Are Both Coordinate Systems in Special Relativity Considered Inertial?

[JM]

So does the result of post 49 apply also to coordinates k, namely 'when anyone clock at rest with k reads a specific time , such as ct =8, then all clocks at rest with k also read '8''?

 

[JM]

May I assume that the answer to post 51 is yes, with the qualifiers given in post 50? If so, can it be concluded that when K is at rest the relation t = T/m holds when comparing the clock at rest at the origin of K with the clock of k that is nearby?

 

[JM]

The point of this is the following:
1, When K is at rest the c lock of k that is near to the origin of K is slow compared to the clock at the origin of K, and
2. When k is at rest the clock at the origin of K is slow compared with the clock of k that is near the origin of K.
The question is whether this result is explainable and accepted using SR, or whether the two results are contradictory and impossible, as some writers suggest?

 

[DrGreg]

The point of this is the following:
1, When K is at rest the c lock of k that is near to the origin of K is slow compared to the clock at the origin of K, and
2. When k is at rest the clock at the origin of K is slow compared with the clock of k that is near the origin of K.
The question is whether this result is explainable and accepted using SR, or whether the two results are contradictory and impossible, as some writers suggest?

It seems contradictory only if you haven't grasped what the relativity of simultaneity is about. To compare two clocks that are separated by a distance, you have to measure both clocks at the same time. Different observers disagree over what "at the same time" means, and there is no definition that everyone can agree on. So that's why one observer says k is slower than K and another observer says K is slower than k. They are both right according to their own definition of simultaneity.

 

[JM]

Thanks for looking in DrGreg,

To compare two clocks that are separated by a distance, you have to measure both clocks at the same time.

I'm not sure what youre referring to here. In my post in both cases 1. and 2. the comparison is between the clock at the origin of K and the clock of k that is immediately next to the origin of K.
I don't think the results are contradictory but I read it stated in a reference as an absolute, and I'm trying to find out how y'all explain it.

JM

 

[Al68]

The point of this is the following:
1, When K is at rest the c lock of k that is near to the origin of K is slow compared to the clock at the origin of K, and
2. When k is at rest the clock at the origin of K is slow compared with the clock of k that is near the origin of K.
The question is whether this result is explainable and accepted using SR, or whether the two results are contradictory and impossible, as some writers suggest?

Those statements are true, but the clocks cannot be local at both the beginning and end of a measurement unless the relative velocity is zero. It makes no sense to say that each clock is slower than the other over a distance of zero and an elapsed time of zero.

So each statement is true for any non-zero time period, meaning that the distance between the clocks must be non-zero either at the beginning or end of the time period.

 

[JM]

Thanks for coming in Al68.

Those statements are true, but the clocks cannot be local at both the beginning and end of a measurement unless the relative velocity is zero. It makes no sense to say that each clock is slower than the other over a distance of zero and an elapsed time of zero.

So each statement is true for any non-zero time period, meaning that the distance between the clocks must be non-zero either at the beginning or end of the time period.

Agreed. The clocks are not together at the beginning of the time period.

The problem considered here is based on Einsteins slow clock calculation, as described in posts 3 and 4 of this thread. The clocks of both K and k are set to zero when the origins of each coincide. Afte a time T the origin of k has moved a distance vT along the X axis of K. At this time there will be a clock of k that is coincident with the clock at the origin of K. The question is how each observer, K and k, will view the comparative time of the two clocks. So are the statements you refer to still true? And how does one work out the answer?

 

[JM]

DrGreg- Specific example-

It seems contradictory only if you haven't grasped what the relativity of simultaneity is about.

I agree that it is important to understand the Relativity of Simultaneity (ROS). But its not clear to me how the principle enters the problem under consideration. I have restated the problem in the just previous post. Would it be possible to describe the solution to this problem so we can all see how its done?

This is not a homework problem.

 

[JesseM]

Thanks for looking in DrGreg,
I'm not sure what youre referring to here. In my post in both cases 1. and 2. the comparison is between the clock at the origin of K and the clock of k that is immediately next to the origin of K.
I don't think the results are contradictory but I read it stated in a reference as an absolute, and I'm trying to find out how y'all explain it.

JM

In order to talk about the rate a clock is ticking in a given frame, you must consider two points on that clock's worldline, and compare the time that elapsed on the clock between those points with the coordinate time between those points in your chosen frame. So even if the first point is the one where K and k are at a common location, the second point on either clock's worldline will have to be one where that clock has moved away from the other clock, and simultaneity issues come into play. For example, if both clocks read 0 when they are at the same position, and in K's frame the k clock reads 0.08 nanoseconds simultaneously with the K clock reading 0.1 nanosecond, then this means that in K's frame the k clock is slowed down by a factor of 0.8; but in k's frame the event of the k clock reading 0.08 nanoseconds is not simultaneous with the K clock reading 0.1 nanosecond, instead it is simultaneous with the K clock reading 0.064 nanoseconds, so in k's frame it is the K clock that is running slow by a factor of 0.8. These simultaneity disagreements are of central importance no matter how tiny the interval between the first reading and the second (even if it is considered to be infinitesimal, so you're talking about the instantaneous rate of ticking between some time t and t + dt).

 

[Al68]

Thanks for coming in Al68.
Agreed. The clocks are not together at the beginning of the time period.

The problem considered here is based on Einsteins slow clock calculation, as described in posts 3 and 4 of this thread. The clocks of both K and k are set to zero when the origins of each coincide. Afte a time T the origin of k has moved a distance vT along the X axis of K. At this time there will be a clock of k that is coincident with the clock at the origin of K. The question is how each observer, K and k, will view the comparative time of the two clocks. So are the statements you refer to still true? And how does one work out the answer?

Yes both statements would be true. Each clock would run slow in the other frame. Notice that relativity of simultaneity now comes into play, so if we define an event (after the origin of K and k coincide) as clock A (at rest in K) reading t, then clock B (at rest in k) will read t2 (less than t) simultaneous (in K) with clock A reading t. But the events of clock A reading t and clock B reading t2 are not simultaneous in k. In k, clock A reading t is simultaneous with clock B reading t3 (more than t).

The situation would be similar if we defined an event prior to the clocks passing. Either way, we have two events, one in which the clocks are local, the other in which they are not.

I notice in your post 3 you say that the twins scenario can be considered to have only two parts: 1. the twins separate, and 2 the twins re-unite.

This is true, but the standard SR equations are not valid in the ship's non-inertial frame in that case, because they are only valid in inertial (unaccelerated) reference frames.

That's why the turnaround is usually treated as a separate event, because then the standard SR equations can be used between each events, since each twin is unaccelerated between the events.

I always recommend Einstein's own twins paradox resolution of 1918, not because it's better or simpler than the rest, but because it directly addresses the point of concern many people have, instead of "dodging" it. And he doesn't bother with the math, he just explains the physics of the situation. And he explains it using the non-inertial frame in which the ship is stationary the whole time.

 

[JesseM]

Agreed. The clocks are not together at the beginning of the time period.

The problem considered here is based on Einsteins slow clock calculation, as described in posts 3 and 4 of this thread. The clocks of both K and k are set to zero when the origins of each coincide. Afte a time T the origin of k has moved a distance vT along the X axis of K. At this time there will be a clock of k that is coincident with the clock at the origin of K. The question is how each observer, K and k, will view the comparative time of the two clocks.

When two clocks coincide at the same point in time and space, all frames always agree on what their respective readings are...otherwise you could get genuinely different physical predictions, like if one clock was set to explode when it reached a different time and different frames disagreed about whether another clock would be next to it when that happened. So both frames agree that the clock at the origin of K is behind the clock of k that it's next to at the moment that the k-clock reads at time T (the clock at the origin of K will read T*\sqrt{1 - v^2/c^2} at this moment). However, in the K frame note that this does not imply the clock at the origin of K was ticking slower than that k-clock, since in the K frame that k-clock would have already shown a time later than zero at the moment the two origins coincided and the clocks at the origins both read zero (again, the relativity of simultaneity).

 

[JM]

According to the definition of simultaneity in their inertial rest frame .

JesseM- What is the definition of simultaneity for clocks at rest in K?

 

[JM]

I always recommend Einstein's own twins paradox resolution of 1918, not because it's better or simpler than the rest, but because it directly addresses the point of concern many people have, instead of "dodging" it. And he doesn't bother with the math, he just explains the physics of the situation. And he explains it using the non-inertial frame in which the ship is stationary the whole time.

Al68-Thanks for your suggestion. Can you give me a more exact reference to the paper of 1918?
Thanks.
Jm

 

[JM]

DrGreg, JesseM, Al68- I appreciate your efforts to respond to my question. But I'm still having a problem putting it all in context. Can you suggest a tutorial on ROS? There seems to be agreement that each observer can see the others clock to be running slow, even when the clocks of K and k are co-located. But there is also a suggestion that when co-located both observers must see the same relation, i.e. if K sees k running slow then k must see K running fast, or did I misunderstand?

I hope for further discussion.
JM

 

[JesseM]

JesseM- What is the definition of simultaneity for clocks at rest in K?

Simultaneity in each inertial frame is defined using the Einstein synchronization convention, which is based on assuming that light moves at the same speed in both directions in that frame. The method Einstein originally gave was that if you have two clocks A and B at rest in the same inertial frame, and clock A sends a light signal to B when A reads T1, and B bounces back the signal when it hits it, then if A reads T2 when the bounced signal hits it, the two clocks are synchronized if B's time when the light hit it was exactly halfway between T1 and T2--i.e. B read a time of T1 + (T1 + T2)/2 when the light hit it. An equivalent procedure which I find a little more intuitive is to set off a flash at the exact midpoint of A and B, and set both clocks to read the same time at the moment the light from the flash reaches them.

With the latter procedure, it's not hard to see why different observer must disagree about simultaneity if they each assume light travels at c in all directions in their own frame. For example, suppose I am on a rocket passing you at high speed, and I want to synchronize clocks at the front and back of the rocket by setting off a flash at the midpoint of the rocket, and setting both clocks to the same time when the light reaches them. In your frame, the front of the rocket is moving away from the point where I set off the flash, while the back of the rocket is moving towards that point, so if you assume the light travels at the same speed in both directions in your frame, naturally you will conclude that the light must hit the back clock before the light hit the front clock, which means if both clocks read the same time when the light hits them, in your frame they must be out-of-sync.

It works out that if two clocks are synchronized and a distance L apart in their own rest frame, then in a frame where they are moving at velocity v (parallel to the axis between them), they will be out-of-sync by vL/c^2. So for example, suppose we have two clocks at rest in k, k0 and k1, which are synchronized and a distance of 20 light-seconds apart in the k frame, with k0 being at the origin and reading a time of 0 seconds at the moment it is next to the clock K0 at the origin of the K frame (which also reads 0 seconds at that moment). Suppose the clock K0 then moves towards k1 at 0.8c as seen in the k frame, so in the k frame it will take a time of 20 light-seconds/0.8c = 25 seconds to reach k1. So, at the time K0 reaches k1, k1 reads 25 seconds, while K0 has been slowed down by time dilation so it only reads 25*\sqrt{1 - 0.8c^2} = 25*0.6 = 15 seconds. But now if we re-analyze the same situation from the perspective of the K frame, things look different--because of length contraction, k0 and k1 are only 0.6*20 = 12 light-seconds apart in this frame, and because of the relativity of simultaneity they are also out-of-sync by vL/c^2 = 0.8c*20 l.s./c^2 = 16 seconds, with the clock at the rear (k1) being ahead of the leading clock (k0) by this amount. So in K's frame, at the moment K0 is next to k0 and both read a time of 0 seconds, k1 is 12 light-seconds away and already reads a time of 16 seconds. Since k1 is approaching K0 at 0.8c, it will take 12/0.8 = 15 seconds for k1 to reach K0 in this frame, meaning K0 will read 15 seconds at the moment k1 passes it, which is what we found before in the other frame. In this frame it is the k clocks that are slowed down by a factor of 0.6 relative to the K clocks, so during these 15 seconds, the clock k1 only ticked forward by 15*0.6 = 9 seconds. But since k1 already had a "head start" of 16 seconds when K0 was passing k0, this means that when k1 passes K0, k1 will read 16 + 9 = 25 seconds, which again matches what we found in the other frame, even though the two frames disagree about whether K0 or k1 was running slower between the event of K0 passing k0 and the event of K0 passing k1.

Can you suggest a tutorial on ROS?

This one is my favorite of the one's I've seen, but if you google "relativity of simultaneity" you should be able to find others.

There seems to be agreement that each observer can see the others clock to be running slow, even when the clocks of K and k are co-located. But there is also a suggestion that when co-located both observers must see the same relation, i.e. if K sees k running slow then k must see K running fast, or did I misunderstand?

You misunderstood that part, observers in each frame always measure clocks in the other frame to be running slow (though they can sometimes see them running fast visually due to the Doppler effect--'measure' refers to how they assign time-coordinates to events on each clock's worldline). As I said in post #59, when talking about the "rate" a clock is ticking you always need at least two measurements at different times, even if the time-interval between these measurements is infinitesimal, so it's potentially misleading to talk about what they measure at the exact instant they are co-located.

 

[JM]

JesseM, Thank you for the clear and complete analysis, and the reference.
At the moment I'm trying to reconcile the results with the results described in posts 3 and 4. Can I get some help on this?

 

[JM]

Re SR dimensions.
In ordinary applications of the equation x = a y, where a is a non-dimensional constant, the units of x and y must agree. If y is in inches then x must be in inches, for example. Is this convention followed in SR also?

 

[JesseM]

JesseM, Thank you for the clear and complete analysis, and the reference.
At the moment I'm trying to reconcile the results with the results described in posts 3 and 4. Can I get some help on this?

Do you mean posts #3 and #4 on the thread (post #4 being by HallsofIvy) or do you mean your own third and fourth posts?

Re SR dimensions.
In ordinary applications of the equation x = a y, where a is a non-dimensional constant, the units of x and y must agree. If y is in inches then x must be in inches, for example. Is this convention followed in SR also?

Yes.

 

[JesseM]

Simultaneity in each inertial frame is defined using the Einstein synchronization convention, which is based on assuming that light moves at the same speed in both directions in that frame. The method Einstein originally gave was that if you have two clocks A and B at rest in the same inertial frame, and clock A sends a light signal to B when A reads T1, and B bounces back the signal when it hits it, then if A reads T2 when the bounced signal hits it, the two clocks are synchronized if B's time when the light hit it was exactly halfway between T1 and T2--i.e. B read a time of T1 + (T1 + T2)/2 when the light hit it. An equivalent procedure which I find a little more intuitive is to set off a flash at the exact midpoint of A and B, and set both clocks to read the same time at the moment the light from the flash reaches them.

By the way, I made a little math error in this paragraph--in order to ensure that "B's time when the light hit it was exactly halfway between T1 and T2", B's time when the light hit it should be T1 + (T2 - T1)/2.

 

[JM]

Do you mean posts #3 and #4 on the thread (post #4 being by HallsofIvy) or do you mean your own third and fourth posts?

Yes.

I meant 3 and 4, where 4 is by Halls.

Following up the 'dimensions' question: Since the clocks of K and k are all set to zero when the axes origins coincide, and time is measured in seconds for both K and k, doesn't that mean that the clocks of K and k are always in synch?

Does an observer ever see his own time dilated or his own dimensions contracted? Doesnt saying that K's time is dilated and his length shortened mean that the view taken is k's?

 

[JesseM]

I meant 3 and 4, where 4 is by Halls.

OK, I already commented on post #3 in my own post #37, and I have no disagreement with what HallsofIvy says in post #4.

Following up the 'dimensions' question: Since the clocks of K and k are all set to zero when the axes origins coincide, and time is measured in seconds for both K and k, doesn't that mean that the clocks of K and k are always in synch?

No, that's where the relativity of simultaneity comes in--the two frames disagree about what distant clocks read "at the same time" that the two clocks at the origins are passing each other and both read 0. In the K frame, all the K clocks read 0 "at the same time" that the clocks at the origin are next to each other reading 0, but all the k clocks are out-of-sync with one another and only the one at the origin reads 0. In the k frame, the reverse is true--all the k clocks read 0 "at the same time" that the clocks at the origin are next to each other reading 0, but all the K clocks are out-of-sync with one another and only the one at the origin reads 0. It may help to take a look at the diagrams I drew up in this thread, showing two rulers with clocks attached at regular intervals moving past one another at relativistic speeds, with all the clocks on a given ruler being synchronized in that ruler's rest frame but out-of-sync in the other ruler's rest frame.

Does an observer ever see his own time dilated or his own dimensions contracted?

No.

Doesnt saying that K's time is dilated and his length shortened mean that the view taken is k's?

Yes, K's time is dilated from the perspective of the k frame, in K's own rest frame his clocks are running normally.

 

[JM]

Yes, K's time is dilated from the perspective of the k frame, in K's own rest frame his clocks are running normally.

I take it that your 'yes' indicates that you agree that saying 'Ks clocks are dilated' means that the analysis is taking ks viewpoint. In that case both of the calculations in your post 65 take ks viewpoint, and that is why you get the same answer for both, and why you dis agree with post 4 that says each observer sees the others clocks as slow.

 

[JM]

No, that's where the relativity of simultaneity comes in--

What I'm saying is that two clocks that are set to zero at the same time and thereafter run at the same rate will always be in synch.
I don't see that ROS has any thing to do with it. I'm not suggesting that each obserrver has to look at the other for this result to be true.

It feels to me that this thread has dead-ended at ROS. Anyone else have any comments?

DrJM

 

[JM]

The explanation for the result t = T/m is not found in the behavior of clocks, but in the 'Fundamental Principle of Relativity'.
Consider the segment of the x-axis of k extending from x=0, where a light source is located, to x=l, where a detector is located. Let a light ray be emitted when the origins of K and k coincide. From ks viewpoint the light ray reaches l at t =l/c. From Ks viewpoint the segment is moving at speed v, so the light ray reaches x =l at T = l/( c - v). So T = t/ (1-v/c). Thus T and t are the coordinates that describe the same event, the arrival of the light ray at x = l, as seen by the respective observers K and k. As measured by clocks that are always in synch.

 

[matheinste]

What I'm saying is that two clocks that are set to zero at the same time and thereafter run at the same rate will always be in synch.
I don't see that ROS has any thing to do with it. I'm not suggesting that each obserrver has to look at the other for this result to be true.

It feels to me that this thread has dead-ended at ROS. Anyone else have any comments?

DrJM

When you set the clocks to zero when the origins are colocated, you do not alter the rate at which time passes for either set of clocks. This difference in the rate of time passing, time dilation, is caused by the relative velocity and the result is that each observer observes others clocks to be running slow compared to his own, and so the others clocks appear desynchronized. Zeroing the clocks when the origins are colocated does not change any of this as the relative motion is still the same. It does not put both sets of clocks permanently in synch, it only sets the accumulated time to zero or, if you like, sets the datum to zero. As the other frames clocks pass by you they appear to you to be progressively more and more out of synch with your own. This is of course a reciprocal effect and applies to both observers.

Does that help? If not when I have a little more time i could perhaps give a clearer explanation. I am sure that JesseM and some other respondants have a better grasp of this than I do but, sometimes another wording helps.

Matheinste.

 

[JesseM]

I take it that your 'yes' indicates that you agree that saying 'Ks clocks are dilated' means that the analysis is taking ks viewpoint.

Yes.

In that case both of the calculations in your post 65 take ks viewpoint, and that is why you get the same answer for both, and why you dis agree with post 4 that says each observer sees the others clocks as slow.

No, I analyzed the same situation from both frames in post 65. Read it again, in the first part I described things from the perspective of the k frame:

It works out that if two clocks are synchronized and a distance L apart in their own rest frame, then in a frame where they are moving at velocity v (parallel to the axis between them), they will be out-of-sync by vL/c^2. So for example, suppose we have two clocks at rest in k, k0 and k1, which are synchronized and a distance of 20 light-seconds apart in the k frame, with k0 being at the origin and reading a time of 0 seconds at the moment it is next to the clock K0 at the origin of the K frame (which also reads 0 seconds at that moment). Suppose the clock K0 then moves towards k1 at 0.8c as seen in the k frame, so in the k frame it will take a time of 20 light-seconds/0.8c = 25 seconds to reach k1. So, at the time K0 reaches k1, k1 reads 25 seconds, while K0 has been slowed down by time dilation so it only reads 25*\sqrt{1 - 0.8c^2} = 25*0.6 = 15 seconds.

Then I analyzed everything again from the perspective of the K frame:

But now if we re-analyze the same situation from the perspective of the K frame, things look different--because of length contraction, k0 and k1 are only 0.6*20 = 12 light-seconds apart in this frame, and because of the relativity of simultaneity they are also out-of-sync by vL/c^2 = 0.8c*20 l.s./c^2 = 16 seconds, with the clock at the rear (k1) being ahead of the leading clock (k0) by this amount. So in K's frame, at the moment K0 is next to k0 and both read a time of 0 seconds, k1 is 12 light-seconds away and already reads a time of 16 seconds. Since k1 is approaching K0 at 0.8c, it will take 12/0.8 = 15 seconds for k1 to reach K0 in this frame, meaning K0 will read 15 seconds at the moment k1 passes it, which is what we found before in the other frame. In this frame it is the k clocks that are slowed down by a factor of 0.6 relative to the K clocks, so during these 15 seconds, the clock k1 only ticked forward by 15*0.6 = 9 seconds. But since k1 already had a "head start" of 16 seconds when K0 was passing k0, this means that when k1 passes K0, k1 will read 16 + 9 = 25 seconds, which again matches what we found in the other frame, even though the two frames disagree about whether K0 or k1 was running slower between the event of K0 passing k0 and the event of K0 passing k1.

You can see that in this second section, it is k's clocks that are running slow ('In this frame it is the k clocks that are slowed down by a factor of 0.6 relative to the K clocks...'), the distance between k's clocks are shrunk ('because of length contraction, k0 and k1 are only 0.6*20 = 12 light-seconds apart in this frame'), and it is the k clocks that are out-of-sync ('because of the relativity of simultaneity they are also out-of-sync by vL/c^2 = 0.8c*20 l.s./c^2 = 16 seconds').

What I'm saying is that two clocks that are set to zero at the same time and thereafter run at the same rate will always be in synch.

What two clocks are you talking about? Are you just talking about the clocks at the origin? I thought you wanted to talk about other clocks at different positions in each frame, like the clock k1 in my example above which passes K0 at a different moment so we can compare the readings on K0 and k1. If you're just talking about the clocks at the origin, k0 and K0 in my example, then it's true that the relativity of simultaneity doesn't enter into it, since all frames agree they both read 0 at the same moment. However, all frames do not agree that they "thereafter run at the same rate", I don't understand where you got that assumption--in any frame where they have different velocities, they run at different rates due to time dilation. For example, in the K frame the k0 clock is running slow, and in the k frame the K0 clock is running slow.

The explanation for the result t = T/m is not found in the behavior of clocks, but in the 'Fundamental Principle of Relativity'.

All time coordinates in any frame are meant to stand for the readings on clocks. Einstein based his coordinate systems on the idea of a physical system of rulers and clocks at rest in a given frame, with the clocks synchronized according to his simultaneity convention; then the position and time coordinates of an event are determined by purely local readings on this system, seeing which marking on the ruler the event was next to as it happened, and what reading was showing on the clock attached to that marking as the event happened.

The equation t = T/m is meaningless unless you specify a particular event which is supposed to have time coordinate t as measured in one frame (i.e. measured using that frame's system of rulers and clocks), while it is supposed to have time coordinate T in the other frame. If T is the time coordinate in the K frame and t is the time coordinate in the k frame, and m is supposed to represent the gamma factor, then this equation t = T/m only works if you are talking about an event on the worldline of the clock k0. On the other hand, if you were talking about an event on the worldline of the clock K0, we would have T = t/m. This was the point I was making in post #37--do you disagree?

Consider the segment of the x-axis of k extending from x=0, where a light source is located, to x=l, where a detector is located. Let a light ray be emitted when the origins of K and k coincide. From ks viewpoint the light ray reaches l at t =l/c.

Yes, that's true in the k frame.

From Ks viewpoint the segment is moving at speed v, so the light ray reaches x =l at T = l/( c - v). So T = t/ (1-v/c).

This is wrong. From K's viewpoint, the segment does not have length l, you're forgetting about length contraction--in the K frame the length of the segment is only l * \sqrt{1 - v^2/c^2}. So it will take a time of T = l * \sqrt{1 - v^2/c^2} / (c - v) for the light to reach the end of this segment in the K frame.

Thus T and t are the coordinates that describe the same event, the arrival of the light ray at x = l, as seen by the respective observers K and k. As measured by clocks that are always in synch.

Again, where do you get the idea that the clocks are always in sync? The clocks at the origin don't stay in sync in either frame because of time dilation. And in SR you wouldn't normally measure the time of the event of the light reaching the end of that segment using clocks at the origin, you need to measure it using different clocks in each frame which are actually at the position of the end of that segment when the light reaches it. Only by using local measurements can you avoid the issue of the delays between when an event actually happens and when the event is seen by the person who is noting the time on his own clock (although of course you can correct for transmission delays if you know the distance, but this will give exactly the same time-coordinates as following the standard procedure of using local measurements on an array of clocks synchronized using the Einstein clock synchronization convention).

 

[JM]

Hit the "Quote" button underneath a post. This incorporates the quoted post, in its entirety, into your post, enclosed in [ quote ] and [ /quote ] tags. (The tags do not actually include any spaces; I added spaces here so the forum software wouldn't interpret them as actual quote tags.)

Thanks, JT, I got it. So, how do I work the multi qoute button?

 

[JM]

25*\sqrt{1 - 0.8c^2}[/tex] = 25*0.6 = 15 seconds. But now if we re-analyze the same situation from the perspective of the K frame, things look different--because of length contraction, k0 and k1 are only 0.6*20 = 12 light-seconds apart in this frame, and because of the relativity of simultaneity they are also out-of-sync by vL/c^2 = 0.8c*20 l.s./c^2 = 16 seconds, with the clock at the rear (k1) being ahead of the leading clock (k0) by this amount. So in K's frame, at the moment K0 is next to k0 and both read a time of 0 seconds, k1 is 12 light-seconds away and already reads a time of 16 seconds. Since k1 is approaching K0 at 0.8c, it will take 12/0.8 = 15 seconds for k1 to reach K0 in this frame, meaning K0 will read 15 seconds at the moment k1 passes it, which is what we found before in the other frame. In this frame it is the k clocks that are slowed down by a factor of 0.6 relative to the K clocks, so during these 15 seconds, the clock k1 only ticked forward by 15*0.6 = 9 seconds. But since k1 already had a "head start" of 16 seconds when K0 was passing k0, this means that when k1 passes K0, k1 will read 16 + 9 = 25 seconds, which again matches what we found in the other frame, even though the two frames disagree about whether K0 or k1 was running slower between the event of K0 passing k0 and the event of K0 passing k1.

1. Where do you get the expressions for out-of-synch, time dilation, and length contraction?If not from Lorentz transforms, then where?
2. How do you know how to manipulate these expressions, regarding sign and magnitude? The reference you gave tells about history but not methods.
3. Is your analysis intended to adress the situation presented in post 3? If so why is your answer different ? Note that post 3 is based on the Lorentz transforms, and that T and t refer to total elapsed time since the origins coincided so two points in time are compared. If not, what situation are you addressing?
4. If you agree with sylas, post 35. that 'Using the Lorentz transformations gives you the correct answers.' then what is the point of using the more-difficult ROS procedure?

 

[jtbell]

how do I work the multi qoute button?

Hit the MULTI QUOTE button on each post that you want to quote from. This highlights the button, but doesn't do anything else immediately. Then hit the QUOTE button on one of those posts. All of the selected posts will be quoted (separately) in your response. Delete unnecessary text and intersperse your comments as usual.

1. Where do you get the expressions for out-of-synch, time dilation, and length contraction? [...]

(just as an example)

Thanks for prodding me to figure out how to do it (had to do a Google search on "vbulletin mulitiquote"). When I've needed to do something like this before, I copied and pasted text, and inserted quote tags by hand for stuff from posts other than the one I used the QUOTE button on.

 

[JM]

Lets review some posts related to 'slow clocks' and see what conclusion result. This thread started with Einsteins coordinate frames K and k:

In his 1905 paper Einstein introduced two coordinate systems which he called K and k. It seems that both systems are inertial because there are no external forces and they are moving at the constant relative speed v.
But I'm not certain, so are they both inertial? Is it correct that either one can be considered the 'stationary' system and the other one the 'moving' system?

and the response was:

Yes and yes.

.

On that basis clock rates were calculated for both frames:

So, with K( X,Y,Z,T) stationary the clocks of k( x,y,z,t) are moving in the + X direction at speed v. Putting X=vT in the transfforms leads to t=T/m, where m is the quantity Einstein calls beta. So t is less than T.
With k being stationary the clocks of K are moving in the -x direction at speed v. Putting x=-vt in the transforms leads to T=t/m, ie T is less than t.
Is there general agreement with this result? Is everyone OK with it?

and the results supported:

Yes, an observer in the K system would take the K system as stationary, K' as moving, and observe time in the K' system, t', as slower than t in the K system. An observer in the K' system would take the K' system as stationary, K as moving, and observe time in the K system, t, as slower than t' in the K' system.

.

Thus, it can be concluded that each observer K or k, when taken to be at rest, observes the other observers clocks to be slow, as viewed from is own rest position.
-----------------------------------------------------------------------------------------
For clarity post 3 can be explained further.
The coordinates of K and k are related by the Lorentz transforms, derived in Einsteins 1905 paper
(1) x = m( X – vT), and ct = m( cT –vX/c). These equations can be inverted to obtain
(2) X = m( x + vt), and cT = m( ct + vx/c). For y,z = Y,Z = 0,0.
The meaning of these equations is that when K observes an event occurring at X,T, then k observes the same event occurring at x, t. Thus each observer is assigning his own coordinates to the event.
These equations can work in either direction. Arbitrary values of X, T can be entered on the right of eqn.( 1) and the corresponding values of x,t determined. This process is referred to as ‘taking the point of view of K’, or ‘taking K to be at rest’. Or similarly, arbitrary values of x,t can be assumed and the corresponding values of X, T determined from eqn. (2). This process is referred to as ‘taking the point of view of k’, or ‘taking k to be at rest’.
These equations have a ‘reciprocal’ property. Specific values of X, T entered into eqn (1) produce specific values of x,t. For example when X, cT = 20, 25 and v =0.8c, eqn (1) produces x,t = 0,15. Entering these values into eqn(2) produces the original values of X, T, namely 20, 25. This process is not the same as ‘taking k to be at rest’ as describe above, because the values of x,t are not arbitrary, but are specifically obtained from eqn (1).
-----------------------------------------------------------------------------------------
There is more, but I think I will post this and follow with another post.

 

[JM]

Comments continued from previous post.
When it was pointed out in Post 37 that post 3 refers to two different events, one occurring at the origin of K and the other at the origin of k, the synchronization procedure of Einstein ( and ROS)was recalled as described in Post 65. This background leads to the following reasoning.

Post 49:

In the 1905 paper Einstein spent much time developing his ideas aboout time. Does his reasoning lead to the result that when anyone clock at rest with K reads a specific time, such as cT= 10, then all the clocks at rest with K also read '10'?

Post 50

According to the definition of simultaneity in their inertial rest frame (and assuming they've been synchronized using the procedure Einstein gave in that paper), yes.

Post 51

So does the result of post 49 apply also to coordinates k, namely 'when anyone clock at rest with k reads a specific time , such as ct =8, then all clocks at rest with k also read '8''?

Post 52

May I assume that the answer to post 51 is yes, with the qualifiers given in post 50? If so, can it be concluded that when K is at rest the relation t = T/m holds when comparing the clock at rest at the origin of K with the clock of k that is nearby?

Post 53

The point of this is the following:
1, When K is at rest the c lock of k that is near to the origin of K is slow compared to the clock at the origin of K, and
2. When k is at rest the clock at the origin of K is slow compared with the clock of k that is near the origin of K.
The question is whether this result is explainable and accepted using SR, or whether the two results are contradictory and impossible, as some writers suggest?

As stated in Post 53 another way, the conclusion is that no matter which two clocks are considered, one of K and one of k, each observer views the other observers clock to be slow. This is a significant result of this discussion, because the conconclusion is not always clear from the the literature. It is a unique property of Special Relativity, embodying , in a way, the Relativity Principle that the laws of physecs must be the same whether referred to either of two inertial coordinate frames. It's also a 'mind bender' when you try to reason it our using 'common sense'.

The result is supported from another viewpoint as follows:
Post 54

It seems contradictory only if you haven't grasped what the relativity of simultaneity is about. To compare two clocks that are separated by a distance, you have to measure both clocks at the same time. Different observers disagree over what "at the same time" means, and there is no definition that everyone can agree on. So that's why one observer says k is slower than K and another observer says K is slower than k. They are both right according to their own definition of simultaneity.

The above presents my understanding of the 'slow clockx' phenomenon. Any comments?

 

[JM]

Comments continued from previous post.
When it was pointed out in Post 37 that post 3 refers to two different events, one occurring at the origin of K and the other at the origin of k, the synchronization procedure of Einstein ( and ROS)was recalled as described in Post 65. This background leads to the following reasoning.

Post 49:

In the 1905 paper Einstein spent much time developing his ideas aboout time. Does his reasoning lead to the result that when anyone clock at rest with K reads a specific time, such as cT= 10, then all the clocks at rest with K also read '10'?

Post 50

According to the definition of simultaneity in their inertial rest frame (and assuming they've been synchronized using the procedure Einstein gave in that paper), yes.

Post 51

So does the result of post 49 apply also to coordinates k, namely 'when anyone clock at rest with k reads a specific time , such as ct =8, then all clocks at rest with k also read '8''?

Post 52

May I assume that the answer to post 51 is yes, with the qualifiers given in post 50? If so, can it be concluded that when K is at rest the relation t = T/m holds when comparing the clock at rest at the origin of K with the clock of k that is nearby?

Post 53

The point of this is the following:
1, When K is at rest the c lock of k that is near to the origin of K is slow compared to the clock at the origin of K, and
2. When k is at rest the clock at the origin of K is slow compared with the clock of k that is near the origin of K.
The question is whether this result is explainable and accepted using SR, or whether the two results are contradictory and impossible, as some writers suggest?

As stated in Post 53 another way, the conclusion is that no matter which two clocks are considered, one of K and one of k, each observer views the other observers clock to be slow. This is a significant result of this discussion, because the conconclusion is not always clear from the the literature. It is a unique property of Special Relativity, embodying , in a way, the Relativity Principle that the laws of physecs must be the same whether referred to either of two inertial coordinate frames. It's also a 'mind bender' when you try to reason it our using 'common sense'.

The result is supported from another viewpoint as follows:
Post 54

It seems contradictory only if you haven't grasped what the relativity of simultaneity is about. To compare two clocks that are separated by a distance, you have to measure both clocks at the same time. Different observers disagree over what "at the same time" means, and there is no definition that everyone can agree on. So that's why one observer says k is slower than K and another observer says K is slower than k. They are both right according to their own definition of simultaneity.

The above presents my understanding of the 'slow clockx' phenomenon. Any comments?

 

[JM]

I want to thank all those who gave their time and effort to contribute to this thread. Their comments help me stretch and sharpen my ideas.
I hope that the lack of comments on my last two posts indicates that there is no important disagreement with the ideas stated there. It would help me to feel that some generally agreed conclusion has been reached.
I note that there have been over 4000 views of this thread, so there must be some interest. Would some of you viewers give your opinions of the thread, did you learn anything, would you like to see more on these subjects? It takes time and effort to carry on these conversations so an encouraging word would help.