Register to reply 
Relativity and The Stopped Clock Paradox 
Share this thread: 
#19
Feb2612, 08:58 AM

PF Gold
P: 4,684




#20
Feb2612, 09:07 AM

P: 169




#21
Feb2612, 09:31 AM

PF Gold
P: 4,684




#22
Feb2612, 10:03 AM

P: 169

I am guessing that you have inadvertently specified that t = 0 at the same moment that t' = 0. But if you do that, the distance 6 μls will not work out properly. In effect, you will be merely converting feet to meters and back again so of course you would get the same results. At t=0, t' ≠ 0. Einstein's t'=0 is an irrelevant mark. He merely needs to measure the change in time from the 6 μls mark. You don't know any history for Einstein's travel, so you can't know nor care what Einstein's clocks might read. I would suggest making them read the same as the station's at that one point of 6 μls (t=14=t') away from the first clock just to clearly see the differences as time moves forward from that point. 


#23
Feb2612, 11:50 AM

Physics
Sci Advisor
PF Gold
P: 6,041




#24
Feb2612, 11:52 AM

P: 140

To answer your questions at the start, everyone will see those clocks reading 18 μs. Since there is nothing you need to do relativity wise, as Einstein's frame of reference never comes into play. All you need to know is that you press the button when it's reading 10 μs. And it takes 4 μs for the pulse to get to the clocks so they stop plus it takes the light 4 μs to get from the clock to the station manager so when the manager actually hits the stop button he is 4 μs behind the clocks.



#25
Feb2612, 12:50 PM

Physics
Sci Advisor
PF Gold
P: 6,041




#26
Feb2612, 04:35 PM

P: 169

This is the fourth time I have had to rewrite this due to a "500 Internal Server Error" so I'm sure I left something out... grrr...
As far as that 6 μls; If we take it that from the station's frame we have 6 between the clock and the train and thus must have 5.196 from the train's POV, then we have presumed that it is the train that is moving. If we take it that the station is moving and that from the train's POV the distance is 5.196, then we have to say that the station's POV is going to be 5.196*.866, yielding 4.5 μls for the station. There has to be a defined common state for both frames somewhere. In the typical scenario when [t,x] = [0,0] then [t',x'] = [0,0]. That was Lorenz' formulation presumption. But we have no idea of Einstein's clock readings and thus they cannot be used as a fact for calculation so directly using Lorenz. If we say that when the button is pressed, the train's Clock is at t'=0, we get different results. In these scenarios we are really only concerned with the change in time readings thus we have to use something else as the common state. It shouldn't matter what Einstein's clocks read at any one moment as long as they remain consistent from there forward. In this scenario, the only thing common that we can use is that distance and if we don't use it, we have a conundrum of not knowing who is moving such as to indicate which clocks are slower. 


#27
Feb2612, 05:28 PM

P: 140

Just as if in the stations frame of reference the train was 10 meter, then in the trains frame of reference the train would be 1/.866 = 11.547 meters. To make it simple, if something is moving in a frame of reference then in that frame of reference that something is length contracted. If instead you look at a frame of reference where that something is not moving then it won't be length contracted in that frame. 


#28
Feb2712, 05:03 AM

PF Gold
P: 4,684

You should also recognize that the times for the events in Einstein's frame are coordinate times on those imaginary clocks that the real Einstein said we needed in order to establish a frame of reference. I never said that Einstein actually carried a clock but even if he did, I agree that what is important is differences in times on that clock. So do you agree that this is a nonissue? 


#29
Feb2712, 05:03 AM

PF Gold
P: 4,684

Please note that Peter is also talking about a station with a clock even though you never described the station, just a station master with no clock and two stopclocks by the tracks. 


#30
Feb2712, 10:57 AM

P: 169

There are only 2 frames. The stationmaster can only measure within his own. Einstein can only measure within his own. In any scenario the travel speed is between the two frames. It never belongs to just one frame. In effect, that is the whole point of relativity. And it is by that thought, that we cannot say that it is "merely" the train moving and thus the station's time is faster. Nor could we have said that it is "merely" the station that is moving and thus the train's time will be faster. We never know who is moving and who isn't. We choose one or the other for sake of calculations at that moment. But those calculations MUST end up with the same results even if we had picked the other as the one moving, because the actual motion is BETWEEN them. That is very fundamental in relativity. I believe that your calculations have translated the original into a different perspective using Lorenz and time, but then merely translated them back using Lorenz and distance, so obviously you will end up with the same answer. If I had not said that it was a train traveling at .5c, but instead had the "two rockets" scenario and one of their clocks read 14, how would you justified that the other clock was slower or faster? The same is true with a given single distance. If one rocket sees it as 6, the other must also see it as 6. 


#31
Feb2712, 11:50 AM

Physics
Sci Advisor
PF Gold
P: 6,041




#32
Feb2712, 11:54 AM

Physics
Sci Advisor
PF Gold
P: 6,041

Doing the calculation in any other frame requires you to calculate, from coordinates of events given in *that* frame, what the coordinate time in the station frame is for those same events. In particular, for the events "photon hits stopclock 1" and "photon hits stopclock 2". I'm still not entirely clear if that is what you are trying to calculate. If it is, then yes, you should get the same answer doing it this way as you get doing it the simple way, using just the coordinates in the station frame. Btw, a very useful tool for these types of problems is a spacetime diagram; has anyone tried to draw one for this scenario? I have a feeling that would help clear up a lot of misunderstandings. 


#33
Feb2712, 11:57 AM

Physics
Sci Advisor
PF Gold
P: 6,041




#34
Feb2712, 12:27 PM

P: 140

Now for the distance it's easy to calculate. If one rocket sees a distance of 6 between two points that are not moving in it's frame of reference, then the other rocket sees that length contracted to 5.176. The problem you are having is that you then you want to say oh well it's length is 5.176 in my frame there for in your frame it's length contracted so it's length is 4.482. The problem with this is that the length you are measuring is moving in your frame of reference. Thus you know that if you switch to the frame of reference where the points aren't moving then you take 5.176 * 1/.866 which gives you 6. Here is another way to look at this. Take 2 points, lets name them A and B. Two frames of reference named 1 and 2. A and B are stationary with respect to each other. In frame 1 A and B are not moving. In frame 2 A and B are moving at .5c. In frame 2 A and B are 1 light year away from each other. How far are they away from each other in frame of reference 1? Well since A and B are moving in frame 2 and are not moving in frame 1. They are 1*(1/.866) away or 1.1547 light years away in frame 1. Now lets take that 1.1547 distance in frame 1, and you want to know how far they are away in frame 2. Since you know A and B are not moving in frame 1, and are moving at .5c in frame 2. This means you take 1.1547 * .866 which means you get 1 light year in frame 2. Which is what we had started with. So we know that in frame 1 A and B are 1.1547 light years away, and in frame 2 A and B are 1 light year away. They are both true, and both equally valid. 


#35
Feb2712, 12:42 PM

PF Gold
P: 4,684

We start with any frame which we consider to be at rest and in which we describe all observers, objects, clocks, whatever. They can be located anywhere and doing anything in terms of moving and/or accelerating. In the first section of Einstein's 1905 paper he says (emphasis his): We can then pick another frame and if we choose to limit ourselves to the standard convention, we assign an arbitrary speed along the xaxis but we fix the directions of the three axes to be the same and the origins (all four coordinates equal zero) of the two frames to be coincident. We don't have to pick a speed that corresponds to the travel speed of an observer or an object in our original frame but there are certain advantages to doing that, mainly that we don't have to then do another step of calculation to determine the Proper Times on the clocks that were moving at a constant speed along the xaxis in the first frame but are stationary in the second frame. So, if you choose the rest frame for the first rocket, then his clock is defined to be running at the same rate as the coordinate time for the frame and his rocket is defined to be the same length as the coordinate distance for the frame but for the second rocket, his clock is calculated to be running at a slower rate than the coordinate time and his rocket is calculated to be shorter than the coordinate distance. The factor gamma which is based on their relative speed determines that amount of change and since both rockets measure the other one to be going the same speed (but opposite directions), then when we switch to the rest frame of the other rocket, all the calculations switch between the two rockets. 


#36
Feb2712, 12:47 PM

PF Gold
P: 4,684




Register to reply 
Related Discussions  
Relativity train and clock synchronization  Special & General Relativity  5  
Matterantimatter ship in GR clock paradox  fuel consumption  Special & General Relativity  1  
Clock synchonization in relativity  Advanced Physics Homework  1  
What is the resolution of the The BugRivet Paradox paradox in special relativity?  Special & General Relativity  4  
Definition of a clock in relativity theory  Special & General Relativity  3 