I Deriving tensor transformation laws

[Daniel_C]

TL;DR Summary

I'm attempting to understand how the definition of a tensor in terms of how it transforms arises from demanding invariance of the tensor when we define it as a multilinear map

Hi, I'm worried I've got a grave misunderstanding. Also, throughout this post, a prime mark (') will indicate the transformed versions of my tensor, coordinates, etc.

I'm going to define a tensor.

$$T^\mu_\nu \partial_\mu \otimes dx^\nu$$

Now I'd like to investigate how the tensor transforms under an arbitrary coordinate transformation where I require that the tensor is invariant under the coordinate transformation. In order to understand how partial-mu transforms, I'm going to apply the chain rule for partial derivatives.

$$\begin{equation} \partial'_\alpha = \frac{\partial}{\partial x'^\alpha} = \frac{\partial x^\mu}{\partial x'^\alpha}\partial_\mu \end{equation}$$

Similarly, I can write down how the basis one-forms transform.

$$\begin{equation} dx'^\alpha = \frac{\partial x'^\alpha}{\partial x^\mu} dx^\mu \end{equation}$$

Therefore, the basis of my tensor transforms like this:

$$\begin{equation} T'^\alpha_\beta \partial'_\alpha \otimes dx'^\beta = T^\mu_\nu \frac{\partial x^\mu}{\partial x'^\alpha} \partial_\mu \otimes \frac{\partial x'^\alpha}{\partial x^\mu} dx^\nu \end{equation}$$

Which mirrors my more familiar definition of a tensor in terms of how it's components transform:

$$\begin{equation} T'^\alpha_\beta = \frac{\partial x'^\alpha}{\partial x^\mu}\frac{\partial x^\nu}{\partial x'^\beta} T^\mu_\nu \end{equation}$$

Is it correct to write down the transformation of the tensor in the way I did in equation (3)? Is this equivalent to equation (4)?

Please help me if I've had a misunderstanding. Thank you very much for any help.

 

[Orodruin]

Summary: I'm attempting to understand how the definition of a tensor in terms of how it transforms arises from demanding invariance of the tensor when we define it as a multilinear map

Is it correct to write down the transformation of the tensor in the way I did in equation (3)? Is this equivalent to equation (4)?

Yes. Yes (the base tensors $\partial_\mu \otimes dx^\nu$ form a complete and linearly independent basis).

Edit: Also note that the tensor product is linear and that the partial derivatives are just numbers that you can move out of it.

 

[Daniel_C]

Yes. Yes (the base tensors $\partial_\mu \otimes dx^\nu$ form a complete and linearly independent basis).

Edit: Also note that the tensor product is linear and that the partial derivatives are just numbers that you can move out of it.

Thanks so much for the quick reply!

 

[Pencilvester]

Therefore, the basis of my tensor transforms like this:

$$T'^\alpha_\beta \partial'_\alpha \otimes dx'^\beta = T^\mu_\nu \frac{\partial x^\mu}{\partial x'^\alpha} \partial_\mu \otimes \frac{\partial x'^\alpha}{\partial x^\mu} dx^\nu$$...
Is it correct to write down the transformation of the tensor in the way I did in equation (3)? Is this equivalent to equation (4)?

Note that the RHS of your equation 3 is not a valid expression as you have four $\mu$ indices. You need to change either the inner two $\mu$’s or the outer two for this to make sense. But even then, if you’re wanting to use equations 1 and 2 and the fact that a tensor remains the same despite a change in basis in order to derive eq. 4, then this might be a more intuitive approach:
$$
\begin{equation*}
\begin{split}
\mathbf T = T^{\bar \alpha}_{\bar \beta} ~ \partial_{\bar \alpha} \otimes dx^{\bar \beta} & = T^{\bar \alpha}_{\bar \beta} (\frac{\partial x^\mu}{\partial \bar x^{\bar \alpha}} \partial_\mu ) \otimes (\frac{\partial \bar x^{\bar \beta}}{\partial x^\nu} dx^\nu) \\
& = (T^{\bar \alpha}_{\bar \beta} \frac{\partial x^\mu}{\partial \bar x^{\bar \alpha}} \frac{\partial \bar x^{\bar \beta}}{\partial x^\nu}) \partial_\mu \otimes dx^\nu
\end{split} \tag{5}
\end{equation*}
$$And because $\mathbf T$ is unchanged by a change in basis, we can say that
$$T^{\bar \alpha}_{\bar \beta} ~ \partial_{\bar \alpha} \otimes dx^{\bar \beta} = T^\mu_\nu ~ \partial_\mu \otimes dx^\nu$$
And therefore
$$T^\mu_\nu = T^{\bar \alpha}_{\bar \beta} \frac{\partial x^\mu}{\partial \bar x^{\bar \alpha}} \frac{\partial \bar x^{\bar \beta}}{\partial x^\nu}$$
We can rearrange like we did in the 2nd line of eq. 5 precisely because of the reason @Orodruin gave in the edit of his post.