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

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I think I fixed my Latex typos.

BTW, because the content of the OP is not very original, the points raised in post #20 also apply to:
http://www.physicsmyths.org.uk/speed-of-light-postulate.htm

 

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I made it to the second sentence on that page and stopped having found an error:

"The coordinates of a light signal" is meaningless. There is no usable frame of reference in which light is still; such a frame is mathematically intractable because any attempt at analysis runs into a divide-by-zero error. The "frame of light" is useful only as a didactic device for explaining relativity to someone who doesn't want to do the math; it has no physical meaning for the reasons given.

 

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You can use any measuring units in each frame + place the respective detectors at an arbitrary distance in these units. Provided neither of the experimenters cheats with his figures, they will in this case find x1/x2=t1/t2 if the speed of light postulate holds.

 

No, they won't, because one experimenter will see everything as x1 and x2', and the other will see everything as x2 and x1'. You're comparing apples and raspberries.

 

Yes, you have two classes of detectors, and thus two classes of detection events. So x2 is not the transformed coordinate of a class 1 event in frame 2 but completely independent of x1. Yet still the speed of light postulate must apply

(1) x1=c*t1 (speed of light in frame 1 is c=x1/t1)
(2) x2=c*t2 (speed of light in frame 2 is c=x2/t2)

 

But the observer in frame 2's observation of the measurement in frame 1 is not necessarily the same as the observer in frame 1's observations of the measurement in frame 1.

This is what you get when you try to compare measurements in two frames without transforming them properly.

 

Experimenter 1 is not measuring x2', only x1. Likewise, experimenter 2 is not measuring x1' only x2. You don't need any coordinate transformation if each experimenter performs an independent measurement.

 

Complete lack of understanding of relativity detected. Raise the deflector shields and arm the photon torpedoes.

And when I speak of relativity in this context, I mean both Galilean and Einsteinian relativity.

 

You will understand relativity if you think more carefully about the points I made.

 

Correction, in light of post #20, you will understand relativity better if you think carefully what A(λ) = B(μ) implies.

That only happens when the event of the family A indicated by parameter λ is exactly the same event as the member of family B indicated by parameter μ. And since all the events of A and B happen just as the flash arrives at the detectors of family A or B, that means we are measuring the same space-time interval between O and A(λ) as between O and B(μ) -- and that space-time interval is a null-interval because light travels at c. So the ratio "q" which you poorly assumed was a free parameter actually tells us how null-intervals transform under a Lorentz transform.

This allows us to decompose a Lorentz transform into the space-time directions of null-rays.

Conventionally, the Lorentz tranform is written in linear algebra as:
\( \begin{pmatrix} t' \\ x' \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{1 - v^2/c^2}} & \frac{v}{c^2\sqrt{1 - v^2/c^2}} \\ \frac{v}{\sqrt{1 - v^2/c^2}} & \frac{1}{\sqrt{1 - v^2/c^2}} \end{pmatrix} \begin{pmatrix} t \\ x \end{pmatrix}\)

But the same transform can also be written as:
\( \begin{pmatrix} t' \\ x' \end{pmatrix} = \begin{pmatrix} \frac{1}{2c} & - \frac{1}{2c} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} \sqrt{\frac{c+v}{c-v}} & 0 \\ 0 & \sqrt{\frac{c-v}{c+v}} \end{pmatrix} \begin{pmatrix} c & -c \\ 1 & 1 \end{pmatrix} \begin{pmatrix} t \\ x \end{pmatrix}\)

(for -c < v < c )

Likewise
\( \begin{pmatrix} \cosh \rho & \frac{1}{c} \sinh \rho \\ c \sinh \rho & \cosh \rho \end{pmatrix} = \begin{pmatrix} \frac{1}{2c} & - \frac{1}{2c} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} e^{\rho} & 0 \\ 0 & e^{-\rho} \end{pmatrix} \begin{pmatrix} c & -c \\ 1 & 1 \end{pmatrix} \)

Learning -- it's fun.

 

You appear to have missed my post #25 above, where I made clear that the class 1 events events are different events from the class 2 events. There is no transformation possible between the two. Each experimenter determines the speed of the light signal independently in his own frame using his own detectors. Experimenter 1 (the light source is at rest with regard to his detector array) measures this

x1=c*t1

Experimenter 2 (the light source is moving with regard to his detector array) measures this

x2=c*t2

from which mathematically follows

x1/x2 = t1/t2


You should forget the about Lorentz transformation (or any transformation for that matter) for a moment. Up to this point it is solely about the application of the speed of light postulate in the above scenario with independent events defining the speed of light, not about Einstein's scenario using the same events (which obviously would require a transformation).

 

So you agree with me that we can talk about the events where the light strikes the detectors which are motionless in frame 1 as the family of events A; and that a specific event of that type can be uniquely indexed (labelled) by its position in coordinates used for frame 1, so a particular event A(λ) happens at position x1=λ and at time t1=| λ | / c. Likewise event B(μ) happens at position x2=μ and time t2= | μ | / c.

This is not true. Because from O, from the point where x1=x2=0, t1=t2=0, comes a pulse of light which hits every location in your 1+1 dimensional space-time at only one time, so the family of all events along the light pulse is a one-dimensional family of events. So A(λ) and B(μ) each parametrize all events along the propagation of the light pulse. So that there must be some absolute correspondence between A(λ) and B(μ) such that for every λ there must be one and only one corresponding μ. Geometrically, this is the meaning of the coordinate-free expression A(λ) = B(μ).

Obviously A(0) = B(0) = O.

Only if you want to forfeit the discussion. If you can't relate events which happen in the same space time to two different coordinate systems, you are not doing relativity.

Therefore there must be a scheme so that x2 and t2 for every event A(λ) can be known. Likewise, there must be a scheme so that x1 and t1 for every event B(μ). Further these schemes must be compatible so that A(λ) = B(μ) means x1_A(λ) = x1_B(μ), t1_A(λ) = t1_B(μ), x2_A(λ) = x2_B(μ), t2_A(λ) = t2_B(μ)

Now I wrote the relation: x2 = (x1 + v t1) / √(1 – v²/c²) , t2 = (t1 + v x1/c²) / √(1 – v²/c²)

But what if it was more general: x2 = R x1 + S t1 ; t2 = T x1 + U t1.
We require objects which are stationary in frame 1 to have velocity v in frame 2, so S = v U
We require objects which have velocity +c in frame 1 to have velocity +c in frame 2, so c² T + ( c – v ) U – c R = 0
We require objects which have velocity –c in frame 1 to have velocity –c in frame 2, so c² T – (c + v) U + c R = 0
Therefore T = v/c² U and R = U.

So the choice of having two different notions of stationary that differ by v, gives us: x2 = U ( x1 + v t1 ) , t2 = U( t1 + (v/c²) x1 )
Already, we have the same-direction velocity composition law: u2 = ( u1 + v ) / ( 1 + u1 v / c² )

There are many arguments that U needs to be 1/√(1 – v²/c²). If we consider the inverse relationship: x1 = V ( x2 – v t2 ) ; x2 = V ( t2 – (v/c²) x1 ) we get V U = 1 / (1 – v²/c²) which for the reason that nothing is special about frame 1 just because you labelled it first means U = 1/√(1 – v²/c²) is sensible.
A stronger argument requires three frames to show that transforming from frame 1 to frame 2 to frame 3 can be done as a single transform from 1 to 3 if U = 1/√(1 – v²/c²).
A higher argument requires det Λ = 1, and so U = 1/√(1 – v²/c²).
Physical experiment is entirely consistent with U = 1/√(1 – v²/c²).

A(λ) happens at position x1=λ and at time t1=| λ | / c which is also is at x2 = U λ ( 1 + (v/c) sgn(λ) ) , t2 = U | λ | / c ( 1 + (v/c) sgn(λ) )
So the requirement that U = 1/√(1 – v²/c²) leads to ratio x1/x2 = t1/t2 = √( ( c – sgn(λ) v ) / ( c + sgn(λ) v ) ) which also shows up in post #31.
Trying to force x1/x2 = 1 is inconsistent, because U would not be constant. So your breezy claim in the OP that "As the value of q is merely a matter of convention, we can therefore, without loss of generality, choose q=1 " is utter nonsense. You not only failed to carry your burden, you wandered into the realm of the barking mad. Your pretentious aping of the language of a mathematical paper is found to be a hollow sham.

Since they are on the same propagating pulse of light, they are necessarily the same class of events, just labelled differently.

Now you are just in naked denial.

That expression only makes any sense if (x1,t1) and (x2,t2) are two different descriptions of the same event. i.e. when A(λ) = B(μ).

If you say you don't care if you are talking about the same event, then q is no constant, but a function of which detectors you choose, x1/x2 = λ/μ. Moreover, if you choose detectors on opposite sides of the origin, then equation (3) doesn't hold, for t1/t2 = |λ/μ|.

Wrong. You have a single spacetime described by two coordinate systems, with a common origin event O, a single family of events generated by a single light cone, and a specification of relative motion, v. So claim now that you are not directly confronting coordinate transformations in equations 7 and 8 is disingenuous.

 

The specific event for the detector at rest in frame 1 is already labelled by the coordinate x1. It is the coordinate at which the detector is placed. Correspondingly x2 in the other frame. So I don't know why you introduce the variables λ and μ here. In fact, your equations above say directly that they are x1 and x2 respectively. So you could as well left it at A(x1) and B(x2)

Assume we send a light signal from the earth to a space probe. The earthbound detector(s) measure the speed of this signal in the earth's frame, and when the signal gets to the space probe, the detectors there measure its speed in the latter's frame. The space probe then send the result of this speed measurement back to earth, so that the speed of light postulate can be confirmed (or not as it may be). There is no correspondence and no transformation between these two measurements required. In fact, there is no correspondence possible as the events related to the earth-based and space-probe-based detectors are absolutely independent (neither of the sides is aware of the events at the other side).

We don't need to do relativity to practically define the speed of light postulate. The former depends on the latter, not the other way around. It would in fact be a circular definition if you would require the Lorentz transformation in order to define the light speed postulate in practice. Einstein did not know anything about the LT either yet when he formulated his postulates.

 

Factually incorrect. Einstein was well aware of Lorenz(1895) which contains the time dilation formula \(t'=t-vx/c^2\). Furthermore, he was also aware that this paper presented the most successful adaptation of Maxwell's equations available at the time he wrote On the Electrodynamics of Moving Bodies, Einstein(1905c). Finally, the paper also used the constancy of the speed of light.

 

Yes, OK he may have known it personally, but for the purpose of his paper he had to formally pretend that he did not know it. After all, he wants to derive it from his postulates. So for the purpose of this discussion you should pretend as well that you don't know the LT. Otherwise we are into circular arguments here.

 

That is silly. You don't have to "pretend" you don't know something in order to show that you have found a way to derive it.

You just have to make sure your derivation does not assume the thing you are trying to derive.

 

Because you were confusing yourself with thinking x1 and x2 were independent. I introduced A(λ) and B(μ) as parametric curves, which allowed me to show A(λ) and B(μ) were necessarily the SAME 1-dimensional curve, and therefore for every λ there existed one and only one value of μ such that A(λ) = B(μ). Every event has an x1, a t1, an x2 and a t2 because events are the elements of the geometry of space-time while coordinates are naming conventions for the elements of space time. Events are more fundamental than coordinates. But when A(λ) = B(μ) we are talking about the SAME event, with the same geometric relationship to event O and so that's the time when the ratio x1/x2 has something to say about how the coordinates (x1,t1) relate to (x2,t2).

Your strange claim to not care if A(λ) = B(μ) would mean you weren't interested in relating coordinates and so your ratios and equations 7 and 8 are unjustified by any rationale. You wouldn't be anywhere the same domain as relativity because you ignored physics and identity. Since you ran off the rails of your own purporting topic, introducing A(λ) and B(μ) was an attempt to show how formalism keeps one on track.

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You originally said there were two classes of detection events, and since an event is a a point of space-time -- a particular time and place -- I pointed out that was not the case for the same 1-dimensional locus parameterized in two ways: A(λ) and B(μ). Did you understand my point? Did you respond meaningfully to it?

Poppycock. For such a comparison x1_A/x2_B may be many orders of magnitude different from 1 because x1_A/x1_B is obviously many orders of magnitude different from 1. The idea that you would think that this ratio would tell you anything about how coordinates (x1,t1) relate to (x2,t2) as you did in equations 7 and 8 is deeply misguided. And you still have not even partially justified your claim that there are two classes of detection events. The motion of the detectors does not change that the detectors only fire when they hit the wavefront of the light cone, so in this 1+1 dimensional spacetime, the locus of all detection events is necessarily 1-dimensional and is completely parameterized by either A(λ) or B(μ) which allows us to compare coordinates for the same event by equating A(λ) and B(μ).

—

LIES! You need the principle of relativity to formulate the speed of light postulate, because only after you assume that OTHER PEOPLES rulers and clocks are just as valid as yours (provided the law of inertia holds) can you ask IF they measure the same speed of light as you do. Special Relativity is what you get when you assume both postulates.

Wrong. In Galilean relativity, different frames would measure different speeds of light. Special Relativity is predicated on both the principle of relativity and the invariance of the speed of light in vacuum.

Incorrect, as others have shown. And the Lorentz transformation was derived from the postulates in 1905, not used to define the postulates.

 

Assuming just the principle of relativity and a few everyday encountered symmetries, you get:

\(x' = \frac{x + vt}{\sqrt{1 - K v^2}} \\ t' = \frac{t + Kvx}{\sqrt{1 - K v^2}} \)​

where K is an unknown constant.

Which leads to 1-dimensional velocity composition law:

\(u' = \frac{u + v}{1 + K v u}\)​

which allows for experimental determination of K. Since 1859 we have had experimental evidence that favors \(K=c^{-2}\) over the Galilean \(K=0\).

 

Yes, exactly, and for this it is best you don't even mention the thing before you actually have derived it. Otherwise, it would not look like a clean deductive derivation (whether that derivation is a correct one is then of course still a different question).