Understanding the Denominator in Differential Descriptions

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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}?
 
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What I mean is, is \partial X^{\mu}\partial X^{\nu} equal to \partial^2 X^2?
 
help1please said:

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##.
 
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help1please said:

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}?
 
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Thanks, was unsure... but I guessed you approach. Thanks again!
 
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?
 
help1please said:
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 Xj.
help1please said:
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 Xj of the partial of ##\phi## with respect to Xj.
help1please said:
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)##
help1please said:
Does anything get squared now?
 
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.
 
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?
 
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  • #10
help1please said:
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:
$$ \frac{\partial^2 \phi}{\partial X_j^2}$$
 
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  • #11
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
help1please said:
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.
help1please said:
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?
 
  • #13
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
Now why can't the latter be expressed using Leibniz rule?
 
  • #15
help1please said:
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.
help1please said:
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.
help1please said:
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
 
  • #16
You said it couldn't, even in the correct format. It has me a bit confused see...
 
  • #17
help1please said:
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.
 
  • #18
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
help1please said:
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.
help1please said:
.. 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.
help1please said:
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)
 
  • #20
So, what you are saying is I need the paranthesis? That is the only difference I see...
 
  • #21
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
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
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
Then why does the derivation above be allowed to collapse it in such a way? (My general relativity equations)
 
  • #25
help1please said:
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.
help1please said:
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?
 
  • #26
Sorry, missed a partial! but thank you.
 
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