hbar\omega_0}{c^2} $$

Now the EM field isn't different. It has not changed; rather, its environment causes it to behave as if it has a mass.

This is how I view mass w.r.t to the Higgs mechanism: all the particles are massless fields until the Higgs turns on. It changes the environment in which the fields propagate, causing a cutoff frequency. At infinite wavelength, there still is finite frequency. With respect to particles, we view that as finite energy at zero momentum: rest mass.


In field theory, each elementary particle is associated with a field of its own. One writes down a Lagrangian (density) $\mathscr{L}$ for the field. When you quantize the theory you obtain particles. The positive coefficient ($\mu^2>0$) of the term in $\mathscr{L}$ which is quadratic in the field becomes proportional to $m^2$ in the relation $E^2=p^2c^2+m^2c^4$ of the particle.

However, sometimes it may so happen that the Lagrangian $\mathscr{L}$ does have a term quadratic in the field but $\mu^2<0$. In the early universe, the Lagrangian Higgs field was such that $\mu^2<0$, and hence cannot be directly associated with the mass of the particle. As the Universe cooled down, $\mu^2$ (being a function of temperature) turned positive and the Lagrangian broke some symmetry called spontaneous breakdown of symmetry with $\mu^2>0$. Since in the Standard Model (SM), the Higgs field is coupled to other particles, its vacuum expectation value gives masses to them (except photon and neutrino).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.