# Identifying a scalar function

We know that a scalar is invariant under rotations. What about a scalar function? Should it also be invariant under rotations? Therefore, under rotation $\phi(x,y,z)$ must be equal to $\phi^\prime(x^\prime,y^\prime , z^\prime)$. Where $(x^\prime,y^\prime , z^\prime)$ is the rotated coordinate system. Does it imply that $$\phi(x,y,z)=x^2+y^2+z^2$$ is the only possible scalar function in three dimension? Can $\phi(x,y,z)=x^2+yz$ be a scalar function?

Good terminology question.

Let's work in some differentiable manifold $M$, our transformation is a smooth map $T: M \to N$. In the case of a rotation $M = N$.

Our $\phi$ is a smooth function $\phi: M \to \mathbb{R}$.

In classical field theory the fact that $\phi$ maps to $\mathbb{R}$ is often expressed by the statement "$\phi$ is a scalar field".

Now the additional demand that $\phi$ is in some sense invariant under $T$ (if $\phi$ is a field it is said that $T$ is a symmetry of the theory) is concretely the requirement:

$$T^*\phi |_{T(p)} = \phi |_p$$ for all $p \in M$, where $*$ denotes the corresponding pullback on a smooth function.

Physicist at times call this property "$\phi$ transforms like a scalar under $T$".

The smooth map $\phi: (x, y, z) \mapsto x^2 + yz$:

• is a scalar valued function
• does not transform like a scalar under rotations.
• I question this a bit. Of course you are technically correct, but I don't think physicists call scalar fields "scalar" when they are invariant under maps. Rather this means that when you have for example a $\phi(x^1,...,x^n)$ field, then if you have some coordinate change expressible as $x^i=x^i(y^1,...y^n)$ then if you plug this directly into the field, it won't change, whereas a tensor field expressed in components would have their components changed in the usual manner. Feb 17, 2015 at 9:13
• Contrast this with a so-called scalar density, which is also a scalar-valued field, but picks up the pullback's determinant as a factor, that is something that is scalar valued, but "does not transform as a scalar". Feb 17, 2015 at 9:14
• Here we have a scalar field - maps to $\mathbb{R}$ don't pick up factors under pullbacks, e.g. see the wiki article linked above.
– zzz
Feb 17, 2015 at 9:32
• Regarding the scalar density - if you and I are thinking about the same thing I think the extra factor comes from transformation of the volume element under the same integral
– zzz
Feb 17, 2015 at 9:43

As usual when dealing with transformations one has to be careful whether they are active or passive. If I understand your question you are implying a passive transformation, which is a mere change of coordinates. In this case all you are doing is changing the way you assign a scalar value to a "vector" of coordinates. Therefore $\phi(\mathbf x) = (\phi\circ\xi)(\mathbf x')$, where $\xi$ is the change of coordinates, i.e. $\mathbf x = \xi(\mathbf x')$. You can now set $\phi' = \phi\circ\xi$ to get the covariance for scalar fields $$\phi(\mathbf x) = \phi'(\mathbf x').$$

To see how this relates to the more general case of tensor fields, where you have the Jacobian of the transformation changing components of tensors, consider that a pseudoscalar (like any tensor density of a certain weight) would transform with a certain power of the determinant of the Jacobian, 1 in this case, hence

$$\phi'(\mathbf x') = \det(D\xi)\phi(\mathbf x).$$

In the active picture you are actually taking the value of the scalar function from a point $p$ to a point $q$ of the manifold and requiring that $\phi(p) = \phi(q)$. This usually happens if $\phi$ has some sort of symmetry, like a spherical potential is invariant under rotation so that you can transport the scalar field along the flow of the transformation (this is well defined in the language of Lie derivatives along vector fields and indeed linked to the notion of symmetries).