# Homework Help: Differential descriptions

1. May 8, 2012

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Suppose you see something like

$$\frac{\partial^2 \phi}{\partial X^{\mu}\partial X^{\nu}}$$

What I am interested here is the denominator. Is the term $$\partial X^{\mu}\partial X^{\nu}$$ simply X taken to the second derivative? I know one can write it like

$$\partial X^{\mu} (\frac{\partial \phi}{\partial X^{\nu}})$$

But what do we calculate when they are lumped together like $$\partial X^{\mu}\partial X^{\nu}$$?

Last edited by a moderator: May 9, 2012
2. May 8, 2012

What I mean is, is $$\partial X^{\mu}\partial X^{\nu}$$ equal to $$\partial^2 X^2$$?

3. May 8, 2012

### Staff: Mentor

I'll take a shot at it. Your notation suggests that $\phi$ is a function of at least two variables: X$\mu$ and X$\nu$

This notation:
$$\frac{\partial^2 \phi}{\partial X^{\mu}\partial X^{\nu}}$$

means

$$\frac{\partial}{\partial X^{\mu}}\left(\frac{\partial \phi}{\partial X^{\nu}}\right)$$

In other words, take the partial of $\phi$ respect to X$\nu$ and then take the partial of that function with respect to X$\mu$.

Last edited: May 9, 2012
4. May 8, 2012

### HallsofIvy

No, it is not. $X^\mu$ and $X^\nu$ are different variables. What have is similar to
$$\frac{\partial^2 \phi}{\partial x\partial y}$$
where x and y are the two variables in the xy-plane. The point is that if you are working in a situation where you have three or four or even more dimensions, it is simpler to write "$X^1$" and "$X^2$" rather than "x" and "y", for example.

Last edited by a moderator: May 9, 2012
5. May 9, 2012

Thanks, was unsure... but I guessed you approach. Thanks again!

6. May 9, 2012

Ok, I have a new question of the same type. Suppose then you might have

$$\frac{\partial \phi}{\partial X^j \partial X^j}$$

This seems to be purporting to the same direction $$X^{j}$$ yes? So how is this interpretated?

Assuming you can bring it out again like so

$$\frac{\partial}{X^j}(\frac{\partial \phi}{ \partial X^j})$$

Does anything get squared now?

7. May 9, 2012

### Staff: Mentor

No, this doesn't make any sense.
It would have to be
$$\frac{\partial^2\phi}{\partial X^j \partial X^j}$$

This is the second partial of $\phi$ with respect to Xj.
It would be interpreted as the partial with respect to Xj of the partial of $\phi$ with respect to Xj.
This isn't quite right.

$$\frac{\partial}{\partial X^j}(\frac{\partial \phi}{ \partial X^j})$$

Compare this to the related Leibniz notation for the second derivative.
$\frac{d^2 y}{dx^2} = \frac{d}{dx} \left(\frac{dy}{dx}\right)$

8. May 9, 2012

Sorry I missed out the $$\partial^2$$. I knew it should have been there...

anyway, this is what I am asking. Just in short, does this $$\partial X^j \partial X^j$$ mean $$\partial X^{j^2}$$?

For

$$\frac{\partial^2 \phi}{\partial X^{\mu} \partial X^{\nu}}$$ you can make it

$$\frac{\partial}{\partial X^{\mu}}(\frac{\partial \phi}{\partial X^{\nu}}$$

why can't I do this for something like

$$\frac{\partial^2 \phi}{\partial X^{j} \partial X^{j}}$$

Is the denominator really $$\partial X^{j^2}$$ because it seem from HallsOfIvy that the denominator in

$$\frac{\partial^2 \phi}{\partial X^{\mu} \partial X^{\nu}}$$

Are seperate variables and so they cannot be squared. With the new case I have given, this is not obvious. You never answered my question very helpfully.

9. May 9, 2012

Let me numb the question down:

HallsofIvy said that

$$\frac{\partial^2 \phi}{\partial X^{\mu} \partial X^{\nu}} = \frac{\partial^2}{\partial X^{\mu}} \frac{\partial \phi} {\partial X^{\nu}}$$

I knew this and it is simple enough. I asked if the product $$X^{\mu}X^{\nu}$$ was some squared value, in which it was $$X^2$$ because they were separate variables. Now I am asking the same of

$$\frac{\partial^2 \phi}{\partial X^{j} \partial X^{j}}$$

Are they separate variables? If not, then can one express it as being squared?

Last edited by a moderator: May 9, 2012
10. May 9, 2012

### Staff: Mentor

No, they're not separate, and yes, you can express that derivative as
$$\frac{\partial^2 \phi}{\partial X^{j2}}$$

That looks pretty messy with the j index as a superscript. That's probably a good reason to write the indexes as subscripts, as X1, X2, and so on.

Doing that, the partial would look like this:
$$\frac{\partial^2 \phi}{\partial X_j^2}$$

Last edited: May 9, 2012
11. May 9, 2012

Right, that's good then. Thank you, one last question. In respect to these $$X_{j}^{2}$$ 's, when can it not be applied to the Leibniz rule?

This has caused a new confusion.

$$\frac{\partial^2 y}{\partial x^2} = \frac{\partial}{\partial x} (\frac{\partial y}{\partial x}$$

especially this part $$\frac{\partial^2 y}{\partial x^2}$$ seems like an identical form of

$$\frac{\partial^2 \phi}{\partial X^{2}_{j}}$$

Thank you

12. May 9, 2012

### Staff: Mentor

He didn't say that. Nor did I. Here's what was said:
$$\frac{\partial^2 \phi}{\partial X^{\mu} \partial X^{\nu}} = \frac{\partial}{\partial X^{\mu}} \frac{\partial \phi} {\partial X^{\nu}}$$

Notice on the right that it is NOT the 2nd partial of the partial - it's the partial of the partial. I removed an exponent of 2 that you had.

13. May 9, 2012

Sorry, I carried that on by a paste.

This has caused a new confusion.

$$\frac{\partial^2 y}{\partial x^2} = \frac{\partial}{\partial x} (\frac{\partial y}{\partial x})$$

especially this part $$\frac{\partial^2 y}{\partial x^2}$$ seems like an identical form of

$$\frac{\partial^2 \phi}{\partial X^{2}_{j}}$$

14. May 9, 2012

Now why can't the latter be expressed using Leibniz rule?

15. May 9, 2012

### Staff: Mentor

As long as each partial is taken with respect to the same independent variable.
Yes. Let's keep to the ordinary derivative notation, since it's easier to type.

$$\frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d^2 y}{dx^2}$$

Working with partial derivatives, it would look pretty much the same. Again, I'm assuming that you're taking the partial with respect to the same variable.

16. May 9, 2012

You said it couldn't, even in the correct format. It has me a bit confused see...

17. May 9, 2012

### Staff: Mentor

I don't understand what you're asking.

18. May 9, 2012

You use some wicked latexing... I have never seen the latex you use... anyway, you qouted the expression I gave you and you said:

This isn't quite right.

$$\frac{\partial}{\partial X^j}(\frac{\partial \phi}{ \partial X^j})$$

Compare this to the related Leibniz notation for the second derivative.
$$\frac{d^2 y}{dx^2} = \frac{d}{dx} \left(\frac{dy}{dx}\right)$$

19. May 9, 2012

### Staff: Mentor

Click any of the things I wrote and you can see how I did it.
This is the same as

$$\frac{\partial^2 \phi}{\partial X_j^2}$$

But you can't collapse the expression if you're doing this:
$$\frac{\partial}{\partial X_j} \frac{\partial \phi}{\partial X_i}$$

Here the partials are with respect to different variables.

20. May 9, 2012

So, what you are saying is I need the paranthesis? That is the only difference I see...

21. May 9, 2012

under closer inspection, I see that the X^i is different, so X^j and X^i are different variables yes? So what you are saying I can't use the L. rule? Yes?

$$\frac{\partial}{\partial X_j} \frac{\partial \phi}{\partial X_i}$$

22. May 9, 2012

If that is true, then this has caused a greater confusion now.

In a general relativistic coursework, we may the equation

$$g^{\mu \nu} \frac{\partial^2 \phi}{\partial X^{\mu}X^{\nu}}=0$$

which is where my first expression arose from.

In certain derivation, it says I can do this:

$$\frac{\partial}{\partial X^{\mu}}(g^{\mu \nu}\frac{\partial \phi}{\partial X^{\nu}})$$

But what I think you are saying... is I can't?

23. May 9, 2012

### Staff: Mentor

Yes, X^j and X^i are different variables, so you can't use the collapsed notation, writing Xj * Xj as Xj2, but that doesn't have anything to do with what you're calling the Leibniz Rule. It's just notation that someone invented to be able to write that partial in a more compact form.

24. May 9, 2012

Then why does the derivation above be allowed to collapse it in such a way? (My general relativity equations)

25. May 9, 2012

### Staff: Mentor

I don't think this is right.
I think it should be
$$g^{\mu \nu} \frac{\partial^2 \phi}{\partial X^{\mu} \partial X^{\nu}}=0$$

Then you could break it up as you have below.