Today Special Relativity dies

[geistkiesel]

Who need SR in this problem?

Do you understand the concept of a distance versus time plot?

Does this look like a drawing of the experiment you describe?

<br /> \begin{picture}(300,240)(0,0)<br /> \put(0,240){\line(1,-3){70}}<br /> \put(70,20){A}<br /> \put(100,240){\line(0,-1){210}}<br /> \put(100,20){M&#039;}<br /> \put(100,240){\line(1,-3){70}}<br /> \put(170,20){M}<br /> \put(200,240){\line(1,-3){70}}<br /> \put(270,20){B}<br /> <br /> \put(0,240){\line(1,0){270}}<br /> \put(280,240){x}<br /> <br /> \put(0,240){\line(0,-1){210}}<br /> \put(0,20){t}<br /> <br /> \put(0,240){\line(1,-1){180}}<br /> \put(200,240){\line(-1,-1){125}}<br /> \end{picture}<br />

__A________________M______|________|_______B__
                   |      |        |
                 M'(t'0)  M'(t'1)  M'(t'3)
                 M'(t'2)     B         A

Right A M and B are stationary points. M the midpoint of photon sources A and B The terms below the line are in the moving frame. The M' represent the actial location of the observer at the indicated times. As you see thee isn't any reason to define loss of simultaneity as it was done.

M' represents the position of the moving observer at various times. We keep a moving observer at M, the midpoint of the photon sources at A and B. Hw ei is motorinzed and moving at -v with respect to the Observer in the mopving frame. We can consider the M'(t'2) the 0 point of the coordinate system in ythe moviong frame. M'(t'0) is when the observer was at the midpoint in the stationary frame. M'(t'1) is the measurement of the B photon. M'(t'2) the measurement of the simultaneous arrival at M'(t'2) of the emitted photons from A and B at the stationary midpoint, all observed by the M'(t'2) observer in the moving frame. M'(t'3) the A photon is detected by the observer.
Who needs SR to look at this problem?

And where did was missing photon?. The one that was left behind when the B photon was emitted first?

 

[Hurkyl]

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 />

 

[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.

Remember the grounded example using his automobiles in motion wrt each other? When adding the observer's velocity the proper length of the automobiles was determined from measurment. Using only the velocity of one of the automobile resultes in a shortening of the car's length. For the same reason you measure a shorter wave length when counting the full wave length increments passing through your eye. Make some calculations.

Janus already knows Galilean relativity.

geistkiesel spouts
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. Get yourself in a higher state of energy and you measure round, from orbit.

 

[ram1024]

http://homepage.sunrise.ch/homepage/schatzer/space-time.html

wow. i found a page that pretty much sums up all of our arguments thus far, and explains my position for "galilean relativity" pretty precisely. (even though i had no idea that's what it was)

it's kind of a long read though, i'll pull some highlight parts that are noteworthy from my side of the fence.

... Briefly, whereas Galilean space-time allows the realization of faster-than-light speeds, at least in principle, Minkowski space-time does not. What is the reason for this difference? In the next sections it is exposed that the key point is the conception of global time, ie. the physical significance of the term simultaneity...

... Because of this absolute time the global notion of past, present and future is the same in all reference frames. If two events are simultaneous in one particular reference frame, this means that they are also simultaneous in all reference frames...

... Although the one-way speed of light is not constant in general (ie. when expressed in an arbitrary reference frame), the mean-speed c of a round-trip is again constant [2], what is in accordance with all experiments (like Michelson-Morley a.s.o.). It should be emphasized again that there has been no experiment which determined the one-way speed of light [3], since this would require the possibility of synchronizing physical clocks by some other means than finite-speed signals...

... Moreover, further arguments can be found which might motivate the reintroduction of (Galilean) absolute time to physics:

First, if there exists a physical absolute time, then the number of fundamental constants is reduced by one, since the (one-way) speed of light is not a constant any longer. This leads to a simplification and a new interpretation of the physical quantities and constants...

...If our universe has a Newtonian background, ie. if there is an absolute time underlying the space-time continuum, then there is no threat on causality by superluminal processes, because time travel and its paradoxes are excluded a priori. And thus, within this framework, faster-than-light travel is possible, at least in principle. ...

 

[quantumdude]

Remember the grounded example using his automobiles in motion wrt each other? When adding the observer's velocity the proper length of the automobiles was determined from measurment. Using only the velocity of one of the automobile resultes in a shortening of the car's length.

You are still confused on the same point that befuddled Grounded. It's not that we have to add the observer's velocity, it's that we have to take the correct velocity of the car with respect to the observer. Using the ground speed to make a length calculation is incorrect in either Galilean relativity or SR. Everyone who knows physics already knows this, and I have already addressed it.

Length contraction is predicted by SR even when you take the correct relative velocity. Furthermore, the length contraction is not equal to the "missing distance" that Grounded calculated. Even he realizes that now.

For the same reason you measure a shorter wave length when counting the full wave length increments passing through your eye.

No, it's not "for the same reason". The wavelength of light is also predicted to change in SR when the correct relative velocity of the source is used.

Make some calculations.

Make some calculations yourself. I performed these exercises a long time ago, and I have moved on to other things. It's time for you to do the same.

geistkiesel spouts
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. Get yourself in a higher state of energy and you measure round, from orbit.

WTF?

 

[Hurkyl]

It should be emphasized again that there has been no experiment which determined the one-way speed of light [3], since this would require the possibility of synchronizing physical clocks by some other means than finite-speed signals...

I learned something today. Assuming http://www.phys.virginia.edu/classes/109N/lectures/spedlite.html is right, the very first measurement of the speed of light was one-way.

First, if there exists a physical absolute time, then the number of fundamental constants is reduced by one, since the (one-way) speed of light is not a constant any longer.

I don't see how this follows.

If our universe has a Newtonian background, ie. if there is an absolute time underlying the space-time continuum, then there is no threat on causality by superluminal processes, because time travel and its paradoxes are excluded a priori. And thus, within this framework, faster-than-light travel is possible, at least in principle.

There's a word for believing something because you want it to be true; gullible.

 

[quantumdude]

Are you telling me, or trying to tell me that SR isn't an imaginary discipline?

There's not a single line of what I wrote that says anything about that, one way or the other. But since you ask, SR is a theoretical discipline. It just so happens that this particular theoretical discipline makes predictions that are agreed with by experiments that aren't imaginary.

And where do you get Psychosis? I know we refer to each other as crazy, but I would like to know where you arrive at psychosis? Just because I or others disagree with you? You read below and tell me thios is the ranting of a mad man as witnessed by his lack of reasoning in his analytic posture.

I already explained why I used the term "psychosis". It is a result not of disagreeing with me, but of confusing the imaginary for the real.

Where have you heard Grounded's theses? Who in your career has said we must add the velocities of the observer into the calculations? And who contiued along in the line taken by grounded in calculating frequency abnd wave length? Who said this and where was she dismemebred?

Grounded is espousing Galilean relativity. Ask me again, and I'll tell you the same.

If you understand Grounded very well then why do you have to resort to some vague reference to the Galilean nature of his theses?

Your mathematical description a hwile back was impressive, as I noted. Clear, straightforward, careful use of proper parameters in the equatons, all of that which is expected of one who knows his theory, which you obviously do.

You are contradicting yourself right here. If you absorbed my mathematical presentation so well, then why do you think my references to Galilean relativity is vague? I spelled Galilean relativity out very clearly, and when I refer to it, I am referring back to that post.

Tom how many times have you pondered the limitations of mathematics to adequately describe physical activity? Just because the math says you can do it doesn't mean it is physically proper.

No kidding. That's why we do experiments. And guess what? The experiments say that Grounded's ideas are not "physically proper".

A train station is stationaary and a train is mobile, moving. To say you can mathematically swap reference frames while being a physical impossibility, then why even broach the matter in analysis? Let me guess because you get the right answer doing it that way?

You are wrong. The observer on the train is perfectly free to say that he is stationary, and that the station is moving away from him.

You mention real experiments and I look them up and see they are flawed with the same erronoeus reasoning that you manifest.

LOL, name a single experiment published in any reputable journal that you've looked up, and why you think it is flawed. Then, I'll explain to you why you are wrong.

It would do you well tom to get yourself into Grounded's shoes.

Again: I've already done this, in this very thread. I have shown many pages ago why the velocity transformation he uses cannot be right.

Also, you might tell me where the photon that follows the first emitted photon in the moving frame is hiding when the first is emitted,. You do rememebr that the photons were emitted simultaneously in the stationary frame don't you?

You seem to be referring to a discussion in which I have not taken part.

The words 'galilean reference frame' is your latest mantra Tom.

Only because I keep getting the same stupid questions about it.

 

[ram1024]

You are wrong. The observer on the train is perfectly free to say that he is stationary, and that the station is moving away from him.

not when light is involved, i have demonstrated this in a past exercise.

because of light's "independent of source" nature, when you change reference frames it creates a fundamental dissimilariy for the situation if perceived as equivalent to the "real" reference frame.

 

[quantumdude]

not when light is involved, i have demonstrated this in a past exercise.

No. Any inertial observer is free to consider himself at rest. If you "demonstrated" otherwise, then you made a mistake.

 

[ram1024]

There's a word for believing something because you want it to be true; gullible.

could say the same thing about your side of the fence with your crazy length contractions, time dialations, and reference based simultaneities.

 

[ram1024]

No. Any inertial observer is free to consider himself at rest. If you "demonstrated" otherwise, then you made a mistake.

well since you HAD to bring it up, here's exactly why

Case #7
Step: 1

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

in this setup, we have but one emitter and one observer. keeping it simple-like. In all cases the emitter is going to emit a pulse of light on the first "frame" of the setup. assume uniform motion (no acceleration).

emitter stays the same place towards the observer.

Step: 2

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

emitter moves towards the observer.


Step: 3

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

emitter moves away from observer.

This is simply a demonstration of what you're saying that light doesn't care what its source does, right? In all 3 cases light would reach the observer at the same time if the first "frame" were synchronized.

now we're going to do what you guys do to things...

Step 4:

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

Step 5:

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

we're going to take the same set ups from above and simply CHANGE the relative motion so that the emitters are stationary and the observer is the one that's moving. this shouldn't change ANYTHING as far as you guys see it right? these cases should be EXACTLY the same as the ones above, we just changed perspective...

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

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

Your steps 1, 2, and 3 seem to take a view from A's frame. But steps 4 and 5 take a view from a frame in which A is moving. Of course that frame will measure different times. It should be no surprize to you by now that time measurements are frame dependent.

that is exactly why picture 2 is NOT the same situation as picture 1. :surprise:

 

[quantumdude]

could say the same thing about your side of the fence with your crazy length contractions, time dialations, and reference based simultaneities.

No, you couldn't say the same thing. We don't believe those things because we want them to be true, we believe them because they are consistent with experimental results. The "crazy" one is the one who refuses to accept reality for what it is.

 

[Hurkyl]

could say the same thing about your side of the fence with your crazy length contractions, time dialations, and reference based simultaneities.

Maybe I'm forgetting; could you point out where I've stated anything about what I (or anyone on my side of the fence) wish was true? Let alone suggesting that this wish is a good reason to believe SR?

 

[quantumdude]

that is exactly why picture 2 is NOT the same situation as picture 1. :surprise:

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:

 

[ram1024]

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

 

[quantumdude]

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?

 

[Hurkyl]

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.

 

[ram1024]

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.

 

[wespe]

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.

 

[ram1024]

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.

 

[Hurkyl]

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)

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

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)

 

[quantumdude]

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.

 

[ram1024]

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)

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

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>

 

[ram1024]

sorry i mean the blue line would.

well whichever... one of them would ;D

 

[Hurkyl]

I drew the wrong picture. it's been fixed; is it right now?

 

[ram1024]

I drew the wrong picture. it's been fixed; is it right now?

lemme try

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

 

[Hurkyl]

Grr, I had the train going the wrong way too, I think... Bah, let me look at picture #1 again to be sure!

 

[ram1024]

<br /> \begin{picture}(400,240)(0,0)<br /> \put(100,210){\textcolor{red}{\line(0,-1){210}}}<br /> \put(110,210){\textcolor{red}{\line(0,-1){210}}}<br /> \put(120,210){\textcolor{red}{\line(0,-1){210}}}<br /> \put(130,210){\textcolor{red}{\line(0,-1){210}}}<br /> \put(140,210){\textcolor{red}{\line(0,-1){210}}}<br /> \put(300,210){\textcolor{red}{\line(0,-1){210}}}<br /> \put(100,210){\textcolor{green}{\line(-1,-1){75}}}<br /> \put(100,210){\textcolor{green}{\line(1,-1){180}}}<br /> \put(300,210){\textcolor{green}{\line(-1,-1){180}}}<br /> \put(300,210){\textcolor{green}{\line(1,-1){75}}}<br /> \put(136.7,210){\textcolor{yellow}{\line(0,-1){210}}}<br /> \put(146.7,210){\textcolor{yellow}{\line(0,-1){210}}}<br /> \put(156.7,210){\textcolor{yellow}{\line(0,-1){210}}}<br /> \put(166.7,210){\textcolor{yellow}{\line(0,-1){210}}}<br /> \put(176.7,210){\textcolor{yellow}{\line(0,-1){210}}}<br /> \put(166.7,210){\textcolor{blue}{\line(1,-3){70}}}<br /> \put(100,210){\line(3,-1){210}}<br /> \end{picture}<br />

HARRRR Lines!

 

[ram1024]

sorry... I'm too easily amused :O

 

[Hurkyl]

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.

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.

 

[Hurkyl]

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?

 

[ram1024]

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.

 

[ram1024]

horizontal cut = now
vertical cut = where

yesh... proceed

 

[wespe]

this picture displays the 'center...stance light travels is the same for a and b.

 

[ram1024]

no idea what you're getting at since i never combined those two

 

[wespe]

no idea what you're getting at since i never combined those two

then why do you say "changing JUST the observer to BE moving changes the distance that light will travel hence change the intercept times." ?

 

[ram1024]

Case #7
Step: 1

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

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

Step: 2

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

Step 4:

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

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


Step: 3

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

Step 5:

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

i move them around so you can do the comparison easier.

 

[ram1024]

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)

 

[wespe]

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..

 

[Hurkyl]

<br /> \begin{picture}(500,240)(0,0)<br /> \put(100,210){\textcolor{red}{\line(0,-1){210}}}<br /> \put(300,210){\textcolor{red}{\line(0,-1){210}}}<br /> \put(100,210){\textcolor{green}{\line(-1,-1){75}}}<br /> \put(100,210){\textcolor{green}{\line(1,-1){180}}}<br /> \put(300,210){\textcolor{green}{\line(-1,-1){180}}}<br /> \put(300,210){\textcolor{green}{\line(1,-1){180}}}<br /> \put(200,210){\textcolor{yellow}{\line(0,-1){210}}}<br /> \put(166.7,210){\textcolor{blue}{\line(1,-3){70}}}<br /> <br /> \put(100,210){\line(3,-1){360}}<br /> \put(325, 135){\circle{4}}<br /> \end{picture}<br />

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!)

 

[ram1024]

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"

 

[ram1024]

<br /> \begin{picture}(500,240)(0,0)<br /> \put(100,210){\textcolor{red}{\line(0,-1){210}}}<br /> \put(300,210){\textcolor{red}{\line(0,-1){210}}}<br /> \put(100,210){\textcolor{green}{\line(-1,-1){75}}}<br /> \put(100,210){\textcolor{green}{\line(1,-1){180}}}<br /> \put(300,210){\textcolor{green}{\line(-1,-1){180}}}<br /> \put(300,210){\textcolor{green}{\line(1,-1){180}}}<br /> \put(200,210){\textcolor{yellow}{\line(0,-1){210}}}<br /> \put(166.7,210){\textcolor{blue}{\line(1,-3){70}}}<br /> <br /> \put(100,210){\line(3,-1){360}}<br /> \put(325, 135){\circle{4}}<br /> \end{picture}<br />

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 :(

 

[ram1024]

ah okay nevermind you said it's a "when" line

 

[ram1024]

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.

 

[Hurkyl]

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.

 

[kurious]

Today special relativity dies...

Kurious:

It will die in space-time...

 

[ram1024]

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.

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"

 

[ram1024]

woops my bad, jdavel suggested space-time diagrams

 

[wespe]

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.

 

[ram1024]

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.