Lorentz Vs. Einstein Who Wins?

[Dale]

the following I copied from a website somewhere

Seems like the website is confused.

 

[John232]

The translations may vary in terms of the text, but the Lorentz transforms are still the Lorentz transforms. The mathematical formulas are surely the same. I don't know where your confusion on this point stems from, but I hope it is clear by now.

I think I do remember them saying that that version of relativity was only used in that version of introductory text. I don't know why they would do that and stir up all this confusion. So are you saying that τ in Einsteins paper was only used to test the experiments and not the reciprocal of the other time clock equations? I think τ in that paper is right, and I saw another version linked to this website four years ago where the other version was in it but their was an issue about translation as well, but as I was saying τ does not equal 1/Δt' as the equations can't be substituded into each other and get the same thing on each side of the equation.

 

[Dale]

You are focusing too much on the symbol used. In OEMB Einstein used tau for coordinate time in system k. Typically we would use t' today. The important part is the set of equations at the very end of section 3, which is the Lorentz transform.

Compare the end of section 3:
http://www.fourmilab.ch/etexts/einstein/specrel/www/

with:
http://en.wikipedia.org/wiki/Lorentz_transformation#Boost_in_the_x-direction

They are the same.

 

[salvestrom]

I think I do remember them saying that that version of relativity was only used in that version of introductory text. I don't know why they would do that and stir up all this confusion. So are you saying that τ in Einsteins paper was only used to test the experiments and not the reciprocal of the other time clock equations? I think τ in that paper is right, and I saw another version linked to this website four years ago where the other version was in it but their was an issue about translation as well, but as I was saying τ does not equal 1/Δt' as the equations can't be substituded into each other and get the same thing on each side of the equation.

The only reciprocal is from the calculation of gamma. 1/sqr rt(1-(v/c)^2). Since t' = t/gamma, then the equation can be written as t.sqr rt(1-(v/c)^2)/1. The division by 1 is obviously redundant and is eliminated.

You need to understand that there are equations for telling us all sorts of things. t'=t/y tells us the time in the primed frame at any given time in the unprimed frame. t=yt' tells us the time in the unprimed frame for any given t', but, and this is very important, it does so from the point of view of the unprimed observer. To know the time in the unprimed frame from the point of view of the primed observer you must use t=t'/y.

The line I have highlighted is absolutely true... because 1/t' is a meaningless calculation that will only tell you something if you use a value of t' that happens to be the reciprocal of the corresponding gamma. Proper time is always 1/1 to the observer that is stationary relative to the clock being observered. When you want to peek at someone else's moving clock you either multiply or divide by gamma. Again, 1/t' is of no meaning in Special relativity, or Lorentz Ether Theory - at least by itself. You'd need to throw in the sqr rt stuff to use it.

 

[NotAName]

Seems like the website is confused.

Lol, probably. Mind elaborating?

 

[John232]

√

The only reciprocal is from the calculation of gamma. 1/sqr rt(1-(v/c)^2). Since t' = t/gamma, then the equation can be written as t.sqr rt(1-(v/c)^2)/1. The division by 1 is obviously redundant and is eliminated.

You need to understand that there are equations for telling us all sorts of things. t'=yt tells us the time in the primed frame at any given time in the unprimed frame. t=yt' tells us the time in the unprimed frame for any given t', but, and this is very important, it does so from the point of view of the unprimed observer. To know the time in the unprimed frame from the point of view of the primed observer you must use t=t'/y.

I don't know what you mean by primed and unprimed frame.
I will do an example.

τ=Δt√(1-v^2/c^2) This gives the proper time in terms of tau. Say an object travels at half the speed of light for 2 secounds.
τ=2√(1-0.5^2/c^2) = 2√3/4

Now solve using the other equation given Δt=Δt/√(1-v^2/c^2)=2/√3/4
then you take the reciprocal and get (√3/4)/2

2√3/4 ≠ (√3/4)/2

 

[m4r35n357]

These two equations do not equal each other. The time variebles are in different locations in the equation? Simply a typo, or is one of the equations more right than another?

The equations DO equal each other; from this and your other comments here it is clear to me that you are either trolling, or WAY out of your depth. In case of the latter I would recommend that you:
1. Learn some mathematics
2. Learn some physics
3. Learn some humility

 

[harrylin]

Lorents and Einstein both made equations to describe time dialation.[..]

Not sure if this has been mentioned, but the expression "Lorentz transformations" refers to the transformations that followed from Lorentz's new theory. Einstein found the same transformations, and thus they were also sometimes called the "transformations of Einstein and Lorentz". These are the transformations that are used in special relativity.

 

[harrylin]

I have seen different translations of this paper and none of them looked exactly the same. I find it odd that it is translated for the proper time, but I have seen versions that where not. The introductory version in first year physics text doesn't say his equation was written this way.

These equations have been written in different ways; that is only a problem if you don't understand their meaning. If you check out other equations such as Maxwell's, you will find the same "problem".

 

[Dale]

Lol, probably. Mind elaborating?

The last two paragraphs you quoted are incorrect:

According to Lorentz, a beam of light traveling to that distant star would take .5 years in the stationary frame and would take .433 years in the moving frame.

According to Einstein, a beam of light traveling to that distant star would take .5 years in the stationary frame and would take .433 years in the moving frame less the movement of the distant star for a total of .288 years.

Lorentz and Einstein use the same math and therefore make the same experimental predictions in all cases.

 

[John232]

I should have made the first post in this forum use Einsteins proper time, it looks like the trolls didn't read all the post of the thread that I made, calling someone a troll doesn't grant you immunity, a troll would just jump into someones thread and make erroneous comments, that doesn't give an answer to anything. If you can't explain something yourself I don't want to hear it. This thread comes from my post #36, if you don't like post #36 then you just don't like this thread.

 

[Dale]

√

I don't know what you mean by primed and unprimed frame.
I will do an example.

τ=Δt√(1-v^2/c^2) This gives the proper time in terms of tau. Say an object travels at half the speed of light for 2 secounds.
τ=2√(1-0.5^2/c^2) = 2√3/4

Now solve using the other equation given Δt=Δt/√(1-v^2/c^2)=2/√3/4
then you take the reciprocal and get (√3/4)/2

2√3/4 ≠ (√3/4)/2

That is why you should always just stick with the Lorentz transform and never use the simplified time dilation equation. If the simplified equation is appropriate then it automatically drops out, and if it is not appropriate then the Lorentz transform still holds.

 

[salvestrom]

I don't know what you mean by primed and unprimed frame.
I will do an example.

τ=Δt√(1-v^2/c^2) This gives the proper time in terms of tau. Say an object travels at half the speed of light for 2 secounds.
τ=2√(1-0.5^2/c^2) = 2√3/4

Now solve using the other equation given Δt=Δt/√(1-v^2/c^2)=2/√3/4
then you take the reciprocal and get (√3/4)/2

2√3/4 ≠ (√3/4)/2

From the top:

You should go find out what they mean then reread my post 34.

Tau IS proper time.

The two equations are not the same thing. They tell you two different pieces of information.

The first equation will tell you the time in the primed frame - the one with the ' - at the same instant as the time - t - that you use.

The second equation tells you the time in the unprimed frame, for any give t', but only from the point of view of the unprimed frame. Until you understand what this sentence means any facts, figures and examples you give will not help you. Noone uses this equation because it's a back to front way to work anything out. You use t'=t/y OR t=t'/y. These equations are for working out the values of t and t' at any particular instant in one frame of reference, from the other frames point of view. They are useful for calculating the placement of various lines on Minkowski diagrams and determining the time gained or lost by a clock used in time dilation experiments.

Finally, the line I highlight bold: You take the reciprocal while calculating gamma, not afterward when using gamma to adjust t'.

Just to help you along, the primed frame is usually the one being treated as moving and the unprimed frame is being treated as stationary. The difference is denoted by '.

 

[NotAName]

The last two paragraphs you quoted are incorrect: Lorentz and Einstein use the same math and therefore make the same experimental predictions in all cases.

No they aren't incorrect. It's the "ether" that is incorrect(problem between the theories). Use the same math but have one preferred frame and you lose some motion.

IE: In SR, Light travels C in both frames. In Lorentz Ether Theory, one of the frames is unaffected by motion or more precisely, in the moving frame, light is slowed by traveling upstream in the ether wind.

It's just a matter of the application of the math. Lorentz originally believed the math was an illusion one experiences in a frame moving with respect to the universal frame of the ether.

So basically some effects are removed if you are in the "true" rest frame. Or another way to see it is that light sort of "travels along with" each frame in SR. It all comes out in the math but in from an old ether theory perspective, the light traveling along with each observer is additive.

The removal of ether and the interesting revelations about the nature of space-time that came along, is what made SR the superior theory otherwise we would have no need of Einstein and would still be using Lorentz's theory instead of just his calculations.

 

[John232]

From the top:

You should go find out what they mean then reread my post 34.

Tau IS proper time.

The two equations are not the same thing. They tell you two different pieces of information.

The first equation will tell you the time in the primed frame - the one with the ' - at the same instant as the time - t - that you use.

The second equation tells you the time in the unprimed frame, for any give t', but only from the point of view of the unprimed frame. Until you understand what this sentence means any facts, figures and examples you give will not help you. Noone uses this equation because it's a back to front way to work anything out. You use t'=t/y OR t=t'/y. These equations are for working out the values of t and t' at any particular instant in one frame of reference, from the other frames point of view. They are useful for calculating the placement of various lines on Minkowski diagrams and determining the time gained or lost by a clock used in time dilation experiments.

Finally, the line I highlight bold: You take the reciprocal while calculating gamma, not afterward when using gamma to adjust t'.

Just to help you along, the primed frame is usually the one being treated as moving and the unprimed frame is being treated as stationary. The difference is denoted by '.

Both equations are supposed to describe both frames. t' is time prime, the moveing frame, t is time in another frame in constant motion. Switching between the two equations you gave sounds insane, they would never work together in a mathmatical setup. I think Dalespam got it when he said you shouldn't use Δt in the lightclock example. I don't think Δt would give you correct values in either frame. It was the simplified equation so they didn't dirty it up with values that include the actual size of the clock and determining its frequency. To use Δt you would have to find the frequency of the clock. If you calaculated the number of times the clock ticked and then divided it by the frequency of the clock you would get the proper time. T'=1/f and Δt'≠f. But you can find tau by just assigning the variebles differently in the same proof since τ=t' you just have to say the observer in motion measures his clock to go straight up and down with t', then the number of times it ticked doesn't matter it would only matter what time he used to measure something to go a certain distance. I don't get why they describes the equations that way to begin with, even if it doesn't change anything I think it just confuses things and is just fail. I would have to say that is how it is taught to solve both of those equations. If you can't take my word for it ask a teacher who does this subject. That is how they will solve it. If you don't take the reciprocal of Δt' you would end up getting the wrong answer that is a greater amount of time than the other observer that uses Δt.

 

[Dale]

No they aren't incorrect.

Yes, they are incorrect. LET and SR make all of the same predictions on the outcome of any possible experiment.

 

[NotAName]

Yes, they are incorrect. LET and SR make all of the same predictions on the outcome of any possible experiment.

That is primarily true, except not about arrival time of light in this one particular instance.

The motion is already part of the equation and therefore adding the motion of the distant planet would be a duplication in the context of LET. In LET, light travels in a medium that is stationary. Because the light travels in the stationary frame alone, it will take .5 years to arrive in the stationary frame and .433 years in the moving frame, period.

The moving frame still cannot detect its own motion, believes light travels at C wrt their frame and believes the distance between them is closing at .5C but measures the distance with different units and time elapses differently and therefore light speed in units per second is also different.

It's just a history thing... you'll probably continue to disagree and insist both theories are the same but all I'm arguing is that ether guides and limits the motion of light with respect to its own frame regardless of observers and their perspectives. Nobody argues that the difference between the theories is ether. That's all I'm saying. It's exactly the same transformation just interpreted differently.

 

[Dale]

That is primarily true, except not about arrival time of light in this one particular instance.

The motion is already part of the equation and therefore adding the motion of the distant planet would be a duplication in the context of LET. In LET, light travels in a medium that is stationary. Because the light travels in the stationary frame alone, it will take .5 years to arrive in the stationary frame and .433 years in the moving frame, period.

This is incorrect. LET uses the exact same formula to transform coordinates from one frame to the other as SR, the Lorentz transform. In this case, the Lorentz transform of (t,x)=(.5,.5) is (t',x')=(.289,.289). There is no addition of motion of the distant planet nor anything else, it is simply a straight transformation of coordinates which applies for both theories. The light ray takes .289 years in the moving frame in both LET and SR.

I'm arguing is that ether guides and limits the motion of light with respect to its own frame regardless of observers and their perspectives. Nobody argues that the difference between the theories is ether. That's all I'm saying. It's exactly the same transformation just interpreted differently.

What you are saying is self-contradictory. If both theories use the same transformation to change between frames then they must get the same answer as to how long the light takes in any given frame. If they get different answers then they are not using the same transform. You cannot have it both ways, it is self contradictory to claim that they can both use the same transform between frames and yet disagree as to how long it takes. The amount of time it takes is completely determined by the transform.

Look, the website you quoted is simply wrong. If that is the source of your learning about SR then it is not surprising that you are confused. I would recommend looking elsewhere for your information.

 

[salvestrom]

Both equations are supposed to describe both frames. t' is time prime, the moveing frame, t is time in another frame in constant motion. Switching between the two equations you gave sounds insane, they would never work together in a mathmatical setup. I think Dalespam got it when he said you shouldn't use Δt in the lightclock example. I don't think Δt would give you correct values in either frame. It was the simplified equation so they didn't dirty it up with values that include the actual size of the clock and determining its frequency. To use Δt you would have to find the frequency of the clock. If you calaculated the number of times the clock ticked and then divided it by the frequency of the clock you would get the proper time. T'=1/f and Δt'≠f. But you can find tau by just assigning the variebles differently in the same proof since τ=t' you just have to say the observer in motion measures his clock to go straight up and down with t', then the number of times it ticked doesn't matter it would only matter what time he used to measure something to go a certain distance. I don't get why they describes the equations that way to begin with, even if it doesn't change anything I think it just confuses things and is just fail. I would have to say that is how it is taught to solve both of those equations. If you can't take my word for it ask a teacher who does this subject. That is how they will solve it. If you don't take the reciprocal of Δt' you would end up getting the wrong answer that is a greater amount of time than the other observer that uses Δt.

When c=300000 and v=200000 then y=1.342 and 1/y=0.745 (this the sqr rt(1-(v/c)^2).

When the time for the unprimed observer is 1s, then the time for the primed observer, according to the unprimed observer at that instant, is t'=t√1-(v/c)^2. t'=0.745. When t' is 1s then the time in the unprimed frame, according to the primed observer is t=t'√1-(v/c)^2. Which is t=0.745. This provides the symmetry of what each observer sees of the others frame of reference.

Your second equation, t=t'y functions like this:

t'=1s then t=1.342s. The value of t here is what the unprimed observer sees on his own clock when that same unprimed observer checks the primed observers clock.

If t=t'y then t/y=t'. t/y is equal to (t√1-(v/c)^2)/1 with the division by 1 being redundant and ignored. Essentially, your two equations go back and forth between the two frames, one telling you the time in t' according to the unprimed observer using his clock time as the starting point, while the other will tell you the time in t, according to the unprimed observer, using the primed observers clock as a starting point.

This is undoubtedly clear as mud. But if nothing else understand this: your two equations both only provide you with information from the point of view of the unprimed observer. When it is 1 second for him, it is 0.745s for the primed observer, but is most definitely never the case that when it is 0.745s on the primed observers clock that the unprimed clock reads 1s (it actually will show 0.556, taken from t/y)

 

[John232]

This is undoubtedly clear as mud. But if nothing else understand this: your two equations both only provide you with information from the point of view of the unprimed observer. When it is 1 second for him, it is 0.745s for the primed observer, but is most definitely never the case that when it is 0.745s on the primed observers clock that the unprimed clock reads 1s (it actually will show 0.556, taken from t/y)

This is like saying that S'=S, but they don't. The whole reason of assigning the prime symbol is to indicate that it is a different frame of reference. Then any value in frame S' would follow to have a prime symbol and any value in S frame would not have a prime symbol. You would then only need to use one of the equations, because it gives the relation to the other frame. An equation should work the same forwards and backwards, if you have 1 sec for a value of time you will only get one answer, if you put that same answer into the other value you should get 1 sec for the first value. Neither frame has to be at rest, they can both be moving but regardless of that the prime frame should never exchange values to the unprimed frame. You can say that the unprimed frame is also in constant motion it doesn't have to read 1 sec for time, then you would only use the equation τ=t√(1-v^2/c^2), where τ is the proper time in the primed frame, then to convert to the unprimed frame you would only need to put in the same value for τ.

The primed frame can't assume values that is unprimed because then it would no longer be a valid frame. So then you can't assume that t<t', you would have to know that t>t' when you assigned the two frames.

 

[salvestrom]

This is like saying that S'=S, but they don't. The whole reason of assigning the prime symbol is to indicate that it is a different frame of reference. Then any value in frame S' would follow to have a prime symbol and any value in S frame would not have a prime symbol. You would then only need to use one of the equations, because it gives the relation to the other frame. An equation should work the same forwards and backwards, if you have 1 sec for a value of time you will only get one answer, if you put that same answer into the other value you should get 1 sec for the first value. Neither frame has to be at rest, they can both be moving but regardless of that the prime frame should never exchange values to the unprimed frame. You can say that the unprimed frame is also in constant motion it doesn't have to read 1 sec for time, then you would only use the equation τ=t√(1-v^2/c^2), where τ is the proper time in the primed frame, then to convert to the unprimed frame you would only need to put in the same value for τ.

The primed frame can't assume values that is unprimed because then it would no longer be a valid frame. So then you can't assume that t<t', you would have to know that t>t' when you assigned the two frames.

t>t' and t'>t. This is the whole point of SR and Lorentz symmetry: t>t' for the unprimed observer and t'>t for the primed observer. Each sees the other's time as passing slower. The symmetry only disappears when clocks get reunited, and even then it depends on how they are brought back together.

The ' is for distinguishing frames. You may find it useful to consider that the frame which is assigned the ' is arbitrary and any given observer will usually, quite biasedly, assign themselves as the unprimed frame.

EDit: Also, using your statement "it's like saying S'=S", what I'm actually saying is: "the difference in time as seen by S' is the same as the difference in time seen by S".

http://img513.imageshack.us/img513/43/spacetimediagram066c.png

I apologise for the scruffyness of the image. The blue lines are constant time for t' and the black lines constant time for t. Note where these lines transect the x=0 line (thick black upright bar) and x'=0 line (thick green, leaning bar). Ignore the mirad red lines that I failed to edit out.

 

[John232]

EDit: Also, using your statement "it's like saying S'=S", what I'm actually saying is: "the difference in time as seen by S' is the same as the difference in time seen by S".

They keep accelerating those particles at the LHC to speeds close to the speed of light, and I don't feel any fatter...

It is hard to imagine a universe that every object can assume the same value S, when they have a corrilated difference from each other to another value S'. It would be like saying that everything experiences an alternate frame that does not go by the same values as it was observed to use. I don't think either theory predicts parrallel universes. I think in order to assume that a value in S' would be equal to a value in S, would have to assume that both frames where always in constant motion to the other frame infinitly into the past. Big Bang Theory predicts acceleration from a moment in the past. The acceleration of galaxies with dark energy has not shown any signs of the effects of SR or LET, but I think it is possible that every object in the universe has accelerated noticeably at some time from one another at some time in the past. Then that assumption wouldn't be valid for any frame in Big Bang Theory.

 

[GeorgeDishman]

I apologise for the scruffyness of the image.

You might find this applet useful:

http://www.reagenix.com/personal/sci/space_time/test.html

 

[salvestrom]

They keep accelerating those particles at the LHC to speeds close to the speed of light, and I don't feel any fatter...

It is hard to imagine a universe that every object can assume the same value S, when they have a corrilated difference from each other to another value S'. It would be like saying that everything experiences an alternate frame that does not go by the same values as it was observed to use. I don't think either theory predicts parrallel universes. I think in order to assume that a value in S' would be equal to a value in S, would have to assume that both frames where always in constant motion to the other frame infinitly into the past. Big Bang Theory predicts acceleration from a moment in the past. The acceleration of galaxies with dark energy has not shown any signs of the effects of SR or LET, but I think it is possible that every object in the universe has accelerated noticeably at some time from one another at some time in the past. Then that assumption wouldn't be valid for any frame in Big Bang Theory.

The value of S never equals S' at the same instant. Only the observered relationship is equal. For v=2/3 c, when S is 1, S' is 0.746 (according to observer S). When S' is 1s, S is 0.746 (according to observer S'). You have to pay attention to who is looking at what. And when.

Galaxies, the Big Bang and parrallel universes need not to be invoked to explain your original question. Or acceleration. These equations are for observers and objects traveling at a constant velocity.

 

[John232]

The value of S never equals S' at the same instant. Only the observered relationship is equal. For v=2/3 c, when S is 1, S' is 0.746 (according to observer S). When S' is 1s, S is 0.746 (according to observer S'). You have to pay attention to who is looking at what. And when.

Galaxies, the Big Bang and parrallel universes need not to be invoked to explain your original question. Or acceleration. These equations are for observers and objects traveling at a constant velocity.

If S≠S' and then S' says that it is S, then S' would be wrong. You would then have to say that there are two different S and two different S' where S≠S in each case. The only way S≠S is if they do not exist in the same equation. The only way you can find that S≠S is if they are two different frames from two different calculations that do not coexist with each other, hence the parrallel universes. I am saying that the transfer to constant motion to acceleration doesn't imply a parrallel universe, t will always be greater than t' unless they where always in constant motion relative to each other infinitly into the past. Then to find out if objects did in fact travel at costant motion relative to each other infinitely into the past you would then have to look into Big Bang Theory. I find it odd that galaxies don't obey SR or LET since they accelerate, but there is no explanation for why this occors or why dark energy affects them this way. Or, there might exist an easy explanation for this problem.

 

[GeorgeDishman]

Or, there might exist an easy explanation for this problem.

There is, SR is a simplified version of GR which is adequate when gravity is negligible in a particular experiment. When you start looking at large scales in cosmology, the speeds that go into SR are often small but gravity governs the overall structure so you have to upgrade to GR.

The most obvious effect of relativity is the cosmological redshift of distant sources, the relativistic Doppler shift due to local speeds is additional to that.

 

[John232]

There is, SR is a simplified version of GR which is adequate when gravity is negligible in a particular experiment. When you start looking at large scales in cosmology, the speeds that go into SR are often small but gravity governs the overall structure so you have to upgrade to GR.

The most obvious effect of relativity is the cosmological redshift of distant sources, the relativistic Doppler shift due to local speeds is additional to that.

Why does GR say that values between galaxies in SR are small?

 

[salvestrom]

If S≠S' and then S' says that it is S, then S' would be wrong. You would then have to say that there are two different S and two different S' where S≠S in each case. The only way S≠S is if they do not exist in the same equation. The only way you can find that S≠S is if they are two different frames from two different calculations that do not coexist with each other, hence the parrallel universes. I am saying that the transfer to constant motion to acceleration doesn't imply a parrallel universe, t will always be greater than t' unless they where always in constant motion relative to each other infinitly into the past. Then to find out if objects did in fact travel at costant motion relative to each other infinitely into the past you would then have to look into Big Bang Theory. I find it odd that galaxies don't obey SR or LET since they accelerate, but there is no explanation for why this occors or why dark energy affects them this way. Or, there might exist an easy explanation for this problem.

The equations are for working out what observers would observe during the period of constant "relative" velocity. During this period both observers describe the other as moving slower through time and having contracted length.

I have no reason, at this time, to consider that absolute velocity and therefore absolute rest do not exist, but I also understand that it is completely impossible to determine if you have attained either. This for me is probably part of why the twin "paradox" occurs, a situation in which one clock does indeed, after the fact, show less time than the other, as you seem to keep insisting they should. But we are not discussing such situations. We are discussing "in the moment" observations of what two observers, seperating at some relative velocity, record of each others frames of reference. Which is exactly the same thing: time is slower for the other fella and his rulers are all foreshortened.

I've basically told you the same thing a dozen times. There's no other way to say it, that I have, so I figure I'm done. Gl with your investigation.

 

[Hurkyl]

There is no way to distinguish Lorentz ether theory and special relativity experimentally because the two formulations will always predict the same results for any experiment. The two formulations only differ in the underlying assumptions used to arrive at the same predictions.
...
One answer is that the underlying assumptions of special relativity are much cleaner, much less ad hoc than those of LET...

I actually prefer to treat the issue by ignoring all issues of pedagogy, and observe that LET is simply equal to SR + Choice of reference frame. And all physical observables are independent of the choice of reference frame. (e.g. by a symmetry argument)

 

[John232]

I've basically told you the same thing a dozen times. There's no other way to say it, that I have, so I figure I'm done. Gl with your investigation.

A paradox is just that, something that can't be solved, if it could it would no longer be a paradox. There is no way to know what twin is really younger as they have traveled only in constant motion relative to each other.

There is no way to know if the true nature of the universe is the same as the nature of mathematics itself. It is always implied that a coordinate plane goes on forever in each direction. There is no way to know for sure if the universe does the same thing. I don't see how a coordinate plane can be described to exist in two different states in terms of S, if S is the whole coordinate plane itself or universe. There would have to be another set of coordinate planes in order to describe it. Maybe the universe did behave this way in the beginning, and the expansion of the galaxies themselves is evidence of that, but once things started accelerating from then on, objects would no longer behave as though they expereinced being at rest as being equally valid. One frame would have to admit its acceleration and would know it could no longer be equal to S. Hopefully, there are not S primes that descibe myself traveling relative to the LHC being smashed to the floor before I even knew what happened.

In theoretical physics it has also been shown that mathmatically a coordinate system can expand at a constant rate in a higher dimension, and then show signs of acceleration in a lower number of dimensions. It could mean that our universe is traveling at a constant speed in a higher dimension. Like the universe is a balloon example, you could blow it up at a constant rate but then two points on the balloon would look to accelerate from each other. But, if constant motion did exist in the past forever and then every object observed a coordinate frame that was different than the coordinate frame they where observered to use, there would have to be two coordinate frames to describe each of these situations. These coordinate frames could no longer have any relation to each other, even after they accelerate from each other. You couldn't use S and S prime to interchange values from each frame from their given equations they would always result into two different discriptions of things that happened. If S prime said that its values of being just S was equally valid, then it would create a different discription of reality. It would be as though each frame was saying that the other frame had accelerated, but it didn't.


It works out this way if you just go by their proper times and not the observed times things happen. They can't have two equally valid proper times. If they had two equally valid proper times it would be like saying there are parrallel universes.