Bulk modulus

Illustration of uniform compression

The bulk modulus ($K$ or $B$) of a substance is a measure of how resistant to compression that substance is. It is defined as the ratio of the infinitesimal pressure increase to the resulting relative decrease of the volume. Other moduli describe the material's response (strain) to other kinds of stress: the shear modulus describes the response to shear, and Young's modulus describes the response to linear stress. For a fluid, only the bulk modulus is meaningful. For a complex anisotropic solid such as wood or paper, these three moduli do not contain enough information to describe its behaviour, and one must use the full generalized Hooke's law.

Definition

The bulk modulus $K>0$ can be formally defined by the equation

$K=-V{\frac {dP}{dV}}$

where $P$ is pressure, $V$ is volume, and $dP/dV$ denotes the derivative of pressure with respect to volume. Considering unit mass,

$K=\rho {\frac {dP}{d\rho }}$

where ρ is density and dP/dρ denotes the derivative of pressure with respect to density (i.e. pressure rate of change with volume). The inverse of the bulk modulus gives a substance's compressibility.

Thermodynamic relation

Strictly speaking, the bulk modulus is a thermodynamic quantity, and in order to specify a bulk modulus it is necessary to specify how the temperature varies during compression: constant-temperature (isothermal $K_{T}$), constant-entropy (isentropic $K_{S}$), and other variations are possible. Such distinctions are especially relevant for gases.

For an ideal gas, the isentropic bulk modulus $K_{S}$ is given by

$K_{S}=\gamma \,p$

and the isothermal bulk modulus $K_{T}$ is given by

$K_{T}=p$

where

γ is the heat capacity ratio

p is the pressure.

When the gas is not ideal, these equations give only an approximation of the bulk modulus. In a fluid, the bulk modulus K and the density ρ determine the speed of sound c (pressure waves), according to the Newton-Laplace formula

$c={\sqrt {\frac {K}{\rho }}}.$

In solids, $K_{S}$ and $K_{T}$ have very similar values. Solids can also sustain transverse waves: for these materials one additional elastic modulus, for example the shear modulus, is needed to determine wave speeds.

Measurement

It is possible to measure the bulk modulus using powder diffraction under applied pressure. It is a property of a fluid which shows its ability to change its volume under its pressure.

Selected values

Approximate bulk modulus (K) for common materials

Material	Bulk modulus in GPa	Bulk modulus in psi
Glass (see also diagram below table)	7001350000000000000♠35 to 7001550000000000000♠55	7006580000000000000♠5.8×106
Steel	7002160000000000000♠160	7007232000000000000♠23.2×106
Diamond (at 4K)	7002443000000000000♠443	7007640000000000000♠64×106

Influences of selected glass component additions on the bulk modulus of a specific base glass.

A material with a bulk modulus of 35 GPa loses one percent of its volume when subjected to an external pressure of 0.35 GPa (~7008350000000000000♠3500 bar).

Approximate bulk modulus (K) for other substances

Water	7009220000000000000♠2.2 GPa (value increases at higher pressures)
Methanol	7008823000000000000♠823 MPa (at 20 °C and 1 Atm)
Air	7005142000000000000♠142 kPa (adiabatic bulk modulus)
Air	7005101000000000000♠101 kPa (constant temperature bulk modulus)
Solid helium	7007500000000000000♠50 MPa (approximate)

Microscopic origin

Interatomic potential and linear elasticity

Interatomic potential and force

Since linear elasticity is a direct result of interatomic interaction, it is related to the extension/compression of bonds. It can then be derived from the interatomic potential for crystalline materials. First, let us examine the potential energy of two interacting atoms. Starting from very far points, they will feel an attraction towards each other. As they approach each other, their potential energy will decrease. On the other hand, when two atoms are very close to each other, their total energy will be very high due to repulsive interaction. Together, these potentials guarantee an interatomic distance that achieves a minimal energy state. This occurs at some distance a₀, where the total force is zero:

$F=-{\partial U \over \partial r}=0$

Where U is interatomic potential and r is the interatomic distance. This means the atoms are in equilibrium.

To extend the two atoms approach into solid, consider a simple model, say, a 1-D array of one element with interatomic distance of a, and the equilibrium distance is a₀. Its potential energy-interatomic distance relationship has similar form as the two atoms case, which reaches minimal at a₀, The Taylor expansion for this is:

$u(a)=u(a_{0})+({\partial u \over \partial r})_{r=a_{0}}(a-a_{0})+{1 \over 2}({\partial ^{2} \over \partial r^{2}}u)_{r=a_{0}}(a-a_{0})^{2}+O((a-a_{0})^{3})$

At equilibrium, the first derivative is 0, so the dominate term is the quadratic one. When displacement is small, the higher order terms should be omitted. The expression becomes:

$u(a)=u(a_{0})+{1 \over 2}({\partial ^{2} \over \partial r^{2}}u)_{r=a_{0}}(a-a_{0})^{2}$

$F(a)=-{\partial u \over \partial r}=({\partial ^{2} \over \partial r^{2}}u)_{r=a_{0}}(a-a_{0})$

Which is clearly linear elasticity.

Note that the derivation is done considering two neighboring atoms, so the Hook’s coefficient is:

$K=a_{0}*{dF \over dr}=a_{0}({\partial ^{2} \over \partial r^{2}}u)_{r=a_{0}}$

This form can be easily extend to 3-D case, with volume per atom(Ω) in place of interatomic distance.

$K=\Omega _{0}({\partial ^{2} \over \partial \Omega ^{2}}u)_{\Omega =\Omega _{0}}$

Relationship with atomic radius

As derived above, the bulk modulus is directly related the interatomic potential and volume per atoms. We can further evaluate the interatomic potential to connect K with other properties. Usually, the interatomic potential can be expressed as a function of distance that has two terms, one term for attraction and another term for repulsion.

$u=-Ar^{-n}+Br^{-m}$

Where A>0 represents the attraction term and B>0 represents repulsion. n and m are usually integral, and m is usually larger than n, which represents short range nature of repulsion. At equilibrium position, u is at it’s minimal, so first order derivative is 0.

$({\partial u \over \partial r})_{r_{0}}=Anr^{-n-1}+-Bmr^{-m-1}=0$

${B \over A}={n \over m}r_{0}^{m-n}$

$u=-Ar^{-n}(1-{B \over A}r^{n-m})=-Ar^{-n}(1-{n \over m}r_{0}^{m-n}r^{n-m})$

when r is close to , recall that the n(usually 1 to 6) is smaller than m(usually 9 to 12), ignore the second term , evaluate the second derivative

$({\partial ^{2} \over \partial r^{2}}u)_{r=a_{0}}=-An(n+1)r_{0}^{-n-2}$

Recall the relationship between r and Ω

$\Omega ={4\pi \over 3}r^{3}$

$({\partial ^{2} \over \partial \Omega ^{2}}u)=({\partial ^{2} \over \partial r^{2}}u)({\partial r \over \partial \Omega })^{2}=({\partial ^{2} \over \partial r^{2}}u)\Omega ^{-4/3}$

$K=\Omega _{0}({\partial ^{2}u \over \partial r^{2}})_{\Omega =\Omega _{0}}\propto r_{0}^{-n-3}$

In many case, such as in metal or ionic material, the attraction force is electrostatic, so n=1, we have

$K\propto r_{0}^{-4}$

This applies to atoms with similar bonding nature. This relationship is verified within alkali metals and many ionic compounds.

Further reading

• De Jong, Maarten; Chen, Wei (2015). "Charting the complete elastic properties of inorganic crystalline compounds". Scientific Data. 2: 150009. Bibcode:2013NatSD...2E0009D. doi:10.1038/sdata.2015.9.

Conversion formulae
Homogeneous isotropic linear elastic materials have their elastic properties uniquely determined by any two moduli among these; thus, given any two, any other of the elastic moduli can be calculated according to these formulas.
	$K=\,$	$E=\,$	$\lambda =\,$	$G=\,$	$\nu =\,$	$M=\,$	Notes
$(K,\,E)$			${\tfrac {3K(3K-E)}{9K-E}}$	${\tfrac {3KE}{9K-E}}$	${\tfrac {3K-E}{6K}}$	${\tfrac {3K(3K+E)}{9K-E}}$
$(K,\,\lambda )$		${\tfrac {9K(K-\lambda )}{3K-\lambda }}$		${\tfrac {3(K-\lambda )}{2}}$	${\tfrac {\lambda }{3K-\lambda }}$	$3K-2\lambda \,$
$(K,\,G)$		${\tfrac {9KG}{3K+G}}$	$K-{\tfrac {2G}{3}}$		${\tfrac {3K-2G}{2(3K+G)}}$	$K+{\tfrac {4G}{3}}$
$(K,\,\nu )$		$3K(1-2\nu )\,$	${\tfrac {3K\nu }{1+\nu }}$	${\tfrac {3K(1-2\nu )}{2(1+\nu )}}$		${\tfrac {3K(1-\nu )}{1+\nu }}$
$(K,\,M)$		${\tfrac {9K(M-K)}{3K+M}}$	${\tfrac {3K-M}{2}}$	${\tfrac {3(M-K)}{4}}$	${\tfrac {3K-M}{3K+M}}$
$(E,\,\lambda )$	${\tfrac {E+3\lambda +R}{6}}$			${\tfrac {E-3\lambda +R}{4}}$	${\tfrac {2\lambda }{E+\lambda +R}}$	${\tfrac {E-\lambda +R}{2}}$	$R={\sqrt {E^{2}+9\lambda ^{2}+2E\lambda }}$
$(E,\,G)$	${\tfrac {EG}{3(3G-E)}}$		${\tfrac {G(E-2G)}{3G-E}}$		${\tfrac {E}{2G}}-1$	${\tfrac {G(4G-E)}{3G-E}}$
$(E,\,\nu )$	${\tfrac {E}{3(1-2\nu )}}$		${\tfrac {E\nu }{(1+\nu )(1-2\nu )}}$	${\tfrac {E}{2(1+\nu )}}$		${\tfrac {E(1-\nu )}{(1+\nu )(1-2\nu )}}$
$(E,\,M)$	${\tfrac {3M-E+S}{6}}$		${\tfrac {M-E+S}{4}}$	${\tfrac {3M+E-S}{8}}$	${\tfrac {E-M+S}{4M}}$		$S=\pm {\sqrt {E^{2}+9M^{2}-10EM}}$ There are two valid solutions. The plus sign leads to $\nu \geq 0$. The minus sign leads to $\nu \leq 0$.
$(\lambda ,\,G)$	$\lambda +{\tfrac {2G}{3}}$	${\tfrac {G(3\lambda +2G)}{\lambda +G}}$			${\tfrac {\lambda }{2(\lambda +G)}}$	$\lambda +2G\,$
$(\lambda ,\,\nu )$	${\tfrac {\lambda (1+\nu )}{3\nu }}$	${\tfrac {\lambda (1+\nu )(1-2\nu )}{\nu }}$		${\tfrac {\lambda (1-2\nu )}{2\nu }}$		${\tfrac {\lambda (1-\nu )}{\nu }}$	Cannot be used when $\nu =0\Leftrightarrow \lambda =0$
$(\lambda ,\,M)$	${\tfrac {M+2\lambda }{3}}$	${\tfrac {(M-\lambda )(M+2\lambda )}{M+\lambda }}$		${\tfrac {M-\lambda }{2}}$	${\tfrac {\lambda }{M+\lambda }}$
$(G,\,\nu )$	${\tfrac {2G(1+\nu )}{3(1-2\nu )}}$	$2G(1+\nu )\,$	${\tfrac {2G\nu }{1-2\nu }}$			${\tfrac {2G(1-\nu )}{1-2\nu }}$
$(G,\,M)$	$M-{\tfrac {4G}{3}}$	${\tfrac {G(3M-4G)}{M-G}}$	$M-2G\,$		${\tfrac {M-2G}{2M-2G}}$
$(\nu ,\,M)$	${\tfrac {M(1+\nu )}{3(1-\nu )}}$	${\tfrac {M(1+\nu )(1-2\nu )}{1-\nu }}$	${\tfrac {M\nu }{1-\nu }}$	${\tfrac {M(1-2\nu )}{2(1-\nu )}}$