Einstein’s clock tower thought experiment question

[MPMin]

7 hours ago, Markus Hanke said:

It’s not that simple. In frame C, the instantaneous tick rate of B is dilated, whereas the distance A-B appears longer for frame B than frame C. We have thus far only concerned ourselves with the final result of the experiment (C accumulating less time than B in total) - if you want to analyse what C visually sees on clock B at every moment of the journey, then things become complicated, because C and B do not share a common notion of simultaneity while there is a spatial separation between them. So you would have to account for relativity of simultaneity as well, and the analysis will lead you to the relativistic Doppler effect.

In actual fact, what you visually see is that the hands on the distant clock advance faster as compared to your own clock - that’s because these intervals “tick out” a total spatial distance that’s different from yours.

The key issue here is that there’s a difference between instantaneous tick rate, and total accumulated time - the instantaneous tick rate of B is dilated with respect to C, but you’re not sharing the same notion of simultaneity, so if you integrate those small infinitesimals, you end up with longer intervals, and thus more overall time passed as seen by you.

It’s like the distant clock projects pictures of itself at you at a steady rate of 1 frame per time unit - but because you are moving towards the clock at high speed, you are encountering each picture at less than one time unit in your own frame, and there will be more of those pictures in total. This is why it visually looks like it’s running fast. But if you were to compare each picture individually to your own clock, using some appropriate concept of relativity of simultaneity, you’d find the instantaneous readings to be dilated with respect to you. The overall times are different, because the overall distance A-B is also different between these frames.

I’m sorry I don’t know how to explain it better - this is neither intuitive nor particularly simple, despite the quite basic scenario. The devil is in the details. Nonetheless, once the maths are done correctly (also not as simple as it might at first seem!), they are found to correspond to what we actually find in the real world.

Thanks again for your reply.

I’m just going to reiterate what you are saying in my layman’s terms to confirm my understanding. If I’m getting it wrong please correct me.

Using the previous scenario where all clocks are synchronised before C departs from A, lets assume C departs from A at 50% speed of light and lets assume acceleration happens instantaneously (just to skip the effects of the acceleration phase and to only consider the constant speed phase). Let’s assume that from A’s frame of reference, C takes 1 min to travel from A to B travelling at 50% speed of light. Let’s also assume all clocks are always within their respective observers frame of reference and to be clear the scenario only describes what each observer sees as per the light that enters the eye.

As soon as C departs A at 50% speed of light, from A’s frame of reference, A sees clock C ticking 50% slower than his own clock at A and observer C from his own frame of reference sees clock A ticking 50% slower than his own clock at C.

Observer C from his own frame of reference never notices any change to the rate of ticking of his own clock. 

From the moment that  C departs A, observer C from his own frame of reference sees clock B ticking 50% faster because C is encountering more light pulses from B as C moves towards B

I’m going to use my intuition to make these next assumptions but i would imagine that, as soon as C left A at 50% speed of light, B would see that C was still stationary at A until C had reached the half way point in terms of distance between A and B. As C passes the half way point, observer A would see that 30 seconds had elapsed on his own clock at A but only 15 seconds had elapsed on clock C. 

I’m still assuming that, At the half way point observer C sees that only 15 seconds has elapsed on his own clock and only 7.5 seconds has elapsed on clock A from C’s frame of reference.

As C passes the half way point, observer B sees C just leaving A, even though C left A 30 seconds ago from A’s frame of reference and 15 seconds ago from C’s frame of reference. Observer C has also seen clock B ticking faster for the past 15 seconds from his own frame of reference. 

When C arrives at B 60 seconds would have elapsed for observer A from their own frame of reference, 30 seconds would have elapsed for observer C from their own from of reference and 30 seconds would have elapsed for observer B at their own frame of reference. 

When i start thinking about what each observer would have seen the times to have been at each stage I can’t resolve the time differences.

 

[Janus]

6 hours ago, MPMin said:

Thanks again for your reply.

I’m just going to reiterate what you are saying in my layman’s terms to confirm my understanding. If I’m getting it wrong please correct me.

Using the previous scenario where all clocks are synchronised before C departs from A, lets assume C departs from A at 50% speed of light and lets assume acceleration happens instantaneously (just to skip the effects of the acceleration phase and to only consider the constant speed phase). Let’s assume that from A’s frame of reference, C takes 1 min to travel from A to B travelling at 50% speed of light. Let’s also assume all clocks are always within their respective observers frame of reference and to be clear the scenario only describes what each observer sees as per the light that enters the eye.

As soon as C departs A at 50% speed of light, from A’s frame of reference, A sees clock C ticking 50% slower than his own clock at A and observer C from his own frame of reference sees clock A ticking 50% slower than his own clock at C.

Observer C from his own frame of reference never notices any change to the rate of ticking of his own clock. 

From the moment that  C departs A, observer C from his own frame of reference sees clock B ticking 50% faster because C is encountering more light pulses from B as C moves towards B

I’m going to use my intuition to make these next assumptions but i would imagine that, as soon as C left A at 50% speed of light, B would see that C was still stationary at A until C had reached the half way point in terms of distance between A and B. As C passes the half way point, observer A would see that 30 seconds had elapsed on his own clock at A but only 15 seconds had elapsed on clock C. 

I’m still assuming that, At the half way point observer C sees that only 15 seconds has elapsed on his own clock and only 7.5 seconds has elapsed on clock A from C’s frame of reference.

As C passes the half way point, observer B sees C just leaving A, even though C left A 30 seconds ago from A’s frame of reference and 15 seconds ago from C’s frame of reference. Observer C has also seen clock B ticking faster for the past 15 seconds from his own frame of reference. 

When C arrives at B 60 seconds would have elapsed for observer A from their own frame of reference, 30 seconds would have elapsed for observer C from their own from of reference and 30 seconds would have elapsed for observer B at their own frame of reference. 

When i start thinking about what each observer would have seen the times to have been at each stage I can’t resolve the time differences.

 

First off: if C is moving at 0.5c, No one would see anyone's else's clocks ticking a a rate of 50% slower or faster than their own. The correct rates are ~57.74% and  ~173.2%. This is the result of accounting for both Doppler shift and time dilation effects.  The combination of these gives what is known as "Relativistic Doppler effect"

What A sees:  C departs at 0.5c, and C's clock is seen as ticking at 57.74% the rate of his own.  It takes C 1 min to reach B. However, since B is 1/2 light min away, it will take A another 1/2 min before he sees this occur.

Thus he sees C's clock recede for 1.5min, while ticking at a rate of 57.74%, and accumulate 0.866 min.  Since the clock at B will be seen, due to propagation delay, to lag behind his by 30 sec during the whole of this interval,  He will see clock B start at -30 sec when C leaves and 1 min upon C arrival.

What B sees:  30 secs after C leaves A, he sees leave A.  Since C will have crossed half the distance by then, it will only take another 30 sec for C to arrive.  During that thirty seconds, he will see clock C tick at a rate 173.2% of his own, and accumulate 0.866 min. (1.732*0.5 = 0.866). His own clock will read 1 min upon arrival of C.

What C sees.  When he leaves A, he sees clock B lagging 30 sec behind clock A, just like A does at this moment  He will see clock B ticking at a rate of 173.2% his own.  By his clock, he will arrive at B 0.866 min later*  and see clock B tick off 0.866 * 1.732 = 1.5 min.  But since he started off seeing clock B reading -30 sec, he sees it as reading 1 min on arrival.

 

*  Why does C say it only takes him 0.866 min to cross 1/2 light min of distance at 0.5c?  Because for him, the distance between A and B is not 1/2 light min, but 0.433 light min.   This is due to length contraction. Since A and B are moving relative to him, he would measure the distance between them as length contracted.   This brings up something that occurs often when people start learning Relativity.  They try and isolate a single effect, such as time dilation, and examine a scenario using only it.  But you need to take all the effects into account to do a proper analysis of it.

[MPMin]

Thanks Janus for your detailed reply.

It’s just occurred to me that defining the synchronisation of the clocks it’s self is a conundrum to me. 
 

I can imagine that if all clocks were in the same proximity that they could all be synchronised together. Let’s assume we synchronise all clocks while they are together and then before the experiment starts, we move clock B and observer B 1 light minute away while leaving Clock A and Clock C begins with their respective observers. Would clock B still be synchronised with clocks A and C after it reaches its position 1 light minute away and does it matter how long it took for clock B to reach its position as to how much the synchronisation changed if at all.

Let’s assume that clock and observer C are due to depart clock A at 10:00:00am in A’s frame of reference. I’m going to take another stab in the dark and assume that if the clocks were all synchronised (without taking into account the effect of moving clock B into position had on the original synchronisation) as observer B is 1 light minute away from clock A, when observer B sees clock A strike 10:00:00, clock B will be seen as displaying 10:01:00 by observer B in his own frame of reference. 
 

If clock A and C show the same time when Clock C departs at 10:00:00am in A and C’s frame of reference, 

When C is half a light minute away, does that mean that from A’s frame of reference that C is half way through its trip to B? Does it mean that from C’s frame of reference, it has already arrived at B? If getting any of this, that should mean that when C arrives at B that, observers C and B will see clock C as being 10:00:30 while Clock B will be 10:01:30 and observers C and B will see that clock A is 10:00:30 from C and B’s frame of reference. Observer A from A’s frame of reference will see that clock C as showing 10:00:15 while seeing clock B to be showing 10:00:30? 

Suspect I’ve got it all wrong though. 
 

 

 

 

 

 

[Markus Hanke]

Janus beat me to it!

Maybe just as well, since I’m actually somewhat confused now - my understanding of the original scenario was that the distance A-B was to be 1 light-minute as seen from frame C, and that clocks were synchronised (taking into account light travel times) to all be seen as showing zero at the instant when C passes A, again as seen from C. My previous answers were based on that. But reading the last few posts, I may not have understood correctly what the OP meant, as the scenario seems to be different now...?

Edited June 1, 2022 by Markus Hanke

[MPMin]

1 hour ago, Markus Hanke said:

Janus beat me to it!

Maybe just as well, since I’m actually somewhat confused now - my understanding of the original scenario was that the distance A-B was to be 1 light-minute as seen from frame C, and that clocks were synchronised (taking into account light travel times) to all be seen as showing zero at the instant when C passes A, again as seen from C. My previous answers were based on that. But reading the last few posts, I may not have understood correctly what the OP meant, as the scenario seems to be different now...?

To be fair to everyone, I’m very confused by many aspects of this thought experiment.

The original thought experiment by Einstein, or should i say the most common version of his thought experiment that I’ve seen only discusses Einstein on a train moving away from a clock towers and what Einstein sees the clock do as he recedes away from the clock. That left too many unknowns for me and i wanted to know to what if there was another clock tower in the opposite direction, left me wondering what Einstein would see the approaching clock do as well as, and, what if Einstein also had a clock with him the whole time what would Einstein’s clock be seen to be doing.

To help my self understand the unknowns Ive elaborated on the thought experiment in attempt to understand all the other possible aspects of the thought experiment. As I’m getting confused with all the relativistic points of view I thought it might help to use times as shown on a clock with reference to some kind of synchronisation to help me understand it more practical terms.

One aspect of the thought experiment is the synchronisation of all the clocks. The thought experiment is intended to be what each observer should see from their own frame of reference when clock C moves from clock A to clock B, where clock B is positioned one light minute away from clock A. With regards to synchronising all the clocks, the only way I can think to synchronise the clocks was to have them all together and synchronising them all from the same frame of reference before moving clock B 1 light minute away before the experiment starts. Perhaps another way to synchronise all the clocks is to firstly position clock B 1 light minute away then observer B sets their own clock being clock B in their own frame of reference to 10:01:00 when observer B sees that clock A strikes 10:00:00 from observers B frame of reference. I suspect that the synchronisation of the clocks would need to be resolved before considering the times shown on the clocks.

To keep things less confusing I would prefer to consider the clocks and their respective observers to be the one and the same entity, so that clock A is also observer A and so on for all clocks, so as to remove any confusion about where the observers are in relation to their own clocks and therefore would consider that what each clock/observer always sees what they see from their own frame of reference. 

I’m going to assume that clock B has been positioned 1 light minute away from clock A. Clock C is positioned at clock A. Clock B synchronises its own clock with clock A by setting his own to 10:01:00 when B sees clock A to display 10:00:00, clock C is synchronised with clock A as clock A and C are right next to each other. From the A’s frame of reference, clock C leaves clock A at 10:00:00 according to clock A, and clock C instantaneously starts travelling towards clock B at 50% speed of light.

If my description makes sense thus far, from B’s perspective, B sees clock B to display 10:01:00 and sees clock A and C both to display 10:00:00 and sees clock C just leave clock A.

From A’s perspective, A sees clock A and C to display 10:00:00 while seeing clock B to display 10:01:00. At that moment clock C leaves at 50% speed of light towards B

At the half way point from A’s frame of reference, A sees clock A to display 10:00:30 while seeing clock C to display 10:00:15 and sees clock B to display 10:01:30

At the half way point from C’s frame of reference, C sees clock A to display 10:00:15 and clock C to display 10:00:07.5 and sees clock B to display 10:00:45

At the half way point from B’s frame of reference, B sees clock A and C to display 10:00:00 and C just leave A and sees his own clock B to display 10:01:00

When C arrives at B from A’s frame of reference, A sees clock A to display 10:01:00, A sees clock C to display 10:00:30 and clock B to display 10:02:00

When C arrives at B from C’s frame of reference, C sees clock A to display 10:00:30, C sees his own clock to display 10:00:15 and clock B to display 10:01:00

When C arrives at B from B’s frame of reference, B sees clock A to display 10:01:00, B sees clock C to display 10:00:15 and his own clock C to display 10:00:30

This all seems wrong but some help would be greatly appreciated please. 

 

 

from here I just can’t resolve what all the other observers should see. I know Janus provided a detailed description but I’m personally unable to equate his description into the practical times each clock would display at each stage being beginning, middle and end from each observers frame of reference. 

[Janus]

11 hours ago, MPMin said:

Thanks Janus for your detailed reply.

It’s just occurred to me that defining the synchronisation of the clocks it’s self is a conundrum to me. 
 

I can imagine that if all clocks were in the same proximity that they could all be synchronised together. Let’s assume we synchronise all clocks while they are together and then before the experiment starts, we move clock B and observer B 1 light minute away while leaving Clock A and Clock C begins with their respective observers. Would clock B still be synchronised with clocks A and C after it reaches its position 1 light minute away and does it matter how long it took for clock B to reach its position as to how much the synchronisation changed if at all.

Let’s assume that clock and observer C are due to depart clock A at 10:00:00am in A’s frame of reference. I’m going to take another stab in the dark and assume that if the clocks were all synchronised (without taking into account the effect of moving clock B into position had on the original synchronisation) as observer B is 1 light minute away from clock A, when observer B sees clock A strike 10:00:00, clock B will be seen as displaying 10:01:00 by observer B in his own frame of reference. 
 

If clock A and C show the same time when Clock C departs at 10:00:00am in A and C’s frame of reference, 

When C is half a light minute away, does that mean that from A’s frame of reference that C is half way through its trip to B? Does it mean that from C’s frame of reference, it has already arrived at B? If getting any of this, that should mean that when C arrives at B that, observers C and B will see clock C as being 10:00:30 while Clock B will be 10:01:30 and observers C and B will see that clock A is 10:00:30 from C and B’s frame of reference. Observer A from A’s frame of reference will see that clock C as showing 10:00:15 while seeing clock B to be showing 10:00:30? 

Suspect I’ve got it all wrong though. 
 

 

 

 

 

 

The synchronization of clocks in SR can be the most difficult concept to grasp when you first start learning the subject. This is mainly because most people come into the subject with preconceived notions about time that they have to lose in order to go forward.

That being said, there are a couple of points we need to go over.

Frames of reference:  You are treating them as being the same as "point of view", such as "A's frame of reference, and "B's frame of reference". This is not what frame of reference means in Relativity.  It is just a choice of coordinate systems. Any object simultaneously exists within an infinite number of frames of reference.  Which one we use when dealing with the object depends on what is most convenient for our purposes at the time. 

 In SR, we deal with inertial frames of reference. Put simply, these are reference frames with constant motions with respect to each other. Again, any object exists simultaneously within an infinite number of these, but it is, in many cases, easiest to work with the one that the object itself is at rest with respect to.  This can be referred to as the "rest frame" of the Object.  And while it will sometimes be called, for example, the "reference frame of A", this is really just shorthand for "the inertial reference frame that A is at rest with respect to".

So let's look at A, B, and C.  At the start of this exercise, they are all at rest with respect with each other, and thus all at rest with respect to the same reference frame. In other words, at this point, there is only one reference frame we need to consider.

Once C begins its trip, A and B remain at rest with respect to the same frame as before, but C is in motion with respect to this frame. C is however at rest with respect to another inertial reference frame, one moving at 0.5c relative to the frame A and B are at rest with respect to.  It has "changed rest frames"  So now we have 2 inertial frames from which we can consider things.

Now, when it comes to synchronizing clocks:  As long as the clocks are at rest with respect to the same reference frame, there is no issue in synchronizing them, no matter how far apart they are.*  If you "see" a clock 1 light min away as lagging behind yours by 1 min.  You know that that the light carrying this info to you left 1 min ago, when your clock read that time.  In other words, 1 min ago both clocks read the same, and as long as nothing has changed for either clock since, they read the same now. This is true for anybody at rest with respect to this frame, no matter where they are located.

However, things change if you consider the clocks from a reference frame in motion with respect to these clocks.

For example, consider the rest frame of C, the instant it starts moving from A to B, It sees clock A read 0, and since there is zero distance between it and A, it knows that Clock A reads 0 at  that moment. It sees clock B reading -30 sec, It does not however conclude that Clock B reads 0 at that moment, but would conclude that it read +15 sec.  In other words, C would conclude that clocks A and B are out of sync by 15 sec, and this 15 sec difference remains throughout the trip.  The upshot, is that while both A and B conclude that their clocks read 15 sec "at the same time", C concludes that A reading 15 sec and B reading 15 sec, occurs at different moments in time.  A and B's "at the same time" is not the same as C's.

So when you ask "When C is half a light minute away, does that mean that from A’s frame of reference that C is half way through its trip to B? Does it mean that from C’s frame of reference, it has already arrived at B?"

You have to be really specific as to what you mean.  When A determines that C is halfway to B, this is when clock A reads 30 sec.  According to A, C's clock will read 0.433 min.  C will say that it is halfway to B when it's clock reads 0.443.  There is no disagreement here.

However, according to C, A does not read 30 sec until C's clock reads 0.577 min. (after he passes the halfway point, but before he arrives at B)

Again we are dealing with the fact that A and B do not always agree as to what occurs "at the same time" . For A, his clock reading 0.5 min, and B's reading 0.433 min occurs at the same time, and for C, A's clock reading 0.5 min and his reading 0.577 min occurs at the same time.

If you mean when A "sees" C arrive at the midpoint, this occurs when A's clock reads 45 sec, at which time, according to A, C is 3/4 of the way to B.

It is really important not to mix up what one "sees", with what one would determine is happening at that moment.  Most times in dealing with Relativity, we don't worry about what is "seen" and instead focus on what is determined.  Put another way, we "cut to the chase" of Relativistic effects without worrying about light propagation delay.

An example of the difference:

For this example, we will assume that C has been already traveling at 0.5c relative to A and B prior to passing A(This avoids complications caused by instantaneous velociyy changes.)

As C passes A, they both "see" Clock B reading 30 sec earlier than Clock A.

Observer A, knowing that A and B have been at rest with respect to each other and 30 light sec apart, concludes that this light left 30 sec ago, and That clock B has ticked off 30 sec since then, and, at that moment reads the same as his own.

But what does C conclude?   We already mentioned that due to length contraction, he would say that B is, at that moment 0.433 light min from A( and himself).  But, the light he is now seeing from B left B before C reached A, and when the distance between himself and B was greater.  Since one of the precepts of SR is that every inertial frame measures light as traveling at c with respect itself, C must assume that the light he sees coming from B as he passes A, left B when they were 0.866 light min(~52 light sec) apart, and it took 52 sec for the light to reach him.   52 sec have ticked off since the light left B and C sees it. This is more than the 30 sec difference that he sees between A and B's clock.  So, as he passes A, he must conclude that the Clock at B already reads sometime later than the clock at A*

So here we have two observers, at the same place, seeing the same thing, but making different(But equally valid) conclusions as to the state of a distant clock at that time.

* Due to time dilation, C would also conclude that Clock B ticks 0.866 as fast as his own, so in his 0.886 min, clock B ticks off 0.75 min or 45 sec.  45-30 = 15, which is the 15 sec difference I mentioned earlier.