Which Train Explodes First in the Time Dilation Paradox?

In summary, the train may know that the timer reaches 2 minutes by some means of communication. The situation is symmetrical, but each person can predict that his timer will reach 2 minutes before the other. The problem is that each person explodes before seeing the other explode.
  • #71
Nugatory said:
"Your frame" is another of those dangerous phrases that can lead to misunderstanding. It is a convenient shorthand for "the frame in which you just coincidentally happen to have zero instantaneous velocity" - and if you try substituting that for the bolded text you may see the problem with saying that you do anything (including seeing) "in a frame".

Events, such as light hitting your retina, don't happen in frames. They just happen; and different frames are just different sets of rules for assigning them space and time coordinates.

Well, I meant to point out that light seen by one observer will not be seen by another, even if the photons are radiated from the same event, each observer sees a different set of photons, and each observer draws his own coordinates (which may or may not coincide with someone else's coordinates), hence, the same light is seen in only one frame of reference, even if it exists in every frame. I guess i should've just said observer instead of frame and save us all from this waste of time.
 
Physics news on Phys.org
  • #72
altergnostic said:
[..] I guess i should've just said observer instead of frame and save us all from this waste of time.
That's better, since anyone can freely choose whatever frame he wants. However it's not really a waste of time, as wrong ways of saying things can propagate misunderstanding.
 
  • #73
altergnostic said:
[..] I say that it is in your own frame because seeing light is always a local event, you can't speak of light in another frame. I mean that regardless of the position and motion of another frame, light is either [..] right in your eyes, or you are not aware of it at all. [..].
altergnostic said:
Well, I meant to point out that light seen by one observer will not be seen by another, even if the photons are radiated from the same event, each observer sees a different set of photons, and each observer draws his own coordinates [..]
In addition to my earlier post of today, this may serve as illustration. Even after eliminating the frame/observer mix-up, you seem to suggest that we can explain SR by using QM (photons), so that an observation by one observer cannot be matched by another observer. If so, then that completely misses the point. SR is based on Maxwell's wave mechanics and shows how the same light wave and even the same phase difference (which can in principle be detected with different detectors) is mapped differently with different reference systems.
Compare the train and station example: one light flash as detected with different reference systems
- http://www.bartleby.com/173/9.html
 
Last edited:
  • #74
i think my brain is going to melt.

everyone has deviated from the original problem. how can person A survive while person B explodes, and vice versa?
 
  • #75
holtto said:
[..] everyone has deviated from the original problem. how can person A survive while person B explodes, and vice versa?
- The problem of the OP was already answered with posts #2, 4, 5 etc.
- Very detailed answers can be found in posts #17, #21.
- References to similar problems and discussions can be found in post #33.

What followed was just a little "after talk". :smile:
 
  • #76
harrylin said:
- The problem of the OP was already answered with posts #2, 4, 5 etc.
- Very detailed answers can be found in posts #17, #21.
- References to similar problems and discussions can be found in post #33.

What followed was just a little "after talk". :smile:
hmmm, only #5 and #21 seem to have some concrete stuff; too bad they had been buried under all the "after talk."I like this paradox, let's call it the "Action Movie Train Scene" paradox.
 
  • #77
so let me guess this straight. the resolution is:

In Adam's frame, Bob dies before Adam dies

In Bob's frame, Adam dies before Bob dies.

In the end, both of them die. Everybody dies.
 
  • #78
holtto said:
so let me guess this straight. the resolution is:

In Adam's [rest] frame, Bob dies before Adam dies

In Bob's [rest] frame, Adam dies before Bob dies.

In the end, both of them die. Everybody dies.
[EDITED] Indeed both of them die (and I added a little precision in [], as explained in post #70).
However - as also illustrated in post #21 - the local clocks must show the same time when they blow up; and as according to each frame's reckoning the distant clock appears to run slower (with synchronous start), this means that "In Adam's [rest] frame, Bob dies after Adam dies."

Anyway, neither of them can see the destruction of the other train. You could put "In the end, both of them die; and neither saw the other die".

PS: The easiest frame to pick is the rest frame of the train station. Using that frame, the symmetry of the situation remains clear and it immediately shows that each train starts exploding before information of the other train's explosion can reach it.
 
Last edited:
  • #79
altergnostic said:
Well, I meant to point out that light seen by one observer will not be seen by another, even if the photons are radiated from the same event, each observer sees a different set of photons
This is all true, it just has nothing to do with reference frames.

altergnostic said:
and each observer draws his own coordinates (which may or may not coincide with someone else's coordinates), hence, the same light is seen in only one frame of reference, even if it exists in every frame.
Suppose that you have a detector on the optic nerve. If the observer sees the light then the detector prints out a piece of paper with the number 1, if the observer does not see the light then the detector prints out a piece of paper with the number 0. If the observer sees the light in one frame of reference there is a 1 on the piece of paper. You are claiming that in some frames there will be a 0 on the piece of paper. That is wrong.

altergnostic said:
I guess i should've just said observer instead of frame and save us all from this waste of time.
Yes, but it is such a common mistake that it deserves a remine
 
  • #80
DaleSpam said:
Suppose that you have a detector on the optic nerve. If the observer sees the light then the detector prints out a piece of paper with the number 1, if the observer does not see the light then the detector prints out a piece of paper with the number 0. If the observer sees the light in one frame of reference there is a 1 on the piece of paper. You are claiming that in some frames there will be a 0 on the piece of paper. That is wrong.

No, that's not what i am saying. you will se the number 1 in a paper if someone else sees a beam of light, but you will not see the beam itself. I am saying that light that gets in your retina doesn't go into anywhere else but your retina. If a photon is in your eye it is not in someone else's eye. If you know about light being seen by another observer, you do only by indirect means, like a paper with the number 1 in your example.
 
  • #81
holtto said:
so let me guess this straight. the resolution is:

In Adam's frame, Bob dies before Adam dies

In Bob's frame, Adam dies before Bob dies.

In the end, both of them die. Everybody dies.

No. In adam's frame, he dies before bob. In bob's, he is the one who dies first.

Both die, but the other is late, light from the explosion that comes from bob takes a while to get to adam, so when it reaches adam, he has already exploded, and vice versa.
 
  • #82
harrylin said:
In addition to my earlier post of today, this may serve as illustration. Even after eliminating the frame/observer mix-up, you seem to suggest that we can explain SR by using QM (photons), so that an observation by one observer cannot be matched by another observer. If so, then that completely misses the point. SR is based on Maxwell's wave mechanics and shows how the same light wave and even the same phase difference (which can in principle be detected with different detectors) is mapped differently with different reference systems.
Compare the train and station example: one light flash as detected with different reference systems
- http://www.bartleby.com/173/9.html

I'm not trying to explain relativity with photons. If you like, you can change photons for a directinalized beam, like a laser. If a blink of light propagates spherically, all observers will eventually see it, but they are not seeing the same patch of light. If i send a laser beam into your eye, you are the only one who will see it. And this has nothing to do with relativity or the topic anymore.
 
  • #83
harrylin said:
That's better, since anyone can freely choose whatever frame he wants. However it's not really a waste of time, as wrong ways of saying things can propagate misunderstanding.

In a thought problem, yes, you can pick any frame you like, but in real life, the observer is always attatched to his particular frame of reference, he can't choose freely nor change frames. Ever.
 
  • #84
altergnostic said:
No, that's not what i am saying. you will se the number 1 in a paper if someone else sees a beam of light, but you will not see the beam itself. I am saying that light that gets in your retina doesn't go into anywhere else but your retina. If a photon is in your eye it is not in someone else's eye. If you know about light being seen by another observer, you do only by indirect means, like a paper with the number 1 in your example.
None of your explanation has anything to do with frames, which is why your statement "the light is seen in only one frame of reference" was wrong. That incorrect statement implies exactly what I described above, which is clearly not what you meant, but is in fact what you said.

The correct statement is "the light is seen by only one observer", which is a true statement regardless of which coordinate system you choose to analyze it in.
 
  • #85
altergnostic said:
In a thought problem, yes, you can pick any frame you like, but in real life, the observer is always attatched to his particular frame of reference, he can't choose freely nor change frames. Ever.
Sure he can. A reference frame is a purely mental construct, a coordinate system, which is just a mathematical bookkeeping convention.

An observer is always free to use a coordinate system in which he is moving. In fact, mentally I suspect that you yourself do this whenever you are driving or playing a sport, you probably mentally use the reference frame of the road or field, even though you are moving in it.

That is all a reference frame is, a mental way to label distances and directions and speeds, you don't need to restrict yourself to one and you can freely choose and change frames for any reason or no reason at all.
 
Last edited:
  • #86
harrylin said:
[..] you seem to suggest that [...], so that an observation by one observer cannot be matched by another observer. If so, then that completely misses the point. SR [..] shows how the same light wave [..] is mapped differently with different reference systems. [..]
altergnostic said:
[..] If a blink of light propagates spherically, all observers will eventually see it, but they are not seeing the same patch of light. If i send a laser beam into your eye, you are the only one who will see it. And this has nothing to do with relativity [..]
That's just what I meant... :rolleyes:
 
  • #87
harrylin said:
For me that's just a space-time coordinate. Regretfully the Wikipedia article only paraphrases the referenced textbooks, so here's a citation of how Alonso and Finn (Fundamental University Physics) define "event" in their chapter on the Lorentz transformation:

"An event is a specific occurrence that happens at a particular point in space and at a particular time".

I won't comment further on this except if it's an issue for bgq.

Also an event/ change-of-state has to generate an em signal to be perceived, otherwise it's a non event.
 
  • #88
The observer's perception corresponds to his position and his local clock. He is his own frame of reference. He can't make an observation from another frame of reference moving relative to himself or use the time of that frames clock.

The fact that 2 observers can't occupy the same spatial position at the same time, requires that their descriptions of an event differ. This was common knowledge thousands of years ago.

If a single photon is emitted, only the observer in line with its path will detect it.
No one else detects it!
If this were not true, then alignments in experiments wouldn't be necessary, and a detector could be placed anywhere in the lab.

If we carry the nit-picking to its conclusion, the 'frame' is another of those ideal concepts that cannot be realized. The first approximation is a set of objects with no relative motion, excluding thermal. In the real world, everything moves.

Remember, definitions are either in terms of other definitions, or circular.
 
  • #89
This is offered as another 'no calculation' demo.
The ops' original scenario has been transferred from trains to space cans (more realistic) with arbitrarily chosen speeds (a generalization). Each can sends a detonation signal when its clock indicates t, to the other can, which has a hidden exlosive device. The hyperbola indicates time t for each can depending on its speed. Signal A to B is path (1,2), signal B to A is path (3,4). No speeds are given because they aren't relevant. If B's speed varies from that of A toward light speed c, it will always send a signal before intercepting the A signal, since the hyperbola and path (1,2) never meet.

https://www.physicsforums.com/attachments/51449
 
Last edited:
  • #90
phyti said:
The observer's perception corresponds to his position and his local clock. He is his own frame of reference. He can't make an observation from another frame of reference moving relative to himself or use the time of that frames clock.

The fact that 2 observers can't occupy the same spatial position at the same time, requires that their descriptions of an event differ. This was common knowledge thousands of years ago.

If a single photon is emitted, only the observer in line with its path will detect it.
No one else detects it!
If this were not true, then alignments in experiments wouldn't be necessary, and a detector could be placed anywhere in the lab.

If we carry the nit-picking to its conclusion, the 'frame' is another of those ideal concepts that cannot be realized. The first approximation is a set of objects with no relative motion, excluding thermal. In the real world, everything moves.

Remember, definitions are either in terms of other definitions, or circular.

That was my point a few posts back, but you articulated it way better. Your reminder that two can't occupy the same place at the same time is all we need to forbid any notion of light seen in another observer's frame. In real life, that light can never be directly a part of our data, we can only know of its existence through secondary effects, like "yeah, i saw that too!"
 
  • #91
phyti said:
The observer's perception corresponds to his position and his local clock. He is his own frame of reference. He can't make an observation from another frame of reference moving relative to himself or use the time of that frames clock.

So, if you are riding a bicycle, you are not entitled to think you are moving; you must insist that the ground is moving? [edit: and if, while riding a bike, I look at a clock tower, I must ignore its time because it is not in my frame.]

More germane to physics, for all these years of colliding particles or ions into stationary targets, physicist were acting incorrectly when they chose to record observations and do calculations in the COM frame rather than the lab frame?
 
Last edited:
  • #92
altergnostic said:
Well, I meant to point out that light seen by one observer will not be seen by another. [..] I guess i should've just said observer instead of frame and save us all from this waste of time.
altergnostic said:
[..] If a blink of light propagates spherically, all observers will eventually see it, but they are not seeing the same patch of light. If i send a laser beam into your eye, you are the only one who will see it. And this has nothing to do with relativity or the topic anymore.
altergnostic said:
[..] Your reminder that two can't occupy the same place at the same time is all we need to forbid any notion of light seen in another observer's frame. In real life, that light can never be directly a part of our data, we can only know of its existence through secondary effects, like "yeah, i saw that too!"
Evidently you recognized that:
- saying "frame" for "a light detector" (or a person who observes) is a mix-up;
- and "the same patch of light" has nothing to do with relativity or this topic;

Thus it's a mystery to me why you continue introducing these things...
 
Last edited:
  • #93
holtto said:
hmmm, only #5 and #21 seem to have some concrete stuff; too bad they had been buried under all the "after talk."


I like this paradox, let's call it the "Action Movie Train Scene" paradox.

I am very happy to know that you like this paradox; however, the "after talk" here is very valuable, I have learned from them much although I am not qualified enough to participate in these discussions.
 
  • #94
phyti said:
The fact that 2 observers can't occupy the same spatial position at the same time, requires that their descriptions of an event differ.
altergnostic said:
That was my point a few posts back, but you articulated it way better. Your reminder that two can't occupy the same place at the same time is all we need to forbid any notion of light seen in another observer's frame.
Welcome to the discussion phyti, you are making the same mistake that altergnostic is making. You can pat each other on the back all you want for being wrong together. The rather trivial fact that two observers cannot occupy the same place at the same time has absolutely nothing whatsoever to do with the question of which reference frames are admissible.

Their description of a given event certainly CAN use the same reference frame, despite the fact that they are in different locations and potentially moving wrt each other. If I get pulled over for going 100 mph then I am perfectly justified in saying that I was going 100 mph, I am not required to say that I was at rest and the officer was going -100 mph.

I am not required to use a reference frame where I am at rest, and I am not required to use a reference frame where I am at the origin.

phyti said:
He can't make an observation from another frame of reference moving relative to himself or use the time of that frames clock.
Yes, he can. GPS satellites do it all the time.

phyti said:
If a single photon is emitted, only the observer in line with its path will detect it.
No one else detects it!
Sure. But that observer detects that photon regardless of what coordinate system you choose to use to analyze the detection. You and altergnostic both seem to confuse the physical observer/detector and the purely mathematical construct of a coordinate system.
 
  • #95
altergnostic: #90
... Your reminder that two can't occupy the same place at the same time is all we need to forbid any notion of light seen in another observer's frame.

That would not be true if the event in common involved multiple photons. In Einsteins train scenario, both observers saw the flashes because they were multiple photon events. My statement was a logical basis for expecting different descriptions of events from 2 different observers. If 10 people circled a statue and each took a photo, each would have a different perspective as evidenced in the photos, yet they would agree there was only 1 statue. That's a simple case with no motion.

In real life, that light can never be directly a part of our data, we can only know of its existence through secondary effects, like "yeah, i saw that too!"
Not sure of what you are saying here, but we perceive the universe indirectly. Light is the messenger and via sensory input, says 'something happened over here', and indicates a direction.
From other posts:
When you discuss A and B observing each others clock running slower than their own, remember each is comparing his current clock event with a clock event in the others past. Does that have any useful meaning?
 
  • #96
PAllen #91
So, if you are riding a bicycle, you are not entitled to think you are moving; you
must insist that the ground is moving? [edit: and if, while riding a bike, I look at a
clock tower, I must ignore its time because it is not in my frame.]

If you read post 88 carefully, it doesn't imply any of what you said. It was just a
response to all the confusing 'frame' talk.
On a bike, on a park bench, it's irrelevant, because all things are moving. SR theory
just says if you are moving, you may choose to assume a pseudo rest frame, and
regardless of which you choose, you use a local clock. In your scenario, which was
not the intended range of application, there would be no significant difference in
using the bikers watch or the clock tower (assuming synchonization).With reference
to the observers frame, A.E. states in the 1905 paper, "It is essential to have time
defined by means of stationary clocks in the stationary system,"

More germane to physics, for all these years of colliding particles or ions into
stationary targets, physicist were acting incorrectly when they chose to record
observations and do calculations in the COM frame rather than the lab
frame?

That's just tranforming from one origin to another, which is what SR does. I don't
see any problem there.
 
  • #97
For those who may not have accepted the geometry based example, and motivated by ghwells example using doppler methods, this is the math for the general case of two observers in relative motion, sending each a signal at a predetermined local time. The signals will arrive at the same local time for each.
https://www.physicsforums.com/attachments/51567
 
Last edited:
  • #98
phyti said:
The signals will arrive at the same local time for each.
Do you think that any reference frame claims otherwise? If not, then no observer is required to use any particular frame to get the correct answer.
 
  • #99
DaleSpam said:
Do you think that any reference frame claims otherwise? If not, then no observer is required to use any particular frame to get the correct answer.

It's intended for bgq, altergnostic, or anyone else who thinks they have to start with a symmetrical situation, i.e. equal speeds in opposite directions.
As for 'frames', I'll wait for a sound/consistent definition, if there is one.
 
  • #100
phyti said:
As for 'frames', I'll wait for a sound/consistent definition, if there is one.
The "official" definition of a frame is given here: http://en.wikipedia.org/wiki/Frame_fields_in_general_relativity

But I usually use the term as though it were synonymous with "coordinate system".

With either definition it should be clear that any observer can use any frame to predict the outcome of any experimental measurement.
 
Last edited:

Similar threads

Replies
14
Views
536
Replies
9
Views
4K
Replies
39
Views
4K
Replies
20
Views
2K
Replies
88
Views
5K
Replies
46
Views
2K
Replies
11
Views
1K
Replies
15
Views
1K
Back
Top