I Inverting the Birch-Murnaghan Equation of State

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Efficient method to compute V at given P for a non-invertible equation of state
Hello,

I have the pressure ([itex]P[/itex])-volume ([itex]V[/itex]) Birch-Murnaghan equation of state coeffcients [itex](V_{0},K_{0}, K^{'}_{0}, K^{''}_{0})[/itex] for a number of different compositions. I'm interested in the volume at very specific pressures only and ideally I would like to compute [itex]V[/itex] for each composition at a given [itex]P[/itex]. My problem is the Birch-Murnaghan equation of state cannot be inverted and yields pressure for a given volume - the opposite of what I need:
[tex]P=3K_{0}f_{E}(1+2f_{E})^{5/2}\left(1+\frac{3}{2}(K^{'}_{0}-4)f_{E}+\frac{3}{2}\left(K_{0}K^{''}_{0}+(K^{'}_{0}-4)(K^{'}_{0}-3)+\frac{35}{9}\right)f^{2}_{E}\right)[/tex]
where [itex]f_{E}=\left[(V_{0}/V)^{2/3}-1\right]/2.[/itex]

So my question is, what would be the best approach to solve [itex]V[/itex] as a function of [itex]P[/itex]?

I was going to calculate [itex]P[/itex] versus [itex]V[/itex], then interpolate the result over a specific [itex]P[/itex] range. But I have a large number of compositions (~100 or so) all with different volumes and equation of state coefficients so this seems quite inefficient, particularly as the range of [itex]V[/itex] to compute [itex]P[/itex] will vary depending on composition

What would be an efficient approach to find [itex]V[/itex] for a given [itex]P[/itex]? Am I forgetting something quite trivial?

Thanks for your help!
 

Charles Link

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I'm puzzled by the first term that contains ## f_E ##. (## 3K_o f_E ##). If ## V \approx V_o ##, then ## f_E \approx 0 ##. Does this mean that ## P \approx 0 ##? ## \\ ##
 
Does this mean that ## P \approx 0 ##? ## \\ ##
Thanks for your reply. Yes at V=V0 the pressure is zero (ambient pressure).
 

Charles Link

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What that means is the equation really reads ## P=P_o+## "what you have as ##P ##". ## \\ ## You may try doing a first order solution of ## f_E ## in terms of ## P-P_o ##, neglecting the ## f_E^2 ## and higher terms. It wouldn't be the perfect solution, but it would be a start. ## \\ ## Alternatively, try something like ## f_E=A(P-P_o)+B(P-P_o)^2+... ##, and see if you can determine ## A ## and ## B ##. (## A ## and ## B ## would come from Taylor series type calculations).
 
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You are "interested in the volume at very specific pressures only". So use the Newton–Raphson method.
 

Charles Link

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A follow-up to post 4: If ## x=At+Bt^2 ##, if my algebra is correct, ##t=\frac{x}{A}-\frac{Bx^2}{A^3} ##, neglecting 3rd order and higher terms.
 

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