Synchronized clocks with respect to rest frame

[mananvpanchal]

Those two diagrams are related by a Lorentz transformation with beta = 0.525. That's the way it is.

The velocity changes don't have to be instantaneous, just shorter than the grid scale.

I'm sorry you're having problems understanding this, but I can't see your difficulties.

That's ok. Don't worry. Thanks Mentz114.

 

[Mentz114]

That's ok. Don't worry. Thanks Mentz114.

I think your problem is with the idea of 'now'. This concept really only has meaning for a particular observer. If the A' observer asks himself 'what is happening to B' now ?' the question has no meaning and leads to weird results, especially if the observers are well separated. The best example is the 'Andromeda paradox', where two observers on Earth with a very small relative velocity apply the idea of 'now' to a very distant place, like the Andromeda galaxy. Using their lines of simultaneity they come up times that are thousands of years apart on Andromeda. So 'now' on a worldline than your own is a frame dependent quantity and so has no useful interpretation.

 

[pervect]

I think your problem is with the idea of 'now'. This concept really only has meaning for a particular observer. If the A' observer asks himself 'what is happening to B' now ?' the question has no meaning and leads to weird results, especially if the observers are well separated. The best example is the 'Andromeda paradox', where two observers on Earth with a very small relative velocity apply the idea of 'now' to a very distant place, like the Andromeda galaxy. Using their lines of simultaneity they come up times that are thousands of years apart on Andromeda. So 'now' on a worldline than your own is a frame dependent quantity and so has no useful interpretation.

I haven't been following the discussion in detail, but I also suspect the problem of "now" is the issue. The other, related issue that may be arising is what "now" means to someone who is making an instantaneous velocity change. I don't think there's any good way of defining the concept of "now" in such a case, frankly. The sudden jump in frames leaves some events without any valid time coordinate, and then there's also the problem of some events having multiple time coordinates.

It may be clearer and make more physical sense if we consider what "now" looks like for an accelerating observer, and take the limit as the acceleration approaches infinity.

This is rather technical, but the end result is rather well known as the "rindler" observer. I'll present the results without explaining how they were arrived at, though time permitting I could probably be convinced to take the effort to draw up the diagrams and post more if there is interest. I should add that I'm sure Mentz already knows this, and he might also be induced to explain it if it's of interest to the OP. I'm not POSITVE this is actually the issue or an issue of interest to the OP, so I'm reluctant to take too much time on it currently.

Anyway, the end result is that there is only a certain region of space-time that can be given valid coordinates, a certain region that the concept of "now" applies to. The edge of this region is bounded by the so-called Rindler horizon, which is an event horizon similar to that formed by a black hole.

The Rindler horizon is at a distance of c^2/g "behind" the accelerating observer. As g approaches infinity, the Rindler horizon gets closer and closer, until in the limit of infinite acceleration, the Rindler coordinates can't cover anything behind the accelerating observer at all.

Since I regard "now" as entirely a mental construct, without much physical significance, this doesn't bother me a whole lot. It's basically mostly a matter of convention - though if you want to set up a coordinate system that obeys Newton's laws, it's an important convention, i.e. you can't both use an arbitrary definition of "now" AND also apply Newton's laws even in a limiting sense.

 

[Dale]

This problems might not so hard for you, but it is actually for me. That is why I am creating problems here...

That and you prefer to argue and come up with really strange ways to tell people that they are wrong rather than try to ask questions and learn. It is not productive, particularly when you know that you are learning a topic.

Btw, you never responded to my post 111.

In general, if you want to describe the behavior of some clock you need to write down a parametric expression for its worldline in some given reference frame. Then, to determine the reading on the clock you use the second formula on this page if the reference frame is inertial:
http://en.wikipedia.org/wiki/Proper_time#In_special_relativity

Or, if it is a non-inertial frame or gravity is involved:
http://en.wikipedia.org/wiki/Proper_time#In_general_relativity

That will give you a complete description of the clock in that frame. If you want to then see the description of the clock in some other frame then you use the Lorentz transform if both frames are inertial or the appropriate coordinate transform otherwise.

That is the general approach for these types of problems.

If you have multiple clocks simply parameterize each one and follow the above, and in order to determine if they are synchronized in some frame simply compare their proper times at some specific coordinate time.

 

[mananvpanchal]

I think your problem is with the idea of 'now'.

Please, understand "see" as "calculate" here.

We might describe this situation like this:
Line of simultaneity of A', B' and C' in unprimed frame moving from bottom to up. There is only one line of simultaneity exist for all three clocks. Now, in unprimed frame all clocks simultaneously change its frame. But, how can we display the simultaneous change in primed frame? We might imagine that three lines of simultaneity created simultaneously in primed frame like three parallel world. This lines of simultaneity is far from each other by time distance. Now, the three lines of simultaneously moving from bottom to up remaining parallel to each other. The events (B' frame change and C' frame change) is already occurred in future of A', but A' cannot see it. The events leaves footprint of own self in spacetime. So, when A's line of simultaneity come to the footprint of the events, A' can see that the events is reoccurring for A'. Here we should not use word "reoccurring", because A' has not seen the events before, it was occurred in future. So, now in primed frame A' still sees in own world that B' has not changed its frame yet. But, B' sees in own world that A' has already changed its frame before himself. The situation is fine uptill now. All events might be occurred already in future of A'. But the events occurs only for one time in A's world. But the symmetry breaks where two frames meet each other. A' might see that events are reoccurring for him in this region (between "p" and "q"). (Possibly reoccurring event is defined as black dot in diagram)

There might be many errors exits in my description, and I expect it to be corrected.

 

[mananvpanchal]

I'll present the results without explaining how they were arrived at, though time permitting I could probably be convinced to take the effort to draw up the diagrams and post more if there is interest.

It would be great help. Thanks.

 

[mananvpanchal]

That and you prefer to argue and come up with really strange ways to tell people that they are wrong rather than try to ask questions and learn. It is not productive, particularly when you know that you are learning a topic.

At starting of this post I had a result in my mind about the scenario. I knew that it might be wrong. So, I have posted my doubt here. I asked that "Am I right? If not then please, give me explanation". I got the answer that "You are wrong", but no explanation. After, this you came with maths. I appreciated your work and was trying to be convinced by your maths that "clocks in synchronized for R, but not for O". But, I found some errors in your maths. I asked about you, and you also came up with solutions.

Btw, you never responded to my post 111.

Can you re-explain your post #42 using subscript with \tau?
But, please this time define clearly:
What is t? What is \tau_a, \tau_o, \tau_b? What is t'?

 

[Mentz114]

mananvpanchal , I'm sorry but I don't understand any of your post#125 except this bit

But, how can we display the simultaneous change in primed frame?

You cannot because the change of velocities is not simultaneous in that frame.
The LT tells me that those 3 events cannot be simultaneous in the primed frame. It is difficult to accept because it is counter-intuitive but it leads to no paradoxes, discontinuities or other unpleasantness.

When you start talking about events 'reoccurring' you've lost me ( and physics). That doesn't happen or someone would have noticed. Similarly there is no point in discussing what goes on in other worldlines because we can only ever see the past of spatially separated observers.

For all I know, the OP might have something important to say, but I can't see what it is and I'm pretty tired of this whole thing.

DaleSpam and Pervect - your posts are right on the button and I hope they help the OP to understand what he/she is trying to fathom.

 

[Dale]

I found some errors in your maths

You found a small inconsistency in notation, not an error. The math is all correct.

Can you re-explain your post #42 using subscript with \tau?
But, please this time define clearly:
What is t? What is \tau_a, \tau_o, \tau_b? What is t'?

Let's analyze the scenario from the OP where at t=0 in the original frame A, B, and M all accelerate instantaneously up to v = .6c and furthermore let's use units where c=1 and where the distance from M to A and from M to B is 1 in the original frame. R's frame is the unprimed frame and t is the coordinate time in the unprimed frame.

So, in R's frame the worldline of A, O, and B are:
r_d=\left(t,x=\begin{cases}<br /> d &amp; \mbox{if } t \lt 0 \\<br /> 0.6 t+d &amp; \mbox{if } t \ge 0 <br /> \end{cases}<br /> ,0,0\right)
where d=-1 for A, d=0 for O, and d=1 for B.

As per the OP, the clocks are initially synchronized in R's frame such that at t=0 they all read 0. So, we can calculate the time displayed on each clock, \tau_d, using the spacetime interval. Solving for t we get:
t=\begin{cases}<br /> \tau_d &amp; \mbox{if } \tau_d \lt 0 \\<br /> 1.25 \tau_d &amp; \mbox{if } \tau_d \ge 0 <br /> \end{cases}

Substituting into the above we get an expression for the worldline parameterized by the proper time on each clock:
r_d=\left(<br /> t=\begin{cases}<br /> \tau_d &amp; \mbox{if } \tau_d \lt 0 \\<br /> 1.25 \tau_d &amp; \mbox{if } \tau_d \ge 0 <br /> \end{cases},<br /> x=\begin{cases}<br /> d &amp; \mbox{if } \tau_d \lt 0 \\<br /> 0.75 \tau_d+d &amp; \mbox{if } \tau_d \ge 0 <br /> \end{cases}<br /> ,0,0\right)

Noting that \tau_d does not depend on d in this frame we see immediately that the clocks remain synchronized in R's frame.

Now, we use the Lorentz transform to boost to the primed frame where O is at rest for \tau_d=t&gt;0. This frame is the unprimed frame and t' is the coordinate time in this frame. We obtain the following expression for the worldline of the clocks in the primed frame, again parameterized by the proper time.
r&#039;_d=\left(<br /> t&#039;=\begin{cases}<br /> 1.25 \tau_d - 0.75 d &amp; \mbox{if } \tau_d \lt 0 \\<br /> \tau_d - 0.75 d &amp; \mbox{if } \tau_d \ge 0 <br /> \end{cases},<br /> x&#039;=\begin{cases}<br /> 1.25 d - 0.75 \tau_d &amp; \mbox{if } \tau_d \lt 0 \\<br /> 1.25 d &amp; \mbox{if } \tau_d \ge 0 <br /> \end{cases}<br /> ,0,0\right)

Noting that \tau_d does depend on d in this frame we see immediately that the clocks are desynchronized in O's frame. It is therefore quite clear the the clocks remain synchronized in the unprimed frame and are not synchronized in the primed frame.

 

[mananvpanchal]

Noting that \tau_d does not depend on d in this frame we see immediately that the clocks remain synchronized in R's frame.

Noting that \tau_d does depend on d in this frame we see immediately that the clocks are desynchronized in O's frame. It is therefore quite clear the the clocks remain synchronized in the unprimed frame and are not synchronized in the primed frame.

Thanks for re-explaining.

Ok, so in unprimed frame \tau_a = \tau_o = \tau_b and in primed frame \tau_a \ne \tau_o \ne \tau_b. Right?

 

[Dale]

Yes.