# I Deriving tensor transformation laws

#### Daniel_C

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.

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

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

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

Therefore, the basis of my tensor transforms like this:

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

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

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

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)?

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#### Orodruin

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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.

"Deriving tensor transformation laws"

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