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A Thermodynamics. Partial derivative tricks.

  1. May 7, 2017 #1
    If we consider function ##z=z(x,y)## then ##dz=(\frac{\partial z}{\partial x})_ydx+(\frac{\partial z}{\partial y})_xdy##. If ##z=const## then ##dz=0##. So,
    [tex](\frac{\partial z}{\partial x})_ydx+(\frac{\partial z}{\partial y})_xdy=0[/tex]
    and from that
    [tex]\frac{dx}{dy}=-\frac{(\frac{\partial z}{\partial y})_x}{(\frac{\partial z}{\partial x})_y}[/tex]
    or
    [tex](\frac{\partial x}{\partial y})_z=-\frac{(\frac{\partial z}{\partial y})_x}{(\frac{\partial z}{\partial x})_y}[/tex]
    My question is is ##z=const## isn't then also ##(\frac{\partial z}{\partial x})_y=0## and ##(\frac{\partial z}{\partial y})_x=0##? This is confusing to me. Could you please answer me that.
     
  2. jcsd
  3. May 7, 2017 #2

    vanhees71

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    Everything you wrote is correct. Your confusion is due to a unfortunate formulation of what's meant by the mathematical operations. What you study is the following:

    You define a function ##x=x(x)## (let me be a bit sloppy; pure mathematicians would kill me for this formulation already ;-)) in an implicit way by using a function ##z=z(x,y)## by setting
    $$z(x,y)=c=\text{const}.$$
    So ##z## is always a function of two independent variables ##x## and ##y##. Setting this function to a constant leads to an equation for ##x## to be solved given the value of the variable ##y##.

    Now taking the partial derivatives of ##z## on both sides of the defining equation makes sense. The total differential on the left-hand side indeed is
    $$\mathrm{d} z=\mathrm{d} x \partial_x z+\mathrm{d} y \partial_{y} z.$$
    Now since you restrict yourself along the line, where ##z(x,y)=c=\text{const}## this means that ##\mathrm{d} z=0##. On the other hand this line can (at least locally) be described by a function ##x=x(y)##, and then you get your formula
    $$\frac{\mathrm{d} x}{\mathrm{d} y}=-\frac{\partial z}{\partial y} \left (\frac{\partial z}{\partial x} \right)^{-1}=-\frac{\partial_y z}{\partial_x z}.$$
    Let's give a very simple example. Let's set
    $$z(x,y)=x y^2.$$
    Now we set
    $$z(x,y)=1=\text{const} \; \Rightarrow\; x y^2=1 \; \Rightarrow x=\frac{1}{y^2}.$$
    Obviously you get
    $$\partial_x z=y^2, \quad \partial_y z=2xy.$$
    Now
    $$\frac{\partial_y z}{\partial_x z}=\frac{2xy}{y^2}=\frac{2x}{y}.$$
    Along the implicitly defined curve you have
    $$\left (\frac{\partial_y z}{\partial_x z} \right)_{x=1/y^2}=\frac{2}{y^3}.$$
    Indeed, calculating the function ##x=1/y^2## directly you get
    $$\frac{\mathrm{d} x}{\mathrm{d} y}=-\frac{2}{y^3}=-\left (\frac{\partial_y z}{\partial_x z} \right )_{x=1/y^2}.$$
     
  4. May 7, 2017 #3
    Thanks a lot. Very useful answer. Could you just explain in this case why you say "at least locally".
     
  5. May 7, 2017 #4
    What you are doing is determining is how x and y have to vary in tandem in order for z to remain constant.
     
  6. May 7, 2017 #5

    vanhees71

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    Take as an example
    $$z(x,y)=x^2+y^2.$$
    Obviously the implicitly defined curve
    $$z(x,y)=R^2=\text{const}$$
    is a circle of radius ##R## around the origin in the ##xy## plane, which cannot be described by a function ##y=y(x)## or ##x=x(y)##, but you can of course do that for two parts of the circle, e.g., one semi-circle in the upper part,
    $$y=\sqrt{R^2-x^2}$$
    or in the lower part
    $$y=-\sqrt{R^2-x^2}.$$
    That's what I meant with "local".
     
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