I am a beginner of relativity. I read Relativity Made Relatively Easy by Andrew M. Steane. At chapter 6.4, p.118, he wrote about the phase velocity of matter wave in the viewpoint of relativity. I totally do not understand how he establish the relationship between particle and phase velocity.

What is the connection of the 'pointer' and motion of particle? and what is the property of 'pointer' here? why did he say the simultaneity of 'pointer is vertical' is lost in other frame?

  • 1
    $\begingroup$ Andrew Steane is a member here, and will probably read this question. If anyone can answer this question with as much detail and clarity, it will be Andrew. $\endgroup$
    – joseph h
    Commented Feb 25, 2022 at 2:15
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    $\begingroup$ Agree, +1 for @josephh. -NN $\endgroup$ Commented Feb 25, 2022 at 5:30

1 Answer 1


Ok so I'm responsible for the book so guess I'd better add some clarification here!

The background is that we have group velocity and phase velocity, and we would like to know what physical interpretation they may have in the case of de Broglie waves. The group velocity is the easier to interpret, since it is the velocity at which a bump in a wavefunction will move along, and this corresponds, in the classical limit, to the velocity of a particle.

The phase velocity is, at first, sight, more puzzling, because it exceeds the speed of light, and in any case the phase for a single wavefunction is not an observable physical property. So one might guess it has no particular physical interpretation. But there is an interpretation in terms of simultaneity. The idea of a 'pointer' is intended as a convenient way to capture the notion of a clock. We imagine that we could furnish each particle with a pointer which behaves like the second-hand of a clock, or something like that. Of course particles such as electrons do not have such things attached to them, but the image is serving as a way to illustrate how the mathematics of quantum theory works, in which these 'pointers' behave like the phase of the wavefunction, which is in general a complex number at any given location.

But let's focus our attention for a moment on a set of classical objects, say little balls made of steel, where each has this clock-hand or pointer attached to it, going round and round. If all the little balls in our illustration are not moving relative to one another, then there is an inertial frame in which they are all at rest. We arrange for their 'pointers' or clock-hands to rotate in synchrony in this frame.

If we now examine that same set of balls from another frame, one moving relative to the first at speed $v$, then the balls are themselves moving at speed $v$ relative to the new frame, and they carry their rotating pointers along with them. As the pointers go round and round, one can notice that they are not in synchrony in the new frame. All those in a line of balls travelling abreast to each each other are in synchrony, but they are out of phase with the ones in the next line of balls behind or ahead. So the events of sets of pointers arriving at vertical (which corresponds to quantum phase arriving at a multiple of $2\pi$) happen at times that make it appear like a mexican wave sweeping down the system. This mexican wave is moving at the phase velocity of the de Broglie waves associated with the steel balls, which is here $c^2/v$.

And to finish, note that the events 'arrive at vertical' for these pointers (or if you prefer, 'arrive at a given multiple of $2\pi$' for the phase of a wavefunction) are events: that is, they are given points in space-time. Those events, which happen one after another in the frame with speed $v$, are simultaneous in the rest frame (because that is how we set the system up in the first place).

Postscript. If you still find this whole observation a bit puzzling, then don't worry about it! It is not very important. It is just a minor observation about the phase of de Broglie waves for your general knowledge as a physicist.

  • $\begingroup$ Prof. Steane, thank you so much! I have a clear understanding about this issue after I see and re-think your interpretation. I am appreciated for your effort writing this book. I learned A LOT from this book. $\endgroup$
    – Hsu Bill
    Commented Mar 1, 2022 at 0:38

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