Consistency of the speed of light

In summary: It is important to note that theories require postulates. While it is theory that the speed of light is constant, for the sake of logical consistency, it is necessary to assume it to be universallly true for the sake of building other theories on it.
  • #71
Hans de Vries said:
A remarkable amount of physics can be extracted from the simple rule that the wave front is always at right angles with the physical velocity, regardless of the reference frame.
Aether said:
Suppose that you are right, doesn't this "simple rule" amount to the choice of Einstein's clock synchronization convention?

In fact, The Lorentz transformations them self can be derived directly from
the single statement that "A wavefront is always at right angles with the
direction of the wave in all reference frames."

(This beauty already laid unrecognized in the M&M experiment)

The example I use here (with the Broglie wave functions) works at any speed
let is be at meters per second or centimeters per second.


Regards, Hans
 
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  • #72
Hans de Vries said:
In fact, The Lorentz transformations them self can be derived directly from the single statement that "A wavefront is always at right angles with the direction of the wave in all reference frames."
Lorentz transformations amount to an arbitrary choice of Einstein's clock synchronization convention. Lorentz symmetry is equally well represented by transformations in which absolute simultaneity is maintained.
 
  • #73
Aether said:
I don't think so, Hans. We actually perceive only dimensionless ratios (http://arxiv.org/abs/hep-th/0208093), and all judgements of simultaneity ultimately depend on one's choice of clock synchronization convention.

Imagine,

Looking at one of the images in my post above sitting right in front of it.
Take a picture from sufficiently far away. The photo will group together all
the space/time points which have equal time in the particular reference frame.
There's no choice here.


Regards, Hans

PS: For two points, say one left and one right of the center, having the same
distance to the viewer. You can take a picture from any distance and you'll
always fetch the same two space/time points togeter. Even though you have
c going at all kinds of angles, with all kinds of different ratios of its x and y
components.
 
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  • #74
Hans de Vries said:
Imagine,

Looking at one of the images in my post above sitting right in front of it.
Take a picture from sufficiently far away. The photo will group together all
the space/time points which have equal time in the particular reference frame.
There's no choice here.


Regards, Hans

PS: For two points, say one left and one right of the center, having the same
distance to the viewer. You can take a picture from any distance and you'll
always fetch the same two space/time points togeter. Even though you have
c going at all kinds of angles, with all kinds of different ratios of its x and y
components.
How do you know that these two points have the same distance to the viewer? What are you assuming about the speed of light? Aren't you assuming that its speed is constant and isotropic?
 
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  • #75
Tom Mattson said:
How's that? Right now, sitting at my desk, I've just observed someone tying his shoe, followed by someone else clicking a mouse. Why do I need a clock synchronization scheme to say which one comes first? And if some other observer goes zipping by at high speed and observes the events in reverse order, why would he need a clock synchronization scheme to say the opposite?
You interpreted what you observed using a set of assumptions including that the speed of light is very fast, and that it is isotropic; right? That is where you adopted Einstein's clock synchronization convention. If you didn't use these assumptions, exactly which ones did you use? If I postulate that the speed of light is a function of the cosine of the angle that your velocity makes with some other velocity, say the CMB rest frame, then clocks are synchronized absolutely and judgments of simultaneity are generally different. You can transform freely between Einstein clock synchronization and absolute clock synchronization, and in doing so you change the directional speed of light.

A dimensionless ratio can not be freely transformed, and is the only truly invariant outcome for any measurement. For example, say that I measure my desk to be 2 meters wide; where is the dimensionless ratio? 2 meters per 1 meter of my meter stick: the ratio of my desk's width to my meter stick's length is 2/1. In any frame, this ratio sticks.

The speed of light is not a dimensionless ratio, and therefore it is not something that you can ever measure without reference to one artificial convention or another!

"When clocks are synchronized according to the Einstein procedure the equality of the velocity of light in two opposite directions is trivial and cannot be the subject of an experiment." - M-S I p.499
 
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  • #76
OK, I've snagged the files (both sources), and will look them over when I get a chance. The .pdf definitely looks readable, I haven't tried the new .doc yet.
 
  • #77
Hans de Vries said:
It is only Special Relativity which can rotate wavefronts, and it does so for both light and matter waves. A Galilean transformation keeps the
wavefronts always directed in the same direction! The mechanism through which Special Relativity manages this is again via the non-simultaneity of time.
There are two things going on here, Hans. First there is Lorentz symmetry, and that is what is actually measurable in experiments. Galilean transformations fail those tests, and I am not implying otherwise.

The second thing that is going on is that Lorentz symmetry is conserved by at least two types of transforms: Lorentz transforms are one type, and they are characterized by postulating that the speed of light c is isotropic in all inertial frames (e.g., SR). However, there is a second type of transform (e.g., LET) where clocks and rods are isotropic in all inertial frames, but the speed of light is not. These two type of transforms are equally valid, and they have nothing whatsoever to do with any experimental result (so far). Interestingly, BOTH theories (SR & LET) would need to be modified upon the detection of a violation of Lorentz symmetry because they are empirically equivalent.

Violations of Lorentz symmetry are empirically measurable in principle, but the conventional choice as to which terms in the transform are constant and which terms are allowed to vary is not. So which one should we use? SR is convenient as long as you don't have a locally preferred frame to use as a "handle", but LET would be better if we ever find a handle.

It is wrong to claim that the constancy of the speed of light is proven by experiment. Does this go for GR as well? I suspect that it does, but I'm still studying GR myself and can't say for sure at this time.
 
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  • #78
If you look at Einstein's original paper, Einstein *assumed* isotropy, which he did not define in great detail, to arrive at his theory of relativity.

Mansouri and Sexyl, from the papers I skimmed (thanks for posting them) are basically exploring the realm of physics of "what happens if one does not assume isotropy", though they don't discuss it in those exact terms. (I personally think their paper would be improved if they did at least mention the term "isotropy", it's got a lot of history).

A perfectly isotropic space-time can be made to appear non-isotropic by the proper (or improper) choice of clock synchronziation methods. So what Mansouri and Sexyl are basically doing is to *not* automatically choice a coordinate system that conforms to the (apparent) isotropy of space-time, by instead considering arbitrary clock synchronizations.

By not assuming isotropy as a given, (as Einstein did), Mansouri and Sexyl's work helps provide a framework for testing it. (Of course I should add that there is currently no evidence that there is any physical aniosotropy in space-time).

Their (M&S) general approach may also be useful in rotating coordinate systems, where the usual assumption of isotropy has issues. While one can always chose not to use rotating coordiantes, they are convenient enough that sometimes it's worth giving up the conveniences of isotropy for the convenience of using rotating coordinates.
 
  • #79
Zapper,
I believe what i said was correct. To throw out a universe with constant time and space, there had to be a reason. Every one accepted a constant time and space as fact. So it was the the MM experiment which led to SR, which led to variable space and variable time. If you have a constant velocity for the observer, then something else must be a variable, Einstein chose space and time. So the whole reason for coming up with SR is the MM experiment. Without the MM experiment, SR was not needed, and thus would never have been accepted in the scientific community.

And yes, relative to the constant speed of light, the observer's speed was 0.
Remember, everyone expected something other than 0, because they were looking for the so called aether. But the experiment was done in air, so relative to the air, the velocity of the interferometer was 0. Not a good experiment.
mike
 
  • #80
Didyoueatpaintchips said:
And yes, relative to the constant speed of light, the observer's speed was 0.
Remember, everyone expected something other than 0, because they were looking for the so called aether. But the experiment was done in air, so relative to the air, the velocity of the interferometer was 0. Not a good experiment.
mike
What difference does it make what its velocity relative to the air is? Neither relativity nor the aether theory would predict your velocity relative to the air would make any difference in terms of the velocity you measure for light.

And "relative to the constant speed of light, the observer's speed was 0" doesn't make sense. You can only talk about your velocity relative to a thing like a car or a light wave, talking about your speed relative to a speed is meaningless. What is my speed relative to the speed of 70 mph? Maybe the idea you're trying to express is that the experimenters originally believed that only an observer whose velocity relative to the aether was 0 would see light moving exactly at c in all directions, whereas relativity predicts that every observer will see that. But this means that every observer has a speed of c relative to light, not a speed of 0 relative to light.
 
  • #81
Didyoueatpaintchips said:
With all these posts no one has stated the reason that SR was needed in the first place. Also how could a forum not include posts from people who don't agree with einstein. This is truly intellectual censorship. No wonder we have had to endure 100 years of relativity, this is what all of the colleges also do, so there are no new ideas.

Welcome to PhysicsForums, Mike.

This forum is intended primarily for mainstream discussions of physics topics. It is not intended for presentation of new theoretical ideas, which are normally presented in the context of peer-reviewed journals. Anyone is free to publish their ideas as they like on their own sites; that is what I do with my work (although most of my stuff is in fact mainstream).

And if you think this forum is devoted to Einstein, you haven't followed the discussions here sufficiently. There are skeptics here, but the presentation should be made within the context of legitimate topics. For example, SR and GR are currently generally accepted theories within the physics community. Therefore, a thread about why SR is wrong is not welcome here. On the other hand, EPR is considered to be outmoded (at least as to the incompleteness of QM) due to Bell's Theorem. Therefore, a critique of Einstein's position on this is acceptable.

Please keep in mind that there are many readers here with varying levels of knowledge. This forum is devoted to those interested in learning more about what is going on in physics and science.

...

Regarding your later post: I wonder why you don't feel the M/M experiment is good. Based on your historical description, how do you see that the speed of the air is an issue? Are you saying that the experiment would only be meaningful in a vacuum? Isn't the air itself also moving "faster" relative to the ether in the direction of the Earth's movement?
 
  • #82
Aether said:
How do you know that these two points have the same distance to the viewer?

You can do that by using a ruler :smile: That's the way we define length.
Under all conditions, even if you might question how we do know that the
ruler is always the same.

Aether said:
What are you assuming about the speed of light? Aren't you assuming that its speed is constant and isotropic?

The example in the postscript shows that you get the same result under
various mixes of cx and cy components.



Regards, Hans
 
  • #83
Aether said:
Violations of Lorentz symmetry are empirically measurable in principle, but the conventional choice as to which terms in the transform are constant and which terms are allowed to vary is not. So which one should we use? SR is convenient as long as you don't have a locally preferred frame to use as a "handle", but LET would be better if we ever find a handle.


Aether,

All Lorentz transforms are equally well valid. They are just different
representations of the same, single, reality. Just different 3D slices from
the single 4D universe.

What are you looking for? A preferred reference frame? what about Einstein's
beloved Machian reference frame, the center of gravity of all mass in the
universe. Or the modern day cosmologist's co-moving frame. more like the
center of mass of the local universe, so there are different co-moving frames
at different places in the universe.

Are you looking for aether? like your name implies. The vacuum is far from
empty. Look at the vacuums from Quantum Field Theory, All the many
different vacua people discuss in Quantum Gravity research. Just never
call it aether. That's a name which is reserved for a substance with a
classical gas like behavior. It has be shown over and over that that's
not the way how it works with incredible accuracy.

Are you maybe looking for the absolute "NOW" ?
Why should two different space/time events separated billions of light years
from each other be connected instaneously by an invisible link? You may
define a mathematical space in any arbitrary way that does so but does
it matter to physics anymore then a statement that two different events
have the same value for x or y? All what matters to physics is how different
events in space time communicate with each other, and they do so from
neighbor point to neighbor point to neighbor point. Two events at different
sides of the universe are completely disconnected.

Special Relativity holds up an illussion of a "NOW" in each reference frame
but in General Relativity there's no global "NOW" any more in any reference
frame.


Regards, Hans
 
  • #84
Hans de Vries said:
What are you looking for? A preferred reference frame? what about Einstein's beloved Machian reference frame, the center of gravity of all mass in the universe. Or the modern day cosmologist's co-moving frame. more like the center of mass of the local universe, so there are different co-moving frames at different places in the universe.
I'm working on a unification theory that makes predictions that are so precise and consistent that I tend to take them seriously, but a locally preferred frame seems to be implied. So, I raised the Aether flag and went forth to test the waters and found that LET is empirically equivalent to SR. Whether or not my own theory pans out, this empirical equivalence of LET and SR is something that I think people should be aware of.

Hans de Vries said:
Are you looking for aether? like your name implies. The vacuum is far from empty. Look at the vacuums from Quantum Field Theory, All the many different vacua people discuss in Quantum Gravity research. Just never call it aether. That's a name which is reserved for a substance with a classical gas like behavior. It has be shown over and over that that's not the way how it works with incredible accuracy.
Aether is used as shorthand for both a rarified gas (which is not how I am using it), and a locally preferred frame (which is how I am using it).

Hans de Vries said:
Are you maybe looking for the absolute "NOW" ? Why should two different space/time events separated billions of light years from each other be connected instaneously by an invisible link? You may define a mathematical space in any arbitrary way that does so but does it matter to physics anymore then a statement that two different events have the same value for x or y? All what matters to physics is how different events in space time communicate with each other, and they do so from neighbor point to neighbor point to neighbor point. Two events at different sides of the universe are completely disconnected.
This is getting close to where I'm going, that events on opposite side of the universe are connected instantaneously. Every event has a temporal coordinate on the farthest edge of the universe, so why shouldn't events be connected out there? There is cosmological evidence of such a link (e.g., the "horizon problem").

Hans de Vries said:
Special Relativity holds up an illussion of a "NOW" in each reference frame but in General Relativity there's no global "NOW" any more in any reference frame.
And, 98% of the matter-energy in the universe is still missing.
 
  • #85
Aether said:
I went forth to test the waters and found that LET is empirically equivalent to SR. Whether or not my own theory pans out, this empirical equivalence of LET and SR is something that I think people should be aware of.


Aether,

Mansouri & Sexl are plainly wrong with the claim that their Lorentz
Ether Theory is empirically equivalent to Special Relativity. There is only
one frame in which it makes some right predictions and that's the frame
in which they did their calculations: The preferred reference frame.
They then erroneously extrapolate that it does work in all the
reference frames.

In the other frames it gives results that are completely different than
those from Special Relativity. The LET is nothing else than a Galilean
transformation with a scaling factor (gamma)

A clear example in where it goes wrong is when you've got two objects,
one moving from left to right and the other from right to left, both with
the same speed. In SR both will have the same Lorentz contraction for
the observer at rest. In their LET however there is no such symmetry.
One object will typically get extended while the other gets contracted.
The only frame where LET and SR give equal results is in their preferred
reference frame.


It's only with the non-simultaneity of SR that two observers both
see each other contracted. With LET one observer will be contracted
and the other will be extended in an asymmetric way.

To see how SR works we can imagine that we instantaneously "freeze"
a bypassing traveler. Walking around him we can now see him "hanging
in the air", indeed being contracted in the direction in he was moving.
The traveler however will complain that his front was stopped first,
before his back was frozen, and argues that this is the reason of his
compressed state.

This now is a symmetric situation. If you'll lend you're "freezing device"
to the traveler then you'll see that you get frozen in the same way:
contracted by non-simultaneity.

With absolute simultaneity this will always lead to a paradox: If A is
contracted in respect with B. Then B is extended relative to A.
It is the non-simultaneity of SR which resolves this paradox.



Regards, Hans


P.S: R.Mansouri and R.U.Sexl
A Test Theory of Special Relativity, I. Simultaneity and Clock Synchronization
General Relativity and Gravitation, Vol 8, No 7 (1977) pp. 497-513
 
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  • #86
Hans de Vries said:
Mansouri & Sexl are plainly wrong with the claim that their Lorentz Ether Theory is empirically equivalent to Special Relativity. There is only
one frame in which it makes some right predictions and that's the frame in which they did their calculations: The preferred reference frame.
Are Kostelecky&Mewes wrong too, http://www.citebase.org/cgi-bin/citations?id=oai:arXiv.org:hep-ph/0205211 [Broken], and everyone else over the past 28 years who has cited M-S without mentioning that they are "plainly wrong" about anything other than a few typos? Otherwise, let's assume that M-S are 100% right.

The asymmetrical length contractions that you see are balanced by an anisotropic speed of light. The point is, Hans, that any difference between the predictions of SR and LET are merely the result of coordinate choice rather than Lorentz symmetry. The two theories are empirically equivalent; any difference that you see is in the interpretation of the measurements and not in the measurements themselves.

Some people here already knew this, and some people here have only just realized this. The fact that some people are still struggling with it proves that it isn't a trivial point, and more care needs to be taken with the teaching of relativity: Local Lorentz symmetry is proven by experiment (up to a point, the search for violations of this symmetry are still ongoing), but special relativity per se is not even a valid subject for experiment.
 
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  • #87
Aether said:
Are Kostelecky&Mewes wrong too, http://www.citebase.org/cgi-bin/citations?id=oai:arXiv.org:hep-ph/0205211 [Broken], and everyone else over the past 28 years who has cited M-S without mentioning that they are "plainly wrong" about anything other than a few typos? Otherwise, let's assume that M-S are 100% right.

This is misleading. The M&S paperer is referred to because of the
parameterization scheme for possible deviations of Special Relativity.
It's only you who uses it to promote your ether theory.

The math of M&S is correct in the preferred frame, not in any other.

Aether said:
Some people here already knew this, and some people here have only just realized this. The fact that some people are still struggling with it proves that it isn't a trivial point, and more care needs to be taken with the teaching of relativity: Local Lorentz symmetry is proven by experiment, but special relativity per se is not even a valid subject for experiment.

Do you at all read my post? do you look at my examples. No you don't

SR is the ABC of physics. Something you have to understand pretty
well before you can start to learn some real physics. The examples I
gave are the simplest it gets in understanding the basic mechanisms in SR
and the simplest way to show that your Ether theory with absolute time
can never work.

Now try to do the math. Try to understand the physics. Don't just rely on
some statement you have found somewhere in a paper. It's now time for you
to prove your ether theory by actually showing how it can account for
these relativistic effects.

1) How can two observers both see the other in a Lorentz contracted state?

2) How can you rotate the deBroglie wavefronts of particles if you go from
one reference frame to another in order to keep them at right angles with
the direction of their speed? How can a single transformation rotate these
wavefront at all kinds of different angles depending on the speed of the
particles?

Let's see if you can do that without non simultaneity.

People here are willing to help others to get ahead. That's why it's called a
Physics Help and Math Help forum. But if there's no response and you just
keep repeating a statement from somewhere then things get pretty useless
after a while. I did the work, the math, the physics, showed you the images
from my simulations.

Now it's up to you.



Regards, Hans.
 
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  • #88
Hans de Vries said:
Mansouri & Sexl are plainly wrong with the claim that their Lorentz Ether Theory is empirically equivalent to Special Relativity. There is only one frame in which it makes some right predictions and that's the frame in which they did their calculations: The preferred reference frame. They then erroneously extrapolate that it does work in all the reference frames.
Don't forget to reset your clocks: "where we have chosen to readjust our clocks according to [tex]f(x,v)=-vx[/tex]" -- Eq. 3.5 M&S-I p. 502

Hans de Vries said:
This is misleading. The M&S paperer is referred to because of the parameterization scheme for possible deviations of Special Relativity. It's only you who uses it to promote your ether theory.

The math of M&S is correct in the preferred frame, not in any other.
You can't legitimately claim that M&S is plainly wrong, that they erroneously extrapolate, and also that I am misrepresenting their papers to promote my own theory. My own theory has nothing to do with this other than it motivates me personally to care about this particular issue.

Hans de Vries said:
Now try to do the math. Try to understand the physics. Don't just rely on some statement you have found somewhere in a paper. It's now time for you to prove your ether theory by actually showing how it can account for these relativistic effects.
M&S and every paper on local Lorentz invariance since then that quotes them is my proof. If they really are "plainly wrong" as you claim, then I'll have to go back to the drawing board. I don't mind working some transformation problems using the M&S papers as a guide to try and show that I am not applying their work improperly. You start by showing the Lorentz transform (Eq. 3.4 from M&S-I p. 501) for any example you choose (M&S restricts their examples to motion along the x-axis, so we will need to agree to do the same), and I will show the corresponding LET transform (Eq. 3.6 from M&S-I p. 502). "This transform is--as far as the prediction of experimental results is concerned--completely equivalent to (3.4)". -- M&S-I p. 502.

Eq. (3.4)
[tex]t=(1-v^2/c_0^2)^{1/2}T-vx[/tex]
[tex]x=(X-vT)/(1-v^2/c_0^2)^{1/2}[/tex]


Eq. (3.6)
[tex]t=(1-v^2/c_0^2)^{1/2}T[/tex]
[tex]x=(X-vT)/(1-v^2/c_0^2)^{1/2}[/tex]

where we have chosen to readjust our clocks according to
Eq. (3.5) [tex]f(x,v)=-vx[/tex]


"We shall investigate here how the results of various experiments, which are usually considered to be tests of special relativity, can be interpreted using [3.6]. The transformation [3.6] is the very relation one would write down if one has to formulate an ether theory in which rods shrink by a factor [tex](1-v^2/c_0^2)^{1/2}[/tex] and clocks slow by a factor [tex](1-v^2/c_0^2)^{1/2}[/tex] when moving with respect to the ether. Note that [3.6] implies the existence of absolute simultaneity since [tex]\Delta T=0[/tex] implies [tex]\Delta t=0[/tex]. We thus arrive at the remarkable result that a theory maintaining absolute simultaneity is equivalent to special relativity." -- M&S-I p. 503
 
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  • #89
Hans de Vries said:
Aether said:
How do you know that these two points have the same distance to the viewer?
You can do that by using a ruler :smile: That's the way we define length.
That's true for an object stationary relative to the observer. For a moving object you must also read a clock, attached to the ruler at the point where the measurement is made.

If you say that two distances are the same, you must mean either they are both constant or else that they were both measured at the same time. So it depends of your definition of simultaneity.
 
  • #90
The one-way speed of light

Aether said:
Eq. (3.6)
[tex]t=(1-v^2)^{1/2}T[/tex]
[tex]x=(X-vT)/(1-v^2)^{1/2}[/tex]

"We shall investigate here how the results of various experiments, which are usually considered to be tests of special relativity, can be interpreted using [3.6]."
That should clear up a lot of confusion. OK. Let's see you compute the one-way speed of light.

Here's the experiment. In an arbitrary frame of reference in the aether model, start with two synchronized clocks side-by-side and slowly transport one of them to any convenient distance D. Then compute D/(t2-t1). The answer better be c. (t1 is the time on the stationary clock when the light pulse is sent. t2 is the time when the light arrives as measured by the slowly transported clock. Take the limit of ultraslow transport for a perfect answer of c).
 
  • #91
Aether said:
You start by showing the Lorentz transform (Eq. 3.4 from M&S-I p. 501) for any example you choose (M&S restricts their examples to motion along the x-axis, so we will need to agree to do the same), and I will show the corresponding LET transform (Eq. 3.6 from M&S-I p. 502). "This transform is--as far as the prediction of experimental results is concerned--completely equivalent to (3.4)". -- M&S-I p. 502.

Eq. (3.4)
[tex]t=(1-v^2)^{1/2}T-vx[/tex]
[tex]x=(X-vT)/(1-v^2)^{1/2}[/tex]


Eq. (3.6)
[tex]t=(1-v^2)^{1/2}T[/tex]
[tex]x=(X-vT)/(1-v^2)^{1/2}[/tex]


where we have chosen to readjust our clocks according to
[tex]f(x,v)=-vx[/tex]

"We thus arrive at the remarkable result that a theory maintaining absolute simultaneity is equivalent to special relativity." -- M&S-I p. 503


Equivalent? Only in the preferred frame.

Note that 3.6 makes both length and speed anisotropic in any other
reference frame besides the preferred frame. Round wheels would be
only really round in the preferred frame.

We at Earth would see al our wheels changing shape in a 24 hours cycle
corresponding with the rotation of the earth.


Regards, Hans
 
  • #92
Perspicacious said:
That should clear up a lot of confusion. OK. Let's see you compute the one-way speed of light.

Here's the experiment. In an arbitrary frame of reference in the aether model, start with two synchronized clocks side-by-side and slowly transport one of them to any convenient distance D. Then compute D/(t2-t1). The answer better be c. (t1 is the time on the stationary clock when the light pulse is sent. t2 is the time when the light arrives as measured by the slowly transported clock. Take the limit of ultraslow transport for a perfect answer of c).
Lorentz transformation:
[tex]t_1=(1-v^2/c_0^2)^{1/2}T_1-vx_1/c_0^2[/tex]
[tex]x_1=(X_1-vT_1)/(1-v^2/c_0^2)^{1/2}[/tex]

[tex]t_2=(1-v^2/c_0^2)^{1/2}T_2-vx_2/c_0^2[/tex]
[tex]x_2=(X_2-vT_2)/(1-v^2/c_0^2)^{1/2}[/tex]

[tex]D/(t_2-t_1)=(x_2-x_1)/(t_2-t_1)=c_0[/tex]

[tex]c(v,\theta)=c_0[/tex] o:)

M&S-I p.511 Eq. (6.16) gives the following result for first order effects when transport synchronization of clocks is used with the Lorentz transformation: [tex]c(\theta)=1-v(1+2\alpha)cos\theta[/tex] where [tex]\alpha =-1/2[/tex] corresponds to perfect Lorentz symmetry.

"In discussing the experiments we need the inverse velocity of light to second order in [tex]v/c[/tex]..." M&S-III p. 810 Eq. (2.1) - [tex]1/c(\theta)=1+(\beta+\delta-1/2)v^2sin^2\theta+(\alpha-\beta+1)v^2[/tex]

LET transformation:
[tex]t_1=(1-v^2/c_0^2)^{1/2}T_1[/tex]
[tex]x_1=(X_1-vT_1)/(1-v^2/c_0^2)^{1/2}[/tex]

[tex]t_2=(1-v^2/c_0^2)^{1/2}T_2[/tex]
[tex]x_2=(X_2-vT_2)/(1-v^2/c_0^2)^{1/2}[/tex]

[tex]D/(t_2-t_1)=(x_2-x_1)/(t_2-t_1)=c_0^2/(c_0+v)[/tex]

[tex]c(v,\theta)=c_0^2/(c_0+v \cdot cos(\theta))[/tex] :cool:

Hint: The speed of light [tex]c_0[/tex] is isotropic in the ether frame: [tex](X_2-X_1)/(T_2-T_1)=c_0[/tex].

Assuming perfect Lorentz symmetry and using the LET transformation I get:
[tex]c_0/c(v,\theta)=1+(v/c_0) \cdot cos(\theta)[/tex]. This is a dimensionless ratio, and as such it is a measurable (e.g., physical) quantity. However, you must synchronize two clocks to make this measurement; and exactly how you choose to do that determines whether the Lorentz transformation or the LET transformation should be applied.


The speed of light is generally anisotropic in LET (e.g., except for within the ether frame), absolute simultaneity is maintained, and this is empirically equivalent to SR. The trick is that the [tex]-vx/c_0^2[/tex] term that is used to maintain a constant speed of light in the Lorentz transformation is used instead to maintain absolute simultaneity in the LET transformation. Both ways are equally valid.
 
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  • #93
Hans de Vries said:
Equivalent? Only in the preferred frame.
Empirically equivalent, yes; independent of the frame. These transforms operate on abstract coordinates, and not on physical objects.

Hans de Vries said:
Note that 3.6 makes both length and speed anisotropic in any other reference frame besides the preferred frame. Round wheels would be only really round in the preferred frame.

We at Earth would see al our wheels changing shape in a 24 hours cycle
corresponding with the rotation of the earth.
Maybe so, but only in the same sense as the arctic circle looks inflated on a Mercator projection. We are talking about mere coordinate systems here, aren't we? My agrument is with taking the attributes of a coordinate system (e.g., constancy of the speed of light) and claiming that it is an emprically proven fact. I do not dispute that for whatever reason one may prefer the Lorentz transform over the LET transform for whatever purpose, but how is this any different than expressing a preference to use spherical polar coordinates over rectangular coordinates in some cases or vice versa? In both cases the two coordinate systems transform into one another quite freely.
 
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  • #94
robphy said:
Different test theories differ in their assumptions about what form the transform equations could reasonably take. There are at present four test theories of SR:
Robertson,Rev. of Mod. Phys. 21, p378 (1949).
Edwards, Am. J. Phys. 31 (1963), p482.
Mansouri and Sexl, Gen. Rel. Grav. 8 (1977), p497, p515, p809.
Zhang, Special Relativity and its Experimental Foundations.
Zhang discusses their interrelationships and presents a unified test theory encompassing the other three, but with a better and more interpretable parameterization. His discussion implies that there will be no more test theories of SR that are not reducible to one of the first three.
I have received my copy of Zhang's book "Special Relativity and its Experimental Foundations", (1997). In the preface he says this: "The key point in Einstein's theory is the postulate concerning the constancy of the (one-way) velocity of light, which contradicts the classical (nonrelativistic) addition law of velocities. The postulate is needed only for constructing well-defined inertial frames of reference or, in other words, only for synchronizing clocks (i.e., defining simultaneity). It is not possible to test the one-way velocity of light because another independent method of clock synchronization has not yet been found...Of course one could use the experiments to yield limits on the parameters in Robertson's transformations but not on the directional parameter q in Edwards' and MS' theories."

On the back cover Zhang says this: "...In particular, the discussions indicate that the one-way speed of light is not observable in the present laboratories...In the third part, variant types of experiments performed up to now are analyzed and compared to the predictions of special relativity. The analyses show that the experiments are tests of the two-way velocity, but not of the one-way velocity, of light."

On page 10 he says: "We have known that there is no instantaneous signal in nature and, therefore, the absolute simultaneity cannot be realized in any laboratory...It is well known that one always use a light signal for the clock synchronization in a laboratory. Therefore Einstein's simultaneity can be directly realized in experiments. We want to stress here that only the two-way speed, but not the one-way speed, of light has been already measured in the experimental measurements, and hence the isotropy of the one-way velocity of light is just a postulate...We shall see from Chap. 6 that a more general postulate, a choice of the anisotropy of the one-way velocity of light, together with the principle of relativity, would give the same physical predictions as Einstein's theory of special relativity."
 
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  • #95
Einstein's second postulate states that the speed of light is constant as viewed from any frame of reference. Most of the books on relativity that I have been reading usually ask the reader to accept that fact because proving it is behind the scope of the book. Can anyone help me understand the actual reason behind the second postulate?

Hi, I'm new to the forum. I didn't read all the responses to your question, but it looks like it wandered away from your original question so I'd like to add a couple of thoughts.

First, if you search the net for "On the Electrodynamics of Moving Bodies" (OEMB for short) you can find Einstein's 1905 paper and read his explanation for the second postulate. Briefly, it can be interpretted this way: When Maxwell's equations (circa 1850) convinced people that light was a wave phenomenon they assumed that the wave must be the motion of some medium. For instance the medium for sound is air. The medium for ocean waves is water, etc.

The name given to the medium for light was aether. A number of experiments were performed over the next half century to observe the properties of aether, the most famous being the Michelson-Morley experiment (you can find Michelson's paper ont he net also). In OEMB Eisntein notes that all the experiments to find the nature or effects of aether failed. He then states his postulate that light propagates in "empty space" at a constant velocity c. Although he doesn't come right out and say it, he is proclaiming that aether doesn't exist (or at least is of no consequence regarding light).

Many people think that this postulate is where he declared that the speed of light is the same for all inertial observers. Actually it is the first postulate that declares this phenomenon. In the first postulate he says that the laws of electrodynamics and optics hold for all inertial observers (inertial means non-accelerating). This means that Maxwell's laws hold. And Maxwell's laws show that the speed of light c is a function of two properites of empty space, the permeability and permitivity of empty space. Since these two properties are constants, then c must be a constant.

In summary then, Einstein made postulates out of what observations of physics seemed to imply; that is, that aether didn't seem to have any measurable effects and that the motion of an observer didn't seem to change the outcome of various physical experiments.
 
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  • #96
cincirob said:
Although he doesn't come right out and say it, he is proclaiming that aether doesn't exist (or at least is of no consequence regarding light).
I just popped into nit pick. In fact he does come right out and say it.

A. Einstein said:
The introduction of a "luminiferous ether'' will prove to be superfluous

http://www.fourmilab.ch/etexts/einstein/specrel/www/
 
  • #97
Quote:
Originally Posted by cincirob
Although he doesn't come right out and say it, he is proclaiming that aether doesn't exist (or at least is of no consequence regarding light).

I just popped into nit pick. In fact he does come right out and say it.


Quote:
Originally Posted by A. Einstein
The introduction of a "luminiferous ether'' will prove to be superfluous
=======================================

You are correct, he does make the comment above. But my comments were directed specifically to the postulates where you must draw the inferrence.
 
  • #98
I have seen some comments in this thread about the Robertson-Mansouri-Sexl framework. I'm not very familiar with the work, but i found this reference to it.

I see that the graphics don't copy so here is the site : http://qom.physik.hu-berlin.de/prl_91_020401_2003.pdf

Modern Michelson-Morley Experiment using Cryogenic Optical Resonators
Holger Mu¨ller,1,2,* Sven Herrmann,1,2 Claus Braxmaier,2 Stephan Schiller,3 and Achim Peters1,†,‡
1Institut fu¨ r Physik, Humboldt-Universita¨t zu Berlin, Hausvogteiplatz 5-7, 10117 Berlin, Germany
2Fachbereich Physik, Universita¨t Konstanz, 78457 Konstanz, Germany
3Institut fu¨ r Experimentalphysik, Heinrich-Heine-Universita¨t Du¨sseldorf, 40225 Du¨sseldorf, Germany
(Received 27 January 2003; published 10 July 2003)
We report on a new test of Lorentz invariance performed by comparing the resonance frequencies of
two orthogonal cryogenic optical resonators subject to Earth’s rotation over 1 yr. For a possible
anisotropy of the speed of light c, we obtain c=c0  2:6  1:7  1015. Within the Robertson-
Mansouri-Sexl (RMS) test theory, this implies an isotropy violation parameter     12
 2:2  1:5  109, about 3 times lower than the best previous result. Within the general extension of the
standard model of particle physics, we extract limits on seven parameters at accuracies down to 1015,
improving the best previous result by about 2 orders of magnitude.
 
  • #99
It may turn out that there is an new theory that explains why the speed of light is the same in all IRF's. This would lead to the same type of question to that theory's foundations. If you understand this you understand the nature of science. It causes one to ask what do I mean by 'why' when I ask a question of science. What am I looking for? What do I expect?
 
  • #100
In the 1860's Maxwell discovered that he could cast the equations, which bear his name, in the form of a standard wave equation. When he did that, he found a combination of physical constants expressed the velocity of these electromagnetic waves. That constant was [itex] \frac 1 { \sqrt{ \mu_0 \epsilon_0}}[/itex]. It is said that he was surprised to find that expression evaluated to a number which was equal to the then accepted value for the speed of light.
So this was a theoretical prediction that the speed of light was a physical constant. The meaning of this was hotly debated for the rest of the century, it implied that electromagnetism behaved differently from massive bodies. As mentioned Michelson and Morley preformed an experiment in an effort to detect the motion of light through the aether. They failed to detect any medium through which light was propagating.
Einstein's postulate that the speed of light was constant to all observers was a result of the failure, over the previous 50 yrs, of physicist trying to prove that it was not. The constancy was predicted theoretically and verified experimentally. Therefore when A.E. wrote down that postulate, it was accepted by the physics community without debate, because it was common knowledge. Einstein tied together the physics of Newton and Electo Magnetism with a simple and very straight forward derivation.
 
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  • #101
Aether said:
Lorentz transformation:
[tex]t_1=(1-v^2/c_0^2)^{1/2}T_1-vx_1/c_0^2[/tex]
[tex]x_1=(X_1-vT_1)/(1-v^2/c_0^2)^{1/2}[/tex]

[tex]t_2=(1-v^2/c_0^2)^{1/2}T_2-vx_2/c_0^2[/tex]
[tex]x_2=(X_2-vT_2)/(1-v^2/c_0^2)^{1/2}[/tex]

[tex]D/(t_2-t_1)=(x_2-x_1)/(t_2-t_1)=c_0[/tex]

[tex]c(v,\theta)=c_0[/tex] o:)

M&S-I p.511 Eq. (6.16) gives the following result for first order effects when transport synchronization of clocks is used with the Lorentz transformation: [tex]c(\theta)=1-v(1+2\alpha)cos\theta[/tex] where [tex]\alpha =-1/2[/tex] corresponds to perfect Lorentz symmetry.

"In discussing the experiments we need the inverse velocity of light to second order in [tex]v/c[/tex]..." M&S-III p. 810 Eq. (2.1) - [tex]1/c(\theta)=1+(\beta+\delta-1/2)v^2sin^2\theta+(\alpha-\beta+1)v^2[/tex]

LET transformation:
[tex]t_1=(1-v^2/c_0^2)^{1/2}T_1[/tex]
[tex]x_1=(X_1-vT_1)/(1-v^2/c_0^2)^{1/2}[/tex]

[tex]t_2=(1-v^2/c_0^2)^{1/2}T_2[/tex]
[tex]x_2=(X_2-vT_2)/(1-v^2/c_0^2)^{1/2}[/tex]

[tex]D/(t_2-t_1)=(x_2-x_1)/(t_2-t_1)=c_0^2/(c_0+v)[/tex]

[tex]c(v,\theta)=c_0^2/(c_0+v \cdot cos(\theta))[/tex] :cool:

Hint: The speed of light [tex]c_0[/tex] is isotropic in the ether frame: [tex](X_2-X_1)/(T_2-T_1)=c_0[/tex].

Assuming perfect Lorentz symmetry and using the LET transformation I get:
[tex]c_0/c(v,\theta)=1+(v/c_0) \cdot cos(\theta)[/tex]. This is a dimensionless ratio, and as such it is a measurable (e.g., physical) quantity. However, you must synchronize two clocks to make this measurement; and exactly how you choose to do that determines whether the Lorentz transformation or the LET transformation should be applied.


The speed of light is generally anisotropic in LET (e.g., except for within the ether frame), absolute simultaneity is maintained, and this is empirically equivalent to SR. The trick is that the [tex]-vx/c_0^2[/tex] term that is used to maintain a constant speed of light in the Lorentz transformation is used instead to maintain absolute simultaneity in the LET transformation. Both ways are equally valid.


Hmmm,

It is clear how you got the second formula but it is not clear at all how you got the first formula [tex]c(v,\theta)=c_0[/tex]. Would you care to show the intermediate steps?
 
  • #102
:
If the two arms of a round-trip interferometer are parallel, then these two terms cancel on subtraction because the two [tex](\delta-\beta+1/2)*[(v/c)sin(\theta)]^2[/tex] terms are both proportional to [tex]sin^2(\theta)[/tex]. These two terms do not cancel in Michelson-Morley or Kennedy-Thorndike experiments because one of the terms is proportional to [tex]sin^2(\theta)[/tex] while the other term is proportional to [tex]cos^2(\theta)[/tex] (when the two arms of the interferometer are orthogonal... but they would cancel if the two arms of the interferometer were made parallel, and that is why they are not made parallel). Gagnon's interferometer is not round-trip so I'm not completely sure that this cancellation of terms applies, but he hasn't explained exactly how he got to Eq. (9)
According to (9) in Gagnon the terms do not cancel.
This has been your challenge from the beginning, to prove that expression (9) is zero. Your explanations have been all over the map, the previous one had to do with delta/beta being zero, the one before had to do with some impossible to correlate transformation thru the Lorentz transforms, most of them had to do with the confusion about the framework (which was incorrectly taken to be SR).

Let's try again:

1. How did you arrive to the expressions
[tex]\phi_1(t)=\phi_1(0)+\omega_1*(2L_1/c)(1/\eta_1+(\delta-\beta+1/2)*[(v/c)sin(\theta)]^2[/tex]

[tex]\phi_2(t)=\phi_2(0)+\omega_2*(2L_1/c)(1/\eta_2+(\delta-\beta+1/2)*[(v/c)sin(\theta)]^2[/tex]

for the two wave guides?

2. Are you aware that the parenses don't match, therefore the expressions are incomprehensible? You will need to correct that.

Look again at [tex]\phi_1(t)=\phi_1(0)+\omega_1*(2L_1/c)(1/\eta_1+(\delta-\beta+1/2)*[(v/c)sin(\theta)]^2[/tex]

3. How does the term [tex]\1\/\eta_1[/tex] "disappear" from your considerations?

4. Where in your calculations do you factor in that the cutoff frequencies for the two waveguides are arranged to be very different? Gagnon seems to make a great deal of the fact that expression (9) is obtained by driving the two waveguides one just above the cutofff and the second one way above the cutoff frequency.
 
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  • #103
clj4 said:
Hmmm,

It is clear how you got the second formula but it is not clear at all how you got the first formula [tex]c(v,\theta)=c_0[/tex]. Would you care to show the intermediate steps?
This is just a statement that in SR one-way light speed is isotropic, but I will examine it anyway.

SR, RMS (Robertson-Mansouri-Sexl text theory), and Lorentz ether theory are all in agreement that the speed of light [tex]c_0[/tex] is defined to be isotropic with respect to at least one particular inertial frame (e.g., within the ether frame [tex]\frac{X_2-X_1}{T_2-T_1}=c_0[/tex] -- actually, all velocites are isotropic within this frame, so this equation really relates just to this example, I suppose that a more general equation like this [tex]\frac{X_2-X_1}{T_2-T_1}=v_0[/tex] is always valid). In SR however, the Lorentz transform relates this particular (arbitrary) inertial frame with every other inertial frame while explicitly preserving light speed, and therefore the speed of light is further defined (within SR) to be isotropic with respect to every other inertial frame as well.

For example, these are (two sets of) the Lorentz transforms for motion along the x-axis:

[tex]t_1=((1-v^2/c_0^2)^{1/2}T_1-vx_1/c_0^2)[/tex],
[tex]x_1=((X_1-vT_1)/(1-v^2/c_0^2)^{1/2})[/tex],
[tex]t_2=((1-v^2/c_0^2)^{1/2}T_2-vx_2/c_0^2)[/tex],
[tex]x_2=((X_2-vT_2)/(1-v^2/c_0^2)^{1/2})[/tex].

Rather than defining light-speed isotropy for all (inertial) frames, I will justify this equation [tex]c(v,\theta)=c_0[/tex] by defining light-speed isotropy only in the ether frame (e.g., [tex]\frac{X_2-X_1}{T_2-T_1}=c_0[/tex]) and then combine this definition with the Lorentz transformations to show that light speed is therefore isotropic in all (inertial) frames [tex]\frac{x_2-x_1}{t_2-t_1}=c_0[/tex]; such a showing is ultimately coordinate-system dependent, and it is definitely not something that could ever be proven by an experiment.

So, to verify that this equation [tex]\frac{x_2-x_1}{t_2-t_1}=c_0[/tex] is true (when [tex]\frac{X_2-X_1}{T_2-T_1}=c_0[/tex]) then simply transform the four coordinates from any arbitrary inertial frame (e.g., use any v you like) into the ether frame using the Lorentz transform equations given above and verify that you always arrive back at this equation [tex]\frac{X_2-X_1}{T_2-T_1}=c_0[/tex] which we have defined above to be true.

Start with:
[tex]\frac{x_2-x_1}{t_2-t_1}=c_0[/tex]

Transform the four space-time coordinates (by substitution) using the Lorentz transforms:
[tex]\frac{((X_2-vT_2)/(1-v^2/c_0^2)^{1/2})-((X_1-vT_1)/(1-v^2/c_0^2)^{1/2})}{((1-v^2/c_0^2)^{1/2}T_2-vx_2/c_0^2)-((1-v^2/c_0^2)^{1/2}T_1-vx_1/c_0^2)}=c_0[/tex]




Now, simply reduce this equation to [tex]\frac{X_2-X_1}{T_2-T_1}=c_0[/tex] to show that [tex]c(v,\theta)=c_0[/tex] is true (this doesn't actually prove that it is true for all [tex]\theta[/tex] because we're only looking at motion along the x-axis).

[tex]\frac{X_2-vT_2-X_1+vT_1}{(1-v^2/c_0^2)T_2-\frac{v(X_2-vT_2)}{c_0^2}-(1-v^2/c_0^2)T_1-\frac{v(X_1-vT_1)}{c_0^2}}=c_0[/tex]

[tex]\frac{(X_2-vT_2-X_1+vT_1)c_0^2}{(1-v^2/c_0^2)T_2c_0^2-v(X_2-vT_2)-(1-v^2/c_0^2)T_1c_0^2-v(X_1-vT_1)}=c_0[/tex]

[tex]\frac{(X_2-vT_2-X_1+vT_1)c_0^2}{( c_0^2-v^2)(T_2-T_1)-v(X_2-vT_2-X_1+vT_1)}=c_0[/tex]

[tex](X_2-vT_2-X_1+vT_1)c_0^2=c_0(( c_0^2-v^2)(T_2-T_1)-v(X_2-vT_2-X_1+vT_1))[/tex]

[tex](X_2-vT_2-X_1+vT_1)c_0^2=c_0( c_0^2-v^2)(T_2-T_1)-v c_0(X_2-vT_2-X_1+vT_1)[/tex]

[tex](X_2-vT_2-X_1+vT_1)(c_0^2+v c_0)=c_0( c_0^2-v^2)(T_2-T_1)[/tex]

[tex](X_2-vT_2-X_1+vT_1)c_0(c_0+v)=c_0( c_0+v)( c_0-v)(T_2-T_1)[/tex]

[tex](X_2-vT_2-X_1+vT_1)c_0=c_0( c_0-v)(T_2-T_1)[/tex]

[tex]X_2-vT_2-X_1+vT_1=c_0(T_2-T_1)-v(T_2-T_1)[/tex]

[tex]X_2-vT_2-X_1+vT_1+vT_2-vT_1=c_0(T_2-T_1)[/tex]

[tex]X_2-X_1=c_0(T_2-T_1)[/tex]

[tex]\frac{X_2-X_1}{T_2-T_1}=c_0 [/tex]
 
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  • #104
Aether said:
This is just a statement that in SR one-way light speed is isotropic, but I will examine it anyway.

SR, RMS (Robertson-Mansouri-Sexl text theory), and Lorentz ether theory are all in agreement that the speed of light [tex]c_0[/tex] is defined to be isotropic with respect to at least one particular inertial frame (e.g., within the ether frame [tex]\frac{X_2-X_1}{T_2-T_1}=c_0[/tex] -- actually, all velocites are isotropic within this frame, so this equation really relates just to this example, I suppose that a more general equation like this [tex]\frac{X_2-X_1}{T_2-T_1}=v_0[/tex] is always valid). In SR however, the Lorentz transform relates this particular (arbitrary) inertial frame with every other inertial frame while explicitly preserving light speed, and therefore the speed of light is further defined (within SR) to be isotropic with respect to every other inertial frame as well.

For example, these are (two sets of) the Lorentz transforms for motion along the x-axis:

[tex]t_1=((1-v^2/c_0^2)^{1/2}T_1-vx_1/c_0^2)[/tex],
[tex]x_1=((X_1-vT_1)/(1-v^2/c_0^2)^{1/2})[/tex],
[tex]t_2=((1-v^2/c_0^2)^{1/2}T_2-vx_2/c_0^2)[/tex],
[tex]x_2=((X_2-vT_2)/(1-v^2/c_0^2)^{1/2})[/tex].

Rather than defining light-speed isotropy for all (inertial) frames, I will justify this equation [tex]c(v,\theta)=c_0[/tex] by defining light-speed isotropy only in the ether frame (e.g., [tex]\frac{X_2-X_1}{T_2-T_1}=c_0[/tex]) and then combine this definition with the Lorentz transformations to show that light speed is therefore isotropic in all (inertial) frames [tex]\frac{x_2-x_1}{t_2-t_1}=c_0[/tex]; such a showing is ultimately coordinate-system dependent, and it is definitely not something that could ever be proven by an experiment.

So, to verify that this equation [tex]\frac{x_2-x_1}{t_2-t_1}=c_0[/tex] is true (when [tex]\frac{X_2-X_1}{T_2-T_1}=c_0[/tex]) then simply transform the four coordinates from any arbitrary inertial frame (e.g., use any v you like) into the ether frame using the Lorentz transform equations given above and verify that you always arrive back at this equation [tex]\frac{X_2-X_1}{T_2-T_1}=c_0[/tex] which we have defined above to be true.

Start with:
[tex]\frac{x_2-x_1}{t_2-t_1}=c_0[/tex]

Transform the four space-time coordinates (by substitution) using the Lorentz transforms:
[tex]\frac{((X_2-vT_2)/(1-v^2/c_0^2)^{1/2})-((X_1-vT_1)/(1-v^2/c_0^2)^{1/2})}{((1-v^2/c_0^2)^{1/2}T_2-vx_2/c_0^2)-((1-v^2/c_0^2)^{1/2}T_1-vx_1/c_0^2)}=c_0[/tex]

Now, simply reduce this equation to [tex]\frac{X_2-X_1}{T_2-T_1}=c_0[/tex] to show that [tex]c(v,\theta)=c_0[/tex] is true (this doesn't actually prove that it is true for all [tex]\theta[/tex] because we're only looking at motion along the x-axis).

Well, not to be nitpicking but you wrote above
[tex]c(v,\theta)=c_0[/tex] for ANY [tex]\theta[/tex]
The following derivation is true only for [tex]\theta=0[/tex]
[tex]\frac{X_2-vT_2-X_1+vT_1}{(1-v^2/c_0^2)T_2-\frac{v(X_2-vT_2)}{c_0^2}-(1-v^2/c_0^2)T_1-\frac{v(X_1-vT_1)}{c_0^2}}=c_0[/tex]

There are some sign errors in the first expression. The correct thing is:

[tex]\frac{X_2-vT_2-X_1+vT_1}{(1-v^2/c_0^2)T_2-\frac{v(X_2-vT_2)}{c_0^2}-(1-v^2/c_0^2)T_1+\frac{v(X_1-vT_1)}{c_0^2}}=c_0[/tex]
You would then get:

[tex]\frac{X_2-vT_2-X_1+vT_1}{T_2-\frac{vX_2}{c_0^2}-T_1+\frac{vX_1}{c_0^2}}=c_0[/tex]

If you divide both the numerator and the denominator by [tex]T_2-T_1[/tex] and you remember that [tex]\frac{X_2-X_1}{T_2-T_1}=c_0[/tex] you indeed get your result.
Very ugly...
 
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  • #105
clj4 said:
Well, not to be nitpicking but you wrote above
[tex]c(v,\theta)=c_0[/tex] for ANY [tex]\theta[/tex]
The following derivation is trur only for [tex]\theta=0[/tex]
That's right. Do you think that it's important to show a more general proof for motion in 3D?

There are some sign errors in the first expression. The correct thing is:

[tex]\frac{X_2-vT_2-X_1+vT_1}{(1-v^2/c_0^2)T_2-\frac{v(X_2-vT_2)}{c_0^2}-(1-v^2/c_0^2)T_1+\frac{v(X_1-vT_1)}{c_0^2}}=c_0[/tex]
You would then get:

[tex]\frac{X_2-vT_2-X_1+vT_1}{T_2-\frac{vX_2}{c_0^2}-T_1+\frac{vX_1}{c_0^2}}=c_0[/tex]

If you divide both the numerator and the denominator by [tex]T_2-T_1[/tex] and you remember that [tex]\frac{X_2-X_1}{T_2-T_1}=c_0[/tex] you indeed get your result.
There is a sign error in the first two equations that I gave, but the rest are OK. The correct equations are:

[tex]\frac{X_2-vT_2-X_1+vT_1}{(1-v^2/c_0^2)T_2-\frac{v(X_2-vT_2)}{c_0^2}-(1-v^2/c_0^2)T_1+\frac{v(X_1-vT_1)}{c_0^2}}=c_0[/tex]
[tex]\frac{(X_2-vT_2-X_1+vT_1)c_0^2}{(1-v^2/c_0^2)c_0^2T_2-v(X_2-vT_2)-(1-v^2/c_0^2)c_0^2T_1+v(X_1-vT_1)}=c_0[/tex]

Very ugly...
What do you mean by this?
 
<h2>1. What is the speed of light?</h2><p>The speed of light is a fundamental constant in physics, denoted by the letter c. It is defined as the speed at which light travels in a vacuum, which is approximately 299,792,458 meters per second.</p><h2>2. Is the speed of light always constant?</h2><p>Yes, according to the theory of relativity, the speed of light is always constant regardless of the observer's frame of reference. This means that no matter how fast an observer is moving or in which direction, they will always measure the speed of light to be the same.</p><h2>3. Can anything travel faster than the speed of light?</h2><p>According to our current understanding of physics, nothing can travel faster than the speed of light. The speed of light is considered to be the cosmic speed limit, and traveling faster than it would violate the laws of physics as we know them.</p><h2>4. How do we know that the speed of light is consistent?</h2><p>Scientists have conducted numerous experiments and observations to confirm the consistency of the speed of light. One of the most famous experiments is the Michelson-Morley experiment, which showed that the speed of light is the same in all directions. Additionally, the laws of physics, such as the theory of relativity, are based on the assumption of the constant speed of light.</p><h2>5. Has the speed of light always been constant?</h2><p>According to our current understanding, the speed of light has always been constant. However, there are some theories that suggest the speed of light may have been different in the early stages of the universe. These theories are still being studied and have not been confirmed.</p>

1. What is the speed of light?

The speed of light is a fundamental constant in physics, denoted by the letter c. It is defined as the speed at which light travels in a vacuum, which is approximately 299,792,458 meters per second.

2. Is the speed of light always constant?

Yes, according to the theory of relativity, the speed of light is always constant regardless of the observer's frame of reference. This means that no matter how fast an observer is moving or in which direction, they will always measure the speed of light to be the same.

3. Can anything travel faster than the speed of light?

According to our current understanding of physics, nothing can travel faster than the speed of light. The speed of light is considered to be the cosmic speed limit, and traveling faster than it would violate the laws of physics as we know them.

4. How do we know that the speed of light is consistent?

Scientists have conducted numerous experiments and observations to confirm the consistency of the speed of light. One of the most famous experiments is the Michelson-Morley experiment, which showed that the speed of light is the same in all directions. Additionally, the laws of physics, such as the theory of relativity, are based on the assumption of the constant speed of light.

5. Has the speed of light always been constant?

According to our current understanding, the speed of light has always been constant. However, there are some theories that suggest the speed of light may have been different in the early stages of the universe. These theories are still being studied and have not been confirmed.

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