Synchronized clocks with respect to rest frame

[Dale]

All clocks is synchronized.
Now at t = \tau = 0
...
When \tau_a = 0, value of t_a=0.75
When \tau_o = 0, value of t_o=0
When \tau_b = 0, value of t_b=-0.75

Please make up your mind. These are two mutually contradictory scenarios so you have to pick only one.

All of our discussion to date has been using the first condition. Changing the scenario at this late date is rather irritating, but I can certainly re-work everything with this new scenario. But you must recognized that it is different and contradictory to the previously discussed scenario.

Here, \tau = t'. Because, clocks is at rest in O's frame. So, we don't need to transform it.

I already gave you a concrete example why the simple fact that a clock is at rest is not sufficient to guarantee that it is synchronized. Your logic is wrong.

 

[mananvpanchal]

When \tau_a = 0, value of t_a=0.75
When \tau_o = 0, value of t_o=0
When \tau_b = 0, value of t_b=-0.75

I know that this calculation creates gap between world line and overlapping of world line. And I know that is strange to you. So, please see below image in which I have tried to show smooth acceleration of three clocks.

We can easily see that for t < 0, the clocks is at rest in R's frame and lines of simultaneity says that the three clocks are synchronized in R's frame.

Now at t = 0 clocks starts accelerating. And at some time t = t_1 the clocks achieves 0.6c speed and after this clocks moving with the constant speed.

Now, you can see lines of simultaneity of O's frame are not parallel during acceleration phase (0 \le t < t1). And lines of simultaneity of O's frame are parallel for t < 0 and for t \ge t_1.

Space axis and time axis is slowly skewed during acceleration. And after achieving constant speed it will not be skewed more. Lines of simultaneity are always parallel to space axis, so it is also skewed with space axis.

Lines of simultaneity of any frame indicates synchronized clocks in that moving frame.

So, lines of simultaneity of O's frame indicates synchronized clocks in O's frame. But, lines of simultaneity of O's frame are not parallel to lines of simultaneity of R's frame. So, O's clocks which is synchronized in O's frame is not synchronized in R's frame.

Now, please see below image in which I have tried to show what you are saying.

You can easily see that it solves the purpose "Clocks of O's frame are synchronized in R's frame".
But, to solve the purpose, the diagram creates some problems.
- Only time axis is skewed. Space axis is not skewed.
- Space axis is not skewed, so space scale of O's frame remains equal to space scale of R's frame. Which should not be.
- Clocks is synchronized in R's frame, but the diagram cannot show how the clocks is not synchronized in O's frame.

Now, please tell me. Do you agree that after finishing acceleration the lines of simultaneity of O's frame are skewed for R's frame?
If you find that I couldn't show your idea in second diagram then please provide me your diagram.

 

[mananvpanchal]

Please make up your mind. These are two mutually contradictory scenarios so you have to pick only one.

All of our discussion to date has been using the first condition. Changing the scenario at this late date is rather irritating, but I can certainly re-work everything with this new scenario. But you must recognized that it is different and contradictory to the previously discussed scenario.

From my original post I am telling that clocks is synchronized for O, but not for R. But, when you posted your maths, I thought that you were telling right. But, I found some errors in your maths.

You are using t = 1.25 \tau to prove that clocks is synchronized for R. Here, you assuming single value of \tau for all three clocks, so you can prove that clocks is synchronized for R.

But when "decreasing desync with increasing \tau" problem created, you came up with idea of \tau_a, \tau_o and \tau_b. So, now you can easily prove that clocks is not synchronized for O with these three values of \tau.

So, now how are you going to prove that clocks is still synchronized for R with these three values of \tau using t = 1.25 \tau?

 

[Dale]

But, I found some errors in your maths.

No, you didn't find any errors. You found that my math disproved your assumptions so you went ahead and assumed your conclusion and stated that therefore the math is wrong. That is a logical fallacy known as "begging the question".

You are using t = 1.25 \tau to prove that clocks is synchronized for R.

I showed you how that was derived. It was derived correctly from the spacetime interval formula. My other results were derived from the Lorentz transform. Those two equations, the spacetime interval and the Lorentz transform, are the fundamental equations of special relativity. If you are not using them then you are not doing special relativity, and in special relativity you can always use them. The fact that they are contradictory to your assumptions proves your assumptions wrong.

Here, you assuming single value of \tau for all three clocks, so you can prove that clocks is synchronized for R.

I didn't assume it, I proved it. The d cancels out.

But when "decreasing desync with increasing \tau" problem created, you came up with idea of \tau_a, \tau_o and \tau_b. So, now you can easily prove that clocks is not synchronized for O with these three values of \tau.

Yes, I made a small mistake in notation, which I quickly corrected immediately upon your pointing it out.

So, now how are you going to prove that clocks is still synchronized for R with these three values of \tau using t = 1.25 \tau?

I already proved it, by deriving (not assuming) the fact that the proper time doesn't depend on d.

 

[Dale]

I know that this calculation creates gap between world line and overlapping of world line. And I know that is strange to you.

Not just strange, it is wrong. It is the most obvious of physical nonsense and should have immediately clued you in that you were wrong. You cannot have clocks magically appearing and disappearing, with duplicates and missing clocks. It is physical nonsense and violates all sorts of conservation laws.

Lines of simultaneity of any frame indicates synchronized clocks in that moving frame.

No, the lines of simultaneity indicate lines of constant coordinate time in the frame. What you need to prove is that the clocks display the same proper time at their respective intersections with one of these lines of simultaneity. That you have never done, but simply assumed.

In fact, your very diagram disproves your assumption. I have only added some dots to your diagram, but not changed the lines. I am assuming (according to the original scenario, not your contradictory scenario) that each clock reads τ=0 at the black dots.

We can easily see that the spacetime intervals between the black dots and the corresponding red dots are the same. That means that the clocks read the same time at the red dots. Since the red dots are on a single plane of simultaneity in the unprimed frame that implies that they are synchronized in the unprimed frame.

We can also easily see that the spacetime intervals between the black dots and the corresponding green dots are not the same. That means that the clocks do not read the same time at the green dots. Since the green dots are on a single plane of simultaneity in the primed frame that implies that they are not synchronized in the primed frame.

Your own graphics disprove your assumptions, as does all of the math.

 

[mananvpanchal]

We can easily see that the spacetime intervals between the black dots and the corresponding red dots are the same. That means that the clocks read the same time at the red dots. Since the red dots are on a single plane of simultaneity in the unprimed frame that implies that they are synchronized in the unprimed frame.

We can also easily see that the spacetime intervals between the black dots and the corresponding green dots are not the same. That means that the clocks do not read the same time at the green dots. Since the green dots are on a single plane of simultaneity in the primed frame that implies that they are not synchronized in the primed frame.

Is this not contradictory?

 

[Dale]

Is this not contradictory?

No, it is not contradictory.

 

[mananvpanchal]

Thanks DaleSpam.

 

[Dale]

You are welcome. Do you understand why it is not contradictory?

 

[mananvpanchal]

I want to clear some points here.

- You are using TD equation to find time readings, that is wrong. I agree that time dilation difference between two readings of each three clocks are same for R. But time readings of each three clocks are not same for R. To find time reading you have to use LT equation.

- You are telling \tau is not dependent of d, and you have proved forcefully that clocks is synchronized for R using TD equation and using single independent value of \tau. But then you have a goal to prove that clocks is not synchronized for O. So, you came up with three values of \tau. But, how could you not realize that if \tau is not dependent of d then how could you get three values of \tau? And if there are three values of \tau exist then how could you prove that clocks is synchronized for R using the TD equation?

- You said my modified diagram (modified by you) shows that clocks is not synchronized for O. Then how does the modified diagram shows that clocks is synchronized for R?

- I cannot understand that why you don't want to understand this: If a clock is at rest in a frame then its proper time (what time interval it has covered in the frame) is equal to co-ordinate time of the frame (because covered space interval is zero in the frame). If a clocks is moving in a frame then proper time (what time interval it has covered in the frame) is not equal to co-ordinate time of the frame (because covered space interval is not zero in the frame). So, \tau is a proper time of train's clocks, the clocks is at rest in O's frame then co-ordinate time of O's frame is equal to proper time of the clocks. And, \tau is a proper time of train's clocks, the clocks is moving in R's frame then co-ordinate time of R's frame is not equal to proper time of the clocks.

- I cannot understand that why you don't want to understand this: Line of simultaneity is the line on which, two events occurred, called simultaneous events. So short hand and long hand of clocks on the line meets at 12 must be simultaneous events on the line of simultaneity. So clocks on the line can be called synchronized. If clocks on line of simultaneity is not synchronized then what is other method you have to prove that two events occurred on the line of simultaneity is simultaneous events?

- Accelerating scenario can be made up using too many small inertial frames changing. My diagram shows how lines of simultaneity of O's frame becomes unparallel to space axis of R's frame (during acceleration and after acceleration). So, clocks on the lines of simultaneity becomes desynchronized for R. But, lines of simultaneity is always parallel to space axis of whatever small inertial frame in which O exist at that time. So, clocks on the lines of simultaneity remain synchronized for O.

- Using LT equation if I pick single value of t, I get three values of \tau, and if I pick single value of \tau, I get three values of t. So, train's clocks is not synchronized for R. If we set two another observer at A and B. The thee A, O and B confirms that train's clocks is synchronized for them, but they all see different value in R's clock.

- You guys explained me that how clocks becomes desync on frame changing: when train is at rest light pulse take equal time to reach to O from both clocks. But after frame change light pulse take different time to reach to O from both clocks (front clock's light beam traveling in opposite direction of train travel and rear clock's light beam traveling in same direction of train travel). But, light speed is same in all direction, so clocks is actually desync.

We can think a experiment on the basis of this. Suppose, there are two more observers at A and B. Train is at rest in R's frame. Clocks firing a pulse to all observers. O gets A and B's "0" pulse at same time but after some time of his own clock reads "0" because of light propagation time. O confirms that all clocks is synchronized. A gets "0" pulse first from O and then from B after some time of his own clock reads "0". A confirms that all clocks is synchronized. B gets "0" pulse first from O and then from A after some time of his own clock reads "0". B confirms that all clocks is synchronized.

Now, train has changed its frame. O gets A and B's "0" pulse at different time but after some time of his own clock reads "0". O confirms that clocks is not synchronized. O confirms that B (front) is ahead of A (rear) (O > B > A). A gets "0" pulse first from O and then from B after some time of his own clock reads "0". Time duration in receiving two pulse is obviously less than what was in rest frame. But A cannot confirms that clocks is not synchronized. A can say that O and B is behind of his own clock (A > O = B). B gets "0" pulse first from O and then from A after some time of his own clock reads "0". Time duration in receiving two pulse is obviously more than what was in rest frame. But B cannot confirms that clocks is not synchronized. B can say that O and A is behind of his own clock (B > O = A).

Can you see the problem here? Three observers in inertial frame is not agree with each other. They are not agree on sequence of desync (which is ahead and which is behind). They are not agree even on the clocks is synchronized or not.

- Earth's motion is accelerating motion around Sun, but we know that Earth's clocks remain synchronized for us. But Earth's clocks would not remain synchronized for Sun.

- I have proved that train's clocks is synchronized for O and not synchronized for R using diagram and maths (LT equation).

 

[Mentz114]

These diagrams might help. They are essentially the same as the one posted by Dalespam in post #95 but show the events from both sets of observers POV.

In the first diagram the stationary frame is the comoving observers A,B,C. The primed observers begin moving away simultaneously in the unprimed observers frame.

The second diagram is the rest frame of the primed observers, and it is clear that their clocks are no longer synchronised because they did not begin their motion simultaneously in their frame.

The diagrams are completely clear and show what is happening. There are no contradictions or paradoxes.

The mistake in your original diagram in post #1 is that you tried to make the moment of departure simultaneous in both frames, which it is not.

 

[mananvpanchal]

@Mentz114

Ok, so A, B, C and A', B', C' is synchronized in ABC's frame (before frame change).
So t_{abc}=t_{a'b'c'} (before frame change).

Now, clocks is changed its frame at t_{abc}=0 simultaneously in ABC's frame and clocks moving with 0.6c speed.

Now, can you calculate for me what is the value of t_{a'}, t_{b'} and t_{c'} at t_{abc}=0? (please, take origin on B's world line)

 

[Mentz114]

@Mentz114

Ok, so A, B, C and A', B', C' is synchronized in ABC's frame (before frame change).
So t_{abc}=t_{a'b'c'} (before frame change).

Now, clocks is changed its frame at t_{abc}=0 simultaneously in ABC's frame and clocks moving with 0.6c speed.

OK. The speed is 0.52c in the diagram below which has the lines of simultaneity. You can read off the answers from this.

Now, can you calculate for me what is the value of t_{a'}, t_{b'} and t_{c'} at t_{abc}=0? (please, take origin on B's world line)

Can you rephrase this in terms of proper times ? t_{abc}=0 is only true in the unprimed frame.

I've run out of time. Try calculating proper times by triangulation from the diagrams using dtau^2 = dt^2-dx^2.

 

[Dale]

- You are using TD equation to find time readings, that is wrong. I agree that time dilation difference between two readings of each three clocks are same for R. But time readings of each three clocks are not same for R. To find time reading you have to use LT equation.

Not only is my use of the equation correct, I proved that it is correct from first principles, the spacetime interval equation.

The Lorentz transform is used to transform from the unprimed frame to the primed frame, it is not useful within a single frame. It is always used, once you have completely specified the problem in one frame, to determine how the problem appears in any other frame.

The LT cannot be used to specify the problem within a single frame. You should have realized that you were making a mistake when your use of the LT contradicted your initial scenario.

- You are telling \tau is not dependent of d, and you have proved forcefully that clocks is synchronized for R using TD equation and using single independent value of \tau. But then you have a goal to prove that clocks is not synchronized for O. So, you came up with three values of \tau. But, how could you not realize that if \tau is not dependent of d then how could you get three values of \tau? And if there are three values of \tau exist then how could you prove that clocks is synchronized for R using the TD equation?

This is a valid point. My apologies for the confusion. There are always 3 values of τ in both frames. So I should have written:
for \tau_d>0
(\tau_d - 0)^2 = (t-0)^2 - ((0.6t+d) - (0 + d))^2
\tau_d = 1.25 t
Where \tau_a, \tau_o, \tau_b are found by substituting the appropriate values of d into the equation.

Is that more clear?

- You said my modified diagram (modified by you) shows that clocks is not synchronized for O. Then how does the modified diagram shows that clocks is synchronized for R?

First, do you agree that the clocks each read 0 at the black dots in your scenario?

If so, then along each worldline simply find points of equal spacetime interval from the black dots and note that they occur simultaneously for R.

If not, then please mark where each clock reads 0 clearly on your diagram.

- I cannot understand that why you don't want to understand this: If a clock is at rest in a frame then its proper time (what time interval it has covered in the frame) is equal to co-ordinate time of the frame (because covered space interval is zero in the frame).

Again, my watch and my desk clock are both at rest in my frame, and they have different readings of proper time. One says 1:19 and the other says 1:21. So it is not possible for them to both be equal to the co-ordinate time. They are not synchronized. This disproves your claim by counterexample.

You are focusing on the fact that they are running at the same rate as coordinate time in my frame. However, that is not relevant to determining synchronization. Two clocks may be running fast or slow and yet be synchronized. They may be running at the correct rate and yet not be synchronized. You are mixing two separate concepts.

- I cannot understand that why you don't want to understand this: Line of simultaneity is the line on which, two events occurred, called simultaneous events. So short hand and long hand of clocks on the line meets at 12 must be simultaneous events on the line of simultaneity. So clocks on the line can be called synchronized. If clocks on line of simultaneity is not synchronized then what is other method you have to prove that two events occurred on the line of simultaneity is simultaneous events?

You are again confusing simultaneity with synchronization. Simply because you can draw a line of simultaneity on a spacetime diagram does not imply that all clocks will read the same on that line of simultaneity unless the clocks are synchronized.

The procedure for synchronization was given by Einstein. In your problem setup you stated that the clocks were initially synchronized in the unprimed frame. Meaning that Einstein's procedure synchronization procedure was carried out initially.

Because the clocks were synchronized in the unprimed frame, and because their velocity profile is the same in the unprimed frame at all times, they remain synchronized in the unprimed frame, as I proved. Because they are synchronized in the unprimed frame, they are not synchronized in the primed frame frame, as I proved.

- Accelerating scenario can be made up using too many small inertial frames changing. My diagram shows how lines of simultaneity of O's frame becomes unparallel to space axis of R's frame (during acceleration and after acceleration). So, clocks on the lines of simultaneity becomes desynchronized for R. But, lines of simultaneity is always parallel to space axis of whatever small inertial frame in which O exist at that time. So, clocks on the lines of simultaneity remain synchronized for O.

No they don't. Look at the gaps between your "sweeping" lines of simultaneity. Do those gaps look like they are marking equal spacetime intervals for each clock? Please answer this question in bold directly.

- I have proved that train's clocks is synchronized for O and not synchronized for R using diagram and maths (LT equation).

No you haven't. You first drew physically impossible diagrams, then you misinterpreted your own diagrams. Your math is incorrect because you tried to use a transform to specify a problem rather than to transform it into a new reference frame.

 

[mananvpanchal]

Can you rephrase this in terms of proper times ? t_{abc}=0 is only true in the unprimed frame.

I've run out of time. Try calculating proper times by triangulation from the diagrams using dtau^2 = dt^2-dx^2.

Suppose, that at some t_a=t_b=t_c=-0.001, value of t'_a=t'_b=t'_c=-0.001 too. But, at t_a=t_b=t_c=0, what should be the value of t_a=?, t_b=?, t_c=?

I think I don't need difference of proper time d\tau and difference of co-ordinate time dt. Please, clarify your point if I have misunderstood this.

 

[mananvpanchal]

This is a valid point. My apologies for the confusion. There are always 3 values of τ in both frames. So I should have written:
for \tau_d>0
(\tau_d - 0)^2 = (t-0)^2 - ((0.6t+d) - (0 + d))^2
\tau_d = 1.25 t
Where \tau_a, \tau_o, \tau_b are found by substituting the appropriate values of d into the equation.

Is that more clear?

No, this is more confusing. If there are three values of \tau, then you should have three values of t using the \tau = 1.25 t equation. Which implies that clocks is not synchronized in R's frame.

No they don't. Look at the gaps between your "sweeping" lines of simultaneity. Do those gaps look like they are marking equal spacetime intervals for each clock? Please answer this question in bold directly.

No, they are not marking equal spacetime intervals for each clock in R's frame.

 

[mananvpanchal]

@Mentz114

Yes, I have to agree that changing the frame of three clocks is simultaneous events for unprimed frame, so it should not be simultaneous for primed frame. From primed frame, first C changes its frame, then B and then A. This is easily understandable with two inertial frames that in one this events is simultaneous, but not in other. But, with this frame change scenario this is very hard for me to understand this. Because, the one by one frame changing of clocks require some time. And the time duration is the cause of desynchronization. So, I want to know how do the clocks changes its time readings in primed frame to be desynchronized?

 

[Mentz114]

Yes, I have to agree that changing the frame of three clocks is simultaneous events for unprimed frame, so it should not be simultaneous for primed frame. From primed frame, first C changes its frame, then B and then A.

I think A', then B' then C' change speed.

So, I want to know how do the clocks changes its time readings in primed frame to be desynchronized?

I can tell you how much, but not how or why. It is not difficult to calculate the proper times on the primed clocks when C' changes speed using the second diagram.

 

[mananvpanchal]

I think A', then B' then C' change speed.

Yes, I forgot that labels are in reverse order in diagram.

I can tell you how much, but not how or why.

That is fine

It is not difficult to calculate the proper times on the primed clocks when C' changes speed using the second diagram.

Should I use LT equation to calculate or some other?

 

[Mentz114]

The simultaneity is not straightforward in this scenario. In the diagram you can see that each of the primed observers will see the other two primed observers changing speed after they have done it.

To calculate proper times use the metric d\tau^2 = dt^2 - dx^2. You can use either diagram because (obviously) proper times will be the same between the same events in either.

I'll try the calculation myself later because I'm going to busy for some hours.

 

[Dale]

No, this is more confusing. If there are three values of \tau, then you should have three values of t using the \tau = 1.25 t equation. Which implies that clocks is not synchronized in R's frame.

Why? The only reason that you would need 3 values of something is if it were the parameter of the 3 worldlines. 3 worldlines -> 3 parameters. The t is coordinate time, not a parameter. The 3 values of τ are the parameters for the 3 worldlines. You don't need to have 6 parameters for 3 worldlines, so why would you need 3 values of τ and also 3 values of t?

You can use t as the parameter along the worldline, but then it gets confusing about whether you are talking about t one of the 3 parameters, or t coordinate time. This is another reason most people like to parameterize by proper time instead. Once you have parameterized the worldlines by τ then there is no need for further parameterization and to do so would only lead to confusion.

In any case, the t on the right hand side of the spacetime interval equation is the coordinate time here.

No, they are not marking equal spacetime intervals for each clock in R's frame.

Then clearly the clocks cannot remain synchronized at those lines of simultaneity.

I mentioned this back in post 49, but I guess I didn't do a good job describing the importance. Let me try to describe it now. Suppose you have two clocks undergoing arbitrary motion, and you draw two spacelike lines intersecting the worldlines of the clocks. Suppose further that you are given that at the first line the clocks both read 0. Then, to determine what each clock reads at the intersection with the next line, all you have to do is calculate the spacetime interval along the clock's path. If it is longer for one of the clocks then that clock will read a larger number. If the lines are lines of simultaneity then the clocks read the same number at the first line of simultaneity (synchronized) but read different numbers at the second line of simultaneity (desynchronized).

 

[mananvpanchal]

The simultaneity is not straightforward in this scenario. In the diagram you can see that each of the primed observers will see the other two primed observers changing speed after they have done it.

No, actually I don't. I see that A' changes frame first, then B' and then C'. They all should agree on sequence of events.

In train lightning example if we put three observers in train and three at platform. They all 6 agree that lightning occurs in platform frame simultaneously and in train frame unsimultaneously (with sequence).

In our example all three clocks A', B' and C' have changed frame simultaneously for unprimed frame, but they haven't for primed frame. They all A, B, C, A', B', C' should agree on simultaneous changing of frame for unprimed frame and sequence of unsimultaneous changing of frame for primed frame.

If they all don't agree with each other then I afraid their is very big confusion going on.

To calculate proper times use the metric d\tau^2 = dt^2 - dx^2. You can use either diagram because (obviously) proper times will be the same between the same events in either.

I really don't know how to calculate using that equation that is why I am asking you.

 

[Mentz114]

OK, I've done the calculation. Refer to the diagram. All quantities are in the primed frame.

I've labelled some points on the worldlines of A' and B'. a is the event where A' changes speed. The elapsed time on A' clock between a->b is ab. The elapsed time on B' clock is pq = ac.
So
<br /> \tau^2_{B&#039;}=\tau^2_{A&#039;}-(bc)^2= \tau^2_{A&#039;}(1-\beta^2) = \tau^2_{A&#039;}/\gamma^2<br />
I could have a made a mistake in my haste, but you can easily check my work now.

If they all don't agree with each other then I afraid their is very big confusion going on.

I'm sorry you find it confusing because taking into account light travel time that is what happens. If they calculate back they get the expected result.

 

[mananvpanchal]

<br /> \tau^2_{B&#039;}=\tau^2_{A&#039;}-(bc)^2= \tau^2_{A&#039;}(1-\beta^2) = \tau^2_{A&#039;}/\gamma^2<br />

Ok, this is fine. But...

Where is my confusion lies:

When A', B' and C' changes its frame simultaneously in unprimed frame at some time t_{a&#039;}=t_{b&#039;}=t_{c&#039;}=0.
Now, in primed frame first A' changes its frame at t_{a&#039;}=t&#039;_{a&#039;}=0. After, some time B' changes its frame at t_{b&#039;}=t&#039;_{b&#039;}=0, at this time A's value is reached to t&#039;_{a&#039;}=t&#039;_{a&#039;1} in primed frame. So, A' and B' is no longer synchronized in primed frame because t&#039;_{b&#039;}=0 and t&#039;_{a&#039;}=t&#039;_{a&#039;1}. A' is ahead of B' in primed frame.
So, far this is ok for me.

Now, confusion starts. I afraid here. Because a moment before changing frame (t_{a&#039;}=-0.000001) A' see same value in B' (t_{b&#039;}=-0.000001) in unprimed frame. So A' sees value < -0.000001 in B' after changing its frame. Because, A' reaches to t&#039;_{a&#039;}=t&#039;_{a&#039;1} when B' have t_{b&#039;}=t&#039;_{b&#039;}=0. So A' sees sudden decreasing change in B' clock's reading after changing its frame.

Please, look at your diagram when A' in unprimed frame at point "a". A' sees value in B' clock at point "q" because line of simultaneity is skewed in unprimed frame. But, after changing frame, line of simultaneity is changed for A'. So now A' sees value in B' clock at point "p". Now, value at "p" < value at "q", so A' sees sudden decreasing change in B' clock's reading after changing its frame.

Am I clear so far?

 

[Mentz114]

Ok, this is fine. But...

Where is my confusion lies:

When A', B' and C' changes its frame simultaneously in unprimed frame at some time t_{a&#039;}=t_{b&#039;}=t_{c&#039;}=0.
Now, in primed frame first A' changes its frame at t_{a&#039;}=t&#039;_{a&#039;}=0. After, some time B' changes its frame at t_{b&#039;}=t&#039;_{b&#039;}=0, at this time A's value is reached to t&#039;_{a&#039;}=t&#039;_{a&#039;1} in primed frame. So, A' and B' is no longer synchronized in primed frame because t&#039;_{b&#039;}=0 and t&#039;_{a&#039;}=t&#039;_{a&#039;1}. A' is ahead of B' in primed frame.
So, far this is ok for me.

Now, confusion starts. I afraid here. Because a moment before changing frame (t_{a&#039;}=-0.000001) A' see same value in B' (t_{b&#039;}=-0.000001) in unprimed frame. So A' sees value < -0.000001 in B' after changing its frame. Because, A' reaches to t&#039;_{a&#039;}=t&#039;_{a&#039;1} when B' have t_{b&#039;}=t&#039;_{b&#039;}=0. So A' sees sudden decreasing change in B' clock's reading after changing its frame.

Please, look at your diagram when A' in unprimed frame at point "a". A' sees value in B' clock at point "q" because line of simultaneity is skewed in unprimed frame. But, after changing frame, line of simultaneity is changed for A'. So now A' sees value in B' clock at point "p". Now, value at "p" < value at "q", so A' sees sudden decreasing change in B' clock's reading after changing its frame.

Am I clear so far?

A' observer cannot see anything until the light reaches them from B' worldline. Doing the calculation means nothing because the two points ( events) involved are not in causal contact. They are outside each others light-cone. It is not a physical number you are calculating.

I don't understand your notation. Please don't use 't'. The only thing I will work in is proper times, denoted by \tau.

 

[mananvpanchal]

A' observer cannot see anything until the light reaches them from B' worldline. Doing the calculation means nothing because the two points ( events) involved are not in causal contact. They are outside each others light-cone. It is not a physical number you are calculating.

I don't understand your notation. Please don't use 't'. The only thing I will work in is proper times, denoted by \tau.

Yes, you are right. I know that that A' actually doesn't see decreasing value of B' clock because of light propagation time. This is just calculation on line of simultaneity.

Thanks

 

[mananvpanchal]

Hello Mentz114

I have a doubt here.

I have added two line of simultaneity of unprimed frame in your diagram. Suppose, two events occurred on A's path and B's path exactly after first line of simultaneity in unprimed frame. The event occurred after "p" on B's path emits light pulse toward A'. We don't know when light pulse is reached to A'. But, we are sure that light pulse is already emitted by the event.

Now, A' changes its frame. After changing the frame, A's line of simultaneity of primed frame becomes horizontal line through "p". So, that event occurred before frame change on B's path already emitted light pulse to A'. Now, the line of simultaneity again reach to the same point.

Would that event be reoccurred for A'?
Does A' see two light pulse (one from event occurred before frame change and other from event occurred after frame change)?
Would the all events between "p" and "q" be reoccurred for A'?

I think instant frame change might be the cause of the problem here. Am I right?

 

[Mentz114]

Hello Mentz114

Would that event be reoccurred for A'?
Does A' see two light pulse (one from event occurred before frame change and other from event occurred after frame change)?
Would the all events between "p" and "q" be reoccurred for A'?

I think instant frame change might be the cause of the problem here. Am I right?

No, the event would not reoccurr. That would be a causal paradox.

There's no problem. I've tried to add regular light pulses from B/B' to A/A'. As you can see, everything is regular. A and A' see a Doppler shift when either B' or A' change speed.

I don't think there is anything to be gained by looking for problems in this scenario because there aren't any. You are making problems by doing a calculation for the unprimed frame in the primed frame.

 

[mananvpanchal]

No, the event would not reoccurr. That would be a causal paradox.

I am sorry, if you guys find that I am creating problems here.

Anyway, I have some other doubts related to instant frame change scenario.

Please, look at this diagram.

We cannot find contracted length between clocks of primed frame for unprimed frame in above digram. Because, rest length in unprimed frame is defines in diagram is about 10, and if we try to find contracted length by seeing the length between clocks at same time in unprimed frame we again get 10.

We can easily find here contracted length between clocks of unprimed frame for primed frame in above digram. Here, we can see that rest length in primed frame is about 12, whereas contracted length in primed frame is about 9.

I think the problem here might be:
In second image we have described frame change events unsimultaneously. That is why the length between vertical world lines in primed frame are more than the length between diagonal world lines in unprimed frame. But in first image the events described are simultaneous. That is why the length between vertical world lines in unprimed frame are equal to the length between diagonal world lines in primed frame.

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

 

[Mentz114]

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 taking a small time compared to the grid scale.

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