Scalar fields transform under a diffeomorphism $x\to\tilde x$ under the rule $\phi(x)\to\tilde \phi (\tilde x) = \phi (x)$. They are often said to be invariant, though only kinda of. For example, if $\phi(t)=\mu t$ and $\tilde t = t^2/\Lambda$, then $\tilde \phi (\tilde t) = \mu \sqrt{\Lambda \tilde t}$ if we want to maintain $\tilde \phi (t^2/\Lambda) = \phi(t)$. So clearly $\phi \ne \tilde \phi$ in the sense of functions; the scalar field, or at least the abstract mapping used to represent it, does transform.

Anyway, I want to ask about arbitrary functions with explicit coordinate-dependence. For example, perhaps you have a time-dependent mass in your Lagrangian:

$$ \mathcal L = \frac{1}{2}\partial_\mu \phi \partial_\nu \phi g^{\mu\nu} - \frac{1}{2}m^2(t)\phi^2. $$

How does the mass transform under a time diffeomorphism? Naively I would say $m(t)\to\tilde m(\tilde t) = m(t)$ since a mass, at least a constant one, is a scalar, but that can't be right. If the mass transforms just like the field, then the full Lagrangian also transforms properly under diffeomorphisms, which I know it doesn't because the explicit time dependence breaks the time diff (energy isn't conserved, etc).

It seems the only sensible transformation law for the mass should be $m(t)\to m(\tilde t)$ meaning it genuinely doesn't transform; I'm simply replacing the time coordinates. Except this seems arbitrary and handwavy. Is there an intuitive reason for me to convince myself that whether a quantity is dynamical (i.e., has an eom like $\phi$ but not $m$) should impact what kind of transformation rule it has?

EDIT to expand on the conserved current issue.

Diff invariance leads to conservation of the gravitational stress-energy tensor. (see page 139 here).

Translational invariance leads to conservation of the canonical stress-energy tensor (Noether's theorem).

For scalars, one stress-energy tensor is conserved if, and only if, the other is as well. (this is claimed in the appendix of this paper but I don't have a proof.)

If $m$ transforms as $\phi$, then $\mathcal L$ is a scalar and $S$ is diff invariant. Thus the gravitational stress-energy tensor is conserved. But the presence of the non-dynamical $m(t)$ in $\mathcal L$ means the canonical stress-energy tensor is not conserved because the Lagrangian has explicit time dependence. This is a contradiction.

So I cannot accept any answer that claims $m$ transforms as $\phi$ unless it resolves this conflict.

  • $\begingroup$ A small remark: the field transforms if given as a function of the coordinates, but if you think of it as a function on the manifold itself, then it is coordinate independent. Not very important to your question, but still. $\endgroup$
    – Javier
    Apr 26, 2017 at 18:45
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    $\begingroup$ I don't want to write an answer because I am missing something right now (it's late at night...). But think about classical mechanics (it's easier), $L = \frac m 2 v^2 - V(x)$. Under reparametrization, we have $V'(x') = V(x(x'))$, so the Lagrangian is "diffeomorphism invariant". But that obviously doesn't mean that spatial translation is a symmetry - to check that, we'd take $V(x) \to V(x + \delta x)$. I think (and this is the part I'm not sure about) the first type of transformation is a kind of gauge symmetry, while the second one (potentially) reveals an actual symmetry of the system. $\endgroup$
    – Noiralef
    Apr 26, 2017 at 20:49
  • $\begingroup$ Is $t \rightarrow t^2$ a diffeomorphism? $\endgroup$
    – image357
    Apr 26, 2017 at 22:16
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    $\begingroup$ If you think of $m(t)$ as of a field $\Phi$ that has gotten a vacuum expectation value, $\langle\Phi\rangle=m(t)$, it should be clear that $m(t)$ must transform as the scalar $\phi$. $\endgroup$
    – TwoBs
    Apr 30, 2017 at 14:19
  • $\begingroup$ If you take a look at Carroll's PDF, you'll see that in the derivation of the gravitational stress-energy tensor he assumes that the equations of motion hold for the fields. But $m$ is a field whose equations of motion we don't care about, so the conservation does not follow, which is what indeed happens. I'll see if I can post this in more detail later. $\endgroup$
    – Javier
    May 3, 2017 at 14:34

2 Answers 2


Short answer: $m$ and $\phi$ transform in the same way, which is the way you call "naive". Their values don't change, you just do the coordinate transformation. The Lagrangian changes functional form but it's a scalar because its value doesn't change; this is not a problem because energy conservation comes from time translation invariance, not diff invariance. The latter doesn't have conservation laws because every theory can be made diff invariant. The (very) long answer follows, with a lot of detail.

Mathematicians and physicists use the word "scalar" in two different ways, I think. This might be a little controversial because people use words differently, but I hope the main point is clear.

In math (and usually in general relativity too) we say that a scalar is a function $f: M \to \mathbb{R}$ defined on your manifold. The value of the function at each point clearly doesn't depend on the coordinates, because coordinates are absent from the definition. For example, take $\mathbb{R}^2$ as our manifold, and let $f(P)$ be the height of $P$ above the x-axis. In cartesian coordinates this is $f(x, y) = y$, but in polar coordinates it's $f(r, \theta) = r \sin \theta$. The form of the function changes, but its value on points does not. $^1$

You might even define functions by how they look in some frame. For example, I'll define $g$ by $g(r, \theta) = r + \cos \theta$. If you want to know how it looks in different frames you just substitute the transformation. For example, we would have $g(x,y) = \sqrt{x^2+y^2} + x/\sqrt{x^2+y^2}$. What we would not do is say $g(x,y) = x + \cos y$: we want to keep the values and change the formula, not the other way around.

In physics (and especially in areas like field theory) we care about the formulas, so we say that something is a scalar if it doesn't change under some transformation. For example, say we're in $\mathbb{R}^2$ and we're interested in functions that are invariant under rotations. Then every such function must be a function of $r = \sqrt{x^2 + y^2}$. This is why in field theory texts you see things like "the only scalar we can form out of $p^\mu$ is $p_\mu p^\mu$". A mathematician would disagree; they'd say that since $p^\mu$ lives in $\mathbb{R}^4$, any function from $\mathbb{R}^4$ to $\mathbb{R}$ is a scalar. The difference, like I said above, is that such a function would have a different formula in different frames.

You say in your comment that the Lagrangian is not always a scalar. This is true in the physics sense but not in the math sense. Let's take your example of $\mathcal{L} = \frac12 g^{00} \partial_\mu \phi \partial^\mu \phi$. If you want this formula to be true in every coordinate system, you'll run into trouble, because fields that are solutions in one coordinate system won't be in another (try an example if you don't believe me). This is unacceptable, because the physics must look the same in every coordinate system, no matter what symmetries we have.

What we can do is first choose some privileged coordinate system, in which $\mathcal{L} = \frac12 g^{00} \partial_\mu \phi \partial^\mu \phi$, and replace the explicit expression for $g^{00}$. For example, if you had the Schwarzschild metric you would put $-1/(1-2M/r)$ in its place. Then, if you want the Lagrangian in a different system of coordinates, you simply decide that it is going to be a scalar, and so all you have to do is transform the coordinates, while not actually changing the value of the function.

$^1$ A mathematician would kill me for writing $f$ both times; technically $f$ is a function from our manifold to $\mathbb{R}$, and the formulas are the compositions of $f$ with coordinate functions. I won't bother with that.

  • $\begingroup$ The Lagrangian isn't always a scalar. $\mathcal L = \frac{1}{2}g^{00} \partial_\mu \phi \partial^\mu \phi $ isn't a scalar, and neither is $ \mathcal L = \frac{1}{2}\partial_\mu A^\nu \partial^\mu A_\nu $. But I'm perfectly allowed to integrate these Lagrangians over spacetime to get actions, it's just that the action isn't invariant under diffs. That's the point of GR, only Lagrangias that are scalars (leading to actions that are diff invariant) are allowed. $\endgroup$
    – Donjon
    Apr 28, 2017 at 13:15
  • $\begingroup$ @Donjon: I rewrote my answer, please let me know if it's better now. It's a bit long because I wanted to make sure I didn't miss anything. $\endgroup$
    – Javier
    Apr 29, 2017 at 23:04
  • $\begingroup$ I still don't see how your conclusion follows. Sure, $m(t)$ is a math scalar, but being a math scalar isn't enough to know its transf law. The determinant $\sqrt g $ is a math scalar as well but transfs non-trivially. Only physics scalars have fixed transf laws, by definition of physics scalar. And I don't think $m(t)$ can be a physics scalar, because then $\mathcal L$ is too. This makes $S$ invariant under diffs, which can't be true (diff invariance leads to conserved stress-energy, but Noether tells us this cannot be the case because $\mathcal L$ has an explicit time dependence). $\endgroup$
    – Donjon
    May 1, 2017 at 9:45
  • $\begingroup$ @Donjon: Let me answer your last point first: it's translation invariance that gives us conserved energy-momentum, not diff invariance. The later is translation invariance but with a position-dependent translation: it's a kind of gauge symmetry, and local gauge symmetries don't have conservation laws. $\endgroup$
    – Javier
    May 1, 2017 at 14:43
  • $\begingroup$ And regarding $\sqrt{g}$, I may have oversimplified a bit, but you have to tell these things apart by context. Given a frame you're right that $\sqrt{|g|}$ is a math scalar, but if you want it to have the same value in every coordinate system you'll break diff invariance, because the measure $d^4 x$ is not invariant. So in this case you have to keep the formula and transform the value, but with the mass it's the other way around. $\endgroup$
    – Javier
    May 1, 2017 at 14:45

Let $x$ be a variable or a set of variables.

A scalar $m$ (it does not matter whether it is a mass field or a scalar field) transforms as following under the diffeomorphism $x \mapsto \bar{x}$:

$m(x) \mapsto m(\bar{x})$.

For a tensor field $T_{\mu_1 \mu_2 \dots \mu_n}$ of rank $n$ (specially, for n=1 a tensor field is a vector field) you have the rule:

$T_{\mu_1 \mu_2 \dots \mu_n}(x) \mapsto \frac{\partial \bar{x}_{\mu_1}}{\partial x_{\nu_1}} \frac{\partial \bar{x}_{\mu_2}}{\partial x_{\nu_2}} \dots \frac{\partial \bar{x}_{\mu_n}}{\partial x_{\nu_n}} T_{\nu_1 \nu_2 \dots \nu_n}(\bar{x})$.

These transformation rules hold in general. You can also define diffeomorphism-invariant derivatives of fields, called the Lie derivative. Under an infinitesimal transformation $\bar{x} = x+ \epsilon$ you have for a scalar

$m(x) \mapsto m(x+\epsilon) = m(x) + \epsilon^\mu\partial_\mu m(x) = m(x) + \mathcal{L}_{\epsilon}m(x)$

it changes by the Lie derivative; tensor fields change also under infinitesimal diffeomorphisms, namely by their Lie derivative.

Note that volume integrals also change under diffeomorphisms $d^{Dimension} \bar{x} = det J d^{Dimension} x$. Here, $J$ is the Jacobi matrix where you derive the variables with bar by the original variables.

  • $\begingroup$ There are also quantities which are not tensors. Position is not a tensor under general coordinate transform. Acceleration (second time derivative of position) also does not transform like a tensor. $\endgroup$
    – lalala
    Apr 26, 2017 at 18:14
  • $\begingroup$ Your second equation is false (the first one too). The components $T_{\nu \dots}(\tilde{x})$ aren't the same functions of $\tilde{x}$ as are the components $T_{\nu \dots}(x)$ of $x$. This isn't the proper tensorial transformation law. You're missing a tilde on the right part, and the indices aren't well placed on the coordinates. $\endgroup$
    – Cham
    Apr 26, 2017 at 21:59

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