Lorentz Vs. Einstein Who Wins?

[John232]

Lorents and Einstein both made equations to describe time dialation. One of these equations was used in quantum mechanics, the other equation was then used in cosomology. These equations then made two new areas in physics. One describes one area of physics and the other describes an area of physics, but neither can be used to describe the other. This has raised a lot of questions to which one may be wrong, or what one is more right than the other. Then I noticed something, these two equations are written to different ways but do not equal each other.

I have seen Einsteins equation written as Δt'=Δt/√(1-v^2/c^2)

I have seen the Lorentz equation written as Δt=Δt'γ
where γ=1/√(1-v^2/c^2)

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?

Then if there are in fact the same equation that brings about two branches of science create mathematics that does not work with each other?

 

[elfmotat]

Lorentz Ether Theory is physically equivalent to Special Relativity. Perhaps you'd care to explain what Δt and Δt' are supposed to represent in the equations you posted.

Lorents and Einstein both made equations to describe time dialation. One of these equations was used in quantum mechanics, the other equation was then used in cosomology. These equations then made two new areas in physics. One describes one area of physics and the other describes an area of physics, but neither can be used to describe the other.

None of this is true.

 

[John232]

ΔWell in Einsteins relativity Δt is the change in time of an observer at rest, and Δt' is the change in time of an object in motion. You get a larger value for dialated time for the object in motion so then you have to take the inverse of the answer to find the proper time.

In Lorentz transformation Δt is the proper time and Δt' is the proper time of a particle in motion. There is no need to take an inverse. But the inverse of one equation is not equal to the inverse of the other equation.

 

[John232]

None of this is true.

I wish it wasn't.

 

[elfmotat]

ΔWell in Einsteins relativity Δt is the change in time of an observer at rest, and Δt' is the change in time of an object in motion. You get a larger value for dialated time for the object in motion so then you have to take the inverse of the answer to find the proper time.

In Lorentz transformation Δt is the proper time and Δt' is the proper time of a particle in motion. There is no need to take an inverse. But the inverse of one equation is not equal to the inverse of the other equation.

Like I said, LET is equivalent to SR; they do not make different predictions. The discrepancy is coming from you misinterpreting things.

Say you're at rest in some reference frame and there's some particle traveling at velocity v down your x-axis. If you measure some time Δt separating two events the particle passes through, then the particle will measure Δt' = Δt√(1-v^2/c^2). LET and SR do not differ on this.

 

[elfmotat]

I wish it wasn't.

It isn't. There's no need for wishes.

 

[John232]

Like I said, LET is equivalent to SR; they do not make different predictions. The discrepancy is coming from you misinterpreting things.

Say you're at rest in some reference frame and there's some particle traveling at velocity v down your x-axis. If you measure some time Δt separating two events the particle passes through, then the particle will measure Δt' = Δt√(1-v^2/c^2). LET and SR do not differ on this.

Put a value into that and then put the same value into the other equation and take the inverse and see what you get, I dare you...

 

[elfmotat]

Put a value into that and then put the same value into the other equation and take the inverse and see what you get, I dare you...

You're not listening to me. The equation Δt' = Δt√(1-v^2/c^2) is the same prediction made by BOTH SR AND LET.

You can call the time you measure Δt' and the particle time Δt if you want, and then the other equation is correct. This doesn't change the meaning.

If you're here because you're confused and want to learn, great. You're coming off as arrogant and argumentative, and you continue to assert your blatantly false ideas.

 

[John232]

http://en.wikipedia.org/wiki/Time_dilation

Just let it roll over for another hundred years when someone learns to do algebra. The time variebles are reversed in this enyclopedia even, in no way shape or form does takeing the inverse exchange variebles in an equation. Nothing in either of these equations can be substituted from anything else in the other. It is even written that way in textbooks that introduce relativity in physics 101. It doesn't make the same prediction, it would be like getting the time variebles missassingned in an equation.

 

[elfmotat]

http://en.wikipedia.org/wiki/Time_dilation

Just let it roll over for another hundred years when someone learns to do algebra. The time variebles are reversed in this enyclopedia even, in no way shape or form does takeing the inverse exchange variebles in an equation. Nothing in either of these equations can be substituted from anything else in the other. It is even written that way in textbooks that introduce relativity in physics 101. It doesn't make the same prediction, it would be like getting the time variebles missassingned in an equation.

I'm done with you. You're not even bothering to try to understand what I said; it seems you'd rather just dig yourself a deeper hole.

 

[Jorriss]

Good thread. I'd read it again. BTW John, they do make equivalent predictions, you're misinterpreting what the theories mean.

 

[John232]

Good thread. I'd read it again. BTW John, they do make equivalent predictions, you're misinterpreting what the theories mean.

I would like to see someone show me how they do. I think they would just end up proving to themselves that they don't. They would only come as close to an answer where you just switched the variables and decided that you could correct it by just takeing the inverse. I don't even know if or where there is any mathmatical rule that says you can do this.

 

[salvestrom]

I have seen Einsteins equation written as Δt'=Δt/√(1-v^2/c^2)

I have seen the Lorentz equation written as Δt=Δt'γ
where γ=1/√(1-v^2/c^2)

You have written the Lorentz equation wrong. The wikipedia page on Lorentz Factors states it as t'=yt. Written like this the two equations are completely identical.

 

[Dale]

Simply a typo, or is one of the equations more right than another?

Yes, just a typo. The equations of Lorentz and Einstein are the same.

 

[Lizwi]

If t' is a moving frame, you must have Dt= gamma*Dt'

 

[John232]

I thought I saw a video in another thread where it was different. Then there was talk about the proper time, that wasn't the same. Where does the equation that elfmotat gave come from? Or is that also a typo, lol.

 

[D H]

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.

So if the two formulations always yield the same results, why is it that special relativity is taught while Lorentz ether theory is largely ignored?

One answer is that the underlying assumptions of special relativity are much cleaner, much less ad hoc than those of LET. Special relativity assumes that the laws of physics are the same in all inertial frames, assumes that the speed of light is the same to all inertial observers, and assumes that spacetime forms a smooth manifold. The assumption of a constant of the speed of light is weird, but that is what Maxwell's equations and what experiment says is the case.

LET assumes the Lorentz transformation and assumes an aether frame in which the electromagnetic waves propagate. This aether frame is undetectable thanks to the Lorentz transformation. This is not physics as there is no way to test this assumption. Another problem is that there is no reason for this aether or aether frame. Physicists in 1905 did not know about quantum mechanics. They assumed that electromagnetic waves were like every other wave phenomena they knew of -- some medium was required to transport the waves. Photons travel just fine through vacuum. The need for this aether disappeared with quantum mechanics.

Yet another problem with LET is that special relativity merges well with quantum mechanics in the form of quantum electrodynamics and generalizes nicely to general relativity. LET does neither.

 

[John232]

Who discovered that you can find the proper time by taking the reciprocal of the number of times a clocks pendelum swings? I find it to be a strange concept as I have never seen it used before in any other method of calculation.

 

[elfmotat]

Who discovered that you can find the proper time by taking the reciprocal of the number of times a clocks pendelum swings? I find it to be a strange concept as I have never seen it used before in any other method of calculation.

You can't. What made you think you could? 1/t doesn't even have units of time - it has units of frequency.

 

[John232]

You can't. What made you think you could? 1/t doesn't even have units of time - it has units of frequency.

In SR you find the proper time of an object traveling at relative speed by taking the reciprocal of the final answer in order to get a smaller value. I was wondering if this is something that works in Galilean relativity or was it introduced in SR? I guess you are thinking of LET and not SR. In SR gamma is derieved by using a light clock. I don't see why a normal clock would work this way if you did not consider time dialation.

 

[John232]

It is strange that you mention the units are in frequency, it is like SR says that the time dilation equation is the frequency of time between two observers.
Why would time only go around one time per secound? And to what? Would this mean something about the wave-like properties of the photon?

 

[elfmotat]

In SR you find the proper time of an object traveling at relative speed by taking the reciprocal of the final answer in order to get a smaller value.

No, you don't. You get the proper time of an object by taking a path integral over its worldline:

\Delta \tau =\int_C \sqrt{-\eta_{\mu \nu}\frac{dx^\mu}{dt}\frac{dx^\nu}{dt}}dt

In the special case that the object is inertial, the integral reduces to:

\Delta \tau = \Delta t \sqrt{1-v^2/c^2}

There are no "reciprocals" involved. None whatsoever. In fact, I'm actually confused as to how you came to your above conclusion.

I was wondering if this is something that works in Galilean relativity or was it introduced in SR?

There're no reciprocals in either.

I guess you are thinking of LET and not SR.

This is the last time I'm going to say it: SR AND LET BOTH HAVE THE SAME EQUATIONS! I really don't know what else I can say to get that through to you.

In SR gamma is derieved by using a light clock.

That's one way of deriving it. There are others as well. For example, time dilation and length contraction fall directly out of the Lorentz Transform.

I don't see why a normal clock would work this way if you did not consider time dialation.

I don't even know what you're asking here.

It is strange that you mention the units are in frequency, it is like SR says that the time dilation equation is the frequency of time between two observers. Why would time only go around one time per secound? And to what?

What does that even mean? It makes absolutely no sense. Time is measured in units of time. Frequency is measured in units of frequency. They are not interchangeable. If you take the reciprocal of a time then it is no longer a time.

Would this mean something about the wave-like properties of the photon?

What? What are you talking about? How does this even slightly relate to the conversation?

 

[John232]

How is it that you have never heard of having to take the reciprocal of the time dilation equation? Everybody just said that SR and LET just use the other equation not tau. You then have to take the reciprocal of an answer from either of those equations to get the final answer for the dialated time. It doesn't say how big the clock is or relate the speed of light to the number of times it would travel across the clock. Then the number of times light would travel across the clock in every calculation would be one time. It is like saying time only goes by once across any size clock. So then instead of the lorentz equation that was a typo in post #1 would be the same equation that you are using for tau. But your equation for tau is not equivelent to the time dilation equations because it doesn't have the same varieble for time and it is not a cycle across a light clock. If you solve using tau and the time dilation equations you get two different answers.

 

[elfmotat]

How is it that you have never heard of having to take the reciprocal of the time dilation equation? Everybody just said that SR and LET just use the other equation not tau. You then have to take the reciprocal of an answer from either of those equations to get the final answer for the dialated time. It doesn't say how big the clock is or relate the speed of light to the number of times it would travel across the clock. Then the number of times light would travel across the clock in every calculation would be one time. It is like saying time only goes by once across any size clock. So then instead of the lorentz equation that was a typo in post #1 would be the same equation that you are using for tau. But your equation for tau is not equivelent to the time dilation equations because it doesn't have the same varieble for time and it is not a cycle across a light clock. If you solve using tau and the time dilation equations you get two different answers.

Now I'm not even sure where your confusion is coming from. You're like a massive aggregation of misconceptions.

First of all, you do realize that you can use different symbols for the same quantity, right? Δτ in my last post is equivalent to Δt' in my other posts. I used τ in my last post because it is the standard symbol for proper time.

Second, I just explained to you how you calculate proper time. There were no "reciprocals" involved. I have realized by now that I need to repeat myself multiple times before you finally acknowledge anything I say, so I'll just go ahead and get it out of the way now: You don't calculate proper time with any reciprocals. You don't calculate proper time with any reciprocals. You don't calculate proper time with any reciprocals.

Also, you know that time dilation applies to any type of clock, don't you? It's not just light clocks.


Honestly, every time you respond I just get more and more frustrated. If I didn't know better I'd guess you were a troll.

 

[John232]

I didn't say that you where calculating tau with reciprocals. If you calculate the proper time without reciprocals with the equation you gave you get a different answer than calculating the proper time with reciprocals in the original time dilation equations like how it is supposed to be done. I feel like I am talking to a brick wall as well because you will not acknowlege that the time dilation equations used in SR and LET are different than the equation that you are sayig is used. I am just telling you how they are different. Calculating the proper time with the equation Δt'=Δt/√(1-v^2/c^2) you have to take the reciprocal, because it is assumed that Δt is the number of times the clock ticks. Tau is supposed to equal 1/Δt', and 1/Δt' is how you find the amount of time dilation from the original equation because it is supposed to be the number of times the clock ticks. It is assumed that Δt is the number of ticks for any given clock so that Δt'≠τ' and τ'=1/Δt'. Where τ is the proper time of what the clock actually says in the same way just t would. Then if you substitute 1/Δt' for tau in your equation you don't end up with the same thing on each side, so the you know that it was solved for incorrectly.

I have thought that t'=t√(1-v^2/c^2) myself, but that is not the equation used in SR and LET. An algebraic proof that doesn't use Δt, but only t in the distance the photon and an object has traveled gives τ or t as I think they should be the same. That is how I know how these equations are different from each other I have worked the proof differently myself and came to the same answer that was different. Because distance does not equal vΔt, in most cases it just equals vt when solving for distance. Then t is not Δt but just the amount of time something would read after reaching a certain distance or the proper time.

Then that raises the question of how does using Δt allow for t to be found by takeing the recirpocal. I don't know that I really agree with it myself. That is why I asked if it was used before in relativity before Einstein, but I think he may have used it because frequency was discovered between these two times. Maybe he thought that frequency could translate Δt this way on his own. I have never seen a proof where the change in time can be related the same way to time in another frame as a frequency. I thought it would settle the issue of this if it was shown to work in Galilean Relativity, but I don't think it does because frequency was not discovered yet so he wouldn't have used it.

I don't think the clock had a timer, I think somehow you would have to take the amount of time that had passed and then use that to determine how many ticks there was on a clock of certain size in order to find the frequency of that clock. But, then the number of times the clocked ticked couldn't always be one. I just don't understand why it would always be one or if it was done this way intentially to go along with results found with frequencies. I don't see how Δt could be the frequency of the clock. But, to get that far you would have to acknowledge that the time dilation equations used that you didn't give involves a reciprocal of the change in time.

 

[Dickfore]

OP, the transformations derived by Einstein in his fundamental work, On the Electrodynamics of Moving Bodies, carry the name "Lorentz transformations", because the same ones were derived by H. A. Lorentz from the requirement that the equations of electrodynamics remain invariant. Einstein used his two postulates only to derive them.

 

[John232]

OP, the transformations derived by Einstein in his fundamental work, On the Electrodynamics of Moving Bodies, carry the name "Lorentz transformations", because the same ones were derived by H. A. Lorentz from the requirement that the equations of electrodynamics remain invariant. Einstein used his two postulates only to derive them.

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.

 

[Dale]

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.

 

[Dickfore]

Equations are not translated, text is. Besides, if you do not believe the translation, you may refer to the original article, which is referenced in the footnotes of the above link.

 

[NotAName]

What I've been told: (which leads me to a question)
Special Relativity differs from Lorentz Ether Theory because of the ether.

In SR, each frame can act as a "universal reference" whereas there is one singular universal reference in Lorentz's derivation. This leads to the following I copied from a website somewhere:


According to Lorentz, a traveller going to a star .5 lightyears away at .5C takes 1 year in the stationary frame but the traveller only records .866 as much time elapsing for a total of .866 years to arrive. Many perspectives were changed for the traveller however: During his travel he believed the point he traveled to was 1.1547 as many units away. He believes light to travel at 1.33 units per second but also still calculates his speed as .5C (because time effects from shortening affect distance inversely leaving only the wind effect visible to in-frame observers).

According to Einstein, a traveller going to a star .5 lightyears away at .5C takes 1 year according to the stationary frame but the traveller only records .866 as much time elapsing for a total of .866 years to arrive. The traveller believes himself to be stationary and that the distant object is approaching at .5C from a distance of .433 lightyears away.

So far there is little effective difference, however:

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.

So the point is that in LET, the motion relative to the universal reference frame is already accounted for in the calculation, whereas in SR, even after the calculation, motion must still be accounted for. (or so I understand)

My question is this: SR leads to two different arrival times for light in the stationary frame: .5 years and (.288 * 1.1547 =) .332 years for the beam of light. Is this part of the lack of simultaneity?

It seems like that since I should only have to apply gamma once after calculating distances:
IE the traveler says he arrives in .866 years, therefore (.866 * 1.1547=)1 year is the simultaneous calculation for my frame.

Yet this somehow seems to not work for light. What did I do wrong and what did I misunderstand?

 

[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.