# Why can no observer measure proper time here?

This is a relatively difficult question I found on a past exam (12% got it right):

Two spacecraft travel in opposite directions, with spacecraft Ajax travelling at a speed of 0.5c and spacecraft Hector travelling at a speed of 0.4c. Both are travelling relative to the inertial frame of the galaxy. The situation is shown below.

A radio signal is emitted by Ajax towards Hector.

How can proper time be measured for the interval between the radio signal being emitted on Ajax and the signal reaching Hector?

A. Use measurements made by the crew on Ajax.
B. Use measurements made by the crew on Hector.
C. Use measurements made by an observer stationary at the point where the signal was emitted.
D. No single observer can measure proper time for this case.

The answer is D. I don't quite understand why, is it because the signal is only one way and there is no way to measure proper time due to there being no singular stationary frame that the event can be measured from? If the signal was reflected back would Ajax be able to measure proper time?

• In this situation, I'm having a problem with the concept of measure. How can the sender of the signal know when it is received? – R.W. Bird Apr 17 at 17:32

In my opinion the question is poorly formulated. Its answer depends a great deal on a subjective or context-dependent choice of terminology, not only on physical facts; it leaves some important points unspecified; and it confuses "being undefined" with "being unmeasurable". These concerns are also expressed in joigus's answer and R.W.Bird's comment. Let me address these points in turn.

First, whether the term "proper time" can be used for a lightlike path is a matter of convention, personal choice, and context. If we are dealing with signals that propagate with speed lower than $$c$$, then the term "proper time" is appropriate and can be calculated or measured (see below) to be larger than zero. If the speed of the signal is increased to have a limit value of $$c$$, the proper time will have a limit value of zero. In such a circumstance it can be useful to simply keep the term "proper time" and say that its value is zero when the signal becomes lightlike, rather than declaring the term to be suddently inapplicable in the limit case. This is a matter of terminological convention, not of physics.

This kind of limit situations appears often in relativity. Consider for example this quote from Gourgoulhon (2012):

As a consequence, in that region, the proper time (of Eulerian observers) between two neighbouring hypersurfaces tends to zero as $$t$$ increases.

In fact, the radio signal could be propagating through rarefied matter, say, and therefore have a speed lower than $$c$$. In this case we can speak of a non-zero proper time of the interval between source and target.

Second, being undefined and being unmeasurable are two very different things, and their distinction is very important in physics. For example, "Regarding electromagnetic field quantities, we take the position that they are not measurable except in vacuo" (Hutter & al 2006), and yet we define them also within matter, where they are not measurable. The unmeasurability leads to several different definitions and formulations (the Chu formulation, the Minkowski formulation, and various others). The definitions are inequivalent, but the formulations are nevertheless equivalent (and many of them are indeed used today) because they lead to the same predictions for the quantities that can be measured. For a discussion of these matters see for example Penfield & Haus (1967) or Hutter & al (2006).

In the present case it could be even argued that we don't even have a problem of definition, see below.

Third, it is not clear (at least from the snippet you quote) what the authors mean by "to measure". The proper time of a timelike path represented by a curve $$C\colon [a,b] \to \text{spacetime}$$ is given by the integral (see eg Misner & al 1973 p. 316) $$\int_a^b \sqrt{\Bigl\lvert \pmb{g}[\dot{C}(t),\dot{C}(t)]\Bigr\rvert}\ \mathrm{d}t \ ,$$ where $$\pmb{g}$$ is the metric.

This integral, which can be generally called "path length", is also defined for lightlike and spacelike paths. There are several ways to measure the path length. For a timelike path it can be directly measured by one clock moving along that path. For spacelike or mixed paths the procedure is a little more involved but still based on observers carrying clocks; see the illuminating discussion in Frankel (1979) ch. 2, or in Landau & Lifshitz (1996) § 84. It's because of the difference between these possibilities that we use the term "proper time" in the timelike case and "proper length" (see eg Misner & al 1973, p. 324) in the spacelike one.

Yet, in all cases the path length can also be measured by any observer (enough close to the region of the path) equipped with a "radar system". Such an observer is basically just measuring and reconstructing spacetime and its 4D geometry in a local neighbourhood, and can therefore compute the lengths of any paths in that neighbourhood. The path length of the radio signal in the question can be measured by any observer with this same method. I would call this a "measurement" as well. After all, we do measure the universe around us, and we rarely (if ever) do so by really sending observers around with clocks. See for example Dautcourt (1983) on this. Komar (1965) could be quoted here:

At any given instant in time an observer "sees" or collects information simultaneously from all events which lie on his past null cone. [...] The events, which we may visualize as stars in the night sky, are not distinguished or located by measurements of distance, but exclusively by measurements of angle and relative intensity of light. If we wish to consider measurements which can extend over some finite time interval we must also include measurements of frequency.

If the signal propagates with speed less than $$c$$, its path length [defined by the integral above] could be directly measured by an observer moving with the signal and carrying a clock. For this reason we call the path length "proper time" in this case. But it could also be measured by any other enough close observer by a system of radar measurements, and all observers would agree on its (non-zero) value. If the signal propagates on a lightlike path, the path length could not be directly measured by the clock-carrying observer, and for this reason we may not want to call it "proper time". But it could still be measured by the radar-system observers, who would agree on a value of zero; in some situations we could agree to still call this "proper time" out of convenience, since its value changes continuously with continuous deformations of signal paths.

I would also counter-propose this (hopefully unrelated) question, taken from Gibson (1964):

(i) Do we, in our schools and colleges, foster the spirit of inquiry, of skepticism, of adventurous thinking, of acquiring experience and reflecting on it? Or do we place a premium on docility, giving major recognition to the ability of the student to return verbatim in examinations that which he has been fed?

References

• I just want to add that I agree with you that the question is ill-defined. One of the reasons is the ambiguity you point out. That's what I meant to say when I wrote "proper time of what exactly?" It's not unheard-of that an exam question is ill-defined, and I'm sure we'll converge to a clarification. – joigus Apr 17 at 18:13
• @joigus Agreed. I'm worried about a tendency to teach only how to use and put together technical terms ("use this word in sentences like this and that", "if they ask you this, reply that"), but without a real understanding of what one is saying... There's a quote by Feynman (from "Surely you're joking..."): I don't know what's the matter with people: they don't learn by understanding; they learn by some other way -- by rote, or something. Their knowledge is so fragile! – pglpm Apr 17 at 18:36

According to Wikipedia,

Proper time can only be defined for timelike paths through spacetime which allow for the construction of an accompanying set of physical rulers and clocks. For lightlike paths, there exists no concept of proper time and it is undefined as the spacetime interval is zero.

The path connecting the 2 events you mentioned i.e. event of radio signal being emitted by Ajax and the event of signal reaching Hector is a light like path. For light like paths, there exists no concept of proper time. It is undefined. This is the reason no observer can measure proper time here, because it is undefined

Proper time is the time experienced by a single clock. A proper time interval between two events is an interval measured by a clock that is present at both of them- the duration of the interval depends on the path taken by the clock, and is longest for a clock that can consider itself to have been stationary between the two events (ie a clock that has not experienced any acceleration).

To measure a proper time interval between the emission of the signal by Ajax and its detection by Hector would require a clock to travel between the two events, which is not possible as clocks cannot travel at the speed of light.

If the signal were reflected back from Hector, then a clock on Ajax could measure the proper time on Ajax between the emission and the return of the signal, but that is quite different to the proper time along the path of the signal.

Joigus points out that the proper time along the path taken by the signal is known to be zero, but I suppose whoever set the exam question took the word 'measured' literally.

• I could just as well say that a photon is a special kind of clock that always measures zero proper time. The wording of the question is really not great. – Brian Bi Apr 18 at 0:20

The right answers are either A, B, or C. The only one that's definitely wrong is D! Any inertial observer can measure proper time. It's only that it happens to be zero for trajectories of objects that move at the speed of light. Pending qualifications on proper time of what exactly?; the signals themselves?, then the answer is zero; of Ajax's and Hector's trajectories? If the latter is the case, then the answer can be worked out from knowledge of velocities as measured from any inertial frame. It's not zero, and corresponds to the time they measure in their respective laboratories by, e.g., using atomic clocks. You can't obviously have an atomic clock sit on a photon and measure time in that way, but you can rest assured the proper time for the photon is zero --so it's well defined.

I'm not saying that the fact that proper time for some motions is identically zero doesn't pose any problems. It does. In general relativity you tackle those by using a technique called "affine parameters". But that's another story. And yes, proper time is well defined in the case described.

• I think there's a typo, from the rest of the sentence you mean "...is D!", and I agree completely – pglpm Apr 17 at 15:01
• If the proper time is zero how can you measure it? The options do not mention "not defined" instead they ask if it can be measured. So isn't D correctly describe the situation? – Lost Apr 17 at 16:34
• You can't measure proper time directly with a clock, because it's a derived quantity. You can measure length with rulers and time with clocks, and then derive proper time from both records. If you think about it, when you measure speed, you're doing a similar thing. Your car registers the number of turns of the wheel, from which the distance is derived, and then uses a clock to derive speeds from that. – joigus Apr 17 at 16:52
• I personally disagree there: proper time is what we measure with a natural clock, and the whole metric can be constructed by observers with such (identically built) clocks. The procedure is for example explained in Frankel (1979): Gravitational Curvature: An Introduction to Einstein's Theory, chapter 2, and in Landau & Lifshitz (1996): The Classical Theory of Fields, §84. – pglpm Apr 17 at 17:06
• Yes, sorry. You're right. And thanks for the correction. For observers moving with $v<c$ in any inertial frame, it would be possible to use clocks taken aboard, so to speak, and then compare records. But not so for $\left\Vert \boldsymbol{v}\right\Vert =c$. – joigus Apr 17 at 17:30