Understanding the Denominator in Differential Descriptions

[help1please]

Homework Statement

Homework Equations

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 $$\partial X^{\mu}\partial X^{\nu}$$?

 

[help1please]

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

 

[Mark44]

Homework Statement

Homework Equations

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 $$\partial X^{\mu}\partial X^{\nu}$$?

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^$\mu$.

 

[HallsofIvy]

Homework Statement

Homework Equations

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?'

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.

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 $$\partial X^{\mu}\partial X^{\nu}$$?

 

[help1please]

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

 

[help1please]

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?

 

[Mark44]

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

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

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 X^j.

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

It would be interpreted as the partial with respect to X^j of the partial of $\phi$ with respect to X^j.

Assuming you can bring it out again like so

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

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

Does anything get squared now?

 

[help1please]

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 separate 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.

 

[help1please]

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?

 

[Mark44]

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?

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

 

[help1please]

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

 

[Mark44]

Let me numb the question down:

HallsofIvy said that

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

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.

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?

 

[help1please]

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

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

 

[help1please]

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

 

[Mark44]

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?

As long as each partial is taken with respect to the same independent variable.

This has caused a new confusion.

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

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.

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

 

[help1please]

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

 

[Mark44]

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

I don't understand what you're asking.

 

[help1please]

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

 

[Mark44]

You use some wicked latexing... I have never seen the latex you use.

Click any of the things I wrote and you can see how I did it.

.. anyway, you qouted the expression I gave you and you said:

This isn't quite right.

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

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.

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

 

[help1please]

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

 

[help1please]

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

 

[help1please]

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?

 

[Mark44]

Yes, X^j and X^i are different variables, so you can't use the collapsed notation, writing X_j * X_j as X_j², 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.

 

[help1please]

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

 

[Mark44]

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

In a general relativistic coursework, we may the equation

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

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.

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?

 

[help1please]

Sorry, missed a partial! but thank you.

 

[Mark44]

Sure, you're welcome!