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Isotherm bulk modulus is defined by $$K=\rho\left(\frac{\partial p}{\partial\rho}\right)_T=-v\left(\frac{\partial p}{\partial v}\right)_T$$ wherein $v=\tfrac{1}{\rho}$ is the specific volume. Studying around the Birch-Murnaghan equation of state and some other related topics I saw everyone talk about the pressure derivative of the isotherm bulk modulus very happily as if it is a function of e.g. pressure, like any other thermodynamic quantity usually is, so that one can calculate its nontrivial pressure derivative for it, even a nontrivial isotherm pressure derivative $\left(\tfrac{\partial}{\partial p}\right)_T$ of it. See for example this link that introduces the quantity $$K'=\left(\frac{\partial K}{\partial p}\right)_T$$ (the reference cited uses $B$ instead of $K$.) But to me its very odd. Let me compute the isotherm pressure derivative of $K$ so you understand why I'm confused: \begin{align} K'&=\left(\frac{\partial K}{\partial p}\right)_T\\ &=\left(\frac{\partial}{\partial p}\right)_T \biggl[\rho\left(\frac{\partial p}{\partial\rho}\right)_T\biggr]\\ &=\left(\frac{\partial\rho}{\partial p}\right)_T\left(\frac{\partial p}{\partial\rho}\right)_T+\rho\left(\frac{\partial^2 p}{\partial\rho\partial p}\right)_T\\ &=1+0\\ &=1 \end{align} A constant, not a function of pressure at all. Oh I don't mean that the Bulk modulus is not an ordinary thermodynamic quantity, but its isotherm pressure derivative can not be function of pressure as the above calculations suggest, yet isentropic pressure dependence could be non-trivial and etc. No reference could I have found to have computed this derivative, everyone who I encountered have only reported to have measured the quantity experimentally. That everyone say it is a function of pressure and experiments validate their idea makes me think I have missed a point in my computations. But there is a greater point that suggests i am wrong, that using the specific-volume-based form of the bulk modulus when is differentiated will result in $-1$ instead of $1$: \begin{align} K'&=\left(\frac{\partial K}{\partial p}\right)_T\\ &=\left(\frac{\partial}{\partial p}\right)_T \biggl[-v\left(\frac{\partial p}{\partial v}\right)_T\biggr]\\ &=-\left(\frac{\partial v}{\partial p}\right)_T\left(\frac{\partial p}{\partial v}\right)_T-v\left(\frac{\partial^2 p}{\partial v\partial p}\right)_T\\ &=-1-0\\ &=-1 \end{align} So that the contradiction $1=-1$ suggests that I must be wrong, but where is the point I am missing? Aren't the derivatives interchangeable in computing $\left(\frac{\partial}{\partial p}\right)_T \left(\frac{\partial p}{\partial\rho}\right)_T$ (I understand the Thermodynamic's underlying geometry can be non-euclidean in different cases) or else?

So the simple question is this: where have I make a mistake in the above calculations?

Regards.

UPDATE. Wrong calculation led into an apparent contradiction, but as could be guessed it was wrong so no contradiction at all. It was a math question which is now resolved, so maybe better to be deleted from ere as it had no physics questioned behind it.

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closed as off-topic by Manishearth Jul 1 '13 at 4:25

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I am confident that your calculation is right (and it is plausible that experimenters overlook this simple mathematical identity as it is easier for them to measure something than to calculate). In fact, your $$K' = 1$$ may be integrated over $p$ to see that $$ K = K_0+p$$ where other arguments really imply that the integration constant $K_0$ has to vanish. So the isothermal bulk modulus of a gas is equal to the pressure. You will find pages that confirm this statement. The bulk modulus only becomes a nontrivial function of the pressure if we consider the adiabatic case (or a different adjective still).

I don't know why you think that the reverted sign when switching from density-based to volume-based formulae is a contradiction.

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  • $\begingroup$ Your first link was interesting, thanks, but that's still strange to me to think more than a century all those giants in science have missed this simple calculation!? That certainly -1=1 is a real contradiction then leads me to conclude I am certainly wrong somewhere. Can you please explain why you think $k'=1$ is not contradicting with $k'=-1$? Cheers $\endgroup$ – owari Jun 28 '13 at 13:24
  • $\begingroup$ Dear @owari, giants of science weren't reviewing random textbooks of basic thermodynamics. It's a subfield of physics that hasn't been on the cutting edge for more than 100 years so the form of many things was "conserved" at a distant past state. I didn't see any calculation deriving $K'=-1$ so I can't address your contradiction. If you define and compute $K'$ with respect to other variables, you get compatible results. Maths is consistent. If you write why you think that $K'=-1$ as well, we may talk. $\endgroup$ – Luboš Motl Jun 28 '13 at 16:08
  • $\begingroup$ The calculation for $K'=-1$ was added, very eager to know your idea about the apparent contradiction, please. I very much believe like you about the consistency of Math, so I should be wrong somewhere! $\endgroup$ – owari Jun 28 '13 at 17:40

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