Error in an argument for classical transverse Doppler effect

Question

I was recently reviewing the classical Doppler effect for intro physics students. One aspect of such is that there's no transverse effect: An observer moving perpendicular to the motion of a plane wave will register no change in frequency. However, my first attempt to construct the argument ran into an issue...

To make things simpler, I'll assume that my source is fixed at $x=-\infty$. Suppose I have an east-moving monochromatic plane wave (frequency $f$, wavelength $\lambda$, and wave velocity $\vec{u}=u \hat{x}=f\lambda\hat{x}$) specified in the medium's rest frame. I introduce an observer moving north with speed $v$.

My first attempt was this. We need to boost to the reference frame of the observer, with coordinates $\vec{x}'=\vec{x}-\vec{v}t$. Such a boost doesn't change the wavelength ($\lambda'=\lambda$) but will modify the wave velocity $\vec{u}=f\lambda \hat{x}$: It transforms as $\vec{u}'=\vec{u}-\vec{v}=u\hat{x}-v\hat{y}$. Hence the wave speed seen by the observer is $u'=\sqrt{u^2+v^2}$, so the frequency as seen by the observer is $$f'=\frac{u'}{\lambda'}=\sqrt{u^2+v^2}{\lambda}=u\sqrt{1+v^2/u^2}.$$ Thus on this argument we would conclude that there is a transverse Doppler effect, albeit of second-order in $v/u$. (The usual Doppler effect is first-order in $v/u$.)

However, this is easily contradicted: Such a boost doesn't affect the horizontal position of the observer, and therefore the wave crests will continue to arrive with their original frequency and speed regardless. So there is no transverse Doppler effect. What, then, is the error in the initial argument?

Answer 1

The key issue is how the wave velocity transforms under Galilean boosts, and the proper response is another question: "Which wave velocity are you referring to?" The point is that there's two distinct concepts of wave velocity: the phase velocity, and the group velocity. For a monochromatic plane wave viewed in a medium's rest frame, the two quantities are identical. But they do not transform identically under Galilean boosts, and so must be distinguished if we work in the observer's rest frame.

For the group velocity $\vec{v}_g$, the transformation law is in fact $\vec{u}'_g=\vec{u}_g-\vec{v}t$. Thus group velocity transforms like a vector under Galilean relativity. This makes sense: A wave group is tracked in the same way we would track any other extended body, and so its velocity should behave accordingly.

However, the phase velocity $v_p=f\lambda$ behaves differently. To clarify this, it's best to work with the plane wave $\cos[2\pi (x/\lambda-ft)]$. Since $(\vec{x}',t')=(\vec{x}-\vec{v} t,t)$ for the observer's coordinates, the plane wave is expressed as $$\cos[2\pi (x/\lambda-ft)]=\cos[2\pi (x'/\lambda-(f-v_x/\lambda)t')].$$ Hence the observer will receive a plane wave with the same wavelength ($\lambda'=\lambda$) but with frequency $$f'=f-v_x\lambda = f(1-v_x/u).$$ This is the expected Doppler effect for a fixed source. Thus the new phase velocity is $$u_p'=f'\lambda'=u-v_x=u-\vec{v}\cdot \hat{x}.$$ That is, the phase velocity is unaffected by boosts which are transverse to the wave propagation---and this is as it should be, because such a boost alters neither crest separation nor the rate at which the crests move horizontally. (This is, after all, the concept underlying phase velocity.) Boosts in the direction of wave propagation, on the other hand, do affect the horizontal velocity of the crests and therefore do produce a Doppler shift.

The argument is then resolved as such: For the purposes of the Doppler effect, what matters is the phase velocity rather than the group velocity. The phase velocity is unaffected by transverse boosts, and there is no classical transverse Doppler effect.

Answer 2

Your construction increases the wavelength as perceived by the observer by exactly the same amount as the increase in velocity. Although the distance between successive peaks along the direction of the wave-travel is still $\lambda$, the observer meets the second wavefront after travelling a distance in the $y$ direction and this has to be added, Pythagoraswise, to $\lambda$ to get $\lambda'$. The ratio is then still the same $f$.