Question

Differentiating in tensor notation.
\[
J^r_{t'}J^{t'}_s=\delta^r_s
\]

Answer 1

How to differentiate the above expression?

Answer 2

\[
\begin{align*}
&\phantom{{}={}}\frac{\partial}{\partial Z^k}\left[J^r_{t'}J^{t'}_s\right]\\
&=\frac{\partial}{\partial Z^k}\left[\frac{\partial Z^r}{\partial Z^{t'}}\left(Z'\left(Z\right)\right)\frac{\partial Z^{t'}}{\partial Z^s}\right]\\
&=\frac{\partial^2 Z^r}{\partial Z^{k'}Z^{t'}}\frac{\partial Z^{k'}}{\partial Z^k}\frac{\partial Z^{t'}}{\partial Z^{s}}+\frac{\partial Z^r}{\partial Z^{t'}}\frac{\partial^2 Z^{t'}}{\partial Z^k Z^s}\\
&=\frac{\partial^2 Z^r}{\partial Z^{k'}Z^{t'}}J^{k'}_k J^{t'}_s+J^r_{t'}\frac{\partial^2 Z^{t'}}{\partial Z^k Z^s}
\end{align*}
\]
What???
@Kainui

Answer 3

*

Answer 4

if \(J_{ik}\) is a cartesian tensor, then we can write these steps:
\[\begin{gathered}
\frac{\partial }{{\partial {x_k}}}\left( {{J_{rt'}}{J_{t's}}} \right) = \frac{\partial }{{\partial {x_k}}}{\delta _{rs}} \hfill \\
\hfill \\
{J_{rt'}}_{,k}\;{J_{t's}} + {J_{rt'}}\;{J_{t's}}_{,k} = 0 \hfill \\
\end{gathered} \]
wherein I have used the Einstein notation, and I have supposed that the right side of the starting equation is the kronecker symbol

Answer 5

What if it isn't?

Answer 6

@Kainui @Kainui Summoning Kainui!

Answer 7

Clearly the right hand side is 0 but I am not sure whether I got the left hand side right.

Answer 8

It is one of the question the lecturer gave in the Tensor Calculus and Calculus of Moving Surfaces playlist.

Answer 9

\[\frac{\partial^2 Z^r}{\partial Z^{k'} \partial Z^{t'}}J^{k'}_k J^{t'}_s+J^r_{t'}\frac{\partial^2 Z^{t'}}{\partial Z^k \partial Z^s}
\]
Yeah that's right, I just fixed one tiny thing with your notation, it's common to put the partial symbol for each term.
Additionally, I believe Grinfeld also ends up defining those in his book just out of convenience as:

\[J^r_{k't'}:=\frac{\partial^2 Z^r}{\partial Z^{k'} \partial Z^{t'}}\]

Answer 10

Yeah I was tired and forgot the partial symbols. Tensor calculus was slightly easier than it looks.

Answer 11

Yeah, at first it's super dense and then it sorta opens up and becomes pretty damn nice and convenient.

Answer 12

I don't really like the fact that tensor notation is really "concrete" though. It is not abstract enough and you kind of lose the big picture. Just look at the correct answer you provided. I don't see why on Earth it is 0.

Answer 13

Oh, why is this true you mean?

\[\frac{\partial}{\partial Z^k} (\delta^i_j)=0\]

Answer 14

Lol I'm not sure what you mean by "It is not abstract enough" xD

Answer 15

At the end of the day, most of this is a more streamlined notation for doing linear algebra that lets you do calculus and other cooler stuff too.

Answer 16

This.
\[\frac{\partial^2 Z^r}{\partial Z^{k'} \partial Z^{t'}}J^{k'}_k J^{t'}_s+J^r_{t'}\frac{\partial^2 Z^{t'}}{\partial Z^k \partial Z^s}=0\]
I don't see any meaning out of this expression and why on Earth the LHS should equal to 0?. That's why I say it is not abstract enough. The notation tells you how exactly to calculate the expression, which is nice, but the detailed instructions on how to calculate it befuddles the (geometric) meaning behind it. Compare the tensor notation to the more restricted normal notation (i.e. \(A\mathbf{x}=\mathbf{b}\)). The normal notation does not tell you how to calculate anything but it is much cleaner. You can clearly see a linear transformation \(A\) is applied to \(\mathbf{x}\). You can do a lot of work without looking at the nitty-gritty details of the matrix in question. Take this idea to the extreme and you get category theory. Here I have no idea what is happening despite I get a method to calculate it and this seems like the complete opposite of category theory, which you know what is happening but you have no idea how to calculate it.

Answer 17

I know pretty much nothing about category theory except that you have arrows and can compose them; I would like to learn more though.

Well I guess it depends on how you are thinking of it. You already knew that the prior equation was equal to the Kronecker delta which is really the identity matrix - a matrix of all only constants.

So since \(J^r_{t'}J^{t'}_s = Const\), then of course its derivative will be zero. But I feel like that probably isn't satisfying enough for you, since maybe you're more bothered by the fact that this thing on the left is equal to a constant or bothered by the fact that the derivative of these Jacobians ends up with the product rule and chain rule looking in the way that it does?

Answer 18

I am bothered by this identity:
\[\frac{\partial^2 Z^r}{\partial Z^{k'} \partial Z^{t'}}J^{k'}_k J^{t'}_s+J^r_{t'}\frac{\partial^2 Z^{t'}}{\partial Z^k \partial Z^s}=0\]Yes it is derived from the \(J^r_{t'}J^{t'}_s=\delta^r_s\). My gripe is that \(J^r_{t'}J^{t'}_s=\delta^r_s\) has a clear interpretation that "changing from a coordinate system to a second one then changing it back to the first one is the same as doing nothing", which clearly makes sense. The identity we derived however admits no clear interpretation. We know that the LHS is equal to 0, we can prove that it is 0 by differentiating \(J^r_{t'}J^{t'}_s=\delta^r_s\), but it is just confusing why it should be the case that LHS = 0.

I actually know nothing about category theory too. I just started reading a e-textbook called "Category Theory - A Gentle Introduction" by Peter Smith from University of Cambridge. Category theory is interesting to me because it avoids all sorts of numerical calculation. I absolutely suck at numerical calculation and I think a primary school student is better at crunching numbers than I do. I am a very lazy person as well. Category theory is highly abstract so you can apply theorems from category theory to multiple different situations. Prove it once and use it multiple times. Saves paper and saves time!

Answer 19

'My gripe is that \(J^r_{t'}J^{t'}_s=\delta^r_s\) has a clear interpretation that "changing from a coordinate system to a second one then changing it back to the first one is the same as doing nothing",'
---

Ahh, so the map \(\delta^r_s\) maps something to itself, which isn't nothing.

The \(change\) in this map is what is \(nothing\).

Translated: the \(derivative\) of this map is \(zero\).

Answer 20

The map \(δ^r_s\) maps something to itself, which isn't nothing, but it is *doing nothing* lol. The change in this map is of course nothing given that it is the identity map, but if someone is given \[ \frac{\partial^2 Z^r}{\partial Z^{k'} \partial Z^{t'}}J^{k'}_k J^{t'}_s+J^r_{t'}\frac{\partial^2 Z^{t'}}{\partial Z^k \partial Z^s}=0\] and this alone, I don't think they can figure out any meaning out of this, except verifying this is true for a few coordinate system.

Answer 21

Well you can tell it's the product rule expanded into something that's easier to calculate, just like

f'g+fg' = (fg)'