1
$\begingroup$

I'm studying general relativity and and learning up on tensors through a lecture series. It says that $\omega_\mu$ represents a 1-form in the $x^\mu$ coordinate system. The coordinate independent 1-form is denoted by $\omega$ as $\omega = \omega_\mu \mathrm{d}x^\mu$. However, if I assume the definition of the 1-form when using two different coordinate systems $x^\mu$ and $x^{\mu'}$, the following is true

$$\omega_{\mu'} = \frac{\partial x^\mu}{\partial x^{\mu'}} \omega_\mu$$

then I can prove the following:

$$\omega_{\mu'} \mathrm{d}x^{\mu'} = \omega_\mu \mathrm{d}x^\mu$$

However, this is how a scalar transforms. So, is $\omega = \omega_\mu \mathrm{d}x^\mu$ a scalar? Or is this just signifying that the quanitity is coordinate independent? If given a generic quanitty that transforms like this, how do I know which one is the case?

$\endgroup$
1
$\begingroup$

Your math is correct. What you have done is confirmed that a 1-form has a coordinate-independent meaning. The fact that $\omega_\mu dx^\mu$ is invariant under coordinate transformations can be used to deduce how the components $\omega_\mu$ of the 1-form must transform: they must transform in such a way that leaves $\omega_\mu dx^\mu$ invariant when $x^\mu$ is transformed. The 1-form is independent of coordinates. The components of the 1-form are not.

Here's a perspective that might help clarify things.

Tensor fields of all types (scalar, vector, 1-form, and so on) can be defined without referring to coordinates at all. Even though using coordinates is still convenient for specific calculations, knowing the coordinate-independent definitions can help clarify how things transform under diffeomorphisms (or under passive coordinate transformations; the difference doesn't really matter in general relativity).

Here are the definitions. Let $M$ denote the underlying smooth manifold. (If we're talking about ordinary spacetime, then $M$ is a four-dimensional smooth manifold.) Also, let $\mathbb{R}$ denote the real numbers.

  • A scalar field is a smooth map from $M$ to $\mathbb{R}$. This one's easy.

  • A vector field is more interesting. To motivate the definition, recall that when working in terms of a coordinate system, a vector field is essentially a linear combination of partial derivatives. Partial derivatives satisfy the "product rule", $\partial_a(fg)=(\partial_a f)g+f(\partial_a g)$. I'm using parentheses here for grouping factors, not to hold the input to a function. With that in mind, here's the coordinate-independent definition of a vector field. If $S$ is the set of scalar fields, then a vector field $v$ is a smooth linear map from $S$ to itself, satisfying $$ v(fg)=(vf)g+f(vg) $$ for all scalar fields $f,g\in S$. In words, the map $v$ must be a derivation. (Saying that $v$ is a "linear" map makes sense because linear combinations of scalar fields are defined in the obvious way.)

  • A 1-form may now be defined simply as a linear map from the set of vector fields to $\mathbb{R}$.

None of these definitions refer to coordinates at all, and that's a good thing, because that means that we can use them to deduce how the components of these various fields transform under coordinate transformations. Tensors are independent of coordinates. The components of tensors depend on which coordinate system we use.

By the way, it's important that none of these definitions depend on the spacetime metric. This is important because the metric field itself is a tensor field, and we don't want our definitions of tensor fields to be circular.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Chiral Anomaly already has a nice answer, but to elaborate on his answer: There are always this tug of war between physicist's and mathematician's way of expressing things. A mathematician would (rightfully) express things in a way that is independent of any coordinate patch on a manifold.

For example, you would read statements like "Let $\omega$ be a section of the cotangent bundle" (i.e. a one form) which doesn't refer to any coordinate patches on the manifold and is a globally defined coordinate system. Though this approach is formally favourable. E.g. otherwise when you are proving something, you would always (very annoyingly) have to show that the thing that you proved doesn't depend on the coordinate chart that you have chosen.

However, this approach is not very favorable for computations, which physicists are more interested in. That's why physicists almost always describe things in local coordinates, so e.g. you would read statements like "The covector $\omega_\mu$ is given as $\omega_\mu = x \, \mathrm d y$. The problem with this approach is choosing different coordinate patches (s. above). The same object that you call a covector $\omega_\mu$ "looks different" in different coordinate patches. To remedy this, physicists have "transformation rules" so that the object $\omega_\mu$ which is strictly speaking defined on a coordinate patch is globally defined.

Note also that the story is very analogous to usual vectors in a vector space. A vector $v=(1,1) \in \mathbb R^2$ can be expressed in standard basis as

$$ v = v^i e_i = 1 e_1 + 1 e_2 \implies v^i = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$$

however if you choose another basis e.g. $b_1 = (1,1)$ and $b_2 = (1,-1)$ the same vector $v$ can be expressed as

$$ v = \tilde v^i b_i = 1 b_1 + 0 b_2 \implies \tilde v^i= \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$

The vector $v$ is still the same thing and has not transformed. However, its components $v^i$ transforms from a given basis to another.

The upshot is the following: The global object $\omega$ as a covector field does not transform and in particular it does not transform as a scalar since this language of "transformation rules" doesn't make sense on this global level. However, if you express $\omega$ in a coordinate patch as $\omega = \omega_\mu \mathrm d x^\mu$, then the components $\omega_\mu$ "transform as a covector" when you switch coordinates.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.