SR Simultaneous Lines Drawn in the Sand

[geistkiesel]

Keep in mind that the principle of relativity is part of classical mechanics; it was not invented by Einstein for the purpose of developing SR.



And to drive home the cosmological point about there being no good point of reference, consider these:

(1) Suppose in your future that we also considered a similar accident occurring on the streets of future earth. In this case, neither car was stationary with respect to the solar system!


(2) The guy who stopped was not at rest with respect to the rest of the universe. And, for instance, you might have been at rest with respect to the Milky Way. The important point is that a particular frame of reference was chosen (in this case, "at rest WRT the solar system"), so the traffic law is defined relative to this choice.


(3) What is "at rest in relation to the rest of the solar system" anyways? Things are moving in all sorts of directions, accelerating all over the place.

Not from the laws of physics do we intuit no absolute reference point but from the sheer technological chore of finding such a point. Even if the best instumentation conceivable were at our disposal and we were able to measure out to 20 billion light years and found a most perfect spot, it is the stellar entiities beyond 20 billion light years that would screw up the perfection. In any event we might perceive a need for a contacted and practical perfect rest point for some perceived need, for some finite duration of time. Ergo to within useful limits I see nio rule or law of physics preventing anyone from determining a relative point that does the job for the situation under consideration. i wouldn't tell any SR theorist about it though, no way. Read the posts, see what you would be subjecting yurself to?

 

[jdavel]

geistkiesel said, "Give us your veyr very best shot at analyzing the situation."

Sorry, the situation you're describing is too complicated to explain in one "shot". And your insistence that SR is wrong would doom to certain failure any attempt of mine to convince you otherwise!

But how about this? I'll explain it one step at a time. As long as you understand, and say that you believe, each step, I'll give you the next one. At the end you'll understand why SR is right and you were wrong.

You can agree to this by starting a new thread on the Relavitity Forum with the title "Why isn't the simultaneity of two events absolute?"

I'll be waiting!

 

[geistkiesel]

Ok, so we have these film strips and after the experiment they will have two little dots on them corresponding to the points A and B when the photons were emitted. In the moving frame we measure the distance between these points using a metre ruler and come up with a length. We have (using my previous notation) that in the stationary frame the distance between A and B is $$2L$$. However, due to Lorentz contraction the stationary observer will observe the moving observer to measure this distance to be $$2L\gamma$$

NB
$$\gamma = \frac{1}{\sqrt(1-\frac{v^2}{c^2})} > 1$$

This is because he sees the moving observer trying to measure a distance of $$2L$$ with a shorter (contracted) measuring stick. So he is observing the moving observer making a measurement larger than $$2L$$.

The moving observer on the other hand sees the distance between A and B contracted. Hence, naively you might think he will measure a length of $$2L/\gamma$$. This cannot be because what each observer sees must be the same (remember that we just had the stationary observer seeing the moving observer measure a longer length), so somehow the moving observer must also measure $$2L\gamma$$. This happens because he does not observe the photon emissions to be simultaneous. Hence there is a little bit of time between the first and the second photon, which makes his measured length larger. This will ensure that his observation coincides exactly with that of the stationary observer.

Matt

No baffledMatt, why do you refer to stationary observers seeing what moving observer do. What is happenming in the physical wold. the language of SR is effectively limited to that used by yourself in this post. The observers perception, and when you take this down to thee eyeball to eyeball level of two observers in dfferent frames observing the same physical result, what do you get? Beside the confounding logic that I just must accept, what physical reality does one come up with? Four eyes seeing the same event, one sees it now and one sees it later, I think this is what you said implicitly at least. The stationary observer, "sees' the photons simultaneously emitted, because it is a given, ok take him to the where it us happeming and show him. Of course the stationary observer cannot observe both events simultabeously, he can only see one at a time. To make it easy on ourselves we postulated the simultaneous emission of photons in the stationsry frme.A photon in one place emits aT t1 and just across a wave length a moving observer sees the same physical 'even' later at time t2, which is also observed by the stationary observer. Me thinks this is a targeted point for SR logic, is it not?

 

[geistkiesel]

ram2048 said, "it's a "Given" part of the experiment that they WERE emitted simultaneously as outlined in Einstein's original set up"

Not quite! They were simultaneous in the stationary frame, but, as it turns out, they weren't simultaneous in the moving frame. That's not an assumption; it's a conclusion based on the assumption that c is the same in both frames.

Here's "Einstein's original set up":

IN THE STATIONARY FRAME THE FLASHES ARE SIMULTANEOUS and are equidistant from the moving observer at the time of the flashes. So, the stationary observer sees light from the two flashes reach the moving observer at different times.

The moving observer agrees, light from the two flashes reached him at different times. Then he looks at the marks that the flashes made on his train, sees they are at equidistant from his seat on the train. Since he knows that c was the same for the light of both flashes, he concludes that THE FLASHES WERE NOT SIMULTANEOUS IN THE MOVING FRAME.

You are 100% absolutely correct. I did not intend to imply differently. In the opening post of this thread I stipulated that SR would predict the events were not simultaneous in bot he frames.

Simulatneity is not measured by your, AE's gedun
jken, it is measured by whether the photons emitted at A and B simultaneously were emitted into the moving frame simultaneoulsy, and the events of the staggered arrival is used as "proof" that the event were not emitted simultaneously in the moving frame. I recognize there is a considerable difference in AE'a trivial example and modern use, but his model is used tioday to emophacize and explain points right?

But look at what AE was using to justify loss of simultabneoty: the reception of photons at different times after the photons were emitted. I see no "relativity" implications here. Take the situation one step farther than AE took. The different arrival times could have been considered by an alert and acute scientist.'" HMMM, two photons different times and places. If the photons were emitted simultaneously and I were moving then I would see one before the other. Or if the photons were emitted at different times I might see them arriving simultaneously at the moving fame or at different times and not know which was emitted first. AE didn't ahve to stop at the naive point in the analysis that SR theorists are quoting this very day, yours not the first.

 

[geistkiesel]

geistkiesel said, "Give us your veyr very best shot at analyzing the situation."

Sorry, the situation you're describing is too complicated to explain in one "shot". And your insistence that SR is wrong would doom to certain failure any attempt of mine to convince you otherwise!

But how about this? I'll explain it one step at a time. As long as you understand, and say that you believe, each step, I'll give you the next one. At the end you'll understand why SR is right and you were wrong.

You can agree to this by starting a new thread on the Relavitity Forum with the title "Why isn't the simultaneity of two events absolute?"

I'll be waiting!

I didn't say Sr was wrong for th e purposes of this thread. I stipulated SR would pedict no simultaneity. What I would ask, is look at the gedunken used as a template for this thread. Discuss in a professsional way the apparent pardox of he observed emission simultaneously being observed nonssimultaneously in the moving frame. Use the measurement condition o of the thread. I've been living SR for a while now, I do not need another proof. I am convicned that you will come up with what eveybody else came up with, Talk about the thread hypo, i f you please,,and if you are able.

 

[geistkiesel]

and I was just about to say exactly the same thing about your link.



Errrm, hello?

*pointing to the last post I made*

Matt

baffledMatt can you point out to me by argument or otherwise whe ehttp://frontiernet.net/~geistkiesel/index_files/ Elaborate or brief, just specifics please. Myself, I thought it original which it will probable remain , who would want to confiscate silly physics? Any hint to a specific flaw(S) would be greatfully appreciated. Thanx G

 

[geistkiesel]

No, if they are sending the same 1 second pulses to each other they will each observe exactly the same thing - either a slowing down of the pulses (red shift) or a speeding up (blue shift) depending whether they are moving toward each other or away from each other.



I still don't see exactly what you mean by this. Could you perhaps give me the dumbed down version of how exactly this is done? It's no good you saying 'using the same physics' since we evidently don't agree on exactly what the physics is!



I can see why you might think that - I remember having exactly these sorts of problems with SR. The way I would always go about solving these problems though was to sit down and just calculate the thing. To illustrate:
Let's make some definitions:
$$x$$ : space coordinate in 'stationary' frame. $$x = 0$$ is the midpoint between A and B, which are located at $$x = -L$$ and $$x = L$$ respectively.

$$v$$ : velocity of the moving observer relative to the stationary observer.

Ok, so now I will calculate what the 'stationary' observer sees. Note I am not going to use any SR as such. We first calculate the location of the point where the 'moving' observer and the light from A coincide. This happens at a time $$t_A$$ after the light is emitted from A and the point where they meet we denote $$d_A$$. By equating the distance the light moves in this time to the distance the moving observer moves we have:

$$vt_A = d_A$$
$$ct_A = d_A + L$$
so
$$\frac{v}{c}(L + d_A) = d_A$$
and
$$d_A = \frac{\frac{v}{c}L}{(1-\frac{v}{c})}$$

Similarly, we can derive $$d_B$$ which is the point where the moving observer meets the light from B
$$vt_B = d_B$$
$$ct_B = (L-d_B)$$
so
$$\frac{v}{c}(L-d_B) = d_B$$
and again
$$d_B = \frac{\frac{v}{c}L}{(1+\frac{v}{c})}$$

Evidently, $$d_A \neq d_B$$ so the photons do not hit the moving observer at the same time/place.



Ok, now repeat the calculation above, but for the moving observer. The only SR result I want you to use is that each observer must observe the same value for the speed of light. If you do this you should see that you need things like time dilation and lack of simultaneity to resolve the observations.

Matt

I ain't gona do your calculations for you. Hurkyl conned me into one of those. But then he complimented me on my calualtions, which showed simple way to determine if thephotons e were whatever. I think he screwed up, by complimenting me on something that effectiely undermined his specific point there. My argument assumes SR. so the easy out is the rationalization for te apparent paradox of the simulatneous emission of photons in a stationary frame turn magically into nonsimultaneity by the mere presen e of an inertial frame other than stationary in the vicinity.

 

[geistkiesel]

baffledMatt, modify my two clocks with the same frequency. The new model was loaded on one ship, Its frequency is 10^10hz. Now they can tell which is which and who is moving slower or faster? assuming they each had acceleration history to share or acceleration history available in some form. I suppose they wouldn't need the different clock speeds then would they?

 

[ram2048]

still seems silly that SR prefers the conclusion that simultaneity is at fault and not personal perspective. it's much more logical to conclude that personal reality is skewed because of movement and calculate backwards to conform to what the rest of the (stationary) universe sees.

but that's one of the things that's a matter of opinion i think. <shrug>

in any case in terms of space and relative fixed points, i think it's imperative that 3 dimensional space be charted in order to be utilized.

when living on Earth it's easy to use relative fixed points for reference, we do it intuitively and almost instinctively daily, driving between the lines, walking on the sidewalk, sitting down on a chair instead of missing it and landing on the floor...

in space you're dealing with far less points of reference AND far greater possible speeds, so everything becomes that much bigger of a problem.

seems only a matter of time before we MUST define a system to all conform to in order to share this real estate.

 

[Doc Al]

Doc Al's error!

In an earlier post in this thread, I made a sloppy statement. I accused poor geistkiesel of assuming that:

(1) The marks on the photo-sensitive strips (caused by the photon emissions) in the moving frame are equally spaced from the point M' (which passed M at the exact moment that the clock at M read t=0): Not true!
(2) That the moving observers detect the photon emissions as happening simultaneously: Not true!

Yes, he made those assumptions--but I had meant to mark the first one as true. That first statement should have read:

(1) The marks on the photo-sensitive strips (caused by the photon emissions) in the moving frame are equally spaced from the point M' (which passed M at the exact moment that the clock at M read t=0): True!​

I had meant to contrast assumption 1 with assumption 2, but I messed it up.

Just for the record, moving frame would record the time and position of the photon emissions (on geistkeisel's photo-sensitive strips) as follows.

In the "stationary" frame (O), photons are emitted at A (x = -L; t = 0) and at B (x = L; t = 0). These flashes are simultaneous in the stationary frame. (I assume that A and B are a distance L from the midpoint M.)

In the moving frame (O'), the photon emissions are recorded on the strips at the following postions and times:

The photon emission from B is recorded at:
$$x' = \gamma L$$
$$t' = -\gamma \frac {vL}{c^2}$$

The photon emission from A is recorded at:
$$x' = -\gamma L$$
$$t' = \gamma \frac {vL}{c^2}$$​

Where, as usual, $\gamma = 1/\sqrt{(1 - \frac{v^2}{c^2})}$.

I apologize for adding any additional confusion to this discussion. (I will add a note to the earlier posts.)

Of course, as I said before, much of this discussion--with geistkiesel's extraneous assumptions--could be avoided by addressing Einstein's actual argument in his train gedanken experiment. Of couse, geistkiesel is unable (or unwilling) to do that.

Note added: Just because that first assumption happens to be true, does not mean that it is justified and can be merely assumed. On the contrary, both assumptions are arbitrary, unjustified, and most importantly extraneous to the simple argument of Einstein. Einstein makes no such assumptions in his demonstration that simultaneity must be relative.

 

[baffledMatt]

baffledMatt can you point out to me by argument or otherwise whe ehttp://frontiernet.net/~geistkiesel/index_files/ Elaborate or brief, just specifics please. Myself, I thought it original which it will probable remain , who would want to confiscate silly physics? Any hint to a specific flaw(S) would be greatfully appreciated. Thanx G

The is no physics in your link! You don't make any calculations so how can I judge it? Then I click on the link to the so-called analysis and there is just a tiny little calculation which has already assumed your point is correct.

I have tried to show you calculations in a way to help you understand how the whole thing works but you reply with:

I ain't gona do your calculations for you. Hurkyl conned me into one of those.

I mean, do you want to be helped? Sorry, but physicists are not philosophers. You have to do calculations to really understand things.

so the easy out is the rationalization for te apparent paradox of the simulatneous emission of photons in a stationary frame turn magically into nonsimultaneity by the mere presen e of an inertial frame other than stationary in the vicinity.

Why is this a paradox? where is the magic? You are the one trying to bring in fairies by demanding something 'is' or 'is not' simultaneous.

No baffledMatt, why do you refer to stationary observers seeing what moving observer do.

This is the crucial point. Imagine I am observing someone moving relative to me. I can watch everything he does, see him measure things with his own measuring devises etc. Now, there must be no discrepancy between the measurements he makes and the ones I observe him to make. It's just that the explanation to why he got the result he did will differ between our frames (I'll say that his measuring stick shrunk wheras he will say that the events were not simultaneous.)

As another example, try this one. Imagine we both have clocks, we know that when they are in the same inertial frame that they tick at exactly the same rate - and they stay like that (they are very good clocks!). Now, you are moving at a velocity v relative to me and at the moment you pass me (ie our x coordinates coincide) we synchronize the clocks - make sure they are telling exactly the same time at that point in time.

Now, you are moving relative to me and so I observe your clock as ticking slightly slower. However, since I am also (relativistically speaking) moving relative to you, you shall also observe my clock as going slower than yours. We each observe each other's clock as going slower. But this is fine, there is no disagreement as such because all I can say is "I observe your clock as slow", I don't know what you might be observing.

Ok, now imagine that you break your clock at time $$T$$ by your clock. This is an event we must both agree on, when your clock had the small hand pointing at $$T$$ you broke it. Now, I observe your clock as ticking slower than mine so my clock reads $$\gamma T$$ at the point when you break it. In fact, as soon as I see your clock stop i stop mine also so we must again both agree on this measurement, we both see $$\gamma T$$ on my clock.

However, now for the 'paradox'. In your frame you were observing my clock as running slower than yours! So you might reason that my clock ought to read $$T / \gamma$$. But it doesn't, it reads $$\gamma T$$ and we can both see this.

Resolve the paradox.

Not quite. As we have an excess of ps-strips, we assure ourselves that whatever shrinking occurs, one strip will be colocated at A and B within a minimum acceptable errror.

But then your establishment of simultaneity will only be accurate to this same error. You say that you will use an excess of strips so this error will in fact be pretty huge. Thus the results will prove nothing.

The numbered ps-strips guarantees the location of mesurements being equal distant from M' established by relected laser measurements.

But how are you performing these measurements?! You are assuming that these are things you can simply measure and will give you your expected result QED. What you are forgetting is that SR is all about measurement! It's a fact that different observers measure different things. But they all observe the same events.

Do you see the subtlety? "The distance between the sources is 2L" is a measurement which observers can disagree on, as is "the emission was simultaneous". But "I measured the distance to be 2L" is an event. All observers will see me make this measurement and observe me getting the result I did. It is only the events which we all agree on.

The frame knows nothing of stationary observers, perceptions or even that an experiment is being conducted.

Yes it does. The frame is trying to measure events which occur in a different frame. I would say that therefore this frame must know a lot about the stationary observer.

Matt

 

[baffledMatt]

in any case in terms of space and relative fixed points, i think it's imperative that 3 dimensional space be charted in order to be utilized.

when living on Earth it's easy to use relative fixed points for reference, we do it intuitively and almost instinctively daily, driving between the lines, walking on the sidewalk, sitting down on a chair instead of missing it and landing on the floor...

in space you're dealing with far less points of reference AND far greater possible speeds, so everything becomes that much bigger of a problem.

seems only a matter of time before we MUST define a system to all conform to in order to share this real estate.

This is then a purely legal issue, but has nothing to do with physics.

Matt

 

[Hurkyl]

to conform to what the rest of the (stationary) universe sees.

The universe of which I'm aware is full of objects traveling every which way, constantly accelerating to different directions, and without any nice, global structure into which everything fits.

So, there are two very problematic things about this idea; why should there be such a thing as a "stationary universe", and how do we tell what it sees?

in any case in terms of space and relative fixed points, i think it's imperative that 3 dimensional space be charted in order to be utilized.

when living on Earth it's easy to use relative fixed points for reference, we do it intuitively and almost instinctively daily, driving between the lines, walking on the sidewalk, sitting down on a chair instead of missing it and landing on the floor...

in space you're dealing with far less points of reference AND far greater possible speeds, so everything becomes that much bigger of a problem.

seems only a matter of time before we MUST define a system to all conform to in order to share this real estate.

And the point is that a reference frame is chosen, and then everything else is defined relative to that.

Relativity does not say that you're not allowed to choose a reference frame as a "standard"; it merely says that any reference frame would suffice.

 

[geistkiesel]

This is then a purely legal issue, but has nothing to do with physics.

Matt

It sounds to me ram2048 is talking about the real estate of space. A need getting our story straight if we are ever going to make some serious ventures into the cosmos.. This is what I read and we all know about perceptions and observations do we not?

 

[geistkiesel]

stationary universes.

The universe of which I'm aware is full of objects traveling every which way, constantly accelerating to different directions, and without any nice, global structure into which everything fits.

So, there are two very problematic things about this idea; why should there be such a thing as a "stationary universe", and how do we tell what it sees?

Why, is a good question, but where and how are more practical questions. For instance the midoint of the surface of an expanding EM sphere is spatially invariant left to its unperturbed own devices.

The midpoint between any two expanding EM spheres invariant and universal. Some stellar objects are (must be) situated such that measurements, using carefully selected the sperical midpoints, as an invariant and absolute location with respect to distant stellar obsject may be exploited.

And the point is that a reference frame is chosen, and then everything else is defined relative to that.

Relativity does not say that you're not allowed to choose a reference frame as a "standard"; it merely says that any reference frame would suffice.

I'm glad Hurkyl said that. I agree, finally, on something.

 

[Hurkyl]

Geistkiesel: here is your experiment drawn as a diagram:

0        0        0
*\       *       /*
* \      *      / *
AZ \     MN    /  BY
AZ  \    MN   /   BY
AZ   \   MN  /    BY
A Z   \  M N/     B Y
A Z    \ M *      B Y
A Z     \M/N      B Y
A  Z     *  N     B  Y
A  Z     M\ N     B  Y
A  Z     M \N     B  Y
A   Z    M  \N    B   Y
A   Z    M   *    B   Y





  Z A    NM     0
  ZA     NM    /*
  ZA     0    / *
  ZA     *   / BY
  0      *  /  BY
  *\    MN /   BY
  * \   MN/   B Y
 AZ  \  M*    B Y
 AZ   \M/N    B Y
 AZ    * N   B  Y
A Z    M\N   B  Y
A Z   M  *   B  Y

Legend:
A, M, B: your A, M, B
Z, N, Y: Your A', M', B'
0: A point where a clock read 0
\, /: photons
*: Multiple things at this point (such as two clocks, or a clock and a photon)

The first diagram is the stationary reference frame. The second diagram is the moving reference frame. Space runs from left to right, and time increases as you go downward.

The first diagram was taken directly from your experimental setup. I placed M and N midway between A/Z and B/Y, and simply drew out the time evolution of the system.

To draw the second diagram, I started with the unambiguous fact that both photons meet precisely when they each meet M. I drew the time evolution backwards, and used the fact that N is midway between Z and Y when M meets N to place where Z and Y should be. I then applied the fact that Z meets A and are both set to zero when the left photon is emitted to finish off the left side of the diagram, and similarly for Y and B. I did, however, have to estimate how N lies relative to M. Any other such choice yields a similar diagram.


Your mistake, as everyone is trying to tell you, is made clear from the diagram. In the moving frame, the clocks are not synchronized; you can see that they are all zero at different times. We see that SR can handle this scenario perfectly well, as long as you don't start with the assumption that synchronization in one frame = synchronization in all frames.




(NOTE: In the second drawing, to keep the diagram small, the lexical distance between A and M is 6 and one-third characters)

 

[geistkiesel]

I still don't see exactly what you mean by this. Could you perhaps give me the dumbed down version of how exactly this is done? It's no good you saying 'using the same physics' since we evidently don't agree on exactly what the physics is!

baffledMatt, whatever the physics used to measure the midpoint in the stationary frame is used in the moving frame.
You can pick you own method.

measure the midpoint of M using some eflected laser scheme relying on time of flight of the beam photons. Ditto for M'.The M' only allows us to guarantee that the number of an an exposed ps-strip near A will have the same number of the exposed ps-strp at B. The scientists calculate shrinking from a known velocity acihieved through repeated acceleration schemes. Approaching A and B from two directions and taking phtographs of the AA' positions, the moving frame can assure itself that wherever the photon expsoes the ps-strip its eflected number at the other end of the frame will be identically exposed. Observers eye-to-eye 1 wave length apart when the photons are emitted must see the same event. If you want to slow down the brain functions in the moving frame ok, by the photons aren't going to play any special games with the moving frame just because it is there. So if brain slowing is the physical ef fect of nonsimultaneity I can understand that.

I can see why you might think that - I remember having exactly these sorts of problems with SR. The way I would always go about solving these problems though was to sit down and just calculate the thing. To illustrate:
Let's make some definitions:
$$x$$ : space coordinate in 'stationary' frame. $$x = 0$$ is the midpoint between A and B, which are located at $$x = -L$$ and $$x = L$$ respectively.

$$v$$ : velocity of the moving observer relative to the stationary observer.

Ok, so now I will calculate what the 'stationary' observer sees. Note I am not going to use any SR as such. We first calculate the location of the point where the 'moving' observer and the light from A coincide. This happens at a time $$t_A$$ after the light is emitted from A and the point where they meet we denote $$d_A$$. By equating the distance the light moves in this time to the distance the moving observer moves we have:

$$vt_A = d_A$$
$$ct_A = d_A + L$$
so
$$\frac{v}{c}(L + d_A) = d_A$$
and
$$d_A = \frac{\frac{v}{c}L}{(1-\frac{v}{c})}$$

Similarly, we can derive $$d_B$$ which is the point where the moving observer meets the light from B
$$vt_B = d_B$$
$$ct_B = (L-d_B)$$
so
$$\frac{v}{c}(L-d_B) = d_B$$
and again
$$d_B = \frac{\frac{v}{c}L}{(1+\frac{v}{c})}$$

Evidently, $$d_A \neq d_B$$ so the photons do not hit the moving observer at the same time/place.



Ok, now repeat the calculation above, but for the moving observer. The only SR result I want you to use is that each observer must observe the same value for the speed of light. If you do this you should see that you need things like time dilation and lack of simultaneity to resolve the observations.

Matt

The problems you used to have were eleiminated using SR. Without SR there is no paradoxes, none. It osunds like a religious convert, really. "My life turned around when I found Godot", or something similar.
Do you want to argue all the experimetnal results proving the postuilates, while I show all the contradictions, exceptions, flaws. You show me my flaws I show you yours. This really isn't a physics discussion, all that much is it?

I know its my fault.

You needn't show this to me I believe it will work as you say. My doing the calculations will not change anything. There is still the matter of the ps-strips being exposed when the photons are emitted. There is no paradox until SR is applied. then we have the big sticker. Mere presence of observers grossly affecting a physical event, by operation of postulate. Is there any language you can use other than that demanded by Sr that can explain the physical manifestation arising from a simultaneous event by operation of the presence eof observers. Can you do this?

If there are no observers on a moving frame observing is there a nonsimultaneous event occurring? I am not asking a question with any implied determinism demanded in the answer, but the inversion of reason and the operation of physical law don't work in SR, at lexst not heer. I discard it.

 

[geistkiesel]

Geistkiesel: here is your experiment drawn as a diagram:

0        0        0
*\       *       /*
* \      *      / *
AZ \     MN    /  BY
AZ  \    MN   /   BY
AZ   \   MN  /    BY
A Z   \  M N/     B Y
A Z    \ M *      B Y
A Z     \M/N      B Y
A  Z     *  N     B  Y
A  Z     M\ N     B  Y
A  Z     M \N     B  Y
A   Z    M  \N    B   Y
A   Z    M   *    B   Y





  Z A    NM     0
  ZA     NM    /*
  ZA     0    / *
  ZA     *   / BY
  0      *  /  BY
  *\    MN /   BY
  * \   MN/   B Y
 AZ  \  M*    B Y
 AZ   \M/N    B Y
 AZ    * N   B  Y
A Z    M\N   B  Y
A Z   M  *   B  Y

Legend:
A, M, B: your A, M, B
Z, N, Y: Your A', M', B'
0: A point where a clock read 0
\, /: photons
*: Multiple things at this point (such as two clocks, or a clock and a photon)

The first diagram is the stationary reference frame. The second diagram is the moving reference frame. Space runs from left to right, and time increases as you go downward.

The first diagram was taken directly from your experimental setup. I placed M and N midway between A/Z and B/Y, and simply drew out the time evolution of the system.

To draw the second diagram, I started with the unambiguous fact that both photons meet precisely when they each meet M. I drew the time evolution backwards, and used the fact that N is midway between Z and Y when M meets N to place where Z and Y should be. I then applied the fact that Z meets A and are both set to zero when the left photon is emitted to finish off the left side of the diagram, and similarly for Y and B. I did, however, have to estimate how N lies relative to M. Any other such choice yields a similar diagram.


Your mistake, as everyone is trying to tell you, is made clear from the diagram. In the moving frame, the clocks are not synchronized; you can see that they are all zero at different times. We see that SR can handle this scenario perfectly well, as long as you don't start with the assumption that synchronization in one frame = synchronization in all frames.




(NOTE: In the second drawing, to keep the diagram small, the lexical distance between A and M is 6 and one-third characters)

You are mistating the thread hypothetical. I have never said the frames were sycnchronized between them. I have stated the opposite. Your statement is a total falsehood, conscious or otherwiise. Your diagram , is a perversion on the original. You are trying to sabotage a thread for your own aggrandizment. Do you get a bonus, or is theprize recognition? Your just another phony, and a liar. You made it up Hurkyl

Take your lexical distances and shove them some place that will do you some good. But all I can predict now, until the last star fades from the sky is that Hurkyl is just another common usless liar and is not to be trusted, ever. I am not answering your crap anymore. Our conversation is over.

 

[russ_watters]

Why don't you cut with the personal attacks and draw out a diagram of your own? Or is intentional obfuscation your goal?

 

[baffledMatt]

The problems you used to have were eleiminated using SR. Without SR there is no paradoxes, none. It osunds like a religious convert, really. "My life turned around when I found Godot", or something similar.
Do you want to argue all the experimetnal results proving the postuilates, while I show all the contradictions, exceptions, flaws. You show me my flaws I show you yours. This really isn't a physics discussion, all that much is it?

The difference between this and religion is that SR is completely logical. You start from the principle of relativity and everything else is deduced from this. So the only thing I got converted to was the principle of relativity - if you want to call that religion then be my guest.

You have not shown us any flaws in SR, only flaws in your interpretation of SR.

There is still the matter of the ps-strips being exposed when the photons are emitted. There is no paradox until SR is applied.

There is no paradox, period! Only gross misunderstandings.

then we have the big sticker. Mere presence of observers grossly affecting a physical event, by operation of postulate. Is there any language you can use other than that demanded by Sr that can explain the physical manifestation arising from a simultaneous event by operation of the presence eof observers. Can you do this?

This depends on your point of view. I do not consider there to be any physical relevance to two things being simultaneous, so the results of SR do not bother me.

Why do you think there is any significance to things being simultaneous? If two events happen at different locations and the same time then there is no physical way the two events are connected - they are outside of each others light cones. So each event has no idea that it is occurring 'at the same time' as another event. Why do you attach so much significance to things which cannot be causally related?

Matt

 

[baffledMatt]

You made it up Hurkyl

Take your lexical distances and shove them some place that will do you some good. But all I can predict now, until the last star fades from the sky is that Hurkyl is just another common usless liar and is not to be trusted, ever. I am not answering your crap anymore. Our conversation is over.

Riiight. All this coming from a guy trying to convince us that his own theory - which is devoid of calculations or even physics for that matter - proves that SR is wrong.

Don't you think it would be better to try and show us exactly why you think Hurkyl is lying, rather than all this immature name calling?

Matt

 

[geistkiesel]

The is no physics in your link! You don't make any calculations so how can I judge it? Then I click on the link to the so-called analysis and there is just a tiny little calculation which has already assumed your point is correct.

I have tried to show you calculations in a way to help you understand how the whole thing works but you reply with:


I mean, do you want to be helped? Sorry, but physicists are not philosophers. You have to do calculations to really understand things.



Why is this a paradox? where is the magic? You are the one trying to bring in fairies by demanding something 'is' or 'is not' simultaneous.



This is the crucial point. Imagine I am observing someone moving relative to me. I can watch everything he does, see him measure things with his own measuring devises etc. Now, there must be no discrepancy between the measurements he makes and the ones I observe him to make. It's just that the explanation to why he got the result he did will differ between our frames (I'll say that his measuring stick shrunk wheras he will say that the events were not simultaneous.)

As another example, try this one. Imagine we both have clocks, we know that when they are in the same inertial frame that they tick at exactly the same rate - and they stay like that (they are very good clocks!). Now, you are moving at a velocity v relative to me and at the moment you pass me (ie our x coordinates coincide) we synchronize the clocks - make sure they are telling exactly the same time at that point in time.

Now, you are moving relative to me and so I observe your clock as ticking slightly slower. However, since I am also (relativistically speaking) moving relative to you, you shall also observe my clock as going slower than yours. We each observe each other's clock as going slower. But this is fine, there is no disagreement as such because all I can say is "I observe your clock as slow", I don't know what you might be observing.

Ok, now imagine that you break your clock at time $$T$$ by your clock. This is an event we must both agree on, when your clock had the small hand pointing at $$T$$ you broke it. Now, I observe your clock as ticking slower than mine so my clock reads $$\gamma T$$ at the point when you break it. In fact, as soon as I see your clock stop i stop mine also so we must again both agree on this measurement, we both see $$\gamma T$$ on my clock.

However, now for the 'paradox'. In your frame you were observing my clock as running slower than yours! So you might reason that my clock ought to read $$T / \gamma$$. But it doesn't, it reads $$\gamma T$$ and we can both see this.

Resolve the paradox.



But then your establishment of simultaneity will only be accurate to this same error. You say that you will use an excess of strips so this error will in fact be pretty huge. Thus the results will prove nothing.



But how are you performing these measurements?! You are assuming that these are things you can simply measure and will give you your expected result QED. What you are forgetting is that SR is all about measurement! It's a fact that different observers measure different things. But they all observe the same events.

Do you see the subtlety? "The distance between the sources is 2L" is a measurement which observers can disagree on, as is "the emission was simultaneous". But "I measured the distance to be 2L" is an event. All observers will see me make this measurement and observe me getting the result I did. It is only the events which we all agree on.



Yes it does. The frame is trying to measure events which occur in a different frame. I would say that therefore this frame must know a lot about the stationary observer.

Matt

why all the talk about clocks we aren' using them? . If the moving frame has a physical entity located a wave length from a stationary entity and these entities are identical ps_strips and a photon is enmitted exactly between them exposing the ls-strips and the same event is occurring on the other end of the frames, the distance betwene the exposed ls-strips is identical at the instant the photons were emitted, lorentz terms notwithstanding.. The exposure, a fraction of a mico second is identically located, like mirror images of each other other. Your SR theory that describe other than this is bogus. Mathematical chicken scratches on a piecxe of paper does not substyitue for physical law and reason. I trust you will always go in peace with your theory.
I realize I have my own perceptions, biases and a personal fault or two, but I don't swlallow SR: it is poison for the mind, the scientific equivalent of fascism.

Explain to me Doc Al and Hurkyl flat out lying and double talking, sonfusin. attemopting to get me of on some wild goose chase. Thee men are your colleagues. I don't fault you for your fdifferences, in fact i don't fault you. You are, to be sure instinctively SR, but I haven't detected the corruption seen in the other two mentioned.

These are "mentors" . I have only seen such peversion and dishnesty in the politics in thios country. I wonder whio is pulling the chain around their necks?

Don't try this at home children, please don't try this at home.

Some say we humans are free thinkers, free to choose our own destiny. Those two mentioned chose the way they think act and work, and corrupt, always to corrupt.
Chroot, pulled a post of mine cruitical to some yoyo named carp or crap. Hedidn't think the post was scientific enough. Jesus christ, Now I cannot edit my own profile or dio any thing except woprk in thi porum. No problem, I just don't appreciate being handkled by a jerk with such insensitivity. The man is arbitary and snot nosed brat.

 

[geistkiesel]

Why don't you cut with the personal attacks and draw out a diagram of your own? Or is intentional obfuscation your goal?

See the first post in the thread. I am not making personal attacks I am reposing obsevations.

Whats the matter,you don't like my threads?

 

[baffledMatt]

If the moving frame has a physical entity located a wave length from a stationary entity and these entities[...]

All the time you are assuming that there are all these physical entities which 'have length X' or 'are at time Y'. How do you know this? How do you know that the distance between A and B is 2L? You measure it of course. But hang on, now you haven't actually determined what this entitie's 'true length' is, you have made a measurement - there is no way you can talk about the 'true length' because there is no measurement free way of determining it. Then what SR tells us is how these measurements will differ between intertial frames.

are identical ps_strips and a photon is enmitted exactly between them exposing the ls-strips and the same event is occurring on the other end of the frames, the distance betwene the exposed ls-strips is identical at the instant the photons were emitted, lorentz terms notwithstanding.. The exposure, a fraction of a mico second is identically located, like mirror images of each other other. Your SR theory that describe other than this is bogus.

You still think that you are determining some true 'real' property using this measurement. What SR tries to tell us is that the only quantities which do in fact have a 'true' property are the invariant ones (such as rest mass).

Until you think very very carefully about the way you are making your measurements you will never understand this.

Mathematical chicken scratches on a piecxe of paper does not substyitue for physical law and reason.

Excuse me, are you a physicist or a philosopher? If you want to talk this way then please move your discussion to the philosophy section of the forum. I'm afraid that mathematics is the only way we know of to build a coherent and consistent model of the world around us.

Your suggestion of using 'physical law and reason' instead of hard mathematics is exactly the kind of thinking that gave us Ptolemy. They reasoned that perfect circles were 'physical law' despite what the mathematics was telling them. Do you really want us to go back to that way of reasoning?

Matt

 

[Doc Al]

last chance

Explain to me Doc Al and Hurkyl flat out lying and double talking, sonfusin. attemopting to get me of on some wild goose chase. Thee men are your colleagues. I don't fault you for your fdifferences, in fact i don't fault you. You are, to be sure instinctively SR, but I haven't detected the corruption seen in the other two mentioned.

Do you even know what the word "lying" means? It means intentionally saying something that isn't true. Can you point to even one post in which Hurkyl or I said something that wasn't true? Never mind intentionally.

These are "mentors" . I have only seen such peversion and dishnesty in the politics in thios country. I wonder whio is pulling the chain around their necks?

You make wild assertions, and when they are questioned all you can do is sputter and fume. In lieu of reason and argument, you make personal attacks. You bring nothing substantive to the table, geistkiestel.

Chroot, pulled a post of mine cruitical to some yoyo named carp or crap. Hedidn't think the post was scientific enough. Jesus christ, Now I cannot edit my own profile or dio any thing except woprk in thi porum. No problem, I just don't appreciate being handkled by a jerk with such insensitivity. The man is arbitary and snot nosed brat.

You spew the same drivel over and over and over again. You've had dozens of chances to make your point. You have always been free to babble away--in TD, where folks can choose to ignore you and you don't disrupt serious discussions.

This is your last chance, geistkiesel. If you care to keep posting here, among adults, then:

No more name calling
No more personal attacks​

 

[geistkiesel]

Riight. All this coming from a guy trying to convince us that his own theory - which is devoid of calculations or even physics for that matter - proves that SR is wrong.

Don't you think it would be better to try and show us exactly why you think Hurkyl is lying, rather than all this immature name calling?

Hurkyl lying: res ipsa loquitor.

Matt

---à motion ----> M' = M
A__________|______|______|______|______B
-------------t1-----t0------t1-------t2
These are the critical positions as follows:

When M’ = M, t0 = 0 and photons emitted from A and B.

The photon from B arrives on moving platform at t1
Photon from A arrives at t2.
Dt = t2 – t1 = 1 unit time.
V = 1 unit tme.
Assume temporarily M’ the midpoint of A and B at t0 = 0.

[Note: baffledMatt: is this assumption what you were telling the world was a rejection of SR from the get go? It was a convenient assumption that is properly tested below? Not quite as you reported it publically, is it, but what the heck, mentor, SR theorist, propagandist, rote robotic believer?].

At t1, when photon from B arrives, the A photon is assumed to be at –t1.
During dt = 1 the photon from A arrives at t2.

During dt = 1 the A photon travel c = t1 + t1 + 1
or rearranging t1 = (c – 1)/2 = k.

If t1 = k the photons emitted simultabneously in both frames. If t1 < k A the A photon emitted first, otherwise B emitted first.

Simple enough? Reasonable enough? fundamental physics.
See the thread for relativistic an observed expansion. similar logic and reason and physics.

 

[geistkiesel]

Do you even know what the word "lying" means? It means intentionally saying something that isn't true. Can you point to even one post in which Hurkyl or I said something that wasn't true? Never mind intentionally.

You make wild assertions, and when they are questioned all you can do is sputter and fume. In lieu of reason and argument, you make personal attacks. You bring nothing substantive to the table, geistkiestel.

You spew the same drivel over and over and over again. You've had dozens of chances to make your point. You have always been free to babble away--in TD, where folks can choose to ignore you and you don't disrupt serious discussions.

This is your last chance, geistkiesel. If you care to keep posting here, among adults, then:

No more name calling
No more personal attacks​

Where is there evidence of maturity and adult behaviour in you people? These men are my colleagues? You are all fired.
You certainly aren't teachers, professors, working to instill curiosity and independence in thinking. Just do it! that is your cfrtyive message. message,
the only message.

I call em like I sees 'em. You want that I hold back, be restrained, give yo a few inches so you can hang some body?

No,.

 

[baffledMatt]

---à motion ----> M' = M
A__________|______|______|______|______B
-------------t1-----t0------t1-------t2
These are the critical positions as follows:

When M’ = M, t0 = 0 and photons emitted from A and B.

The photon from B arrives on moving platform at t1
Photon from A arrives at t2.
Dt = t2 – t1 = 1 unit time.
V = 1 unit tme.
Assume temporarily M’ the midpoint of A and B at t0 = 0.

Do you see that by defining dt = 1 and v = 1 you have already specified all the distances in the problem?

For instance, (using my notation again, note that your t1 and t2 are my $$t_B$$ and $$t_A$$).

$$v (t_B - t_A) = 1 = d_A - d_B = \frac{v}{c} L \left(\frac{1}{1-\frac{v}{c}} - \frac{1}{1+\frac{v}{c}}\right)$$

using the fact that v = 1 and $$t_B - t_A = 1$$.

so:

$$L = \frac{c^2 - 1}{2}$$.

Now we can calculate the location of the photon from A when the photon from B hits M' (i.e. at t1)

$$d = ct1 - L = ct1 - \frac{c^2-1}{2} \neq - ct1$$

hence:

At t1, when photon from B arrives, the A photon is assumed to be at –t1.
During dt = 1 the photon from A arrives at t2.

you are no longer free to make this assumption - it is wrong.

[Note: baffledMatt: is this assumption what you were telling the world was a rejection of SR from the get go? It was a convenient assumption that is properly tested below? Not quite as you reported it publically, is it, but what the heck, mentor, SR theorist, propagandist, rote robotic believer?].

I have no problem with the midpoint as such. The problem is your assumption that the measured distance of A and B from this midpoint will be the same in each frame. So, in each frame they will indeed see M and M' exactly in between A and B, but if you ask each observer to measure the distance between the two they will disagree.

Matt

 

[geistkiesel]

Do you see that by defining dt = 1 and v = 1 you have already specified all the distances in the problem?

I don't think so. 1 is just another symbol. The velocity has to be something I just mae it 1 for convenience. When I make calulation testing t1 I would have to use a eal functional value for he velocity

For instance, (using my notation again, note that your t1 and t2 are my $$t_B$$ and $$t_A$$).

$$v (t_B - t_A) = 1 = d_A - d_B = \frac{v}{c} L \left(\frac{1}{1-\frac{v}{c}} - \frac{1}{1+\frac{v}{c}}\right)$$

using the fact that v = 1 and $$t_B - t_A = 1$$.

so:

$$L = \frac{c^2 - 1}{2}$$.

Now we can calculate the location of the photon from A when the photon from B hits M' (i.e. at t1)

$$d = ct1 - L = ct1 - \frac{c^2-1}{2} \neq - ct1$$

hence:



you are no longer free to make this assumption - it is wrong.

How do you calculate A if you don't know if it was emitted simultaneosusly or when it emitted?


Yes I am free to do this as I haven't made any tests yet. I merely continue with the assumption that when M'=M that the photons were emitted simultaneously, which happens to be what happened in this hypo. I make the same claim against you for arbitrarily inserting SR constraints into the problem.

Continuing wih the assumption that the photons were emitted simultaeously when B is detected thenm M = M' at t = 0 is the midpoint hence -t1 the location of the oncoming A photon. The photon reaches t2 during dt = t2 - t1 = t'" if this is more satisfying and let v = v. The distance c moved to arrive at t2 is c(t") = t1(v) + t1(v) + t"(v) or 2(t1v) = ct" - t"(v) or t1v = (ct" - t"v)/2

or t1 = (ct" - t"v)/2v =(t"/2v)(c -v) = k. Now I ask the same question:
Is t1 = k? If so the photons were emitted simultaeously. If t1 < k, the photon left A first, otherwise the photon left B first.
Why did you use the expressions 1 + v/c amd 1 -v/c? was this necessary?

I got, t1 = (ct" - t"v)/2v =(t"/2v)(c -v)

I have no problem with the midpoint as such. The problem is your assumption that the measured distance of A and B from this midpoint will be the same in each frame. So, in each frame they will indeed see M and M' exactly in between A and B, but if you ask each observer to measure the distance between the two they will disagree.

Look again isn't there only the implication that that M' was the midpoint, not necessaily of A and B but of the wave fronts of the photons emitted at somewhere?
This is the divegence point. I say your injection of SR to properly describe the midpoint assumes SR here, does it not?, and therefore, assume they
would not still maintain the charateristic of having the midpoints M = M'? Once you've done this the problem has been hijacked by SR, perhaps hijacked is a tad strong , but you know what I mean.

I think if you looked again I really only assuned the M =M' was the midpoint of the photon wave front, and there is no inference of the A and B location,
Maintaining steadfastly in the moving frame I get an expression that looks the same as yours up to a point.

Well then I just won't ask each observer to measure anything as I see no benefit as this time for anything to measure. Why and what do you measure here?
Now when my expession is rolled out it is aleady to be asked to undergo the test: t1 ?= (t"/2v).(c - v) .

I suspect you don't like the c- v?

 

[baffledMatt]

I don't think so. 1 is just another symbol. The velocity has to be something I just mae it 1 for convenience. When I make calulation testing t1 I would have to use a eal functional value for he velocity

No, by specifying v=1 and dt=1 you are setting the time and velocity. distance is velocity multiplied by time, ergo you have also specified distance. Ipso facto your subsequent analysis is erroneous.

How do you calculate A if you don't know if it was emitted simultaneosusly or when it emitted?

I know that they were simultaneous in the stationary frame because that's how the situation was set up. However, this is not to say that they will be simultaneous in the moving frame.

I make the same claim against you for arbitrarily inserting SR constraints into the problem.

I am not inserting SR arbitrarily. I am using results which follow directly from the principle of relativity. I could just as well go through each and every calculation from first principles, but I will still get the same result.

The photon reaches t2 during dt = t2 - t1 = t'" if this is more satisfying and let v = v. The distance c moved to arrive at t2 is c(t") = t1(v) + t1(v) + t"(v) or 2(t1v) = ct" - t"(v) or t1v = (ct" - t"v)/2

Ok, what have you done here. You have equated ct'', which is the distance a photon travels in time t'' to (t1 + t1 + t'')*v, which is the distance the observer moving at velocity v moves in time t1 + t2. What makes you think that these two quantities are equal?! I'm afraid that what you have done here is complete and utter nonsense.

I suspect you don't like the c- v?

This is the least of my concerns!

Matt

 

[baffledMatt]

I personally think I am something rare and special [...]

Well, we can't help what you personally think about yourself, but that still doesn't warrant all this hostility.

Besides, if they have been rude to you surely it would be a better show of character not to stoop to their level?

Matt

 

[geistkiesel]

Well, we can't help what you personally think about yourself, but that still doesn't warrant all this hostility.

Besides, if they have been rude to you surely it would be a better show of character not to stoop to their level?

Matt

whta else do you want

 

[Hurkyl]

Some say we humans are free thinkers

I always find it terribly ironic that those who talk most about "free thought" tend to be those who are least likely to consider that others might actually have a good reason to disagree.

If t1 = k the photons emitted simultabneously in both frames.

I find it curious that you conclude this, because everything up until this statement has involved only a single frame...

 

[baffledMatt]

whta else do you want

Politeness would be a good start.

Being civil is not a sign of weakness. And if you stop with all the vulgarity people might be willing to listen more carefully to what you are saying.

Matt

 

[geistkiesel]

I always find it terribly ironic that those who talk most about "free thought" tend to be those who are least likely to consider that others might actually have a good reason to disagree.




I find it curious that you conclude this, because everything up until this statement has involved only a single frame...

t1 = (ct" - t"v)/2v =(t"/2v)(c -v) = k. Yes it involved calculation in the moving frame. To get to this point an assumpotion was made that the photons were emitted simultaneously in the moving frame. Having gotten this far the next step is to test the assumption. If the mesured t1 = k the assumpion was true, if t`< k then the A photon was emitted first, otherwise the B photon was emitted first. All the calculations were performed in the moving frame. So what does "I find it curious ..." mean? Do you find the math, the assumptions, the conclusion erroenous, or am I supposed to guess what you mean? If you have a specific objection I suggest you state it as clearly as you are able so we don't have to read each others minds.
Thee were nio SR assumptions expresssed in the derivaion of the expression. as no SR implications were presented in the problem, without the presentaion being in a contrived mode.