Today Special Relativity dies

[ram1024]

correct me if I'm wrong in saying the following.

SR believes that aether travels the speed and direction of the observer.
SR believes that light travels through Aether at approximately 300,000 km/s (true value defined by Maxwell Equation)

 

[ram1024]

What's the difference?? You're still only going to be finding the speed of Buoy #2 relative to #1 or vice versa!

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

you're afraid to do it, i don't blame you. it makes you confused doesn't it?

if photons are emitted simultaneously for the first frame and hit both observers at the same time <according to what you've told me> then a photon has traveled both one distance and another completely different distance AT THE SAME TIME.

do photons have a probabilistic existence?

 

[Alkatran]

if photons are emitted simultaneously for the first frame and hit both observers at the same time <according to what you've told me> then a photon has traveled both one distance and another completely different distance AT THE SAME TIME.

Listen VERY CAREFULLY: IF the photons are emited simultaneously for the first frame they are NOT for the other!

Your problems don't confuse me at all, they are all the same thing drawn slightly differently. Stop refusing to learn the basics and pick up a book.

 

[ram1024]

Listen VERY CAREFULLY: IF the photons are emited simultaneously for the first frame they are NOT for the other!

Your problems don't confuse me at all, they are all the same thing drawn slightly differently. Stop refusing to learn the basics and pick up a book.

what are you saying, that makes no sense.

Step 1: photons emitted simultaneously

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

Step 2: motion

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

Step 3: more motion

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

Step 4: final step <interception>

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

Light from Observer 2's LEFT Emitter would have had to have traveled MORE distance in the SAME amount of time to meet observer 2 in the center of HIS moving frame this is compared to light from Observer1's LEFT emitter which would have traveled to the center of observer 1's STATIONARY frame.

yet in both cases light was EMITTED from the same spot at the SAME TIME.

light traveling 2 different distances at the same interval = NOT constant.

 

[Alkatran]

How is this any different than any other of your cases? Here, I'll draw ANOTHER picture... sigh...

 

[ram1024]

Lemme make it easier for you. instead of 4 emitters we have 2 explosions, one on each side.

[u]¤¤¤¤¤                            (o)                            ¤¤¤¤¤[/u]
[u]¤¤¤¤¤                            (o)                            ¤¤¤¤¤[/u]

[u]¤¤¤¤¤                            (o)                            ¤¤¤¤¤[/u]
[u]¤¤¤¤¤                                (o)                        ¤¤¤¤¤[/u]

[u]¤¤¤¤¤                            (o)                            ¤¤¤¤¤[/u]
[u]¤¤¤¤¤                                    (o)                    ¤¤¤¤¤[/u]

[u]¤¤¤¤¤                            (o)                            ¤¤¤¤¤[/u]
[u]¤¤¤¤¤                                        (o)                ¤¤¤¤¤[/u]

light is independant of source right? it doesn't matter if i don't move the explosions with the second observer.

this should yield exactly the same results right? observer 1 and observer 2 both witness the photons simultaneously, right?

 

[Alkatran]

Look at this:

Now, HERE'S WHAT's HAPPENING:
Top Graph: The observer who isn't moving in the picture.
Bottom graph: Second observer, duh

Now, it looks like the graphs contradict each other, until you remember that, gasp*, they don't have the same time, or distance.

http://www.vbforums.com/attachment.php?s=&postid=1728412

 

[Michael F. Dmitriyev]

Your discussion follows beyond relativity, guys.
The light is excluded from this theory. Its speed is absolute initially.
The theory of relativity examines movement of objects concerning light but not on the contrary.
Relativity can be applied only for objects having mass, because the mass is a source of this phenomenon.

Michael

 

[ram1024]

Look at this:

Now, HERE'S WHAT's HAPPENING:
Top Graph: The observer who isn't moving in the picture.
Bottom graph: Second observer, duh

Now, it looks like the graphs contradict each other, until you remember that, gasp*, they don't have the same time, or distance.

http://www.vbforums.com/attachment.php?s=&postid=1728412

now put a stationary observer at the point where 2 intercepts light simultaneously from HIS 2 sources.

Observer 3 is co-located at that point where Observer 2 gets hit simultaneously.

Observer 3 MUST get hit by the photons simultaneously as well.

Observer 3 is NOT centered between the 2 "emitters" during the time of emission, yet BECAUSE of observer 2 he DOES get hit simultaneously by light from 2 differing distances?

Ooooh it's like MAGIC. just by having someone LOOK at the light and running BACKWARDS really FAST we can make it intercept objects between us and the light SOONER!

i think you guys are smoking and not sharing...

 

[Alkatran]

now put a stationary observer at the point where 2 intercepts light simultaneously from HIS 2 sources.

Observer 3 is co-located at that point where Observer 2 gets hit simultaneously.

Observer 3 MUST get hit by the photons simultaneously as well.

Observer 3 is NOT centered between the 2 "emitters" during the time of emission, yet BECAUSE of observer 2 he DOES get hit simultaneously by light from 2 differing distances?

Ooooh it's like MAGIC. just by having someone LOOK at the light and running BACKWARDS really FAST we can make it intercept objects between us and the light SOONER!

i think you guys are smoking and not sharing...

You know what? I'm done in this thread. It's obvious you don't want to learn.

As for someone running backwards and having light intercept faster: That's BECAUSE HE REACHES THAT POINT IN SPACETIME SOONER. There, happy? Oh wait, you aren't going to read and understand a thing I wrote.

 

[ram1024]

you're the one who isn't understanding the consequences of your own theories.

you refuse to admit that light traveling the same relative speed for any observer causes light to travel different distances for all observers, (Even stationary ones!).

1. Observer 1 stationary in the center getting hit simultaneously
2. Observer 2 moving and STILL getting hit simultaneously
3. Observer 3 stationary and co-located to the position that observer 2 is hit simultaneously, getting hit simultaneously.

this leads to the obvious and clear conclusion, that it doesn't matter where you are between two light sources, they will hit you at the same time. all you need is someone from the middle to travel to your location?

You know what? I'm done in this thread. It's obvious you don't want to learn.

You don't even understand your own theory, how do you expect to teach anyone?

 

[Doc Al]

Lemme make it easier for you. instead of 4 emitters we have 2 explosions, one on each side.

[u]¤¤¤¤¤                            (o)                            ¤¤¤¤¤[/u]
[u]¤¤¤¤¤                            (o)                            ¤¤¤¤¤[/u]

[u]¤¤¤¤¤                            (o)                            ¤¤¤¤¤[/u]
[u]¤¤¤¤¤                                (o)                        ¤¤¤¤¤[/u]

[u]¤¤¤¤¤                            (o)                            ¤¤¤¤¤[/u]
[u]¤¤¤¤¤                                    (o)                    ¤¤¤¤¤[/u]

[u]¤¤¤¤¤                            (o)                            ¤¤¤¤¤[/u]
[u]¤¤¤¤¤                                        (o)                ¤¤¤¤¤[/u]

Your 4 emitters are all fixed with respect to O1. (So you really only need 2; the extra 2 add nothing.) Do those explosions occur simultaneously? According to who? O1 or O2? They don't agree.

light is independant of source right? it doesn't matter if i don't move the explosions with the second observer.

Yes, light speed is independent of the source. But what matters is when the explosions occur. If they are simultaneous in the O1 frame, they cannot be simultaneous in the O2 frame.

this should yield exactly the same results right? observer 1 and observer 2 both witness the photons simultaneously, right?

Wrong. Obviously O1 sees O2 traveling towards the oncoming light: so O1 knows that O2 will detect the oncoming light before he does. Both O1 and O2 agree that O2 detects the light before O1.

 

[Alkatran]

Yes, I'll have to apologize for my faulty picture. I was confused for a minute that the emition was simultaneous in both frames, which it can't be.

 

[Janus]

Maybe this will get the point across:
http://www.geocities.com/janus58.geo/simultaneous.html

 

[Michael F. Dmitriyev]

Your dispute about not existing thing.
The relativity of speed of light is absurdity.

 

[HallsofIvy]

"relativity of speed of light "? The whole point of relativity is that the speed of light is the one thing that is not relative!

 

[Michael F. Dmitriyev]

"relativity of speed of light "? The whole point of relativity is that the speed of light is the one thing that is not relative!

I have said more definitely:
- It is absurdity.
So?

 

[Alkatran]

I like Michael's argument style:
It's right 'cause I said so!

 

[ram1024]

Your 4 emitters are all fixed with respect to O1. (So you really only need 2; the extra 2 add nothing.) Do those explosions occur simultaneously? According to who? O1 or O2? They don't agree.

fine, O1 then. but doesn't O2 AGREE that O1 receives photons simultaneously?

Wrong. Obviously O1 sees O2 traveling towards the oncoming light: so O1 knows that O2 will detect the oncoming light before he does. Both O1 and O2 agree that O2 detects the light before O1.

Good. so i set my clocks when I'm "dead in space" so that emitted photons will hit me simultaneously. then whatever relative motion i make, light will NOT hit me simultaneously because i would be moving to intercept it, correct?

 

[Alkatran]

fine, O1 then. but doesn't O2 AGREE that O1 receives photons simultaneously?



Good. so i set my clocks when I'm "dead in space" so that emitted photons will hit me simultaneously. then whatever relative motion i make, light will NOT hit me simultaneously because i would be moving to intercept it, correct?

Yes, he will agree he received them simultaneously. But only because he perceived the photons as being emitted non-simultaneously. All events at the same position and time are simultaneous to all observers. If there's a difference in position or time they won't be (for all)

 

[wespe]

light is independant of source right? it doesn't matter if i don't move the explosions with the second observer.

this should yield exactly the same results right? observer 1 and observer 2 both witness the photons simultaneously, right?

This reveals you still haven't understood a single thing. Otherwise you wouldn't make this wrong guess.

If the explosions occurred simultaneously in the "moving" frame, observer2 would see them at the same time.

Observer2 didn't see them at the same time, therefore the explosions didn't occur simultaneously in that frame.

Logically, if a then b, if not b then not a.

What problem is there? Is it that the explosions did occur simultaneously in the "stationary" frame? But Observer1 did see them at the same time. You just have to understand that simultaneity is relative, instead of coming up with different examples for the same thing.

 

[geistkiesel]

This reveals you still haven't understood a single thing. Otherwise you wouldn't make this wrong guess.

If the explosions occurred simultaneously in the "moving" frame, observer2 would see them at the same time.

Observer2 didn't see them at the same time, therefore the explosions didn't occur simultaneously in that frame.

Logically, if a then b, if not b then not a.

What problem is there? Is it that the explosions did occur simultaneously in the "stationary" frame? But Observer1 did see them at the same time. You just have to understand that simultaneity is relative, instead of coming up with different examples for the same thing.

Simultaneity is just a word. How do you prove simultaneity without just repeating the mantra that it comes from special relativity. How do you prove simultaneity? Surely it is more than sheer belief, is it?

 

[geistkiesel]

Your discussion follows beyond relativity, guys.
The light is excluded from this theory. Its speed is absolute initially.
The theory of relativity examines movement of objects concerning light but not on the contrary.
Relativity can be applied only for objects having mass, because the mass is a source of this phenomenon.

Michael

Michael, How do you prove simultaneity, briefly? This is what they are discussing isn't it?
Does Einstein's stationary - platform/moving - train prove it? or is this just a definition?
Thanx, Geistkiesel.

 

[geistkiesel]

You know what? I'm done in this thread. It's obvious you don't want to learn.

As for someone running backwards and having light intercept faster: That's BECAUSE HE REACHES THAT POINT IN SPACETIME SOONER. There, happy? Oh wait, you aren't going to read and understand a thing I wrote.

It is all to do with simultaneity right? Other than merely state that it is SR, how do you prove simultaneity? Briefly. Is Einstein's definition enough?

I am asking around about this question,
thanx, geistkiesel.

PS how do you create and post your attachments?
Thanx again.

 

[wespe]

Simultaneity is just a word. How do you prove simultaneity without just repeating the mantra that it comes from special relativity. How do you prove simultaneity? Surely it is more than sheer belief, is it?

Einstein's method to determine simultaneity is the most logical one possible. Consider two events separated by a distance. How else can you prove that they happened at the same time? Be realistic; you can't see or know everything instantly (like in the thought experiments). Your best option is to assign a midpoint observer to be the judge, and use light signals. If this will make more sense, think of the midpoint observer actually watching both events at the same time (using mirrors). If he sees the events occur at the same time, the events are simultaneous.

I know, of course, you will now say that if the midpoint observer is moving, he can't be the judge. You know, of course, I will say that he considers himself stationary and also the emission points of light stationary. Therefore the only requirement is that the observer was at the midpoint when the events occurred and his relative speed to something else doesn't matter. Then, of course, you won't agree, just like the hundred times you were told this before. OK, but please discuss properly, don't start with your unintelligible essays again please. Quote what you don't agree with and provide your reasoning.

 

[geistkiesel]

There is no need to "include the observer", because the observer is at rest in his own frame. All the observer has to do is record the time and place of the emission, and the time and place of detection. Take the ratio of Δx and Δt, and you get c.



There is no error. The rate at which a photon approaches me is the distance I measure it to cover divided by the time it took to do it.

Excuse the late rehash, but I've been thinking about this.
Yes but it approaching you means you areapproaching it. If from the instant the Obsever was at M until measuring B at t1v, the photon has traveled ct1. You are eliminating the relative velocity of the observer, naturally you will calculate C. See below for a corected vesion of your statement.

When the obsever on the moving frame (MF) detects the oncming B photon ahe notes the time in her frame. T1 is measured from zero when she crossed hrough the midpoint of A and B in the stationary Frame (SF). Simlarly T2 measues he time the A photon arrives from behind. Using the fact that the photon from A at T1 is equidistant to M and must cover c(T2 -T1) space during the train moving from T1v to T2v or v(T2 -T1), which is T1v + T1v + (T2 - T1)v we arive at T2 = T1(C + v)/(C - v).

You must mean that the distance the train moved to the B detection spot was T1v, during which time the light moved from B to T1C.
The place of emission of the B photon to the right of M is L - T1v = T1C and then L = t1(c + V), right? And from the rear, from symmetry considerations, the distance the photon travel from T 1 to T2, when detected, is, C(T2 - T1) = L + t1v + v(T2 - T1), or L = C(T2 - T1) -vT1 - v(T2 - T1), or T2 = T1(C + v)/(c - v).

The velocity of C then in terms of v, T1 and T2 is simply
c= v(t1 + T2)/( T2 - T1)

Another way to say all this:

Maintain a series of photon detectors along the train and measure when the A and B photons arrive simultaneously at M. What ever that time happens to be it is a singularly measured time. If the photon detector in the SF and the MF are separated by one wavelength then the photons are measured (after reflecting thrug a distance of one wavelength) as simultaneously being emitted from the SF into the MF, correct? We find then that the relative velocity C is just = v(t1 + T2)?t2 - T1).

Janus already knows Galilean relativity.

Maybe Janus should be nstructed in the art of including the osbervers velocity, speed.

quoted by geistkiesel in a snit of impatience..
OK smart guy, measure the flat of the planet on the surface,. You will never be able to measure round. The ships coming to port over the horizon are measurements at a distance. Within limits the planet is measured flat man. Get used to it.

Tom,
Take a look at my post to "Has anyone seen the lost simultaneity?" #55. It is much more complete.

 

[geistkiesel]

Oh wait, is (A, M, B) supposed to be the "stationary" letters?

Okay, does this diagram look like your experiment?

<br /> \begin{picture}(240,240)(0,0)<br /> \put(0,240){\line(0,-1){210}}<br /> \put(0,20){A,t}<br /> \put(100,240){\line(0,-1){210}}<br /> \put(100,20){M}<br /> \put(200,240){\line(0,-1){210}}<br /> \put(200,20){B}<br /> \put(100,240){\line(1,-3){70}}<br /> \put(170,20){M&#039;}<br /> \put(0,240){\line(1,-1){180}}<br /> \put(200,240){\line(-1,-1){125}}<br /> <br /> \put(0,240){\line(1,0){210}}<br /> \put(220,230){x}<br /> \end{picture}<br />

Hurkyl, I may have overlooked answering you. Yes the drawing here is correct. First the B photon is detected in the moving Frame then the A detected This appears to be your drawing. As this condition is really a test of simultaneity we needn't introduce any SR at this point correct? It will, must, show up in the analysis. (See Has anyone seen the lost simultaneity post # 55).

During he time the photon traveled to its detection point it traveled cT1 while the train traveled vt1, During the time the A photon traveled to the detection point is v(t2 - t1). From synmmetry the A photon was at -t1v when the B photon was detected. Shortening the algebra we get that, for t1 = 1, C = 1, and v = .1,
t2 = t1(C + V)/(C -V) = 19, or that C = V(t2 + t1)/(t2 - t1) = 1.

Who needs SR to get the right answer?

Don't tell me you are going to bootsrap a reply using time dilation etc. You should be able, you must be able, to get it from this analysis. The biggest problem in SR is the very long reach of assuming that because the different imes that the observer measured the arrival of the A and B photons, you jump to a loss of simultabneity, discarding absolute time, insering time dilation and mass shrinking.

let us put an observer that measures the simultaneous arrival of the A and B photons at M (arriving into reflecting mirrors reflecting to M and M' at the same time). Just put your slanted line for the world events located at M and M = M'. The spatial difference in detectors is one wavelength of the photons. Are you still going to insist that the moving frame observer will detect the photon arriving on the moving frame at different times?

Or said another way,

Are events simultaneous in the stationary frame not simultaneous in the moving frame? Where the events, of course, is the simultaneous emission of photons in the stationary frme at A and B.

 

[geistkiesel]

It would appear (with a gentle nudge from Hurkyl) that you answered your own question. Why couldn't you have just read up on it like I (we) suggested instead of guessing and getting combative when you guessed wrong? Its simpler, faster, and less painful for all of us. And this is why you are having such a hard time learning these things. Its human nature to not want to be wrong and so I understand your frustration. You are trying to learn by a process that ensures that for you to learn something you must first make a guess and be wrong (and accept that you were wrong). On this forum it manifests itself as you making guesses, us trying to help you, and you responding with the natural human instinct of getting defensive/combative. Why put yourself through that? Just learn the right way the first time. Don't guess.

Yes, can be tough to find credible info on the web. We can help. But if you don't trust us, its not hard to figure out on your own what is credible and what isn't.

Russ_watters, you have to be one obnoxious snob, talking like ou do in yopur trust you? Is this your argument or are you the reincarnation of Ptolemy, are you? If so, please excuse the 'snob' reference, I can understand your personal position.

 

[geistkiesel]

There is no need to "include the observer", because the observer is at rest in his own frame. All the observer has to do is record the time and place of the emission, and the time and place of detection. Take the ratio of Δx and Δt, and you get c.



There is no error. The rate at which a photon approaches me is the distance I measure it to cover divided by the time it took to do it.

Excuse the late rehash, but I've been thinking about this.
Yes but it approaching you means you areapproaching it. If from the instant the Obsever was at M until measuring B at t1v, the photon has traveled ct1. You are eliminating the relative velocity of the observer, naturally you will calculate C. See below for a corected vesion of your statement.

When the obsever on the moving frame (MF) detects the oncming B photon ahe notes the time in her frame. T1 is measured from zero when she crossed hrough the midpoint of A and B in the stationary Frame (SF). Simlarly T2 measues he time the A photon arrives from behind. Using the fact that the photon from A at T1 is equidistant to M and must cover c(T2 -T1) space during the train moving from T1v to T2v or v(T2 -T1), which is T1v + T1v + (T2 - T1)v we arive at T2 = T1(C + v)/(C - v).

You must mean that the distance the train moved to the B detection spot was T1v, during which time the light moved from B to T1C.
The place of emission of the B photon to the right of M is L - T1v = T1C and then L = t1(c + V), right? And from the rear, from symmetry considerations, the distance the photon travel from T 1 to T2, when detected, is, C(T2 - T1) = L + t1v + v(T2 - T1), or L = C(T2 - T1) -vT1 - v(T2 - T1), or T2 = T1(C + v)/(c - v).

The velocity of C then in terms of v, T1 and T2 is simply
c= v(t1 + T2)/( T2 - T1)

Another way to say all this:

Maintain a series of photon detectors along the train and measure when the A and B photons arrive simultaneously at M. What ever that time happens to be it is a singularly measured time. If the photon detector in the SF and the MF are separated by one wavelength then the photons are measured (after reflecting thrug a distance of one wavelength) as simultaneously being emitted from the SF into the MF, correct? We find then that the relative velocity C is just = v(t1 + T2)?t2 - T1).

Janus already knows Galilean relativity.

Maybe Janus should be nstructed in the art of including the osbervers velocity, speed.

quoted by geistkiesel in a snit of impatience..
OK smart guy, measure the flat of the planet on the surface,. You will never be able to measure round. The ships coming to port over the horizon are measurements at a distance. Within limits the planet is measured flat man. Get used to it.

Tom,
Take a look at my post to "Has anyone seen the lost simultaneity?" #55. It is much more complete.

 

[geistkiesel]

Einstein's method to determine simultaneity is the most logical one possible. Consider two events separated by a distance. How else can you prove that they happened at the same time? Be realistic; you can't see or know everything instantly (like in the thought experiments). Your best option is to assign a midpoint observer to be the judge, and use light signals. If this will make more sense, think of the midpoint observer actually watching both events at the same time (using mirrors). If he sees the events occur at the same time, the events are simultaneous.

I know, of course, you will now say that if the midpoint observer is moving, he can't be the judge. You know, of course, I will say that he considers himself stationary and also the emission points of light stationary. Therefore the only requirement is that the observer was at the midpoint when the events occurred and his relative speed to something else doesn't matter. Then, of course, you won't agree, just like the hundred times you were told this before. OK, but please discuss properly, don't start with your unintelligible essays again please. Quote what you don't agree with and provide your reasoning.

You took the words right out of my typing fingers. See the post #55 in "Has anyonme seen the lost simultaneity?'

I put mirrors at M, the midpoint, in the stationary frame, reflecting into the moving train just as the detecters in the moving frame pass the midpoint M when the photons arrive simultaneously from A and B sources. The photons are immediately directed into the moving frame and immediately detected after a distance traveled of one wavelength of the photon light.

To say it gently: "The photons emitted simultaneously in the stationary frame are simultaneously emitted in the moving frame."

I demand that the midpoint observer be the judge, absolutely.
For a detailed solution to your conundrum wespe check out my #55. and try to keep your mind open. Unintelligble? posts? only to those unable to reason past their robotic limits.

I always recognized that the B photon is detected before the A photon, I have just been arguing that the naive conclusions of Einstein, and yourself for instance, are just that, naive.

 

[ram1024]

Einstein's method to determine simultaneity is the most logical one possible. Consider two events separated by a distance. How else can you prove that they happened at the same time? Be realistic; you can't see or know everything instantly (like in the thought experiments). Your best option is to assign a midpoint observer to be the judge, and use light signals. If this will make more sense, think of the midpoint observer actually watching both events at the same time (using mirrors). If he sees the events occur at the same time, the events are simultaneous

but he's not a midpoint observer if he moves relative to the sources. you can't possibly tell me that an observer who moves to intercept photons AT the midpoint can call himself stationary and believe he was always at the midpoint the whole time. measure the distances, it's that simple

 

[geistkiesel]

but he's not a midpoint observer if he moves relative to the sources. you can't possibly tell me that an observer who moves to intercept photons AT the midpoint can call himself stationary and believe he was always at the midpoint the whole time. measure the distances, it's that simple

Ram. In einstein's train the single observer O' is the focus of the argument, only!.

What about the passengers that are sitting adjacent to the midpoint just as the photons from A and B arrive there, after the intercepion of the B photon by O' but before O' detects the A photon? For these passengers the photons are detected at the very instant the photons arrive at the midpoint. Hence, these passengers do observe the simultaneous arrival of the photons in the movng frame.

But according to SR all the other passengers "must. therefore" conclude something different. Something for everybody I suppose.

I wonder, did the passenegrs on Einstein's train argue this out among themselves?

 

[wespe]

but he's not a midpoint observer if he moves relative to the sources. you can't possibly tell me that an observer who moves to intercept photons AT the midpoint

He doesn't move to intercept the photons at the midpoint of the stationary frame. He remains at the midpoint in his frame. He also doesn't see the events occurring at the same time, therefore concludes they did not occur simultaneously in his frame.

can call himself stationary and believe he was always at the midpoint the whole time. measure the distances, it's that simple

In the moving frame, he is stationary and always at the midpoint of the two events. Explained below.

First, don't think of the moving frame consisting only the observer. Extend it so that it covers the locations the events occured. It's a train with an observer in it, sufficiently long to include both events.

Don't think of the events belonging to the stationary frame. The events also occur in the train frame. That's why lightenings were used in the gedanken. They occur in an instant and they can leave burning marks in both frames.

The midpoint observers in both frames are always the same distance from the marks in their frame. That's how we can say the observer in the moving frame is and always was at the midpoint of the two events. That's why both can be the judge in determining simultaneity in their frame.

Of course, the burning marks in the two frames don't remain co-located after the lightenings strike (and both were never co-located at the same time in both frames, but that will be the conclusion). Since the lightenings don't belong to either frame, and speed of light is independent of its source, the relative motion between the frames is irrelevant to the events.

 

[ram1024]

that's ridiculous and wrong.

the observer did NOT start out in the middle of the two events, he moved to BE at the midpoint WHEN the light hit.

<talking about the train gedunken experiment portrayed in the animated gifs>

even in HIS frame he was not in the midpoint of the two emitters. you're basically telling me that via length contraction NOTHING in the universe moves EXCEPT light

 

[wespe]

<talking about the train gedunken experiment portrayed in the animated gifs>

Post #425 which you replied in #431 was not talking about animated gif. Do you even care to read anything others write?

 

[pallidin]

This is a great discussion, but I have a problem with it.
"Thought experiments" are of great potential value to be sure, but often entail the concept of "perfect" circumstances.
More often than not, these "perfect" circumstances exclude or inappropriately minimize important influences associated with reality; often rendering experimental validation impossible, so what are we left with?
In all fairness, thought experiments should be considered a valued commodity in the scientific community, as most foundations of development began with this.

 

[ram1024]

if i was replying to that post i would have quoted it. i am replying to your most recent post, of course

 

[wespe]

if i was replying to that post i would have quoted it. i am replying to your most recent post, of course

Assuming this was a reply to post #435 (why don't you press the quote button?)

Here's the history:

#422 by Geistkiesel (asks "how do you prove simultaneity?")
#425 by Wespe (talks about Einstein's gedanken to prove simultaneity)
#431 by Ram1024 (asks something about an animated gif without mentioning it)
#433 by Wespe (replies thinking of Einstein's gedanken)
#434 by Ram1024 (says it's wrong according to the animated gif)
#435 by Wespe (says he wasn't talking about animated gif)
#437 by Ram1024 (quoted above)

 

[ram1024]

doesn't matter if you were talking about the animated gif or NOT

you made a statement that it fallacious applied to previously accepted SR arguments "In the moving frame, he is stationary and always at the midpoint of the two events." in the case of:

http://home.teleport.com/~parvey/train1.gif
http://home.teleport.com/~parvey/train2.gif

 

[Alkatran]

doesn't matter if you were talking about the animated gif or NOT

you made a statement that it fallacious applied to previously accepted SR arguments "In the moving frame, he is stationary and always at the midpoint of the two events." in the case of:

http://home.teleport.com/~parvey/train1.gif
http://home.teleport.com/~parvey/train2.gif

"OH MY GOD"

I know I said I would stay away from this thread but this is unimaginable!

The midpoint of the two events is the midpoint of the two expanding spheres/circles of LIGHT! Light moves at c for all observers! The mid point is different depending on frame FOR THE LIGHT!

He was talking about the midpoint between the EMITTERS.

 

[ram1024]

but in this case he is NOT always in the midpoint between the two emitters.

OR the light sources for that matter.

but he still receives simultaneous light. if he assumes that the light he receives simultaneously came at him from the same distances THEN light speed would be constant for him as a stationary observer. simultaneous reception for him is NOT simultaneous emission only because you guys have no idea how to measure things when you switch frames.

 

[Alkatran]

but in this case he is NOT always in the midpoint between the two emitters.

OR the light sources for that matter.

but he still receives simultaneous light. if he assumes that the light he receives simultaneously came at him from the same distances THEN light speed would be constant for him as a stationary observer. simultaneous reception for him is NOT simultaneous emission only because you guys have no idea how to measure things when you switch frames.

Thanks for the insult, I'm sure everyone here "can't measure things".

Look, just face it, you have no idea what you're talking about. All your arguments are the same thing.

 

[ram1024]

Thanks for the insult, I'm sure everyone here "can't measure things".

Look, just face it, you have no idea what you're talking about. All your arguments are the same thing.

the imperfection is yours.

you guys bring forth an argument and when i prove it doesn't and CAN'T work that way, you say I'm wrong without saying why I'm wrong.

why am i wrong then?

 

[Alkatran]

First of all, how can we say you're wrong when you won't even tell us your theory?

Secondly, relativity applies (remarkably accurately) to reality.

 

[ram1024]

K I've figured out exactly where you went wrong with this "light speed constant to the observer" stuff

i don't even need the data anymore.

in every case you're taking the data with the OBSERVER as stationary and it doesn't matter where the light source is moving because you're "ASSUMING" the sources are stationary with the observer BECAUSE you believe light to NOT be dependant on source.

Essentially you're saying that MOVING, the SOURCE will always MOVE with the observer. <reverse frame>

well of course you're always going to measure constant light speed if you use that definition to measure it. it's like saying a car can only move at 5 miles an hour towards me, but i cannot make any motion towards it.

Light speed is only constant BECAUSE you're assuming a stationary observer and applying your measurements to that frame <with a stationary source>

i have already proven WHY this does not accurately reflect reality when something is indeed moving.

 

[Alkatran]

K I've figured out exactly where you went wrong with this "light speed constant to the observer" stuff

i don't even need the data anymore.

in every case you're taking the data with the OBSERVER as stationary and it doesn't matter where the light source is moving because you're "ASSUMING" the sources are stationary with the observer BECAUSE you believe light to NOT be dependant on source.

Essentially you're saying that MOVING, the SOURCE will always MOVE with the observer. <reverse frame>

well of course you're always going to measure constant light speed if you use that definition to measure it. it's like saying a car can only move at 5 miles an hour towards me, but i cannot make any motion towards it.

Light speed is only constant BECAUSE you're assuming a stationary observer and applying your measurements to that frame <with a stationary source>

i have already proven WHY this does not accurately reflect reality when something is indeed moving.

Congradulations! You've just realized that light is at c relative to yourself, meaning that, technicly, everyone is stationary relative to light traveling at c.

Don't think too hard, wouldn't want to go down the right path.

It's like saying that if you start moving towards the car (moving away from you) it speeds up.

 

[ram1024]

Congradulations! You've just realized that light is at c relative to yourself, meaning that, technicly, everyone is stationary relative to light traveling at c.

Don't think too hard, wouldn't want to go down the right path.

It's like saying that if you start moving towards the car (moving away from you) it speeds up.

exactly... it DOES speed up. but not relative to you, because YOU ARE STATIONARY.

it's an exercise in futility. like catching those motes in your eye... you track them and they move away,,,

of course i have already proven why this reasoning is false, it helps now that i understand why you think this way, so i can better explain it to others who may have the same misconceptions.

please go back and read why this is NOT reality.

 

[Alkatran]

exactly... it DOES speed up. but not relative to you, because YOU ARE STATIONARY.

it's an exercise in futility. like catching those motes in your eye... you track them and they move away,,,

of course i have already proven why this reasoning is false, it helps now that i understand why you think this way, so i can better explain it to others who may have the same misconceptions.

please go back and read why this is NOT reality.

I've read through this entire thread and I can tell you that it's amazing you've somehow managed to avoid learning the entire time. You seemed so close with the space-time graphs... but alas..

Acceleration has to do with general relativity, and this thread is about special relativity, so let's just ignore that, alright?

You see, the main concept you need to grasp is this: If observer A is moving relative to observer B, events separated by distance that are simultaenous for A will not be for B, and vice versa.

That's it. That's all you need to accept, no, CONSIDER, before the learning really begins.

 

[ram1024]

wrong.

events that APPEAR to be simultaneous for a stationary observer will not APPEAR to be simultaneous for a moving observer.

but that's a GIVEN in ANY relativity. I've never disputed THAT.

but you guys are given to the notion that if something APPEARS to be simultaneous then it IS simultaneous, which is quite illogical considering the multitudes of situations where things contradict that logic.

Light can move at a constant calculated speed and still not screw up anything that you guys hold precious.

the whole relative light speed is a miscalculation. that is all

 

[Alkatran]

wrong.

events that APPEAR to be simultaneous for a stationary observer will not APPEAR to be simultaneous for a moving observer.

but that's a GIVEN in ANY relativity. I've never disputed THAT.

but you guys are given to the notion that if something APPEARS to be simultaneous then it IS simultaneous, which is quite illogical considering the multitudes of situations where things contradict that logic.

Light can move at a constant calculated speed and still not screw up anything that you guys hold precious.

the whole relative light speed is a miscalculation. that is all

What in the world are you talking about?
How can something not be simultaneous but "APPEAR" to be simultaneous? If you include a few calculations based on relativity, things that are simultaneous for you are "simultaneous" for me. But simultanity is frame-dependant (because it has to be at the same time FOR YOU).

Please give an example where something "appears" to be simultaneous but isn't.