# Special Relativity/Waves

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## Applications of Special Relativity

In this chapter we continue the study of special relativity by applying the ideas developed in the previous chapter to the study of waves.

First, we shall show how to describe waves in the context of spacetime. We then see how waves which have no preferred reference frame (such as that of a medium supporting them) are constrained by special relativity to have a dispersion relation of a particular form. This dispersion relation turns out to be that of the relativistic matter waves of quantum mechanics.

Second, we shall investigate the Doppler shift phenomenon, in which the frequency of a wave takes on different values in different coordinate systems.

Third, we shall show how to add velocities in a relativistically consistent manner. This will also prove useful when we come to discuss particle behaviour in special relativity.

A new mathematical idea will be presented in the context of relativistic waves, namely the spacetime vector or four-vector. Writing the laws of physics totally in terms of relativistic scalars and four-vectors ensures that they will be valid in all inertial reference frames.

### Waves in Spacetime

Waves in Spacetime

We now look at the characteristics of waves in spacetime. Recall that a wave in one space dimension can be represented by

${\displaystyle A(x,t)=A_{0}\sin(kx-\omega t)\,}$

where ${\displaystyle A_{0}}$ is the (constant) amplitude of the wave, ${\displaystyle k}$ is the wavenumber, and ${\displaystyle \omega }$ is the angular frequency, and that the quantity ${\displaystyle \phi =kx-\omega t}$ is called the phase of the wave. For a wave in three space dimensions, the wave is represented in a similar way,

${\displaystyle A(\mathbf {x} ,t)=A_{0}\sin(\mathbf {k} \cdot \mathbf {x} -\omega t)}$

where ${\displaystyle \mathbf {x} }$ is now the position vector and ${\displaystyle \mathbf {k} }$ is the wave vector. The magnitude of the wave vector, ${\displaystyle |\mathbf {k} |=k}$ is just the wavenumber of the wave and the direction of this vector indicates the direction the wave is moving. The phase of the wave in this case is ${\displaystyle \phi =\mathbf {k} \cdot \mathbf {x} -\omega t}$.

Figure 5.1: Sketch of wave fronts for a wave in spacetime. The large arrow is the associated wave four-vector, which has slope ${\displaystyle \omega /ck}$. The slope of the wave fronts is the inverse, ${\displaystyle ck/\omega }$.

In the one-dimensional case ${\displaystyle \phi =kx-\omega t}$. A wave front has constant phase ${\displaystyle \phi }$, so solving this equation for ${\displaystyle t}$ and multiplying by ${\displaystyle c}$, the speed of light in a vacuum, gives us an equation for the world line of a wave front:

${\displaystyle ct={\frac {ckx}{\omega }}-{\frac {c\phi }{\omega }}={\frac {cx}{u_{p}}}-{\frac {c\phi }{\omega }}\quad {\mbox{(wave front).}}}$

The slope of the world line in a spacetime diagram is the coefficient of ${\displaystyle x}$, or ${\displaystyle c/u_{p}}$, where ${\displaystyle u_{p}=\omega /k}$ is the phase speed.

### An application of relativity to waves

Returning to the phase of a wave, we immediately see that

${\displaystyle \phi =\mathbf {k} \cdot \mathbf {x} -\omega t=\mathbf {k} \cdot \mathbf {x} -(\omega /c)(ct)={\underline {k}}\cdot {\underline {x}}.}$

Thus, a compact way to write a wave is ${\displaystyle A({\underline {x}})=\sin({\underline {k}}\cdot {\underline {x}}).}$(6.8)

Since ${\displaystyle {\underline {x}}}$ is known to be a four-vector and since the phase of a wave is known to be a scalar independent of reference frame, it follows that ${\displaystyle {\underline {k}}}$ is indeed a four-vector rather than just a set of numbers. Thus, the square of the length of the wave four-vector must also be a scalar independent of reference frame

${\displaystyle {\underline {k}}\cdot {\underline {k}}=\mathbf {k} \cdot \mathbf {k} -\omega ^{2}/c^{2}=const.}$

Resolution of a four-vector into components in two different reference frames.

Let us review precisely what this means. As this figure shows, we can resolve a position four-vector x into components in two different reference frames, e.g (X,T) and (X′,T′), but these are just different ways of writing the same vector.

This is exactly the same as the way a three-vector has different components in a rotated frame.

Similarly, just as a three-vector has the same magnititude in all frames, so does a 4-vector; i.e,

${\displaystyle X^{2}-c^{2}T^{2}=X^{\prime 2}-c^{2}T^{\prime 2}}$

Applying this to the wave four-vector, we infer that

${\displaystyle k^{2}-c^{-2}\omega ^{2}=k^{\prime 2}-c^{-2}\omega ^{\prime 2}}$

where the unprimed and primed values of k and ω refer to the components of the wave four-vector in two different reference frames.

Up to now, this argument applies to any wave. However, waves can be divided into two categories, those for which a special reference frame exists, and those for which there is no such special frame.

As an example of the former, sound waves look simplest in the reference frame in which the gas carrying the sound is stationary. The same is true of light propagating through a material medium with an index of refraction not equal to unity. In both cases the speed of the wave is the same in all directions only in the frame in which the material medium is stationary.

If there is no material medium, then there is no unambiguous way of finding a special frame so the waves must fall into the second category. This includes all waves in a vacuum, such as light.

In this case the following argument can be made. An observer moving with respect to waves of frequency ω and wave number k sees waves of frequency ω′ and wave number k′. If the observer can tell in any way that they come from a source moving with respect to them, then they can use this to identify a special frame for those waves, so the waves must look just like ones from a stationary source of frequency ω

This forces us to conclude that for such waves

${\displaystyle \omega ^{2}=c^{2}k^{2}+\mu ^{2}\,}$

where μ is a constant. All waves in a vacuum must have this form, a much more restricted choice than in classical physics.

In classical physics, ω and k for light are related by

${\displaystyle \omega =ck\,}$

In relativistic physics, we've seen that for waves with no special reference frame, such as light, ω and k are related by

${\displaystyle \omega ^{2}=c^{2}k^{2}+\mu ^{2}\,}$

If μ=0 then the relativistic equation reduces to the classical, so we can assume that, for light, μ does equal zero.

This means that light does not have a mimimum frequency.

If μ is not zero then the wave being described are dispersive. The phase speed is

${\displaystyle u_{p}={\frac {\omega }{k}}={\sqrt {c^{2}+{\frac {\mu ^{2}}{k^{2}}}}}}$

This phase speed always exceeds c, which at first may seem like an unphysical conclusion. However, the group velocity of the wave is

${\displaystyle u_{g}={\frac {d\omega }{dk}}={\frac {kc^{2}}{\sqrt {k^{2}c^{2}+\mu ^{2}}}}={\frac {kc^{2}}{\omega }}={\frac {c^{2}}{u_{p}}}}$

which is always less than c. Since wave packets and hence signals propagate at the group velocity, waves of this type are physically reasonable even though the phase speed exceeds the speed of light.

Another interesting property of such waves is that the wave four-vector is parallel to the world line of a wave packet in spacetime. This is easily shown by the following argument.

The spacelike component of a wave four-vector is k, while the timelike component is ω/c . The slope of the four-vector on a spacetime diagram is therefore ω/kc. However, the slope of the world line of a wave packet moving with group velocity is c/ug, which is also ω/kc .

Note that when we have k is zero we have ω=μ. In this case the group velocity of the wave is zero. For this reason we sometimes call μ the rest frequency of the wave.

### The Doppler Effect

You have probably heard how the pitch of a train horn changes as it passes you. When the train is approaching, the pitch or frequency is higher than when it is moving away from you. This is called the Doppler effect. A similar, but distinct effect occurs if you are moving past a source of sound. If a stationary whistle is blowing, the pitch perceived from a moving car is higher while moving toward the source than when moving away. The first case thus has a moving source, while the second case has a moving observer.

In this section we will compute the Doppler effect as it applies to light moving through a vacuum. The figure below shows the geometry for computing the time between wave fronts of light for a stationary and a moving reference frame.

• The time between wavefronts for the stationary observer, in the stationary frame, is ${\displaystyle T}$.
• The time between wavefronts for the moving observer, in the stationary frame, is ${\displaystyle T^{\prime }}$.
• The time between wavefronts for the moving observer, in the stationary frame, is ${\displaystyle \tau }$.

Since the world lines of the wave fronts have a slope of unity, the sides of the shaded triangle both have the same value, C. If the observer is moving at velocity ${\displaystyle v}$ , the slope of the observer's world line is ${\displaystyle c/v}$, i.e

${\displaystyle {\frac {c}{v}}={\frac {cT+X}{X}}}$

Solving this for X and substituting in give ${\displaystyle cT^{\prime }=cT+X}$ gives

${\displaystyle T^{\prime }={\frac {T}{1-{\frac {v}{c}}}}\quad (1)}$

In classical physics ${\displaystyle T^{\prime }}$ and ${\displaystyle \tau }$ are the same so this formula, as it stands, leads directly to the classical Doppler shift for a moving observer.

However, in relativity ${\displaystyle T^{\prime }}$ and ${\displaystyle \tau }$ are different. We can use the Lorentz transformation to correct for this.

The second wavefront passes the moving observer at (${\displaystyle vT^{\prime }}$, ${\displaystyle cT^{\prime }}$) in the stationary observers frame, but at (0,${\displaystyle c\tau }$) in its own frame. The Lorentz transform tells us that.

${\displaystyle c\tau =\gamma \left(cT'-{\frac {v}{c}}vT'\right)=cT'\gamma \left(1-{\frac {v^{2}}{c^{2}}}\right)=cT'/\gamma }$

Substituting in equation (1) gives

${\displaystyle {\begin{matrix}\tau &=&T{\frac {\sqrt {1-v^{2}/c^{2}}}{1-v/c}}\\&=&T{\sqrt {\frac {(1-v/c)(1+v/c)}{(1-v/c)^{2}}}}\\&=&T{\sqrt {\frac {c+v}{c-v}}}\end{matrix}}}$

From this we infer the relativistic Doppler shift formula for light in a vacuum:

${\displaystyle \nu ^{\prime }=\nu {\sqrt {\frac {c-v}{c+v}}}}$

since frequency is inversely proportional to time.

We could go on to determine the Doppler shift resulting from a moving source. However, by the principle of relativity, the laws of physics should be the same in the reference frame in which the observer is stationary and the source is moving. Furthermore, the speed of light is still c in that frame. Therefore, the problem of a stationary observer and a moving source is conceptually the same as the problem of a moving observer and a stationary source when the wave is moving at speed c.

This is unlike the case for, say, sound waves, where the stationary observer and the stationary source yield different formulas for the Doppler shift.