# Starting from an expression of E(V) and P(V) for the Birch-Murnaghan's equation of state, is there a way of obtaining an expression for E(P)?

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:

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

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

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

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

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.

• You could try to manually eliminate $V$. Jun 18, 2016 at 12:54
• Whoops ... comment deleted. Jun 18, 2016 at 13:08
• @Karlo In the hypothetical case you were to eliminate $V$ from Eq. 1, you would not end up with an expression of $E(P)$ Jun 18, 2016 at 13:52
• @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 $E=E(P)$. Jun 18, 2016 at 15:09
• @LubošMotl There is no possible way to describe a combination of Equations 1,3 that have no $V$ Jun 18, 2016 at 15:50

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.
• Does that mean that $E(P)$ is simply $E(P) = PV + E_{0}+\frac{9V_{0}B_{0}}{16}\left \{ \left [ \left ( \frac{V_{0}}{V} \right )^\frac{2}{3} -1 \right ]^3 B_{0}^{'}+\left [ \left ( \frac{V_{0}}{V} \right )^{\frac{2}{3}}-1 \right ]^{2}\left [ 6-4\left (\frac{V_0}{V} \right )^{\frac{2}{3}} \right ]\right \}$ ? Jun 19, 2016 at 14:47
• @DavidC. Yes, but there is the problem that $E(V,S)$ is convex in a small interval and may not be what you need. Jun 19, 2016 at 15:03
• @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. Jun 19, 2016 at 15:07