Length contraction applet - have they got it wrong?

[FranzDiCoccio]

Hi,

I was browsing the internet for interactive simulations illustrating special relativity concepts. It seems to me that those in this http://kcvs.ca/concrete/visualizations/special-relativity are mostly nice and clear, although not very "chrome-friendly" (I have to use firefox to play their swf files).

I'm having some trouble with their visualization of http://www.kcvs.ca/site/projects/physics_files/specialRelativity/synchClocks/synchClocks.swf, though.
There are several things that seem really strange to me.

1. the behavior of the "photon" in the moving light clock seems wrong. As far as I understand the photon in the rocket should move always as fast as the ones in the lab frame. So I'd expect that it takes a longer time to go from the rear of the rocket to its front, and a shorter time in the opposite direction. Instead it takes the same time for both paths.
2. In order to do the above, the photon slows down when traveling towards the rear end of the rocket, like there was a Galilean composition of its velocity and that of the rocket. That feels really strange.
3. When the factor $\sqrt{1-v^2/c^2}$ is punched in the clock length field, the traveling clock is synchronized with the stationary ones. But then I do not understand the meaning of the rescaled length. I think that, in order to be synchronized with the stationary clocks, the traveling clock should be seen as even shorter. That is, the factor should be $1-v^2/c^2$.

I say all this because in the "standard" visualization one looks at a "vertical" traveling light clock, whose mirrors are parallel to the rocket velocity, as those in the lab frame. The traveling clock is not synchronized with those in the lab, but slower (because, from the lab point of view, light takes more time to get to the mirror and back).
Now I can think of putting a second clock on the rocket that is identical to the first, but orthogonal to it. I believe that the departure and arrival of the photons in both clocks can be made simultaneous (the two "photon guns" can be put very close to one another). So both clock tick simultaneously (in the rocket frame _and_ in the lab frame).
This means that the "horizontal" clock should be seen as shorter than the "vertical" clock, by a factor $\sqrt{1-v^2/c^2}$.
Now if the second clock has to tick simultaneously with those in the lab, it should be _really_ shorter (as opposed to just contracted).

Perhaps the authors of the simulation are focusing on the _real_ length of the clock, and not on its length as it appears from the lab. That is, they are choosing not to represent length contraction.
That would probably work, although I find it a bit confusing.

Anyway, I'm pretty sure that the photon in the traveling clock does not hit the mirror at the same time as the stationary clocks, and does not slow down, as depicted in the app. Since the rocket is traveling, it takes more time to reach the front of the rocket. This time is compensated by the shorter time on the way back. That is, the overall journey of the photon should be synchronized, not each leg.

Any insight on this is really appreciated.

Thanks a lot
Franz

 

[stevendaryl]

I went to that page, and I couldn't make any sense of it.

 

[FranzDiCoccio]

I went to that page, and I couldn't make any sense of it.

Hi stevendaryl,

thanks for your reply. So do you agree with me that there is something strange in their simulation?

It's a shame, because their apps look nice. I would have made different choices in other apps too, but the one about synchronization is really confusing in my opinion.

 

[stevendaryl]

Hi stevendaryl,

thanks for your reply. So do you agree with me that there is something strange in their simulation?

It's a shame, because their apps look nice. I would have made different choices in other apps too, but the one about synchronization is really confusing in my opinion.

Maybe it was my browser not working correctly, but I couldn't get the length adjustment to do anything at all.

 

[FranzDiCoccio]

I was able to make it work in firefox. Chrome and Safari just download the swf file.

In firefox you should be able to change the clock length. You can do that only when the clock is not moving (hit rewind). You might need to delete the number that's in the field by default (this is not very clear, because they do not show a blinking cursor when you're in the field).

Anyway, even if you cannot reduce the clock's length, you should be able to see what I mean, if you're able to play the simulation.
Irrespective of the speed, the "photon" takes the same time going forward and backward, and it slows down on the second leg.
This seems extermely weird.

 

[itfitmewelltoo]

Hi,

I was browsing the internet for interactive simulations illustrating special relativity concepts. It seems to me that those in this http://kcvs.ca/concrete/visualizations/special-relativity are mostly nice and clear, although not very "chrome-friendly" (I have to use firefox to play their swf files).

I'm having some trouble with their visualization of http://www.kcvs.ca/site/projects/physics_files/specialRelativity/synchClocks/synchClocks.swf, though.
There are several things that seem really strange to me.

1. the behavior of the "photon" in the moving light clock seems wrong. As far as I understand the photon in the rocket should move always as fast as the ones in the lab frame. So I'd expect that it takes a longer time to go from the rear of the rocket to its front, and a shorter time in the opposite direction. Instead it takes the same time for both paths.
2. In order to do the above, the photon slows down when traveling towards the rear end of the rocket, like there was a Galilean composition of its velocity and that of the rocket. That feels really strange.
3. When the factor $\sqrt{1-v^2/c^2}$ is punched in the clock length field, the traveling clock is synchronized with the stationary ones. But then I do not understand the meaning of the rescaled length. I think that, in order to be synchronized with the stationary clocks, the traveling clock should be seen as even shorter. That is, the factor should be $1-v^2/c^2$.

I say all this because in the "standard" visualization one looks at a "vertical" traveling light clock, whose mirrors are parallel to the rocket velocity, as those in the lab frame. The traveling clock is not synchronized with those in the lab, but slower (because, from the lab point of view, light takes more time to get to the mirror and back).
Now I can think of putting a second clock on the rocket that is identical to the first, but orthogonal to it. I believe that the departure and arrival of the photons in both clocks can be made simultaneous (the two "photon guns" can be put very close to one another). So both clock tick simultaneously (in the rocket frame _and_ in the lab frame).
This means that the "horizontal" clock should be seen as shorter than the "vertical" clock, by a factor $\sqrt{1-v^2/c^2}$.
Now if the second clock has to tick simultaneously with those in the lab, it should be _really_ shorter (as opposed to just contracted).

Perhaps the authors of the simulation are focusing on the _real_ length of the clock, and not on its length as it appears from the lab. That is, they are choosing not to represent length contraction.
That would probably work, although I find it a bit confusing.

Anyway, I'm pretty sure that the photon in the traveling clock does not hit the mirror at the same time as the stationary clocks, and does not slow down, as depicted in the app. Since the rocket is traveling, it takes more time to reach the front of the rocket. This time is compensated by the shorter time on the way back. That is, the overall journey of the photon should be synchronized, not each leg.

Any insight on this is really appreciated.

Thanks a lot
Franz

I have the applet open in chrome and it is working fine.

Going to your point (2), yes, the simulation is, unfortunately, limited in it's ability to depict what is going on.

You are correct. In our frame, and that of the stationary vertical clocks, the light pulse of the horizontal clock must be exactly the speed of light in both the forward and reverse direction as we view it. And it isn't in the applet. As I have set it, with the speed at .74 and clock length of .67, the reverse pulse stops and waits for the rear mirror to catch up to it.

The reason for this is not entirely clear to me at this time. It may simply be that the underlying math driving the simulation is not precise but rather a more artistic depiction.

It is true to say that the relative speed between the mirrors and light pulses are classical in addition. I can refer to Einstein's paper, part I, section 3. The derivation of his first differential equation begins with the quantities of c+v and c-v. Specifically, the ratios of x'/(c-v) and x'/(c+v) are distance divided by velocity which is, of course, the time of the pulses going back and forward, in the stationary frame of reference. The depiction of the moving clock as horizontal is consistent with Einstein's derivation.

One possible issue that the applet may suffer from has is that is that it is down-scaled from near light speed to something we can actually watch. If you can imagine, at 75% of light speed, the clock and pulse will have traveled quite a long way before the forward mirror is struck and then on the reverse pulse, the rear mirror will be hit almost instantaneously. Consider the ratios of x'/(c-v) and x'/(c+v). Plug in some realistic values for x' and v=.75c. The difference between forward and reverse are clear.

Yep, so there it is.

 

[Pencilvester]

I'm having some trouble with their visualization of http://www.kcvs.ca/site/projects/physics_files/specialRelativity/synchClocks/synchClocks.swf, though.

I think this is simply an illustration of the fact that if you have multiple synchronized light clocks at rest in frame K with lengths of 1 as measured in frame K, and a rocket moving with speed $v$ in frame K, and a light clock aboard the rocket with length $\gamma^{-1}$ as measured in the rocket's FOR, and we call the event where the rocket passes one of the clocks at rest in K as $t = t' = 0$, then the subsequent number of ticks on the light clock aboard the rocket will equal the number of subsequent ticks on the light clock at rest in K that the rocket happens to be next to. I don't think the orientation of the clocks makes any significant difference to this fact. I do question the necessity of having this concept depicted visually, though.

 

[Pencilvester]

I think this is simply an illustration of the fact that...

Maybe put more simply: a tick on a clock moving with speed $v$ lasts $\gamma$ ticks on an identical clock at rest (this is basic time dilation). If you design the moving clock to tick more quickly than it's (previously) identical counterpart so that when they're at rest relative to one another, a tick lasts $\gamma^{-1}$ ticks on the "good" clock, then (when moving at speed $v$) the moving clock will not appear to show any time dilation. I'm being a little sloppy with my phrasing, but I think in the context of this thread there shouldn't be confusion.

 

[FranzDiCoccio]

Hi,
thanks to both of you for your insight. I had no time to think carefully about them, because I'm off to work now. I'll possibly ask some questions later.

Anyway, I think that the applet would look ok if the traveling clock was "vertical". In that case it would tick simultaneously with the static ones (as seen from the lab) if its real length was shorter by $\gamma^{-1}$. Both legs of the photon journey would take the same time, as in the lab.

If the same traveling clock was "horizontal", as in the applet, it would be also synchronized but

1. it would appear even shorter, due to length contraction
2. the two legs of the photon journey would not take the same time, and hence the syncronization would apply only to the "round trip" from the source and back to it. The photon in the traveling clock would not hit the traveling mirror at the same time as in the static clocks. This is possible because of the relativity of simultaneity.

I agree with Pencilvester that there is no real reason for turning the clock horizontal. Even worse, that choice would involve the two effects I mention.
I am not really sure what frame of reference the authors have in mind for the traveling clock. In my opinion everything should be "seen" from the lab frame, assuming that's possible. But that's what one posits in the first place.

 

[itfitmewelltoo]

Every time I look at a different set up of the problem, there is always some new little thing that stands out that I hadn't seen before. And I'm left again with another nagging little issue. Really? So the backward going pulse, from the rest frame, ticks faster then the forward going pulse? How does that work?

Einstein's setup defines synchronized with this setup. If I'm reading it right, this is exactly his point. Observers moving with the moving frame would find that clocks at each mirror are not synchronized. (He doesn't define the mirror and bouncy pulse as the clock. His clocks are separate.) They are synchronized with respect to the stationary system, at the point where they happen to be in the stationary system. That is what the applet shows. Those two vertical clocks are synchronized. The horizontal "clock" in the applet is synced to the vertical clocks. The pulse hits each mirror in sync. Something has to give in the example and that is that observers at each mirror would find that their clocks aren't in sync.

I'm pretty sure that is right. I'll check back. There is something that bothers me in my mind.

 

[itfitmewelltoo]

Hi,

Anyway, I think that the applet would look ok if the traveling clock was "vertical".

Vertical clocks is how I solved for the Lorenz transformation. It has a visual advantage in that the moving pulse follows a sawtooth path. The geometry is that of right triangles with a^2 = b^2 + c^2. With the equating the proper variables and some manipulation, the Lorenz form of dilation and contraction can be had.

Visually, it can be better seen that internal to the moving frame, time has to be experienced as slower. After all, in the stationary frame, the light has a longer path. In the moving frame, that path is shorter. Time is always [light path]/c for the clock internal to the frame in which it is stationary. But for the moving clock from the stationary frame, that light path by an amount determined by it's velocity. Something has to give and at least one of these is time being experienced as slower internal to the moving frame.

You'll have to forgive me as the mathematical derivation has been long forgotten. The moving clock pulse, from the rest frame, doesn't have to backtrack. It is always "chasing" the next mirror.

 

[Pencilvester]

Anyway, I think that the applet would look ok if the traveling clock was "vertical".

The simulation does separate the rocket FOR with that horizontal line through the middle of the screen. So the fact that it shows the rocket’s clock’s photon taking the same amount of time on the emission and reflection legs of the journey is consistent with what the rocket observer sees, as is the length of the clock, regardless of how they orient it. I think the confusion is arising because of the fact that they show that clock moving, whereas the rocket observer typically wouldn’t choose a frame for himself in which he is moving in these types of setups. I think they do this simply to emphasize that you need at least two clocks (spacially separated) in the lab frame to measure the rate at which the rocket clock is ticking, and you need the rocket to travel from one of those clocks to the other. But again, this does seem like an odd concept to want to represent visually, especially in the way they do.

 

[itfitmewelltoo]

Maybe it was my browser not working correctly, but I couldn't get the length adjustment to do anything at all.

You have to click "Enter". I found that first clicking rewind, followed by entering a new clock length in the box, clicking enter, and lastly starting it worked out.

 

[itfitmewelltoo]

I'm highlighting a particular point. It's buried in a longer comment and seems worth repeating. Re the comment "there is no real reason for turning the clock horizontal.", The setup in the applet is consistent with the setup that Einstein uses in his paper. He defines time with a horizontal light clock of this nature. Two standard clocks at each end provide a definition of synchronous. The two clock are synchronous if the forward trip time is equal to the reverse trip time according to the standard clock readings when the pulse reaches that mirror, Tb-Ta = Ta'-Tb. I found this clearer when considering the results of the two not being synchronous. If the clock at the front runs slower than that at the back, then the forward and return trips are of different lengths of time by the readings given by these clocks. Imagine that the applet stationary vertical clocks are not in sync.

This becomes important in the derivation of LC and TD. The next setup is such that the stationary frame has a continuum of clocks on the x axis. The two clocks in the moving FOR are then taken to be constantly in sync with the stationary clocks to the specific clock at the place where the moving clock is at that moment.

The applet presents only the two stationary clock necessary for this. (There is also a reason the stationary clocks are shown as vertical but not necessarily.)

So, as we've noted, the length of time that the pulse in the moving clock takes is much different on the forward and return trips. The conclusion then is that for the moving observer, the moving clocks at each end of the light clock are not synchronized.

I'm pretty sure this is the setup that allows Einstein to equate the two FORs and derive LC and TD.

 

[stevendaryl]

This might be the point of the app:

Three facts about "light clocks" that together sort of imply time dilation and length contraction

A light clock is conceptually a pair of parallel mirrors with a pulse of light bouncing between them. Let $T$ be the period, or the time it takes for the light to bounce from one mirror to the other and back to the first. Using the fact that light always travels at speed c, regardless of the speed of the source, we find for a light clock of length $L$:

1. For a clock at rest, $T = \frac{2L}{c}$
2. For a clock moving at speed $v$ parallel to the line connecting the mirrors, $T = \frac{1}{1-\frac{v^2}{c^2}} \frac{2L}{c}$
3. For a clock moving at speed $v$ perpendicularly to the line connecting the mirrors, $T = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \frac{2L}{c}$

From 2 & 3, if you have a pair of moving light clocks, one oriented one way and one oriented the other way, the only way for them to stay synchronized (to have the same period) is if the one that is oriented parallel to the motion has length $\sqrt{1-\frac{v^2}{c^2}}$ times as big as the one oriented perpendicular to the motion. To complete the argument about time dilation, you need to assume that the light clock oriented perpendicular to the motion experiences no length contraction.

 

[Janus]

Every time I look at a different set up of the problem, there is always some new little thing that stands out that I hadn't seen before. And I'm left again with another nagging little issue. Really? So the backward going pulse, from the rest frame, ticks faster then the forward going pulse? How does that work?

Einstein's setup defines synchronized with this setup. If I'm reading it right, this is exactly his point. Observers moving with the moving frame would find that clocks at each mirror are not synchronized. (He doesn't define the mirror and bouncy pulse as the clock. His clocks are separate.) They are synchronized with respect to the stationary system, at the point where they happen to be in the stationary system. That is what the applet shows. Those two vertical clocks are synchronized. The horizontal "clock" in the applet is synced to the vertical clocks. The pulse hits each mirror in sync. Something has to give in the example and that is that observers at each mirror would find that their clocks aren't in sync.

I'm pretty sure that is right. I'll check back. There is something that bothers me in my mind.

Here's a couple of animations that I made to show the relation ship between time dilation and length contraction. Each clock and mirror set up has two alignments vertical and horizontal. The animations are drawn in the frame of one clock. First we have how it would look without length contraction. We start both with both set of mirrors at the same spot. That way, we can use the expanding circle to give a reference for the light pulses.

As we can see the right moving pulse for the "stationary" setup hits the right mirror and returns to the left mirror, before the pulse in the moving setup has even caught up to its right mirror. With the vertical pulses, the stationary setup up makes 2 complete round trips while the with the moving setup, only 1 round trip is made. (the relative speed between set ups is 0.866, so this makes the time dilation factor 2) The other thing to note is that for the moving clock, the vertical and horizontal pulses are out of sync with each other.

Next we show what happens when we include length contraction and we reduce the horizontal distance between the moving mirrors by 1/2 (remember, someone moving with the mirrors would still measure their distance as being the same as that of the vertical set.

The right moving pulse for the moving set still take almost the same amount of time to reach the right mirror as the stationary clock does to complete two round trips, but the return leg trip is much shorter and it ends up making one complete round trip for the two by the stationary clock, this matches the time dilation factor, and more importantly puts the round trips of the vertical and horizontal pulses back in sync with each other.

 

[FranzDiCoccio]

Hi itfitmewelltoo,

I'm back from work. I'm commenting your answers below.

I have the applet open in chrome and it is working fine.

Huh! I am not able to do that. I looked up in the internet and lots of people are having the same issue with swf. Probably it has to do with my version of chrome, which is too up-to-date. I think that google is actively dumping flash.

It is true to say that the relative speed between the mirrors and light pulses are classical in addition. I can refer to Einstein's paper, part I, section 3. The derivation of his first differential equation begins with the quantities of c+v and c-v. Specifically, the ratios of x'/(c-v) and x'/(c+v) are distance divided by velocity which is, of course, the time of the pulses going back and forward, in the stationary frame of reference. The depiction of the moving clock as horizontal is consistent with Einstein's derivation.

What you're saying sounds strange. I do not think that is a classical addition of velocities (composition). Those are the velocities of light and of the rocket in the f.o.r. of the lab. There is no reason for adding them. It's simply that on one leg the photon is chasing the front mirror, on the other leg the photon and the back mirror are heading towards each other.

One possible issue that the applet may suffer from has is that is that it is down-scaled from near light speed to something we can actually watch. If you can imagine, at 75% of light speed, the clock and pulse will have traveled quite a long way before the forward mirror is struck and then on the reverse pulse, the rear mirror will be hit almost instantaneously. Consider the ratios of x'/(c-v) and x'/(c+v). Plug in some realistic values for x' and v=.75c. The difference between forward and reverse are clear.

Sure, but someone in the rest frame (lab) would not see the photon (almost) stopping on the way back. I see no reason for that.
It cannot be related to the fact that light travels very fast. Any representation should work fine assuming that the speed of light is slow, provided that it's faster than anything else.

I have to admit that I am not sure of what you're saying in your subsequent answers. It's not a criticism, it's only that I do not understand the points you're making.
Perhaps it's because I'm not a native English speaker and I'm missing some subtleties of the language.

Do you think that the applet makes sense, or it has something weird? I still think that it's weird and confusing...

 

[FranzDiCoccio]

The right moving pulse for the moving set still take almost the same amount of time to reach the right mirror as the stationary clock does to complete two round trips, but the return leg trip is much shorter and it ends up making one complete round trip for the two by the stationary clock, this matches the time dilation factor, and more importantly puts the round trips of the vertical and horizontal pulses back in sync with each other.

Hi Janus,

I'm happy because this is what I said in my first message. I believe your movies show what I tried to describe in words. Length contraction would put the horizontal clock in sync with the vertical clock on the rocket, not with that in the lab.

Do you see any interpretation where the applet I'm pointing to makes sense?

 

[FranzDiCoccio]

The simulation does separate the rocket FOR with that horizontal line through the middle of the screen. So the fact that it shows the rocket’s clock’s photon taking the same amount of time on the emission and reflection legs of the journey is consistent with what the rocket observer sees, as is the length of the clock, regardless of how they orient it. I think the confusion is arising because of the fact that they show that clock moving, whereas the rocket observer typically wouldn’t choose a frame for himself in which he is moving in these types of setups. I think they do this simply to emphasize that you need at least two clocks (spacially separated) in the lab frame to measure the rate at which the rocket clock is ticking, and you need the rocket to travel from one of those clocks to the other. But again, this does seem like an odd concept to want to represent visually, especially in the way they do.

No wait. I see your point that "rocket frame" leads to think that this is what someone on the rocket would see. But that would raise more questions.
For instance, why a shorter clock would work ok? Wouldn't it run faster for the observer on the rocket?
And an observer on the rocket would see the clocks in the lab run slower. If they wanted to make their clock synchronized with those in the lab they should make them longer, right?
Also, why the rocket would move in the rocket frame?
Really confusing.

I'd stick with the usual point of view that we're looking at the FOR of the rocket as seen from the lab.

 

[FranzDiCoccio]

1. For a clock at rest, $T = \frac{2L}{c}$
2. For a clock moving at speed $v$ parallel to the line connecting the mirrors, $T = \frac{1}{1-\frac{v^2}{c^2}} \frac{2L}{c}$
3. For a clock moving at speed $v$ perpendicularly to the line connecting the mirrors, $T = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \frac{2L}{c}$

From 2 & 3, if you have a pair of moving light clocks, one oriented one way and one oriented the other way, the only way for them to stay synchronized (to have the same period) is if the one that is oriented parallel to the motion has length $\sqrt{1-\frac{v^2}{c^2}}$ times as big as the one oriented perpendicular to the motion. To complete the argument about time dilation, you need to assume that the light clock oriented perpendicular to the motion experiences no length contraction.

Hi stevendaryl,
this is basically what I say in my first message, perhaps not as schematically.

This might be the point of the app:

But how?
Are you suggesting that they are showing only the "real" length of the clock and not the effect of length contraction seen from the lab?
This still does not account for the weird motion of the photon.
And it also confuses me, because on the one hand we see the real length of the clock (in the rocket FOR), on the other hand we hear its dilated tick (in the lab FOR).
Confusing...

[edit]

I now realize that what I said in my OP came out sligthly wrong.
Of course the factor to be punched in is $\gamma^{-1}$, because this is factor to obtain the _real length_ of a clock on the rocket that would be synchronized with the stationary ones.
What I was tryng to say is that the clock should appear shorter by a factor $\gamma^{-2}$, because of the further $\gamma^{-1}$ factor coming from the length contraction.

So, if I was to create an app like the one I mentioned, I would show the traveling clock both

• in the rocket FOR, where it's stationary, (really) shorter, runs faster and the photon takes an equal time going back and forth between the mirrors
• and in the lab FOR, where it moves, it appears even shorter, it is synchronized with the static clocks, and the photon does not take the same time going either way (this last feature is true irrespective of the clock length or its synchronization with the stationary clocks).

 

[Pencilvester]

why a shorter clock would work ok? Wouldn't it run faster for the observer on the rocket?

Yes. See post #8.

And an observer on the rocket would see the clocks in the lab run slower.

Yes, but the rocket’s observer would not measure the lab’s clocks to be synchronized with each other. This means that although the rocketeer will see each lab clock run more slowly than his own, if set up correctly (meaning if the rocketeer uses a light clock shortened by the right amount so that it’s ticks are just the right amount of fast), then the rocketeer will read the same time on his fast clock as he reads on the lab clock that he happens to be passing at that moment (but different times on all the rest of the lab clocks).

Also, why the rocket would move in the rocket frame?
Really confusing.

This is exactly what I considered a potential source of confusion.

I'd stick with the usual point of view that we're looking at the FOR of the rocket as seen from the lab.

“as seen from the lab” automatically implies the lab’s FOR.

 

[FranzDiCoccio]

Hi Pencilvester,
I think I was a bit confused the phrasing in your posts. So, as far as I understand you're stressing the synchronisation between the clocks. I see.
Let me try to summarize:

• seen from the lab frame the (shortened) clock on the rocket would always tick "simultaneously" with those in the lab (here by "tick" I mean the photon round trip, not each leg). However it would appear moving and shortened by a further $\gamma^{-1}$, so it would appear $\gamma^{-2}$ shorter than a "standard" clock.
• seen from the rocket the same clock would not move, be (not appear) shorter than a standard clock by a factor $\gamma^{-1}$. Also it would tick faster than a standard clock, and its ticking would be symmetric. In general it would not be synchronized with the stationary clocks, which would be seen as slower. However, the clock in the rocket would agree with the first stationary clock at $t=0$, and with the second stationary clock at the time $t=T$ when it gets there. This is possible because the stationary clocks are not synchronized with each other, as seen from the rocket.

Ok, I think that this might work.

I still find the applet really misleading. As I say, in order to highlight what we've been discussing, I would represent the traveling clock both in the lab frame and in the rocket frame. But that would probably be too much.

I have read again the "Instructions" in the applet. I am not a native English speaker, but it seems to me that what we said implies that one can "keep the rocket clock synchronized with the lab clocks" only when the former is seen from the lab. Otherwise the sinchronization happens at two events only.
Therefore they should adjust the clock length to account for the length contraction (which they mention but do not represent) and adjust the photon behavior, that is wrong from every point of view.
That is, they should do something along the lines of what Janus showed (but theirs would be "interactive", meaning that one can adjust the rocket speed and the length of the clock).

 

[Pencilvester]

Let seen from the lab frame the (shortened) clock on the rocket would always tick "simultaneously" with those in the lab (here by "tick" I mean the photon round trip, not each leg)

Yes.

However it would appear moving and shortened by a further $\gamma^{-1}$, so it would appear $\gamma^{-2}$ shorter than a "standard" clock.

This depends on how you orient the rocket’s clock, which shouldn’t be an emphasis of the simulation, so why they decided to depict the clock as oriented with its length along the direction of motion instead of orthogonal, like the lab clocks are, is beyond me. But yes, if you assume the rocket clock’s length is oriented along its direction of motion, then your statement is correct.

seen from the rocket the same clock would not move, be (not appear) shorter than a standard clock by a factor $\gamma^{-1}$. Also it would tick faster than a standard clock, and its ticking would be symmetric.

Yes.

In general it would not be synchronized with the stationary clocks, which would be seen as slower. However, the clock in the rocket would agree with the first stationary clock at $t=0$, and with the second stationary clock at the time $t=T$ when it gets there. This is possible because the stationary clocks are not synchronized with each other, as seen from the rocket.

Yes.

I still find the applet really misleading.

Agreed.

one can "keep the rocket clock synchronized with the lab clocks" only when the former is seen from the lab. Otherwise the sinchronization happens at two events only.

For any number of synchronized clocks in the lab FOR, all possible observers, including the rocketeer (assuming he does not change the speed or direction of the rocket) will always see the rocket clock read the same time as the co-located lab clock (the lab clock that is in the same location in spacetime as the rocket clock; i.e. the lab clock that the rocket is passing). In general, everyone (other than the lab techs) will say that all other clocks are out of sync. And to the lab techs, all clocks in the whole scenario are synchronized.

 

[FranzDiCoccio]

Yes.
For any number of synchronized clocks in the lab FOR, all possible observers, including the rocketeer (assuming he does not change the speed or direction of the rocket) will always see the rocket clock read the same time as the co-located lab clock (the lab clock that is in the same location in spacetime as the rocket clock; i.e. the lab clock that the rocket is passing). In general, everyone (other than the lab techs) will say that all other clocks are out of sync. And to the lab techs, all clocks in the whole scenario are synchronized.

You're referring to one scenario in which instead of two "stationary" clocks at the "ends" of the journey (as in the app) there is a succession of identical stationary clocks synchronized with one another (in the lab frame).
As the rocket passes "over" one of the stationary clocks, the clock on the rocket would agree with that clock, and that clock only.
Am I correct?

I think I recently came across an app depicting that. I found that a bit overwhelming though. Far too many clocks to be readable.

Anyway, what's all this fuss of having a clock instantaneously synchronized with the "stationary" clocks?
I think I'm losing the perspective here. Why is this important? Such a clock would be pretty useless for the people on the rocket, since it is wrong (goes too fast).
Is this a way of checking whether the "stationary" clocks are synchronized?

Wouldn't be easier to measure the distance between the clocks (in the rest frame) and send synchronization signals between clocks (of course taking into account the journey time?)

That is: I have a second clock at a distance L from the first one. I can program the first clock so that it sends a signal to the second at $t=0$, and the second so that it starts running when it receives the signal. Of course it should start from $t=L/c$. That way I would be sure that the clocks are synchronized, right?

 

[Pencilvester]

Why is this important? Such a clock would be pretty useless for the people on the rocket, since it is wrong (goes too fast).

I’m more familiar with the theory side of things as opposed to the applied side, but as far as a use, I think this sort of synchronization is used in GPS satellites (although, I believe they have to take GR effects into account as well). In this case, there’s no person actually aboard the satellite who would need a clock that tells them the amount of time that has actually passed on the satellite, so it doesn’t matter that the clock runs a little fast (or slow—I can’t remember if SR or GR effects predominate in a GPS satellite).

Is this a way of checking whether the "stationary" clocks are synchronized?

Nope. See above for an example of a practical use.

I have a second clock at a distance L from the first one. I can program the first clock so that it sends a signal to the second at $t=0$, and the second so that it starts running when it receives the signal. Of course it should start from $t=L/c$. That way I would be sure that the clocks are synchronized, right?

If clock 1 sends a light signal at its own $t = 0$, and clock 2 (which I am assuming is at rest relative to clock 1) starts ticking from 0 when it receives the light signal, then I’m afraid those clocks will be out of sync for all observers.

 

[Pencilvester]

Of course it should start from t=L/c.

Ok, I just now realized what you meant there, so scratch my previous answer:

I’m afraid those clocks will be out of sync for all observers.

Yes, you are correct in saying that those clocks would be synchronized in the lab's FOR, but I don't think the applet tells us much of anything about how to synchronize clocks at relative rest to one another. But now that we're on that subject, it would probably just be a little simpler to stand exactly halfway between the two lab clocks, and then use a light-based remote to start both clocks at the same time (emit 2 photons simultaneously- one directed to each clock, and have each clock set to start at t=0 when it receives a photon).