They say that the Eulerian finite strain is necessary and useful to consider HP compression and construct proper equation of states [Wiki].

However, it is not clear (at least, it was absolutely unclear for me.) how useful the Eulerian finite strain is.

Therefore, let me argue in this page why the Eulerian strain is useful.

**Eulerian scheme**

The unstrained state before compression are expressed by the coordinates in the strained state after compression.

**Lagrangian scheme**

The strained state after compression are expressed by the coordinates in the unstrained state before compression.

We adopt the Eulerian scheme. Then, why was the Lagrangian scheme discarded? In order to answer this question, let me construct an equation of state in the Lagrangian scheme.

Let us consider a cube with an edge length of *X*_{0} and volume *V*_{0} = *X*_{0}^{3} in an unstrained state. This cube is compressed to an
edge length of *X* and volume *V* = *X*^{3} (See the figure below). The change of the edge length is expressed by *u*.

Namely,

(1)

(2)

(3)

This setting is identical to the case of the Eulerian scheme.

As is mentioned above, we argue change in square of the edge length by compression. However, unlike the Eulerian scheme, we express it not by *X* but by *X*_{0} in the
Lagrangian scheme.

(4)

Similar to the Eulerian scheme, we have a proportionality between *X*_{0} and *u*.

(5)

Then we have:

(6)

We define the Lagrangian finite strain as:

(7)

Similar to the Eulerian scheme,

(8)

(9)

The Lagrangian finite strain is expressed by the volume ratio as:

(10)

We have a convenient Lagrangian finite strain as:

(11)

Then, let us construct the 2^{nd}-order equation of state based on the Lagrangian finite strain, which is called the Lagrangian EOS here.

Like the 2^{nd}-order Birch-Murnaghan EOS (Eulerian EOS), we first assume that the Helmholtz free energy increases with square of the Lagrangian finite strain as:

(12)

where *f*_{L} is the Lagrangian finite strain.

The formula of *P* in the Lagrangian scheme is identical to the Eulerian scheme.

(13)

However, the *V* derivative of the finite strain is different:

(14)

By substituting Eqs. (11) and (14) into Eq. (13), we have

(15)

Then, we have the 2^{nd}-order Lagrangian equation of state as:

(16)

Let us compare the Eulerian (2^{nd}-order Birch-Murnaghan) and Lagrangian equations of state using
MgSiO_{3} bridgimanite as a working matter. Siogeikin *et al*. [2004] obtained *K*_{S0} = 253(3) GPa by Brillouin scattering. Here we
assume *K*_{T0} = *K*_{S0}. Tange *et al*. [2012] measured volumes of MgSiO_{3} bridgmanite at pressures of 28 to 98 GPa and ambient
temperature. Pressures obtained by the Eulerian (BM Eq. 8) and Lagrangian (Eq. 16) EOS's based on the above
*K*_{T0}. are compared with MgO pressures in the diagram below.

Fig. 1.

This diagram demonstrates that, although the Eulerian EOS (2. BM-EOS) gives correct pressures only with knowledge of
*K*_{T0}, the Lagrangian EOS significantly underestimates pressures. In other words, the Eulerian
EOS contains the property that matter becomes less compressible at high pressures, whereas the Lagrangian EOS does not. The reason for this difference
is that the *V* derivative of the finite strain increases with inverse proportion to the 5/3 power of *V* in the Eulerian scheme (BM2 Eq. 6), whereas it increases with inverse proportion to only the 1/3 power of *V* in the Lagrangian scheme (this page, Eq. 14). This fact is
also seen by their volume exponents in the two EOS's (BM2 Eq. 8 and this page Eq. 16). The exponents are more close to 0 in the Lagrangian scheme (-1/3 and
1/3) than the Eulerian scheme (7/3 and 5/3).

Let us discuss further more why the Eulerian pressure increases more rapidly than the Lagrangian pressure. As is shown in Eq. (3), the pressure is proportional to the product of finite strain and its volume derivative. Fig. 2 shows increase of the Eulerian and Lagrangian finite strains with volume decrease.

Fig. 2

As is seen, the Eulerian strain more rapidly increases than the Lagrangian strain.

Why like this? Eulerian and Lagrangian finite strains are both differences between the 2/3 power of volume ratio and unity. In the case of Eulerian strain, the ratio is the unstrained volume to the strained volume, and therefore the Eulerian strain can be infinitely large with decreasing volume, whereas the Lagrangian strain is limited to unity.

However, the more essential reason is the volume derivatives of the finite strain, which are shown in Fig. 3.

Fig. 3

The volume derivative of the Eulerian strain dramatically increases with volume decrease, whereas that of the Lagrangian strain remains nearly constant. For these reasons, the Eulerian pressure increases more greatly than the Lagrangian pressure.

Next, let us construct the 3^{rd}-order Lagrangian EOS. Just like the 3^{rd}-order Eulerian EOS, increase in Helmholtz free energy is expanded into the third power of *f*_{L} as:

(17)

The differentiation of Eq (17) by *V* leads to pressure:

(18)

By differentiating Eq. (14), the second *V* derivative of *f*_{L} is given by

(19)

Unlike the Eulerian scheme, the coefficient of the right side does not contain the factor of 5.

At *P* = 0, we have:

(20)

By differentiating Eq. (18) by *V* twice, we have the same equation with Eq. (10) in the Eulerian scheme. Eq. (3) in the derivation of the parameter "a" in the BM2.-EOS and Eq. (20) is substituted into this equation, and we have:

(21)

In the Eulerian scheme, the constant term in the bracket is not 3 but 15.

The second *V* derivative of *P* at *P* = 0 is:

(22)

By equating Eq. (21) and (22), and substituting Eq (5) in the BM2-EOS page into it, we have:

(23)

By simplifying Eq. (23), we have

(24)

Unlike the Eulerian scheme, Eq. (24) has no constant term of -4.

In the exactly same way as the Eulerian scheme, we have the 3rd-order EOS in the Lagrangian scheme:

(25)

One of the important difference of the Lagrangian EOS from the 3^{rd}-order Eulerian EOS is absence of constant
term of -4 to *K*_{T0}'. This means that the 3^{rd}-order Lagrangian EOS becomes equivalent to the 2^{nd}-order EOS
(Eq. 16) when ** K_{T0}' = 0**. In the case of the Eulerian scheme, the
3

Let us compare the 3^{rd}-order EOS in the Eulerian and Lagrangian schemes (Fig. 2). Let us set
*K*_{T0} = 254 GPa and *K*_{T0}' = 4.

Fig. 4

3.E: 3rd-order Eulerian EOS

3.L: 3rd-order Lagrangian EOS

2.L: 2nd-order Lagrangian EOS

The 3^{rd}-order Lagrangian EOS becomes less compressible especially at high compression than the 2^{nd}-order
Lagrangian EOS. However, it is still more compressible than the 3^{rd}-order Eulerian EOS. In order that the Lagrangian EOS has similar
(*V*/*V*_{0})-*P* relations with the Eulerian EOS, *K*_{T0}' should be around 6. We can imagine that the EOS's in these two schemes become more similar at higher order.

The purpose of EOS in geophysics is to make it possible to *evaluate density at high pressures with a small number of information*. The fantastic point of the
**Birch-Murnaghan EOS** is that we can estimate density of matter with high accuracy **only by** **K_{T0}**! If we adopt the Lagrangian scheme, we cannot properly evaluate
density at HP, even though we know both of

It is suggested that the Eulerian strain is more reasonable than the Lagrangian strain due to the following reason. As *V* goes to zero, *F* must go to infinity. As *V* goes
to zero, the Eulerian strain goes to infinity, and therefore *F* goes to infinity. On the other hand, the Lagrangian strain goes to unity only, and *F* does not go to infinity.

It seems that the arguments in this page were issues in the early 20^{th} century or even earlier. I do not find this kind of argument in any current literature. Therefore, I am not sure
whether the argument in this page is correct or not. If you find any problem in this page, please let me know
it.