Are Both Coordinate Systems in Special Relativity Considered Inertial?

[JM]

Al68 JesseM

OOOHKaY? I'm really trying to find out whether there'a anything at all that we can agree on, so help out here.

I sense that you haven't read the first four posts of this thread. So I think these are yes/no questions.
Have you read these posts?
Is there any thing in them that you don't understand?

 

[sylas]

Al68 JesseM

OOOHKaY? I'm really trying to find out whether there'a anything at all that we can agree on, so help out here.

I sense that you haven't read the first four posts of this thread. So I think these are yes/no questions.
Have you read these posts?
Is there any thing in them that you don't understand?

I am not sure that you have ever really grasped the fact that "k" is a not a frame for the clock as it moves in polygons, or circles.

That error was clearest in [post=2264139]msg #11[/post], where you said

What I am saying is that k has an equal right to consider himself to be stationary! Thus , if K sees k moving away, turning around, and returning, then k, considering himself to be stationary,sees K moving away, turning around and returning.

The earlier posts in the thread are fine as worded, but only if you recognize that "k" refers to an inertial frame for an observer or clock in straight line unaccelerated motion. If, however, "what you were saying" in those earlier posts was predicated on k being somehow associated with the clock moving in polygons, or a closed circle, then the answers are all different. In fact, a clock that moves and returns does not have an "equal right" to consider itself stationary, and k is not a frame for motions in any kind of curve.

Cheers -- sylas

 

[JesseM]

I sense that you haven't read the first four posts of this thread. So I think these are yes/no questions.
Have you read these posts?
Is there any thing in them that you don't understand?

Of course I read them, and I already pointed out my disagreement with your fourth post (post #7) in post #8 (specifically the part where you said T=t/m in the k frame, which I took to mean that you were claiming the k frame would predict the inertial clock B would be behind the accelerated clock A when they met). Again, if we use the physical situation as Einstein described it, with the two clocks being initially synchronized in the K frame where both clocks were at rest before A accelerated, then even if we analyze things from the perspective of the k frame where both clocks were moving and then A came to rest after it accelerated, this frame will still predict that A will be behind B when they meet, in spite of the fact that this frame predicts A was ticking faster than B after they accelerated. As I said, the reason is that the two clocks were not initially synchronized in this frame, B had a significant "head start" at the time A was accelerated and came to rest in this frame. Do you have trouble understanding this analysis, or do you disagree with it in any way?

 

[JM]

Al68, Sylas: Thanks for your comments.

These threads diverge so quickly and far that I've tried to start at the very simplest, basic ideas of Einsteins relativity. In my first post I identified K and k as Einsteins original coordinates, he even calls them K and k. They are both inertial, as agreed by Mentor George Jones. The connection between K,k and the curved/ segmented path is so unclear that I don't want to try to connect them.

I am trying to find something that we can agree on as a starting place for our discussion. I propose that we put our blinders on and focus on the first four posts of this thread. I don't want to assume anything, and it would encourage me greatly to hear a 'yes' in agreement with these posts..( I almost fell off my chair when I read Jones 'yes and yes'!)

 

[sylas]

Al68, Sylas: Thanks for your comments.

These threads diverge so quickly and far that I've tried to start at the very simplest, basic ideas of Einsteins relativity. In my first post I identified K and k as Einsteins original coordinates, he even calls them K and k. They are both inertial, as agreed by Mentor George Jones. The connection between K,k and the curved/ segmented path is so unclear that I don't want to try to connect them.

I am trying to find something that we can agree on as a starting place for our discussion. I propose that we put our blinders on and focus on the first four posts of this thread. I don't want to assume anything, and it would encourage me greatly to hear a 'yes' in agreement with these posts..( I almost fell off my chair when I read Jones 'yes and yes'!)

George's reponse (yes, and yes) was to your message #1. That's quite unexceptional. I don't know why you found it surprising.

Your message #3 is poorly expressed. You refer to co-ordinates t,x,y,z and T,X,Y,Z; and show t=T/m. The m here is the gamma factor, and this is a description of the time dilation effect.

But think about it. t and T are not variables denoting a single value, but co-ordinates, which range over many values. When you write t = T/m, you are comparing a whole range of events that are simultaneous for one observer... but NOT the other. If you switch to the other frame, you get T = t/m. But this is involving a different notion of simultaneity.

Specifically. For observer in the k frame, the other clock is moving at v and reads T/m simultaneously with their own clock reading t.

Equivalently, for the observer in the K frame, the first clock is the one moving at v, and it reads t/m simulaneously with their own clock reading T.

By writing t=T/m, and then also T=t/m, you are not in fact writing two inconsistent equations. You are writing the relations for what one clock reads simultaneously with the other; for two different notions of simultaneity... depending on the frame.

This seems to be the difficulty everyone has with relativity. It nearly always comes down to failing to appreciate that simultaneity depends on the frame being used. Einstein's paper does it correctly. Using the Lorentz transformations gives you the correct answers.

Cheers -- sylas

 

[DrGreg]

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?

All correct.

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.

To make this clear, I would spell out when x = 0, X = vT and t = T/m.

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.

To make this clear, I would spell out when X = 0, x = -vt and T = t/m.

The bits in bold are important, and make it clear that the two paragraphs apply to two different situations. It's not possible for x and X to both be zero (except for the special case when t and T are both zero).

 

[JesseM]

JM, I have no problems with your first post. As for the second, you say:

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?

I agree with this, but just to be clear, you understand that you're talking about two different events here right? X=vT would be some event on the worldline of the clock at rest in k, so in t=T/m, t would be the time on this k-clock and T would be the time in the K frame of the event of the k-clock reading t. On the other hand, x=-vt would be some event on the worldline of the clock at rest in K, so in T=t/m, T would be the time on this K-clock and t would be the time in the k frame of the event of the K-clock reading T.

 

[JM]

Sylas, DrGreg, JesseM:
Thank you for your responses, they are positive and encouraging. There is a lot to think about in them, I want to take some time to understand and prepare appropriate responnses. So stay tuned.

 

[JM]

Sylas writes: "Your message #3 is poorly expressed."
My posts may be abbreviated, with details added as necessary.

And "...this is a description of the time dilation effect."
This particular expression of time dilation has the advantages of its direct derivation from the transforms, and its symmetry. For convenience we might refer to it as 'double slow clocks', with each half called a 'slow clock'.

 

[JM]

But think about it. t and T are not variables denoting a single value, but co-ordinates, which range over many values. When you write t = T/m, you are comparing a whole range of events that are simultaneous for one observer... but NOT the other. If you switch to the other frame, you get T = t/m. But this is involving a different notion of simultaneity.

Specifically. For observer in the k frame, the other clock is moving at v and reads T/m simultaneously with their own clock reading t.

Equivalently, for the observer in the K frame, the first clock is the one moving at v, and it reads t/m simulaneously with their own clock reading T.


Cheers -- sylas

I agree with this analysis.
The basic statement of The Relativity of Simultanaity (ROS) is ' A number of events that are simultaneous in one coordinate system are not necessarily simultaneous in any other coordinate system.' That doesn't seem too difficult to understand. But the application of this principle to a particular situation is not so easy to see.
The book 'Special Relativity' by A. P. French includes good descriptions in the sections 'The Relativity of Simultaneity' p.74, and Relativity is Truly Relative, p.111. The use of ROS there provides unique insights into the behavior of clocks, and develops the same resulst as given in posts 3 and 4.

 

[JM]

By writing t=T/m, and then also T=t/m, you are not in fact writing two inconsistent equations. You are writing the relations for what one clock reads simultaneously with the other; for two different notions of simultaneity... depending on the frame.


Cheers -- sylas

This idea seems to be related to the process of designating a coordinate/observer as being 'stationary', or 'the reference system'. This is important and related to the question of how many events there are. To be discussed.

 

[JM]

All correct.

To make this clear, I would spell out when x = 0, X = vT and t = T/m.

To make this clear, I would spell out when X = 0, x = -vt and T = t/m.

The bits in bold are important, and make it clear that the two paragraphs apply to two different situations. It's not possible for x and X to both be zero (except for the special case when t and T are both zero).

Agreed. In his analysis of the slow clock Einstein specified that the expression X=vT describes the position of the clock at x=0 of k.

 

[JM]

I agree with this, but just to be clear, you understand that you're talking about two different events here right? X=vT would be some event on the worldline of the clock at rest in k, so in t=T/m, t would be the time on this k-clock and T would be the time in the K frame of the event of the k-clock reading t. On the other hand, x=-vt would be some event on the worldline of the clock at rest in K, so in T=t/m, T would be the time on this K-clock and t would be the time in the k frame of the event of the K-clock reading T.

I see what JesseM is driving at here, and in the context of world lines I agree.
The description of the physical situation under posts 3/4 needs to be clarified. To begin, there is no known location of absolute rest, and only relative motions have meaning. So these posts describe only one physical event, namely the two coordinates/observers are separating at speed v. The two results T=t/m and t=T/m come from the process of designating one or the other as 'stationary', or equivalently taking the viewpoint of one observer or the other as each views the same event. Isn't the world line approach consistent with this idea?
Comments?

 

[JM]

Slow Clocks.

The idea expressed in post 4,that each obsercer 'sees' the others clock running slow, has been considered berore, for example by G Builder, Phil. Sci., Apr. 1959, p. 135-144. He considers this idea paradoxical because, he says, the two results t=T/m and T=t/m are contradictories, i.e. if one is true the other must be false. On the other hand both results are considered here to be true. Explanation is foound in understanding of the meaning of the terms.

Builders analysis indicates that he views the calculations to apply directly to the time reading of the moving clock. He says that ...one of the clocks must be retarded relative to the other. But in deriving the time delay from the Lorentz trasnforms,Einstein clearly states that '...the time marked by the (moving) clock (viewed in the stationary system)is slow...' Emphasis added. Builder has apparently ignored this qualifying phrase, as is commonly done even nowadays. But there is a sugnificant difference between 'the time on the clock' and 'the time perceived be the stationary viewer'. Thus each observer can correctly state that the others clock appears to run slow. An example sometimes used to illustrate this point is the difference between the actual height of a person standing at some distance away and the reduced height perceived by each of the facing observers.

Agreement on these ideas would remove a significant source of confusion for me, and I hope for others.

 

[JM]

The Twins again

In the paper cited above,Builder considers the clock/twin paradox in the form of a thought exaperiment. In the nomenclature of this thread, he pictures a clock k, originally at the origin of inertial coordinates K , that moves at speed v along the X axis of K, stops, and returns to the origin of K. He considers several methods for calculating the time retardation when the clock k arrives back at the origin, and concludes that all methods give the same result, t =T/m. He also notes that the asymmetry of the predicted retardation is obviously consistent with the dynamical asymmetry involved in the experiment itself, which requires that k should be subjected to accelerations while K remained at rest. His analysis is of spcial intest because it is representative of the ideas pesented by many others.

continued next post

 

[JM]

Twins, cont.

So, let's consider whether there are other thought experiments that can be develpped from Einsteins description of he's clock paradox, par. 4 of the 1905 paper. Note that the clock paradox involves only the clocks A and B, and that no connection is made between A,B and K,k. So we are on our own to make the connecrion. Builder has evidently chosen to identify the stationary B with coordinates K, and specifies the path of A as the back-and=forth motion of k along the X axis.

It is equally corect to identify the stationary B with coordinates k, and the path of A as a back-and-forth motion of K along the negative x axis. The analysis methods Builder uses are equally applicable to this second experiment, with the result T = t/m.

cont,

 

[JM]

Twins, concluded

The presence of acceleration at the turn around may be part of Builders experiment,, but it is not part od Einseteins, who diesn't mention accelerations. Accelaerations are not necessary anyway because they don;t affect the calculated time retardation, alnd because designation of which coordinate is stationsary is enough to specify the experimental conditions.

For the reasons given in post 43 the twins journey can be considered as one event with two parts, namely 1. the twins separate, and 2 the twins re-unite. Builders experiment takes Ks perspetive, who 'sees' k recede and then retrn. The second experiment takes ks perspective, who 'sees' K recede and then return. Thus when the twins re-unite each 'sees' the others clock to be running slow.

So that's all that I meant to say in this thread. Is it 'QED' or 'back to the drawing board'?
What do y'all think?

 

[JM]

I agree with this, but just to be clear, you understand that you're talking about two different events here right? .

Right. When K is at rest the expression X=vT identifies the clock at rest with K that is adjacent to the origin of the k system. So this event takes place at the origin of k. When k is at rest the expression x = -vt identifies the clock at rest with k that is located at the origin of the K system. So this event takes place at the origin of K. Since the origins are at different places the events are separate.

I believe that Silas and DrGreg were making the same point about there being two different events.

OK?

 

[JM]

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'?

 

[JesseM]

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'?

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.

 

[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.