Reciprocal Theorem proof in 3 variables

Click For Summary

Homework Help Overview

The problem involves proving a relationship between partial derivatives of a function w, which depends on three variables: x, y, and z. The specific statement to prove is that the partial derivative of w with respect to x, holding y and z constant, is equal to the reciprocal of the partial derivative of x with respect to w, also holding y and z constant.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the operations involved in differentiating a function of three variables and question whether the problem is trivial or significant. Some express uncertainty about the correctness of the operations and the implications of treating certain variables as constants.

Discussion Status

There is an ongoing examination of the operations performed by the original poster, with some participants suggesting that the problem may be simpler than it appears. Others emphasize the importance of the theorem in thermodynamics and the need for careful attention to detail in the formulation of the problem.

Contextual Notes

Participants note potential issues with the transcription of the problem, including the correctness of subscripts and the presence of any minus signs. There is also mention of the context in which this theorem is presented in thermodynamics, suggesting that it may have deeper implications than initially perceived.

RJLiberator
Gold Member
Messages
1,094
Reaction score
63

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}
 
Physics news on Phys.org
:olduhh:
Isn't this a triviality, tantamount to saying a = 1/(1/a) ?
 
  • Like
Likes   Reactions: 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.
 
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.
 
  • Like
Likes   Reactions: 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.
 
Are you sure you have copied it right, have the subscripts right, there is no minus sign?

Things like that are important in thermodynamics.
 
  • Like
Likes   Reactions: 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.
 
Essentially I said that since y, z are constant, this is the two-dimensional case..
 
  • Like
Likes   Reactions: RJLiberator
RJLiberator said:
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:
RJLiberator said:
...
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}dzNow, ...
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 ##​
 
  • Like
Likes   Reactions: RJLiberator
  • #10
Yes, that was a typo. Thank you for checking my operations.
 
  • #11
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.
 
  • Like
Likes   Reactions: RJLiberator

Similar threads

Replies
4
Views
1K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K