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So, I'm currently teaching myself the basics of tensors, and one of the definitions I continually run into for Contravariant and Covariant tensors is that they transform according to $\bar A^i = \frac{\partial \bar x^i}{\partial x^j} A^j$ and $\bar A^i = \frac{\partial x^j}{\partial \bar x^i} A^j$ respectively, where the bars indicate the new coordinate system. Specifically, one of the texts I'm using is A.J. McConnell's Applications of Tensor Analysis, in which he describes that an arbitrary tensor, $a_{np}^m$, has the transformation $\bar a_{st}^r = c_m^r \gamma_s^n \gamma_t^p a_{np}^m$, where he has only defined $c$ as a linear transformation, and $\gamma$ as the cofactor of $c$ divided by the determinant, $|c|$. He then goes on to say that upon contraction of $r$ and $t$, that $c_m^r \gamma_r^p = \delta_m^p$.

Firstly, I assume that $c_m^r$ and $\gamma_r^p$ can, in at least some instances (pls correct me if I'm wrong), be viewed as $c_m^r = \frac{\partial \bar x^r}{\partial x^m}$, and $\gamma_r^p = \frac{\partial x^p}{\partial \bar x^r}$, based on my previously mentioned definitions.

My question, given that my assumption is correct, is why does $\frac{\partial \bar x^r}{\partial x^m} \frac{\partial x^p}{\partial \bar x^r} = \delta_m^p$ ? Or, at least, why does the case $m \neq p$ lead to $\frac{\partial \bar x^r}{\partial x^m} \frac{\partial x^p}{\partial \bar x^r} = 0$. The other case makes sense to me.

Moreover, in what circumstances does this apply? For all coordinate transformations? Or just a subset (i.e. orthonormal, orthogonal, non-curvilinear)? At the very least, non-linear coordinate transformations wouldn't necessarily have this condition, right?

Any help would be greatly appreciated!

Person Person
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  • Thinking about it component wise won't be helpful. Instead, think of the two partial derivatives being evaluated at a specific point. Then the partials are the usual change of basis matrices from linear algebra. The two partials you have describe the opposite change of bases, so they are inverses of each other, hence their product is the identity matrix. Since this logic applies at every point, then the contraction of the two tensors gets you the delta. – Ninad Munshi Nov 13 '19 at 22:21
  • I see, thanks! Since this logic only applies to linear transformations though, are tensors useful for non-linear transformations? Say from Cartesian to Spherical? – Person Person Nov 13 '19 at 23:10
  • No this logic is applicable to tensors. That's why I said when you plug in a point, you are dealing with a change of basis between vector spaces (the vector spaces being the tangent space at that point). – Ninad Munshi Nov 14 '19 at 00:03
  • Right, no, I get it applies to tensors. My question was more along the lines of: Say I wanted to transform a tensor from one coordinate system to another. If the transformation between the two coordinate systems is linear (i.e. each coordinate of the new system is a linear combination of the old ones), the tensor transformation is easily defined as describee above. However, what happens in the case when the transformation is non-linear? Say from Cartesian to Spherical? I don't think this has been answered here (though maybe I'm wrong). – Person Person Nov 14 '19 at 00:25
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    The point I was trying to make is that the partial derivatives act locally, and they are always linear because they are operators on the tangent space of a point. – Ninad Munshi Nov 14 '19 at 00:48
  • Oh, I see what you're saying. Thanks! – Person Person Nov 14 '19 at 16:06

1 Answers1

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The expression you are interested in obeys the chain rule.

$$ \sum_{r}\frac{\partial \bar{x}^r }{\partial x^m } \frac{\partial x^p }{\partial \bar{x}^r } = \frac{\partial x^p }{\partial x^m} $$

Now consider $\frac{\partial x^p}{\partial x^m}$. This expression will equal $1$ if $p=m$ because $x^p$ and $x^m$ would refer to the same variable (the partial derivative of a variable with respect to itself is $1$). Then if $p\neq m$ the expression will equal zero because $x^p$ and $x^m$ would refer to different variables (the partial derivative of a variable with respect to another variable is $0$).

These observations mean that $\frac{\partial x^p}{\partial x^m} = \delta^p_m$.

We conclude that,

$$ \boxed{\sum_{r}\frac{\partial \bar{x}^r }{\partial x^m } \frac{\partial x^p }{\partial \bar{x}^r } = \delta^p_m }$$

$c$ and $\gamma$ are equal to the partial derivative expressions that you identified. Not that by definition $\gamma$ is the inverse of $c$ (the cofactor matrix divided by the determinant is the inverse matrix).

These rules would apply to all smooth coordinate transformations from a manifold onto itself. If the coordinate transformation is very non-smooth to the point where it doesn't have a well defined derivative then you will definitely run into problems for sure.

Spencer
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  • Thanks for the answer! Quick question though. You said "the partial derivative of a variable with respect to another variable is 0." However, doesn't this assume an orthogonal system? If I were working with a transformation involving skew coordinate systems, wouldn't that not necessarily be true? Or am I missing something? – Person Person Nov 17 '19 at 22:15
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    The statement it true whether the coordinates are skew or orthogonal. Consider the coordinates $u=x+2y$ and $v=x+3y$. $\partial / \partial u = (\partial x / \partial u) \partial / \partial x + (\partial y/\partial u) \partial / \partial y = 3 \partial / \partial x - \partial / \partial y $. Now consider $\partial u / \partial v = 3 \partial v / \partial x - \partial v / \partial y = 0 $. Note that $u$ and $v$ form skew coordinates, but the partial derivative of $v$ with respect to $u$ is still $0$. – Spencer Nov 18 '19 at 01:11
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    You will learn that partial derivatives transform covariantly and coordinates transform contravariantly. This means that their transformation matrices are inverses. The inverse relationship ensures that "the partial derivative of one variable with respect to another variable is $0$" will be true in all coordinate systems. – Spencer Nov 18 '19 at 01:14