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Cyclic Rule

  1. Sep 20, 2009 #1
    I dont quite understand why cyclic rule works (from Pchem)

    (del x/ del y)_z = part of x with respect to y, hold z constant

    I dont know why is it negative 1?

    del x/ del y)_z * del y/ del z)_x * del z/ del x)_y = -1
  2. jcsd
  3. Sep 20, 2009 #2
    You could derive it. Not sure if there is any other way to imagine it
    [tex]\mathrm{d}z=\left.\frac{\partial z}{\partial x}\right)_y\mathrm{d}x+\left.\frac{\partial z}{\partial y}\right)_x\mathrm{d}y[/tex]
    Now to get [itex]\left.\frac{\partial x}{\partial y}\right)_z[/itex] from the above equation just find the fraction [itex]\frac{\mathrm{d}x}{\mathrm{d}y}[/itex] under the condition that [itex]\mathrm{d}z=0[/itex], i.e.
    [tex]\left.\frac{\partial x}{\partial y}\right)_z=\left.\frac{\mathrm{d} x}{\mathrm{d} y}\right)_{\mathrm{d}z=0}=-\frac{\left.\partial z/\partial y\right)_x}{\left.\partial z/\partial x\right)_y}[/tex]
    The minus sign comes from rearranging the first equation.

    [tex]\left.\frac{\partial z}{\partial y}\right)_x=\frac{1}{\left.\frac{\partial y}{\partial z}\right)_x}[/tex]
    [tex]\left.\frac{\partial z}{\partial y}\right)_x\left.\frac{\partial y}{\partial z}\right)_x=1[/tex]
    The apparent cancellation is only possible since both derivatives have the same variables kept constant!
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