Simultaneity in SR

[Grimble]

Simultaneity in SR.

1.The coordinate scales for an Inertial Frame of Reference are Proper time and Proper distance.
2.Using the Lorentz Transformations, any IFoR can calculate, from its own measurements, that the proper units of other IFoRs have the same magnitude as those proper units in its own IFoR.
3.If two rigid bodies, say Einstein's train(CD) and embankment(AB), with identical proper lengths, pass one another and we consider two Space-time events; their front ends, A & C aligning and their back ends, B & D aligning, then those two events will be simultaneous in their respective IFoRs.

Reasoning.
1.Coordinate scales
i.Proper time is the time recorded on an inertial clock.
ii.The length of the train measured in the train frame is the proper length of the train (AB)
iii.The length of the embankment , measured in the frame of the embankment is the proper length of the embankment (CD).
2.Proper units in different IFoRs are of equal magnitude.
1.They would be measured so from any other IFoR that transformed their measurements
3.Simultaneity.
i.We have two bodies passing one another.
ii.The time-like space-time interval between the two events ds²= (cdt² – (dx² + dy² + dz²) where dt = 0 is the same for each body as dy = 0, dz = 0 in both cases and dx is the proper distance between AB and betwee A' and B'

[macaw]

3.If two rigid bodies, with identical proper lengths, pass one another and we consider two Space-time events; their front ends, A & A' aligning and their back ends, B & B' aligning, then those two events will be simultaneous in their respective IFoRs.

Q1: Which are the two events ?
Q2: What makes you think that , if A and A' are aligned, B and B' will also align? SR length contraction says that , as viewed from the unprimed frame, A'B'<AB. SR also says that , as viewed from the primed frame AB<A'B'. So, it appears that the two rods do not align in either frame.

[AstroRockHunter]

Simultaneity in SR.

1.The coordinate scales for an Inertial Frame of Reference are Proper time and Proper distance.
2.Using the Lorentz Transformations, any IFoR can calculate, from its own measurements, that the proper units of other IFoRs have the same magnitude as those proper units in its own IFoR.
3.If two rigid bodies, with identical proper lengths, pass one another and we consider two Space-time events; their front ends, A & A' aligning and their back ends, B & B' aligning, then those two events will be simultaneous in their respective IFoRs.

Reasoning.
1.Coordinate scales
i.Proper time is the time recorded on an inertial clock.
ii.The length of the train measured in the train frame is the proper length of the train (AB)
iii.The length of the embankment , measured in the frame of the embankment is the proper length of the embankment (CD).
2.Proper units in different IFoRs are of equal magnitude.
1.They would be measured so from any other IFoR that transformed their measurements
3.Simultaneity.
i.We have two bodies passing one another.
ii.The time-like space-time interval between the two events ds²= (cdt² – (dx² + dy² + dz²) where dt = 0 is the same for each body as dy = 0, dz = 0 in both cases and dx is the proper distance between AB and betwee A' and B'

Obviously, you copied/pasted your conclusions from somewhere, as you first use rigid bodies but your 'proof' uses a train and an embankment. This is the first inconsistently of your post.

Next, you fail to specify if events

A & A' aligning and their back ends, B & B' aligning

happen at the same time T.

Then, you fail to whether the direction of travel of the two frames is parallel to the length of the rods or perpendicular to the length of the rods.

Could you please clear these up?

[Grimble]

Hello again! Let's make sure we keep this away from tutoring or we may upset someone?

Q1: Which are the two events ?

A & A' aligning and their back ends, B & B' aligning

Q2: What makes you think that , if A and A' are aligned, B and B' will also align? SR length contraction says that , as viewed from the unprimed frame, A'B'<AB. SR also says that , as viewed from the primed frame AB<A'B'. So, it appears that the two rods do not align in either frame.

This is where we are talking at cross purposes; I am only concerned with how it appears to each party, from that same party's point of view. So there will be no question of TD or LC as they are only effects perceived by a moving observer.
No, I am concerned about whether the observer on the train sees the two events as simultaneous. What he sees is light from the lightning, reflected by mirrors passing down the train.
The distance A'B' on the train is the same proper (invariant) length as AB the proper(invariant) length of the embankment.
Therefore if A&B are simultaneous for the embankment A'&B' must be simultaneous for the train.

This has nothing to do with what either observer will calculate that the other observer will measure.

[Grimble]

Obviously, you copied/pasted your conclusions from somewhere, as you first use rigid bodies but your 'proof' uses a train and an embankment. This is the first inconsistently of your post.

Next, you fail to specify if events happen at the same time T.

Then, you fail to whether the direction of travel of the two frames is parallel to the length of the rods or perpendicular to the length of the rods.

Could you please clear these up?

Sorry! and yes it was part copied and it led to inconsistencies as you so rightly pointed out.

The scenario was initially based on Einstein's train and embankment here

A & B are also the points where the lightning strikes and are given as simultaneous.

All movement is parallel not perpendicular.

How's that?

Grimble

[macaw]

Hello again! Let's make sure we keep this away from tutoring or we may upset someone?



This is where we are talking at cross purposes; I am only concerned with how it appears to each party, from that same party's point of view. So there will be no question of TD or LC as they are only effects perceived by a moving observer.

Yet, if A coincides with A', B cannot coincide with B' due to length contraction. Frames F and F' are in motion wrt each other.

No, I am concerned about whether the observer on the train sees the two events as simultaneous.

Once again:

Q1: What two events?

What he sees is light from the lightning, reflected by mirrors passing down the train.

Q3: What light? What lightning?

The points A' and B' are the same proper (invariant) length as the AB the proper(invariant) length of the embankment.
Therefore if A&B are simultaneous for the embankment A'&B' must be simultaneous for the train.

A,A',B,B' are points , therefore they are NOT events.

This has nothing to do with what either observer will calculate that the other observer will measure.

It also has nothing to do with mainstream science, as we know it.

[Kwalish Kid]

This is where we are talking at cross purposes; I am only concerned with how it appears to each party, from that same party's point of view. So there will be no question of TD or LC as they are only effects perceived by a moving observer.

This is simply not true. Any observer moving in any way whatsoever can use whatever reference frame they wish to refer to events. And they must be able to determine, from a description in one frame, what happens in another frame.

What you are doing is demanding that there is no consistency between the description in one frame and the description in another frame.

[AstroRockHunter]

Sorry! and yes it was part copied and it led to inconsistencies as you so rightly pointed out.

The scenario was initially based on Einstein's train and embankment here

A & B are also the points where the lightning strikes and are given as simultaneous.

All movement is parallel not perpendicular.

How's that?

Grimble

Bold mine.

You realize that with the highlighted statement, you have fundementally changed your scenerio?

[Grimble]

Yet, if A coincides with A', B cannot coincide with B' due to length contraction. Frames F and F' are in motion wrt each other.

Once again, Length contraction only occurs when one frame is viewed from another moving with respect to the first.
That is not the case I am considering.
I am looking at the where the events occur, and how this relates to the embankment and how it relates to the train.

Once again:

Q1: What two events?

The front of the train, A', passing the beginning of the embankment A and the rear of the train, B', passing the end of the embankment B.

Q3: What light? What lightning?

Sorry, I was referring to the Einstein lightning which helps to fix the events and tells us that the two events are simultaneous from the embankment.

A,A',B,B' are points , therefore they are NOT events.

A passing A', and B passing B' are the events.

It also has nothing to do with mainstream science, as we know it.

Take away the moving observers and what can be seen from the other frame of reference and consider what is happening in each frame of reference as seen by the observer in that same frame of reference and you must agree that LC does not come into it in any way.
It has all to do with the proper times and lengths experienced within the two frames of reference. How each of them experiences the two events, Each, on their own with no reference to the other.
Is the experience of one comparable to the other?

[Grimble]

This is simply not true. Any observer moving in any way whatsoever can use whatever reference frame they wish to refer to events. And they must be able to determine, from a description in one frame, what happens in another frame.

What you are doing is demanding that there is no consistency between the description in one frame and the description in another frame.

Not at all.
Of course one can calculate how the other will experience events.
And they must be able to determine, from a description in one frame, what they will see, that is how they will experience, what happens in another frame.

What I am saying is that the two events, A' passing A and B' passing B are two spacetime events. Invariant wherever they are viewed from.
And that as AB is equal to A' B', those being the two proper lengths from the embankment and the train, then the time-like separation for one must be equal to the time-like separation for the other.

i.e. ds^2 = (cdt)^2-(dx^2+dy^2+dz^2) = (cdt')^2-(dx'^2+dy'^2+dz'^2) and as
(dx^2+dy^2+dz^2) = (dx'^2+dy'^2+dz'^2), dt must = dt'

[macaw]

Once again, Length contraction only occurs when one frame is viewed from another moving with respect to the first.
That is not the case I am considering.
I am looking at the where the events occur, and how this relates to the embankment and how it relates to the train.

Then you don't need two frames. One is enough.

The front of the train, A', passing the beginning of the embankment A and the rear of the train, B', passing the end of the embankment B.

If this is the case, you need to take length contraction and RoS into account. So, you need to make up your mind. Either way, A,A',B,B' are not events, they are locations in the two frames of reference. You need to correct that.

Q3: What light? What lightning?

Sorry, I was referring to the Einstein lightning which helps to fix the events and tells us that the two events are simultaneous from the embankment.

You are mixing up scenarios and you keep bringing in new stuff. You need to go back and describe the scenario from scratch. You also need to understand that if you have more than one frame, you need to consider the full spectrum of relativistic effects. That includes length contraction. You can't pick and choose.

A passing A', and B passing B' are the events.

A,A',B,B' are points.

Q5: Try to define an event using its mathematical description.

Take away the moving observers and what can be seen from the other frame of reference and consider what is happening in each frame of reference as seen by the observer in that same frame of reference and you must agree that LC does not come into it in any way.

Then you only need one frame. Make up your mind:

Q6: does the scenario have two frames or one frame only?

It has all to do with the proper times and lengths experienced within the two frames of reference. How each of them experiences the two events, Each, on their own with no reference to the other.
Is the experience of one comparable to the other?

You need to figure it out. Before you do that, you need to clean up the descripton of the scenario.

[macaw]

i.e. ds^2 = (cdt)^2-(dx^2+dy^2+dz^2) = (cdt')^2-(dx'^2+dy'^2+dz'^2) and as
(dx^2+dy^2+dz^2) = (dx'^2+dy'^2+dz'^2), dt must = dt'

What you said here is correct but has nothing to do with your scenario. You need to clean up the descrition of your scenario and you will get it to align with the math. BTW, your conclusion is absolutely trivial, it says: if I send a light signal from midway of the train and if I send another light signal from midway of the embankment, both signals will reach the respective ends of the train and of the embankment simultaneously in their respective frames.

Q7: What happens if I only send one signal, from the center of the train when its center point coincides with the center of the embakment? will it reach the ends of the embakment simultaneously as viewed from the embankment frame?

Q8: same question as judged from the train frame. Calculate the answer.

[Strange]

Once again, Length contraction only occurs when one frame is viewed from another moving with respect to the first.
That is not the case I am considering.
I am looking at the where the events occur, and how this relates to the embankment and how it relates to the train.

But you have two frames of reference: train and embankment. Which are you using, and how are you comparing the lengths between these two frames of reference? You can't just say that they each measure them as X and therefore they are both the same because at that point you are comparing things in two different frames of reference. At which point you need to take into account contraction and all the rest.

[macaw]

But you have two frames of reference: train and embankment. Which are you using, and how are you comparing the lengths between these two frames of reference? You can't just say that they each measure them as X and therefore they are both the same because at that point you are comparing things in two different frames of reference. At which point you need to take into account contraction and all the rest.

Absolutely right, let's see how Grimble answers Q7 and Q8.

[Strange]

Absolutely right, let's see how Grimble answers Q7 and Q8.

I know you like to rely on the math, but I think that when the basic assumptions or logic are fundamentally flawed it sometimes helps to fall back to something cruder

[macaw]

I know you like to rely on the math, but I think that when the basic assumptions or logic are fundamentally flawed it sometimes helps to fall back to something cruder

Math is absolute, it leaves no room for ambiguity. I prefer it when it comes to nailing the ATM claims.

[grav]

Not at all.
Of course one can calculate how the other will experience events.
And they must be able to determine, from a description in one frame, what they will see, that is how they will experience, what happens in another frame.

What I am saying is that the two events, A' passing A and B' passing B are two spacetime events. Invariant wherever they are viewed from.
And that as AB is equal to A' B', those being the two proper lengths from the embankment and the train, then the time-like separation for one must be equal to the time-like separation for the other.

i.e. ds^2 = (cdt)^2-(dx^2+dy^2+dz^2) = (cdt')^2-(dx'^2+dy'^2+dz'^2) and as
(dx^2+dy^2+dz^2) = (dx'^2+dy'^2+dz'^2), dt must = dt'

But dx^2 + dy^2 + dz^2 does not equal dx'^2 + dy'^2 + dz'^2 in Relativity and so neither does dt = dt'. To see this, let's try an experiment. We are the embankment observer and see the train pass with the same length d as the embankment as we measure them, so the back of the train reaches the left side of the embankment simultaneously with the front of the train reaching the right side of the embankment. Okay, now let's say that a pulse of light is sent from the back of the train to the front of the train.

1) How much time t_forward will pass according to the embankment observer for the pulse to travel from the back of the train to the front?

2) How much time t_back will pass for a pulse of light to travel from the front of the train to the back according to the embankment observer?

Remember that star aberration shows that light always travels at the same speed from point A to point B at the same speed regardless of the motion of the source and that this speed has been measured isotropically at c by several experiments that test Relativity, and also that the Ives-Stilwell exeriment has disproved the ballistic theory of light travelling with the extra speed of the source. The light will always travel at c to the embankment observer, then, and if you wish, you can consider that just one particular preferred frame where light happens to travel isotropically at c in every direction regardless of the motion of the source, even if you don't think it does for any other.

3) Now, after gaining those two times for the light to travel each way, ignoring that any time dilation takes place between the frames for the moment and so still using the embankment times directly, how much would the clock in the back of the train need to be set forward to have the same time of travel for the light from the back to the front as to travel from the front to the back when taking the difference between the readings on the clocks at the front and back when the light passes?

4) If everything from pulses of light to tennis balls that are thrown in the same way from the back to the front of the train in the same time as from the front to the back, then do you think that the passengers of the train would consider their own clocks in the back and front of the train to be synchronized to each other?

[macaw]

But dx^2 + dy^2 + dz^2 does not equal dx'^2 + dy'^2 + dz'^2

In the case of Grimble's scenario the proper lenghts of the embankment and of the train are the same so he's right on this one.

[grav]

In the case of Grimble's scenario the proper lenghts of the embankment and of the train are the same so he's right on this one.

dx and dx' in this context are not two distances such as the length of the train and of the embankment that are measured from the same frame, or of the proper distances of each as measured in their own frames, but of the same distance that is measured from two different frames.

[macaw]

dx and dx' in this context are not two distances such as the length of the train and of the embankment that are measured from the same frame, or of the proper distances of each as measured in their own frames, but of the same distance that is measured from two different frames.

One more time, in Grimble's scenario they represent the proper lengths of the train and of the embankment and he takes them as being equal, by definition.

[grav]

One more time, in Grimble's scenario they represent the proper lengths of the train and of the embankment and he takes them as being equal, by definition.
You need to read carefully.

I did read it. He says that A meets C simultaneously with B meeting D at each end of the train and each side of the embankment in each of the two frames, so dt = 0 and dt' = 0. But if these two events occur simultaneously in the embankment frame, then they cannot occur simultaneously in the train's frame, so dt = 0 in the embankment frame while dt' <> 0 in the train's frame in that case. The reason is that if the train is the same length as the embankment in the embankment frame, then the train is longer than the embankment in the train's frame, so A meets C first at the front of the train, then some time dt' passes according to the train's clocks while the embankment moves past at some relative speed, then B meets D at the back of the train.

[macaw]

I did read it. He says that A meets C simultaneously with B meeting D at each end of the train and each side of the embankment in each of the two frames, so dt = 0 and dt' = 0.

This is a different issue. Grimble is right about the proper length of the train being equal to the proper length of the embankment. We will get to his errors in terms of understanding RoS when he answers Q7 and Q8, unless you would like to answer for him.

But if these two events occur simultaneously in the embankment frame, then they cannot occur simultaneously in the train's frame, so dt = 0 in the embankment frame while dt' <> 0 in the train's frame in that case. The reason is that if the train is the same length as the embankment in the embankment frame, then the train is longer than the embankment in the train's frame,

No, this is definitely not the reason. The center of the train coincides to the center of the embankment when Grimble's flashes are emitted. So, why don't you calculate the answers to Q7 and Q8?

[grav]

This is a different issue. Grimble is right about the proper length of the train being equal to the proper length of the embankment.

Not from the link he gave. If the length of the train and embankment have the same length d from the frame of the embankment, then the proper length of the embankment is d while the proper length of the train is d / sqrt[1 - (v/c)^2] and the train observer also measures the embankment at a length of sqrt[1 - (v/c)^2] d.

No, this is definitely not the reason. The center of the train coincides to the center of the embankment when Grimle's flashes are emitted.

Only according to the embankment frame. We are considering the events of the front and back of the train lining up with each side of the embankment, which will only occur simultaneously from the embankment frame while the centers also coincide. If we also have lightning strikes at each end of the train and each side of the embankment simultaneously in the embankment frame, then from the frame of the train, since the train is longer than the embankment according to that frame, then the furthest side of the embankment will coincide at the front of the train when the first lightning strike occurs in the same place, then the embankment will pass the train until the centers of each line up after a time of t according to the train's clocks, then at 2t on the train's clocks, the closest side of the embankment will reach the back of the train where the second lightning strike will occur at that time according to the train observer.

[macaw]

then the proper length of the embankment is d while the proper length of the train is d / sqrt[1 - (v/c)^2].

No, you do not understand the notion attached to "proper".

Only according to the embankment frame. We are considering the events of the front and back of the train lining up with each side of the embankment, which will only occur simultaneously from the embankment frame while the centers also coincide. If we also have lightning strikes at each end of the train and each side of the embankment simultaneously in the embankment frame, then from the frame of the train, since the train is longer than the embankment according to that frame, then the furthest side of the embankment will coincide at the front of the train when the first lightning strike occurs in the same place, then the embankment will pass the train until the centers of each line up after a time of t according to the train's clocks, then at 2t on the train's clocks, the closest side of the embankment will reach the back of the train where the second lightning strike will occur at that time according to the train observer.

Please try to answer Q7 and Q8, would you?

[grav]

No, you do not understand the notion attached to "proper".

The proper frame is just the frame of the observer. The proper time is the time measured by that observer's own clock and any other clock that is stationary within the same frame since they are synchronized to that observer's own clock according to the perspective of the observer's frame and the proper distance is the length that is measured of an object that is stationary to the observer in the same frame.

Please try to answer Q7 and Q8, would you?

Fine. Q7) Yes, they will reach each end of the embankment simultaneously at t_embankment = (d / 2) / c if the embankment and train have a length of d as measured by the embankment observer. Q8) They will reach each end of the train simultaneously in a time of t_train = (y d / 2) / c. From the frame of the train when the embankment moves past at v, whereas y = 1 / sqrt[1 - (v/c)^2], the light will reach the furthest side of the embankment toward the front of the train first at a time of t_furthest = ((d / 2) / y) / (c + v) and the closest side toward the back of the train at a time of t_closest = ((d / 2) / y) / (c - v).

[macaw]

Fine. Q7) Yes, they will reach each end of the embankment simultaneously at t_embankment = (d / 2) / c if the embankment and train have a length of d as measured by the embankment observer.

Wrong answer, the signals have been sent from the center of the train, remember?

Q8) They will reach each end of the train simultaneously

Yes.

in a time of t_train = (y d / 2) / c. From the frame of the train when the embankment moves past at v, whereas y = 1 / sqrt[1 - (v/c)^2],

nope , the signals have been sent from the center of the train , remember? And you keep getting the length contraction backwards.

[grav]

Wrong answer, the signals have been sent from the center of the train, remember?

Right, and the the center of the train coincides with the center of the embankment, so the according to the embankment observer, the signals just travel half the distance d to each side of the embankment at c, so in a time of t_embankment = (d / 2) / c.

nope , the signals have been sent from the center of the train , remember? And you keep getting the length contraction backwards.

Right, from the center of the train when it coincides with the center of the embankment, which to the train observer the embankment has a length of d / y, so (d / 2) / y from the center to one edge, and one signal travels at c toward the front of the train and the furthest edge of the embankment at c while the furthest edge of the embankment also travels past the train at v toward the signal at the same time, so c t_furthest = (d / 2) / y - v t_furthest, t_furthest = ((d / 2) / y) / (c + v). The other signal travels toward the back of the train at c while the closest edge of the embankment travels past the train at v away from the signal, so c t_closest = (d / 2) / y + v t_closest, t_closest = ((d / 2) / y) / (c - v). The embankment observer says that the signals struck each side of the embankment simultaneously, but due to the relativity of simultaneity, the train observer says the signals struck the furthest side of the embankment first, then the closest side at a time of t = ((d / 2) / y) / (c - v) - ((d / 2) / y) / (c + v) = ((d / 2) / y) [(c + v) - (c - v)] / [(c - v) (c + v)] = d v / (y (c^2 - v^2)) = d v / (c^2 sqrt(1 - (v/c)^2)) later.

[north]

SR , GR , is about perspective , fundamenally

but to the object(s) perspective doesn't matter in the end

[RussT]

Originally Posted by Grimble
3.If two rigid bodies, with identical proper lengths, pass one another and we consider two Space-time events; their front ends, A & A' aligning and their back ends, B & B' aligning, then those two events will be simultaneous in their respective IFoRs.

OrginallyPosted by macaw Post #2]
Q1: Which are the two events ?
Q2: What makes you think that , if A and A' are aligned, B and B' will also align? SR length contraction says that , as viewed from the unprimed frame, A'B'<AB. SR also says that , as viewed from the primed frame AB<A'B'. So, it appears that the two rods do not align in either frame.

In the case of Grimble's scenario the proper lenghts of the embankment and of the train are the same so he's right on this one.

Sorry, macaw, but you were the one that caused this in your post #2.

But, you are correct that grav is switching where the lightning strikes are coming from and going to!

BUT...

Orginally pPosted by Grimble
A passing A', and B passing B' are the events.

A,A',B,B' are points.

BUT...regardless of where the lightning strikes are coming from or going to...

IF the distance in the separation between A and A' is 0 and the separation between B and B' is 0, so that dt=0 for BOTH...

Then that is a "Singularity"

[grav]

But, you are correct that grav is switching where the lightning strikes are coming from and going to!

The lightning strikes aren't considered to be coming from or going to anywhere. They are just two events that occur at the front of the train and furthest side of the embankment when those two points coincide and at the back of the train and closest side of the embankment when those two points coincide.

BUT...regardless of where the lightning strikes are coming from or going to...

IF the distance in the separation between A and A' is 0 and the separation between B and B' is 0, so that dt=0 for BOTH...

Then that is a "Singularity"

That's true. The only way to make dt = dt' = 0 would be to make dx = dx' = 0 as well, so the train and embankment would have no length (for the separation AB and A'B', though) and the events would occur simultaneously in the same place in both frames in that case.