# Tensors (summation convention)

1. Sep 11, 2010

### bremvil

Hi everyone,

I recently started a course on continuum mechanics. It started with the mathematical background of transforming tensors with contravariant and/or covariant indices. There is one thing I don't understand and it should be really straight forward. I hope you can give me a hint.

http://s35.photobucket.com/albums/d174/Brasempje/?action=view&current=tensor.jpg

In equation 5 on the page that I linked above I do not see how I can get from term 3 to term 4. I end up with 3 kronecker delta's instead of just 1. Since there is a double index 'm' summation can be performed and it should lead to the same result as when you use a 'shortcut'. I can show what I do using 'latex' notation, with ^ = superscript, _ = subscript, d = kronecker delta, t = theta

'term 3' in Equation (5) reads:
(dt^i/dx^m) * (dx^m/dt^j)

in case I decide to do summation this equation will turn into:
(dt^i/dx^1) * (dx^1/dt^j) + (dt^i/dx^2) * (dx^2/dt^j) +
(dt^i/dx^3) * (dx^3/dt^j)

Each component of vector x is a function of all three components of vector theta. And each component of vector theta is a function of all three components of vector x. By the chain rule the last expression would become

dt^i/dt^j + dt^i/dt^j + dt^i/dt^j

this is:
d^i_j + d^i_j + d^i_j = 3 * d^i_j

so in case I decide to do the summation I end up with something different than I
would expect. 3 kronecker delta's instead of 1! Is there any objection to using a sum in this
case?

with kind regards,

Bremvil

2. Sep 11, 2010

### Fredrik

Staff Emeritus
That equality is just the chain rule. It doesn't have anything to do with tensors or properties of the Kronecker delta.

When $g:\mathbb R^n\rightarrow\mathbb R^n$ and $f:\mathbb R^n\rightarrow\mathbb R$, I like to write the chain rule like this:

$$(f\circ g)_{,i}(x)=f_{,j}(g(x)) g^j{}_{,i}(x)$$

Here ",i" denotes partial derivative with respect to the ith variable, and $g^j$ is the jth component of the function g. If $f:\mathbb R^n\rightarrow\mathbb R^n$, we have

$$(f\circ g)^i(x)=(f(g(x))^i=f^i(g(x))=f^i\circ g(x)$$

so

$$\delta^i_j=(f\circ f^{-1})^i{}_{,j}(x)=(f^i\circ (f^{-1}))_{,j}(x)$$

Define $g=f^{-1}$ to unclutter the notation somewhat. Then the above is

$$=(f^i\circ g)_{,j}(x)=f^i{}_{,k}(g(x))g^k{}_{,j}(x)$$

and...uhh...I can't explain why this is written in the form

$$\frac{\partial A^i}{\partial B^k}\frac{\partial B^k}{\partial A_j}$$

without explaining partial derivatives with respect to a coordinate system. (Edit: Actually I can. See the comments at the end of the post). On a manifold, expressions like f(x+h) don't work, because in general, addition isn't defined for points in the manifold. This is why we have to use a coordinate system to define partial derivatives. A coordinate system is a function $x:U\rightarrow R^n$, where U is an open subset that contains the point p at which we want to define the partial derivative. If f is a function from the manifold to the real numbers, we define

$$\frac{\partial f}{\partial x^i}(p)=(f\circ x^{-1})_{,i}(x(p))$$

In particular, if y is another coordinate system,

$$\frac{\partial y^j}{\partial x^i}(p)=(y^j\circ x^{-1})_{,i}(x(p))$$

The f above is a coordinate change function, i.e. an expression of the form $A\circ B^{-1}$, where A and B are coordinate systems. So we have

$$\delta^i_j=f^i{}_{,k}(g(x))g^k{}_{,j}(x)=(A^i\circ B^{-1})_{,k}(g(x))(B^k\circ A^{-1}){}_{,j}(x)=\frac{\partial A^i}{\partial B^k}(A^{-1}(x))\frac{\partial B^k}{\partial A_j}(A^{-1}(x))$$

Edit: It turned out to be easier than I thought to explain the notation. No techniques from differential geometry are needed.

$$\delta^i_j=f^i{}_{,k}(g(x))g^k{}_{,j}(x)=f^i{}_{,k}(g(x))g^k{}_{,j}(f(g(x)))=\frac{\partial f^i}{\partial g^k}(g(x))\frac{\partial g^k}{\partial f_j}(f(g(x))$$

Last edited: Sep 11, 2010
3. Sep 11, 2010

### bremvil

Dear Fredrik,

Thanks for your reply. I read through it carefully but I find it quite difficult, my math background is not as strong as yours. So I'm still not really there yet. In your explanation you started with:

$$(f\circ g)_{,i}(x)=f_{,j}(g(x)) g^j{}_{,i}(x)$$

but this step is basically my entire problem! The index j appears twice which would mean summation right. I will try to write down my original problem with latex. Could you instead maybe tell where I am making the error?

http://s35.photobucket.com/albums/d174/Brasempje/?action=view&current=tensor.jpg
equation 5, the problem is in step from term 3 to term 4.

$$x_i = x_i(\theta_1 ,\theta_2 , \theta_3)$$

$$\theta_i = \theta_i(x_1 ,x_2 , x_3)$$

If I decide to apply summation over 'index m' in the equation below I get:

$$\frac{\partial \theta^i}{\partial x^m}\frac{\partial x^m}{\partial \theta^j} = \frac{\partial \theta^i}{\partial x^1}\frac{\partial x^1}{\partial \theta^j} + \frac{\partial \theta^i}{\partial x^2}\frac{\partial x^2}{\partial \theta^j} + \frac{\partial \theta^i}{\partial x^3}\frac{\partial x^3}{\partial \theta^j}$$

by the chain rule I get

$$\frac{\partial \theta^i}{\partial \theta^j} + \frac{\partial \theta^i}{\partial \theta^j} + \frac{\partial \theta^i}{\partial \theta^j} = \delta^i_j + \delta^i_j + \delta^i_j = 3\delta^i_j$$

So I get 3 times the delta function instead of only 1 delta function. I guess I should not do summation for some reason, but I don't understand why. There is a double 'index m' so a summation should be justified. In every part of the book 'classical and computational solid mechanics' the presence of a double index means summation.

4. Sep 11, 2010

### Fredrik

Staff Emeritus
The first line is correct, but that's not the chain rule. By the chain rule, what you have on the upper right is equal to

$$\frac{\partial\theta^i}{\partial\theta_j}$$

(once, not three times).

5. Sep 11, 2010

### bremvil

So basically you are saying that the term on the top right:

$$\frac{\partial \theta^i}{\partial x^m}\frac{\partial x^m}{\partial \theta^j} = \frac{\partial \theta^i}{\partial x^1}\frac{\partial x^1}{\partial \theta^j} + \frac{\partial \theta^i}{\partial x^2}\frac{\partial x^2}{\partial \theta^j} + \frac{\partial \theta^i}{\partial x^3}\frac{\partial x^3}{\partial \theta^j}$$

equals a single delta function? The way I interpret it, each term within the expression above is a single delta function.

6. Sep 11, 2010

### Fredrik

Staff Emeritus
You're not applying the chain rule correctly. Another example:

$$\frac{d}{dx}f(g(x),h(x))=\frac{\partial f}{\partial g}\frac{dg}{dx}+\frac{\partial f}{\partial h}\frac{dh}{dx}\neq \frac{df}{dx}+\frac{df}{dx}$$

7. Sep 11, 2010

### bremvil

I finally see it! Thanks a lot.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook