Scalar Fields

Suppose I have a hot cup of coffee described by a temperature field $\phi(\bs{x})$, and I slide the cup by a displacement $\bs{a}$ on my desk. If physics is going to be of any use to us, then the total heat content of the coffee better not have changed. Why must this be?

Because space is homogeneous — there is no preferred or privileged position. Mathematically, the energy $F[\phi]$ of the coffee cup is invariant under actions from the translation group $\mathbb{R}^3$. How do we describe the transformed temperature field of this displaced coffee?

Evidently, the transformed temperature field $\phi'$ now has support in a different region of space. This is a whole new function, it is an entirely different configuration of the coffee cup. The function $\phi$ is not invariant under translations, unlike the energy. This new function is defined by

$$\begin{align*} \phi'(\bs{x} + \bs{a}) = \phi(\bs{x}). \end{align*}$$

This is the most straightforward way to describe the transformation, however, it is customary (and also useful) to describe the transformed field from its perspective. Letting $\bs{x}' = \bs{x} + \bs{a}$, we get

In general, if $g$ is an element of a group $G$, then the transformed scalar field $g \cdot \phi$ is

If the inverse group element is ever confusing, just remember that it comes from expressing the field and points in terms of the new reference frame as in Eq. (1). This is called a pullback — we define the transformed field $\phi'$ by pulling the evaluation point $\bs{x}$ back to its original position $g^{-1}\bs{x}$ and evaluate the old function there.

Vector Fields

Now, let’s consider a bar magnet whose local magnetization is described by a vector field $\bs{m}(\bs{x})$. In the absence of any external field, I can rotate this magnet by rotation matrix $R$, and the energy of the magnetic field should not change. Why must this be?

Because in addition to being homogeneous, space is also isotropic — there is no preferred or privileged direction. That means the energy of the bar magnet $F[\bs{m}]$ is not only invariant under translations $\mathbb{R}^3$, but also under the group of rotations $O(3)$. Together, these transformations make up the Euclidean group $E(3)$, the fundamental group of symmetries for flat three-dimensional space. aside For relativistic systems, the fundamental group of symmetries is the Poincaré group $P(1,3)$, which is the combination of spacetime translations $\mathbb{R}^{(1,3)}$ and spacetime rotations $O(1,3)$. Let’s see how our magnetic field transforms under a rotation:

From the diagram above, we can see that the transformed field $\bs{m}'$ is described by,

$$\begin{align*} \bs{m}'(R\bs{x}) = R\bs{m}(\bs{x}). \end{align*}$$

Letting $\bs{x}' = R \bs{x}$, we can again express this as a pullback function,

so that the rotated field is defined entirely in terms of the untransformed field. This reveals an important distinction between symmetry transformations on scalar fields (Eq. (2)) and on vector fields: for vector fields, one must transform both the inputs and the outputs. From Eq. (3), it may appear that we should just transform the vector field opposite how we transform the base space, however, we can come up with an immediate counterexample.

Consider now a general rigid-body transformation $\bs{x}' = R\bs{x} + \bs{a}$. Translating the magnet should certainly not change the direction of the field, so only the rotation operator should act on $\bs{m}$, giving $\bs{m}'(\bs{x}') = R \bs{m}(R^{-1}(\bs{x}' - \bs{a})).$ So, we see that the transformation acting on the vector field $\bs{m}$ is not simply the rigid body motion $g$. The proper transformation is given by the Jacobian of the rigid body motion $J(\bs{x}) = R$,

$$\bs{m}'(\bs{x}) = J(g^{-1}\bs{x}) \bs{m}(g^{-1}\bs{x}).$$

The Jacobian may vary over space for more general coordinate maps, but it is constant in this simple example.

Derivation: Why the Jacobian shows up

Covector Fields

Recall

Recall that a covector $\omega_{\bs{x}}$ is simply a linear map specified at a point $\bs{x}$ that eats a vector and returns a scalar. For instance, the bra’s in quantum mechanics are covectors, or also a force field used to determine the amount of work done $\omega_x = \langle F(x) | \cdot \rangle$.

A covector field $\omega$ must transform in such a way that its contraction with a vector is unchanged, since that is just a scalar,

$$\omega_{\bs{x}'}'(\bs{v}'(\bs{x}')) = \omega_{\bs{x}}(\bs{v}(\bs{x})).$$

For example, you can think of $\omega$ as returning the amount of work done in a given direction vector $\bs{v}$ at a point $\bs{x}$. Since we know how vectors transform, $\bs{v}' = J\bs{v}$, the covector transformation rule follows immediately¹,

This is equivalent to

in index-free form.

Tensor Fields

Now that we know how vectors and covectors transform, we can write down the transformation rule for arbitrary $(p,q)$-tensors. For a general group element $g\in G$, a tensor field $T$ transforms as

$$(g \cdot T)(g\bs{x}) = \rho(J(\bs{x})) T(\bs{x})$$

where

$$\rho(J) = J^{\otimes p} \otimes ({J^{-T}}) ^{\otimes q}$$

is a representation of $GL(n)$ on the space where $T$ lives. This simply says that the transformation rule for a tensor involves a copy of $J$ for each upper index and a copy of $J^{-T}$ for each lower index. As an example, let us consider the conductivity tensor that governs Ohm’s law $j^i = \sigma^i{}_j E^j$. This is a $(1,1)$-tensor that transforms as

$$\sigma'^i{}_j = J^i{}_m (J^{-1})^n{}_{j} \sigma^{m}{}_n.$$

Since $(1,1)$ tensors transform by similarity transformations, their eigenvalues and trace are invariant. This is why the principal axes of conductivity are meaningful physical quantities.

Footnotes

1. Suppressing the base space evaluation points to avoid subscript hell ↩