8
$\begingroup$

Like curvature of a geometrical shape, is there any purely mathematical explanation for what is mass?

Or is it absolutely a physical quantity in which case this question has no meaning?

I thought that mass might be a physical quantity but that it can be described in terms of something more fundamental i.e can be derived from another.

$\endgroup$
3
  • 4
    $\begingroup$ Does this answer your question? Physical meaning of the Casimir operators of Poincarè algebra $\endgroup$ – my2cts Jun 8 '20 at 21:02
  • $\begingroup$ If you go deeper, you can always find energy in place of mass. For example, in QFT, the mass of the elementary particles (and all of their "compounds") comes from a) the energy of the oscillations in the corresponding quantum field and b) the interaction energy between the fields. But I'm not sure if that's in the spirit of your question; in modern physics (20th century+), mass is usually thought of as the energy of a system; once you account for all the energies, there's no "intrinsic mass" left over. But whether that's physically important or just a matter of formalism... $\endgroup$ – Luaan Jun 9 '20 at 13:42
  • 1
    $\begingroup$ @my2cts Mass being a Casimir invariant of the Poincaré algebra is one of the multiple satisfactory answers given here. But the question itself is not a duplicate of the other, since the two are coming from a very different standpoint - one is about the actual physical meaning of a specific algebraic operation and this one about how to describe/derive a physical quantity in purely mathematical/geometrical terms $\endgroup$ – Yash Sharma Jun 9 '20 at 17:25
17
$\begingroup$

At the moment mass is one of the axiomatically defined quantities in the MKS (meter, kilogram, second) system. As with axioms in mathematics, other units can and have been defined and then the (MKS) units become derivative.

The theoretical models of physics use mathematics with its axioms, and in addition impose additional axioms and axiomatic statements to connect the mathematical solutions to observations and data and to predict new situations. In particle physics it is the standard model.

Since ancient times there are two schools of thought, the Platonic, which can be encapsulated into "mathematics creates reality", and the pragmatic, which can be summed into "mathematics models observations and data".

In my observation, many theorists and theoretically inclined physicists belong to the Platonic school. Usually experimental physicists belong to the pragmatic, because they are aware how one generation's physics theory becomes an emergent one from the current theory in the next generation.

If the platonists are correct and there exists a Theory Of Everything (TOE) and the masses of the elementary particles in the standard model of particle physics come out as predictions of the TOE, the answer to your question will be "yes".

At the moment the pragmatists are on the ascendant so the answer is that no, there isn't any prediction of elementary masses in the current models of physics.

$\endgroup$
5
  • $\begingroup$ OK, I now understand the current definition. But since we have candidates for TOE namely String theory or the Mathematical Universe Hypothesis (MUH), even though they are far from being a complete theory. Do they contain any hint about how mass can be defined or predicted? $\endgroup$ – Yash Sharma Jun 8 '20 at 12:05
  • $\begingroup$ If a TOE turns out to be a string theory, the masses of the elementary particles will be the excitation energies of the elementary string.. As for the MUH that I just searched for, in the article in Wikipedia it sounds like the platonic case i describe above. arxiv.org/abs/0704.0646 There is no concrete predictive mathematics ( as there is in string theory). It is a platonic hypothetically existing theory. $\endgroup$ – anna v Jun 8 '20 at 13:15
  • $\begingroup$ This answer severely misrepresents Platonism which holds pretty clearly to dualism (or even trialism). Platonism holds that mathematics exists independently of physical reality. "Mathematics creates reality" is something else entirely and only the most radical platonists might even consider it. "Mathematics models observations and data" is the official perspective of mathematics and since about half of mathematicians are Platonists (as am I), it seems unlikely that they would adopt an perspective that so directly contradicts there own understanding and interpretation of mathematics. $\endgroup$ – RBarryYoung Jun 9 '20 at 21:05
  • $\begingroup$ @RBarryYoung This answer is from a physicist in a physics blog, not a philosophy or mathematics one, and it is describin the state of physics theories. Maybe it is the Pythagorean view of platonism that dominates it, but certainly most physics theorist in my opinion believe that mathematics creates reality. You should hear them on the reality of the fields in the standard model.of particle physics. $\endgroup$ – anna v Jun 10 '20 at 3:20
  • $\begingroup$ @annav That does not make it OK to mischaracterize the metaphysics of Platonism so extensively. Further your whole answer was based on this misrepresentation, so it calls your whole answer into question. I assume that accuracy and correctness are still important to physicists even in a physics forum. $\endgroup$ – RBarryYoung Jun 10 '20 at 3:32
15
$\begingroup$

Yes, mass has a geometrical explanation. The mass of a system is the magnitude or “length” of its energy-momentum four-vector $p=(E,p_x,p_y,p_z)$, using the Minkowski metric $\text{diag}(1,-1,-1,-1)$ of four-dimensional spacetime. This is the standard interpretation of

$$m^2=p\cdot p=E^2-p_x^2-p_y^2-p_z^2$$

in units where $c=1$.

$\endgroup$
2
  • 1
    $\begingroup$ Isn't this somewhat circular? Does the four-vector have a definition that isn't rooted in a pre-defined notion of mass? $\endgroup$ – chepner Jun 8 '20 at 22:16
  • 9
    $\begingroup$ @chepner No. Energy and momentum can be understood as conserved quantities associated with temporal and spatial translations via Noether’s Theorem, without reference to mass. And you can have energy and momentum without mass, such as in electromagnetic fields. $\endgroup$ – G. Smith Jun 8 '20 at 23:26
10
$\begingroup$

Yes; you can derive mass from the representation theory of the Lie group ${\rm Spin}(3,1)$.

It's a long story, but I think I can usefully summarize it. If you need some areas further expanded, please say so in the comments.

First, a definition. The Lie group ${\rm SO}(3,1)$ describes the world that we live in according to special relativity. It is the local gauge group of space-time, so it's the exact meaning of the geometry in your question. The Lie group ${\rm Spin}(3,1)$ is the double cover of ${\rm SO}(3,1)$, and it's simply connected [note 1].

How does representation theory show up? Imagine you do an experiment, with some inputs $x$ and you get some output $y=f(x)$. The results should be invariant under the gauge group $G$, which might be ${\rm SO}(3)$ if we're non-relativistic or ${\rm SO}(3,1)$ if we're using special relativity. It also includes translations by ${\mathbb R}^3$ or ${\mathbb R}^4$ but these are less important to the story of mass. Anyway, invariance under $G$ means that $f(g x)=g f(x)$ for $g\in G$. That means that $f$ is actually a representation of $G$. So a fundamental particle is actually described by a representation of ${\rm SO}(3,1)$.

So why ${\rm Spin}(3,1)$? In physics, we note that we need to include so-called spin representations of ${\rm SO}(3,1)$, by which we mean representations of the double cover ${\rm Spin}(3,1)$. Phenomenologically, this is certainly true, because fermions are described by a spin representation, and they exist!

Okay, so what is the representation theory of ${\rm Spin}(3,1)$? Let's use the sign convention $(+,-,-,-)$. There are 3 types of representation, corresponding to a 4-momentum $p_i=(p_0, p_1, p_2, p_3)$ [note 2] with a squared-length $p^2=p_0^2-p_1^2-p_2^2-p_3^2$ for which $p^2<0$, $p^2=0$ or $p^2>0$. In each case, we look at the subgroup $H_p<G$ which fixes $p_i$, we find the representations of $H_p$, and we induce these representations back up to $G$. This is called Mackey theory. For $p^2<0$, we get tachyons, which may or may not be physical, but in any case we won't discuss this further. For $p^2=0$ (but $p_i$ is not the zero 4-vector), we get massless light-like particles with $H_p\cong{\rm SO}(2)$ [note 3]. The representation theory of ${\rm SO}(2)$ describes polarization. But the important case for our purposes is $p^2>0$. Here, $H_p\cong{\rm Spin}(3)$, which is the double cover of ${\rm SO}(3)$ [note 4]. The Lie group ${\rm Spin}(3)$ has one irreducible representation in each dimension. If a particle corresponds to an irreducible representation of dimension $m$, we say that the particle has spin $(m-1)/2$. If $m$ is odd (so the spin is an integer) we have a boson, otherwise we have a fermion.

We've gone on a bit of a journey into representation theory, but the important take-away for our purposes is that the representation corresponding to a massive particle depends fundamentally on the 4-momentum $p_i$ through its squared-length $p^2>0$. But the length is its mass. The 4-momentum is $p_i=(E, {\bf p})$, where $E$ is the (combined rest-mass and kinetic) energy and ${\bf p}=(p_1, p_2, p_3)$ is the non-relativistic momentum. In the rest-frame of the particle, the 4-momentum is $(E,{\bf 0})$, so $p=E$. Now the possibly most well-known equation in physics, $E=m c^2$, finishes the story and tells you that a fundamental particle must have a well-defined mass.

Notes

  • [note 1]: You may see ${\rm Spin}(3,1)$ described as ${\rm SL}(2,{\mathbb C})$. These groups are isomorphic, but I find it more helpful for this story to regard this as a coincidence.
  • [note 2]: Physicists actually write $p^i$ rather than $p_i$, but the meaning is essentially the same.
  • [note 3]: Actually the double cover ${\rm Spin}(2)$ of ${\rm SO}(2)$, but they are isomorphic, so it's not too important.
  • [note 4]: Again, ${\rm Spin}(3)$ is often described as ${\rm SU}(2)$, but for our purposes this is best regarded as a coincidence.

Further reading

  • Gerald B. Folland, Quantum Field Theory: A Tourist Guide for Mathematicians
$\endgroup$
2
  • 1
    $\begingroup$ This might not be a valid question and please correct if that is so - Can the study of Lie group Spin(3,1) be directly used to determine/predict or at least rule out which masses will be instantiated in Nature? Or will it always be that we first conduct the experiment and then describe the found mass of a particle using group Spin(3,1)? $\endgroup$ – Yash Sharma Jun 9 '20 at 16:15
  • $\begingroup$ The question is valid and excellent, but the answer is no, this approach tells us nothing about the mass except that it’s well-defined - we still need to fix the fundamental masses with measurements. $\endgroup$ – Adam Chalcraft Jun 10 '20 at 6:05
3
$\begingroup$

This question can be seen as a duplicate of https://physics.stackexchange.com/questions/434664/physical-meaning-of-the-casimir-operators-of-poincarè-algebra .

Mass or rather $m^2$ is a Casimir invariant of the Poincaré algebra. The second invariant is spin , or rather $-m^2 s(s+1)$.

$\endgroup$
0
1
$\begingroup$

In basic physics, fundamental quantities are defined in terms of how they are measured: length with a ruler, time with a clock, and mass with an inertial balance. (I'm thinking it would make more sense to measure force with a spring scale, and then define mass with Newton's Law.)

$\endgroup$
0
$\begingroup$

In Newtonian Mechanics, mass of the $i^{th}$ particle in a system of $n$ particles can be defined as the unique constant $c_i$ which leaves the quantity $\sum c_i\vec{v_i}$ conserved over time, in the absence of external forces on the system.

As a constant times a conserved quantity is another conserved quantity, the above defines mass upto multiplication by a constant (which makes sense as the magnitude of mass is measured in terms of a freely chosen unit). Though I'm not sure if this definition (or some altered version of it) applies to more advanced physics like quantum physics and relativity.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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