# Why is this term not zero?

1. Jul 25, 2010

### Nick R

PSI is a relative scalar field of weight w in the sense
that it transforms between the hatted and unhatted coordinate systems
according to,

$$\bar{\psi}=J^{w}\psi$$

Where J is the jacobian. According to the book, the ordinary partial
derivative of a relative scalar field is not a relative tensor field
because,

$$\frac{\partial\bar{\psi}}{\partial\bar{x}^{j}}=J^{w}\frac{\partial x^{h}}{\partial\bar{x}^{j}}\frac{\partial\psi}{\partial\bar{x}^{h}}+wJ^{w-1}\frac{\partial J}{\partial\bar{x}^{j}}\psi$$

(this transformation law is not the tranformation law of any sort
of relative scalar field unless the second term on the right it zero)

Later the book asserts the following relation relation (note the LHS
below is a factor in the second term on the RHS above):

$$\frac{\partial J}{\partial\bar{x}^{j}}=J\frac{\partial^{2}x^{h}}{\partial\bar{x}^{j}\partial\bar{x}^{l}}\frac{\partial\bar{x}^{l}}{\partial x^{h}}$$

... Why isn't this zero always??? Invoking the chain rule $$\sum\frac{\partial a}{\partial x^{h}}\frac{\partial x^{h}}{\partial b}=\frac{\partial a}{\partial b}$$
gives (einstein notation is being used - sums are implicit),

$$\frac{\partial^{2}x^{h}}{\partial\bar{x}^{j}\partial\bar{x}^{l}}\frac{\partial\bar{x}^{l}}{\partial x^{h}}=\frac{\partial^{2}x^{h}}{\partial\bar{x}^{j}\partial x^{h}}=\frac{\partial}{\partial\bar{x}^{j}}\left(\frac{\partial x^{h}}{\partial x^{h}}\right)=\frac{\partial}{\partial\bar{x}^{j}}\left(n\right)=0$$

This there something wrong in what is written immediatly above?

2. Jul 25, 2010

### Mute

$$\frac{\partial^{2}x^{h}}{\partial\bar{x}^{j}\partial\bar{x}^{l}}\frac{\partial\bar{x}^{l}}{\partial x^{h}} = \frac{\partial}{\partial \bar{x}^j}\left(\frac{\partial x^{h}}{\partial\bar{x}^l}\right)\frac{\partial\bar{x}^{l}}{\partial x^{h}} \neq \frac{\partial}{\partial \bar{x}^j}\left(\frac{\partial x^{h}}{\partial\bar{x}^l}\frac{\partial\bar{x}^{l}}{\partial x^{h}}\right) = \frac{\partial}{\partial \bar{x}^j}\left(\frac{\partial x^{h}}{\partial x^{h}}\right) = 0$$

The problem, as shown, is that you were implicitly moving a term inside the derivative when it wasn't constant, giving you the erroneous result (unless that term just so happened to be constant). What you did was the same as having two orthogonal vectors $\mathbf{u}(t)$ and $\mathbf{v}(t)$ so that $\mathbf{u}(t) \cdot \mathbf{v}(t) = 0$ for all t, and then concluding that $d\mathbf{u}/dt \cdot \mathbf{v} = 0$. This is hopefully obviously not correct; it was probably the einstein notation which obscured this error in the original question.

3. Jul 26, 2010

### Nick R

That isn't quite what I did - I should have posted exactly how I did the first step so now I will. In fact, I will show two different approaches to get the same thing (that this term is zero).

What is the implication if this term is zero??? It means that the ordinary partial derivative of a relative scalar is a relative tensor (i.e. the "covariant derivative" of a relative scalar is just the ordinary partial derivative). This would not be hard to believe for me, because the covariant derivative of a scalar is just the ordinary partial derivative.

The only thing I'm assuming here is that the coordinates are independent in the sense that

$$\frac{\partial x^{h}}{\partial x^{k}}=\delta_{k}^{h}$$

Which seems to be something they do throughout this entire book (Tensors, Differential Forms, and Variational Principles by Lovelock and Rund - right now I'm working in ch. 4 at around page 106).

Consider the following:

$$0=\frac{\partial}{\partial\bar{x}^{j}}\left(\frac{\partial x^{h}}{\partial\bar{x}^{l}}\frac{\partial\bar{x}^{l}}{\partial x^{h}}\right)=\frac{\partial^{2}x^{h}}{\partial\bar{x}^{j}\partial\bar{x}^{l}}\frac{\partial\bar{x}^{l}}{\partial x^{h}}+\frac{\partial^{2}\bar{x}^{l}}{\partial\bar{x}^{j}\partial x^{h}}\frac{\partial x^{h}}{\partial\bar{x}^{l}}$$

$$\frac{\partial^{2}x^{h}}{\partial\bar{x}^{j}\partial\bar{x}^{l}}\frac{\partial\bar{x}^{l}}{\partial x^{h}}=-\frac{\partial^{2}\bar{x}^{l}}{\partial\bar{x}^{j}\partial x^{h}}\frac{\partial x^{h}}{\partial\bar{x}^{l}}$$

Assuming the coordinates are independent,

$$-\frac{\partial^{2}\bar{x}^{l}}{\partial\bar{x}^{j}\partial x^{h}}\frac{\partial x^{h}}{\partial\bar{x}^{l}}=-\left[\frac{\partial}{\partial x^{h}}\left(\delta_{j}^{l}\right)\right]\frac{\partial x^{h}}{\partial\bar{x}^{l}}=0$$

Therefore,

$$\frac{\partial^{2}x^{h}}{\partial\bar{x}^{j}\partial\bar{x}^{l}}\frac{\partial\bar{x}^{l}}{\partial x^{h}}=0$$

Here is ANOTHER approach to show the same thing:

$$\frac{\partial^{2}x^{h}}{\partial\bar{x}^{j}\partial\bar{x}^{l}}\frac{\partial\bar{x}^{l}}{\partial x^{h}}=\frac{\partial}{\partial\bar{x}^{l}}\left(\frac{\partial x^{h}}{\partial\bar{x}^{j}}\right)\frac{\partial\bar{x}^{l}}{\partial x^{h}}$$

Now define

$$a_{j}^{h}\equiv\frac{\partial x^{h}}{\partial\bar{x}^{j}}$$

So we can rewrite the expression as,

$$\frac{\partial a_{j}^{h}}{\partial\bar{x}^{l}}\frac{\partial\bar{x}^{l}}{\partial x^{h}}$$

Invoking the chain rule,

$$=\frac{\partial a_{j}^{h}}{\partial x^{h}}=\frac{\partial^{2}x^{h}}{\partial\bar{x}^{j}\partial x^{h}}=0$$

Now if these things are true, we could conclude that

$$\frac{\partial J}{\partial\bar{x}^{j}}=J\frac{\partial^{2}x^{h}}{\partial\bar{x}^{j}\partial\bar{x}^{l}}\frac{\partial\bar{x}^{l}}{\partial x^{h}}=0$$

Because the jacobian J is not an indexed quantity

Therefore,

$$\frac{\partial\bar{\psi}}{\partial\bar{x}^{j}}=J^{w}\frac{\partial x^{h}}{\partial\bar{x}^{j}}\frac{\partial\psi}{\partial x^{h}}+wJ^{w-1}\frac{\partial J}{\partial\bar{x}^{j}}\psi$$

$$\frac{\partial\bar{\psi}}{\partial\bar{x}^{j}}=J^{w}\frac{\partial x^{h}}{\partial\bar{x}^{j}}\frac{\partial\psi}{\partial x^{h}}$$

So the ordinary partial derivative of a relative scalar (a quantity that transforms according to $$\bar{\psi}=J^{w}\psi$$) is a relative tensor of type (0,1).

This is contrary to what the book says! The book takes this term to be non-zero, writes it in terms of an affine connection, does a step or two of algebraic manipulation and defines covariant differentiation of a relative scalar field to be

$$\psi_{|h}=\frac{\partial\psi}{\partial x^{h}}-w\Gamma_{hl}^{l}\psi$$

Which is not the ordinary partial derivative...

4. Jul 26, 2010

### Mute

You cannot invoke the chain rule there. $a^h_j$ is a function of the $\bar{x}_\ell$, NOT the $x_\ell$. To invoke the chain rule in that expression $a^h_j(\bar{x})$ must be expressed in terms of x via a transformation

$$a^h_j(\bar{x}) = \frac{\partial \bar{x}^h(x)}{\partial x^\ell} \frac{\partial \bar{x}_j(x)}{\partial x_k} \tilde{a}^\ell_k(x)$$.

This is now expressed as a function of x, and so if you take the derivative of it with respect to $\bar{x}^i$ you can use the chain rule. Before you just had a function of $\bar{x}$ and were taking a derivative with respect to $\bar{x}^i$, so there was no chain rule to be done.

You have a similar problem here:

The first term on the right of the first = sign is a derivative of a function of $\bar{x}$ times a function of $x$. So, you really have

$$0=\frac{\partial}{\partial\bar{x}^{j}}\left(\frac{\partial x^{h}}{\partial\bar{x}^{l}}\frac{\partial\bar{x}^{l}}{\partial x^{h}}\right) = \frac{\partial^2 \bar{x}^h(\bar{x})}{\partial \bar{x}^j\partial \bar{x}^\ell} \frac{\partial \bar{x}^\ell(x)}{\partial x^h} + \frac{\partial x^h(\bar{x})}{\partial \bar{x}^\ell} \frac{\partial}{\partial \bar{x}^j}\left(\frac{\partial \bar{x}^\ell(x)}{\partial x^h}\right)$$

and you have to use the chain rule on this last term:
$$\frac{\partial}{\partial \bar{x}^j}\left(\frac{\partial \bar{x}^\ell(x)}{\partial x^h}\right) = \frac{\partial^2\bar{x}^\ell(x)}{\partial x^l \partial x^h} \frac{\partial x^k(\bar{x})}{\partial \bar{x}^j}$$

So, you see that your expressions weren't quite correct because you didn't take into account what each thing being differentiated was actually a function of.

Last edited: Jul 26, 2010
5. Jul 26, 2010

Thanks.