How does this proof demonstrate the concept of time dilation?

[Stephanus]

[..]Now according to A, he sees the photon going to the top and coming back to the bottom(with velocity c) of compartment after hitting the top. He looks at his clock when it has returned and the digital clock clearly shows 2 sec.
Now according to B, he sees the same photon going towards the top(with velocity c) but not perpendicular to the bottom of compartment and rather at a certain angle(we can calculate this angle easily),clearly we can see that to reach the top the photon has to travel a bigger distance according to B and lesser distance according to A.
Now comes my question sir,
B also has a digital clock which is identical to that of A's, so when A's clock shows 1 sec (the instant when according to A the photon has hit the top) B's clock must also read 1 sec and it's just that according to him(B) photon will take somewhat more time to reach the top as the distance it has to travel is more as compared to A's.

No!. B clock's does not read 1!
B's clock will read...

(w.r.t whom the light beam clock moves at v=c/2)

$t=\sqrt{1^2 + (\frac{1}{2})^2} = \sqrt{1.25} = 1.118$
Because...

B will see the light travels at 1.118 * 300000 = 335,410.1966 km. So, for B, the light won't take 1 second to reach the top, but 1.118 sec.
For A? Of course 1 second.

Now according to B, he sees the same photon going towards the top(with velocity c) but not perpendicular to the bottom of compartment and rather at a certain angle(we can calculate this angle easily),clearly we can see that to reach the top the photon has to travel a bigger distance according to B and lesser distance according to A.

Now you should stop right here!

Now comes my question sir,

STOP
Longer distance according to B, shorter distance according to A, and if their clock show the same, than for B the light travels more than c.
Remember:

The two postulates of special relativity are covariance (inertial frames have the same physics) and universality of c regardless of relative motion.

So the clock couldn't be the same. Only the velocity is invariant!

 

[PeterDonis]

when A's clock shows 1 sec (the instant when according to A the photon has hit the top) B's clock must also read 1 sec

Ah, I see; what you're missing isn't time dilation, it's relativity of simultaneity. B's clock is identical in construction to A's, but it is not at rest relative to A's, so "the instant when according to A the photon has hit the top" only has meaning for A, not for B. In other words, the word "when" here means "when, according to A's rest frame", which is only relevant when talking about the tick rate of A's clock, not B's clock. It does not mean "when" in any absolute sense, because there's no such thing in relativity. Your argument assumes that there is, so your argument is based on a false premise.

what I am saying here is that both the clock's show same time at every instant(i'm saying on the basis that both are identical) but it's just that in B's frame of reference somewhat more time will be taken due to the increment in the distance the photon has to travel to reach to the top.

This can't be right, because it amounts to saying that B's clock does not read off time in B's frame of reference. The speed of the photon is the same in all reference frames, so the photon has to take more time to travel a longer distance. But that means B's clock must read that longer time; otherwise B's clock is not a valid clock, because a valid clock always reads off time in its rest frame. B's clock must read off time in B's rest frame; it can't read off time in A's rest frame.

Your error, again, is in the statement "both clocks show same time at every instant". "Every instant" has no absolute meaning; it is different in each frame of reference, so "every instant" according to A is different from "every instant" according to B.

 

[universal_101]

Now comes my question sir,
B also has a digital clock which is identical to that of A's, so when A's clock shows 1 sec (the instant when according to A the photon has hit the top) B's clock must also read 1 sec and it's just that according to him(B) photon will take somewhat more time to reach the top as the distance it has to travel is more as compared to A's.So what I am saying here is that both the clock's show same time at every instant(i'm saying on the basis that both are identical) but it's just that in B's frame of reference somewhat more time will be taken due to the increment in the distance the photon has to travel to reach to the top.
So by my argument when A's clock shows 2 sec. B's clock too reads 2 sec. and it's just that the photon for A by this time has completed his 1 round journey whereas for B it's yet to reach the bottom and when both clocks show 3.40 seconds(2*gamma) then B sees the photon hitting the bottom .
So my conclusion of the whole eg. is :
The increment in length in B's frame has just caused the total time interval to differ although the rate of digital clock ticking is same.

I really like your view on this, for it is very natural to think this way, OK let me try to show the differences between the two views(of yours and others).

Your view of the scenario is based on the absolute 'time' and variable 'speed of light', that's how you can easily explain the observation by 'A' and 'B', whereas similarly, others have the view of variable 'time' and absolute 'speed of light'.

You can easily realize that the difference is because of different interpretation, your version of explanation says the time rate is same for the two observers, it's just that light has to travel a longer distance in frame of B(which implicitly means different light speed for different observers). While the other view is about the light speed being same for the two observers, and implying that the time rate itself changes(that is instead of rate of time being same, the instant of light beam striking the ceiling or the floor is same in both frames, which means the light journey in two frames represent the ticking clock, implying a different time rate).

 

[Stephanus]

Your view of the scenario is based on the absolute 'time' and variable 'speed of light', that's how you can easily explain the observation by 'A' and 'B', whereas similarly, others have the view of variable 'time' and absolute 'speed of light'.

Now this one I like

 

[georgir]

Now comes my question sir,
B also has a digital clock which is identical to that of A's, so when A's clock shows 1 sec (the instant when according to A the photon has hit the top) B's clock must also read 1 sec...

And here you already messed up your reasoning. Digital or light or sand or biological, all clocks are affected by time dilation. They may be identical, but they are not moving identically so they will not show 1 sec at the same time.

 

[Raman Choudhary]

Okay universal_101 I think you have very well quote my mistake in very simple words(really thank u for appreciating the view though it clearly stands against the very theory i am struggling to learn)but could you please explain to me that by simply calculating the time observed by different observors how can we show that for one observor the clock is ticking slow . I mean we haven't showed what there clocks read when suddenly asked to reveal the readings. We have just calculated the time intervals taken, which to me doesn't sound in anyway related to the different ticking rates.I hope you will again through some light on this using your simple qoutes.

 

[PeterDonis]

We have just calculated the time intervals taken which to me doesn't sound in anyway related to the different ticking rates.

The time intervals are the ticking rates. That's what they mean, physically. Any calculation in a scientific theory has to be based on some correspondence between the numbers in the calculation and actual physical observations. In the case of relativity, the calculated time intervals correspond to the actual physically observed readings on clocks. If they didn't, the theory would be useless.

 

[Raman Choudhary]

I really like your view on this, for it is very natural to think this way, OK let me try to show the differences between the two views(of yours and others).

Your view of the scenario is based on the absolute 'time' and variable 'speed of light', that's how you can easily explain the observation by 'A' and 'B', whereas similarly, others have the view of variable 'time' and absolute 'speed of light'.

You can easily realize that the difference is because of different interpretation, your version of explanation says the time rate is same for the two observers, it's just that light has to travel a longer distance in frame of B(which implicitly means different light speed for different observers). While the other view is about the light speed being same for the two observers, and implying that the time rate itself changes(that is instead of rate of time being same, the instant of light beam striking the ceiling or the floor is same in both frames, which means the light journey in two frames represent the ticking clock, implying a different time rate).

Okay universal_101 I think you have very well quote my mistake in very simple words(really thank u for appreciating the view though it clearly stands against the very theory i am struggling to learn)but could you please explain to me that by simply calculating the time observed by different observors how can we show that for one observor the clock is ticking slow . I mean we haven't showed what there clocks read when suddenly asked to reveal the readings. We have just calculated the time intervals taken, which to me doesn't sound in anyway related to the different ticking rates.I hope you will again throw some light on this using your simple qoutes.

 

[PeterDonis]

I mean we haven't showed what there clocks read when suddenly asked to reveal the readings.

Yes we have. Please read my post #37, and stop and think very carefully about what you are saying. When we calculate the time intervals taken, we are calculating what the clocks will read. That is how a scientific theory works; you aren't just calculating abstract numbers, you are calculating numbers that tell you the results of direct observations. Otherwise, as I said in my last post, the theory would be useless.

 

[universal_101]

Okay universal_101 I think you have very well quote my mistake in very simple words(really thank u for appreciating the view though it clearly stands against the very theory i am struggling to learn)but could you please explain to me that by simply calculating the time observed by different observers how can we show that for one observer the clock is ticking slow .

Happy to help!

First of all you should ponder upon the idea that this light clock doesn't result in same calculated time difference, for different orientation of the light clock, that is, it is not a general case(try doing the same calculation with light clock being horizontal). So it is not a very good analogy by any means.

But anyways, let's consider this special case(where light clock is vertical), and your question about how can we show that for one observer the clock is ticking slow ?
This question is the central conflict since the beginning of these kind of theories, between principle of relativity(which implies similarity/reciprocity between the observers A and B) and time dilation in relativity, that is, anyone of the observer A or B can claim that it's the other one which is moving, which brings up your question.

There are several resolutions to this conflict in mainstream physics, like, the one who accelerates will be the one who's time will be dilated, or the one which takes longer 'space-time path' will be the one who's time will be dilated.

I mean we haven't showed what their clocks read when suddenly asked to reveal the readings. We have just calculated the time intervals taken, which to me doesn't sound in anyway related to the different ticking rates.I hope you will again throw some light on this using your simple quotes.

No we haven't, but what the other person's clock reads depend on who is observing whom. And the concept of suddenly, or at the same instant, alone, has no meaning in special relativity(you need more information to ask this question in special relativity), because the time rate itself is considered different for the two observers moving relative to each other. Therefore, according to special relativity, you should also specify from which reference frame you want to ask this question.

I hope you understand the situation, there's NO absolute time in special relativity, so the question what the two clocks of A and B reads at any instant is incomplete and requires more information, like, from which reference frame.

 

[PeterDonis]

No we haven't

If you mean we haven't calculated "what their clocks will read", this is incorrect; we have. See my previous post.

what the other person's clock reads depend on who is observing whom

Careful. You are coming close to saying that what a given clock reads at a given event is frame-dependent, which is not correct. I think what you mean here is that what the other person's clock reads "at the same time" as one person's clock (assuming the two clocks are spatially separated) depends on who is observing whom, i.e., which reference frame we are using. But this is because which pairs of spatially separated events happen "at the same time" is frame-dependent. Once you've picked a particular event, i.e., a particular point in spacetime, the reading of a clock located at that event is an invariant; it doesn't depend on which frame you are using.

 

[universal_101]

If you mean we haven't calculated "what their clocks will read", this is incorrect; we have. See my previous post.

I was trying to explain that there is NO meaning of suddenly without specifying a reference frame.

Careful. You are coming close to saying that what a given clock reads at a given event is frame-dependent, which is not correct. I think what you mean here is that what the other person's clock reads "at the same time" as one person's clock (assuming the two clocks are spatially separated) depends on who is observing whom, i.e., which reference frame we are using. But this is because which pairs of spatially separated events happen "at the same time" is frame-dependent. Once you've picked a particular event, i.e., a particular point in spacetime, the reading of a clock located at that event is an invariant; it doesn't depend on which frame you are using.

I never said anything about any spacetime event(I feel the concept of spacetime is far less unintuitive, compared to basic one-to-one treatment of things from the perspective of the different observers). So, yes, I mean what you think I mean.

 

[PeterDonis]

I never said anything about any spacetime event(I feel the concept of spacetime is far less unintuitive, compared to basic one-to-one treatment of things from the perspective of the different observers).

But the basic, one-to-one treatment of things from the perspective of different observers requires the concept of "spacetime event", because the events are what coordinates are attached to, and transforming from one observer's perspective to another means transforming the coordinates while keeping the events and their invariant relationships constant. It may not be as intuitive to visualize what you're doing in terms of geometry in a flat spacetime (though I think that's more a matter of training your intuition than anything else), but you can't avoid the concept of spacetime event.

 

[exmarine]

B and A see each other's clocks run slow

You guys are the experts, but ? Here is a paragraph from Einstein's 1905 paper, bottom of page 10:

From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by t*beta^2/2 (up to magnitudes of fourth and higher order), t being the time occupied by the journey from A to B.

So B sees A run slow. Please tell me how A can then see B run slow? (And don't the GPS clocks now verify Einstein's take on this?)

 

[cpsinkule]

You guys are the experts, but ? Here is a paragraph from Einstein's 1905 paper, bottom of page 10:

From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by t*beta^2/2 (up to magnitudes of fourth and higher order), t being the time occupied by the journey from A to B.

So B sees A run slow. Please tell me how A can then see B run slow? (And don't the GPS clocks now verify Einstein's take on this?)

Clock A is no longer an inertial observer if he is accelerated to velocity v and then decelerated to compare at B, the symmetry is broken when an observer accelerates, and thus both observers can tell who was actually moving.

 

[PeterDonis]

So B sees A run slow. Please tell me how A can then see B run slow?

Because of relativity of simultaneity. Work out the Lorentz transformations for A and B both inertial and moving relative to each other, and you'll see.

don't the GPS clocks now verify Einstein's take on this?

They verify Einstein's theories of SR and GR, but that doesn't mean what you think it means. B and A each see the other's clocks running slow if they are both inertial observers in relative motion in flat spacetime. The GPS clocks and Earthbound clocks are in curved spacetime, and things don't work the same in curved spacetime.

 

[Janus]

You guys are the experts, but ? Here is a paragraph from Einstein's 1905 paper, bottom of page 10:

From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by t*beta^2/2 (up to magnitudes of fourth and higher order), t being the time occupied by the journey from A to B.

So B sees A run slow. Please tell me how A can then see B run slow? (And don't the GPS clocks now verify Einstein's take on this?)

When A goes from being stationary with respect to B to moving with respect to B, A has to accelerate. During this period of acceleration, according to A, B will run fast. How fast depends on the magnitude of the acceleration and the distance to B. After the acceleration is over, B will run slow according to A, However, it will have enough time during the acceleration, that When A and B meet up, B will still be ahead of A. Now to be clear, this is what A would determine is happening to B, not what A visually sees happening to B. Because of the propagation delay of light and Doppler shift, they will see the following:

For simplicity, we will assume near instantaneous acceleration for A.

A starts off seeing B as being behind his clock by an amount determined by the distance between them. (if they are 1 light hour apart, A will see B read 1 hr earlier than A reads, because the light arriving at A left B 1 hr ago.) The clocks run at the same rate
After instantaneously accelerating to some speed with respect to B, B starts off as reading an hour behind B, but because of Relativistic Doppler shift will see B running faster than A up to the point where A and B meet and A come to rest with respect to B. The faster running rate will more than make up for the 1 hr lag B started with.

B, on the other hand sees this: Like A, it starts off seeing A running 1 hr behind. But unlike A, it does not see a Doppler Shift from A the instant A starts moving. B doesn't see a Doppler shift until 1 hr after A started moving. (A saw the Shift immediately because it is the one that accelerated, B on the other hand must wait an hr for the light from A to reach him before he sees A start to move and sees the Doppler Shift. ) This means that by the time B starts to see A run fast, A has already covered a good part of the distance between Its start point and B. ( If A was moving at 90% of c, it would only be 6 light min from B by the time B even saw that A started moving, meaning that B only sees A running fast for 6 2/3 min before A arrives.) The time that B sees A running fast is not enough to make up for the 1 hr lag.

So the upshot is that even though A and B each see each others clocks running fast by the same rate, A sees B running fast for a longer time than B sees A running fast, with the end result being that both agree that A will be behind B when they meet up.

GPS clocks are a different matter as they are moving in circular orbits and circular motion is accelerated motion. This puts the GPS clock continuously in an accelerated frame, in which you can't use the simple forms of the SR transformations.

 

[FactChecker]

cpsinkule said: ↑
B and A see each other's clocks run slow.

HOW?

Don't forget that when A looks at B's clocks, he is not looking at the same clock in the same position from one second to the next. He is looking at widely separated clocks that B says are synchronized. A looks at one B clock at one second and at the next second, he is looking at another B clock in the direction that A has traveled in that second. A is looking at B clocks that B swears he has synchronized. A has to take B's word that B's clocks are synchronized in B's physical world. A and B just disagree on what "simultaneous" and "synchronized" means. The reason for their disagreement is symmetric for both A and B. Each one thinks that the other one is wrong and that the clocks are wrong in a similar manor in his forward direction of motion. So each one thinks that the other one's clocks are synchronized wrong and "running slow" as he moves from the clock at one location to the "synchronized" clock at the next location.

 

[PeterDonis]

GPS clocks are a different matter as they are moving in circular orbits and circular motion is accelerated motion. This puts the GPS clock continuously in an accelerated frame, in which you can't use the simple forms of the SR transformations.

It's worth noting that here you are using "accelerated" in the sense of coordinate acceleration, not proper acceleration. The GPS satellites are in free-fall orbits. That's possible because spacetime around the Earth is curved, not flat; so it's not as simple as just not being able to use the simple forms of SR transformations. You can define an "Earth-centered inertial" (ECI) frame in which the GPS satellite orbits are indeed "accelerated" in the sense of coordinate acceleration; but this frame is not an SR inertial frame.

 

[exmarine]

Thanks for all the replies. But this one might be in error?

Don't forget that when A looks at B's clocks, he is not looking at the same clock in the same position from one second to the next. He is looking at widely separated clocks that B says are synchronized. A looks at one B clock at one second and at the next second, he is looking at another B clock in the direction that A has traveled in that second. A is looking at B clocks that B swears he has synchronized. A has to take B's word that B's clocks are synchronized in B's physical world. A and B just disagree on what "simultaneous" and "synchronized" means. The reason for their disagreement is symmetric for both A and B. Each one thinks that the other one is wrong and that the clocks are wrong in a similar manor in his forward direction of motion. So each one thinks that the other one's clocks are synchronized wrong and "running slow" as he moves from the clock at one location to the "synchronized" clock at the next location.

There once was a prominent SRT critic named Dingle (?) whose argument rested on exactly this, i.e., NOT watching a SINGLE clock in the other reference frame. At least that's what I remember. I finally convinced myself that he was wrong because of that. You can probably find his argument somewhere on the internet to check for yourself. And that's also why I mentioned the GPS clocks in connection with Einstein's paragraph. They show that the time-dilation is "real" for one of the clocks, and not just symmetric "appearances".

 

[FactChecker]

They show that the time-dilation is "real" for one of the clocks, and not just symmetric "appearances".

I didn't mean to imply that the time dilation was not real. Both are real and observable by the other coordinate system, but not within it's own coordinate system. That is why the speed of light is the same for all inertial coordinate systems. I was just trying to explain how both A and B could both think that the other's clock was going slower. It is not possible to compare the same clocks side-by-side over any length of time because the clocks move relative to each other. You can only compare separated clocks that have been synchronized in each coordinate system.

I talked about A observing different clocks in B as time progressed, but the argument could be made that A can watch the same clock in B but at different locations in his own coordinate system. The clocks at the different A locations are synchronized in A's coordinate system, and the result is the same.

 

[Markus Hanke]

So B sees A run slow. Please tell me how A can then see B run slow?

Imagine two people standing on a flat plane very far apart. Person A sees person B somewhere in the distance, and perceives him to be very small. Person B looks at person A in the distance, and also perceives him to be very small. The laws of optical perspective are perfectly symmetric in this scenario, and I'm sure no one would argue that there is any kind of paradox arising from this. It's just common sense - both of these observers are right, in their own frames of reference. There is no contradiction.

The same is true for inertial frames - to get from frame A to frame B, you perform a hyperbolic rotation about some angle in space-time ( i.e. a Lorentz transformation ). To get back from frame B to frame A, you perform another rotation about the same angle. And again, there is no mystery or paradox to this - it's a perfectly symmetric scenario, just like the simple thought experiment about the observer in the distance. In many ways, going from one frame to another is really just a change in perspective - choosing a different point of view in space-time, if you will - and one that is identical to its inverse operation.