1
$\begingroup$

I am studying "Introducing Einstein's Relativity" (D'Inverno, Oxford University Press) and I am trying to understand Fig.2.12, page 23. There it goes:

fIG.2.12, PAGE 23

Observer B sends:

  • a light ray towards Q at time t(R) which bounces back at t(U)

  • a light ray towards P at time t(S) which bounces back at t(V)

Observer B concludes that P and Q are not simultaneous, but that anyhow they are equidistant (because light takes equal time intervals RU and SV to bounce back).

The doubt I have is about the need for active illumination of targets for which we want to measure the distance.

Clearly, our observer B is using a radar, or a flashlight, or some other sort of emitting device. I can see that using a radar is radically different from using a binocular, and that telescopes require parallax considerations in order to calculate distances.

But if active illumination is the only way to produce distance estimations then in Einstein's train carriage experiment the observer on the train (who's only receiving light beams) won't be able to judge his own position inside the carriage. Is this correct?

$\endgroup$
  • 1
    $\begingroup$ This question does not show any research effort; it is unclear or not useful. Love that.... $\endgroup$ – Marco Faustinelli Mar 24 '17 at 16:25
1
$\begingroup$

if active illumination is the only way to produce distance estimations

This requirement (cmp. Einstein 1905) stands in the context of thought-experiments, for the purpose of definition of geometric relations (such as distance ratios) and the evaluation of systematic uncertainties of estimations which might be obtained by other means.

in Einstein's train carriage experiment the observer on the train (who's only receiving light beams)

It is true that the description of Einstein's train carriage experiment (definition of "simultaneity", 1917) concentrates on participant M observing signal (flash) indications of participants A and B (and not the other way(s) around).

But there is explicitly required the determination to identify M as "mid-point" {between} A and B. Arguably this necessitates (for the purposes of unambiguous definition) mutual observations of A, B, and M, namely:

(1) for each signal indication by A, that
- A saw that B saw that A had signalled, in coincidence with
- A having seen that M saw that A saw that M saw that A had signalled,

(2) for each signal indication by B, that
- B saw that A saw that B had signalled, in coincidence with
- B having seen that M saw that B saw that M saw that B had signalled, and

(3) for each signal indication by M, that
- M saw that A saw that M had signalled, in coincidence with
- M having seen that B saw that B had signalled.

(And satisfying these conditions may still not suffice to identify M uniquely as "mid-point between" A and B; additional conditions involving certain additional participants may to be satisfied.)

So:

[...] Is this correct?

Yes.

$\endgroup$
1
$\begingroup$

The entire theory of special relativity is built upon the axiom that light speed is constant to any observer in any reference frame. All the rest is just consequences of this assumption.

So yes, Einstein's passenger will see the train like any other train passenger sees their train.

$\endgroup$
  • $\begingroup$ hear! hear! Absolutely! $\endgroup$ – ZeroTheHero Mar 24 '17 at 23:50
1
$\begingroup$

Radar methods are not the only way to establish position and time coordinates. One can use Einstein's rods and clocks. For inertial motion in Special Relativity, they lead to identical coordinate assignments. Some folks (like me) prefer radar methods because they are simpler to analyze and practical. I cannot rely on a really long ruler made of some material extended into space. I think the point is this: provide an operational definition of your measurements---don't presume that some everyday-intuitive notion from Galilean physics carries over.

$\endgroup$
  • $\begingroup$ When I place these words aside the ones you wrote for my previous question, things start to shape up. Thank you for your perseverance :-) $\endgroup$ – Marco Faustinelli Mar 24 '17 at 16:29
1
$\begingroup$

There is nothing about relativity that requires active illumination. Indeed, there is nothing that requires light at all. It is in some sense unfortunate, from a conceptual point of view, that there is anything that actually moves at c, i.e., it is unfortunate for the understanding of relativity that light exists. Sure, it was part of the history of the discovery of relativity, and we use it to great value in experiments, but conceptually, if there was no such thing as light, we could still understand relativity just fine, perhaps even better-- because we would then never fall under the misconception that relativity has something to do with a trick of light. Relativity is about the geometry of spacetime, and any valid means of measuring times and distances will encounter it, even if light is never used at all. It is equivalent to having light that always moves at the same speed in any frame, but the postulates of relativity can be stated without the existence of light. If we do it that way, then the postulates could be used to derive the result that if there existed a massless particle, it would always move at c in any frame, but we could have those postulates and not have any massless particle. An example would be the twin paradox-- even if there were not light, a spaceship moving at close to c would encounter the twin effect.

$\endgroup$
  • $\begingroup$ It is unfortunate for the understanding of relativity that light exists. Relativity is about the geometry of spacetime. I'll try to use these words as my guiding principle. Let's see where they'll lead me. Hopefully where you and robphy are :-) Thank you. $\endgroup$ – Marco Faustinelli Mar 24 '17 at 16:32
  • $\begingroup$ I disagree with some of your statements. The central postulate of special relativity is that light travels at the same speed in any inertial frame. @Thriveth is completely correct in pointing out that everything else - the Lorentz transformations, the addition of velocities, the geometrical structure underlying SR and everything else - follow from this. $\endgroup$ – ZeroTheHero Mar 24 '17 at 23:49
  • $\begingroup$ It is not logically correct to say that since all those things follow from that one postulate, that this implies SR must be postulated in that way. Nor does it logically follow that light even needs to exist, in order to have SR. There are many equivalent ways to postulate SR, some require no reference to light. Hence, you have not actually refuted anything I said. Essentially, SR is identical to the Lorentz transformation, and light need not exist to have that, nor even to discover it, but it makes it easier to discover-- but it also makes a lot of people think SR is a trick of light. $\endgroup$ – Ken G Mar 25 '17 at 1:41
  • $\begingroup$ You are right, lightspeed is not the essence of SR. Neither is the Lorenz transformation. The essence of SR is the postulate that any experiment is going to yield identical results in any inertial frame, which is a generalization of Galileo's principle. But the assumption of constant light speed is a consequence of this, and the Lorenz transformation is a consequence of the constant light speed. It is true nothing keeps you from developing the Lorenz transformation without light. But without light, what is it but a mathematical function among an infinity of functions? $\endgroup$ – Thriveth Mar 25 '17 at 3:27
  • $\begingroup$ It is not enough that inertial observers get the same physics, there remains to determine what that physics is. That's where the Lorentz transformation comes in. Even without light, the Lorentz transformation is still the way we find we need to map one observer's inertial coordinates into another's, it gives the correct invariants. All one needs to discover this is observers that can move at high relative speeds. Of course, we actually don't get observers moving at high speeds, but we do have elementary particles and so on, so we get the notion of proper time without any need for light. $\endgroup$ – Ken G Mar 25 '17 at 7:59

Not the answer you're looking for? Browse other questions tagged or ask your own question.