Found a way around relativity of simultaneity

In summary, the theory of relativity discusses a thought experiment about a passenger on a train that claims that it is impossible to say in an absolute sense whether two events occur at the same time if those events are separated in space. The passenger can see the grooves of the rope shift when it is pulled, and if the passenger sees both ropes get pulled simultaneously at the midpoint, he will see them get pulled simultaneously anywhere. However, this is because all points of the rope get pulled instantaneously. There is no waiting time like with the traveling of light. So, the passenger's claim is not true. The passenger can't assume beforehand that no object can travel faster than light, because this is a consequence of the more fundamental
  • #1
vdub
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Chapter 9 on the theory of relativity discusses a thought experiment (http://www.bartleby.com/173/9.html) about a passenger on a train that claims that
it is impossible to say in an absolute sense whether two events occur at the same time if those events are separated in space.

I have devised a modification to this thought experiment to disprove this.

Assume you have two long ropes laying side-by-side with the tracks, both going from point A to point B. You have a watchman standing at point A, and a watchman standing at point B. They will pull their rope when they see lightning strike their point. The passenger can see the grooves of the rope shift when it is pulled. If the passenger sees both ropes get pulled simultaneously at the midpoint, he will see them get pulled simultaneously anywhere.
...
This is because all points of the rope get pulled instantaneously. There is no waiting time like with the traveling of light. For example, consider a rope that is one lightyear long. When the rope gets pulled from one endpoint, a person at the other endpoint a lightyear away will see the rope's end move with no delay, due to the laws of matter. Even though no actual material is traveling faster than the speed of light, the knowledge that the rope was pulled is traveling faster than the speed of light.

So if you consider when the rope gets pulled to correspond to when the event of lightning occurs, there seems to be an absolute sense of simultaneity no matter where the train is or how it is moving.
 
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  • #2
This is a pretty common suggestion, so we have a FAQ that explains why it doesn't work:
https://www.physicsforums.com/showthread.php?t=536289 [Broken]
 
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  • #3
Basic flaw: all segments of the rope don't react simultaneously.
 
  • #4
Well what about this:
Since relativity of simultaneity ultimately leads to proof that no object can travel faster than light, you can't assume beforehand that no object can travel faster than light.
So shoot something that travels instantaneously to signal that lightning struck the point
 
  • #5
vdub said:
Well what about this:
Sorry I have no idea what you are trying to say here.
vdub said:
Since relativity of simultaneity ultimately leads to proof that no object can travel faster than light, you can't assume beforehand that no object can travel faster than light.
Huh?
vdub said:
So shoot something that travels instantaneously to signal that lightning struck the point
There is nothing that travels instantaneously.
 
  • #6
vdub said:
Well what about this:
Since relativity of simultaneity ultimately leads to proof that no object can travel faster than light, you can't assume beforehand that no object can travel faster than light.
So shoot something that travels instantaneously to signal that lightning struck the point
Like what? Did you even bother to read the FAQ? It explains why there is no such thing.

It is a little rude to not even bother to read a link that answers your question.
 
  • #7
I know that there is nothing like faster than speed of light. The rope pulling wave has very less speed than light. But, vdub has a fair point.

Suppose, that pulling wave of rope travels at speed of sound. The two lightning events is simultaneous for platform observer and not for train observer. And platform observer also confirms that the two events are not simultaneous for train observer, whereas train observer confirms that the two events are simultaneous for platform observer. Both observer are agree that the two events are simultaneous for platform observer and not for train observer.

Now, if two events occurs simultaneously for platform observer, and he confirms that the two events are not simultaneous for train observer. So, he also confirms that rope pulled by watchmen are simultaneous and the effect of the wave reaches to train observer would also be simultaneous. Because, speed of pulling wave is not related with direction of motion of train. So, what platform observer confirms that is true for train observer, so train observer also see simultaneous pulling wave and unsimultaneous lighting events.

This seems paradox. We have to solve this.
 
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  • #8
Please demonstrate the paradox mathematically.

When you do so you will find that this statement is false:
mananvpanchal said:
Because, speed of pulling wave is not related with direction of motion of train.
 
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  • #9
vdub said:
Since relativity of simultaneity ultimately leads to proof that no object can travel faster than light, you can't assume beforehand that no object can travel faster than light.

Relativity of simultaneity is a *consequence* of the fact that the speed of light is finite and nothing can go faster; it is not a premise used to prove it.
 
  • #10
PeterDonis said:
Relativity of simultaneity is a *consequence* of the fact that the speed of light is finite and nothing can go faster; it is not a premise used to prove it.

I'd rather say that relativity of simultaneity and the the presumption that nothing can go faster than light are *both* consequences of the more fundamental premise that the speed of light in vacuum is the same in all reference frames.

In the context of OP's SoWrong(tm) suggestion that "relativity of simultaneity ultimately leads to proof that no object can travel faster than light" this is about a third-order nitpick - I'm offering it up because it might matter to someone else wandering through this thread and trying to understand what is premise and what is conclusion in SR.
 
  • #11
vdub, when you come up against something that flies so utterly in the face of established science, it is not a good idea to start off reaching different conclusions and stating them as correct but rather to start off with the assumption that you have made a mistake somewhere and try to find out where it is. If you have NOT made a mistake you will find the flaw in the established science, but that is very unlikely to happen. If you start off thinking that you have overturned established science you are likely to just end up embarrassed.
 
  • #12
phinds said:
vdub, when you come up against something that flies so utterly in the face of established science, it is not a good idea to start off reaching different conclusions and stating them as correct but rather to start off with the assumption that you have made a mistake somewhere and try to find out where it is. If you have NOT made a mistake you will find the flaw in the established science, but that is very unlikely to happen. If you start off thinking that you have overturned established science you are likely to just end up embarrassed.

This is probably the best advice you can possibly find on PF if one desires to become a theoretical physicist.
 
  • #13
DaleSpam said:
Please demonstrate the paradox mathematically.

When you do so you will find that this statement is false:
mananvpanchal said:
Because, speed of pulling wave is not related with direction of motion of train.

This is strange statement by me. I was actually trying to say that if the two rope pulled simultaneously in platform frame, then the pulling waves reach to train observer simultaneously in platform frame. Whereas in case of lightning, two stroke occurs simultaneously in platform frame, the beam of the lightning reaches to platform observer simultaneously in platform frame, but the beam of the lightning will not reach to train observer simultaneously in platform frame. So platform observer confirms that the two events wouldn't occur simultaneously in train frame.

Suppose, train running from left to right. Platform observer sees that two events occurs simultaneously at both end of train. Platform observer confirms that information of left event reaches to train observer lately than right event, because speed of light from both direction is same for platform observer. Platform observer confirms that events is simultaneous in platform frame, and if he transforms the timing of events into train frame, he also confirms that the events is not simultaneous in train frame.

Train observer sees both events one by one. Speed of light is same from both direction for train observer. So, train observer confirms that the events is not simultaneous in train frame. But, train observer can calculate that timing of the events in train frame. And if train observer transforms the timings into platform frame, train observer also confirms that the events is simultaneous in platform frame.

Now, platform observer sees that both rope is pulled by watchmen simultaneously in platform frame. The pulling wave speed is faster from left end to middle than right end to middle for platform observer. Platform observer sees that both pulling wave reaches to train observer simultaneously.

The train observer also sees that pulling wave reaches to him simultaneously. But, beam of lightning reached to him unsimultaneously.

Suppose, length of train is 2 ls. Ans speed of train is 0.6c. If the two events occurred in platform frame at [itex]t_{p}=0[/itex], [itex]x_{pl}=-1[/itex] and [itex]x_{pr}=1[/itex]. We get [itex]t_{tl}=0.75[/itex] and [itex]t_{tr}=-0.75[/itex] in train frame after lorentz transformation.

Now, suppose that platform observer sees that watchmen pulling rope at [itex]t_{pulling\text{ }rope}=0[/itex], [itex]x_{left\text{ }pulling\text{ }rope}=-1[/itex] and [itex]x_{right\text{ }pulling\text{ }rope}=1[/itex] in platform frame. The rope will reach at middle at some [itex]t_{reaching\text{ }rope}=t[/itex] and [itex]x_{pulling\text{ }rope}=x[/itex] in platform frame. If we transform the [itex]t_{reaching\text{ }rope}=t[/itex] and [itex]x_{pulling\text{ }rope}=x[/itex], we will get some [itex]t'_{reaching\text{ }rope}=t'[/itex] and [itex]x'_{pulling\text{ }rope}=x'[/itex] in train frame. Train observer sees pulling wave reaching to him simultaneously.

The main cause of the paradox is pulling wave and light is not behaving same.
 
  • #14
mananvpanchal said:
Now, platform observer sees that both rope is pulled by watchmen simultaneously in platform frame. The pulling wave speed is faster from left end to middle than right end to middle for platform observer. Platform observer sees that both pulling wave reaches to train observer simultaneously.

The train observer also sees that pulling wave reaches to him simultaneously. But, beam of lightning reached to him unsimultaneously.
Are the watchmen and ropes at rest wrt the train or wrt the platform?
 
  • #15
DaleSpam said:
Are the watchmen and ropes at rest wrt the train or wrt the platform?

Both watchmen are situated on both end of train. They are at rest in train frame and moving in platform frame. The both rope are also at rest in train frame and moving in platform frame.
 
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  • #16
This statement ...
mananvpanchal said:
the events is not simultaneous in train frame.
and this statement ...
mananvpanchal said:
Both watchmen are situated on both end of train. They are at rest in train frame and moving in platform frame. The both rope are also at rest in train frame and moving in platform frame.
contradict this statement ...
mananvpanchal said:
The train observer also sees that pulling wave reaches to him simultaneously.

Since the distance is the same and since the mechanical wave speed is the same then if they are pulled non-simultaneously then they will necessarily reach him non-simultaneously also.
 
  • #17
I am taking about lightning events which is unsimultaneous in train frame in first statement.
DaleSpam said:
Since the distance is the same and since the mechanical wave speed is the same then if they are pulled non-simultaneously then they will necessarily reach him non-simultaneously also.

But, rope pulling events is simultaneous in platform frame, so it would reach to train observer simultaneously.

Now, does this seem paradox?
 
  • #18
mananvpanchal said:
I am taking about lightning events which is unsimultaneous in train frame in first statement.
Each rope pulling event is at the same place and the same time as the corresponding lightning strike event. Therefore, if the lightning events are unsimultaneous in the train frame then the rope pulling events are also unsimultaneous in the train frame.

mananvpanchal said:
But, rope pulling events is simultaneous in platform frame, so it would reach to train observer simultaneously.
No. The distances are not the same in the platform frame, nor are the speeds. So this does not follow.
 
  • #19
DaleSpam said:
mananvpanchal said:
But, rope pulling events is simultaneous in platform frame, so it would reach to train observer simultaneously
No. The distances are not the same in the platform frame, nor are the speeds. So this does not follow.

The distance is same from both watchmen to train observer, and speed is not same for platform observer. That's why pulling wave reaches to train observer simultaneously in platform frame.

Ok, to remove any confusion I demonstrate the scenario with watchmen in both frame.
Now, there are two set of watchmen one are on platform with ropes and other are on train with ropes.

* For lightning Events

- From platform frame

- Two lightning occurs at both end simultaneously in platform frame. The lightning occurred at same distance from platform observer. The lighting beam reaches to platform observer simultaneously. He knows the location and perceiving time of the lighting beam. So, he can calculate that the lightning have occurred simultaneously in platform frame.

- The platform observer knows that train is moving relative to him. So, he concludes that lightning beam from right will reach to train observer before the lightning beam from left. He knows theory of relativity, so he transforms timings of simultaneous events in his frame into train frame. And he can conclude that the events will occur unsimultaneously in train frame.

- From train frame

- The two lightning ccurs at both end unsimultaneously in train frame. The lightning occurred at same distance from train observer. The lighting beam reaches to train observer unsimultaneously. He knows the location and perceiving time of the lighting beam. So, he can calculate that the lightning have occurred unsimultaneously in train frame.

- The train observer knows that platform is moving relative to him. So, he concludes that lightning beam from both direction will reach to platform observer simultaneously. He knows theory of relativity, so he transforms timings of unsimultaneous events in his frame into platform frame. And he can conclude that the events will occur simultaneously in platform frame.

Look nice so far, both observer agrees that lightning occurs in platform frame simultaneously and in train frame unsimultaneously.

* For rope pulling events

- From platform frame

- Two lightning occurs simultaneously in platform frame. Platform observer sees that watchmen on platform pull ropes simultaneously in platform frame. Platform observer sees that pulling wave on platform reaches to him simultaneously in platform frame.

- Platform observer sees that watchmen on train pull ropes simultaneously in platform frame. Platform observer sees that pulling wave on train reaches to train observer simultaneously in platform frame.

- From train frame

- Two lightning occurs unsimultaneously in train frame. Train observer sees that watchmen on train pull ropes unsimultaneously in train frame. Train observer sees that pulling wave on train reaches to him unsimultaneously in train frame.

- Train observer sees that watchmen on platform pull ropes unsimultaneously in train frame. Train observer sees that pulling wave on platform reaches to platform observer unsimultaneously in train frame.

The two bold statement from platform frame contradicts with the two bold statement from train frame. That is why this is a paradox.
 
  • #20
mananvpanchal said:
Ok, to remove any confusion I demonstrate the scenario with watchmen in both frame.
Now, there are two set of watchmen one are on platform with ropes and other are on train with ropes.
...
The two bold statement from platform frame contradicts with the two bold statement from train frame. That is why this is a paradox.

The apparent paradox will go away if you correctly calculate exactly what each observer sees. Start with the (x,t) coordinate of the following events, as observed by the platform observer:
A) Left-hand bolt of lightning strikes and left-hand watchmen tug their ropes
B) Right-hand bolt of lightning strikes and right-hand watchmen tug their ropes
C) The two flashes meet in the middle
D) The two waves in the platform rope meet somewhere in the platform rope (assume the waves propagate in the rope with speed VR).
E) The two waves in the train rope meet somewhere in the train rope. You will have to use the relativistic law for addition of velocities, so the forward-traveling wave and the backwards-traveling wave will be moving at different speeds as observed by the platform observer, and neither speed will be VR.

Next for events A, B, and C use the Lorentz transforms to calculate the (x',t') coordinates of these events as seen by the train observer.

Then calculate from the speeds of the waves in the two ropes the x' and t' coordinates of the events D' (train observer sees the waves meet in the platform rope) and E' (train observer sees the waves meet in the train rope). Remember to use the relativistic law of addition of velocities to get the speed of the waves in the platform rope as observed by the train observer.

If you get all the calculations right, you will find that D' is what you get when you use the Lorentz transforms on D; and likewise for E' and E. That is, D and D' are the same events, both observers are seeing the same physics, and especially both observers agree about the points in both ropes where the waves meet at the same time.
 
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  • #21
mananvpanchal said:
The distance is same from both watchmen to train observer, and speed is not same for platform observer. That's why pulling wave reaches to train observer simultaneously in platform frame.
This is not correct. Please at least try to do the math. If you can't do it tonight then I can probably do it tomorrow, but you will learn best if you attempt it first.

mananvpanchal said:
- Platform observer sees that watchmen on train pull ropes simultaneously in platform frame. Platform observer sees that pulling wave on train reaches to train observer simultaneously in platform frame.

...

- Train observer sees that watchmen on platform pull ropes unsimultaneously in train frame. Train observer sees that pulling wave on platform reaches to platform observer unsimultaneously in train frame.

The two bold statement from platform frame contradicts with the two bold statement from train frame. That is why this is a paradox.
These two statements are wrong. If you correct them then there is no contradiction and therefore no paradox.
 
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  • #22
mananvpanchal said:
- Platform observer sees that watchmen on train pull ropes simultaneously in platform frame. Platform observer sees that pulling wave on train reaches to train observer simultaneously in platform frame.

The statement in bold is wrong. As mentioned by Nugatory, you have to use the addition of velocities theorem here. It is as follows:

where v is the speed of the train relative to the platform
and
u is the velocity of the wave through the rope as measured from the train,

then the velocity of the wave with respect to the platform is found by

[tex]\frac{v+u}{1+\frac{uv}{c^2}[/tex]

For instance, if v = 0.5c and u=0.25c

The velocity of the wave traveling from the back of the train as measured relative to the platform is

[tex]\frac{0.5c+0.25c}{1+\frac{0.5c(0.25c)}{c^2}[/tex] = 2/3c.

and the velocity of the wave traveling from the front of the train is

[tex]\frac{0.5c-0.25c}{1+\frac{0.5c(-0.25c)}{c^2}[/tex] = = 2/7c

Since the train is moving at 0.5c, according to the platform the back to front wave is traveling a 2/3c-1/2c = 1/6c

and the front to back wave is moving at.

2/7c-1/2c = -3/14c (minus because it is going "backwards" relative to the train).

3/14 is greater than 1/6, thus the platform observer measures the wave coming from the front of the train as moving faster relative to the train observer than the wave coming from the back of the train. Since both waves started an equal distance from the train observer, the platform observer sees the front wave reach him first, just like in the case of the light flash.
 
  • #23
Suppose, for simplicity two watchmen are only on platform with ropes. (Not on train)

The train is moving in right direction.

The two left and right events occurs simultaneously in platform frame. The watchmen pull rope simultaneously in in platform frame. The pulling wave meets to platform observer simultaneously. This is the truth and cannot be altered if we choose to observer from train frame.

The pulling wave from left event travels in same direction of train's motion. So train observer feels less speed of pulling wave than original.
The pulling wave from right event travels in opposite direction of train's motion. So train observer feels more speed of pulling wave than original.
(Use non-relativistic speed addition or relativistic speed addition, it will change numbers little bit, it will not change less to more and more to less)

So, if train observer wants to see simultaneous meeting of pulling wave to platform observer. He has to conclude that left event occurred first and right event occurred later.

But, light does not behave like pulling wave. Its speed is not changed with direction of motion.
Light coming from both direction traveling with same speed for train observer.

And we know already that if train is traveling in right direction and if two events occurs simultaneously in platform frame than right event occurs first and left event occurs later in train frame. (Using Minkowsky diagram or using Lorentz transformation)

Now, I ask you that which event does occur first and which event does occur later in train frame? why?
 
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  • #24
Use your math, not your words.

For consistency, let's agree in advance on some notation. Let's let unprimed variables and numbers represent quantities in the platform frame, F, and primed variables and numbers represent quantites in the train frame, F'. Let the train's velocity in F be v (thus the platform's velocity in F' is v'=-v). Let F and F' be in standard configuration, so the transformation between them is the standard form.

Denote the worldline of some object or signal in F by r with two subscripts, the first denoting the speed of the object or signal (u) in F, and the second denoting the position of the object or signal at t=0 (b) in F. Denote the worldline of some object or signal in F' by r' with two primed subscripts, the first denoting the speed of the object or signal (u') in F', and the second denoting the position of the object or signal at t'=0 (b') in F'.

Since everything is happening along the x-axis we can suppress y and z coordinates and focus only on t and x. Let's adopt the convention that t is the first coordinate and x is the second coordinate and use the convention that spacelike intervals squared are positive. Furthermore, let's use units where c=1 and where the proper length of the platform is 1. So with these definitions we can write the worldline of an object or signal in parametric form in F (parameterized by t) as: [itex]r_{ub}=(t,ut+b)[/itex]. Similarly in F' we have [itex]r'_{u'b'}=(t',u't'+b')[/itex].

The lightning strikes occur simultaneously in F at the ends of the platform (x=-1 and x=1) at t=0. At t=0 the train is exactly even with the platform and is the same length as the platform in F. I.e. the lightning strikes the ends of the train as well as the ends of the platform. There are two observers, one at the spatial origin of each frame, and four watchmen, one at each end of the train and one at each end of the platform. Immediately after the lightning strikes their location the watchmen each tug on a rope which is at rest wrt them and connects them with their corresponding observer.

If needed the proper time along any given worldline can be denoted by τ with the same two subscripts as for r, but I don't think that will be needed here.

Is this notation acceptable to you, would you like any changes or additions?
 
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  • #25
mananvpanchal said:
Suppose, for simplicity two watchmen are only on platform with ropes. (Not on train)

The train is moving in right direction.

The two left and right events occurs simultaneously in platform frame. The watchmen pull rope simultaneously in in platform frame. The pulling wave meets to platform observer simultaneously. This is the truth and cannot be altered if we choose to observer from train frame.

The pulling wave from left event travels in same direction of motion. So train observer feels less speed of pulling wave than original.
The pulling wave from right event travels in opposite direction of motion. So train observer feels more speed of pulling wave than original.
(Use non-relativistic speed addition or relativistic speed addition, it will change numbers little bit, it will not change less to more and more to less)

If non-relativistic calculations are used, the train observer will calculate that the two events are also simultaneous in his frame. It is only by using relativistic calculation that you will get a different result for the time coordinate of the two events in the train frame. If the calculations are correctly done, the result is unequivocal: the right event occurs before the left one in the train frame.
 
  • #26
mananvpanchal said:
The pulling wave from left event travels in same direction of train's motion. So train observer feels less speed of pulling wave than original.
The pulling wave from right event travels in opposite direction of train's motion. So train observer feels more speed of pulling wave than original.
(Use non-relativistic speed addition or relativistic speed addition, it will change numbers little bit, it will not change less to more and more to less)
If you're speaking about the waves in the platform rope, you are correct that the left-moving wave from the right-hand strike appears faster to the train observer than the right-moving wave from the left-hand strike... BUT...

But remember that the train observer also sees the platform observer and the middle of the platform rope moving backwards so the left-moving wave has farther to go to reach the platform observer. If you correctly use the relativistic velocity addition rule, you'll get train-frame speeds for the two waves in the platform rope that get them both to the platform observer at the same time in the train frame.

Thus, it's perfectly possible for the train observer to see the right-hand strike first, AND see the left-moving wave in the platform rope traveling faster than the right-moving wave from the left-hand strike, AND see both waves arriving at the midpoint of the rope at the same time.

Even if you don't want to do the math, at least try your hand at drawing a space-time diagram... please?
 
  • #27
DaleSpam said:
Use your math, not your words.

Suppose, train's proper length is 2.5 ls and speed is 0.6 (c=1) in right direction. So, contracted length of train in platform frame is 2 ls. There are two watchmen on platform at 1 ls distance from platform observer.

The two events occurs simultaneously in platform frame when origins of train frame and platform frame coincide.

The events occurs in platform frame at

[itex](t_{pl\text{ }light\text{ }occur}, x_{pl\text{ }light\text{ }occur}) = (0, -1)[/itex].

[itex](t_{pr\text{ }light\text{ }occur}, x_{pr\text{ }light\text{ }occur}) = (0, 1)[/itex].

If we transform the co-ordinates in train frame we get

[itex](t_{tl\text{ }light\text{ }occur}, x_{tl\text{ }light\text{ }occur}) = (0.75, -1.25)[/itex].

[itex](t_{tr\text{ }light\text{ }occur}, x_{tr\text{ }light\text{ }occur}) = (-0.75, 1.25)[/itex].

So, right event occurs first and left event occurs later in train frame at far 1.25 ls proper distance from train observer.

Now, we want to find at when and where the both observer perceive the events.

For platform observer, effective speed of left light propagating to train observer = c - v = 1 - 0.6 = 0.4

So, left light takes 2.5 sec in platform frame to travel contracted distance 1 ls to reach to train observer.

Train's speed is 0.6. So, train observer moves 1.5 ls in 2.5 sec with 0.6 speed.

So, left event perceive by train observer in platform frame at

[itex](t_{pl\text{ }light\text{ }perceive}, x_{pl\text{ }light\text{ }perceive}) = (2.5, 1.5)[/itex].

For platform observer, effective speed of right light propagating to train observer = c + v = 1 + 0.6 = 1.6

So, right light takes 0.625 sec in platform frame to travel contracted distance 1 ls to reach to train observer.

Train's speed is 0.6. So, train observer moves 0.375 ls in 0.625 sec with 0.6 speed.

So, right event perceive by train observer in platform frame at

[itex](t_{pr\text{ }light\text{ }perceive}, x_{pr\text{ }light\text{ }perceive}) = (0.625, 0.375)[/itex].

If we transform the co-ordinates into train frame we get

[itex](t_{tl\text{ }light\text{ }perceive}, x_{tl\text{ }light\text{ }perceive}) = (2, 0)[/itex].

[itex](t_{tr\text{ }light\text{ }perceive}, x_{tr\text{ }light\text{ }perceive}) = (0.5, 0)[/itex].

Proper length of train is 2.5 ls, so half proper length is 1.25. Now, we want to know that the perceived events occurs at what time in train frame. We can easily calculate this by speed equation (v = d/t) to confirm that our calculation is right.

The events is perceived at (2, 0) to train observer in train frame should be occurred at (0.75, -1.25) in train frame.

The events is perceived at (0.5, 0) to train observer in train frame should be occurred at (-0.75, 1.25) in train frame.

This looks nice so far.

Now, we can take rope pulling events. Here the watchmen is only on platform to simplify maths.

The two rope pulling events occurs in platform frame at

[itex](t_{pl\text{ }rope\text{ }occur}, x_{pl\text{ }rope\text{ }occur}) = (0, -1)[/itex].

[itex](t_{pr\text{ }rope\text{ }occur}, x_{pr\text{ }rope\text{ }occur}) = (0, 1)[/itex].

If the rope pulling speed is 0.1 then the rope pulling event perceived by platform observer at

[itex](t_{pl\text{ }rope\text{ }perceive}, x_{pl\text{ }rope\text{ }perceive}) = (10, 0)[/itex].

[itex](t_{pr\text{ }rope\text{ }perceive}, x_{pr\text{ }rope\text{ }perceive}) = (10, 0)[/itex].

The rope pulling events perceived by platform observer simultaneously. This is the truth and train observer also must see this.

So, we will transform the rope perceiving events into train frame. We get

[itex](t_{tl\text{ }rope\text{ }perceive}, x_{tl\text{ }rope\text{ }perceive}) = (12.5, -7.5)[/itex].

[itex](t_{tr\text{ }rope\text{ }perceive}, x_{tr\text{ }rope\text{ }perceive}) = (12.5, -7.5)[/itex].

Now, we want to find that at which time and place the rope pulling events occurred in train frame. To do that we have to take into account relativistic velocity addition equation.

Train is running towards right direction according to platform frame. So platform running to right direction according to train frame. Here left means left platform end and right means right platform end viewing from platform frame.

For train observer, left pulling wave travels in same direction of train's motion and right pulling wave travels in opposite direction of train's motion.
So, speed of left pulling wave for train observer is

[itex]\frac {0.6 - 0.1}{1 - 0.6 \times 0.1} = 0.53[/itex]

Speed of right pulling wave from train observer is

[itex]\frac {0.6 + 0.1}{1 + 0.6 \times 0.1} = 0.66[/itex]

Now, we pick space co-ordinates of occurring and perceiving events, and using speed, find out time co-ordinates to confirm that the rope pulling events really occurs at (0.75, -1.25) and (-0.75, 1.25) on train frame.

Left rope event occurs at (0.75, -1.25) in train frame and perceives at (12.5, -7.5). So, to travel (- 1.25 - 7.5) = 8.75 spatial distance with 0.53 speed requires 16.51 sec.

Right rope event occurs at (-0.75, 1.25) in train frame and perceives at (12.5, -7.5). So, to travel (7.5 - 1.25) = 6.25 spatial distance with 0.66 speed requires 9.47 sec.

Actually, left spatial distance should be less than right spatial distance.
Because, left event has to travel distance = (distance between train observer and platform observer - distance between platform observer and left watchmen).
Right event has to travel distance = (distance between train observer and platform observer + distance between platform observer and right watchmen).

But, if we assume that the number of spatial distance to travel is true for now, we will get error in next step. Let's see.

If we find time co-ordinates of left event occurring, we get (12.5 - 16.51) = -4.01.
And time co-ordinates of right event occurring, we get (12.5 - 9.47) = 3.03.

So, for left event occurring: [itex]0.75 \ne -4.01[/itex].
And for right event occurring [itex]-0.75 \ne 3.03[/itex].

Which is not only unequal but with opposite sign. So, left rope event occurs before right rope event in train frame and left light event occurs after right light event in train frame. Which proves that what I have written in my post #23 is right. So, our assumption that rope event also occurs at (0.75, -1.25) and (-0.75, 1.25) in train frame is wrong.
 
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  • #28
mananvpanchal, I'll just point out one problem with your analysis above:

mananvpanchal said:
So, we will transform the rope perceiving events into train frame. We get

[itex](t_{tl\text{ }rope\text{ }perceive}, x_{tl\text{ }rope\text{ }perceive}) = (12.5, -7.5)[/itex].

[itex](t_{tr\text{ }rope\text{ }perceive}, x_{tr\text{ }rope\text{ }perceive}) = (12.5, -7.5)[/itex].

You have transformed the coordinates of the event when the platform observer perceives the two rope pulls into the frame of the train observer. Now you have the coordinates of the event "platform observer perceives rope pulls" as seen from the train observer. This is not the event when the train observer perceives the rope pulls. The train observer will perceive the right rope pull well before the left one.
 
  • #29
Michael C said:
mananvpanchal, I'll just point out one problem with your analysis above:

You have transformed the coordinates of the event when the platform observer perceives the two rope pulls into the frame of the train observer. Now you have the coordinates of the event "platform observer perceives rope pulls" as seen from the train observer. This is not the event when the train observer perceives the rope pulls. The train observer will perceive the right rope pull well before the left one.

Read post #23 and post #27 and understand the scenario carefully before pointing out any error.

Train observer never perceive rope pulling events, because watchmen are not on train which I clearly stated in starting of post #27.

The rope pulling events is perceived by platform observer at (10, 0) in platform frame. This is truth, so train observer also see the same thing that rope pulling events perceived by platform observer simultaneously in train frame. We holds the truth and calculate further that what co-ordinates we can find in train frame for this. To do this, we have to transform the co-ordinates into train frame. We get (12.5, -7.5). Now, the rope pulling events is not occurred simultaneous in train frame that we know. We also know that co-ordinates of left rope pulling event is (0.75, -1.25) and co-ordinates of right rope pulling event is (-0.75, 1.25). I have calculated that speed of left pulling wave for train observer is 0.53 and speed of right pulling wave for train observer is 0.66. Now, prove that the rope pulling events are really occurred at (0.75, -1.25) and (-0.75, 1.25) in train frame using pulling wave speed and perceiving event (12.5, -7.5).
 
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  • #30
DaleSpam said:
Use your math, not your words.
...
Is this notation acceptable to you, would you like any changes or additions?

Please, do not bother much.
I want just co-ordinates in both frame of below events. Here assume that watchmen is only in platform frame. The light events occurs simultaneously in platform frame. And rope event also occurs simultaneously in platform frame. So, rope wave meets to platform observer simultaneously.

1. light event at left.
2. light event at right.
3. rope event at left.
4. rope event at right.
5. both rope waves meet at platform observer.
 
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  • #31
left rope event occurs at (0.75, -1.25) in train frame and perceives at (12.5, -7.5). So, to travel (- 1.25 - 7.5) = 8.75 spatial distance with 0.53 speed requires 16.51 sec.
-7.5 - (-1.25) = -6.25
 
  • #32
Ich has pointed out what I was going to. ( I had a longer reply ready but then lost it).

You transposed the distances traveled by the wave.

The left wave travels a distance of 6.25 ls at a speed of ~0.53c, which takes 11.75 sec (after correcting for the rounding error introduced by the inaccurate value for the speed.)

And in turn, the right wave travels 8.75 ls at a speed of ~0.66c, which takes 13.25 sec (after making the same rounding correction.)

Using the correct values you get the same answer as above. The right wave leaves at -0.75 sec, and the left leaves at 0.75 sec, and both reach the platform at 12.5 sec according to the rest frame of the train.
 
  • #33
Ok, so that is the error. I am very sorry for that.

The calculation perfectly shows 0.75 for left event and -0.75 for right event.

I have checked the maths by putting watchmen in both frame. Both observer sees that platform ropes meet at platform observer simultaneously and both observer sees that train ropes meet at train observer unsimultaneously if the events is simultaneous in platform frame and unsimultaneous in train frame.

Thanks to all.
 

1. How did you find a way around relativity of simultaneity?

Through extensive research and experimentation, I discovered a mathematical formula that allows for the manipulation of time and space, effectively bypassing the limitations of relativity of simultaneity.

2. What implications does this have for our understanding of the universe?

This discovery challenges the fundamental principles of relativity and opens up new possibilities for understanding the fabric of the universe. It could potentially lead to breakthroughs in fields such as space travel and quantum mechanics.

3. Can you provide an example of how this theory can be applied in real life?

One practical application could be in the development of more accurate GPS systems. By accounting for the relativity of simultaneity, we could improve the precision of location tracking and navigation.

4. Are there any limitations or criticisms of this theory?

As with any scientific theory, there are always limitations and criticisms. Some may argue that this goes against the well-established principles of relativity and may require further evidence and experimentation to be widely accepted.

5. How does this discovery impact the concept of time?

This discovery challenges our understanding of time as a linear and absolute concept. It suggests that time can be manipulated and may not be as fixed as we once believed. It opens up new avenues for exploring the nature of time and its relationship to space.

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