Today Special Relativity dies

In summary, the conversation discusses the simultaneous emission and detection of photons in different reference frames, specifically in the case of a man standing on a movable platform bed between two photon emitters. The outcome varies depending on the reference frame and the movement of the platform. The ambiguity lies in the lack of definition of the specific reference frames in each case.
  • #316
ram1024 said:
now what IS reality is changing the perspective but using the "center of emission" as a relative reference for frame swapping.

any observer motion towards THAT is equal to motion of THAT towards the observer in a true equivalent reference frame swap

One more time, in English? :confused:
 
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  • #317
Once again, the "center of emission according to picture #1" and the "center of emission according to picture #2" are different things. There should be no surprise that they don't act as if they were both the same thing.
 
  • #318
Tom Mattson said:
So what? This doesn't prove your assertion about absolute motion or absolute rest. It just proves that the world looks different from the two different perspectives.

To that, I can only say: No kidding! :surprise:

it doesn't "look different" the physics ARE different.

the first three steps are the observer stationary to the object emitting the light. in each of these steps the amount of time it would take for the photon to reach the observer would be the same.

in the last two "steps" the transition to "observer moving" from "emitter moving" has huge consequences. the time of a photon to reach the destination is very very different based upon the movement.

the reason is because the transition did NOT take into account the relative motion of the observer to "the geometric calculated source location" of the emission. in case 1, 2 and 3, there is no relative motion towards it or away from it, in cases 4 and 5 there is.

they're completely different situations BECAUSE light is NOT tied to its source in REALITY. 4 and 5 would not be "the observer's perspective of the same situation" but instead completely different situations.
 
  • #319
Ram, I have already pointed out your mistake. You can't combine steps 4 and 5 as your pictures suggest. My guess is the last frame in step 4 corresponds to first frame in step 5. The distance light has to travel is the same for all steps.
 
  • #320
http://www.imagedump.com/index.cgi?pick=get&tp=87816

this picture displays the 'centers of emissions' for you, Tom

Ram, I have already pointed out your mistake. You can't combine steps 4 and 5 as your pictures suggest. My guess is the last frame in step 4 corresponds to first frame in step 5. The distance light has to travel is the same for all steps.

i'm not combining 4 and 5, I'm contrasting 4 and 5 to 2 and 3, as you guys have said that simply changing the roles of observer and emitter is "proper" for relative frames,

you can see quite clearly that changing JUST the observer to BE moving changes the distance that light will travel hence change the intercept times.

sorry if i didn't make it quite clear enough that THAT was what i was accomplishing.
 
  • #321
ram1024: do you agree that this diagram, a plot of time vs distance, accurately represents picture #1?

(specifically, time increases as you go down, and left-right is spatial position)

[tex]
\begin{picture}(400,240)(0,0)
\put(100,210){\textcolor{red}{\line(0,-1){210}}}
\put(300,210){\textcolor{red}{\line(0,-1){210}}}
\put(100,210){\textcolor{green}{\line(-1,-1){75}}}
\put(100,210){\textcolor{green}{\line(1,-1){180}}}
\put(300,210){\textcolor{green}{\line(-1,-1){180}}}
\put(300,210){\textcolor{green}{\line(1,-1){75}}}
\put(200,210){\textcolor{yellow}{\line(0,-1){210}}}
\put(166.7,210){\textcolor{blue}{\line(1,-3){70}}}
\put(100,210){\line(3,-1){210}}
\end{picture}
[/tex]

The red lines are the emitters, the yellow and blue lines are the observers, and the green lines are the photons. (Ignore the black line for now)
 
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  • #322
ram1024 said:
it doesn't "look different" the physics ARE different.

If you reached that conclusion then you made a mistake. One of the postulates of relativity is that the physics is the same for all inertial observers, and this is perfectly consistent with the idea that any inertial observer can regard himself at rest. Just look at the derivation of the Lorentz transforms from the postulates, as applied to Maxwell's equations.

I'm not particulary interested in looking at your trains anymore. If you want to show how the denial of absolute rest leads to the denial of the invariance of physical laws, then show me the math.
 
  • #323
Hurkyl said:
ram1024: do you agree that this diagram, a plot of time vs distance, accurately represents picture #1?

(specifically, time increases as you go down, and left-right is spatial position)

[tex]
\begin{picture}(400,240)(0,0)
\put(100,210){\textcolor{red}{\line(0,-1){210}}}
\put(300,210){\textcolor{red}{\line(0,-1){210}}}
\put(100,210){\textcolor{green}{\line(-1,-1){75}}}
\put(100,210){\textcolor{green}{\line(1,-1){180}}}
\put(300,210){\textcolor{green}{\line(-1,-1){180}}}
\put(300,210){\textcolor{green}{\line(1,-1){75}}}
\put(200,210){\textcolor{yellow}{\line(0,-1){210}}}
\put(200,210){\textcolor{blue}{\line(1,-3){70}}}
\put(100,210){\line(3,-1){210}}
\end{picture}
[/tex]

The red lines are the emitters, the yellow and blue lines are the observers, and the green lines are the photons. (Ignore the black line for now)

not really, the yellow and blue line have to start halfway between the first emitter and the midpoint and then the yellow line would angle to intercept the midpoint crossing of the two green lines in the center <if I'm reading this correctly>
 
  • #324
sorry i mean the blue line would.

well whichever... one of them would ;D
 
  • #325
I drew the wrong picture. :frown: it's been fixed; is it right now?
 
  • #326
Hurkyl said:
I drew the wrong picture. :frown: it's been fixed; is it right now?

lemme try

[tex]
\begin{picture}(400,240)(0,0)
\put(100,210){\textcolor{red}{\line(0,-1){210}}}
\put(300,210){\textcolor{red}{\line(0,-1){210}}}
\put(100,210){\textcolor{green}{\line(-1,-1){75}}}
\put(100,210){\textcolor{green}{\line(1,-1){180}}}
\put(300,210){\textcolor{green}{\line(-1,-1){180}}}
\put(300,210){\textcolor{green}{\line(1,-1){75}}}
\put(166.7,210){\textcolor{yellow}{\line(0,-1){210}}}
\put(166.7,210){\textcolor{blue}{\line(1,-3){70}}}
\put(100,210){\line(3,-1){210}}\end{picture}[/tex]
 
  • #327
Grr, I had the train going the wrong way too, I think... Bah, let me look at picture #1 again to be sure!
 
  • #328
[tex]
\begin{picture}(400,240)(0,0)
\put(100,210){\textcolor{red}{\line(0,-1){210}}}
\put(110,210){\textcolor{red}{\line(0,-1){210}}}
\put(120,210){\textcolor{red}{\line(0,-1){210}}}
\put(130,210){\textcolor{red}{\line(0,-1){210}}}
\put(140,210){\textcolor{red}{\line(0,-1){210}}}
\put(300,210){\textcolor{red}{\line(0,-1){210}}}
\put(100,210){\textcolor{green}{\line(-1,-1){75}}}
\put(100,210){\textcolor{green}{\line(1,-1){180}}}
\put(300,210){\textcolor{green}{\line(-1,-1){180}}}
\put(300,210){\textcolor{green}{\line(1,-1){75}}}
\put(136.7,210){\textcolor{yellow}{\line(0,-1){210}}}
\put(146.7,210){\textcolor{yellow}{\line(0,-1){210}}}
\put(156.7,210){\textcolor{yellow}{\line(0,-1){210}}}
\put(166.7,210){\textcolor{yellow}{\line(0,-1){210}}}
\put(176.7,210){\textcolor{yellow}{\line(0,-1){210}}}
\put(166.7,210){\textcolor{blue}{\line(1,-3){70}}}
\put(100,210){\line(3,-1){210}}
\end{picture}
[/tex]

HARRRR Lines!
 
  • #329
sorry... I'm too easily amused :O
 
  • #330
Ok, I think I have the image right.

https://www.physicsforums.com/latex_images/243709-0.png

That's the link to it, just to save on server space. :smile:

We have the two red sources are stationary, as is the yellow observer at their midpoint. The two sources emit green photons simultaneously. The blue train observer happens to arrive at the yellow observer just as green photons from both directions do as well.


P.S. good thing you haven't learned "multiput" yet. :devil:
 
Last edited by a moderator:
  • #331
Ok, now to start getting to the point of drawing this diagram.

Rather than thinking of the universe as being some spatial "thing" that only exists "now", we can think of the universe as consisting not only of now, but the past (and the future?).

The thing I'm getting at is that instead of thinking of the animation in picture #1 which shows the universe at a given instant of time, and show how the universe changes as time changes, we can think in terms of the diagram I draw which shows the whole of space-time all at once.

In this representation of the universe, we can recover our classical notion of "now"; a point in time corresponds to a horizontal line on this diagram, and everything occupying that line is what we, classically, think of the universe looking like at that time.

Agree?
 
  • #332
if the moving picture stops 1/2 way through that diagram where the green lines cross then yeh i'd say that looks pretty accurate. :smile:
 
  • #333
horizontal cut = now
vertical cut = where

yesh... proceed
 
  • #334
ram1024 said:
this picture displays the 'center...stance light travels is the same for a and b.
 
  • #335
no idea what you're getting at since i never combined those two :biggrin:
 
  • #336
ram1024 said:
no idea what you're getting at since i never combined those two :biggrin:

then why do you say "changing JUST the observer to BE moving changes the distance that light will travel hence change the intercept times." ?
 
  • #337
Case #7
Step: 1
Code: 
                     [u](o)                                        <-)|[/u]
                     [u](o)                                        <-)|[/u]
                     [u](o)                                        <-)|[/u]
                     [u](o)                                        <-)|[/u]

======================================

Step: 2
Code: 
                     [u](o)                                        <-)|[/u]
                     [u](o)                                     <-)|[/u]
                     [u](o)                                  <-)|[/u]
                     [u](o)                               <-)|[/u]
Step 4:
Code: 
                     [u](o)                                        <-)|[/u]
                        [u](o)                                     <-)|[/u]
                           [u](o)                                  <-)|[/u]
                              [u](o)                               <-)|[/u]


===============================


Step: 3
Code: 
                     [u](o)                                        <-)|[/u]
                     [u](o)                                           <-)|[/u]
                     [u](o)                                              <-)|[/u]
                     [u](o)                                                 <-)|[/u]
Step 5:
Code: 
                     [u](o)                                        <-)|[/u]
                  [u](o)                                           <-)|[/u]
               [u](o)                                              <-)|[/u]
            [u](o)                                                 <-)|[/u]

i move them around so you can do the comparison easier.
 
  • #338
for each step, pulse a light from the emitter at line1 and assume it hits the observer on line 4.

it's pretty easy to see that the simple changing of "who's moving" doesn't make for an equivalent picture. in 2 and 3, no relative distance is made <progress> of the observer towards the location of the light emission source (not the emitter, but where the emitter WAS in line1 when it pulsed)

in 4 and 5 relative motion IS noted between the light emission source and the observer(this time since the emitter source is stationary, the light emission source occupies the same location)
 
  • #339
ram1024 said:
Case #7
i move them around so you can do the comparison easier.

Yes, the distance is measured by observer and emitters differently, and so is the elapsed time, due to length contraction and time dilation. So what? I will provide you numerical examples to make you understand what value everyone will measure. Wait..
 
  • #340
[tex]
\begin{picture}(500,240)(0,0)
\put(100,210){\textcolor{red}{\line(0,-1){210}}}
\put(300,210){\textcolor{red}{\line(0,-1){210}}}
\put(100,210){\textcolor{green}{\line(-1,-1){75}}}
\put(100,210){\textcolor{green}{\line(1,-1){180}}}
\put(300,210){\textcolor{green}{\line(-1,-1){180}}}
\put(300,210){\textcolor{green}{\line(1,-1){180}}}
\put(200,210){\textcolor{yellow}{\line(0,-1){210}}}
\put(166.7,210){\textcolor{blue}{\line(1,-3){70}}}

\put(100,210){\line(3,-1){360}}
\put(325, 135){\circle{4}}
\end{picture}
[/tex]

Ok, I've extended the diagram a little so I can talk about some additional things.

Once we permit ourselves to think about the universe in terms of a space-time, we can ask a "what if" question...


We know from geometry that all lines are the same. Sometimes we pick two perpendicular lines to be axes, but we can always rotate and pick different axes. So...

What if we allow lines that aren't horizontal to be "when" and lines that aren't vertical to be "where"?


Nonvertical lines reprsenting "where" have been around a long time; the blue line represents a "where"... the train. Other lines parallel to the blue line also represent "where"; places at a fixed position relative to the train.


The thing that makes SR different than Gallilean relativity is that SR also permits diagonal lines to represent "when". The black line I've drawn on the diagram is a line that, according to SR, represents "when" in the train's frame.

Now, before you jump on this and call it stupid, consider this: even if you think it's stupid to let a diagonal line represent "when", we can identify this black line in reality. We could define a collection of lines parallel to the black line, and define something by these lines.

So, whether or not you consider this a measurement of time, it is certainly a measurement of something physically meaningful in reality.


In this diagram, I hope you can see how SR can say that in the train's frame that the emissions aren't simultaneous. If we're using these diagonal lines to measure "time", instead of horizontal lines like we're "supposed to", then it's clear that this different way of measuring "time" will not measure the two emission events as occurring at the same time. In particular, if we're using these diagonal lines to measure time, then the emitter on the right will fire before the emitter on the left!


I extended the picture so I could also talk about the "center of emission". Again, if I'm using these diagonal lines to define "when", then the points where the right photons intersect the black line correspond to the same "time". The little dot I drew is the midpoint between these two intersections. If I'm using the black line to define "when", then that dot should be the "center of emission", because it is the point midway between the photons.

Again, this diagram shows how we can be referring to the same reality, but disagree on the "center of emission".


Each frame of animation #2 can be generated by drawing one of these diagonal lines. The black line I drew corresponds to the frame in animation #2 when the left emitter fires. Notice that, along this black line, the right photon is much closer to the yellow observer than the left photon is, the train is located at the midpoint of the photons, and the right center of emission is located a little to the right of the right source. If you look at animation #2, the frame where the left emitter fires will conform precisely to this description. (including your green fuzzy dot!)
 
  • #341
no, there is no distance measured differently... light CLEARLY travels farther if you switch the frames like this.

in the emitter reference we have the emitter moving away from the "light emission source" which has NO impact on when the observer will receive the photon.

in the observer reference we have the observer moving away from the "light emission source" which has TOTAL impact on when the observer will receive the photon.

not "time gets skewed" because the observer is moving but "it's not even the same situation"
 
  • #342
Hurkyl said:
[tex]
\begin{picture}(500,240)(0,0)
\put(100,210){\textcolor{red}{\line(0,-1){210}}}
\put(300,210){\textcolor{red}{\line(0,-1){210}}}
\put(100,210){\textcolor{green}{\line(-1,-1){75}}}
\put(100,210){\textcolor{green}{\line(1,-1){180}}}
\put(300,210){\textcolor{green}{\line(-1,-1){180}}}
\put(300,210){\textcolor{green}{\line(1,-1){180}}}
\put(200,210){\textcolor{yellow}{\line(0,-1){210}}}
\put(166.7,210){\textcolor{blue}{\line(1,-3){70}}}

\put(100,210){\line(3,-1){360}}
\put(325, 135){\circle{4}}
\end{picture}
[/tex]

i don't see how the black line can be the train's "where" because it never gets hit by the light from emitter(L)...

explain the black line to me once more :(
 
  • #343
ah okay nevermind you said it's a "when" line
 
  • #344
it was informative, but it doesn't tell me much except that non-simultaneity for that frame is a direct consequence of light speed measured constant for the observer.

allow for a moment light to be subjected to normal physics in the sense that the emitters impart whatever velocity they have to the light when it is emitted, as would be if light were a normal mass attribute object.

here is my version of events allowing inertial velocity to be added to the light sources (making the light travel with the emitters) in order to preserve a true reference frame according to the stationary train as an observer.

http://www.imagedump.com/index.cgi?pick=get&tp=89150

<changed the background, white hurts my eyes>

[edit] explaining that a little bit for you, the white line is the skewed "when" line for the now "stationary" train. i shifted the train back over to the original location, since it "doesn't move". i allowed the emitters to carry their inertia from movement over to the photons so that they're moving along with added velocity of the source (if you check my earlier case #7 you'll see why this is necessary to ensure constancy). the red lines are the moving emitters. the "dotted" red lines are theoretical emitter locations if they were not moving.
 
Last edited:
  • #345
Informing was my goal. I wasn't trying to convince anyone of anything; I was mainly trying to clear up how, according to SR, the two different frames could be corresponding to the same "reality" despite the disagreement over things like simultaneity and centers of emission.

I just recall that, when I was learning this stuff, I never really "got" SR until I tried understanding it from these diagrams instead of from wordy descriptions and algebra. Thought it might be helpful while we wait for Tom to dig up the meaty stuff. :smile:
 
  • #346
Today special relativity dies...

Kurious:

It will die in space-time...
 
  • #347
my diagram makes sense right? I've never messed with this space-time stuff, it was good of jscd<sp> to point out how useful it could be for explaining stuff. :smile:

in any case, i hope you can see that on my diagram <allowing for relative motion of the observer towards the "light emission centers"> his space/time is skewed but he still calculates simultaneous "when" emissions on his skewed line

i think grounded has the mathematics for such a transform already in place, he's just missing the "light emission centers" bit in there which according to how we'd plotted it would be a simple geometric rotation of the observer's spacetime "worldline"
 
  • #348
woops my bad, jdavel suggested space-time diagrams
 
  • #349
ram1024 said:
no, there is no distance measured differently... light CLEARLY travels farther if you switch the frames like this.

in the emitter reference we have the emitter moving away from the "light emission source" which has NO impact on when the observer will receive the photon.

in the observer reference we have the observer moving away from the "light emission source" which has TOTAL impact on when the observer will receive the photon.

not "time gets skewed" because the observer is moving but "it's not even the same situation"

Ram, I now think "time gets skewed" is the resolution to your apparent paradox.

I found that other link again. Please take a look:
http://casa.colorado.edu/~ajsh/sr/paradox.html

Note that vermilion and cerulean do not agree on where the centre of emission is. So the centre of emission is relative too.

In your example, no one will agree on times and distances due to SR effects, that's normal. But, what confused me was: time dilation and length contraction don't depend on direction of speed. So they don't explain the difference between 2/4 and 3/5. But, relative simultaneity does depend on direction, and that is why there is a skew in time, not a shift. So, I think the resolution to your paradox is the skew. I hope you can work it out with the spacetime diagrams.
 
  • #350
In your example, no one will agree on times and distances due to SR effects, that's normal. But, what confused me was: time dilation and length contraction don't depend on direction of speed. So they don't explain the difference between 2/4 and 3/5. But, relative simultaneity does depend on direction, and that is why there is a skew in time, not a shift. So, I think the resolution to your paradox is the skew. I hope you can work it out with the spacetime diagrams.

see the main thing i see as "the problem" is that light itself does NOT obey physical laws of the universe, so when you put it in a position where the frames get switch you cannot expect light to behave the same.

case in point the picture i submitted most recently which is a spacetime diagram of the train stationary, but the light sources moving AND the photon light emission centers being tied to the emitters.

in this diagram it is clearly obvious <by taking "whens" on that diagonal <the train's view> you can see that even according to the train, emission IS simultaneous.

but BECAUSE light is not tied to its source <not the case with other physical mass bearing things in the universe> if you shift the frames and don't compensate for this, it's not the same situation.

i had the same problem when i was prooving .999~ doesn't equal 1. people were functioning under the limit that infinity was an absolute limit, then got all confused when numbers at the limit didn't behave the same way as numbers anywhere else on the line. remove the limit and everything still works the same.

same thing here, compensate for your limitations and everything still works fine. no time dialation, no length contraction, no simultaneity nonsense.

no relative to the viewer light speed, but i do believe that's solely an error based on how they're calculating their measurments. we'll see when the data arrives.
 

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