Relativity and The Stopped Clock Paradox


by James S Saint
Tags: paradox, relativity, stopped clock
ghwellsjr
ghwellsjr is online now
#19
Feb26-12, 08:58 AM
PF Gold
P: 4,525
Quote Quote by James S Saint View Post
This is a critical part. The rest we can hash out.

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 calculation.
Of course it's symmetrical but you didn't discuss the length of the train or have any other events on the train. If you had, then I could show you how in the station frame, the length of the train is length contracted and from the train frame the length of the station is contracted. If you would have established a symmetrical scenario, then you would see that everything is symmetrical.
James S Saint
James S Saint is offline
#20
Feb26-12, 09:07 AM
P: 169
Quote Quote by ghwellsjr View Post
Of course it's symmetrical but you didn't discuss the length of the train or have any other events on the train. If you had, then I could show you how in the station frame, the length of the train is length contracted and from the train frame the length of the station is contracted. If you would have established a symmetrical scenario, then you would see that everything is symmetrical.
I understand that the other lengths are all CONTRACTED, not lengthened. But do we agree that at the button press moment, the train is at 6 μls for both frames?
ghwellsjr
ghwellsjr is online now
#21
Feb26-12, 09:31 AM
PF Gold
P: 4,525
Quote Quote by James S Saint View Post
I understand that the other lengths are all CONTRACTED, not lengthened. But do we agree that at the button press moment, the train is at 6 μls for both frames?
No and I didn't say any length was lengthened. I showed that the length between Einstein and the button at the moment of the press was 10 in the station frame and 8.660 in Einstein's frame.
James S Saint
James S Saint is offline
#22
Feb26-12, 10:03 AM
P: 169
Quote Quote by ghwellsjr View Post
No and I didn't say any length was lengthened. I showed that the length between Einstein and the button at the moment of the press was 10 in the station frame and 8.660 in Einstein's frame.
So what are you specifying as common between them so as to make any calculation at all?
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.

Quote Quote by ghwellsjr View Post
[6.351,11.547] Station-master when the stop-button is pressed.
That 11.547 is a problem since the stationmaster, at the same instant, sees the distance as only 10. As you have just now said, it definitely has to be only 8.66.
PeterDonis
PeterDonis is offline
#23
Feb26-12, 11:50 AM
Physics
Sci Advisor
PF Gold
P: 5,507
Quote Quote by James S Saint View Post
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 don't understand why a calculation from the train's frame is even relevant. Read carefully what I posted: you are asking *what time the stop-clocks actually flash once they are stopped*; in other words, you are asking about the actual digits visible on the stop-clocks' LEDs, or which points on the stop-clocks' faces the clock hands point to. That observable depends *only* on the "time" as measured in the stop-clocks' rest frame, since that's what determines the time actually displayed by the stop-clocks; the train frame is irrelevant. The *time coordinates* assigned in the train frame to the events "stop-clock 1 receives a photon and stops" and "stop-clock 2 receives a photon and stops" will be different, but that's not what you are asking for.
darkhorror
darkhorror is offline
#24
Feb26-12, 11:52 AM
P: 141
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.

Quote Quote by James S Saint View Post
I understand that the other lengths are all CONTRACTED, not lengthened. But do we agree that at the button press moment, the train is at 6 μls for both frames?
That is not possible. I assumed that the information given at the beginning was from the stations frame of reference. In the stations frame of reference you have a point on the track that is 6 μls away from the first stop-clock. This is when the button is pushed. In Einstein's frame of reference that distance is length contracted by .866 so in Einstein's frame of reference that distance is .866*6 = 5.196 μls away. Also in Einstein's frame of reference the distance between the stop button and the clocks is 4*.866 = 3.464 μls. Edit: but since in that frame those points are moving it takes differnet times.
PeterDonis
PeterDonis is offline
#25
Feb26-12, 12:50 PM
Physics
Sci Advisor
PF Gold
P: 5,507
Quote Quote by ghwellsjr View Post
Now we use the Lorentz Transform to convert the coordinates of these events into Einstein's rest frame.
This version of the LT is only valid if the two frames have a common origin: in other words, if Einstein is just passing the station-master (assuming the station-master is at the spatial origin, x = 0, in the station frame) when both Einstein's and the station's clocks read 0. I don't think that's true of the OP's scenario, so all the LT formulas become more complicated since they have to include the offset of one frame's origin relative to the other. As far as I can see, all of the calculations in this thread need to be re-done because of this.
James S Saint
James S Saint is offline
#26
Feb26-12, 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...

Quote Quote by PeterDonis View Post
This version of the LT is only valid if the two frames have a common origin: in other words, if Einstein is just passing the station-master (assuming the station-master is at the spatial origin, x = 0, in the station frame) when both Einstein's and the station's clocks read 0. I don't think that's true of the OP's scenario, so all the LT formulas become more complicated since they have to include the offset of one frame's origin relative to the other. As far as I can see, all of the calculations in this thread need to be re-done because of this.
There you go. The presumption that when t=0, t'=0 corrupts the calculations.

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.
darkhorror
darkhorror is offline
#27
Feb26-12, 05:28 PM
P: 141
Quote Quote by James S Saint View Post
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...


There you go. The presumption that when t=0, t'=0 corrupts the calculations.

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.
From the stations frame of reference the train IS moving. Just like from the trains frame of reference the station IS moving. In the train's frame of reference it know's that the station is moving, thus it knows that when it looks at the station it's looking at a station that is length contracted. Thus to figure out the length of the station in it's own frame of reference if it's length is 5.196 in the train's. You take 5.196*1/.866 = 6 μls is the distance in the station's frame of reference.

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.
ghwellsjr
ghwellsjr is online now
#28
Feb27-12, 05:03 AM
PF Gold
P: 4,525
Quote Quote by James S Saint View Post
Quote Quote by ghwellsjr View Post
No and I didn't say any length was lengthened. I showed that the length between Einstein and the button at the moment of the press was 10 in the station frame and 8.660 in Einstein's frame.
So what are you specifying as common between them so as to make any calculation at all?
I am guessing that you have inadvertently specified that t = 0 at the same moment that t' = 0.
It wasn't inadvertent--it was purposeful. Of course the origins of the two reference frames have to coincide in order to use the standard configuration of the Lorentz Transformation. And it's not just t=t'=0, it's also x=x'=0 and y=y'=0 and z=z'=0.
Quote Quote by James S Saint View Post
But if you do that, the distance 6 μls will not work out properly.
What's not working out properly? I have no idea what your concern is. What are you saying that I'm doing that is not working out properly and what do you say is the right thing to do to make it work out properly?
Quote Quote by James S Saint View Post
In effect, you will be merely converting feet to meters and back again so of course you would get the same results.
Please show me the calculations that I'm doing that you don't agree with.
Quote Quote by James S Saint View Post
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.
Of course I don't know all these things because you won't tell me. I asked you to provide details of your scenario but you refused. When I provided some, you agreed that they were OK, but now you don't like them.

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 non-issue?
Quote Quote by James S Saint View Post
Quote Quote by ghwellsjr View Post
[6.351,11.547] Station-master when the stop-button is pressed.
That 11.547 is a problem since the stationmaster, at the same instant, sees the distance as only 10. As you have just now said, it definitely has to be only 8.66.
If you go back and look at the listing of events in the station frame (post #7) you will see four of them for different places when the stop-button is pressed. Notice that they all share the same time coordinate of 14. This is a requirement in any Frame of Reference when you want to see how far apart different things are--you have to find events in which the time coordinates are all the same. The fact that when these four events are transformed into Einstein's frame they end up with different time coordinates is demonstrating the relativity of simultaneity--simultaneous events in one frame may not be simultaneous in another frame. That's why I had to do some extra computations to figure out what the various lengths were in Einstein's frame.
ghwellsjr
ghwellsjr is online now
#29
Feb27-12, 05:03 AM
PF Gold
P: 4,525
Quote Quote by James S Saint View Post
Quote Quote by PeterDonis View Post
This version of the LT is only valid if the two frames have a common origin: in other words, if Einstein is just passing the station-master (assuming the station-master is at the spatial origin, x = 0, in the station frame) when both Einstein's and the station's clocks read 0. I don't think that's true of the OP's scenario, so all the LT formulas become more complicated since they have to include the offset of one frame's origin relative to the other. As far as I can see, all of the calculations in this thread need to be re-done because of this.
There you go. The presumption that when t=0, t'=0 corrupts the calculations.
What are you talking about? Peter just said that this version of the LT (he's talking about the standard configuration) is only valid if the two frames have a common origin which means t=0 and t'=0 (and the same for the x, y, and z coordinates as I mention on the previous thread). He then went on to suggest one way to do this but realized that wouldn't fit your scenario and then said there needed to be an offset between the two frame's origins. Well I accomplished the same thing by providing an offset between the station masters location and the station frame's origin. If you had provide details that you cared about when I asked for them, we could have avoided all these issues that you are now raising.

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 stop-clocks by the tracks.
Quote Quote by James S Saint View Post
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.
No, there is a distance in the station frame that you said was 6 μls. That distance when measured by Einstein is 5.196. It's actual length is frame dependent. In the station frame, it's 6, in Einstein's frame it's 5.196. When Einstein measures it in either frame (or any other frame), he gets 5.196. When the station master measures it in either frame (or any other frame), he gets 6. Nobody measures it as 4.5 although there are frames of reference in which it is 4.5 but not one of the ones we have considered so far.
Quote Quote by James S Saint View Post
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.
I think this issue has already been dealt with. Just keep in mind that the times for events that got transformed are coordinate times on imaginary clocks and if you want to give Einstein a real clock, you can add or subtract an offset to relate it to the coordinate clocks.
Quote Quote by James S Saint View Post
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.
We know who is moving. You told us. But we can transform your description based on the rest frame of the station into any other frame of our own choosing and it will be just as valid as the one you started with.
James S Saint
James S Saint is offline
#30
Feb27-12, 10:57 AM
P: 169
Quote Quote by ghwellsjr View Post
No, there is a distance in the station frame that you said was 6 μls. That distance when measured by Einstein is 5.196. It's actual length is frame dependent. In the station frame, it's 6, in Einstein's frame it's 5.196. When Einstein measures it in either frame (or any other frame), he gets 5.196. When the station master measures it in either frame (or any other frame), he gets 6. Nobody measures it as 4.5 although there are frames of reference in which it is 4.5 but not one of the ones we have considered so far.

.
.

We know who is moving. You told us. But we can transform your description based on the rest frame of the station into any other frame of our own choosing and it will be just as valid as the one you started with.
Now I am seeing a common misunderstanding.

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.
PeterDonis
PeterDonis is offline
#31
Feb27-12, 11:50 AM
Physics
Sci Advisor
PF Gold
P: 5,507
Quote Quote by ghwellsjr View Post
Well I accomplished the same thing by providing an offset between the station masters location and the station frame's origin.
Ah, ok, this would take care of it. So basically you are defining the origin of the station frame (t=0, x=0, y=0, z=0) as the event where Einstein is at some point *other* than the station master's location; and at that event, the station's clock reading *and* Einstein's clock reading are both taken to be zero?
PeterDonis
PeterDonis is offline
#32
Feb27-12, 11:54 AM
Physics
Sci Advisor
PF Gold
P: 5,507
Quote Quote by James S Saint View Post
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.
This is true: you should be able to calculate what the stop-clocks' displayed times will be in any frame. Doing it in the station frame is simplest because the clocks are at rest in that frame, so the time they will display corresponds to coordinate time in that frame, and that's easy to calculate from the statement of the problem.

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 stop-clock 1" and "photon hits stop-clock 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.
PeterDonis
PeterDonis is offline
#33
Feb27-12, 11:57 AM
Physics
Sci Advisor
PF Gold
P: 5,507
Quote Quote by James S Saint View Post
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.
Well, since your original statement of the problem only gave coordinates in the station frame, how else are we to proceed? The ultimate answer we want is the coordinate time in the station frame for two specific events, since that's what the stop-clocks will display. Taking the roundabout route of transforming into Einstein's frame, seeing how things look from his perspective, then transforming back, is not necessary to calculate the answer; as I've said before, that can be done entirely within the station frame. But it does illustrate why the answer must be the same regardless of whose frame is used to compute it.
darkhorror
darkhorror is offline
#34
Feb27-12, 12:27 PM
P: 141
Quote Quote by James S Saint View Post
Now I am seeing a common misunderstanding.

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.
But we know exactly who is moving and who isn't moving. In the train's frame of reference the station is moving. In the stations frame of reference the train is moving. Who is moving and who isn't moving just depends on the frame of reference you choose. Both are equally valid.

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.
No that is not right at all, Both rockets would see the other's clock moving slower than their own. Just as the train sees the stations clock moving slower than it's own, and the station see's the train's clock moving slower than it's own.

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.
ghwellsjr
ghwellsjr is online now
#35
Feb27-12, 12:42 PM
PF Gold
P: 4,525
Quote Quote by James S Saint View Post
Now I am seeing a common misunderstanding.

There are only 2 frames.
There are an infinite number of frames, all equally valid for describing, illustrating, analyzing, etc any given scenario. You have arbitrarily chosen to focus on two of them and you didn't even specify them, you left that up to me and when I did it, you first accepted them but now you are rejecting them.
Quote Quote by James S Saint View Post
The stationmaster can only measure within his own. Einstein can only measure within his own.
No, the stationmaster does not need a frame in which to make measurements, neither does Einstein. We use frames to help us analyze in thought problems what the stationmaster would measure and what Einstein would measure if they were actually carrying out the scenario. But we only need one frame to do this and it doesn't matter which one it is, except of course that the calculations can be very easy using one frame and very difficult using a different frame, as evidenced by my posts on this thread. Since you have described the scenario in a general way from the POV in which the station is at rest, that is the easiest one in which to do a rigorous specification of a frame and events in it and then to do the calculations, because we don't need to do any transformations. We can analyze exactly what the stationmaster would measure (without him using any concept of a frame--he just measures) and exactly what Einstein would measure (without him using any concept of a frame--he just measures).
Quote Quote by James S Saint View Post
In any scenario the travel speed is between the two frames.
Not if we don't want it to be.

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):
If a material point is at rest relatively to this system of co-ordinates, its position can be defined relatively thereto by the employment of rigid standards of measurement and the methods of Euclidean geometry, and can be expressed in Cartesian co-ordinates.

If we wish to describe the motion of a material point, we give the values of its co-ordinates as functions of the time.
I don't know why you think a single frame doesn't allow motion of objects or observers in it.

We can then pick another frame and if we choose to limit ourselves to the standard convention, we assign an arbitrary speed along the x-axis 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 x-axis in the first frame but are stationary in the second frame.
Quote Quote by James S Saint View Post
It never belongs to just one frame.
Nothing belongs to just one frame, not even an observer at rest in the frame. Everything is in all frames...all of the infinity of equally valid frames.
Quote Quote by James S Saint View Post
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.
What you are describing is the situation prior to Einstein's Special Relativity. All we could know is what each observer could see and measure. But no observer can tell what time is on another moving observer's clock, only what he can see after the image of that clock is transmitted to him at the speed of light. Einstein's Special Relativity allows us to arbitrarily assign coordinate times to distant locations and define what time is on another moving observer's clock at any particular coordinate time. But these coordinate times are dependent on the particular frame we decide to use and will change when we pick a different one in motion to the first one.
Quote Quote by James S Saint View Post
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.
Of course we will get the same answer when we are calculating what each observer sees and measures or what a clock displays when a photon hits it and things like that. But the arbitrary coordinate times and distances are just that--arbitrary and can be different in each frame.
Quote Quote by James S Saint View Post
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.
If you want to talk about two identical rockets traveling toward or away from each other, then we have an exactly symmetrical situation. Each one sees and measures exactly what the other one sees and measures and this is independent of any frame we use to describe and analyze the situation. Each one sees the other ones clock running slower and the other ones rocket as being shorter. But without selecting a frame, neither one can say (nor can we say) what time is on the other ones clock corresponding to any time on their own. Nor can they say how far away either one is at any given time on their own clock. We need SR for that (or some other equivalent theory). And depending on which frame you select, you can get different answers for those coordinates,.especially those that are remote from each observer.

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.
ghwellsjr
ghwellsjr is online now
#36
Feb27-12, 12:47 PM
PF Gold
P: 4,525
Quote Quote by PeterDonis View Post
Quote Quote by ghwellsjr View Post
Well I accomplished the same thing by providing an offset between the station masters location and the station frame's origin.
Ah, ok, this would take care of it. So basically you are defining the origin of the station frame (t=0, x=0, y=0, z=0) as the event where Einstein is at some point *other* than the station master's location; and at that event, the station's clock reading *and* Einstein's clock reading are both taken to be zero?
Sort of, but please note, James did not specify a station clock, especially one located with the station master. I did it the way I did so that the coordinate clocks for the station frame would have the same time on them as the two stop-clocks located remotely from the station master.


Register to reply

Related Discussions
relativity train and clock synchronization Special & General Relativity 5
matter-antimatter 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 Bug-Rivet Paradox paradox in special relativity? Special & General Relativity 4
definition of a clock in relativity theory Special & General Relativity 3