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From Misner, Thorne and Wheeler, page 115.

0-Form or Scalar, $f$

An example in the context of 3-space and Newtonian physics is temperature $T\left(x,y,z\right),$ and in the context of spacetime, a scalar potential, $\phi\left(t,x,y,z\right).$

I'm trying to think of an example of such a scalar potential. Is there one? Electrostatic potential is the time component of the electromagnetic 4-vector potential, so it's really a vector with 0-valued space components.

Steven Thomas Hatton
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  • Scalar potentials play a crucial role in nuclear physics. See my answer here: https://physics.stackexchange.com/questions/288357/why-is-nuclear-force-spin-dependent/288499#288499 and here: https://physics.stackexchange.com/questions/292913/why-is-the-density-of-the-fermi-gas-in-a-neutron-star-not-changing-the-potential/293523#293523 – Lewis Miller Feb 11 '22 at 03:32

3 Answers3

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Within the Standard Model, the simplest model of the Higgs field is a multiplet of Lorentz scalar fields. This multiplet does have a non-trivial transformation under an underlying gauge group of the Standard Model; but under Lorentz transformations, the Higgs field is invariant, as all good Lorentz scalars should be.

Of course, the Higgs field is not a "potential" in the sense that a "potential" is a field whose derivative is a physically observable field; so if you're strictly looking for a scalar potential this isn't what you're looking for. But to the best of my knowledge it is the only fundamental scalar field we know to date.

Michael Seifert
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  • This answer is not entirely in my field of expertise, so if I've garbled anything above, feel free to send me corrections, amendments, and execrations. – Michael Seifert Feb 11 '22 at 03:15
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If you have a vector or tensor field, then you can get a scalar field by contraction.

Examples:

$J^\mu$ = 4-flux of some quantity. Scalar field: $\rho = \sqrt{J^\mu J_\mu}/c$. Interpretation: proper density.

$k^\mu$ = 4-wave vector; $x^\mu$ = 4-position. Scalar field: $\phi = k^\mu x_\mu$. Interpretation: phase of a plane wave.

Electromagnetic field tensor $F^{\mu\nu}$. Scalar fields: $F^\mu_\mu$ and $F^{\mu\nu} F_{\mu \nu}$ and $F^{\mu \nu} \tilde{F}_{\mu \nu}$. The first of these is zero, the second is $2(E^2 - c^2 B^2)/c^2$ and the third is $4 {\bf E} \cdot {\bf B}$.

The above are scalar fields, though not normally called 'potentials' because their gradient does not relate to a force. However we can introduce a potential which is by definition a scalar invariant, and then consider the gradient to be a 4-force. We thus obtain $$ f^\mu = - \partial^\mu \phi. $$ Such a 4-force is not the electromagnetic force, but it can be used to construct simple models of the strong force.

Andrew Steane
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    Why should we consider those to be potentials? – Steven Thomas Hatton Feb 11 '22 at 03:53
  • @StevenThomasHatton I may have misunderstood your question, but since you mentioned temperature I thought you were just thinking of anything scalar and invariant and a continuous field. I modified the answer a bit. – Andrew Steane Feb 11 '22 at 09:24
  • I didn't ask the question well. I should have specified "classical physics". And even excluded variational dynamics. I figure there must be some way of concocting a Lagrangian with a 4-potential that is "physically meaningful". But, IMO, that is a different mathematical space, which is "less real" than the mathematical space of classical general relativity. I wasn't even considering QFT. And since I haven't looked at QFT for quite some time, I'm not in a position to assess any of the answers involving it. It may be a year or more before I am prepared to accept an answer. – Steven Thomas Hatton Feb 14 '22 at 20:14
  • @StevenThomasHatton The added remark I made about strong force but construed either as a classical force or as a contribution to a quantum field. – Andrew Steane Feb 14 '22 at 20:36
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There are two different things one can mean by a potential.

The first is in the sense of a gauge field whose derivatives (in some combination) give the field-strength tensor. For example, the electromagnetic potential $A^\mu(t,\vec{x})$ as you mention. One can certainly write down a theory with a scalar gauge field $B(t,\vec{x})$, however, such a gauge field does not appear in the standard model of particle physics. But, there is nothing wrong with writing down such a theory. See, for example, this paper where the $U(1)$ symmetry is gauged using a scalar gauge field.

The second is in the sense of the potential terms, i.e., the interaction/self-interaction terms in the Lagrangian density. For example, we say things like "the shape of the Higgs potential looks like the logo of a famous StackExchange site". All such potential terms are always scalar because a Lagrangian is not allowed to be charged.