Partial derivatives are an important mathematical tool in a number of physics disciplines, most notably field theories (e.g. electricity & magnetism and general relativity) and in thermodynamics.<\/p>\n\n\n\n
However, working with partial derivatives are always a bit tricky and teaching students about them is usually fraught with difficulties. So it was to my pleasant surprise that I found a really nice discussion of how to derive the various ‘classic’ rules cleanly presented in Classical and Statistical Thermodynamics<\/em> by Ashley H. Carter. <\/p>\n\n\n\n My presentation here is strongly influenced and closely follows her presentation in Appendix A, although I’ve added on a bit in the theoretical flow and I’ve also provided explicit examples in terms of the standard paraboloid found in freshman calculus. <\/p>\n\n\n\n Assume a function of 3 variables can be expressed as $$f(x,y,z) = 0$$. This equation can be viewed as a constrain equation linking the values of the variables such that two variables are independent. That means that we can (at least locally) solve for:<\/p>\n\n\n\n Focus on the first and second forms (other pairings will follow a simple relabeling of the variables). The corresponding differentials are<\/p>\n\n\n\n \\[ dx = \\left( \\frac{\\partial x}{\\partial y} \\right)_z dy + \\left( \\frac{\\partial x}{\\partial z} \\right)_y dz \\; \\]<\/p>\n\n\n\n and<\/p>\n\n\n\n \\[ dy = \\left( \\frac{\\partial y}{\\partial x} \\right)_z dx + \\left( \\frac{\\partial y}{\\partial z} \\right)_x dz \\; .\\]<\/p>\n\n\n\n Now substitute the expansion of $$dy$$ into the expansion of $$dx$$<\/p>\n\n\n\n \\[ dx = \\left(\\frac{\\partial x}{\\partial y}\\right)_z \\left[ \\left(\\frac{\\partial y}{\\partial x}\\right)_z dx + \\left(\\frac{\\partial y}{\\partial z}\\right)_x dz \\right] + \\left(\\frac{\\partial x}{\\partial z}\\right)_y dz \\; ,\\]<\/p>\n\n\n\n which simplifies to<\/p>\n\n\n\n \\[ dx = \\left(\\frac{\\partial x}{\\partial y}\\right)_z \\left(\\frac{\\partial y}{\\partial x}\\right)_z dx + \\left[ \\left(\\frac{\\partial x}{\\partial y}\\right)_z \\left(\\frac{\\partial y}{\\partial z}\\right)_x + \\left(\\frac{\\partial x}{\\partial z}\\right)_y \\right] dz \\; .\\]<\/p>\n\n\n\n Putting it all together gives<\/p>\n\n\n\n \\[ \\left[ 1 – \\left(\\frac{\\partial x}{\\partial y}\\right)_z \\left(\\frac{\\partial y}{\\partial x}\\right)_z \\right] dx – \\left[ \\left(\\frac{\\partial x}{\\partial y}\\right)_z \\left(\\frac{\\partial y}{\\partial z}\\right)_x + \\left(\\frac{\\partial x}{\\partial z}\\right)_y \\right] dz = 0 \\; .\\]<\/p>\n\n\n\n Since $$dx$$ and $$dz$$ are independent, each differential can be set to zero independently, giving one of the classic identities.<\/p>\n\n\n\n First set $$dz = 0$$ to get<\/p>\n\n\n\n \\[ \\left(\\frac{\\partial x}{\\partial y}\\right)_z = 1\/ \\left(\\frac{\\partial y}{\\partial x}\\right)_z \\; , \\]<\/p>\n\n\n\n which is called the reciprocal rule.<\/p>\n\n\n\n Next, setting $$dx = 0$$ yields<\/p>\n\n\n\n \\[ \\left(\\frac{\\partial x}{\\partial y}\\right)_z \\left(\\frac{\\partial y}{\\partial z}\\right)_x = \\; – \\; \\left(\\frac{\\partial x}{\\partial z}\\right)_y \\; , \\]<\/p>\n\n\n\n which is called the fraction rule.<\/p>\n\n\n\n The manipulations are complete when using the reciprocal rule in the fraction rule and simplify to get<\/p>\n\n\n\n \\[ \\left(\\frac{\\partial x}{\\partial y}\\right)_z \\left(\\frac{\\partial y}{\\partial z}\\right)_x \\left(\\frac{\\partial z}{\\partial x}\\right)_y = -1 \\; , \\]<\/p>\n\n\n\n which is called the cyclic rule.<\/p>\n\n\n\n Let\u2019s take a look at these relationships in action. Consider the implicit definition of the paraboloid<\/p>\n\n\n\n \\[ x^2 + y^2 – z = 0 \\; . \\]<\/p>\n\n\n\n As mentioned earlier, this equation can be considered as a constraint equation that selects out a value for any one of the three variables given the other two. In other words, we can imagine a look up table where we select a value of $$x$$ and $$y$$, we rummage through the table to find a row with both values and then we scan to the right to find the allowed value of $$z$$ that makes it satisfy the implicit equation.<\/p>\n\n\n\n How do you construct this table; not at a finite set of points but functionally so that it works at any point? It is natural and easy to determine $$z$$ given $$x$$ and $$y$$ by simply rewriting the implicit equation as<\/p>\n\n\n\n \\[ z(x,y) = x^2 + y^2 \\; .\\]<\/p>\n\n\n\n However, it isn\u2019t as easy to express $$x$$ or $$y$$ as functions of the remaining two variables because of the two possible signs that result from taking the square root. We need to have four functional relationships<\/p>\n\n\n\n \\[ x_p(y,z) = \\sqrt{ z \\; – \\; y^2 } \\; ,\\]<\/p>\n\n\n\n \\[ x_n(y,z) = \\; – \\; \\sqrt{z \\; – \\; y^2} \\; , \\]<\/p>\n\n\n\n \\[ y_p(x,z) = \\sqrt{z \\; – \\; x^2} \\; , \\]<\/p>\n\n\n\n and<\/p>\n\n\n\n \\[ y_n(x,z) = \\; – \\; \\sqrt{z \\; – \\; x^2} \\; , \\]<\/p>\n\n\n\n depending on the particular combination of whether $$x$$ is positive or negative and whether $$y$$ is also positive or negative. In the language of differential geometry, we have a 5 charts in our atlas.<\/p>\n\n\n\n We are now in position to try the various relations derived above. For example, let’s examine the reciprocal relation in the first quadrant of the $$x$$-$$y$$ plane. We need to use $$x_p$$ as our local chart.<\/p>\n\n\n\n \\[ \\left(\\frac{\\partial x_p}{\\partial z}\\right)_y = \\frac{1}{2} \\frac{1}{\\sqrt{z – y^2}} \\; \\]<\/p>\n\n\n\n or once we recognize the denominator as $$x_p$$ <\/p>\n\n\n\n \\[\\left(\\frac{\\partial x_p}{\\partial z}\\right)_y = \\frac{1}{2 x_p} \\; . \\]<\/p>\n\n\n\n The ‘reciprocal’ partial derivative is<\/p>\n\n\n\n \\[ \\left( \\frac{\\partial z}{\\partial x} \\right)_y = 2 x = 2 x_p \\; ,\\]<\/p>\n\n\n\n where there is no need for the $$z$$-chart to distinguish between positive and negative values of $$x$$. As expected the derivatives are reciprocals of each other.<\/p>\n\n\n\n Next, let’s test the fraction rule. For fun, this time let’s test it in the 2nd quadrant in the $$x$$-$$y$$ plane ($$x < 0$$ and $$y > 0$$). Calculating the partial derivatives on the left-hand side yields<\/p>\n\n\n\n \\[ \\left(\\frac{\\partial x_n }{\\partial y_p } \\right)_z = \\frac{y_p}{\\sqrt{z – y_p^2}} = \\; – \\; \\frac{y_p}{x_n} \\; \\]<\/p>\n\n\n\n and<\/p>\n\n\n\n \\[ \\left(\\frac{\\partial y_p }{\\partial z} \\right)_{x_n} = \\frac{1}{2\\sqrt{z-x_n^2}} = \\frac{1}{2 y_p} \\; . \\]<\/p>\n\n\n\n It is a simple matter to verify that<\/p>\n\n\n\n \\[ \\left(\\frac{\\partial x_n }{\\partial y_p} \\right)_z \\left(\\frac{\\partial y_p }{\\partial z} \\right)_{x_n} = -\\frac{1}{2 x_n} \\; \\]<\/p>\n\n\n\n is identical to <\/p>\n\n\n\n \\[ – \\left(\\frac{\\partial x_n }{\\partial z} \\right)_{y_p} = \\frac{1}{2 \\sqrt{z – y_p^2} } = -\\frac{1}{2 x_n} \\; .\\]<\/p>\n\n\n\n Finally, for the cyclic rule, let’s go into the 4th quadrant in the $$x$$-$$y$$ plane ($$x>0$$ and $$y<0$$). Taking each partial derivative in turns yields<\/p>\n\n\n\n \\[ \\left(\\frac{\\partial x_p }{\\partial y_n} \\right)_{z} = -\\frac{y_n}{\\sqrt{z-y_n^2}} = -\\frac{y_n}{x_p} \\; ,\\]<\/p>\n\n\n\n \\[ \\left(\\frac{\\partial y_n }{\\partial z} \\right)_{x_p} = -\\frac{1}{2 \\sqrt{z-x_p^2} } = \\frac{1}{2 y_n} \\; ,\\]<\/p>\n\n\n\n and <\/p>\n\n\n\n \\[ \\left(\\frac{\\partial z }{\\partial x_p} \\right)_{y_n} = 2 x_p \\; .\\]<\/p>\n\n\n\n Multiplying these terms in order gives<\/p>\n\n\n\n \\[ -\\frac{y_n}{x_p} \\frac{1}{2 y_n} 2 x_p = -1 \\; .\\]<\/p>\n\n\n\n Nice, neat, and more than partially done.<\/p>\n\n\n\n \n","protected":false},"excerpt":{"rendered":" Partial derivatives are an important mathematical tool in a number of physics disciplines, most notably field theories (e.g. electricity & magnetism and general relativity) and in thermodynamics. 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