Wave equations YouTube

Covariant components

In this course, we have defined the covariant components, $X_\mu$, of a four-vector in terms of their contravariant counterparts, $X^\mu$, $$X_\mu = \eta_{\mu \nu} X^\nu$$ where $$\eta_{\mu \nu} = \begin{bmatrix}1 & 0 & 0 & 0\\0 & -1 & 0 & 0\\ 0 &0 & -1 & 0\\0 & 0 & 0 & -1\end{bmatrix}$$

We haven't yet considered how the covariant components of a tensor (of any rank) transform, but we can also do this using contravariant components.

The archetypal contravariant four-vector is a displacement, $\Delta X^\mu$, which transforms like $$\Delta X^{\mu'} = L_\nu^{\mu'} \Delta X^\nu$$ where $L_\nu^\mu$ is a matrix representing the Lorentz transformation from the primed frame to the unprimed frame. But, if $\Delta X^\mu$ is a small displacement, then we can also write $$\Delta X^{\mu'} = \frac{\partial X^{\mu'}}{\partial X^\nu} \Delta X^\nu$$

Thus, a Lorentz transformation matrix can be considered as a matrix of partial derivatives of the components of one frame with respect to the coordinates of another frame (called the Jacobian matrix of the transformation), $$L_\nu^{\mu'} = \frac{\partial X^{\mu'}}{\partial X^\nu}$$

For completeness,
The inverse Lorentz transformation - in our case from the umprimed frame to the primed frame - is simply $$\Delta X^\nu = \frac{\partial X^\nu}{\partial X^{\mu'}} \Delta X^{\mu'} = \tilde{L}_{\mu'}^\nu \Delta X^{\mu'}$$ where the tilde (~) is just to emphasise that this is the inverse.

To confirm that that this is the inverse, note that $$\tilde{L}_{\mu'}^\nu L_\rho^{\mu'} = \frac{\partial X^\nu}{\partial X^{\mu'}} \frac{\partial X^{\mu'}}{\partial X^\rho} = \delta_\rho^\nu = \left\{\begin{matrix}1 &\text{ if } \rho = \nu\\0 &\text{ otherwise }\end{matrix}\right.$$ which are the components of the identity matrix.

Nb. Using the current notation, where we distinguish primed frames from unprimed frames, we could actually leave off the tilde, since the primes on the indices do determine which frame which are transforming from and which frame we are transforming to.

Now, the proper length of any four-vector, $X_\mu$, is a scalar, and hence invariant, quantity. Thus $$X_{\mu'} X^{\mu'} = X_\nu X^\nu = X_\nu \left(\tilde{L}_{\mu'}^\nu X^{\mu'}\right) = \tilde{L}_{\mu'}^\nu X_\nu X^{\mu'}$$ which we can write as $$\left(X_{\mu'} - \tilde{L}_{\mu'}^\nu X_\nu \right) X^{\mu'} = 0$$

Since $X^{\mu'}$ is an arbitrary choice (and we can choose $X^{\mu'} \neq 0$) we must have $$X_{\mu'} - \tilde{L}_{\mu'}^\nu X_\nu = 0$$ or $$X_{\mu'} = \tilde{L}_{\mu'}^\nu X_\nu$$

Thus, the covariant components of a four-vector (or any tensor) transform inversely to the contravariant components.

Covariant differentiation

A scalar field, $\phi$, is a function of all four space-time variables, $$\phi = \phi(t,x,y,z) = \phi(X^\mu)$$

In terms of tensors, a scalar is a rank-zero tensor, meaning that it is invariant under Lorentz transformations. But, we can still imagine that the field changes over space and time.

For example, we could consider the change in the field, $\Delta \phi$, over a small displacement, $\Delta X^\mu$, in space-time, $$\Delta \phi = \frac{\partial \phi}{\partial X^\mu} \Delta X^\mu$$

Since $\Delta \phi$ is a scalar, and $\Delta X^\mu$ is a contravariant four-vector, then it must be the case that $$\frac{\partial \phi}{\partial X^\mu} \rightarrow \left(\frac{\partial \phi}{\partial t},\frac{\partial \phi}{\partial x},\frac{\partial \phi}{\partial y},\frac{\partial \phi}{\partial z}\right)$$ is a covariant four-vector. To emphasise this, we often use notation that clearly places the index is in the covariant position, $$\partial_\mu \phi \equiv \frac{\partial \phi}{\partial X^\mu}$$

This notation also allows to write the contravariant components of the derivative easily as $$\partial^\mu \phi = \eta^{\mu \nu} \partial_\nu \phi \rightarrow \left(\frac{\partial \phi}{\partial t},-\frac{\partial \phi}{\partial x},-\frac{\partial \phi}{\partial y},-\frac{\partial \phi}{\partial z}\right)$$

We can think of covariant differentiation as an operation, where the operator $$\partial_\mu = \frac{\partial}{\partial X^\mu}$$ increases the covariant rank of its argument by one. Thus, given a four-vector, $V^\mu$, then $$\partial_\nu V^\mu = T_\nu^\mu$$ is a mixed tensor with one covariant index, along with the contravariant index from the original four-vector.

Since the components of $\partial_\mu$ are covariant, they transform inversely to $\mathrm{d} X^\mu$, $$\partial_{\mu'} = \tilde{L}_{\mu'}^\nu \partial_\nu$$

The wave equation

Any equations written strictly in terms of scalars will automatically be the same in all reference frames.

Wave equations are of this form and interesting wave equations, Maxwell's equations for example, always involve $2^\text{nd}$ derivatives of scalar fields.

We can write the $2^\text{nd}$ derivavtives in terms of the covariant derivative operator, using the fact that the contraction, $$\partial^\mu \partial_\mu = \partial_\mu \partial^\mu \rightarrow \frac{\partial^2 }{\partial t^2} - \frac{\partial^2 }{\partial x^2} - \frac{\partial^2 }{\partial y^2} - \frac{\partial^2 }{\partial z^2}$$ is a scalar.

Nb. If we set $c \neq 1$, then $c$ only appears in the coordinate form, $\partial^\mu \partial_\mu \rightarrow \frac{1}{c^2}\frac{\partial^2 }{\partial t^2} - \frac{\partial^2 }{\partial x^2} - \frac{\partial^2 }{\partial y^2} - \frac{\partial^2 }{\partial z^2}$.

The general form of the wave equation (for our purposes) is then, $$\partial^\mu \partial_\mu \phi = f(\phi)$$ where

  • $\phi$ is a scalar field
  • $f(\phi)$ is a function of the scalar field, $\phi$, hence is a scalar
  • $\partial^\mu \partial_\mu$ is a scalar

So we can immediately see that the equation will have exactly the same form in all reference frames.

The form of the wave equation of particular interest to us is where the right-hand side is zero, $$\frac{\partial^2 \phi}{\partial t^2} - \frac{\partial^2 \phi}{\partial x^2} - \frac{\partial^2 \phi}{\partial y^2} - \frac{\partial^2 \phi}{\partial z^2} = 0$$

Any solution to the equation will always have to depend on time - otherwise nothing will be moving - and at least one component of space. In fact, for us, we only need to consider solutions of one spatial direction, $x$ say, $$\frac{\partial^2 \phi}{\partial t^2} - \frac{\partial^2 \phi}{\partial x^2} = 0$$

Nb. If we were considering waves that travelled at velocities, $v$, other than the speed of light (for example, sound waves) then we would need to reflect this in the equation, which would then look like $\frac{1}{v^2}\frac{\partial^2 \phi}{\partial t^2} - \frac{\partial^2 \phi}{\partial x^2} = 0$ but we are only interested in electromagnetic waves, which do travel at the speed of light.

Solutions to the wave equation

The general solution to the one-spatial dimensional version of the electromagnetic wave equation, $$\frac{\partial^2 \phi}{\partial t^2} - \frac{\partial^2 \phi}{\partial x^2} = 0$$ has the form $$\phi = \phi(t, x) = F(x \pm t)$$ where $F$ could any real function. To show this, write $u = x \pm t$, then $$\begin{align*}\frac{\partial^2 \phi}{\partial t^2} - \frac{\partial^2 \phi}{\partial x^2} &= \frac{\partial }{\partial t}\left(\frac{\partial F}{\partial t}\right) - \frac{\partial}{\partial x} \left(\frac{\partial F}{\partial x}\right)\\&= \frac{\partial}{\partial t} \left(\frac{\mathrm{d} F}{\mathrm{d} u} \frac{\partial u}{\partial t} \right)- \frac{\partial}{\partial x} \left(\frac{\mathrm{d} F}{\mathrm{d} u} \frac{\partial u}{\partial x}\right)\\&= (\pm 1)\frac{\partial}{\partial t} \left(\frac{\mathrm{d} F}{\mathrm{d} u} \right) - (1) \frac{\partial}{\partial x} \left(\frac{\mathrm{d} F}{\mathrm{d} u} \right)\\&= (\pm 1)\frac{\mathrm{d}^2 F}{\mathrm{d} u^2} \frac{\partial u}{\partial t} - \frac{\mathrm{d}^2 F}{\mathrm{d} u^2} \frac{\partial u}{\partial x}\\&= (\pm 1)^2\frac{\mathrm{d}^2 F}{\mathrm{d} u^2} - \frac{\mathrm{d}^2 F}{\mathrm{d} u^2}\\&= \frac{\mathrm{d}^2 F}{\mathrm{d} u^2} - \frac{\mathrm{d}^2 F}{\mathrm{d} u^2}\\&= 0\end{align*}$$

$F(x - t)$ represent waves moving at the speed of light in the positive $x$-direction and $F(x + t)$ represent waves moving at the speed of light in the negative $x$-direction

We could also show that solutions to the wave equation are additive, in that if $\phi, \psi$ are both solutions, then so is $\phi + \psi$.

There is absolutely no restriction on $F$ at all - it could be the function of a straight line, for example - but we are interested in periodic functions, where there exists a number $a$, such that for all $u$, $$F(u + a) = F(u)$$

The archetypal solutions are the sinusoidal functions, sine and cosine. For example, $$\phi(t,x) = \cos k(x \pm t)$$ where $k$ is called the wave number and is related to both the wavelength $\lambda$, (the distance between crests of the wave in space) and the period, $T$, (the interval between crests of the wave in time).

Nb. We are talking about three periods here: the period of the $\cos$ function ($2\pi$), the period of the wave $T$ and the wavelength $\lambda$, which is a period in space.

The wavelength and the period of the wave are related by the speed, $v$, of the wave, $$\lambda = vT$$ but with the speed of light set to be $c \equiv 1$, they are equal, $$\lambda = T$$

Figure 7.4 - Wave number and wavelength

Figure 7.4 - Wave number and wavelength

If we fix $t = 0$ say, then, by definition of the wavelength, we want $$\cos k(x + \lambda) = \cos kx$$ but we also know that $$\cos (kx + 2\pi) = \cos kx$$ which means that $$k \lambda = 2\pi$$ or $$k = \frac{2\pi}{\lambda}$$

Thus, the wave number is inversely proportional to the wavelength (and the period) of an electromagnetic wave. Larger wavelength implies smaller wave number and vice versa.