Today Special Relativity dies

[wespe]

in the Einstein Gedanken the observer moves towards an emitter <lightning> and away from another, starting at the midpoint.

the "remains there in his frame" is wrong. As per Case #7.

Let me clarify.

The "moving" observer in Einstein Gedanken is M'. When the lightenings strike, M' is at the midpoint according to the embankement. M' then moves to the right, according to the embankement.

Now, according to M', the lightenings don't strike simultaneously. So, when looking from M' frame, we can't actually say "M' was at the midpoint when the lightenings stroke", because that's not a single instant. But we can say "M' was at the midpoint of the locations where the lightenings stroke". Those locations are where the burning marks are made on the train (not the embankement, we are in the train frame now). Then, M' does not move anywhere (in his own frame, according to himself), thus remains at the midpoint of the events the whole time.

About your case #7 (post#249, took some time to find), all observers remain at the same distance from the location of any emitted photons (in accordance to what I wrote above). In other words, the location any event never changes *within a frame*. The event has happened at an instant and its location cannot be carried with the source of the event or whatever.

 

[ram1024]

Now, according to M', the lightenings don't strike simultaneously. So, when looking from M' frame, we can't actually say "M' was at the midpoint when the lightenings stroke", because that's not a single instant. But we can say "M' was at the midpoint of the locations where the lightenings stroke". Those locations are where the burning marks are made on the train (not the embankement, we are in the train frame now). Then, M' does not move anywhere (in his own frame, according to himself), thus remains at the midpoint of the events the whole time.

if he was at the midpoint of the "events" and light speed is constant relative to him, it stands to reason that HE views the light as striking simultaneously as well, does it not? he would not lose simultaneity in this case...

About your case #7 (post#249, took some time to find), all observers remain at the same distance from the location of any emitted photons (in accordance to what I wrote above). In other words, the location any event never changes *within a frame*. The event has happened at an instant and its location cannot be carried with the source of the event or whatever.

but the law cannot be frame inviolate as i have outlined, the times for light to reach an observer from "an event" is different if the observer is moving than if the emitter is moving. the emitter does not impart relative change in motion of an "event" but the observer DOES.

in other words, the light is NOT tied to it's SOURCE in an event, it IS however tied to an Observer.

Make it so Observers are both Observers AND Emitters. Make one of them move and the other one stationary. ping the emitters simultaneously in a "midpoint" frame. One of them <the moving one> will be hit with the opposing emitter's photon first <by virtue of closing a distance towards the emitter>

now switch "frames" make the moving one stationary. all of a sudden the OTHER emitter (now the moving one) gets hit by the opposing photon first.

Holy Rusted Metal, Batman

Paradox

 

[ram1024]

still working on the gif for this one... but i figured the explanation should be good enough to get the point across for the time being

stupid animated gif programs not being free :O

 

[wespe]

if he was at the midpoint of the "events" and light speed is constant relative to him, it stands to reason that HE views the light as striking simultaneously as well, does it not? he would not lose simultaneity in this case...

Yes "he was at the midpoint of the events" and yes, "light speed is constant relative to him". Therefore, yes, IF he saw lights striking simultaneously, he would rightfully conclude the events occurred simultaneously. But he does NOT see them simultaneously, therefore concludes they were not simultaneous events in his own frame. You may think "why not change the observation rather than the conclusion. How are we sure he does not see them at the same time?". Because that's what the stationary frame sees: that the moving observer meets the lights at different points so he can't have possibly seen them at the same time. I could think of a paradox if this was not the case.

but the law cannot be frame inviolate as i have outlined,

I have stated my objection to that above so let's not continue until that's clear.

 

[ram1024]

Yes "he was at the midpoint of the events" and yes, "light speed is constant relative to him". Therefore, yes, IF he saw lights striking simultaneously, he would rightfully conclude the events occurred simultaneously. But he does NOT see them simultaneously

Einstein's Gedanken right? observer moving FROM the center of the light points AT THE INSTANT that lights are fired simultaneously in the stationary frame. Make that observer stationary. Lights are STILL fired in the first instant of the experiment when the distance between observers and is equal on both sides, light does not move with sources (in this case lightning doesn't leave a "source" to move with anyways). observer does not move either. midpoint DOES move.

conclusion? observer receives light simultaneously.

I have stated my objection to that above so let's not continue until that's clear.

and my continuation should prove beyond a doubt that my reasoning in case #7 holds true. Postulate: "Motion of light is independant of the motion of the source, but DEPENDANT on the motion of the observer"

 

[Alkatran]

if he was at the midpoint of the "events" and light speed is constant relative to him, it stands to reason that HE views the light as striking simultaneously as well, does it not? he would not lose simultaneity in this case...



but the law cannot be frame inviolate as i have outlined, the times for light to reach an observer from "an event" is different if the observer is moving than if the emitter is moving. the emitter does not impart relative change in motion of an "event" but the observer DOES.

in other words, the light is NOT tied to it's SOURCE in an event, it IS however tied to an Observer.

Make it so Observers are both Observers AND Emitters. Make one of them move and the other one stationary. ping the emitters simultaneously in a "midpoint" frame. One of them <the moving one> will be hit with the opposing emitter's photon first <by virtue of closing a distance towards the emitter>

now switch "frames" make the moving one stationary. all of a sudden the OTHER emitter (now the moving one) gets hit by the opposing photon first.

Holy Rusted Metal, Batman

Paradox

Oh, I get what you mean now. No, that's not right. They actually do go the same distance... here allow me to explain:

Observer 1 is stationary relative to the emitter, observer 2 is moving towards it.

The Emitter releases a photon at 0.
Observer 1 receives this photon at t (in M), after Observer 2. Observer 2 receives it at t' (in M'), before Observer 1.

Now, what distances have these photons travelled? Well, they've traveled the same distance in each frame (t in M and t' in M') because you don't measure the distance between the emitter and the observer when he receives the light, because the speed of the emitter doesn't affect the photon's speed!

This means you need to use the emitter's original location >IN THAT FRAME<.
In M' (the moving frame) the emitter's original location is behind the emitter's location in M. So the distances are equal.

 

[ram1024]

i have no idea what you mean by that. But i just woke up so I'm slow starting :D

according to SR, both frames are equally valid. we KNOW from experimentation <i'm going to assume SOMEONE did Einstein's Gedanken before> that an observer moving towards a light receives photons before a person stationary at the same start location.

but it doesn't work in this case. if you switch frames so that the moving observer is "the stationary one" then the stationary observer in reality (now the moving one) gets hit by photon first. How can you determine who gets hit first in this case?

 

[jcsd]

No if you switch frames ram the observer who was stationary in the frame in your first paragraphy is now moving away from the light source in the frame in your second parargraph, consequently in this frame also it is this obsrever who receives the light last.

 

[ram1024]

nah we've moved on to Case #8 now, jcsd

lemme diagram it for you

Case #8

     [u](o)                    |                    (o)[/u]
     [u](o)                    |                  (o)[/u]
     [u](o)                    |                (o)[/u]
     [u](o)                    |              (o)[/u]

The observers in this case are both Observers AND Emitters. at the onset of the experiment, ObserverA and ObserverB shine a light towards each other. Simultaneity is verified by an observer at the midpoint of the starting locations of both observers. ObserverB then moves towards ObserverA. Predict who receives a photon first.

     [u](o)                    |                    (o)[/u]
       [u](o)                    |                  (o)[/u]
         [u](o)                    |                (o)[/u]
           [u](o)                    |              (o)[/u]

now all we've done is taken that same experiment from ObserverB's point of view AS IF he was the stationary one. in this case the reverse result is shown, ObserverA receives his photon first. It is also interesting to note that the midpoint observer no longer receives photons simultaneously in this view, further proving that frame shifting is wrong for this type of experiment. How can something absolute such as simultaneity AT A POINT be violated by merely changing how you view a situation?

 

[Alkatran]

nah we've moved on to Case #8 now, jcsd

lemme diagram it for you

Case #8

     [u](o)                    |                    (o)[/u]
     [u](o)                    |                  (o)[/u]
     [u](o)                    |                (o)[/u]
     [u](o)                    |              (o)[/u]

The observers in this case are both Observers AND Emitters. at the onset of the experiment, ObserverA and ObserverB shine a light towards each other. Simultaneity is verified by an observer at the midpoint of the starting locations of both observers. ObserverB then moves towards ObserverA. Predict who receives a photon first.

     [u](o)                    |                    (o)[/u]
       [u](o)                    |                  (o)[/u]
         [u](o)                    |                (o)[/u]
           [u](o)                    |              (o)[/u]

now all we've done is taken that same experiment from ObserverB's point of view AS IF he was the stationary one. in this case the reverse result is shown, ObserverA receives his photon first. It is also interesting to note that the midpoint observer no longer receives photons simultaneously in this view, further proving that frame shifting is wrong for this type of experiment. How can something absolute such as simultaneity AT A POINT be violated by merely changing how you view a situation?

They don't agree on the order because the two events are separated by distance.
I told you you weren't reading my posts.

 

[wespe]

Einstein's Gedanken right? observer moving FROM the center of the light points AT THE INSTANT that lights are fired simultaneously in the stationary frame. Make that observer stationary. Lights are STILL fired in the first instant of the experiment when the distance between observers and is equal on both sides, light does not move with sources (in this case lightning doesn't leave a "source" to move with anyways). observer does not move either. midpoint DOES move.
conclusion? observer receives light simultaneously.

Einstein's gedanken corresponds to your case#3. The moving observer does not see the lightenings at the same time (since they occurred simultaneously in the stationary frame, they can't have occurred simultaneously in the train frame). I am 100% sure this is SR's prediction. Plus, if he saw them at the same time, there would be a paradox, because the non-moving midpoint observer sees them at the same time at another point in space.

But if you make the moving observer stationary, of course he sees the lightenings at the same time, because he would be in the stationary frame now. That would correspond to case#1.

I don't understand why you say "midpoint DOES move". The midpoint of events never move within any frame, because the location of events never move within a frame. Location of an event in a frame may look like moving from another frame (like burning marks on the train as seen from the embankement), but they would move at the same speed as the train.

Postulate: "Motion of light is independant of the motion of the source, but DEPENDANT on the motion of the observer"

For the hundreth time, there is no motion of the observer in the observer's frame, observer is at rest wrt itself. Of course separation speed changes when looked from another frame, but that's irrelevant. What changes in a frame is the simultaneity of events, not how light moves. That's the whole point you are missing.

 

[Eyesaw]

Yes it does matter, if he is at the midpoint he will detect both simultaneously, if he's not he won't.

Example: the two following gifs show what happens according to the frame the emitters are in and the frame the man is in for the same situation. (the actual animation deals with a railway car moving along a track, but we will just assume that the track represents the platform. In this situation we have to men, one that stays stationary to the platform and one that moves with it.

We further stipulate that the moving man is next to the stationary man when the light is first detected by either.

The first animation shows things from the perspective of the emitters and the stationary man:
http://home.teleport.com/~parvey/train1.gif

The photons expand at c in two spherical fronts that reach the mid point at the same time by the observation of both men (both men see the photons arrive at the same time.)

The second animation shows what happens according to the man movig relative to the platform/track.

Since the speed of light is invarient for all observers, he also must see photons exapand outward from the point of emmission as a sphere in his frame. But from his perspective, the emitters do not stay at the the point of emmission. Therefore, in order for him to detect the emissions from both emitters at the same time, and at the same time as the man stationary to the platform, in his frame, the emiiters do not emit simultaneously, but one emits after the other.

http://home.teleport.com/~parvey/train2.gif

The reason things have to be this way is that the two men, being at the same point at the same time according to both of them, and both seeing the flash from the emitters arrive simultaneously is a spacetime event that is invarient and must be agreed upon by everyone.

I guess you dropped the football a lot in high school huh? . When a quarterback throws a ball to a "stationary" receiver and the football is intercepted by a running safety as they collide, you must think the ball was thrown at a different time for the receiver than the safety, that's why one guy misses it right?

 

[ram1024]

They don't agree on the order because the two events are separated by distance.
I told you you weren't reading my posts.

read the case. simultaneity is verified by a midpoint observer. in one "view" the midpoint observer DOES see the events simultaneously. in the other "view" he doesn't. but it's the SAME CASE. we're not doing the experiment twice.

simultaneity at a point is INVIOLATE.

you're obviously not reading MY posts. I'll let you ruminate on that subject for a while.

For the hundreth time, there is no motion of the observer in the observer's frame, observer is at rest wrt itself. Of course separation speed changes when looked from another frame, but that's irrelevant. What changes in a frame is the simultaneity of events, not how light moves. That's the whole point you are missing.

you seem to be dodging case #8, wespe. please give it a try

 

[Alkatran]

read the case. simultaneity is verified by a midpoint observer. in one "view" the midpoint observer DOES see the events simultaneously. in the other "view" he doesn't. but it's the SAME CASE. we're not doing the experiment twice.

simultaneity at a point is INVIOLATE.

you're obviously not reading MY posts. I'll let you ruminate on that subject for a while.



you seem to be dodging case #8, wespe. please give it a try

Oh, sorry, I forgot to tell you AGAIN that the photons aren't emitted simultaneously in the moving frame, making the middle observer receive them at the same time.

And this is because the two events (the emitions) are SEPERATED BY DISTANCE.

 

[wespe]

you seem to be dodging case #8, wespe. please give it a try

You have dodged cases #1 to #7 and some 400 replies. Please go back and read all the thread again.

OK, case #8, involves acceleration. ObserverB first shines light and then starts moving, therefore switching to a different frame. So he can't declare himself stationary all the time (or a pseudo gravity field must be added). Similarly to the Twins Paradox, see:

http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

So, explaining case #8 as is requires invoking General relativity, which I can't. To avoid this, assume ObserverB never moves but an inertial ObserverC was passing near ObserverB as the light was shined. Then, midpoint always receives the lights at the same time. Only for ObserverC they were not emitted simultaneously, just as with the previous cases.

 

[ram1024]

You have dodged cases #1 to #7 and some 400 replies. Please go back and read all the thread again.

i'm the one making the cases, i don't have to answer them. I'm finding out what SR thinks of the situations. i already KNOW what I think of the situations...

So, explaining case #8 as is requires invoking General relativity, which I can't. To avoid this, assume ObserverB never moves but an inertial ObserverC was passing near ObserverB as the light was shined. Then, midpoint always receives the lights at the same time. Only for ObserverC they were not emitted simultaneously, just as with the previous cases.

fine make ObserverB an inertial viewer. don't need to add more observers into the mess.

ObserverB moving towards ObserverA shines a light in such a fashion that the light he shines and the light ObserverA shines hits the midpoint viewer simultaneously.

switch view frames so ObserverB is NOT a moving observer.

Midpoint Observer and ObserverA have to move in this frame. Midpoint viewer AND ObserverA both do not receive photons in the correct time frame in this "view"

please resolve this paradox.

what was once simultaneous interception for the midpoint viewer is no longer such. just by changing "views"

the obvious conclusion is changing views does NOT work when dealing with light BECAUSE light is observer motion dependant

QED

 

[Eyesaw]

Einstein's gedanken corresponds to your case#3. The moving observer does not see the lightenings at the same time (since they occurred simultaneously in the stationary frame, they can't have occurred simultaneously in the train frame). I am 100% sure this is SR's prediction. Plus, if he saw them at the same time, there would be a paradox, because the non-moving midpoint observer sees them at the same time at another point in space.

In Janus's animated presentation, she argues (and you agreed) that the moving observer observes the lightning flashes simultaneously (at M')because they were not emitted simultaneously in the moving frame. If I read you correctly, you are saying here that the moving observer does not receive the lightning flashes simultaneously (at M') because they were not emitted simultaneously in the moving frame. Sounds like a contradiction here- care to explain?

In particular, I'd like to address your statement here : "The moving observer does not see the lightenings at the same time (since they occurred simultaneously in the stationary frame, they can't have occurred simultaneously in the train frame)." : Only if you are silly enough to use the front and rear of the train as the distance the two photons from events A and B traveled to the moving observer in lieu of the actual places where the events occurred as marked by the burned wood on the tracks to deduce the time of events. And only if you are naive enough to think the sequence of actual events happening is equivalent to the sequence of detection of light signals from those events.

 

[ram1024]

Eyesaw: they define simultaneity at a distance different from normal people. For them something at a distance only happens "at the same time" if it can be witnessed by an observer in the center of "the events" "at the same time"

thus the source of 30 pages of confusion :D

finally have the whole thing nailed down as a simple postulate error involving frame changes. pretty sure all "relative light speed to the observer" calculations were done as observer stationary frame, which is wrong.

 

[Eyesaw]

Eyesaw: they define simultaneity at a distance different from normal people. For them something at a distance only happens "at the same time" if it can be witnessed by an observer in the center of "the events" "at the same time"

thus the source of 30 pages of confusion :D

finally have the whole thing nailed down as a simple postulate error involving frame changes. pretty sure all "relative light speed to the observer" calculations were done as observer stationary frame, which is wrong.

I think you are right.

 

[ram1024]

well thank you, it's nice to have a favorable vote once in a while :D

 

[wespe]

i'm the one making the cases, i don't have to answer them. I'm finding out what SR thinks of the situations. i already KNOW what I think of the situations...

Well you aren't just asking questions. You are presenting arguments and drawing conclusions. When you are answered with a counter argument and you don't have anything left to say, you simply change the case while you should be accepting that your initial argument was incorrect. Therefore you aren't learning anything. That's what I meant by dodging. Please do read http://www.ephilosopher.com/Sections-article36-page1.html

fine make ObserverB an inertial viewer. don't need to add more observers into the mess.

ObserverB moving towards ObserverA shines a light in such a fashion that the light he shines and the light ObserverA shines hits the midpoint viewer simultaneously.

switch view frames so ObserverB is NOT a moving observer.

Midpoint Observer and ObserverA have to move in this frame.

It's fine up to this point.

Midpoint viewer AND ObserverA both do not receive photons in the correct time frame in this "view"

You draw this conclusion without showing me the reasoning. It is a wrong conclusion. The setup is that midpoint receives both photons at the same time. Switching the view won't change this fact. Why did you think so? Please explain.

please resolve this paradox.

No paradox. From observerB's perspective, ObserverA doesn't shine light simultaneously with ObserverB. ObserverB shines light at a certain time such that the midpoint receives the photons at the same time. I could show you how this is so (requires adding another observer in ObserverB frame), but first tell me what made you think:

what was once simultaneous interception for the midpoint viewer is no longer such. just by changing "views"

My guess is that you didn't think; you just used your intuition.

 

[wespe]

In Janus's animated presentation, she argues (and you agreed) that the moving observer observes the lightning flashes simultaneously (at M')because they were not emitted simultaneously in the moving frame. If I read you correctly, you are saying here that the moving observer does not receive the lightning flashes simultaneously (at M') because they were not emitted simultaneously in the moving frame. Sounds like a contradiction here- care to explain?

Well, there were different cases discussed in this thread. One of them is an animated gif which shows the moving observer seeing the flashes at the same time. In that gif, the observer is not at the midpoint. The one you quoted is different; the observer is at the midpoint and does not see the flashes at the same time.

In particular, I'd like to address your statement here : "The moving observer does not see the lightenings at the same time (since they occurred simultaneously in the stationary frame, they can't have occurred simultaneously in the train frame)." : Only if you are silly enough to use the front and rear of the train as the distance the two photons from events A and B traveled to the moving observer in lieu of the actual places where the events occurred as marked by the burned wood on the tracks to deduce the time of events.

So your suggestion is :the train observer must prefer the burned wood on the tracks over the burn marks on the train, as the location of events. Why? What is special about the tracks?

And only if you are naive enough to think the sequence of actual events happening is equivalent to the sequence of detection of light signals from those events.

Speed of light is constant relative to all observers (do you disagree at this point?). Then, if light from both events travel the same distance (that is, the observer is at the midpoint), the order of detection of light signals can be used to conclude about the sequence of events. Why do you think not?

 

[ram1024]

Fine, you want my observations this time, you can have them

1. before the frame shift, ObserverA is stationary and the Midpoint Observer is stationary. ObserverB is moving towards ObserverA <and the Midpoint Observer>

2. In this "Frame" Lights are emitted and progress is made by ObserverB Towards the opposing source of light. no progress is made by ObserverA towards HIS opposing light source. Light for ObserverB will be intercepted before ObserverA. the midpoint observer being stationary and at the midpoint of the emission sources will receive the emissions simultaneously.

3. Frames are switched. ObserverB is now stationary. ObserverA and Midpoint Observer move to preserve "reality" and "relative movements". in this frame, ObserverB makes no motion towards the light sources. ObserverA and Midpoint Observer DO make progress. The Midpoint Observer now progressing in a direction towards a light source will no longer receive photons simultaneously in this case, as he is no longer stationary at the midpoint of the two light sources at the time of emission and reception.

No paradox. From observerB's perspective, ObserverA doesn't shine light simultaneously with ObserverB. ObserverB shines light at a certain time such that the midpoint receives the photons at the same time. I could show you how this is so (requires adding another observer in ObserverB frame), but first tell me what made you think:

please do show me how the midpoint observer can still receive photons simultaneously in this case. i can almost guarantee you'll come to the same conclusion / revelation that I've been holding out of your reach for 30 pages.

 

[ram1024]

oh and remember to preserve the speed of light relative to the midpoint observer ;D

 

[wespe]

Fine, you want my observations this time, you can have them

1. before the frame shift, ObserverA is stationary and the Midpoint Observer is stationary. ObserverB is moving towards ObserverA <and the Midpoint Observer>

2. In this "Frame" Lights are emitted and progress is made by ObserverB Towards the opposing source of light. no progress is made by ObserverA towards HIS opposing light source. Light for ObserverB will be intercepted before ObserverA. the midpoint observer being stationary and at the midpoint of the emission sources will receive the emissions simultaneously.

3. Frames are switched. ObserverB is now stationary. ObserverA and Midpoint Observer move to preserve "reality" and "relative movements". in this frame, ObserverB makes no motion towards the light sources. ObserverA and Midpoint Observer DO make progress.The Midpoint Observer now progressing in a direction towards a light source

fine up to this point

will no longer receive photons simultaneously in this case,

this is your wrong conclusion. your reasoning is:

as he is no longer stationary at the midpoint of the two light sources at the time of emission and reception.

true, no longer at the midpoint, but you also assume that in ObserverB frame, ObserverA emitted light simultaneously with ObserverB. Not true. According to ObserverB, ObserverA emitted light before ObserverB, such that the moved mid-observer received the lights at the same time.

please do show me how the midpoint observer can still receive photons simultaneously in this case. i can almost guarantee you'll come to the same conclusion / revelation that I've been holding out of your reach for 30 pages.

I explained above how midpoint observer still receives photons simultaneously in ObserverB frame. Now I am expecting you to present your counter-argument. No more cases, enough is enough.

oh and remember to preserve the speed of light relative to the midpoint observer ;D

From ObserverB's perspective, the SEPARATION speed between two light signals and mid-observer are of course different than c. But from mid-observer's perspective, relative speed of light signals are still c.

 

[Eyesaw]

Originally Posted by Eyesaw
In particular, I'd like to address your statement here : "The moving observer does not see the lightenings at the same time (since they occurred simultaneously in the stationary frame, they can't have occurred simultaneously in the train frame)." : Only if you are silly enough to use the front and rear of the train as the distance the two photons from events A and B traveled to the moving observer in lieu of the actual places where the events occurred as marked by the burned wood on the tracks to deduce the time of events.

So your suggestion is :the train observer must prefer the burned wood on the tracks over the burn marks on the train, as the location of events. Why? What is special about the tracks?

Actually my suggestion is that you try running from the burn marks on the track until you catch up to the observer inside the train and then tell us if there is a difference between running just from the burn marks on the train to the middle of the train.




Quote:
Originally Posted by Eyesaw
And only if you are naive enough to think the sequence of actual events happening is equivalent to the sequence of detection of light signals from those events.

Speed of light is constant relative to all observers (do you disagree at this point?). Then, if light from both events travel the same distance (that is, the observer is at the midpoint), the order of detection of light signals can be used to conclude about the sequence of events. Why do you think not?

Because if a frog and a duck were on the track at the time and place of simultaneous lightning flashes A&B (according to the embankment observer *snort*) , they would both be fried, despite what the train observer using SR's wrong assumptions calculate.

 

[wespe]

Actually my suggestion is that you try running from the burn marks on the track until you catch up to the observer inside the train and then tell us if there is a difference between running just from the burn marks on the train to the middle of the train.

I don't understand your point. What kind of difference? Elaborate please..

Because if a frog and a duck were on the track at the time and place of simultaneous lightning flashes A&B (according to the embankment observer *snort*) , they would both be fried, despite what the train observer using SR's wrong assumptions calculate.

Yes, both frog and duck are fried at the same time in the stationary frame. And, both frog and duck will be fried in the train frame, despite at different times. But their frying at different times doesn't mean you can prevent the frog or duck from frying after the other one is fried. That would require sending a message faster than light, which means going back in time (does bring up paradoxes, but FTL is not possible according to SR).

 

[Doc Al]

case #8 -- not so great

You still seem to think that all you need to do to switch a diagram from one frame to another is just slide things over so that the pictures line up with the other observer. If things were that easy then everyone would understand relativity!

You keep ignoring simultaneity, length contraction, and other relativistic effects.

nah we've moved on to Case #8 now, jcsd

lemme diagram it for you

Case #8

     [u](o)                    |                    (o)[/u]
     [u](o)                    |                  (o)[/u]
     [u](o)                    |                (o)[/u]
     [u](o)                    |              (o)[/u]

The observers in this case are both Observers AND Emitters. at the onset of the experiment, ObserverA and ObserverB shine a light towards each other. Simultaneity is verified by an observer at the midpoint of the starting locations of both observers.

So the midpoint is in the frame of Observer A. And the lights are switched on simultaneously according to Observer A, but not Observer B. The diagram shows the view from the Observer A frame.

ObserverB then moves towards ObserverA. Predict who receives a photon first.

According to the A frame clocks, Observer B detects a photon first. But so what? The two events--(1) Observer B detects a photon and (2) Observer A detects a photon--are not causally connected and happen at different locations. Their time order is frame dependent.

     [u](o)                    |                    (o)[/u]
       [u](o)                    |                  (o)[/u]
         [u](o)                    |                (o)[/u]
           [u](o)                    |              (o)[/u]

now all we've done is taken that same experiment from ObserverB's point of view AS IF he was the stationary one.

You think you are showing the view from B's view, but you're not drawing it accurately. So the diagram is useless for drawing any meaningul conclusions. An accurate diagram would show length contraction, time dilation, and most importantly the lack of simultaneity of the two light flashes.

in this case the reverse result is shown, ObserverA receives his photon first. It is also interesting to note that the midpoint observer no longer receives photons simultaneously in this view, further proving that frame shifting is wrong for this type of experiment.

All this demonstrates is that you don't know how to draw the diagram from B's viewpoint. It's not easy.

How can something absolute such as simultaneity AT A POINT be violated by merely changing how you view a situation?

I'm glad you realize that that would make no sense! But the problem lies with your analysis, not with relativity. Both frames agree that the photons arrive at M simultaneously.

 

[Alkatran]

Well, RAM. You've had a few different people say exactly the same thing for an entire page. That's how special relativity sees it. It seems like you have your "own" special relativity that you're trying to disprove to prove your own theory.

You can't change an existing theory to disprove it.

 

[ram1024]

Been wracking my brain for two days trying to find SOME way to get you guys to understand me.

i may have stumbled upon the answer, however, so try and hear me out once more.

How does one measure distance? it is a measurement based upon a constant "movement" over an allotted "time"

speed is the direct proponent of this calculation including both distance AND time.

now when you say the speed of light is constant relative to the observer, that's all well and good, but only if the observer is stationary. stationary in respect to WHAT you may ask. stationary in respect to THE SPEED OF LIGHT says SR. The measurement "method" of SR is to assume that the observer is ALWAYS stationary, and since any "motion" of a light source has no effect, light will always be stationary as well.

think of how anything else moves through the universe however. If something IS moving, then it covers a distance TOWARDS something else.

Hence if i have 3 positions on a line A, B and C. and a person at point C throws a ball to point B while A moves to point B to reach that point the same time the ball reaches there.

the relative velocity of the ball, to ParticipantA is the velocity of the distance C-B over the time it took the ball to get from C-B PLUS the velocity of the distance A-B divided by the time it took ParticipantA to get from A-B. basically A-B/t + B-C/t = A-C/2t

Even taking ParticipantA as stationary, now ParticipantB has a velocity of B-A and the ball has a velocity C-B making the ball's velocity towards A, A-C/2t

now the problem with relativity is you have to abandon this "stationary observer" method of measurement, or account for the problem caused by light being unbound to its source. The problem you run into is as follows.

A is making progress to a point B, reaching there at the same time as a photon from C. The relative velocity of light for A in THIS case is of course easily measured as above. A covers a distance A-B/t and the photon covers a distance C-B/t. the relative velocity of the photon is of course A-C (total distance covered)/2t (total time elapsed)

case1
http://www.imagedump.com/index.cgi?pick=get&tp=93222

now move to make A a stationary observer.

I'll present to you two pictures and let you guess which is right. which is wrong. which is SR prediction.

case2
http://www.imagedump.com/index.cgi?pick=get&tp=93223

case3
http://www.imagedump.com/index.cgi?pick=get&tp=93224

Now looking at these two pictures you can note a few things.

Case2 asserts that A is stationary, it does NOT move, which means B must move relatively to its position and light from C intercepts them both at that location.

Case3 asserts THE SAME THING, except it asserts A is already IN B's position. well then where is B? obviously B has to move relatively to get to A so we move it over to maintain consistancy. C, which is not an object, but light from a source does NOT move, so lo and behold the measurement of light from C to A is the same as it was from C to B in the previous case.

Case3 ignores a very real distance, that is a distance covered from A-B in the time it took light to cover the distance C-B. Case 3 is an incorrect measurement of the speed of light.

 

[Alkatran]

Judging entirely by sight, the first one can't be true because light is exceding c according to the observer. (Assuming the yellow line is a photon)

So, by elimination, the second one is true. But I fail to see your point.

Are you going to give another argument about the observers being in two places at once? Because I've already told you it's because of their "when" line being skewed.

 

[ram1024]

by "second one" do you mean case3 ?

Are you going to give another argument about the observers being in two places at once? Because I've already told you it's because of their "when" line being skewed.

no, not at all. in case3 the setup applied was that there has to exist relative gains between Participants A and B so if A is already in position B <to keep lightspeed constant> then logic dictates B is closer to C than the original picture. but we know this is NOT the case, so the picture is wrong.

 

[Alkatran]

What in the world are you talking about? In ALL the pictures B is ALWAYS closer to C than A.

 

[ram1024]

BC in case1 is a different distance than BC in case3

simply

 

[Alkatran]

That would be because of *gasp* length contraction. Due to the "when" line being skewed WHICH I ALREADY SAID!