Naked Science Forum
On the Lighter Side => New Theories => Topic started by: Jaaanosik on 17/01/2024 19:09:42
-
[Mod note: This was split off of here: https://thenakedscientists.com/forum/index.php?topic=86033.0
The OP of that topic did post two videos, but neither of the videos mention anything relevant to a discussion about the train car and platform being discussed here.]
Here is a demonstration of what went wrong in the OP video.
It is impossible to align the following 4 events due to disagreement on the simultaneity between the inertial frames.
Imaging a 'space train car' and a 'space platform'.
(https://i.imgur.com/5II0hJY.png)
Figure 1: The train car inertial reference system S′.
(https://i.imgur.com/Mne73CP.png)
Figure 4: The platform inertial reference system S.
The front of the train F' is the origin of the train S'. The back of the train B' is -6.9282cs' back.
The P is location of the platform observer and origin of the platform frame.
When we assign time to locations we have events.
We can have time synchronized on the train S' for the F', B' events.
We can have time synchronized on the platform S for P, B events.
The events F', B', P, B cannot be aligned though.
The disagreement on the simultaneity does now allow that.
Then the invariance of the space-time intervals of a light round-trip defines stationary preferred frame with the slowest time.
-
Then the invariance of the space-time intervals of a light round-trip defines stationary preferred frame with the slowest time.
The light round-trip can happen only in one grid of inertial observers with synchronized clocks.
The other frame observes the emission and reception at two different locations.
(https://i.imgur.com/HAMsg8X.png)
Figure 5: The light emission on the train car and the reflection at the other side of the train car.
The thought experiment works with x, y, x', y' axis and time t, t'.
(https://i.imgur.com/nsc5ezu.png)
Figure 6: The light absorption back at the front observer in the train car inertial grid.
(https://i.imgur.com/hI6de1z.png)
Figure 7: The train light round-trip observed from the platform inertial reference frame.
(https://i.imgur.com/9MWaUXo.png)
Figure 8: The space time diagram for the light round-trip for the platform inertial grid.
The blue arrows represent the light.
(https://i.imgur.com/zS1Fyn2.png)
Figure 9: The red photon is reflected back to the platform observer P for the platform light round-trip.
(https://i.imgur.com/9R9UpOS.png)
Figure 10: The second red photon in the train grid of inertial observers at time t'=2s'.
Comparing Figure 9 and Figure 10.
Question 1: Why there is a disagreement on the y, y' position of the red photon?
-
...
Then the invariance of the space-time intervals of a light round-trip defines stationary preferred frame with the slowest time.
This comment would be false if it made any sense, but it's not even wrong. The spacetime interval of light is always zero. That fact does not in any way define a preferred reference frame.
(https://i.imgur.com/PJnpeQs.png)
Even though the light space-time interval is equal 0 there is a space separation that light crossed.
To say ds'^2=ds^2 Lorentz transformation has to be applied to get those zeros.
How do we do that for Figure 10?
(https://i.imgur.com/9R9UpOS.png)
-
The events F', B', P, B cannot be aligned though
It is unclear what you think it means for events to be 'aligned'. You mean they are not the same events? At time zero, apparently F' and B are the same event, and B' and P are not.
I glean this not from this post, but from figures 7 & 9 in the subsequent post.
Then the invariance of the space-time intervals of a light round-trip defines stationary preferred frame with the slowest time.
This comment would be false if it made any sense, but it's not even wrong. The spacetime interval of light is always zero. That fact does not in any way define a preferred reference frame.
I split this into a new topic since all this discussion of a train car is unrelated to the twins discussion.
Then the invariance of the space-time intervals of a light round-trip defines stationary preferred frame with the slowest time.
You repeat this, but don't back it. Nothing you post suggests this.
The light round-trip can happen only in one grid of inertial observers with synchronized clocks.
The other frame observes the emission and reception at two different locations.
The fact that emission at reception events are at the same point in space in only a limited set of frames is not in any way invalidate the physics going on in the frames that assign different spatial locations to the exact same two events. The difference is merely one of abstract assignment of coordinates to the events, not one of different physics.
Figure 5: The light emission on the train car and the reflection at the other side of the train car.
The thought experiment works with x, y, x', y' axis and time t, t'.
Figures 5-10 seem to depict a light pulse that is emitted at event F'/B and is sent to mirrors at the opposite side of the train car, which reflect the pulse two ways, one back to the front of the car where the pulse was emitted, and one to the opposite corner at the rear of the car.
Figure 8 is pretty baffling, but is apparently trying to show 3 dimensions at once. It is unclear what the grayed regions are meant to represent. Neither are the car or the platform worldlines since neither is present at the origin.
Comparing Figure 9 and Figure 10.
Question 1: Why there is a disagreement on the y, y' position of the red photon?
It's a light pulse, not a photon. There's no reason not to keep things classical.
The text image you included in post 2 makes the same mistake.
There isn't a disagreement. The two figures depict the exact same thing, but from different frames. They are both space diagrams (no time axis), not spacetime diagrams like figure 8 is apparently trying to be.
My question: What has any of this to do with the opening line, that a preferred frame somehow gets defined by all this? It is for that unbacked assertion that this topic has been moved to new theories.
Even though the light space-time interval is equal 0 there is a space separation that light crossed.
Yes, but the interval is invariant. The space separation (and thus the time to get from one to the other) is frame dependent.
To say ds'^2=ds^2
I don't know what these are. It looks like you're multiplying distance by the name of a frame, which makes no sense at all. If you're somehow asserting that the distance traveled by light should be the same from one frame to the next, that would be wrong.
Lorentz transformation has to be applied to get those zeros.
I don't know what you mean by this either.
How do we do that for Figure 10?
How do we do what? Fig 10 shows both pulses, from the train PoV. It shows the blue one reflected back to the emission location, taking time 2 to go back and forth the width of 1. The red pulse goes length 1 to get to the mirror, and 7 to go all the way to the opposite corner. No Lorentz transform is needed to see that, or to compute (using the coordinates of all events in that frame) the interval of zero for either light path. The transform is useful to go between figures 9 and 10, which shows the same events, but in different frames.
The bold part, instead of delta I typed d.
I am going to try: Δs'2=Δs2
OK, part of the problem is that you're using nearly the same variable for different things.
You use S and S' as the name of the frames but lowercase s and s' for some unspecified duration of time, or the name of your time coordinate, or possibly one unit of time like it's different from one frame to the next. You have a little γ' in the upper right of the picture, totally unexplained.
There is just γ, the factor of change between the two frames. There is no γ'.
About your statement with the deltas above, this seems very wrong since it seems to suggest that some change in time (what, between two unspecified events??), or the square of those changes in time, would be equal in two different frames, which is very unlikely.
But you persist in not explaining that assertion, so I can only presume it's something you're just making up.
How much time elapsed in the platform frame when the red photon crossed 1cs' in the train car frame?
Here you use s' like a unit, but a unit that is different than the s unit used in the S frame (what, minutes?), and by multiplying by c you change it into a distance. So if s' means a second, you're asking how much time elapses in S' for a light pulse (not a photon) to cross the distance of one light second in S', which is pretty obviously 1 second. Do you not know this? Am I missing something that actually would make this a non-trivial question?
The red pulse is going to need 8 time units (assuming natural units in your diagram) to go from emitter at the lower right to the far detector at the lower left. Surely you can figure that out yourself. The total distance in S' is 8, so it takes time 8 to make the trip. We don't need relativity at all to see that.
Maybe you're asking a different question, but then I cannot figure out what it is.
You're supposed to be showing how your assertion in the early posts about the preferred frame is concluded from all this, and you're making no attempt at doing so. Should I then lock the topic before I have to play 20 questions or are you going to show how all this playing with the train car demonstrates that assertion?
-
...
To say ds'^2=ds^2
I don't know what these are. It looks like you're multiplying distance by the name of a frame, which makes no sense at all. If you're somehow asserting that the distance traveled by light should be the same from one frame to the next, that would be wrong.
Lorentz transformation has to be applied to get those zeros.
I don't know what you mean by this either.
How do we do that for Figure 10?
How do we do what? Fig 10 shows both pulses, from the train PoV. It shows the blue one reflected back to the emission location, taking time 2 to go back and forth the width of 1. The red pulse goes length 1 to get to the mirror, and 7 to go all the way to the opposite corner. No Lorentz transform is needed to see that, or to compute (using the coordinates of all events in that frame) the interval of zero for either light path. The transform is useful to go between figures 9 and 10, which shows the same events, but in different frames.
The bold part, instead of delta I typed d.
I am going to try: Δs'2=Δs2
What do you know, it works.
Figure 10:
(https://i.imgur.com/9R9UpOS.png)
How much time elapsed in the platform frame when the red photon crossed 1cs' in the train car frame?
-
The bold part, instead of delta I typed d.
I am going to try: Δs'2=Δs2
OK, part of the problem is that you're using nearly the same variable for different things.
You use S and S' as the name of the frames but lowercase s and s' for some unspecified duration of time, or the name of your time coordinate, or possibly one unit of time like it's different from one frame to the next. You have a little γ' in the upper right of the picture, totally unexplained.
There is just γ, the factor of change between the two frames. There is no γ'.
S, S' are inertial reference systems.
Δs2, Δs'2 are space-time intervals.
Space coordinates x,y belong to S, space coordinates x',y' belong to S'.
There is only one \gamma γ=2.
About your statement with the deltas above, this seems very wrong since it seems to suggest that some change in time (what, between two unspecified events??), or the square of those changes in time, would be equal in two different frames, which is very unlikely.
But you persist in not explaining that assertion, so I can only presume it's something you're just making up.
I did not make anything up.
(https://i.imgur.com/gKny6xH.png)
(https://i.imgur.com/NwAUdPB.png)
How much time elapsed in the platform frame when the red photon crossed 1cs' in the train car frame?
Here you use s' like a unit, but a unit that is different than the s unit used in the S frame (what, minutes?), and by multiplying by c you change it into a distance. So if s' means a second, you're asking how much time elapses in S' for a light pulse (not a photon) to cross the distance of one light second in S', which is pretty obviously 1 second. Do you not know this? Am I missing something that actually would make this a non-trivial question?
Time 1s', one second in train car is not 1s on the platform.
The red photon reflects from the half mirror to the back of the train car at 2s on the platform.
The blue photon reflected to the front of the train car gets back to F' at 2s' of the train car and that corresponds with 4s on the platform.
The event F'[2,0,0]=[2s',0cs',0cs'] is 'aligned' with the event P[4,3.46,0]=[4s,3.46cs,0cs]
I am not even asking what coordinates are the end of the red arrow, the photon position, only time observed on the platform.
The red pulse is going to need 8 time units (assuming natural units in your diagram) to go from emitter at the lower right to the far detector at the lower left. Surely you can figure that out yourself. The total distance in S' is 8, so it takes time 8 to make the trip. We don't need relativity at all to see that.
Maybe you're asking a different question, but then I cannot figure out what it is.
You're supposed to be showing how your assertion in the early posts about the preferred frame is concluded from all this, and you're making no attempt at doing so. Should I then lock the topic before I have to play 20 questions or are you going to show how all this playing with the train car demonstrates that assertion?
I used the word preferred to point out that if there is a light round-trip in a reference frame, then this frame is 'preferred' because its time is the shortest.
Every other moving inertial reference frame has to agree that the round-trip takes more time in any other moving reference frame because it is not a round-trip in those frames. There is a space separation and that's adding the additional time.
The round-trip rest frame has that attribute of time being the shortest/slowest.
The round-trip P[0,0,0]->P[2,3.46,1]->P[4,0,0] takes 8s' in the train car frame but only 4s in the platform frame.
That's the problem.
-
this frame is 'preferred' because its time is the shortest.
If you consider that a personal preference, then that's fine. It doesn't make that frame any more correct than the others, though. All frames are equally correct.
-
S, S' are inertial reference systems.
Space coordinates x,y belong to S, space coordinates x',y' belong to S'.
I already figured those out.
Δs2, Δs'2 are space-time intervals.
Nowhere was this stated before. OK, so I guessed wrong.
What interval? I mean, an interval is between two events (or along a specific worldline), but you've not identified those events.
Yes, the interval will be frame independent, so they'll equal each other.
Time 1s', one second in train car is not 1s on the platform.
Now you're using s to be a second (or some other unit of time). You just said it was an interval, which is not a time. Using the same symbol for at least three different things is not being clear.
I am now once again presuming s means seconds. Maybe it means seasons (making it a really wide train car).
I presume we're not talking about intervals that are also s.
The red photon reflects from the half mirror to the back of the train car at 2s on the platform.
Both pulses (not photons) reflect from their respective mirrors at t=2 seconds in S, yes. I'm using t for time here, since using s is way too confusing.
Both pulses reach their respective receptors simultaneously in S at t=4 sec since both pulses travel the same distance.
Now we're switching to the S' frame:
The event F'[2,0,0]=[2s',0cs',0cs'] is 'aligned' with the event P[4,3.46,0]=[4s,3.46cs,0cs]
No. Event F was specified as having coordinates [0,0,0] in S'. Event P is similarly [0,0,0] in S.and these are not the same event. So you're contradicting yourself.
I would agree that the event [2,0,0] in S' and [4,0,0] in S (not the coordinates you gave) are the same event. 'Aligned' is not a term I've seen used. This is the receptor event of the blue pulse. You've not given this event a unique name, so I'll simply call it BR for blue receptor event, as opposed to RR red receptor on the right, and PR for pulse reflection event where the mirrors get hit, and PE (pulse emission) which is the event called both B and F in the OP.
I am not even asking what coordinates are the end of the red arrow, the photon position, only time observed on the platform.
RR is [4, 03.46, 0] in S and [8, -6.92, 0] in S'. You show the red arrow going all the way to RR in figure 9 but only 1/7th of the way there in figure 10, so you're not consistent with drawing the same red arrow in both those figures. The head of those two arrows is thus not the same event.
I used the word preferred to point out that if there is a light round-trip in a reference frame, then this frame is 'preferred' because its time is the shortest.
That doesn't make any frame preferred since the path will be shortest in any frame where the emitter and receptor are at the same spatial location, which can be arranged in any frame. The claim is no more remarkable that noting that he speed of the train is lowest (zero) in S', therefore S' is special.
Also, your red pulse takes less time in S, but even less in some other frame, making S' very much not preferred by your funny definition.
A preferred frame would be some test that would demonstrate a way to objectively determine which frame is the preferred one, not which frame in which some arbitrary thing is stationary, which is what you're doing here.
The round-trip rest frame has that attribute of time being the shortest/slowest.
But now do the same thing with a mirror stationary in S, opposite the platform 1cs away. That pulse comes back to the emitter after 2s only in the S frame and takes longer in any other. That trick can be played with any frame, thus there's no objective frame that behaves any different than any other.
Or keep the mirrors on the train, but put them at location x'=-0.933,y'=1 instead of 0, 1,. Now the platform observer still records the shortest round trip time.
-
this frame is 'preferred' because its time is the shortest.
If you consider that a personal preference, then that's fine. It doesn't make that frame any more correct than the others, though. All frames are equally correct.
Well, there are some issues, let's discuss...
-
...
Time 1s', one second in train car is not 1s on the platform.
Now you're using s to be a second (or some other unit of time). You just said it was an interval, which is not a time. Using the same symbol for at least three different things is not being clear.
The character 's' means seconds. The character " ' " means seconds or cs (light-second) belongs to S' frame - train car.
If " ' " is not there then it is the S - platform frame.
I am now once again presuming s means seconds. Maybe it means seasons (making it a really wide train car).
I presume we're not talking about intervals that are also s.
The red photon reflects from the half mirror to the back of the train car at 2s on the platform.
Both pulses (not photons) reflect from their respective mirrors at t=2 seconds in S, yes. I'm using t for time here, since using s is way too confusing.
Both pulses reach their respective receptors simultaneously in S at t=4 sec since both pulses travel the same distance.
Pulse, fine, I had wave packet (photon) as particle in mind.
t=2s in S is t'=1s' in S' when we are looking from S to S'.
Now we're switching to the S' frame:
The event F'[2,0,0]=[2s',0cs',0cs'] is 'aligned' with the event P[4,3.46,0]=[4s,3.46cs,0cs]
No. Event F was specified as having coordinates [0,0,0] in S'. Event P is similarly [0,0,0] in S.and these are not the same event. So you're contradicting yourself.
There is some logic to this 'madness'. :D
The character P with no " ' " means a 'point' in platform frame S.
Two types: location or event.
P[0,0]=[0cs,0cs] is a location, origin of the platform frame at coordinates x=0cs, y=0cs.
P[0,0] is just name.
P[0,0,0]=[0s,0cs,0cs] is an event at the origin at time t=0s.
The event first coordinate is time. I suggest dropping 's, cs', they were there to point the difference between events and locations.
If coordinates have just two numbers [0,0] it is a location [x,y].
If coordinates have have three numbers [0,0,0] it is an event [t,x,y].
F'[2,0,0] is just a name of an event in the train car frame at [2',0',0'].
I suggest dropping " ' " as well, the name determines the frame units of measure.
We have to remember t'=γt and x'=γx.
P[4,3.46,0] is a name of an event at t=4s and x=3.46cs, y=0cs.
P[4,3.46,0] is aligned with F'[2,0,0] because the front of the train car is at x=3.46cs.
...
So far, I'll respond to the rest later.
[/quote]
-
Well, there are some issues, let's discuss...
Such as?
-
Well, there are some issues, let's discuss...
Such as?
To start, even though we do analysis in the Special Relativity, any time a change of a grid of inertial observers is applied it involves an acceleration.
The simple comment that we are going to say one grid of inertial observers sees/observes something and we would like to say that another grid of inertial observers sees something else, this change of observers has to undergo acceleration analysis.
-
...
I would agree that the event [2,0,0] in S' and [4,0,0] in S (not the coordinates you gave) are the same event. 'Aligned' is not a term I've seen used. This is the receptor event of the blue pulse. You've not given this event a unique name, so I'll simply call it BR for blue receptor event, as opposed to RR red receptor on the right, and PR for pulse reflection event where the mirrors get hit, and PE (pulse emission) which is the event called both B and F in the OP.
The alignment of events is problematic for two or more grids of inertial observers.
Only one event can have coordinates 'aligned', 'next to each other'. Any other events happening 'at the same time' are misaligned because of disagreement on the simultaneity.
Clock desynchronization
I am not even asking what coordinates are the end of the red arrow, the photon position, only time observed on the platform.
RR is [4, 03.46, 0] in S and [8, -6.92, 0] in S'. You show the red arrow going all the way to RR in figure 9 but only 1/7th of the way there in figure 10, so you're not consistent with drawing the same red arrow in both those figures. The head of those two arrows is thus not the same event.
This is the problem of switching the inertial grids. Figure 9 is platform frame and Figure 10 is train car frame.
The photon is supposed to be at y=0 at t=4s in the platform frame.
But the photon is supposed to be at y'=6/7 at t'=2s' in the train car frame.
What is y, y' position of the photon? It comes back to the question who is looking?
I used the word preferred to point out that if there is a light round-trip in a reference frame, then this frame is 'preferred' because its time is the shortest.
That doesn't make any frame preferred since the path will be shortest in any frame where the emitter and receptor are at the same spatial location, which can be arranged in any frame. The claim is no more remarkable that noting that he speed of the train is lowest (zero) in S', therefore S' is special.
Also, your red pulse takes less time in S, but even less in some other frame, making S' very much not preferred by your funny definition.
A preferred frame would be some test that would demonstrate a way to objectively determine which frame is the preferred one, not which frame in which some arbitrary thing is stationary, which is what you're doing here.
The round-trip rest frame has that attribute of time being the shortest/slowest.
But now do the same thing with a mirror stationary in S, opposite the platform 1cs away. That pulse comes back to the emitter after 2s only in the S frame and takes longer in any other. That trick can be played with any frame, thus there's no objective frame that behaves any different than any other.
Or keep the mirrors on the train, but put them at location x'=-0.933,y'=1 instead of 0, 1,. Now the platform observer still records the shortest round trip time.
This is the contradiction of the special relativity.
The objective reality does not care about our flawed SR.
How does the reality work?
The logical conclusion is that if we compare all the inertial frames then there has to be one that resolves all the contradictions, simultaneity, causality, ... all the possible issues.
-
To start, even though we do analysis in the Special Relativity, any time a change of a grid of inertial observers is applied it involves an acceleration.
The simple comment that we are going to say one grid of inertial observers sees/observes something and we would like to say that another grid of inertial observers sees something else, this change of observers has to undergo acceleration analysis.
Says who? Nothing is actually accelerating. All you are doing is changing the perspective from which you are viewing the situation.
-
To start, even though we do analysis in the Special Relativity, any time a change of a grid of inertial observers is applied it involves an acceleration.
The simple comment that we are going to say one grid of inertial observers sees/observes something and we would like to say that another grid of inertial observers sees something else, this change of observers has to undergo acceleration analysis.
Says who? Nothing is actually accelerating. All you are doing is changing the perspective from which you are viewing the situation.
I say so. :D
Seriously, the thought experiment I introduced demonstrates (eventually when we get through it) how 'changing the perspective' involves acceleration.
(https://i.imgur.com/zS1Fyn2.png)
Figure 9: The red photon is reflected back to the platform observer P for the platform light round-trip.
(https://i.imgur.com/9R9UpOS.png)
Figure 10: The second red photon in the train grid of inertial observers at time t'=2s'.
Changing the perspective as shown in Figure 9 and Figure 10, suddenly y!=y'.
Is that correct? Do you agree?
-
Is that correct? Do you agree?
If you're talking about a particular object changing the reference frame that it is in, then yes, we can say that's an acceleration. If you are talking about switching reference frames by looking at the same situation from two different viewpoints, then all you've done is a mental exercise. Nothing is actually, physically, accelerated.
-
Is that correct? Do you agree?
If you're talking about a particular object changing the reference frame that it is in, then yes, we can say that's an acceleration. If you are talking about switching reference frames by looking at the same situation from two different viewpoints, then all you've done is a mental exercise. Nothing is actually, physically, accelerated.
Let us assume just changing of viewpoints, as you say mental exercise.
Changing the perspective as shown in Figure 9 and Figure 10, suddenly y≠y'.
Do you agree y≠y'?
-
Do you agree y≠y'?
Yes, but no acceleration has occurred. Acceleration requires a physical change, not a conceptual one.
-
Do you agree y≠y'?
Yes, but no acceleration has occurred. Acceleration requires a physical change, not a conceptual one.
If y≠y' then does it mean the Lorentz transformation is 'broken'?
-
If y≠y' then does it mean the Lorentz transformation is 'broken'?
Why would it be?
-
If y≠y' then does it mean the Lorentz transformation is 'broken'?
Why would it be?
(https://i.imgur.com/O8N0PX5.png)
The Lorentz transformation says y=y'.
(https://i.imgur.com/hI6de1z.png)
Figure 7: The train light round-trip observed from the platform inertial reference frame.
It is obvious how this is the true for the train light round-trip for any chosen time in t=0s and t=4s platform time interval.
(https://i.imgur.com/9R9UpOS.png)
Figure 10: The second red photon in the train grid of inertial observers at time t'=2s'.
We see y≠y' for the platform light round-trip between t>2s and t<=4s.
The proper time between two inertial grids cannot be synchronized.
The Lorentz transformation inherently determines who is looking at a specific event.
-
We see y≠y'
I must be missing something because it looks like y=1 and y'=1 on your illustrations.
-
We see y≠y'
I must be missing something because it looks like y=1 and y'=1 on your illustrations.
That's true for the event of reflection.
I wrote: We see y≠y' for the platform light round-trip between t>2s and t<=4s.
-
Here is a space-time diagram to ponder.
(https://i.imgur.com/Mv3rKS8.png)
Figure 11: Train car space-time diagram for the train time t'=2s'.
Just to demonstrate the problem. The event of reception at the platform observer does not exist for the train grid of inertial observers yet. It is in the future.
There are some additional arrows, just ignore them at the moment.
-
Changing the perspective as shown in Figure 9 and Figure 10, suddenly y!=y'.
Is that correct? Do you agree?
In all your diagrams, for any event anywhere, the y coordinate of that event in S is idential to the y' coordinate in S'. So your assertion of y != y' (assuming that y and y' are the coordinate of some event) is unfounded, and your descriptions do not show any such thing.
When Kryptid agreed that y != y', that was when y and y' were the names of the axes in their respective frames, and indeed, the axes are not the same from one frame to the next.
But then you changed y and y' to mean a coordinate of some event, in which case they are the same for any event between frames which have a zero velocity difference component in the y direction.
(https://i.imgur.com/hI6de1z.png)
Figure 7: The train light round-trip observed from the platform inertial reference frame.
Fig 7 does not show a round trip in S. It shows a light pulse starting at one end of the platform and ending at the other. It takes 4 seconds (in S) to do this.
(https://i.imgur.com/9R9UpOS.png)
Figure 10: The second red photon in the train grid of inertial observers at time t'=2s'.
OK, for some reason you've taken a special interest in this unlabeled pulse event where the red arrow ends. The y coordinate in S of that event is identical to the y' coordinate in S' of that same event. I cannot fathom why you take such an interest in that particular event.
We see y≠y' for the platform light round-trip between t>2s and t<=4s.
Between t>2s and t <= 4s, no round trip is taken in S. There is no y or y' for any worldline in either frame, so no y to equal y' or not. A line is not an event, and thus has no single set of coordinates.
The proper time between two inertial grids cannot be synchronized.
Grids don't have a proper time. If you're saying that clocks stationary in different respective inertial frames cannot be synchronized over time, then SR agrees with that.
The Lorentz transformation inherently determines who is looking at a specific event.
The LT makes no mention of observers. It works just fine in the complete absence of observers.
Here is a space-time diagram to ponder.
(https://i.imgur.com/Mv3rKS8.png)
Figure 11: Train car space-time diagram for the train time t'=2s'.
Nice picture! The lower ~horizontal grey bit is the train at t'=2 seconds in S', yes. The other grey bar seems to be the platform position events corresponding to t=4s in S
Just to demonstrate the problem. The event of reception at the platform observer does not exist for the train grid of inertial observers yet. It is in the future.
SR does not posit a past/present/future ontology for events. Doing so would contradict the first premise of SR. So any mention of these is meaningless.
The event of reception at the platform observer(s) do very much exist in your picture. You label at least one of them and everything.
And y = y' for every single event in the diagram.
If your issue is that the reception events at the two ends of the platform that are simultaneous in the S frame are not simultaneous in the S' frame, that is indeed the case, as per relativity of simultaneity. You seem not to have identified any issue with SR at all. Your issue seem to be related to an attempt to drag an additional premise presentism into a theory that has a premise (both premises actually) that would be in contradiction with presentism. But no argument based on presentism has been part of any of the discussion before the immediate prior post.
-
SR does not posit a past/present/future ontology for events. Doing so would contradict the first premise of SR. So any mention of these is meaningless.
The event of reception at the platform observer(s) do very much exist in your picture. You label at least one of them and everything.
What do you mean by past/present/future ontology for events?
What is the first premise of SR?
-
SR does not posit a past/present/future ontology for events. Doing so would contradict the first premise of SR. So any mention of these is meaningless.
The event of reception at the platform observer(s) do very much exist in your picture. You label at least one of them and everything.
The past/present/future is clearly defined by the space-time interval.
(https://i.imgur.com/PJnpeQs.png)