The ol' Ladder-in-the-Garage thought experiment in a different guise?

Hi Guest. Yes and no. It may be similar, but it has a few more twists to it. Watch this space...

Jorrie

Hi Jorrie... Nice puzzle... And so damn difficult... ...

I suspect that (if we are standing on the plate), maybe, we should take into account the time expansion of the outside device too...

I mean that the time interval that a photodiode doesn't receive any photon (as we see this from the plate) is actually much smaller for someone standing on the device's reference system... Hence, (from the device's point of view) the plate doesn't "have enough time" to cut more than one beam as it passes through the device... ... (???)...

The result will depend on which frame of reference one moves. Is an airplane flying thru the atmosphere, or is the atmosphere simply moving past the airplane as the Earth rotates.

If my understanding is correct, the length does not actually change, it only appears to from the observers point of view, therefore, the plate will cut the same amount of beams regardless of it's relative speed, so there's only one outcome.

This is difficult. I'm resisting the urge to search the web for an answer.

I don't believe there would be two different outcomes depending on frame of reference.

It strikes me that the question is talking about simultaneous photodiode detection and that what is simultaneous in one frame of reference is not necessarily simultaneous in another, maybe due to time dilation, but I just can't figure out a way to make that line of thought work.

I'm at a loss. I'm looking forward to the answer.

Hi Roger, you wrote: "It strikes me that the question is talking about simultaneous photodiode detection and that what is simultaneous in one frame of reference is not necessarily simultaneous in another, ..."

This at the core of the problem. To start off with, how do we know that two photo-detectors are detecting light (or no light) at the same time if they are spatially separated?

The simplest way I can think of is to put an ideal (= incredibly fast) latching AND gate^[1] precisely mid-way between them, with identical wiring from the two detectors, to equalize signal delays. Do that for every adjacent pair of detectors and feed the latch outputs to a multi-input OR gate setup, driving the alarm. With the plate slowly moving, it will always break at least two beams and set the latches progressively.

Now the question is: if you reset the latches and move the plate at 0.88c, will it ever trigger any latch? I'm asking this, because there are no clock synchronizations required - the setup does not care what the clocks at the detectors read, or the clocks at the two ends of the plate, for that matter. Either two beams are broken at the same time for this setup, or they are not.

This boils down to asking: is Lorentz contraction "real", or is it just a conventional thing based on the convention used for clock synchronization? Tough question!

Jorrie

[1] Assuming that the photo-diode signals go low when they receive light. Anyway, the AND gate must set the latch when both beams of the pair are broken.

For the purpose of the question Jorrie, rather than concern ourselves with AND gates and OR gates (to determine simultaneous disruption of two photodiodes), can't we just simply say the information is passed by signal at the speed of light?

I'm thinking like this. The plate is traveling at .88c and disrupts photodiode #1 beam. Immediately photodiode #1 begins transmitting a signal to photodiode #2 that it has been disrupted (signal travels at c). This signal continues to be transmitted until the plate moves on and no longer disrupts the signal to photodiode #1. At that point photodiode #1 stops signaling photodiode #2. However some signal is still on route. If photodiode #2 gets disrupted while its still recieving the signal from photodiode #1 (that its disrupted), then the alarm will go off, if the signal from photodiode #1 stops arriving at photodiode #2 before photodiode #2 is disrupted then no alarm will sound when the plate disrupts photodiode #2.

Since the plate is contracted to only a little less than "a" and needs only move a small fraction of "a" (travelling at .88c) to block photodiode #2, whereas the signal must travel all of the distance "a" (travelling at c) to communicate its status (no longer disrupted) with photodiode #2, I'm thinking now that the alarm will definitely go off, even though the plate is only big enough to block one beam at a time.

I really should have thought of the devices having to communicate to each other. It makes all the difference in the world in the problem.

Hi Roger, yea, one can send the signals by light (rather than by wiring) and simplify the situation somewhat. It does change the dynamics of the problem somewhat. The speed of light in vacuum is constant for all inertial frames, but the speed of signals in wires are only constant in the frame of the wires. In frames moving relative to the wire frame, the speeds of those signals are functions of the relative velocities (by the relativistic velocity addition formula).

BTW, the way that you proposed is not compatible with the puzzle, demanding simultaneous detection in the detector frame, which your proposal does not provide. The logic must be halfway between the sensors, or you will have to build in fancy delay-circuitry in one line. And you still need the AND function for every pair of adjacent detectors…

Hope you get well soon - and that this damn puzzle does not add to the flu headaches!

Jorrie

Jorrie,

Thanks for the good wishes. The puzzle has been a nice diversion.

I agree about the detector set up. Instead, lets just say the alarm sits in between the two photodiodes and is signaled when it's neighboring photodiodes are disrupted.

If you don't mind waiting one more day to post an answer, I will do a more detailed math answer tomorrow to see if what I'm thinking (thanks to your explanation) works or doesn't.

Roger

Hi Roger. Will love it to see your solution first and 'uncontaminated'!

Jorrie

Assume that the time interval that the plate will cut the beams is Δt₁ for the device's reference frame and Δt₂ for the plate's reference frame... Then we have Δt₂>Δt₁ due to the time expansion of the device's reference frame as it's seen from the plate's reference frame... Assume, also, that the length between the beams is a₁ for the device's reference frame and a₂ for the plate's reference frame... Then we have a₁>a₂ due to the length contraction of the device's system (frame) as it's seen from the plate's reference frame... So if you are standing on the plate: you see a longer Δt and a shorter a , or, in other words, you see that the plate is cutting the beams in the same rate as it's seen from the device's point of view (no alarm will sound)...

Am I right???... ...

Hi again George.

It is not very clear what you mean by the Δt's in: "Assume that the time interval that the plate will cut the beams is Δt₁ for the device's reference frame and Δt₂ for the plate's reference frame..."

It seems like it is the time that one beam is cut by the plate(?), but then the argument makes no sense. If it is the time between interrupting successive beams, then your Δt₂>Δt₁ is the wrong way around. For a 10 ft separation and v/c=0.88, I get Δt₁ = 10/0.88 = 11.36ns and Δt₂ = 4.75/0.88 = 5.34ns.

So I do not follow your solution. Time dilation does not really come into it, only simultaneity.

Jorrie

Simultaneity has to do with the "relative time flow"... So the "time dilation" is, obviously, related with the simultaneity... If you stand on the plate and see what is happening to the device you see the device's "time flow" slowing down: two events separated by Δt₁ in the device's frame are now seen - from the parspective of the plate's frame - to be separated by a longer time interval Δt₂ ... So, Δt₂>Δt₁ ... Put it in other words: if two events are happening simultaneously in the device's frame, they are not happening simultaneously in the plate's frame (due to the relative time dilation of the device's frame, as it is observered by the plate's frame)... I don't get your objection... Am I wrong somewhere???... ...

Hi George. You said: "Simultaneity has to do with the "relative time flow"... So the "time dilation" is, obviously, related with the simultaneity..."

I may perhaps be proven wrong, but I think they are not directly related. Time dilation's formula is: dt² = dtau²[1-v²/c²], while the equation for synchronization offset is: Δt = vL/c, where L is the length in the direction of motion. Lorentz contraction and synchronization offset are more directly related in the sense that: (vL/c)² + L²[1-v²/c²] = L².

"... if two events are happening simultaneously in the device's frame, they are not happening simultaneously in the plate's frame (due to the relative time dilation of the device's frame, as it is observed by the plate's frame)... "

No, I don't think this is due to time dilation; but simply due to synchronization offset, vL/c. Another argument against this is that if two events are simultaneous in one frame, they are simultaneous only in that frame and always non-simultaneous in all other frames. Time dilation, on the other hand, is a reciprocal thing. Each frame view all other frames' time as running slower. Hence the two concepts are not compatible.

Still, I do not see your solution working out!

Jorrie

Thanks Jorrie... It seems you are right...

Good morning ,maybe in the reception a delay of distance ,or the fact electron frequency are slowing believing g.but an electron can't be static with a little energy.g=.5

Henri Poincare:Because the Time Scanner never produces faster than light speeds, it also perfectly handles the relativistic law of speed addition. So it can accelerate the moving gear below again and again and it will never reach the speed of light.

There is no preferred frame of reference any more. This is clearly identical to Einstein's 1905 Special Relativity and it should be emphasized that it was published in 1901.I will still be a Lorentz disciple .....I can not wait for the awser...phil

At this point if I do not acknowledge Roger Experiment it mean will go with the answer and stick with it.But the Fact are Roger get to the most NEAR,as even those who get the best EQUIPMENT,INSTRUMENTATION,better degree,get far or near problem.When I look at fact science relied on complex,imaginary,base 2,..on...and on...algorithm,rapid variation,ZERO some TIME,(SRTS).I know and a bit disappointed REGARDING the fact all of you guy's play with this stuff ...it is like when a Kid go to the park.if I look like an IDIOT it is because in this website:cr4.globalspec.com Roger prove and personally I don't CARE on which theory....c,straight line,delay of tools,of nature,of some convention protecting the PASS because they scare one day will manipulate those beautiful masterpiece's Theory.We mite use it for or own stuff.ROGER prove is quick and accurate way to set fast and quick this problem.Now of coarse this still a problem,but ROGER experiment confirm the fact of a lack some where and for sure more than one.and the frequency just fit into this result.ROGER as ALL who try confirm,near or far the lack of .....................................?I wish I was their .phil

I had a feeling once about mathematics -- that I saw it all. Depth beyond depth was revealed to me -- the Byss and the Abyss. I saw -- as one might see the transit of Venus or even the Lord Mayor's Show -- a quantity passing through infinity and changing its sign from plus to minus. I saw exactly why it happened and why tergiversation was inevitable -- but it was after dinner and I let it go.

• --Winston Churchill

Coherent Control of Vacuum Squeezing in the Gravitational-Wave Detection Band

Henning Vahlbruch, Simon Chelkowski, Boris Hage, Alexander Franzen, Karsten Danzmann, and Roman Schnabel

Max-Planck-Institut für Gravitationsphysik (Albert-Einstein-Institut) and Institut für Gravitationsphysik, Universität Hannover, Callinstraße 38, 30167 Hannover, Germany

(Received 5 April 2006; published 6 July 2006)

Towards a better Æther Theory.

Present-day measurement standards intimately link light speed, by way of atomic resonance clocking, to the definitions of time, length, motion and energy, despite the fact that light remains an magnetic resonance effect produced and propagated by atomic events in matter. Despite accepting invalid Nineteenth Century notions about time, clocks do not rule Nature, nor does light, nor does humanity's pet scientific theories. When applied to an object, the important factors defining an apparent force involve time, reaction rate, from reception to propagation, through matter's response changing position, alignment, activity, structure, and magnetic field strength. The applied force only becomes effective when compatible and exerted through the correct time frame as an energy-form with an appropriate energy level needed to affect or produce an effect in matter. Compatibility means that the matter must be in a state of activity that accepts and responds to that energy-form. Consequently, energy-forms may be absorbed, converted, stored, released, propagated, or rejected.

Jorrie, Does this puzzle have a definite answer that will be posted, or is this just a thinking game? At this point it looks like the 'twins paradox', which I don't know if it has ever been settled.

StandardsGuy,

I'm not sure if Jorrie intended his post to indicate a solution, but my post is intended to be. I'll do the math when I have time but I'm pretty sure the "trick" to the problem is the fact that the LEDs themselves have to communicate with each other, or a third point (alarm) in order to indicate that they both are disrupted and that this communication isn't instantaneous but occurs at speed of light.

This communication between sensors results in the two scenarios producing the same result. The trick of the question is, you can't just take one part of relativity (contraction) and ignore the other part (speed limit on the transfer of information). Words can be to ambiguous though, when I have time, I'll try to do the math (I'm pretty run down with the flu right now, just about the worst flu I've ever had (hell, I never get the flu)), but in one of my more healthy moments I'll try to do this with math.

Sorry you're under the weather Roger. Get well soon!

We need your input.

-John

Hi S.

Yes, the puzzle has a 'standard answer' if you like (and we will get to that), but, as Roger has pointed out, there are some technical issues to be taken care of.

I don't quite agree with Roger's experimental setup - I think it violates the original puzzle conditions. Will reply directly to that post.

BTW, the 'twin paradox' also has a standard solution showing that it is no paradox...

Jorrie

Ok, I've tried to do the math and make it work but I can't, not with the alarms sitting in between the photodiodes (at a/2). Note, for both scenarios I've placed alarms in between the photodiodes. The alarms can detect any photodiode interruption (each photodiode has a distinct frequency. The signal from the photodiode to the alarm is assumed to travel at c.

Here's what I calculated:

Scenario #1 - 2.1a Plate is moving (at .88c) while photodiodes spaced "a" distance apart remain still.

Contracted length of the Plate = .9975a
Time dilation from the plate = 2.11T_detectors
Alarm distance from photodiode detector = a/2

So only one blocked beam at a time in this scenario.

Scenario #2 - 2.1a plate standing still as photodiodes spaced "a" distance apart move by at .88c.

Contracted distance between photodiodes = .475a
Time dilation for the photodiodes = 2.11T_plate
Alarm distance from photodiode detector = .2375a

At first glance it seems like 4 blocked photodiodes simultaneously.

Jorrie, I just don't know.

I tried to figure how time dilation would help, but it doesn't seem like it does. I've looked at the signal lag, but with the alarms in the middle, there isn't any. I'm looking for the two scenario's to be equivalent since they are the same way of looking at the same thing but I'm not getting that. I look forward to reading your explanation and seeing where I'm going wrong here.

Roger

Hi Roger,

Your sums seem correct: I got similar results before. A naive 'At first glance it seems like 4 blocked photodiodes simultaneously', does not work, as you said. Also, you are right that time dilation does not solve the issue, so what is it?

We all know that the sounding of the alarm here cannot depend on who is watching (this isn't the quantum physical, Schrödinger's Cat type, thank goodness!), or on who is considered stationary. The alarm will only sound if both are static and not when there is 0.88c of relative movement. But why so?

I'll delay the 'official solution' for another day to see if yourself or someone else can come up with something.

Jorrie

Hi Roger. You asked for a Minkowski spacetime diagram. It's more difficult to draw than the Epstein or Loedel, so I've used a copy out of my eBook and annotated it slightly for this puzzle; the speed is different here (v/c=0.4), but the principles are the same.

The green bar at the bottom, at time zero, is shown just before it interrupts LED1, while it has already stopped interrupting LED0, i.e., it's actually in-between LEDs. As you can see, lengths are interpreted quite differently in Minkowski. The moving bar's proper length is 1.09 units, and although it 'looks' longer than that in the x', ct' axes, it is in fact shorter than 1 unit when projected onto the x,ct axes (the projection happens parallel to the x'=1 line). The higher bars and LEDs are just snapshots at later times.

Lines parallel to the x-axis are lines of simultaneity for the LED frame and lines parallel to the x'-axis are lines of simultaneity for the plate frame. I guess the best way forward is if you ask me about the aspects that you feel uncomfortable with.

Jorrie

Oops! I've made a silly mistake in the Minkowski spacetime diagram that I posted above.

I thought of leaving it as-is for a "test", but that's probably counter-productive in this case. It should look like on the right, with the green bar progressing against the ct axis at the same rate as the red dots. This represents the correct time dilation as observed from the LED reference frame.

Jorrie

PS: GK, this does perhaps provide another answer to your problem: although the green bar cuts the light beams later due to time dilation, it makes no difference as to how many beams are cut at any time.

-J

As Roger suspected, the answer lurks in simultaneity issues and also in a bit of vagueness in the original wording, i.e.:

"… the LED separation must be contracted to less than a/2 and more than two beams (up to 4) should be broken at a time, with a resulting alarm."

It did not prompt the reader to the fact that the LED frame contraction is only relative to the plate, which is now chosen as the reference frame, while the decisions on sounding an alarm are still made in the LED frame. The simultaneity (or rather synchronization) status is best summarized on an Epstein-space-propertime-diagram. It makes the Lorentz contraction of the (green) moving plate very clear relative to the (blue) LED-frame. It also shows the synchronization offsets between the two frames (vL/c ns).

The blue time values are how the blue LED frame observes the sync-offsets of clocks at the ends of the plate. The green values are how plate observers view the sync-offsets of the various detector pairs. Plate observers might say: "yea, no alarm sounds, because the detectors are not sampled at the same time. Sample them simultaneously according to our definition and the plate breaks more than two beams at a time and there should be an alarm." However, savvy plate observers will realize why the LED frame's observers synchronized the detectors the way they did, i.e., putting logic midway between every two adjacent detectors and trigger the alarm whenever two 'interrupt lines' are high simultaneously. It is the 'logical way'.

There are however other (perhaps less logical) ways, which we can discuss once the dust has settled on this solution…

Jorrie

PS: Roger, I will draw a Minkowski spacetime diagram for a later post.

Hi Jorrie,

I don't agree with your solution.

"How can the same experiment yield two contradictory results, depending on the inertial frame of reference chosen?"

You didn't address this in your "solution".

Some snippets from Wikipedia:

Isaac Newton embraced the Principle of Relativity of Galileo Galilei, which states that the fundamental laws of physics are the same in all inertial frames. The term Galilean invariance today usually refers to this principle as applied to Newtonian mechanics, that is, Newton's laws hold in all inertial frames. In this context it is sometimes called Newtonian relativity. It states that all uniform motion is relative, and that there is no absolute and well-defined state of rest (no privileged reference frames).

And from Wikipedia under Time Dilation:

Time dilation is the phenomenon whereby an observer finds that another's clock which is physically identical to their own is ticking at a slower rate as measured by their own clock. This is often taken to mean that time has "slowed down" for the other clock, but that is only true in the context of the observer's frame of reference... In special relativity, the time dilation effect is reciprocal: as observed from the point of view of any two clocks which are in motion with respect to each other, it will be the other party's clock that is time dilated. (This presumes that the relative motion of both parties is uniform; that is, they do not accelerate with respect to one another during the course of the observations.) (emphasis is mine)

What is true for time is true for length. So as I said in my first post, there is only one outcome.

S

Hi S, you said I did not address the original question in the puzzle and you may be right - I only showed how the answer can be deduced.

Likewise, your: "If my understanding is correct, the length does not actually change, it only appears to from the observers point of view, therefore, the plate will cut the same amount of beams regardless of it's relative speed, so there's only one outcome." also doesn't answer the question! There is only one outcome, but which one?

Does "the same amount of beams" mean that the plate always cuts 2 beams? If so, your answer is wrong, I'm afraid! The reasoning for this is in my solution post.

Regards,

Jorrie

Due to reciprocity both the plate and the frame are shortened equally (if at all). The outcome is the same as if the is plate is moving slowly. You chose the frame to be the reference frame. That is "illegal" because there is no privileged reference frame. At least that is the way I see it.

S

Hi S.

The relativistically correct answer to the puzzle is that at v << 0.88c, the alarm will sound because the plate breaks more than two beams at a time. At v=0.88c, no alarm will sound because the plate can only break one beam at a time. If this is wrong, the Einstein's relativity theory must be wrong.

You wrote: "Due to reciprocity both the plate and the frame are shortened equally (if at all). The outcome is the same as if the is plate is moving slowly."

The problem is that simultaneity is not really a reciprocal thing – the figure in post #21 is supposed to show that clearly.

You said: "You chose the frame to be the reference frame. That is "illegal" because there is no privileged reference frame."

One can choose any inertial frame to do your analysis from. In this case, we stuck the sensors onto the LED frame and they measure things as the LED frame 'observes' – hence the LED frame is the "preferred frame" here. If we also stuck sensors onto the plate (all along its length) and did the same comparison, the plate alarm would always have sounded, irrespective of its speed relative to the LED frame. The outcomes here differ, depending on who does the observation, i.e., sensing the number of beams broken simultaneously, because simultaneity is relative.

Food for thought…

Regards,

Jorrie

Hi Jorrie,

"If this is wrong, the Einstein's relativity theory must be wrong."

I don't think the theory is wrong, only the application of it.

"The problem is that simultaneity is not really a reciprocal thing – the figure in post #21 is supposed to show that clearly."

It shows that the LED frame is contracted from 10.5 feet to 10 feet from the plates point of view. You could have drawn another from the frames point of view and showed it to be contracted to 10' from the frames point of view. Wikipedia says that time dilation is reciprocal (as I pointed out in post 33). It follows, then, by logic that simultaneity is also reciprocal. If your definition says it isn't then it's wrong, no matter how logical it seems.

You can crank out all the math you want, but it's an exercise in futility.

S

Hi S, you wrote:

"It shows that the LED frame is contracted from 10.5 feet to 10 feet from the plates point of view. You could have drawn another from the frames point of view and showed it to be contracted to 10' from the frames point of view."

What you stated here is correct as far as it goes, but it is exactly the 'catch' of this puzzle! The LEDs and detectors are all in the LED frame and they do not care about how the plate frame "sees" things. In their frame, according to their simultaneity, the plate contracts enough at 0.88c to not disrupt two beams at any time, so they trigger the alarm. I think Einstein would have agreed.

"Wikipedia says that time dilation is reciprocal (as I pointed out in post 33). It follows, then, by logic that simultaneity is also reciprocal. If your definition says it isn't then it's wrong, no matter how logical it seems."

I'm sorry S, but you're somewhat off the mark here. Time dilation and simultaneity actually have nothing to do with each other, apart from the fact that both are relativistic effects. A cursory glance at the applicable formulae shows that. Time dilation depends on sqrt(1-v^2/c^2) and simultaneity on vL/c, where L is the distance between the observers. Clearly no direct relationship exists; further, one is dependent on the sign of v and the other one not.

Regards,

Jorrie

Hi Jorrie,

This 'debate' is getting old. If you are right, then it probably means that I am not smart enough to understand it, and that would be sad , but not impossible to believe. The drawing didn't clearly tell me anything about simultaneity. It would be nice if other people would speak up and say if they understand it and if they agree with it.

I told you that it was improper to pick a reference frame, and you countered: "One can choose any inertial frame to do your analysis from." So here's the deal: If you pick the plate instead of the frame for your analysis, and get the same solution, then you win (and I will take your word for it). If you get a different solution, then I win. In that case you can e-mail me instead of posting the results here if you wish.

Regards,

S

Hi S, you wrote:

"If you are right, then it probably means that I am not smart enough to understand it, and that would be sad , but not impossible to believe.""

No, I guess it's sad that I did not convey the all-important concept of simultaneity properly! Maybe I should describe the figure more clearly.

The blue x-axis represents t=0 for the LED frame. All lines parallel to it are lines of simultaneity for the blue frame, i.e., having the same time coordinate. All lines perpendicular to it are lines of co-location, i.e., having the same x-coordinate.

The green (plate) line represents t=0 for the plate frame, i.e., it is the slope of all lines of simultaneity for the plate frame. It is clear that plate and LED observers do not agree on simultaneity. The offsets that I showed (9.24 ns and 8.8 ns) are the so-called clock synchronization offsets caused by the difference in simultaneity, given by Lv/c, where L is the distance between clocks and v the relative speed.

What it means is that if there were clocks at the two ends of the plate, synchronized in the plate frame, the LED frame would observe them to be out of sync by twice 9.24 ns (18.48 ns). Likewise, observers with synchronized clocks riding on the plate would observe the LED frame's clocks to be out of sync by 8.8 ns each way from the center, but in the opposite sense. I hope this clears the basic meaning of simultaneity for this scenario.

Now on to your 'deal': from the plate's perspective, according to its definition of simultaneity, it is correct that the LED spacing drops to less than 5 ft, while the plate itself remains at 21ft long, apparently breaking at least 4 beams. The issue is that the logic is still in the LED frame, making timing measurements and logic decisions based on the LED frame's definition of simultaneity. Accordingly, it measures the moving plate as less than 10 ft long and hence never breaking more than 1 beam, not sounding an alarm.

If you somehow transmit the detector outputs to the plate frame and allow the plate's clocks and logic to decide on the number of beams broken simultaneously, then the situation would have been be different (the alarm would have sounded). It is then obviously a different 'experiment'.

The bottom line is that one can choose any inertial frame as your reference, but then you must make all measurements and decisions in that frame, otherwise the results can be paradoxical.

Let me know if this still does not settle it.

Regards,

Jorrie

Hi Jorrie,

I appreciate the time you took to explain the drawing more clearly, however you didn't quite take my challenge, did you? I, also, am apparently not making my point clearly, so I'll try one more time. Instead of the challenge, lets do a thought experiment.

First, the frame is stationary to an arbitrary reference. For convenience, lets use the CMB. The plate is in motion relative to the CMB as .88c as you describe with all hardware as you say. Next, we'll make the plate stationary to the CMB and put the frame in motion in the opposite direction at .88c, with no change in the hardware or definitions. Since there is no absolute motion, it's really the same experiment as the first time, so it has the same outcome. I think Einstein would agree.

Regards,

S

Hi S, you wrote:

"Since there is no absolute motion, it's really the same experiment as the first time, so it has the same outcome. I think Einstein would agree."

So did I, way back in post 20 and 21! The 'puzzle' was put in the 'inverse' to prompt thought, but the answer is simple. When zero relative velocity, the alarm sounds, while at 0.88c relative velocity, no alarm, irrespective of from which frame you do the analysis.

So we were miss-communicating now for some time, although I think your post #5, referred to in post #33, through me off the track, because your answer was not quite right there…

Regards,

Jorrie