# Are signal fronts in a beam not at rest to each other?

I'd like to investigate how the notion of "mutual rest" might be applied consistently, but distinctively, in the following thought experiment:

Consider a light source ("$A$") which directs a beam towards a receiver ("$B$"). Source $A$ is modulating the beam, emitting discrete recognizable signals in rapid succession (i.e. with short durations from stating one particular recognizable signal modulation until stating the next; compared to $A$'s duration from stating one particular recognizable signal modulation until observing the corresponding echo from receiver $B$).

Is there some definite, operational sense of "mutual rest" whereby in this case it could be said

• that source $A$ and receiver $B$ had been "at rest to each other" ("throughout the experiment", if applicable); but

• that a successive pair of signal fronts had not been "at rest to each other" ("while both were on their way")

?

• This question has been asked in reaction to the statement "there is no reference frame moving with speed *c* relative to another reference frame" in this answer and, as it turned out, also in anticipation of statements such as "An object moving at $c$ does not have a rest frame" in this more recent answer. – user12262 Mar 29 '14 at 9:00

## 2 Answers

In relativity, in order for something to be at rest with respect to something else, both of them must be particles with nonzero mass in their respective rest frames, which are represented by time-like four-vectors. Saying that two photons are at rest one w.r.t another does not make sense, because there exists no inertial frame in which a photon can be at rest, photons are represented by null-like four-vectors. An inertial frame is the coordinate system of an inertial observer, who is time-like by definition. This is a reformulation of the axiom that the maximum speed of a particle is the speed of light.

Wave fronts are not particles, they are the parts of a wave with equal phase, traveling at the speed of light. As such, saying one is at rest w.r.t another does not make sense at all. You could say that they move at the same speed of propagation, though.

• auxsvr: "in order for something to be at rest with respect to something else, both of them must be [...] represented by time-like four-vectors." -- However: being (separately) "represented by time-like four-vectors" is apparently not sufficient for two "something"s to be at rest wrt. each other. A satisfactory answer should point out which condition is sufficient for determining "mutual rest", and derive explicitly that such a condition cannot be satisfied by "somethings which are (separately) represented by null-like four-vectors". (contd.) – user12262 Mar 30 '14 at 8:08
• auxsvr: "An inertial frame is the coordinate system of [...]" -- The argument should be made without referring to any coordinates; because otherwise it would just beg the question how such coordinates might have been assigned to the participants (and their individual indications, or even their coincidence events) in the first place. "Wave fronts [...] are the parts of a wave with equal phase" -- Not really; cmp. en.wikipedia.org/wiki/Front_velocity "here exists no inertial frame in which a photon can be at rest" -- Why not e.g. a google.com/search?q="light+cone+frame"? – user12262 Mar 30 '14 at 8:18
• @user12262 Relativity is a physical theory and inertial frames are an integral part of it, because that's how we do measurements. If there cannot exist an observer that observes something at rest, saying it is at rest does not make sense. – auxsvr Mar 30 '14 at 8:25
• @user12262 I'm not certain how the link to front velocity proves me wrong. Also, I checked the light-cone frame definition in Zwiebach, A first course in string theory and it has nothing to do with an inertial frame of the photon, it is just a change of coordinates, not a Lorentz transformation. – auxsvr Mar 30 '14 at 8:54
• auxsvr: "[...] because that's how we do measurements" -- Explicate how "we (time-like guys) do measurements" (hint: google.com/#q=%22chronogeometry%22 ) and prove that therefore "null-like objects" are incapable of "doing likewise" (and consequently they're indeed not called "observers"). "the light-cone frame [...] has nothing to do with an inertial frame of the photon" -- Correct (given the justification I keep asking about). Though only looking at coordinates seems inconclusive: e.g. the set "$\{ \forall \theta \in \mathbb R: (\theta, 0, 0, 0) \}$". – user12262 Mar 30 '14 at 15:01

A necessary condition for mutual rest is mutual rigidity; and a necessary condition for mutual rigidity is according to Synge (Relativity: The General Theory, p. 115) operationally ("chrono-geometrically") defined as

[...] sending photons from one [...] to another, and receiving back the scattered or reflected photons. The criterion for rigidity is that the elapsed time [duration of the signal source] from emission to return [...] should be constant.

Any two distinct signal fronts in a beam are however not exchanging any photons (as signal fronts) between each other. Consequently they cannot be said having been rigid to each other; nor therefore having been at rest to each other.

Remark:
Synge's more complete, sufficient definition refers to:

[...] sending photons from one timelike curve to another [...]

• You're confused. The point is that there must exist an observer to measure something at rest, it is not that the wave fronts do not exchange photons that makes what you ask not make sense at all. The measurement process includes sending and receiving photons, but what use would that be even if photons could do that, since a photon cannot measure the distance? To make a measurement you'll need at least a clock and a device to send and receive photons, how can a photon on its own do that? Even Synge says so, the curves must be time-like, why do you keep ignoring this? – auxsvr Apr 5 '14 at 16:37
• auxsvr: "You're confused." -- Well, at least one of us seems confused. "it is not that the wave fronts do not exchange photons that makes what you ask not make sense" -- The sheer formal logic of my argumentation seems stringent enough, for starters. (And terminology-wise I rather stick to "signal front".) So your point is(?) 1.: "you'll need at least a clock" -- The possibility to assign parametrizations to a signal front (a.k.a. "null-trajectory") has apparently been considered. And 2.: ... [to be contd.] – user12262 Apr 5 '14 at 18:14
• auxsvr: "and a device to send and receive photons" -- For each event in which any one front had a part there's an entire "light cone" of events such that, at least kinematically, photons from the "past half-cone" could've been received by the front, or photons could've been sent from it to the "future half-cone". Certain is only: There's no such "signal round trip" between successive fronts of the same beam. As it happens: that's sufficient to answer my question. "Even Synge says so, the curves must be time-like" -- The "Remark" even quoted it. I just didn't have to use that subclause. – user12262 Apr 5 '14 at 18:25
• You still don't understand what I'm telling you. Synge writes about measuring the elapsed time between sending and receiving a photon, how do you measure that on a photon? – auxsvr Apr 5 '14 at 18:30
• auxsvr: "Synge writes about measuring the elapsed time between sending and receiving a photon" -- To be precise (ABAIU): the elapsed time between sending a photon and (in general separately) receiving not just some other photon(s), but the scattered or reflected one(s), in response to what had been sent. "how do you measure that on a photon?" -- My point is: this questions doesn't even arise; because there is no pair of events (one "sending" and a/the corresponding one "receiving") identifiable to begin with, for which to try to evaluate "elapsed time". Otherwise ...[contd.] – user12262 Apr 5 '14 at 19:59