Reciprocal Theorem proof in 3 variables

[RJLiberator]

Homework Statement

w is a function of three variables x, y, and z. Prove that
$$\frac{\partial w}{\partial x}_{y,z} = \frac{1}{\frac{\partial x}{\partial w}}_{y,z}$$

Homework Equations

The Attempt at a Solution

w=w(x,y,z)

$$dx = \frac{\partial w}{\partial x}_{y,z}dx +\frac{\partial w}{\partial y}_{x,z}dy+\frac{\partial w}{\partial z}_{x,y}dz$$

Now, we need to use the trick to showcase one of the variables as a function of the others.

$$dx=\frac{\partial x}{dw}_{y,z}dw+\frac{\partial x}{dy}_{z}dy+\frac{\partial x}{dz}_{y}dz$$

Now, we plug in dx from the second equation into the first and we let dy =0, dz = 0, dx =/= 0.

We get that $$\frac{\partial w}{\partial x}_{y,z}\frac{\partial x}{\partial w}_{y,z}=1$$

And so the result follows that
$$\frac{\partial w}{\partial x}_{y,z} = \frac{1}{\frac{\partial x}{\partial w}}_{y,z}$$

 

[epenguin]

Isn't this a triviality, tantamount to saying a = 1/(1/a) ?

 

[RJLiberator]

Well, if it is... I must be able to prove it, apparently.

What do you think of my operations? I am not sure if I operated everything correctly as differentiating by parts with 3 variables is quite confusing.

 

[epenguin]

I don't know what is required of you - seems a bit strange to me. I mean it appears just that simple algebraic fact I stated.

If there is any calculus in it, you can ignore the ∂ 's, since two variables are constant on both sides this is just ordinary differential coefficients so the problem if you have copied it out right is just asking to prove

dw/dx = 1/(dx/dw).

If that is considered to requiring proof, it is about slopes, so just consider a straight line, for generality not through the origin, of w plotted against x. and then what it is if you switched the variables around (e.g. You rotated the page and then flipped it over so you end with W increasing horizontally, and X vertically.

 

[RJLiberator]

Hm. You make a good point stating that this is trivial, however, my book depicts this as an important theorem to prove in thermodynamics.

In the first chapter, they go through a proof of the reciprocal theorem on page 13. This can actually be viewed on Amazon preview here: https://www.amazon.com/dp/0521274567/?tag=pfamazon01-20

Now, the question I have to answer is how to solve it in three dimensions.

 

[epenguin]

Are you sure you have copied it right, have the subscripts right, there is no minus sign?

Things like that are important in thermodynamics.

 

[RJLiberator]

Yup. I've checked multiple times to make sure what I have in the first post of this thread question and the book is the same. It does seem odd, but the preview in Amazon goes through the proof o the 2-dimensional one.

 

[epenguin]

Essentially I said that since y, z are constant, this is the two-dimensional case..

 

[SammyS]

Well, if it is... I must be able to prove it, apparently.

What do you think of my operations? I am not sure if I operated everything correctly as differentiating by parts with 3 variables is quite confusing.

It looks good, except for the following typo:

You wrote:

...
3. The Attempt at a Solution
w=w(x,y,z)$$dx = \frac{\partial w}{\partial x}_{y,z}dx +\frac{\partial w}{\partial y}_{x,z}dy+\frac{\partial w}{\partial z}_{x,y}dz$$Now, ...

I think you intended to write:

$\displaystyle dw = \frac{\partial w}{\partial x}_{y,z}dx +\frac{\partial w}{\partial y}_{x,z}dy+\frac{\partial w}{\partial z}_{x,y}dz$​

 

[RJLiberator]

Yes, that was a typo. Thank you for checking my operations.

 

[epenguin]

I hope things are becoming clearer. That last formula is okay - note however that in your first formula of #1 you have y, z constant, that is in your last formula dy = 0. dz = 0 so this leads you back to a triviaity. But in other cases you will get into real and important useful stuff.

Looking back I do seem to remember I myself had some doubts when I first saw anybody use 1/(dx/dy) = dy/dx without justification. It seemed like a superficial carryover of algebra into calculus, something you maybe couldn't necessarily do without justification. But anyway I think I've indicated to you how to to justify it. Note that dy/dx is the slope of the tangent line that goes through the point (x, y) after which you only need to consider this line, i.e. your arguments don't need to get in the into any hoo-ha about limits.