Today Special Relativity dies

[Alkatran]

Alright! I fixed it.

The green and blue lines are the two different observers.
Blue is the stationary (relative to emitter) observer and green is the moving
The top frame is blue's frame and the bottom is green's
The thick lines are the opposite observer's frame
(you see what they mean by skewed?)

 

[ram1024]

i said already, don't use attachment, they don't get approved. we'll never be able to see them

host it on imagedump or something :|

 

[Alkatran]

You're kidding me? You can't see any attachments until they're approved? Even pictures?! (Can you even put damaging stuff in jpeg's? I mean that will be run)

Anyways, I modified the picture a bit, pretty much the same except the thick lines are boxes now.

http://www.vbforums.com/attachment.php?s=&postid=1726678

Red Dot: The buoy
Blue Dot: photon heading towards buoy from emitter
Green Dot: photon heading towards buoy from emitter
http://www.vbforums.com/attachment.php?s=&postid=1726680
I'd also like to point out that this picture is "impossible" because it's drawn as if the buoy were moving relative to it and yet light doesn't move normally (for no speed). It's as if we are observing a frame from a a different frame.

 

[ram1024]

have NO idea what I'm looking at in your first diagram.

in the second one it just seems to prove that it SHOULD work. in order for the lights to simultaneously meet in the center the light from blue travels at 1m/s towards red which travels at 1m/s towards it (1+1=2) and light from green travels at 3m/s but red travels away from it at 1m/s so (3-1=2) light is traveling at 2m/s <lightspeed> relative to the observer <red>

but wait a minute. that means light was really <to no observer> traveling at 3 in one instance and 1 in another instance... what gives?

just to reiterate, the space-buoy does NOT work, but that is not why it does NOT work :D

 

[Alkatran]

Do you notice how the light intersects at the buoy (this is a spacetime diagram) and is always equi-distant from it on both sides? That's why ti doesn't work.

The first is also a space time diagram. First ignore the large square (dark blue/dark green) and look. The black line (going straight up in the first, since it is unmoving) is the emitter, red is light (same in both), green is one observer and blue is another.

The dark squares are the other frames. (rotate your head to look at them like that and they look similar to the other diagram, but it is not an accurate representation)

 

[ram1024]

the space buoy picture as you have drawn it is not how SR sees the events.

if it was then it would prove the inconsistancy of light speed, 3 m/s from the left and 1m/s from the right.

SR asserts that the left light would be emitted first, then the right light emitted later, so that they both meet when red dot reaches 1. but even doing so, red has to be moving away from green at SOME speed so light from the direction of green will always be traveling at "greater than light speed" in ANY case where motion of the apparatus is assumed.

SR elites have yet to attack my space-buoy because they see the paradox that occurs when the theory is applied to it.

gotta goto work today :( laters!

 

[wespe]

Ram, your pictures are still based on absolute time and space. They give the illusion that everything is happening in parallel in both frames. If you could examine #385, I have shown this is not so.

Specifically, your bad assumption is:

we're putting two observers next to each other. a flash of light is emitted at the same time from emitters same distance from them. <this can be the same emitter so don't worry about synchronicity>

Even if there is only one emitter, you still have to worry about length contraction. The two observers don't measure the distance the same. Therefore light cannot be emitted at the same time for both observers.

 

[Alkatran]

Did you even read my post explaining that we're watching one frame from another? If we were watching the non-moving frame, THAT'S when there woul dbe loss of simultanity because it IS simultaneous in the moving frame. geez

In fact, for any speed of the buoy you get the same thing in the buoy frame: The light is emitted at the same time and received at the same time. All you'll prove is the buoy is at rest relative to itself (duh).

 

[ram1024]

In fact, for any speed of the buoy you get the same thing in the buoy frame: The light is emitted at the same time and received at the same time. All you'll prove is the buoy is at rest relative to itself (duh).

how so? in your very own picture you have light approaching from the rear at 3m/s <measured> and from the front at 1m/s <measured>

put a stationary observer at the final interception point. light from the left covers a REAL distance of 3 to get to him from the left and a REAL distance of 1 to get to him from the right. there is no length contraction craziness, these would be real measurements to the stationary viewer. both of these observers would be hit by the same photons at the same time <occupying the same space> so there can be no discrepancies as far as where and when the lights were emitted.

Even if there is only one emitter, you still have to worry about length contraction. The two observers don't measure the distance the same. Therefore light cannot be emitted at the same time for both observers.

length contraction is bunk. working on a picture

 

[Alkatran]

Ugh, did you even comprehend what I wrote? Here, picture this: The same picture, except the buoy doesn't move (since we are in its frame!) and light approaches from both sides at the same speed, intercepts at the same speed.

There, that's how the buoy sees it, it won't work.

 

[wespe]

length contraction is bunk.

This is not an argument.

working on a picture

Have fun.

 

[ram1024]

Ugh, did you even comprehend what I wrote? Here, picture this: The same picture, except the buoy doesn't move (since we are in its frame!) and light approaches from both sides at the same speed, intercepts at the same speed.

There, that's how the buoy sees it, it won't work.

are you saying that's how the buoy sees it. no matter what?

because picture 2 buoys. one stationary and one moving away relative to the other.supposeing we use the same measurements as the other instance 2m/s lightspeed and the second buoy is moving away at 1m/s. they buoys start at the same place and fire at the same time.

determine the the reception times for photons of the second buoy to hit buoy 1 and vice versa.

 

[geistkiesel]

Ram, your pictures are still based on absolute time and space. They give the illusion that everything is happening in parallel in both frames. If you could examine #385, I have shown this is not so.

Specifically, your bad assumption is:



Even if there is only one emitter, you still have to worry about length contraction. The two observers don't measure the distance the same. Therefore light cannot be emitted at the same time for both observers.

Length contraction comes directly from simultaneity concepts that form the basis of relativity and light propagation. In this sense simultaneity precedes Sr. You have no loss of simultaneity, you have no SR, In the case where two photons are emitted simultaneously from A and B and meet at the midpoint M (of A and B) which has deflecting mirrors (\/) that immediately deflect both photons into the passing moving frame there is no loss of simultaneity. Einsteins definition the "events simultaneous with respect to the stationary frame are not simultaneous with respect to the moving frame." must be discarded as contrary to physical law and experimental results..

       M
  A -->\/<--B
       ||
       ||
    ##########  Detectors in the moving frame --> motion
     A'M'B'

We assume the photons move only a few wave lengths before being detected in the moving frame. Einstein's example which says: The observers ". . .must, therfore, come to the conclusion that . . ." the photons were not emitted simultaneouisly in the moving frame. This is because the B photon was detected before th A photon caught up from the rear of the frame in AE's original gedunken, which is obviously flawed. It is a difficult point to get accrioss, but one ought not take dogma so seriously. This definition was described by AE as a "natural definition" that operated to discard absolute space and all the rest.

How are the photons not going to be observed emitted simultaneously in the moving frame? Except by operation of the definition?

 

[Alkatran]

are you saying that's how the buoy sees it. no matter what?

because picture 2 buoys. one stationary and one moving away relative to the other.supposeing we use the same measurements as the other instance 2m/s lightspeed and the second buoy is moving away at 1m/s. they buoys start at the same place and fire at the same time.

determine the the reception times for photons of the second buoy to hit buoy 1 and vice versa.

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

 

[Doc Al]

those wacky diagrams

After all these hundreds of posts, it seems like very little progress has been made.

so in the moving observer frame, the center of emission is stationary

in the stationary observer frame the center of emission is moving.

that would be consistant right?

Wrong. The "center of emission" is stationary in every frame. This is a direct consequence of the invariant speed of light.

but it's not so according to the way you guys are calculating it. you have it stationary in the moving frame and stationary in the stationary frame.

That's correct.

you say it's not a real thing, but it has to be real enough in such that light emitted from that point will spread "at light speed" in all directions. that is a given.

The center of emission is not a "thing". It is just the place where the source was (with respect to an observer) when the light was flashed. The source may move on (depending upon what frame we are describing) but the center of emission never does: it is always fixed in the observer's frame.

let's analyze that further.

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

we're putting two observers next to each other. a flash of light is emitted at the same time from emitters same distance from them. <this can be the same emitter so don't worry about synchronicity>

I presume the source is in the "stationary" frame (the O1 frame). Again, saying that the flash is emitted "at the same time" is ambiguous. You certainly realize by now that only the O1 frame would observe the flash being emitted at the same time as the two observers pass each other. O2 would disagree. Simultaneity is key to understanding how to relate the observations made in different frames (along with length contraction and time dilation).

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

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

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

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

one of the observers moves towards the emitters. it's guaranteed he gets hit before the other observer due to less distance light has to travel. correct?

Both observers agree that the light hits O2 before it hits O1.

now we swap frames

OK, now I presume those diagrams are your attempt to illustrate the same events from three different frames. But your diagrams ignore length contraction. And since they don't show the center of emission, or where the pulse is at the moment of observation, or the extent of length contraction, it is difficult to draw any meaningful conclusions.

You pick three frames: O1, O2, and a third in which observers O1 and O2 are moving apart (at the same speed I presume?). Let's call that third observer O3. OK, so what?

these are the three relative frames i managed to come up with.

the first defining stationary as the point between O1 and O2, having each other move away at same speed. the emitters have to close some of the distance.

OK, this is the one I call O3. If the emitter moves with O1, then of course it will move in the O3 frame. So what?

the second being O2 stationary, so O1 must move away and the emitters must close the distances

Yeah, yeah. What do you mean "emitter must close the distance"? It's moving! More interesting would be to show (1) the location of the center of emission (2) the location of the light pulse at this moment (3) the length contraction of the O1 - emitter distance that a moving observer would find.

the third being O1 stationary and O2 moving. this is exactly the same case as outlined in the first example.

OK. And the point is?

let's define a length that light travels at a certain time to be light's speed in this scenario.

Huh? I hope you realize by now that light travels at the speed c in every frame!

Your ensuing discussion of length (or something) I cannot decipher, but you manage to conclude that some frames don't work with SR. Well, do it over! All inertial frames are equally valid in SR.

 

[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!

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

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

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

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