
#1
Feb2312, 01:10 PM

P: 169

Might there be anyone here who knows the math to work this one out?
Stopped Clock Paradox Simplified "The speed of light is the same for all observers." Scenario The stationmaster has set two synchronized stopclocks along side the track. The stationmaster placed the clocks exactly 8 μls apart. Exactly half way between the clocks is a stopbutton. The stopbutton functions by flashing a photon toward each stopclock, stopping them. At the exact moment the clocks stop, they light up flashing the time they stopped. Einstein is approaching the station in his train traveling at 0.5c. The stationmaster pressed the stopbutton when the train was 6 μls from the first clock. The stationmaster noted that the clocks read exactly 10 μs when he pressed the button. Questions What time will Einstein see flashed from each clock and why? What time will the stationmaster see flashed from each clock and why? What time will the clocks later show as the time they actually stopped? From Einstein’s POV; The station and clocks are moving toward him. After the button is pressed, one clock is moving toward the stopbutton’s flash. The other is moving away. Thus it seems that the clocks must stop at different time readings unless they are out of sync. And though they would be out of sync as perceived by Einstein, the math doesn't seem to work out such that both of them could agree with the actual stopped reading as revealed later. 



#2
Feb2312, 02:04 PM

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I guess we are to surmise that the stationmaster is colocated with the stop button because he can see the same time on the stop clocks before they are stopped and flash the stopped time, correct?
If so, then it will take 4 μs for the photons to reach the stop clocks at which point they will read 14 μs, and that is the answer to all your questions. You sure threw in a lot of extraneous junk that has nothing to do with your questions. EDIT: Opps, I didn't do this right. When the stationmaster sees the clocks reading 10 μs, they will actually be reading 14 μs. It will then take 4 μs for the photons to get to the stopclocks at which point they will be reading 18 μs and that is the answer to all your questions. 



#3
Feb2312, 05:28 PM

P: 169

"extraneous junk"?
Really? You did notice it was a relativity question? You showed the obvious and trivial. Care to show the math involved in Einstein's POV? 



#4
Feb2312, 05:38 PM

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Relativity and The Stopped Clock Paradox 



#5
Feb2312, 06:52 PM

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Where is the stationmaster located? How does the stationmaster know when to press the stop button? Can the stationmaster read the time on the stopclocks before they are stopped? Am I supposed to assume that the distance the train is from the first clock is in the rest frame of the station? Where is Einstein located in the train? Why did you ask your three questions as if they could have different answers? Ditto what jtbell asked: what math are you using that creates a paradox? The proper way to deal with problems like this is to define in detail all the events in one frame of reference and then transform them into another frame of reference using the Lorentz Transform. It's guaranteed to get the correct answer without any fear of a paradox. So if you can provide the information for the details of the events in the station frame, I will show you the numbers in Einstein's frame. 



#6
Feb2312, 07:21 PM

P: 169

Okay;




#7
Feb2412, 01:04 PM

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Since there are enumerable ways in which to select a frame and since you didn't specify one, I will define one here:
It is a frame in which the station is at rest. Along the xaxis we have the following items. (Distance units are in μls and time units are in μs.) At x=0, we have Einstein at the start of the scenario. At x=7, we have the location of Einstein when the stopbutton is pressed, at t=14. Note that he is traveling at 0.5c. At x=13, we have the first stopclock. Note that it is 6 μls from where Einstein was when the stopbutton is pressed. At x=17, we have the location of the stationmaster and the stopbutton. At x=21, we have the second stopclock. Note that there is 8 μls between the two stopclocks and the stopbutton is midway between them. I believe this adheres to your description of the scenario. So now we can define some significant events. I will use the nomenclature of [t,x]. Since Einstein is traveling at 0.5c, all of his events will have the xcoordinate equal to onehalf of the tcoordinate. [0,0] Einstein at the start of the scenario. [0,17] Stationmaster at the start of the scenario. [14,7] Einstein when the stopbutton is pressed. [14,17] Stationmaster when the stopbutton is pressed. [14,13] First stopclock when the stopbutton is pressed. [14,21] Second stopclock when the stopbutton is pressed. [18,9] Einstein when the photons reach the stopclocks. [18,13] First stopclock when photon hits it. [20.667,10.333] Einstein first sees image of first stopclock reading 18. [18,21] Second stopclock when photon hits it. [26,13] Einstein first sees image of second stopclock reading 18. [26,13] Einstein colocated with first stopclock (when it would have read 26). [42,21] Einstein colocated with second stopclock (when it would have read 42). I think a little explanation is in order for how I arrived at the events for which Einstein sees the two stopclocks reading 18. It based on the fact that light travels twice as fast as Einstein, therefore Einstein will cover 1/3 of the distance while the image covers 2/3. The distance between Einstein and the first stopclock is 139 = 4 so Einstein will have traveled 4/3 or 1.333 beyond 9 where he was when the photon stopped the first clock which equals 10.333. The distance between Einstein and the second stopclock is 219 = 12 so Einstein will have traveled 12/3 or 4 beyond 9 where he was when the photon stopped the second clock which equals 13. Now we use the Lorentz Transform to convert the coordinates of these events into Einstein's rest frame. First we need to calculate gamma for beta equal to onehalf: γ = 1/√(1β^{2}) = 1/√(10.5^{2}) = 1/√(10.25) = 1/√(0.75) = 1/0.866 = 1.1547 Now we use the version of the Lorentz Transform that applies to units where c=1 and distances are in μls and time is in μs: t' = γ(txβ) x' = γ(xtβ) The first event [0,0] transforms into the same coordinates so I will show the calculation for the second event and the rest of the events I show just the final results: t' = γ(txβ) = 1.1547(017*0.5) = 1.1547(8.5) = 9.815 x' = γ(xtβ) = 1.1547(170*0.5) = 1.1547(17) = 19.630 [0,0] Einstein at the start of the scenario. [9.815,19.630] Stationmaster at the start of the scenario. [12.124,0] Einstein when the stopbutton is pressed. [6.351,11.547] Stationmaster when the stopbutton is pressed. [8.660,6.928] First stopclock when the stopbutton is pressed. [4.041,16.166] Second stopclock when the stopbutton is pressed. [15.588,0] Einstein when the photons reach the stopclocks. [13.279,4.619] First stopclock when photon hits it. [17.898,0] Einstein first sees image of first stopclock reading 18. [8.660,13.856] Second stopclock when photon hits it. [22.517,0] Einstein first sees image of second stopclock reading 18. [22.517,0] Einstein colocated with first stopclock (when it would have read 26). [36.373,0] Einstein colocated with second stopclock (when it would have read 42). You will note that all of the events corresponding to Einstein have an xcoordinate of 0 since he is at rest in this frame. The first thing to notice is the time and distance that each photon has to travel in Einstein's rest frame. For the first stopclock, the event of the start of the photon is [6.351,11.547] and the arrival of the photon at the first stopclock is [13.279,4.619]. If we take the difference between the components of these two events, we get [6.928,6.928] which shows us that the photon traveled a distance of 6.928 μls in 6.928 μs which means it traveled at c in Einstein's frame. Doing the same thing for the second stopclock we use the same start event [6.351,11.547] but the arrival of the photon at the second stopclock is [8.660,13.856] for a difference of [2.309,2.309], again showing that this photon traveled at c in Einstein's frame but has a shorter distance to travel and took less time, as you noted it would in your first post. The second thing to notice is regarding the stopped times on the two stopclocks. First off, you have to be aware that the times in the events are coordinate times so the events pertaining to the stopclocks will not show you the times on the stopclocks. One way to figure out from Einstein's frame what time will be on the stopclocks, is to see what time would have been on the stopclocks when Einstein is colocated with them and then determine Einstein's coordinate time deltas from when the photons hit the clocks and then use the time dilation factor to determine the proper time delta on those clocks. It sounds complicated and it is complicated but it will give us the same answer that we got in the station rest frame and that's all that matters. So for the first stopclock, we note that in Einstein's frame, he is colocated with it at [22.517,0] when it would have read 26 (the same as the coordinate time in its rest frame). The time at which the the photon hits the stopclock in Einstein's frame is 13.279. The delta is 9.238. Dividing this by gamma we get 8. Subtracting 8 from 26 we get 18, the flashed time on the first stopclock. And for the second stopclock, we note that in Einstein's frame, he is colocated with it at [36.373,0] when it would have read 42 (the same as the coordinate time in its rest frame). The time at which the the photon hits the stopclock in Einstein's frame is 8.660. The delta is 27.713. Dividing this by gamma we get 24. Subtracting 24 from 42 we get 18, the flashed time on the second stopclock. Did I cover all your concerns? Do you still think there is a paradox? 



#8
Feb2412, 03:56 PM

P: 169

I am going to have to take this a little at a time, especially since you seem to be adding unnecessary elements. But we'll see..
So, I have to assume that you have x=0 at t=0 (which wasn't intended, but I think is okay). At t=0; xE (x location of Einstein) = 0 At t=14 xE = 7 So at that point, the button gets pressed (t=14), thus Button Press == (14,7)_{S} (station frame) I will get to the rest shortly.. 



#9
Feb2512, 01:57 AM

P: 169

But wait. The station sees Einstein at a distance of 17 from the button at t=0 as per prior post. But now Einstein sees the button at a different distance of 19.630? A length expansion? Why would Einstein see the button further away than the button would see Einstein? I think the notation conversion might need a little explaining. The stop button is pressed when Einstein sees the stopclocks reading 12.124. At that moment, the stationmaster/button is 11.547 away. Where does the 6.351 come from? Need some help here with your notation changeover. 



#10
Feb2512, 11:57 AM

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#11
Feb2512, 01:58 PM

P: 169

..and btw, my method for getting to the conundrum is much simpler. I am looking for anything that would indicate what my error might be. 



#12
Feb2512, 04:25 PM

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In the station frame, we assume that there are synchronized clocks at every possible location. These are coordinate clocks and remain fixed at coordinate locations. The time on a coordinate clock plus the x, y, z coordinates for the location define an event. I have ignored the y and z coordinates because they are zero. I described several events and gave their coordinates in the station frame. Then I used the Lorentz Transform to see what the coordinates in Einstein's frame would be. His Frame of Reference is created in the same way as the station frame, by setting up synchronized clocks at every possible location but these clocks and locations are stationary with respect to Einstein but they are all moving with respect to the station's Frame of Reference. All the clocks and locations are different in the two frames and that's why we need to use the Lorentz Transform to get from one to the other. We have to keep them straight. We have to make the calculations of the type I showed you to see what Einstein will actually see during the scenario but the main point is that when doing these calculations based on Einstein's frame, we will get the same answers that we get from the station frame, except that it is generally a lot more work, as should be obvious by now. So to answer your question, Einstein's "coordinate clocks" are stationary with respect to him and all have the same time on them and all run at the same rate, so he can tell what any of them has on them by looking at the one that is next to him at location zero. On the other hand, the other clocks are moving with respect to him so they run at a different rate and don't have the same time on them so he can't know what time is on any one of them unless he does some calculations. All he knows is the time on the clock that is nearest him as it goes flying by him at onehalf the speed of light. All other clocks, he has to wait for an image of it to reach him at which point it will have a different time on it and be in a different location. t' = γ(txβ) = 1.1547(1417*0.5) = 1.1547(148.5) = 1.1547(5.5) = 6.351 Is that clear? 



#13
Feb2512, 08:15 PM

P: 143

pictures. that is all....




#14
Feb2612, 12:21 AM

P: 169

We have t as the station time, and we have t' as Einstein's time. We have [t,x]s and, we have [t',x']e [14,17]s = [6.351,x']t for the moment and distance of stoptime at distance x'. x' = γ(xtβ) = 1.1547(1714*0.5) = 1.1547(177) = 1.1547(10) = 11.547. thus; [14,17]s = [6.351,11.547]e The original stipulation was that when the button was pressed, the train was only 10 from the button from the station POV. Now we are talking about a 17 and an 11.547. I accepted that you took it that the stationmaster saw "10" and thus pressed the button, which wasn't the actual stipulation, but we can work with that. So now we have the stopclock at 14 when the button is pressed. Now the stipulation was also that the button was pressed when the train was 6 from the first clock, or 10 from the button. If again, you assume that the stationmaster must perceive the train to be at 6 from the first clock, then it would take 10 more for him to perceive that (complicating it further) and leading to the assessment that the button was pressed when the train was 5 closer to the button, making it 5 away. The button is pressed at t=10 and Einstein is only x=10 away (from station POV) or x=5 if you assume the perception issue with the stationmaster. Either way, I am not seeing how you are now coming up with him being 11.547 away from his POV. He cannot be perceiving himself any further way than the stationmaster perceives him. You seem to be saying that the station sees Einstein 10 away from the button when Einstein sees the button 11.547 away from himself. Those distances can't be different. 



#15
Feb2612, 12:52 AM

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(1) The readings shown by the stopclocks must be the same for all observers, since they are stopped by definite events (photons hitting them after traveling from a common starting event) and the time they flash after they are stopped is the time of those definite events in a particular frame (the clocks' rest frame). So the answers to all three of the questions above must be the same. (Note that Einstein, who is moving relative to the stopclocks and the stop button, will assign different *coordinate times* to the events of photons hitting each stopclock, but that's not what the above questions ask forthey ask specifically for the time "flashed from each clock", not the coordinate time assigned to the event "photon hits clock".) (2) Since the stopclocks and the stop button are all at rest with respect to each other, and each stopclock is specified to be the same distance (4 uls) from the stop button in the frame in which they are mutually at rest (this isn't explicitly stated in the OP but I can't see any other way of making sense of the OP), then the time shown by both stopclocks must be the same, and must be 4 us after the time the stop button is pushed (again, "time" being in the frame in which the button and both clocks are mutually at rest). 



#16
Feb2612, 01:48 AM

P: 169

Obviously the issues of relativity are the issue. The question proposes that you calculate the simple case for the station frame. And it also proposes that you calculate from the train's frame. Of course it is expected that the times would be the same. But I have yet to see that both calculations work out to be the same. I haven't finished with ghwellsjr's argument yet, but it appears that he is simply reverse calculating from the presumed station's POV so as to make the answer come out right. In the long run, I'm pretty sure that fails. But thus far, it feels like a shell game of switching notations around. We are still working on it. 



#17
Feb2612, 03:15 AM

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You should not think of the events in Einstein's frame as belonging exclusively to Einstein or associated with him. They are merely coordinates in his established Frame of Reference. So any event anywhere and at any time in the station frame also has coordinates in Einstein's frame but that isn't meant to imply that they are located where Einstein is. They can be in front of him or behind him. Also again, I made no assumption with regard to the stationmaster perceiving the arrival of the train 10 μls away from him. I was just following your statement that he doesn't know when to press the button. So any of your comments regarding timing of events different from what I outlined in post #7 should not be brought up now. You agreed to them. But you are making a mistake when you say that the button is pressed at t=10. You earlier stated "So now we have the stopclock at 14 when the button is pressed" so why are you changing your mind now? If you want to determine how far away Einstein was from the button when it was pressed, one way to do this is to start with the coordinate time of when Einstein arrived at the button. I worked this out near the end of post #10 and it's 29.445. Now you subtract from this the coordinate time of when the button was pressed: 29.44512.124=17.321. Since Einstein is traveling at 0.5c we can calculate the distance as 17.321*0.5=8.660. So when you say that the distance the station sees Einstein away from the button cannot be different than the distance Einstein sees himself from the button, you are wrong. This is another fundamental concept of Special Relativity, that distances are relative, specifically, length contracted. The distance of 10 in the station frame becomes 10/γ or 10/1.1547 which equals 8.66. 



#18
Feb2612, 08:47 AM

P: 169

It seems that if neither party can know that they are moving, then you have symmetry between their perceptions. There can be no preference involved. All we have (now) is when he pressed the button the train was 6 μls away from the first clock. If point A sees point B to be 6 μls away, how can point B see point A to be a different distance away? The speed of travel is the same for both of them. One can't have preference such as to be shorter or longer regardless of any clocks. And any calculation that would apply to one frame would equally apply to the other's. It could have equally been stated that the first stop clock was 6 μls away from Einstein when the button was pressed. That one number is the only thing actually affixed in the scenario. Everything else is a frame dependent calculation. You have to have something with which to start that is common for both frames. It makes more sense to me to make that first clock x = 0 = x'. 


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