3.4.1 The linear chain of harmonic oscillators
Figure 3.2: Diagram of a one-dimensional elastic medium modeled as a chain of harmonic oscillators. The displacements of the masses (dark blocks) are given in terms of a vector function u (z)
.
So far, we have only considered the harmonic oscillator as a localised source. To see how such a disturbance may be propagated throughout space, we extend our treatment to consider many small masses coupled together by ideal springs. To simplify our analysis, we shall restrict our-selves to the one-dimensional case (and neglect external forces such as gravity). A diagram of such a system is shown in Fig. 3.2.
The mass of an infinitesimal section of the coupled system may be given as m = µzdz, where µz is the mass per unit length. We may also write the spring constant K as the tension Tz per unit length, K = Tz/dz. Allowing the displacement at a point z to be given by the vector quantity u (z) and applying Newton’s second law to the element at z , we then have
µzdz∂2u (z)
∂t2 = Tz
dz[u (z + dz) − u (z)] − Tz
dz [u (z) − u (z − dz)] , (3.11) so
∂2u (z + dz)
∂t2 = Tz µz
u (z + 2dz) − 2u (z + dz) + u (z)
dz2 . (3.12)
Taking the limit dz → 0, we get
3.4. THE WAVE EQUATION 53
This may equation may be generalised to three dimensions, yielding
∂2u
is a solution of Eq. (3.14). If f maintains its shape as it propagates, then we see that v must be the velocity with which this wave profile propagates in the positive or negative z-direction. This is known as the phase velocity . Now let us put
ξ = z ± vt, (3.16)
so that, by the chain rule,
∂
Hence, substituting Eq. (3.15) into Eq. (3.14), we have v2∂2f
∂ξ2 = T µ
∂2f
∂ξ2, (3.22)
which gives
v = T µ
1/2
. (3.23)
We may therefore write the wave equation in the more general form
∇2u = 1 v2
∂2u
∂t2. (3.24)
This is the general form of the wave equation for a linear medium. Note that, although was derived for the oscillations of an elastic medium, the particu-lar physics of the system have been washed out, leaving just the abstract mathematical form. This equation may then be used to to find the wave solutions for diverse physical systems, so long as the harmonic oscillator remains a valid model. In Chapter 6, we shall find that the same equation governs the electromagnetic oscillations of light in a linear medium.
3.4.3 The polarisation
The vectorial nature of the displacement implies that that u may be in the direction of the propagation (longitudinal) or transverse to it. This consti-tutes the polarisation of the medium. In general, there are various cases to consider, although we shall find later that not all applied to the optic field.
• Unpolarised
means that the displacement is randomly orientated in space.
• Longitudinal polarisation
In this case, the displacement is in the propagation direction . As an example, sound waves in fluids and solids are longitudinal. However, electromagnetic waves generally are not.
• Transverse polarisation
In this case, the displacement is perpendicular to the propagation direction. For example, surface waves on a liquid are transverse.
Electromagnetic waves (constituting light) are generally transverse.
Moreover, the transverse polarisation may be – linearly polarised
in which the displacement always lies in the same plane, or – elliptically polarised
in which the polarisation rotates around the direction of propa-gation.
3.4. THE WAVE EQUATION 55 We shall defer detailed discussion of the polarisation of optical waves until Chapter 7. For the time being, we shall assume for simplicity that we have linear polarisation (although this is not the most general case). Taking the direction of propagation to be in the z axis, the polarisation may then be specified by a column vector
u0= |u0| respec-tively, and θ is the angle the polarisation vector makes with the x-axis.
3.4.4 Linearity of the wave equation Linear differential equations
Equation (3.24) is an example of a linear differential equation. This cate-gorization has a more general, abstract meaning than just that the forces depend only on linear displacements. Rather, it implies that the principle of linear superposition applies to the solutions. That is, if we can find two solutions f and g of a linear equation, then the function h = af + bg, where aand b are scalars, is also a solution.
It is straightforward to show that Eq. (3.24) observes this rule. Let us suppose that the solutions we have found are f (r, t) and g (r, t). Then, inserting f (r, t) into Eq. (3.24), we have
∇2f − 1 v2
∂2f
∂t2 = 0 (3.26)
and similarly for g (r, t). Let us now substitute
h (r, t) = af (r, t) + bg (r, t) (3.27) into Eq. (3.24). This gives
∇2h − 1 which shows that h (r, t) is also a solution.
Linear operators
More generally, a linear differential equation does not involve products of the derivatives of the solutions. This includes the ‘zero’-th derivative of a solution (i.e. the solution itself). For instance if E is a solution for a
dynamical system, then E · E is a (second order) non-linear term. Such terms would not appear in a linear differential equation.
In general, we may characterise a linear differential equation by saying that every term within it is a linear operator . Such an operator is charac-terised by the fact that it is distributive over addition. For instance, if L (u) is a linear operator, then
L (f + g) = L (g) + L (f ) . (3.29)
On the other hand, suppose we had
L (u) = u2. (3.32)
Now
L (f + g) = (f + g)2 = f2+ 2f g + g2 6= L (f ) + L (g) . (3.33) Hence, in this case, L (u) is nonlinear . As another example, consider
L (u) = udu
dx. (3.34)
In this case, we have
L (f + g) = (f + g) df
so again, L (u) is nonlinear.
This discussion is of relevance to optics since the response of a material medium to a very intense optical field may include nonlinear terms. This is the subject matter of nonlinear optics, which is beyond the scope of this course. It is therefore an important classification of a given medium that it is linear to good approximation, by which we mean that the dynamics of light propagation within it may be modelled by linear operators.