Light Clock Problem

DaleSpam

Why are we looking for T? You already agreed that it cannot affect the operation of the clock in any way. It is irrelevant.

altergnostic

But there is! The beam is our object of analysis here, and it is going at c in the primed frame. We are not seeing it directly, so we need the signal from event B to know its speed.

Do it like the train and embankment problem and simply ditch the angle. You are given the primed distance between AB and the primed time for event B, you don't need anything else.



I aknowledge that you found a setup that allows us to observe the beam, and you did it only by sending light from the to the detector, so I give you that. But that also proves what I have been saying, that you can't simply take local or primed numbers for the beam and use them as unprimed data. The beam has since acted like an object subject to relative velocities, it is just like a moving train that we see with light reflected from it. The light between the train and embankment acts the same as light between beam and detector, and it is this light that we directly observe that we know moves at c. It is this light that brings us the coordinates for the beam.

Answered: with a new setup. Current setup is incomplete.

Actually this hasn't been answered at all, but I already concluded that light that is moving in any direction other than directly at us doesn't have to follow any constancy, since we can't be considered neither source nor observer in that case.
Clearly the answer is yes, otherwise we wouldn't have even mentioned scattering or signaling.
You showed it can be done with a rigorous setup, but I still contend the numbers are different from current diagrams.

That is a conclusion I came to only in recent posts (I actually learned a lot from your scattering setup) so it is not like I changed the subject, I only went where the discussion led me. You refuse to do the analysis with the set of givens I proposed, which are very realistic. And now you claim that substituting the beam with a prijectile is overly complex when I showed that it isn't, you don't even have to take any angle into account since I gave you the distance AB measured locally so you can put the x axis directly in line with it, and I also gave primed times so you can calculate the speed.

DaleSpam

Besides the funny self-contradiction, it is one of the postulates of SR that it is going at c in all frames. We know its speed according to SR, and alternative theories that disagree with SR are not permitted here.

harrylin

OK, I think that we are getting somewhere:
A bullet doesn't have the same speed in both frames; I'll stick to the topic instead, which allows much better clarity.
I already illustrated that it is not necessarily to use that light for the time data. So, if you mean that it is light or radio waves or other means that brings us the clock synchronisation, then I completely agree.
In post #8 I gave you the answer (in the link). However, you did not comment on it.
[EDIT: In fact, my answer took care of the spatial rotation misconception]
Of course, we narrowed down the problem as stemming from a calculation error - either made by "all" textbooks and students, or made by you. My purpose was (and still is) to get rid of the bug, after which you can be an asset to this forum - and you will feel better.
Thanks - such feedback is helpful to remain motivated with this voluntary work!
OK. I will next look into your calculation example with the light beams, and give my analysis.

altergnostic

My premise has nothing to do with the operation of the clock, it has to do with apparent effects. Like doppler and such things.

Because T is the time shown in the unprimed frame when the observer receives the signal from B, which is different from the primed time, and which is the time that determines the perceived speed of the beam from the observer at the origin. It is the time the beam seems to take to cross the distance AB. It is far from irrelevant.

There's no contradiction whatsoever. Observed/detected light is going c in all frames. Einstein never once thought about how light would look like at a distance since it seemed ridiculous to do so, since we can't have any knowledge of undetected light. He and everyone else always addressed this issue by placing another observer in the path of light, but that says nothing about the observed behavior of light receding from the observer (I really believe that taking harrylin setup of scattering light from the beam's path plus my suggestion that we take this scattered light and bring it to one single point - the observer - is a novelty).
How do you measure the speed of the train? You take the distance it covers over the time it takes to cover it. This velocity is not the same as marked on the trains own speedometer, for instance. You are on the train and it marks .5c on the speedometer. You cross 1 light second, from your perspective, in 2 seconds.
An observer at the embankment would see the train cross that distance in 2 seconds + the time it takes light to reach back to him from that distance (which would be the time marked on the observers watch at the moment he sees the train at that distance). The speed will seem to be slower relative to the speed as seen on the onboard speedometer!

I have a diagram on this from a discussion I had a few months ago:


But I'm afraid this will divert us from the topic, maybe this needs a thread itself.

PeterDonis

My previous analysis was correct, it just didn't include the extra light signal from the mirror back to the observer. Altergnostic apparently still thinks I made a mistake (which I didn't); I'll address his issue in a separate post.

You are correct that it doesn't matter how the x and y axes are oriented; either choice leads to the same physics. I switched it in my revised analysis only because I didn't see the point of getting into a protracted wrangle about it when I can do the analysis with altergnostic's preferred axis orientation and still show that there is no problem.

altergnostic

This is precisely my point!!!!! The same is true for the beam!

That's almost what I mean. I take that the sync has been done previously (at the origin) and the light actually brings time information, in this particular case. We have, at the origin, the same time marked both on the observer's watch as the light clock's time. The light clock departs. The observer will see the next second on the light clock at the moment he sees the signal, but the time marked on his watch when the signal reaches him is no t'B=1s, it is t'B + the time it takes light to cross BA.

I guess I forgot about that post. But there's no disagreement here.

You are welcome! That was ingenious indeed.

I started this post certain that the observer couldn't possibly trace the beam (and from current diagrams he really can't - you are the only one who made this possible so far in all the discussions I have participated, I really respect you for this). My last diagram was only possible because of this insight, I am only afraid that my conclusions are so far off current accepted diagrams that I won't get any credit for this. I don't care if I am wrong, really, I just want a really rigorous analysis, not just more of the same assumption that the beam must be travelling at the speed of light for this particular observer because it always travels at c for all observers - but I insist this is not your common observer, it is a non-observer, if you like. He is not observing the beam directly. The speed of light must apply to light that reaches him, since that is the light he detects going at c directly. The beam's velocity has to be calculated from that, just like you said above, the bullet has a different velocity in each frame, and so must this beam. I think this is even more consistent with such experiments than the opposite assumption that the beam must seem to move at c in this situation. Than we would have to explain why should the beam act differently from the bullet, since the setup is fundamentally the same.

Thanks!

DaleSpam

You have explicitly stated multiple times that the standard analysis of the operation of a light clock is incorrect or incomplete. Do you now agree that the standard analysis is both a correct and complete analysis of the operation of the clock? If not, then please state exactly your objection to the standard analysis of the operation of the clock which you agree has nothing to do with any signals sent to any observers.

Note, the standard analysis only purports to be an analysis of the operation of the clock and not an analysis of the subsequent transmission of the results to any observer. Such considerations are irrelevant to the operation, as you have already agreed.

You are factually wrong on this count. The postulate of SR is that all light pulses travel at c, regardless of whether they are detected or not. Please read the postulates and note that they do not mention anything about detection. The postulate is that light travels at c, end of story, no distinction whatsoever between light going towards or away from any observer or about the detectability of the light. Get your facts straight.

altergnostic

Sorry Peter, but I have to insist that you reconsider this carefully.
I insist that you have to align the x axis with AB and completely ditch the speed of the light clock in the former x axis, since it is actually causing a distraction here.
Please note that the observer at the origin doesn't even need to know that speed to calculate anything:
He is given the distance AB
He knows the primed times, since he has synchronized his watch with the light clock back at the origin. He knows that the reflection at B occurs at t'=1s.
He receives the signal event from B at T=t'+ the time it took for the light signal to cross BA and reach him - that is the time he actually sees the signal, and is the time he observes the beam to reach the point B.

From the given distances and the observed times he can only observe that the beam crossed AB in T seconds, and then subtract the time it takes light to cross BA to get the primed time - which will be 1s again.

This seems all extremely consistent, and in accordance with what we would expect to have by replacing the beam with a projectile. If a projectile hits B at t'=2s, the observer will observe it crossing AB at T=t' + the time it took light to cross BA, and then subtract the time it takes light to cross BA to get the primed time - which will be 2s again.

By this method, the primed velocity will make the primed distance smaller than AB (length contraction), and the calculated distance will become y', which again is fully consistent with the setup. Like so:
Observed speed of the beam = AB/T = 1.12/2.12 =0.528c
Transformed time of event B = t'= 1s

We know that light travels at c locally from many many experiments, so to get the distance travelled by light in the primed frame, we just have to figure out the distance light crossed in 1 second (the primed time) = 1lightsecond = y'

Do you see my reasoning? You may not agree with it for some reason, but it is very consistent, and all the numbers add up. I can't see any error on it, if you can, please do point it out, since I am not here to defend any preconceived opinion, I'm just going where logic and math has taken me and sharing this with you for open analysis.

And just as a footnote, I don't think this contradicts any postulates nor disproves or changes SR in any way. It is simply a setup that hasn't been thought of yet, as far as I know (and I have to share credit for this with harrylin, even if he disagrees with my analysis). Observed light still travels at c. It also travels at c relative to the source. It is undetected light, or indirectly observed light, that I conclude doesn't need to follow the constant, but this is not your ordinary light. It is not the light that enters any known equation as c. This is light moving at a distance. Not a single light equation takes c to be light at a distance, everytime you have c in an equation, that c is referring strictly to light that reaches the observer, which is the light you do transforms with.
That's the importance of the signal arriving from B: it is with this light that we observe the beam, and it is with this light that we must do transforms, just like in any other setup.
To give the speed of light both for this incoming light and for the beam, as seen in the unprimed frame, is similar to believing that the speed of the man walking inside a moving train is the same for both primed (the train) and unprimed (embankment) frames.
If you still think the beam will be observed to go at c from A to B as observed from the observer at the origin A, than you have to explain why wouldn't the speed of a projectile (in place of the the beam) be the same in the primed and unprimed frame as well.

Please consider this carefully and think about this with no prejudice (I mean no harm to SR )

PeterDonis

It isn't a mistake (if there's any mistake involved here it's yours, in not understanding how relative motion is modeled in SR).

Yes, but in order to get "numbers for the beam", you have to model the motion of the light clock as well. Otherwise you can't enforce the constraint that the beam has to be co-located with particular parts of the light clock at particular events (the source/detector at events D0 and D2, and the mirror at event B1a). Without that constraint in place any analysis of the beam's motion is meaningless.

I didn't *assume* that the beam was moving at c; I *proved* that it moves at c, by enforcing the constraint I referred to above. Please read my analysis again, carefully. What I did was compute the locations of the light clock's source/detector at events D0 and D2, and compute the location of the mirror at event B1a, using only facts about the light clock's motion relative to the observer. I then showed that the spacetime interval between events D0 and B1a, and between events B1a and D2, is of zero length; this proves that the light pulse travels at c between those pairs of events. I did this in both frames to confirm that the interval is invariant (as it should be).

I also, once I understood that there was a second light signal involved (from event B1b to A2), did the same kind of computation for that light signal: compute the location of the observer at event A2 (we already know the mirror's location at event B1b because event B1b is co-located with event B1a), and confirm that the spacetime interval between events B1b and A2 is of zero length, again in both frames.

Sure, by all means let's do the analysis as many ways as you like; the answer will remain the same.

We now have the x axis oriented in such a way that the event coordinates in the unprimed frame are as follows:

A0 = D0 = (0, 0, 0)

B1a = B1b = (t1, x1, 0)

A2 = (t2, 0, 0)

D2 = (t2, x2, y2)

Our objective is to find t1, x1, t2, x2, and y2, and then compute the spacetime intervals (A0 to B1a), (B1b to A2), and (B1b to D2), and verify that they are all zero.

The first thing to note is that the time coordinates are unchanged from my previous analysis; i.e., we still have t1 = 1.16 and t2 = 2.32.

The second thing to note is that x1 must be given by the sum of the squares of the x and y coordinates for events B1a/B1b in my previous analysis (since it's the same spatial distance, we've just rotated the axes to put that distance all along one axis); i.e., we must have

[tex]x_1 = \sqrt{ 1 + ( 0.58 )^2 } = 1.16[/tex]

The third thing to note is that the sum of the squares of x2 and y2 must equal the square of the y coordinate of event D2 in my previous analysis (again, the same spatial distance from the origin, just with the axes rotated); i.e., we must have

[tex]x_2^2 + y_2^2 = ( 1.16 )^2 = x_1^2[/tex]

This tells us something very important: the distances AB, BD, and AD in your diagram are all *equal*. (We could have seen this directly by noting that the light clock, which moves half as fast as the light pulse, covers distance AD in the same time as the light pulse covers distance AB + BD; therefore AB + BD = 2AD, which combined with AB = BD gives AB = BD = AD.) This means that the angle between lines AB and AD is 60 degrees (since it's an angle of an equilateral triangle), so we must have

[tex]\frac{y_2}{x_2} = tan (60 deg) = \sqrt{3}[/tex]

Substituting this into the equation above gives

[tex]x_2^2 + 3 x_2^2 = x_1^2[/tex]

or

[tex]2 x_2 = x_1[/tex]

Now we can confirm that the spacetime intervals I mentioned are zero. First, we have t1 = x1, so the interval (A0 to B1a) is obviously zero. Similarly, we have t2 - t1 = t1 = x1, so the interval (B1b to A2) is obviously zero. Finally, the interval (B1a to D2) is given by

[tex](t_2 - t_1)^2 - (x_2 - x_1)^2 - (y_2 - 0)^2 = ( 1.16 )^2 - ( x_2^2 + y_2^2 ) - x_1^2 + 2 x_2 x_1 = ( 1.16 )^2 - ( 1.16 )^2 - x_1^2 + x_1^2 = 0[/tex]

As I said, the answer remains the same.

These numbers are from different frames; trying to mix them in formulas is going to give you meaningless answers.

Oh, so by Vab you meant the speed of light, not the speed of the clock? Fine, but your calculation of it still makes no sense; once again, you can't mix numbers from different frames and expect to get meaningful answers. See above.

PeterDonis

The second part of your diagram (the "TRAIN FRAME" part) is incorrect; in the train frame the blinker is motionless, so all of the space coordinates should be marked as zero.

Before starting one, I strongly advise you to review a good basic SR textbook, such as Taylor & Wheeler.

altergnostic

That is the train frame measuring its velocity against the ground, just like you do in your car. It is the speed marked on its apeedometer.

PeterDonis

Done; see my previous post (actually the last but one from this one). The answer remains the same.

But he can't use the primed times along with unprimed distances to calculate anything meaningful. To correctly calculate a speed you must use the distance and time from the same frame. The "distance AB" that he is given is in the unprimed frame. That distance is *not* perpendicular to the relative motion of the light clock and the observer in the unprimed frame (it is in the *primed* frame, but it is *not* in the unprimed frame--perhaps this is one of the points where you are confused). So the observer *can't* combine it with a time in the primed frame to get a meaningful answer, because of length contraction.

Nope, this mixes numbers from different frames again.

The velocity of an object moving at less than c *does* change when you change frames. The velocity of light does not. Please review a good basic relativity textbook, paying particular attention to the formula for relativistic velocity addition; you will note that that formula always gives c for the velocity of a light beam.

Read my latest analysis again, carefully; also read my comments above, carefully. Your error is that you are mixing numbers from different frames and expecting them to give you meaningful answers. Also, you may be mistakenly assuming that the distance AB is the same in both frames; it's not, for the reason I gave above.

Then you don't understand SR very well.

This is all confused. All light travels at c in SR, in any inertial frame. If you read my latest analysis, you will see that the light beam traveling from B1a to D2 travels at c, even though it is not "observed" (both events are "at a distance" from the observer).

PeterDonis

But the ground is moving in the opposite direction, and the blinker is not moving with it. A correct "TRAIN FRAME" diagram would have the blinker on the right, motionless, and points on ground moving to the left. Then you would have to calculate *which* points on the ground would be at which spatial coordinates in the train frame at which times.

A better tool for this job, IMO, would be a proper spacetime diagram, since that directly shows both space and time relationships and allows you to draw actual worldlines so you can see how they are related. If you aren't familiar with spacetime diagrams, I would recommend learning them; they really help to remove confusion for scenarios like this one.

altergnostic

Agreed. But still it is not hard to see, from the bottom diagram, that the speed of the ground measured from the train (the speed on its speedometer, to simplify things) will not be the same speed as that observed by the man at the origin. This actually compromises gamma a bit, since it is hard to decide which velocity to plug into it.

But please, let's leave this be for now, since we will surely divert from the topic. If you feel this deserves attention and want to discuss it further, I will gladly start a thread on this, but let's please not digress here, since this thread is already dense enough as it is. Deal?

DaleSpam

No, it isn't. The invariance of the speed of light is a postulate of SR. If that is precisely your point then your point is WRONG.

PeterDonis

You're going to have to explain how this is "not hard to see", because I don't see it. The only thing that changes about the velocity from frame to frame is the sign: in the ground frame, the train moves at +v, and in the train frame, the ground moves at -v.

I suspect you are confused because you think length contraction affects the velocity from frame to frame. It doesn't, because length contraction and time dilation *both* come into play, and their effects cancel when applied to the relative velocity. If you don't see how that works, read through my analysis again, carefully; for example, look at how the coordinates of events A2 and D2 transform between the primed and the unprimed frame, and note that the time and space coordinates change in concert to keep the relative velocity the same.

[Edit: To expand on this a little more, events D0 and D2 both lie on the worldline of the light clock source/detector, which is equivalent to the "train"; and events A0 and A2 both lie on the worldline of the observer, which is equivalent to the "ground". Events A0 and D0 are co-located at the origin, so their coordinates drop out of the analysis, and we can use the coordinates of event D2 in the unprimed frame, and A2 in the primed frame, to compute the relative velocity. In the unprimed frame, event D2 has coordinates (t2, x2, y2), and the relative velocity is given by sqrt(x2^2 + y2^2) / t2, which works out to 0.5. In the primed frame, event A2 has coordinates (t2', x2', y2'), and the relative velocity is given by sqrt(x'2^2 + y2'^2) / t2', which works out to 0.5. Then we just have to remember to set the sign appropriately, based on the sign of (x2, y2) or (x2', y2'), which gives v = +0.5 in the unprimed frame and v = -0.5 in the primed frame.]

I don't see how starting another thread will help. This is an absolutely fundamental point that underlies the "problem" you posed in the OP, and all of my analysis, and indeed all of SR. If you think it's wrong, you basically think SR is wrong; and if you can't back your claim up cogently (which you haven't so far) and you won't abandon it, any further discussion of your claim is basically out of bounds here on PF, since we don't discuss personal speculations that contradict SR. Moving the discussion to another thread won't change that.