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... QuoteThen 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.
Then the invariance of the space-time intervals of a light round-trip defines stationary preferred frame with the slowest time.
The events F', B', P, B cannot be aligned though
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.
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'.
Comparing Figure 9 and Figure 10.Question 1: Why there is a disagreement on the y, y' position of the red photon?
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?
The bold part, instead of delta I typed d.I am going to try: Δs'2=Δs2
How much time elapsed in the platform frame when the red photon crossed 1cs' in the train car frame?
...QuoteTo say ds'^2=ds^2I 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.QuoteLorentz transformation has to be applied to get those zeros.I don't know what you mean by this either.QuoteHow 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.
Quote from: Jaaanosik on 23/01/2024 16:03:23The bold part, instead of delta I typed d.I am going to try: Δs'2=Δs2OK, 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.
QuoteHow 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?
this frame is 'preferred' because its time is the shortest.
S, S' are inertial reference systems.Space coordinates x,y belong to S, space coordinates x',y' belong to S'.
Δs2, Δs'2 are space-time intervals.
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 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.
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.
The round-trip rest frame has that attribute of time being the shortest/slowest.
Quote from: Jaaanosik on 23/01/2024 18:34:39this 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.
...QuoteTime 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.QuoteThe 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:QuoteThe 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.
Well, there are some issues, let's discuss...
Quote from: Jaaanosik on 24/01/2024 21:45:24Well, there are some issues, let's discuss...Such as?
...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.
QuoteI 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.
QuoteI 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.QuoteThe 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.
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.
Quote from: Jaaanosik on 25/01/2024 21:28:06To 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.
Is that correct? Do you agree?
Quote from: Jaaanosik on 26/01/2024 18:27:59Is 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.
Do you agree y≠y'?
Quote from: Jaaanosik on 30/01/2024 12:38:12Do 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'?