The discussion centers on the implications of Special Relativity (SR) regarding the detection of photons emitted simultaneously from two sources in different reference frames. In Case #1, both the stationary observer and the moving platform agree that the photons are detected simultaneously, making it true. In Case #2, while the stationary observer sees the emissions as simultaneous, the moving observer does not, leading to a false conclusion for the moving frame. Case #3 highlights that the moving observer detects the photons at different times due to their motion towards one emitter, reinforcing the relativity of simultaneity. The conversation emphasizes the importance of clearly defining reference frames to avoid confusion in understanding the outcomes of these scenarios.
#91
wespe
202
0
ram1024 said:
we can ditch the acceleration and say train 1 was running along the track at 5ft/s the whole time and when it hit the point where both trains were aligned nose-to-nose, THEN the experiment begins
No, you can't do that. Because, there is length contraction. The ends of two trains will not meet at the same time, each one will see the other contracted. (If you make one of the trains longer, still no, becaue length contraction is mutual)
I have to admit I may not be able to answer your question if it involves acceleration. I don't know much GR. Maybe this is a good time for me to start GR.
As has already come up, #4 has not been formulated in a way consistent with SR.
In other words, you're assuming the answer before you "do" the experiment.
If you accelerate the train, you run into the problem of no rigid bodies, and if both trains had been moving with different constant velocitys (possibly zero), then it cannot be the case that the two setups are identical (particularly the same proper length) and that the endpoints can coincide simultaneously in either of their two rest frames.
Furthermore, yet again I can only assume that when you say "simultaneous" you mean in the frame you draw pictures (which I'll now call the "picture frame")
Anyways, if I assume you do something like accept there's a rigid body problem, or allow one train to be longer than the other, or something or other, I deduce that both observers will observe the right photon before the left photon. If you can get the two observers along the same worldline, then they will detect the right photon simultaneously, and similarly for the left photon.
This is consistent with my previous answers where, in both case #2 and #3 (under the assumption that "simultaneously" was measured in the picture frame), I stated that the observer would not detect both photons simultaneously.
#94
wespe
202
0
ram1024 said:
damn, bummer then
teach me while you're at it, wespe. contractions perplex me
However, if you consider that after the acceleration is over, train1 will be inertial again, and length contraction must have already occured. (you can't reject length contraction at this point. If you do, you have to also reject the given SR answers). That would explain some things. Anyway, if expert people here consider your question worthy, you will get your answer quickly. My answer will be delayed since acceleration is involved.
Take care.
#95
geistkiesel
538
1
ram1024 said:
Mark it on your calendars, people. Let us begin.
Case #1:
Code:
[u]|(-> (o) <-)|[/u]
man standing on a movable platform bed. at the EXACT center between two photon emitters. SR concludes that the simultaneously emitted photons from the two emitters will be detected by the observer at the same exact time.
ADDENDUM:the photon emitters are tied to precise atomic clocks
these clocks are perfectly aligned and synchronized and in all cases they move within the same inertial frame so they can stay calibrated.
Man is moving on platform towards an emitter. SR concludes that photons are NOT detected at the exact same time.
(True / False) ?
once we square these we'll move on to stage 2.
!. True, the photon detected simultaneously.
2. False. The photons are detected sequentially. Light is measured at c in all inertial frames. The fact that the photon source is moving or not is irelevant. The observer is moving to the right toward a photon moving to the left. The left moving photon has a shorter distance to travel than does the left photon, therefore the times of flight of both photons is different. SR should predict sequential detection and that the photons were not emitted simulaneously in both stationary and moving frames.
3. same as 2). The photons will move at the speed of light. The left and right moving photons move at the same speed (from whatever frame measured). As the photon moving left has a shorter disance to cover befor meeting the observer moving right this photon will be detected first.
In both 2 and 3 SR will predict the photons were not emitted simultaneoulsy in the moving frame.
#96
wespe
202
0
Hurkyl said:
(under the assumption that "simultaneously" was measured in the picture frame)
I was under the impression that "simultaneously" was measured in the platform/emitters frame. (from post #14). Maybe ram1024 would like to make that clear.
#97
O Great One
98
0
Correct
ram1024 said:
actually if they're giving me a hard time it's probably because i deserve it :D
but I'm ridiculously patient and not really functioning under a deadline so meh, it doesn't matter :D
you have pointed out something interesting though with
that. say a man standing on a platform halfway to halfway (crap i better make a picture)
Code:
[u]|(-> (o) (o) <-)|[/u]
observer1 observer2
at the time of emission, observer 2 is at the exact center. but at the exact moment the photons are emitted he starts running left and ends up at the position where observer one is. according to what you said he gets hit by photons from both sides simultaneously. but observer 1 just standing there doesn't. he gets hit by photon to the left first and then by photon to the right.
This is correct. Let's say that I'm 10 miles from somebody who then runs directly towards me at a speed always relative to me at 5 m.p.h. No matter what I do after he starts running towards me, he will reach me in 2 hours. If I start moving away at 20 m.p.h. relative to the earth, he will then be moving towards me at 25 m.p.h. relative to the earth, so that he maintains his constant speed of 5 m.p.h. relative to me. If light always moves towards us at the same constant speed relative to us, then the time it takes for light to reach us depends on the distance at the time of emission.
#98
wespe
202
0
O Great One said:
If light always moves towards us at the same constant speed relative to us, then the time it takes for light to reach us depends on the distance at the time of emission.
Yes, light always moves towards us at the same constant speed relative to us. And the distance is the same when lights were emitted. So the time it takes for light to reach us are the same. BUT, according to the moving observer, they were not emitted at the same time, so he detects them at different times. [or you can say: since he detects them at different times, they were not emitted at the same time according to him] That's the whole point of relative simultaneity.
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#99
geistkiesel
538
1
wespe said:
Do you realize you wasted people's time by not specifying this in the first place? OK, in that case, #1 and #2 are true.
But, as Janus says, #3 is still unclear: where is the man when he detects the photons?
Number 3. is as clear as the rest. Once tghe photons have been emitted, whether from a moving platform or not, the motion of the sources is irelevant. The moving source may measure the speed of light, or he can simply look at his photon detectors. He is moving tot he right towarda photon moving tot he left. The time of flight of both photons is different.The left moving photon meet the observer before the right moving photon.
AS long as some v > 0 is specified the moving observer will always detect the photons in sequence. The measured delta t between measurements will surely differ from those measured by a stationary observer, and the distance traveled by the photons will be measured differently between the two observers, but this does not change the reality that the time of flight of both photons is different. The distance from L to Midpont is the same as R to the midpoint in both frames, though the frame to frame measurement number comparison may differ.
SR does not say that the right moving photon will move faster than the left moving photon in any of the cases. This would have to be the conclusion if the observer ever saw the detected photons simultaneously in 2 and 3. SR merely says the photons were not emitted simultaneously in both frames.
As long as some v > 0 is specified the moving observer will always detect the photons in sequence. It seems we have an absolute frame implied. In Grounded's examples, the measured photons are emasured against a stationary source, as the speed of light is invariant with the speed of the source, correct? Therefore meassuring the observers velocity wrt the wave lengths ppasing through the observers position, the observers speed wrt the photon source speed = 0 can be determined.
#100
O Great One
98
0
wespe said:
Yes, light always moves towards us at the same constant speed relative to us. And the distance is the same when lights were emitted. So the time it takes for light to reach us are the same. BUT, according to the moving observer, they were not emitted at the same time, so he detects them at different times. [or you can say: since he detects them at different times, they were not emitted at the same time according to him] That's the whole point of relative simultaneity.
OK. So the guy moving to the left concludes that the light on the left was emitted before the light on the right. Let's say that there was a guy running to the right then he would have to conclude that the light on the right was emitted before the light on the left. BOTH CAN'T BE TRUE.
You are aware that GPS receivers use SR to determine their exact location relative to the satellites? That's why you need to have 4 sat's in view. You triangulate your position with three, but you need the fourth to solve for the SR clock bias.
Now that, I did not know. Thanks.
And hey, don't throw out the reciever quite yet - I'm sure SR will be replaced with another theory that is mathematically exactly the same, if any of the anti-relativity guys ever get around to working out the math.
#102
ram1024
301
0
EEK :surprise:
was trying to stay on target and only get opinions using SR in this thread... i guess that's all out the window now
that's an assumption. I'll elaborate when we get to the second stage. need hurkyl russ or tom to confirm (True, True, True) on the first 3 cases.
Sorry - today was Father's Day, which meant I was having lunch with my parents/grandparents, then playing golf with my dad. I'm not quite at my computer 24/7.
#104
ram1024
301
0
oh crap today was father's day?
#105
ram1024
301
0
OK. So the guy moving to the left concludes that the light on the left was emitted before the light on the right. Let's say that there was a guy running to the right then he would have to conclude that the light on the right was emitted before the light on the left. BOTH CAN'T BE TRUE.
hehe yeh it's things like this that make me a skeptic :D
good work spotting it though, Great One. apparently it's not easy to spot :|
#106
wespe
202
0
O Great One said:
OK. So the guy moving to the left concludes that the light on the left was emitted before the light on the right. Let's say that there was a guy running to the right then he would have to conclude that the light on the right was emitted before the light on the left. BOTH CAN'T BE TRUE.
Both must be true. Given the constant speed of light, there is no other explanation (well, if you have one...). Look at it this way: simultaneity is defined by what the midpoint observer concludes in this setup. Since there can be any number or midpoint observers moving at different speeds, each one concludes differently, and each of them are correct according to themselves. That's relative simultaneity, as opposed to absolute (one that everyone agrees on) simultaneity. If you have a better definition of simultaneity, I'd like to hear that. Of course, defining simultaneity has consequences. It must be consistent with what you define as time, and with casuality. Note that all observers still agree that events occurring at a single point in space are absolutely simultaneous. Only events separated by distance causes disagreement. But this disagreement does not yield any casuality paradoxes, since SR also limits all speeds to be below c. Considering the speeds involved in our daily lives, it has little impact and not a big deal really (except when you need very high precision and stuff)..
Take care.
#107
wespe
202
0
ram1024 said:
hehe yeh it's things like this that make me a skeptic :D
good work spotting it though, Great One. apparently it's not easy to spot :|
Ram1024, that's funny, but, did you read #93 and #96? Hurkyl still thinks simultaneity was measured in "picture frame". Maybe you should make it clear before wasting any more of his time.
#108
ram1024
301
0
the consequences are way too far-fetched to believe in.
Length Contractions, Time Dialation, No Absolute Time, No Absolute Space...
it's like alice in wonderland on your guys' side of the fence :D
i'll define my version some time tomorrow i think. sure would like more people to toss in their opinions first of all, though.
it's intriguing that people who know SR still can't agree what the answers are to these.
the consequences are way too far-fetched to believe in.
Length Contractions, Time Dialation, No Absolute Time, No Absolute Space...
it's like alice in wonderland on your guys' side of the fence :D
i'll define my version some time tomorrow i think. sure would like more people to toss in their opinions first of all, though.
it's intriguing that people who know SR still can't agree what the answers are to these.
Clearly, this approach isn't working. Perhaps you'd be more inclined to accept SR if you started looking at what the evidence actually shows. Real experiments showing these "alice in wonderland" concepts like time dilation and length contractions.
How about GPS satellites? Do you know how they keep their clocks synchronized?
How about speed of light measurements: do you know that they all give the same answer?
#110
ram1024
301
0
Hurkyl has unfortunately come up with the "correct" answers for #2 #3 and #4, but I'm not sure he acquired them through SR since he didn't really elaborate.
if he has indeed come to that conclusion based on SR then i am pretty much SOL for refuting it since that's what my relativity would say as well
was trying to hide him under the mattress or something so i could get a bit further along in the discussion, but you just HAD to go and point him out
#111
geistkiesel
538
1
ram1024 said:
Hurkyl has unfortunately come up with the "correct" answers for #2 #3 and #4, but I'm not sure he acquired them through SR since he didn't really elaborate.
if he has indeed come to that conclusion based on SR then i am pretty much SOL for refuting it since that's what my relativity would say as well
was trying to hide him under the mattress or something so i could get a bit further along in the discussion, but you just HAD to go and point him out
you gave up too much., too soon. try this one out for size.
All moving frame values are non-primed with the exception of M’, the consistent location of the observer O in the moving frame.
At no time is there an inference that M’ was at the midpoint of the A and B photons emitted in the stationary frame.
To demonstrate the following:
Einstein’s moving train calculation indicating when the oncoming B photon is detected at t1 the A photon was located at a position consistent with –t1. Said in other words, as t1 is determined from t0 which locates M’ at t0, the A and B were equidistant to M’(t0) when t = t1.
Proof:
A moving observer located at M’ on a moving frame passes through the midpoint M of photon sources located at A and B in the stationary frame just as A and B emit photons. M’ is moving along a line connecting A and B, toward B.
At this instant the moving source t = t0. Later the moving observer detects the photon from B at t1, and later the photon from A at t2. The observer has measured her velocity wrt the stationary frame as v. Determine the position of the A photon at tx in terms of t0, t1, t2, and v when the B photon was detected at t1.
The photon from A must reach the position of M’ when t = t2. Therefore, the distance traveled by the A photon during Δt = t2 – t1, is Δtc. This is equal to the distance cΔt = vΔt + vt1 + vtx . Now we rearrange somewhat to arrive at, vtx = vΔt – cΔt + –vt1. Now as vΔt - cΔt is just -vtx - vt1
vtx = -vtx - vt1 – vt1
2tx = -2t1
tx = -t1
Therefore, in the moving frame the photon from A and the photon from B were equidistant from M’(t0) at t1.
#112
O Great One
98
0
OK. So the guy moving to the left concludes that the light on the left was emitted before the light on the right. Let's say that there was a guy running to the right then he would have to conclude that the light on the right was emitted before the light on the left. BOTH CAN'T BE TRUE.
wespe said:
Both must be true. Given the constant speed of light, there is no other explanation (well, if you have one...). Look at it this way: simultaneity is defined by what the midpoint observer concludes in this setup. Since there can be any number or midpoint observers moving at different speeds, each one concludes differently, and each of them are correct according to themselves. That's relative simultaneity, as opposed to absolute (one that everyone agrees on) simultaneity. If you have a better definition of simultaneity, I'd like to hear that. Of course, defining simultaneity has consequences. It must be consistent with what you define as time, and with casuality. Note that all observers still agree that events occurring at a single point in space are absolutely simultaneous. Only events separated by distance causes disagreement. But this disagreement does not yield any casuality paradoxes, since SR also limits all speeds to be below c. Considering the speeds involved in our daily lives, it has little impact and not a big deal really (except when you need very high precision and stuff)..
Take care.
It is logically impossible that both statements are true. Let's imagine that the platform is balanced like a teeter-totter and when the light is emitted the platform tips in that direction. So the guy running towards the left thinks that the platform tips to the left and then to the right, and the guy running towards the right thinks the platform tips to the right and then to the left. Try to explain how they are both correct!
#113
ram1024
301
0
actually it all suits my purposes. i don't care to retute SR as long as it sees things MY way ;D
#114
geistkiesel
538
1
Hurkyl said:
As has already come up, #4 has not been formulated in a way consistent with SR.
In other words, you're assuming the answer before you "do" the experiment.
If you accelerate the train, you run into the problem of no rigid bodies, and if both trains had been moving with different constant velocitys (possibly zero), then it cannot be the case that the two setups are identical (particularly the same proper length) and that the endpoints can coincide simultaneously in either of their two rest frames.
Furthermore, yet again I can only assume that when you say "simultaneous" you mean in the frame you draw pictures (which I'll now call the "picture frame")
Anyways, if I assume you do something like accept there's a rigid body problem, or allow one train to be longer than the other, or something or other, I deduce that both observers will observe the right photon before the left photon. If you can get the two observers along the same worldline, then they will detect the right photon simultaneously, and similarly for the left photon.
This is consistent with my previous answers where, in both case #2 and #3 (under the assumption that "simultaneously" was measured in the picture frame), I stated that the observer would not detect both photons simultaneously.
Hurkyl, the following is similar to the problem you had me solving a while back. This should be slam dunk easy for you.
All moving frame values are non-primed with the exception of M’, the consistent location of the observer O in the moving frame.
At no time is there an inference that M’ was at the midpoint of the A and B photons emitted in the stationary frame.
To demonstrate the following:
Einstein’s moving train calculation indicating when the oncoming B photon is detected at t1 the A photon was located at a position consistent with –t1. Said in other words, as t1 is determined from t0 which locates M’ at t0, the A and B were equidistant to M’(t0) when t = t1.
Proof:
A moving observer located at M’ on a moving frame passes through the midpoint M of photon sources located at A and B in the stationary frame just as A and B emit photons. M’ is moving along a line connecting A and B, toward B.
At this instant the moving source t = t0. Later the moving observer detects the photon from B at t1, and later the photon from A at t2. The observer has measured her velocity wrt the stationary frame as v. Determine the position of the A photon at tx in terms of t0, t1, t2, and v when the B photon was detected at t1.
The photon from A must reach the position of M’ when t = t2. Therefore, the distance traveled by the A photon during Δt = t2 – t1, is Δtc. This is equal to the distance cΔt = vΔt + vt1 + vtx . Now we rearrange somewhat to arrive at, vtx = vΔt – cΔt + –vt1. Now as vΔt - cΔt is just -vtx - vt1
vtx = -vtx - vt1 – vt1
2tx = -2t1
tx = -t1
Therefore, in the moving frame the photon from A and the photon from B were equidistant from M’(t0) at t1.
#115
wespe
202
0
O Great One said:
It is logically impossible that both statements are true. Let's imagine that the platform is balanced like a teeter-totter and when the light is emitted the platform tips in that direction. So the guy running towards the left thinks that the platform tips to the left and then to the right, and the guy running towards the right thinks the platform tips to the right and then to the left. Try to explain how they are both correct!
That's because there is no totally rigid objects in reality. Imagine that the platform is several light years long. When one side goes down, this event cannot have an immediate effect on the midpoint or the other side. Even if the change propagates at light speed, it has to travel for several years to reach the midpoint or to make the other side go up. Therefore, what the observer sees is more up-to-date than what he feels under his feet. Was this what you meant?
#116
O Great One
98
0
OK. Let's try again. Let's say that when the light is emitted, the emitter blows up in a huge explosion. Somebody is standing off in the distance observing this. He either observes the explosion on the left and then the explosion on the right or he observes the explosion on the right and then on the left.
That will depend on how that person is moving. It is the original situation all over again. If he is moving to the left, he will get first the light from the left, and since he measures the speed of that light to be the same as that from the right, he is entitled to deduce that the blast from the left occurred first.
Same happens if the person in the distance is moving to the right.
#118
ram1024
301
0
the emitter exploding is a good one, Great. i was tempted to use explosion as emitters but was afraid someone would say something crazy like "the explosion imparts and inertial vectored thrust quotient based on non-linear cohesian and sub temporal disjunction, and therefore the light particles get immersed in an endothermic tesla-radiation that elevates their states to 15th dimensional particles. SR predicts that these particles exist everywhere at once. Good Game, ram1024, YOU LOSE"
#119
geistkiesel
538
1
Hurkyl said:
In this case, then, indeed, (true, true, true).
However, I would like to comment on other things you have said:
This will not work. There is no such thing as a rigid body, so you cannot use that as a way to keep things in sync.
In fact, if you sync the clocks while they're stationary (in the background frame), then accelerate the platform, the clocks cannot be synchronized in their rest frame, nor in the background frame.
This is easy to see in the background frame; due to length contraction, one of the clocks must have been displaced more than the other.
Which cannot happen, according to SR, unless the situation is trivial; either the two clocks are at the same place, or two frames are the same.
(for simplicity, I'm speaking in one spatial dimension)
I'm lagging somewhat behind the course of the conversation (darned TV!) but I'm going to post this anyways.
Can't you rotate the clocks 90 degrees, accelerate and then rotate back to align as before acceleration?
#120
geistkiesel
538
1
Tom Mattson said:
A better way to look at it is the way JCSD described. Nothing "causes" them to be unsynchronized in other frames, they simply are unsynchronized. There is no way that the postulates of SR can hold and for simultaneity to be absolute. So the question is, Do the postulates hold?
Experimentation has answered with an emphatic "YES".
No, you are expected to accept it because the evidence[/color] says so. And even if you don't accept it, surely you must be able to accept that simply assuming that SR is false does not disprove it.
And why not? Experimentation is the final court of appeals in science. If you aren't open to that, then there is no hope for you.
That's correct, because nothing was "done" to the clocks, period. It's not as though some invisible agent resets the clocks so that they are out of synch when a moving observer passes by. It is just a simple consequence of the fact that, in our universe, the laws of physics and the speed of light are the same for everyone.
But we can[/color] discern that the clocks don't tick at the same rate for all observers.
Tom, Can't you rotate the clocks 90 degrees transverse to the accelerated direction and when velocity is achieved rotate the clocks -90 degrees?
