# What does it mean for a law to be fundamental?

I was recently reading up about Coulomb's law and Gauss law and several sources seemed to state that the Gauss law was more "fundamental" than Coulomb's law even though one is deducible from the other, which got me thinking: what does it even mean for a law/theorem to be more fundamental?

• For example in this case, Coloumb's law refers to charges and so implies that electricity only exists when these are involved. Gauss law on the other hand makes use of the concept of electric field per se, which turns out to exist even when there are no charges at all. So Gauss law has broader meaning (and, in fact, is one of the four maxwell equations..) – pp.ch.te Jun 29 '17 at 18:32

Roughly speaking, one law is more fundamental than another if it explains it. (There's no guarantee any law is "fundamental", in the sense of there being nothing even more fundamental; maybe all laws have a deeper explanation, but at any given time our knowledge is finite.)

The most obvious guess at what it means for $$A$$ to explain $$B$$ is that $$B$$ is deducible from $$A$$, but if the deduction works both ways this doesn't reveal which is more fundamental, which is the point you've hit on. (If you want to get technical, in the philosophy of science the simple definition of explanation that I just critiqued is the deductive-nomological model.) Indeed, we can obtain Coulomb's law as a special case of Gauss's law, or Gauss's law from Coulomb's law by linearity.

More fundamental claims provide deeper insight. Gauss's law is more fundamental in the sense that from Maxwell's equations we obtain a vector-calculus description of the electromagnetic fields that works for arbitrary charge distributions. It is at this point that the fields $$\vec{E},\,\vec{B}$$ become related in a theory that unifies them. Unification is typically a sign of deeper insight in physics, whereas Coulomb's law speaks only of $$\vec{E}$$.

From Maxwell's equations emerge Lorentz-invariant wave equations that ultimately inspired special relativity. If we rewrite $$\vec{E},\,\vec{B}$$ in terms of $$\vec{A},\,\phi$$ (which unite into $$A^\mu$$ relativistically), we reduce Gauss's law to $$\nabla^2\phi=-\frac{\rho}{\epsilon_0}$$. But a manifestly relativistic formalism gives an even deeper understanding of electromagnetism, far beyond anything Coulomb imagined.

At this point we wonder only where $$A^\mu$$ comes from. Scalar electrodynamics explains this in terms of local symmetries of a scalar field; this provides an even more fundamental exposition. (We could go further, but you see my point.)

Gauss's law is more fundamental in a few ways:

It is applicable in more situations:

A version of Gauss's Law that involves the vector potential is still valid in quantum field theory, regardless of choice of gauge, whereas Coulomb's Law only arises after choosing the Coulomb gauge, $\nabla\cdot\vec{A}=0$.

It requires fewer physical assumptions:

Gauss's Law is essentially just the divergence theorem, which requires no physical assumptions as it is a mathematical statement based on the structure of $\mathbb{R}^3$. Gauss's Law then defines the charge as the divergence of the electric field*, scaled by an arbitrary constant. Meanwhile, Coulomb's Law essentially starts by assuming an interaction potential, which is a physical assumption, and defines charge based on this physical assumption. The two can be put on equal footing if you assume that neither the charge nor the electric field is moving, but this is also a physical assumption.

It has broader meaning in general:

Coulomb's Law is a statement about the forces and/or electric fields generated in the presence of charge. Gauss's Law, in contrast, is a statement about the behavior of the electric field in general, regardless of whether or not charge is present. As such, Gauss's Law still gives useful statements in the context of electromagnetic radiation in a vacuum, while Coulomb's Law gives only a vacuous answer.

*Here we assume the existence of the electric field. This is not problematic since even QFT treats it as fundamental.

• "It requires fewer physical assumptions" is a valid statement only if you count the static nature of things as an assumption. That is the only extra assumption that we need for Coulomb's law. If you can summarize the physical content of assumptions that go into the Gauss's law in one sentence and can't do it for the Coulomb's law then it doesn't mean it really takes in fewer assumptions than the Coulomb's law. – Feynmans Out for Grumpy Cat Jun 29 '17 at 18:45
• Coulomb's law is also a statement about the electric field only. One CAN certainly assert that the electric field - in the absence of any test charges whatsoever - can be determined by the Coulomb's law if the situation is static. – Feynmans Out for Grumpy Cat Jun 29 '17 at 18:47
• @Dvij Edited to reflect this. – probably_someone Jun 29 '17 at 18:49
• Still, Coulomb's law is valid without the presence of test charges just as much as Gauss's law. It is the dynamic nature of the situation that distinguishes their validity - not the presence of charges. – Feynmans Out for Grumpy Cat Jun 29 '17 at 18:51
• @probably_someone Do not forget, though, that Gauß law requires the fields to vanish at infinity (and this is in turn a pretty strong physical assumption on the electric field). – gented Jun 30 '17 at 11:58

Roughly speaking, a more general validity can be thought of as being more fundamental.

In the case of the comparison between the Gauss's law and the Coulomb's law, both are exactly equivalent in the static cases but Gauss's law is a valid law in a generic situation as well. Thus, Gauss's law can be considered as more fundamental than the Coulomb's law.

In other words, the Coulomb's law, $$\vec{E}(\vec{r})=\dfrac{1}{4\pi\epsilon_0}\displaystyle\int_{\text{space}}\dfrac{\vec{r}-\vec{r'}}{|\vec{r}-\vec{r'}|^3}{\rho(\vec{r'})} d^3\vec{r'} \tag{1}$$

is a valid expression for the electric field at a point $\vec{r}$ due to a charge distribution $\rho(\vec{r'})$ only in a static situation.

But, the Gauss's law, $$\nabla\cdot\vec{E}(\vec{r})=\dfrac{\rho(\vec{r})}{\epsilon_0} \tag{2}$$

is one of the general Maxwell's equations and is always valid - both in the static as well as the dynamic case.

In the static case, $(2)$ implies $(1)$ (and vice-versa) and thus, are equivalent. But, in the general case, $(1)$ doesn't hold while the $(2)$ does - making $(2)$ more fundamental than $(1)$. So, as I said, in a rough sense, we call something a more fundamental feature of the laws of physics if the feature survives more generalizations.

## protected by Qmechanic♦Jun 29 '17 at 19:23

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