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I have a function, an equation of state named the Birch-Murnaghan's equation of state, in which the Energy ( $E$ ) is a function of the Volume ($V$), where $E_{0}$, $V_{0}$, $B_{0}$ and $B_{0}^{'}$ are parameters that can be obtained from the fit:

\begin{equation} \label{BM-EOS-Energy-Volume} E\left ( V \right )=E_{0}+\frac{9V_{0}B_{0}}{16}\left \{ \left [ \left ( \frac{V_{0}}{V} \right )^\frac{2}{3} -1 \right ]^3 B_{0}^{'}+\left [ \left ( \frac{V_{0}}{V} \right )^{\frac{2}{3}}-1 \right ]^{2}\left [ 6-4\left (\frac{V_0}{V} \right )^{\frac{2}{3}} \right ]\right \}, \tag{1} \end{equation}

The pressure P may be written as a function of the volume V as:

\begin{equation} \label{BM-EOS-Energy-Volume-derivative} P\left ( V \right )=-\left ( \frac{\partial E}{\partial V} \right )_{S}, \tag{2} \end{equation}

so that we can get the expression of $P\left ( V \right )$ : \begin{equation} \label{BM-EOS-Pressure-Volume} P\left ( V \right )=\frac{3B_{0}}{2}\left [ \left ( \frac{V_{0}}{V} \right )^\frac{7}{3} - \left ( \frac{V_{0}}{V} \right )^\frac{5}{3}\right ]\left \{ 1+\frac{3}{4}\left ( B_{0}^{'}-4 \right )\left [ \left ( \frac{V_{0}}{V} \right )^\frac{2}{3}-1 \right ] \right \} \tag{3} \end{equation}

Now, I would like to get an expression of $E(P)$, but I do not find the way to accomplish that goal starting from the expression of $E(V)$ (Equation 1) and $P(V)$ (Equation 3).

I would appreciate if you could help me.

EDIT:

I have data of $E$ vs $V$. After plotting, I have fitted this data using Eq. (1).

I have transformed each $V$ data to $P$ using Eq. (3).

I have now plotted $P$ vs $V$, and fitted using Eq. (3).

Now I have plotted $E$ vs $P$ and I need to fit this data using a function. And this function would have to be the expression of $E(P)$, which I don't know a possible way of working it out.

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  • $\begingroup$ You could try to manually eliminate $V$. $\endgroup$
    – Karlo
    Jun 18, 2016 at 12:54
  • $\begingroup$ Whoops ... comment deleted. $\endgroup$
    – garyp
    Jun 18, 2016 at 13:08
  • $\begingroup$ @Karlo In the hypothetical case you were to eliminate $V$ from Eq. 1, you would not end up with an expression of $E(P)$ $\endgroup$
    – DavidC.
    Jun 18, 2016 at 13:52
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    $\begingroup$ @DavidC - you misunderstood Karlo's obvious recipe. You clearly can't eliminate $V$ just from equation 1. You need to elimininate $V$ from the set of equations 1,3. These are two equations involving $E,P,V$, if you eliminate $V$, i.e. find a combination of 1,3 that has no $V$, you will have an equation with $E,P$ only which is an implicit (or explicit, if you are lucky) prescription for $E=E(P)$. $\endgroup$ Jun 18, 2016 at 15:09
  • $\begingroup$ @LubošMotl There is no possible way to describe a combination of Equations 1,3 that have no $V$ $\endgroup$
    – DavidC.
    Jun 18, 2016 at 15:50

1 Answer 1

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This is the purpose of the Legendre transform, which lies at the core of hamiltonian mechanics. You have a function $E(V,S)$, which is convex for certain interval of $V$, and want to find $E'(P,S)$, where $P \equiv -\left.\frac{\partial E}{\partial V}\right|_S$. This function is given by the Legendre transform of $E$: $$E'(P,S) = PV + E(V,S)$$ for the interval of $V$ in which $E$ is convex.

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  • $\begingroup$ Does that mean that $E(P)$ is simply $E(P) = PV + E_{0}+\frac{9V_{0}B_{0}}{16}\left \{ \left [ \left ( \frac{V_{0}}{V} \right )^\frac{2}{3} -1 \right ]^3 B_{0}^{'}+\left [ \left ( \frac{V_{0}}{V} \right )^{\frac{2}{3}}-1 \right ]^{2}\left [ 6-4\left (\frac{V_0}{V} \right )^{\frac{2}{3}} \right ]\right \}$ ? $\endgroup$
    – DavidC.
    Jun 19, 2016 at 14:47
  • $\begingroup$ @DavidC. Yes, but there is the problem that $E(V,S)$ is convex in a small interval and may not be what you need. $\endgroup$
    – auxsvr
    Jun 19, 2016 at 15:03
  • $\begingroup$ @DavidC. Also, I know that many physicists (me included) abuse notation and write $E(P)$, when in fact they mean "a function that takes values of pressure and returns energy", but in mathematical terms $E(P)$ means $E(V)$ if you replace $V\mapsto P$, which doesn't make sense and may confuse you. $\endgroup$
    – auxsvr
    Jun 19, 2016 at 15:07

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