# The Light Speed Postulate and its Interpretation in Derivations of the Lorentz Transformation

Discussion in 'Pseudoscience' started by tsmid, Apr 24, 2016.

1. ### tsmidRegistered Senior Member

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368
• This is pseudomathematics and pseudoscience. It apes the form of math and science without doing more than confirm the writer's biases against 20th century.
I. INTRODUCTION

The invariance of the speed of light in different reference frames lies at the heart of the derivation of the Lorentz transformation and thus the theory of Special Relativity. The derivation typically assumes two given events associated with the propagation of a light signal, and then looks for a linear transformation of the space and time coordinates of these events that would also correspond to the same speed of light c in a different reference frame moving with a velocity v relatively to the first. In the following it is shown that such a transformation can not exist without violating the speed of light postulate applied to an independent measurement of the speed of light in the second reference frame.

II. PROOF

Version 1

We designate the two reference frames with the indices 1 and 2 respectively and assume that in each frame we have a separate stationary array of light detectors. Let a light signal be emitted by a light source at the origin of both reference frames, the light source being at rest in frame 1 but moving with velocity v in frame 2. Assuming the detectors to be equipped with synchronized clocks and the light signal emitted at zero system time in each frame, the detector locations and associated detection times are then given by

(1) x1=c*t1
(2) x2=c*t2

where (2) follows from the speed of light postulate.
We have formally allowed different time units here, but use of the same numerical value for c implies obviously

(3) x1/x2 =t1/t2 =q

where q is some constant. As the value of q is merely a matter of convention, we can therefore, without loss of generality, choose q=1 and thus set

(4) t1 =t2 =t

and instead of (1) and (2) we have then

(5) x1=c*t
(6) x2=c*t

Note that these timings, despite being identical here due to the chosen convention, are as such completely independent of each other. Frame 2 obtains its data completely independently of the data obtained in frame 1. Equations (5) and (6) define the speed of light postulate therefore without the need of transforming the frame 1 data into frame 2. If we wanted to do the latter in addition, we would require that the coordinate transformed location of x1 coincides with x2, i.e. using (5) and (6) we would get

(7) x1'=x2=c*t=x1

It is obvious that this does not allow any velocity dependent transformation of the form

(8) x1'=a*(x1-b*t)

unless a=1 and b=0 (note that (7) requires that x1' is reflected about the origin whenever x1 is reflected about the origin.).

Version 2

In contrast to version 1, we do not take any conventions at all about the time units used in the two reference frames, so we retain the initial equations

(9) x1=c*t1
(10) x2=c*t2

Again, the corresponding detection events in both equations are completely independent of each other (the data in one frame can be obtained without having any knowledge of those obtained in the other frame). This system of equations consists thus of two independent variables (the locations of the detectors x1 and x2) and two dependent variables (the corresponding detection times t1 and t2). If we would require that the event (x2,t2) can be represented as a coordinate transformation of the event (x1,t1), we would have the additional constraints

(11) x2=x1'=f(x1,t1,v)
(12) t2=t1'=g(x1,t1,v)

where f and g are the transformation functions of the arguments.
However, (11) would turn the independent variable x2 into a dependent variable, and (12) would, via the inversion of the equation, turn the independent variable x1 into a dependent variable. There would thus be no independent variable left in the system of equations (9) and (10). This would be unacceptable both mathematically and physically.

III. DISCUSSION

As is evident from both of the above considered scenarios, any velocity dependent transformation is inconsistent with the speed of light postulate, Such a transformation is only applicable for material (massive) objects, for which the speed is not invariant in different reference frames. For those, independent measurements in the two reference frames (moving with velocity v relatively to each other) would instead of (5) and (6) yield

(13) x1=u*t
(14) x2=(u-v)*t

i.e.

(15) x2=x1-v*t

and thus a transformation equation of the form (8) follows naturally.
In case of a propagating light signal, the problem is of course not the equation (8) but (7) where x1' is assumed to be co-located with x2. If we don’t make this assumption, we can without any issues take for instance

(16) x1'=x1-v*t

as the speed of light postulate is independently satisfied by (5) and (6).
And this would of course also remove the impossible situation of having a system of equations without any independent variables (as argued in version 2 above).

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3. ### Xelasnave.1947Valued Senior Member

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Set up an actual experiment and gather some real data maybe?

Alex

5. ### SchneibsterRegistered Member

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390
You have combined coordinates from two different frames without transforming them; x1 is only valid in frame 1, in frame 2 it will have a different value. When you divided one by the other in equation (3) is where you made your first error. You can't do that; you haven't transformed x2 into its coordinates in frame 1. You will actually have two different values for both x1 and x2: their values in frame 1, and their values in frame 2. When you put them into the same term you have to use either their values in frame 1 or their values in frame 2. You can't mix them. It's like dividing unicorns by basilisks.

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7. ### tsmidRegistered Senior Member

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368
x1 and x2 are not transforms of each other here, they are by definitions independent locations (the locations of the stationary detectors in each frame). And if the speed of light postulate holds, then we must have the independent equations

(1) x1=c*t1
(2) x2=c*t2

Requiring that x1 and x2 are also coordinate transforms of each other would mean that x2=x1'.

8. ### Farsight

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3,492
It doesn't. See Einstein talking about it here:

"However, the writer of these lines is of the opinion that the theory of relativity is still in need of a generalization, in the sense that the principle of the constancy of the velocity of light is to be abandoned".

Or see this:

"I arrived at the result that the velocity of light is not to be regarded as independent of the gravitational potential. Thus the principle of the constancy of the velocity of light is incompatible with the equivalence hypothesis".

Also see this Baez article:

"That the speed of light depends on position when measured by a non-inertial observer is a fact routinely used by laser gyroscopes that form the core of some inertial navigation systems. These gyroscopes send light around a closed loop, and if the loop rotates, an observer riding on the loop will measure light to travel more slowly when it traverses the loop in one direction than when it traverses the loop in the opposite direction. This is known as the Sagnac Effect..."

"Einstein talked about the speed of light changing in his new theory. In the English translation of his 1920 book "Relativity: the special and general theory" he wrote: "according to the general theory of relativity, the law of the constancy of the velocity [Einstein clearly means speed here, since velocity (a vector) is not in keeping with the rest of his sentence] of light in vacuo, which constitutes one of the two fundamental assumptions in the special theory of relativity [...] cannot claim any unlimited validity. A curvature of rays of light can only take place when the velocity [speed] of propagation of light varies with position" . This difference in speeds is precisely that referred to above by ceiling and floor observers."

9. ### tsmidRegistered Senior Member

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368
Farsight
I am not sure what you are trying to say here, and why you are bringing in General Relativity. The invariance of the speed of light is an observational fact that exists before Relativity. It is only that there are different ways of defining this invariance, either by using the same events in both reference frame (like Einstein did) or by using different events (albeit relating to the same light signal).

10. ### SchneibsterRegistered Member

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390
Errr, I don't think you're getting it. You can't define a coordinate in one frame, then divide it by a coordinate in another. And that's what you did in (3), which I notice you never mentioned.

For this to even work, you'd have to have

(1a) x1=c*t1
(1b) x'1 = c*t'1
(2a) x2 = c*t2
(2b) x'2 = c*t'2

Then 3 would become:
(3a) x1/x'2 = t1/t'2 = q1
(3b)x'1/x2 = t'1/t2 = q2

but you can't have q1 = q2, at which point your "proof" dies because you can't say t1 =t2 either, as you did in (4). Like I said, you can't divide unicorns by manticores and expect to get anything meaningful.

Last edited: Apr 24, 2016
11. ### Farsight

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3,492
Because it subsumes Special Relativity.

The speed of light varies in the room you're in. If it didn't, light wouldn't curve and your pencil wouldn't fall down. If you placed NIST optical clocks at different elevations then plotted their different readings, your plot would map out the speed of light at different elevations. There's no time flowing inside those clocks. The higher clock goes faster because the speed of light is greater.

12. ### tsmidRegistered Senior Member

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368
If the observer in frame 1 has his detector at the fixed location x1 registering the light signal at time t1 (assuming the light is emitted from the origin at time t1=0), and independently of that, the observer in frame 2 has his detector at the fixed location x2 registering the light signal at time t2 (assuming the light is emitted from the origin at time t2=0), then, even if the observers are initially not aware of the measurement in the other frame, they can afterwards compare their figures, and if the light speed postulate holds, they must find x1/x2=t1/t2.
I don't understand why you disagree with this.

Last edited: Apr 24, 2016
13. ### SchneibsterRegistered Member

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390
All fine, but 1 ≠ 2 and therefore t1 ≠ t2. If you're going to start by talking about t1, you have to use t'1 to talk about t2 in frame 1. Otherwise you can't do any mathematical operation that has even mathematical meaning, far less physical meaning. The relationship of two frames is arbitrary.

Because you're mixing similes.

Last edited: Apr 24, 2016
14. ### tsmidRegistered Senior Member

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368

According to

(1) x1=c*t1
(2) x2=c*t2

t1 and t2 depend on x1 and x2 respectively. If t1 ≠ t2 then x1 ≠ x2. As (1) and (2) relate to different events (not the same event as in Einstein's assumption), we don't have to relate to t1 in any way to defined t2. The two times are completely unrelated. If you want to relate them by a coordinate transformation, then you would need the additional constraint

(2a) x2=x1'

15. ### SchneibsterRegistered Member

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390
Yep.

And if they're completely unrelated then you can't put them in the same equation, much less the same term.

Nope. x2 = f(x'1). f is the transform.

16. ### tsmidRegistered Senior Member

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368
What you makes you think that? If I measure my height as h1 and you measure independently your height as h2, then we can do anything with these figures

h1>h2
h1<h2
h1=h2
h1/h2 =q

or anything else you want.
The situation is in no way different here if (x1,t1) and (x2,t2) relate to different, unrelated, events.

17. ### SchneibsterRegistered Member

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390
Math.

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21,803
Rubbish.
The speed of light never varies, period! The apparent coordinate speed by definition may vary, but this is simply because light has travelled further.

With regards to your clock rubbish, you have been continually refuted and derided on that aspect many times, but you keep coming back with the same nonsense.
A clock higher up in the atmosphere, or from the surface of a planet, most certainly does go faster, just as you and I would age faster, because it/you are further from the center of gravity, and gravity slows down time, both physically and biologically.
https://en.wikipedia.org/wiki/Gravitational_time_dilation
Gravitational time dilation is a form of time dilation, an actual difference of elapsed timebetween two events as measured by observers situated at varying distances from a gravitating mass. The weaker thegravitational potential (the farther the clock is from the source of gravitation), the faster time passes.
""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""

Certainly not as you interpret or presumably is noted in your pop science book, and most certainly not as it is viewed and accepted by mainstream.
Perhaps even Schmelzer may correct you on that misunderstanding.

19. ### Confused2Registered Senior Member

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503
If it were left to me I'd have two identical rockets A and B. A is heading nose first towards B and B is orientated at 90 degrees to the the path path of A, effectively heading towards A 'sideways'. In each rocket a pulse of light is fired up the rocket (of length l) taking a time l/c. It is then shown that all observations are consistent with a constant c. It might be nice (later) to have arbitrary orientations but the simplest case should illuminate any problems in principle while keeping the practice reasonably simple.
-C2.

20. ### SchneibsterRegistered Member

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390
But we're using different measuring sticks; yours is 8 inches and mine is 12. We can't just go ahead and add your measurement with your stick of your height to my measurement with my stick of my height. We both gotta use the same length stick, or we gotta multiply your measurements by 2/3 or mine by 3/2. Otherwise, we're transferring coordinates from one frame to another without transforming them.

21. ### przyksquishyValued Senior Member

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3,173
Referring to Version 2:
Based on my understanding of your thought experiment, this step in the reasoning makes no sense to me:
Coordinate transformations in general relate different coordinates that different observers* would attribute to the same event ("point") in space and time. So if (x1, t1) are the coordinates of an event according to observer 1 and (x2, t2) are the coordinates of an event according to observer 2, then it only makes sense to relate these by a coordinate transformation if you believe that these events are the same or that they happen to coincide (both observers agree that the events occur at the same place and at the same time).

In your thought experiment, t1 is the time at which a detector (call it "detector 1") located at x1 detects a light pulse according to observer 1, and t2 the time at which a different detector ("detector 2") detects a light pulse according to observer 2. The transformed coordinates t1' = g(x1, t1, v) and x1' = f(x1, t1, v) have nothing a priori to do with t2 and x2. Here, t1' would be the time at which detector 1 detects the light pulse according to observer 2. Detector 1 is (presumably) at rest according to observer 1, so it would be moving according to observer 2, and x1' is where it is (temporarily) located according to observer 2 when it detects its pulse. Invariance of c here only means that x1' and t1' should be related by x1' = c t1' if x1 and t1 are related by x1 = c t1.

If you assert that x2 = x1' and that t2 = t1' then you are claiming that the detection events happen to coincide, i.e., that detectors 1 and 2 happen to be located at the same place at the same time when they both detect the light pulse. In other words, you are yourself introducing an assumption that contradicts your own earlier claim that these events are unrelated.

----------

*This is a figure of speech. Coordinate systems don't in general need to be associated to observers.

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22. ### Confused2Registered Senior Member

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503
Bit of a test here:-
(fail)

23. ### rpennerFully WiredValued Senior Member

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4,833
That means a signal moving at the speed of light in vacuum, c, can move between any two space-time events in one frame if and only if it can move between the same two space-time events as described in another frame. Or algebraically: $c^2 (t_2 -t_1)^2 - (\vec{x}_2 - \vec{x}_1)^2 = 0 \Leftrightarrow c^2 (t'_2 -t'_1)^2 - (\vec{x}'_2 - \vec{x}'_1)^2 = 0$.

Well if $t'(t, \vec{x}) = \frac{1}{\sqrt{1-v^2/c^2}} ( t - c^{-2} \vec{v} \cdot \vec{x} ) + T, \vec{x}'(t, \vec{x}) = \vec{x} + \left( \frac{1}{\sqrt{1-v^2/c^2}} - 1 \right) v^{-2} ( \vec{v} \cdot \vec{x} ) \vec{v} - \frac{1}{\sqrt{1-v^2/c^2}} t \vec{v} + \vec{X}$ , I hope you will recognize that as linear in t and $\vec{x}$. Then:
$c^2 (t'_2 -t'_1)^2 - (\vec{x}'_2 - \vec{x}'_1)^2 = c^2 (t_2 -t_1)^2 - (\vec{x}_2 - \vec{x}_1)^2$

Since a frame isn't a physical thing, you have made a conceptual error by saying there could be separate arrays of light detectors. The light detectors have to co-exist in both frames. So you have two classes of light detectors: those stationary in frame 1 and those stationary in frame 2.

Event O (x1=0, t1=0; x2=0, t2=0). x2 = (x1 + v t1) / √(1 – v²/c²) , t2 = (t1 + v x1/c²) / √(1 – v²/c²)

Which class of detectors?
how synchronized?
Wrong.
You have parametrized (labeled) your array of detectors at rest in frame 1 with their frame-1 x-coordinate, which needs a name, λ. This choice parametrizes arrival events at the frame-1 detectors as:
(x1_A = λ, t1_A = |λ|/c ; x2_A = (λ + (v/c) |λ|) / √(1 – v²/c²) , t2 = (|λ|/c + v λ/c²) / √(1 – v²/c²))
Similarly, treating the frame-2 detectors, we need a new family of event, B, and a new parameter, μ.
(x1_B = (μ – (v/c) |μ|) / √(1 – v²/c²), t1_B = (|μ|/c – v μ/c²) / √(1 – v²/c²); x2_B = μ, t2 _B= |μ|/c)
Since both the A events and the B events happen along the same pulse of light, it follows that every event A(λ) is co-incident with an event B(μ). Since |v| < c we have:

A(λ) = B(μ) ⇔ λ = μ ( 1 – sgn(μ) v/c ) / √(1 – v²/c²) AND μ = λ ( 1 + sgn(λ) v/c ) / √(1 – v²/c²) AND sgn(μ) = sgn(λ)

q is not constant, but depends on the sign of λ

x1_A/x2_A = √(1 – v²/c²) / (1 + sgn(λ) (v/c) ) = √( (c - sgn(λ) v ) / (c + sgn(λ) v ) )
t1_A/t2_A = √( (c - sgn(λ) v ) / (c + sgn(λ) v ) )
x1_B/x2_B = (1 – sgn(μ) (v/c) ) / √(1 – v²/c²) = √( (c - sgn(μ) v ) / (c + sgn(μ) v ) ) = √( (c - sgn(λ) v ) / (c + sgn(λ) v ) ) ; when A(λ) = B(μ)
t1_B/t2_B = √( (c - sgn(λ) v ) / (c + sgn(λ) v ) ) ; when A(λ) = B(μ)

FALSE.
FALSE. Such a choice of q=1 leads to complete loss of generality.
Such a choice requires v=0 (or λ=0), and so you are no longer talking about DIFFERENT standards of rest in two DIFFERENT inertial coordinate systems.

From the choice you made of q=1, you get v=0, t1=t2, x1=x2 and so equations (5), (6), (7), (8) are trivially true and say nothing about transformations between DIFFERENT coordinate systems.

since eqn (9) (10) are redundant to (1)(2), you are wasting space.

FALSE. Conceptually wrong from start to finish. That's why I bothered to introduce λ to parametrize A, and μ to parametrize B.

FALSE. One was given above and used throughout.

FALSE. Completely misconceptualizes the need for relativity.

equations (13-15) recapitulate Galilean relativity, which is incompatible with electromagnetic phenomena and therefore reality.

Last edited: Apr 25, 2016