In classical mechanics, a scalar field is characterised by the fact that its value at a particular point must be invariant under rotations and reflections of coordinates. That is, one requires that $\phi'(x')=\phi(x)$, where a point, $x'$ in the new coordinate system could be related to a point, $x$ in the old one by either a rotation, $x'=Rx$ (where $R$ is a rotation matrix), or a reflection, $x'=-x$.

Then, in special relativity, one requires that a scalar field must be invariant under Poincaré transformations, i.e. under Lorentz transformations and space-time translations, $x'=\Lambda x+a$, such that $\phi'(\Lambda x+a)=\phi(x)$.

However, when one considers general relativity one is confronted with more general coordinate transformations. In this case, how does a scalar field transform under general coordinate transformations? Does one still require that it transforms trivially, i.e. such that $\phi'(x')=\phi(x)$?


2 Answers 2


Since you are going into GR I believe some elaboration might be good. In General Relativity spacetime is a smooth four dimensional Lorentzian manifold $(M,g)$.

Being a manifold means that $M$ is one Hausdorff topological space (don't mind too much about this by now) and that there is a collection of pairs $\mathcal{A}=\{(x_\lambda,U_\lambda)\}$ where $U_\lambda\subset M$ is open, $M$ is the union of all the $U_\lambda$ and $x_\lambda : U_\lambda \to \mathbb{R}^4$ is a coordinate system: i.e., it assigns to each point $p\in U_\lambda\subset M$ the coordinates


There is the additional condition that given two coordinate systems $(x,U)$ and $(y,V)$ on $M$ if their domains overlap $U\cap V \neq \emptyset$ then $T_{y,x}=x\circ y^{-1}$ and $T_{x,y}=y\circ x^{-1}$ are both $C^\infty$ functions on $\mathbb{R}^4$. These are respectively the map that transforms $y$ coordinates in $x$ coordinates and vice versa.

In summary: being a topological space gives the notion of continuity and open sets, and the existence of these functions with the overlap condition, ensures you have local coordinates which are smoothly put together to make up the whole spacetime.

So in GR we have a clear separation between the points and their coordinates. The points lie in $M$, the coordinates lie in $\mathbb{R}^4$ and what brings them together are the coordinate systems as defined above.

Now, a real scalar field is a $C^\infty$ function $\phi : M\to \mathbb{R}$. Nothing more to say about it.

How can it be? Where the transformation law is gone? Well, I haven't introduced coordinates above. I've defined the object intrinsicaly to spacetime $M$.

Now let $p\in M$ and a coordinate system $(x,U)$ around $p$ be given, namely $p\in U$. Then, we can localy, on $U$, resolve $\phi$ in coordinates. Since $x : U\to \mathbb{R}^4$ we have that $x^{-1} : \mathbb{R}^4\to U$ takes coordinates to the corresponding point.

The composite function $\phi_{(x)}=\phi \circ x^{-1} : \mathbb{R}^4\to \mathbb{R}$ is the coordinate expression of $\phi$. Notice that it takes some coordinate values and gives $\phi$ on the point corresponding to them.

Now the issue with coordinate systems is that they are not unique nor there is any prefered one. Imagine I pick another coordinate system $(y,V)$ with $p\in U\cap V$. Then we can also get the coordinate expression $\phi_{(y)}=\phi\circ y^{-1}$.

Now focus on the overlap. You shall notice that the following happens:

$$\phi_{(y)}=\phi\circ y^{-1}=(\phi\circ x^{-1})\circ x \circ y^{-1}=\phi_{(x)}\circ (x\circ y^{-1})=\phi_{(y)}\circ T_{y,x}$$

Obviously $T_{x,y} = (T_{y,x})^{-1}$ Thus you finaly conclude that

$$\phi_{(y)}\circ T_{x,y} = \phi_{(x)}$$

This is precisely what you wrote, take a time to understand that.

When $\phi : M\to \mathbb{R}$ is already defined as a function on $M$ then this condition is satisfied as a result.

But generaly you want to define it from coordinates, because it is easier and more intuitive. Then this becomes a compatibility condition to ensure that the object you are defining is actually a well defined function on $M$, i.e., it doesn't depend on the coordinates themselves.


Yes, that is the definition of a scalar field in a theory with general covariance.


where $x^{\prime}$ is the coordinate in the new coordinate system corresponding to a given $x$ in the original coordinate system. i.e. $x$ and $x^{\prime}$ are, in general, different values representing the same point on the manifold.

  • $\begingroup$ Why does one not simply demand this from the start in classical mechanics (CM) then? In CM one is taught that a scalar should be rotationally invariant (since it has no directional dependence), but at the end of the day, a scalar field is simply a function that maps points to numbers, so one would expect that it should be invariant under general coordinate transformations?! ... $\endgroup$
    – Will
    Jul 27, 2016 at 13:49
  • $\begingroup$ Also, should one interpret $\phi'(x')=\phi(x)$ as "saying" that the coordinate representation of the scalar field is different in the two different coordinate systems (given by $\phi$ in one and $\phi'$ in the other), but since $x$ and $x'$ are different coordinates for the same point, we require that $\phi'$ evaluated at $x'$ should be equal to $\phi$ evaluated at $x$, i.e. $\phi'(x')=\phi(x)$?! $\endgroup$
    – Will
    Jul 27, 2016 at 13:49
  • $\begingroup$ Yep, scalar fields are just functions on a manifold, so they transform in the simplest possible way under changes of coordinate. (Writing it out in coordinates only complicates matters.) $\endgroup$
    – user1504
    Jul 27, 2016 at 15:44
  • $\begingroup$ I have edited my answer as a response to your second comment. See this physics.stackexchange.com/questions/263954/… for discussion related to your first comment. $\endgroup$ Jul 27, 2016 at 17:38
  • 1
    $\begingroup$ @Blazej, the question was about a scalar 'field' in particular $\endgroup$ Jul 28, 2016 at 2:36

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