Acoustic theory is the field relating to mathematical description of
sound waves. It is derived from
fluid dynamics. See
acoustics for the
engineering approach.
The propagation of sound waves in a fluid (such as air) can be modeled by an equation of motion (conservation of momentum) and an equation of continuity (conservation of mass). With some simplifications, in particular constant density, they can be given as follows:
begin{align}
rho_0 frac{partial mathbf{v}}{partial t} + nabla p & = 0 qquad text{(Momentum balance)}
frac{partial p}{partial t} + kappa~nabla cdot mathbf{v} & = 0 qquad text{(Mass balance)}
end{align}
where
is the acoustic pressure and
is the acoustic fluid velocity vector,
is the vector of spatial coordinates
,
is the time,
is the static mass density of the medium and
is the
bulk modulus of the medium. The bulk modulus can be expressed in terms of the density and the speed of sound in the medium (
) as
The acoustic wave equation is a combination of these two sets of balance equations and can be expressed as
cfrac{partial^2 mathbf{v}}{partial t^2} - c_0^2~nabla^2mathbf{v} = 0
qquad text{or} qquad
cfrac{partial^2 p}{partial t^2} - c_0^2~nabla^2 p = 0
The
acoustic wave equation (and the mass and momentum balance equations) are often expressed in terms of a
scalar potential where
. In that case the
acoustic wave equation is written as
cfrac{partial^2 varphi}{partial t^2} - c_0^2~nabla^2 varphi = 0
and the momentum balance and mass balance are expressed as
p + rho_0~cfrac{partialvarphi}{partial t} = 0 ~;~~
rho + cfrac{rho_0}{c_0^2}~cfrac{partialvarphi}{partial t} = 0 ~.
Derivation of the governing equations
The derivations of the above equations for waves in an acoustic medium are given below.
Conservation of momentum
The equations for the
conservation of linear momentum for a fluid medium are
rho left(frac{partial mathbf{v}}{partial t} + mathbf{v} cdot nabla mathbf{v}right) = -nabla p + nabla cdotboldsymbol{s} + rhomathbf{b}
where
is the body force per unit mass,
is the pressure, and
is the deviatoric stress. If
is the Cauchy stress, then
p := -tfrac{1}{3}~text{tr}(boldsymbol{sigma}) ~;~~
boldsymbol{s} := boldsymbol{sigma} + p~boldsymbol{mathit{1}}
where
is the rank-2 identity tensor.
We make several assumptions to derive the momentum balance equation for an acoustic medium. These assumptions and the resulting forms of the momentum equations are outlined below.
Assumption 1: Newtonian fluid
In acoustics, the fluid medium is assumed to be
Newtonian. For a Newtonian fluid, the deviatoric stress tensor is related to the velocity by
lambda~(nabla cdot mathbf{v})~boldsymbol{mathit{1}}
where
is the shear
viscosity and
is the
bulk viscosity.
Therefore, the divergence of is given by
begin{align}
nablacdotboldsymbol{s} equiv cfrac{partial s_{ij}}{partial x_i} & =
mu left[cfrac{partial}{partial x_i}left(cfrac{partial v_i}{partial x_j}+cfrac{partial v_j}{partial x_i}right)right] + lambda~left[cfrac{partial}{partial x_i}left(cfrac{partial v_k}{partial x_k}right)right]delta_{ij}
& = mu~cfrac{partial^2 v_i}{partial x_i partial x_j} + mu~cfrac{partial^2 v_j}{partial x_ipartial x_i} + lambda~cfrac{partial^2 v_k}{partial x_kpartial x_j}
& = (mu + lambda)~cfrac{partial^2 v_i}{partial x_i partial x_j} + mu~cfrac{partial^2 v_j}{partial x_i^2}
& equiv (mu + lambda)~nabla(nablacdotmathbf{v}) + mu~nabla^2mathbf{v} ~.
end{align}
Using the identity
, we have
nablacdotboldsymbol{s} = (2mu + lambda)~nabla(nablacdotmathbf{v}) -
mu~nablatimesnablatimesmathbf{v}~.
The equations for the conservation of momentum may then be written as
rho left(frac{partial mathbf{v}}{partial t} + mathbf{v} cdot nabla mathbf{v}right) = -nabla p + (2mu + lambda)~nabla(nablacdotmathbf{v}) -
mu~nablatimesnablatimesmathbf{v} + rhomathbf{b}
Assumption 2: Irrotational flow
For most acoustics problems we assume that the flow is irrotational, that is, the
vorticity is zero. In that case
nablatimesmathbf{v} = 0
and the momentum equation reduces to
rho left(frac{partial mathbf{v}}{partial t} + mathbf{v} cdot nabla mathbf{v}right) = -nabla p + (2mu + lambda)~nabla(nablacdotmathbf{v}) + rhomathbf{b}
Assumption 3: No body forces
Another frequently made assumption is that effect of body forces on the fluid medium is negligible. The momentum equation then further simplifies to
rho left(frac{partial mathbf{v}}{partial t} + mathbf{v} cdot nabla mathbf{v}right) = -nabla p + (2mu + lambda)~nabla(nablacdotmathbf{v})
Assumption 4: No viscous forces
Additionally, if we assume that there are no viscous forces in the medium (the bulk and shear viscosities are zero), the momentum equation takes the form
rho left(frac{partial mathbf{v}}{partial t} + mathbf{v} cdot nabla mathbf{v}right) = -nabla p
Assumption 5: Small disturbances
An important simplifying assumption for acoustic waves is that the amplitude of the disturbance of the field quantities is small. This assumption leads to the linear or small signal acoustic wave equation. Then we can express the variables as the sum of the (time averaged) mean field (
) that varies in space and a small fluctuating field (
) that varies in space and time. That is
p = langle prangle + tilde{p} ~;~~
rho = langlerhorangle + tilde{rho} ~;~~
mathbf{v} = langlemathbf{v}rangle + tilde{mathbf{v}}
and
cfrac{partiallangle p rangle}{partial t} = 0 ~;~~ cfrac{partiallangle rho rangle}{partial t} = 0 ~;~~
cfrac{partiallangle mathbf{v} rangle}{partial t} = mathbf{0} ~.
Then the momentum equation can be expressed as
left[langlerhorangle+tilde{rho}right] left[frac{partialtilde{mathbf{v}}}{partial t} + left[langlemathbf{v}rangle+tilde{mathbf{v}}right] cdot nabla left[langlemathbf{v}rangle+tilde{mathbf{v}}right]right] = -nabla left[langle prangle+tilde{p}right]
Since the fluctuations are assumed to be small, products of the fluctuation terms can be neglected (to first order) and we have
begin{align}
langlerhorangle~frac{partialtilde{mathbf{v}}}{partial t} & +
left[langlerhorangle+tilde{rho}right]left[langlemathbf{v}ranglecdotnabla langlemathbf{v}rangleright]+
langlerhorangleleft[langlemathbf{v}ranglecdotnablatilde{mathbf{v}} +
tilde{mathbf{v}}cdotnablalanglemathbf{v}rangleright]
& = -nabla left[langle prangle+tilde{p}right]
end{align}
Assumption 6: Homogeneous medium
Next we assume that the medium is homogeneous; in the sense that the time averaged variables
and
have zero gradients, i.e.,
nablalangle p rangle = 0 ~;~~ nablalangle rho rangle = 0 ~.
The momentum equation then becomes
langlerhorangle~frac{partialtilde{mathbf{v}}}{partial t} +
left[langlerhorangle+tilde{rho}right]left[langlemathbf{v}ranglecdotnabla langlemathbf{v}rangleright]+
langlerhorangleleft[langlemathbf{v}ranglecdotnablatilde{mathbf{v}} +
tilde{mathbf{v}}cdotnablalanglemathbf{v}rangleright]
= -nablatilde{p}
Assumption 7: Medium at rest
At this stage we assume that the medium is at rest which implies that the mean velocity is zero, i.e.
. Then the balance of momentum reduces to
langlerhorangle~frac{partialtilde{mathbf{v}}}{partial t} = -nablatilde{p}
Dropping the tildes and using
, we get the commonly used form of the acoustic momentum equation
rho_0~frac{partialmathbf{v}}{partial t} + nabla p = 0 ~.
Conservation of mass
The equation for the
conservation of mass in a fluid volume (without any mass sources or sinks) is given by
where
is the mass density of the fluid and
is the fluid velocity.
The equation for the conservation of mass for an acoustic medium can also be derived in a manner similar to that used for the conservation of momentum.
Assumption 1: Small disturbances
From the assumption of small disturbances we have
p = langle prangle + tilde{p} ~;~~
rho = langlerhorangle + tilde{rho} ~;~~
mathbf{v} = langlemathbf{v}rangle + tilde{mathbf{v}}
and
cfrac{partiallangle p rangle}{partial t} = 0 ~;~~ cfrac{partiallangle rho rangle}{partial t} = 0 ~;~~
cfrac{partiallangle mathbf{v} rangle}{partial t} = mathbf{0} ~.
Then the mass balance equation can be written as
frac{partialtilde{rho}}{partial t} +
left[langlerhorangle+tilde{rho}right]nabla cdotleft[langlemathbf{v}rangle+tilde{mathbf{v}}right] +
nablaleft[langlerhorangle+tilde{rho}right]cdot left[langlemathbf{v}rangle+tilde{mathbf{v}}right]= 0
If we neglect higher than first order terms in the fluctuations, the mass balance equation becomes
frac{partialtilde{rho}}{partial t} +
left[langlerhorangle+tilde{rho}right]nabla cdotlanglemathbf{v}rangle+
langlerhoranglenablacdottilde{mathbf{v}} +
nablaleft[langlerhorangle+tilde{rho}right]cdotlanglemathbf{v}rangle+
nablalanglerhoranglecdottilde{mathbf{v}}= 0
Assumption 2: Homogeneous medium
Next we assume that the medium is homogeneous, i.e.,
nablalangle rho rangle = 0 ~.
Then the mass balance equation takes the form
frac{partialtilde{rho}}{partial t} +
left[langlerhorangle+tilde{rho}right]nabla cdotlanglemathbf{v}rangle+
langlerhoranglenablacdottilde{mathbf{v}} +
nablatilde{rho}cdotlanglemathbf{v}rangle
= 0
Assumption 3: Medium at rest
At this stage we assume that the medium is at rest, i.e.,
. Then the mass balance equation can be expressed as
frac{partialtilde{rho}}{partial t} +
langlerhoranglenablacdottilde{mathbf{v}} = 0
Assumption 4: Ideal gas, adiabatic, reversible
In order to close the system of equations we need an
equation of state for the pressure. To do that we assume that the medium is an
ideal gas and all acoustic waves compress the medium in an
adiabatic and
reversible manner. The equation of state can then be expressed in the form of the differential equation:
cfrac{dp}{drho} = cfrac{gamma~p}{rho} ~;~~ gamma := cfrac{c_p}{c_v} ~;~~ c^2 = cfrac{gamma~p}{rho} ~.
where
is the
specific heat at constant pressure,
is the
specific heat at constant volume, and
is the wave speed. The value of
is 1.4 if the acoustic medium is air.
For small disturbances
cfrac{dp}{drho} approx cfrac{tilde{p}}{tilde{rho}} ~;~~
cfrac{p}{rho} approx cfrac{langle p rangle}{langle rho rangle} ~;~~
c^2 approx c_0^2 = cfrac{gamma~langle prangle}{langle rho rangle} ~.
where
is the
speed of sound in the medium.
Therefore,
cfrac{tilde{p}}{tilde{rho}} = gamma~cfrac{langle p rangle}{langle rho rangle}
= c_0^2 qquad implies qquad
cfrac{partialtilde{p}}{partial t} = c_0^2 cfrac{partialtilde{rho}}{partial t}
The balance of mass can then be written as
cfrac{1}{c_0^2}frac{partialtilde{p}}{partial t} +
langlerhoranglenablacdottilde{mathbf{v}} = 0
Dropping the tildes and defining
gives us the commonly used expression for the balance of mass in an acoustic medium:
frac{partial p}{partial t} + rho_0~c_0^2~nablacdotmathbf{v} = 0 ~.
Governing equations in cylindrical coordinates
If we use a
cylindrical coordinate system with
basis vectors , then the
gradient of
and the
divergence of
are given by
begin{align}
nabla p & = cfrac{partial p}{partial r}~mathbf{e}_r + cfrac{1}{r}~cfrac{partial p}{partial theta}~mathbf{e}_theta + cfrac{partial p}{partial z}~mathbf{e}_z
nablacdotmathbf{v} & = cfrac{partial v_r}{partial r} + cfrac{1}{r}left(cfrac{partial v_theta}{partial theta} + v_rright) + cfrac{partial v_z}{partial z}
end{align}
where the
velocity has been expressed as
.
The equations for the conservation of momentum may then be written as
rho_0~left[cfrac{partial v_r}{partial t}~mathbf{e}_r+cfrac{partial v_theta}{partial t}~mathbf{e}_theta+cfrac{partial v_z}{partial t}~mathbf{e}_zright] +
cfrac{partial p}{partial r}~mathbf{e}_r + cfrac{1}{r}~cfrac{partial p}{partial theta}~mathbf{e}_theta + cfrac{partial p}{partial z}~mathbf{e}_z = 0
In terms of components, these three equations for the
conservation of momentum in
cylindrical coordinates are
rho_0~cfrac{partial v_r}{partial t} + cfrac{partial p}{partial r} = 0 ~;~~
rho_0~cfrac{partial v_theta}{partial t} + cfrac{1}{r}~cfrac{partial p}{partial theta} = 0 ~;~~
rho_0~cfrac{partial v_z}{partial t} + cfrac{partial p}{partial z} = 0 ~.
The equation for the conservation of mass can similarly be written in cylindrical coordinates as
cfrac{partial p}{partial t} + kappaleft[cfrac{partial v_r}{partial r} + cfrac{1}{r}left(cfrac{partial v_theta}{partial theta} + v_rright) + cfrac{partial v_z}{partial z}right] = 0 ~.
Time harmonic acoustic equations in cylindrical coordinates
The acoustic equations for the
conservation of momentum and the
conservation of mass are often expressed in time
harmonic form (at fixed
frequency). In that case, the pressures and the velocity are assumed to be time harmonic functions of the form
p(mathbf{x}, t) = hat{p}(mathbf{x})~e^{-iomega t} ~;~~
mathbf{v}(mathbf{x}, t) = hat{mathbf{v}}(mathbf{x})~e^{-iomega t} ~;~~ i := sqrt{-1}
where
is the frequency. Substitution of these expressions into the governing equations in
cylindrical coordinates gives us the fixed frequency form of the
conservation of momentum
cfrac{partialhat{p}}{partial r} = iomega~rho_0~hat{v}_r ~;~~
cfrac{1}{r}~cfrac{partialhat{p}}{partial theta} = iomega~rho_0~hat{v}_theta ~;~~
cfrac{partialhat{p}}{partial z} = iomega~rho_0~hat{v}_z
and the fixed frequency form of the
conservation of mass
cfrac{iomega hat{p}}{kappa} = cfrac{partial hat{v}_r}{partial r} + cfrac{1}{r}left(cfrac{partial hat{v}_theta}{partial theta} + hat{v}_rright) + cfrac{partial hat{v}_z}{partial z} ~.
Special case: No z-dependence
In the special case where the field quantities are independent of the z-coordinate we can eliminate
to get
frac{partial^2 p}{partial r^2} + frac{1}{r}frac{partial p}{partial r} +
frac{1}{r^2}~frac{partial^2 p}{partialtheta^2} + frac{omega^2rho_0}{kappa}~p = 0
Assuming that the solution of this equation can be written as
p(r,theta) = R(r)~Q(theta)
we can write the partial differential equation as
cfrac{r^2}{R}~cfrac{d^2R}{dr^2} + cfrac{r}{R}~cfrac{dR}{dr} + cfrac{r^2omega^2rho_0}{kappa} = -cfrac{1}{Q}~cfrac{d^2Q}{dtheta^2}
The left hand side is not a function of
while the right hand side is not a function of
. Hence,
r^2~cfrac{d^2R}{dr^2} + r~cfrac{dR}{dr} + cfrac{r^2omega^2rho_0}{kappa}~R = alpha^2~R ~;~~ cfrac{d^2Q}{dtheta^2} = -alpha^2~Q
where
is a constant. Using the substitution
tilde{r} leftarrow left(omegasqrt{cfrac{rho_0}{kappa}}right) r = k~r
we have
tilde{r}^2~cfrac{d^2R}{dtilde{r}^2} + tilde{r}~cfrac{dR}{dtilde{r}} + (tilde{r}^2-alpha^2)~R = 0 ~;~~ cfrac{d^2Q}{dtheta^2} = -alpha^2~Q
The equation on the left is the
Bessel equation which has the general solution
R(r) = A_alpha~J_alpha(k~r) + B_alpha~J_{-alpha}(k~r)
where
is the cylindrical
Bessel function of the first kind and
are undetermined constants. The equation on the right has the general solution
Q(theta) = C_alpha~e^{ialphatheta} + D_alpha~e^{-ialphatheta}
where
are undetermined constants. Then the solution of the acoustic wave equation is
p(r,theta) = left[A_alpha~J_alpha(k~r) + B_alpha~J_{-alpha}(k~r)right]left(C_alpha~e^{ialphatheta} + D_alpha~e^{-ialphatheta}right)
Boundary conditions are needed at this stage to determine
and the other undetermined constants.
References
See also