Refutation of Special Relativity for Dummies

[Perpetual Student]

Look my post #177. The units in vx/c^2 clearly reduce to simply s, which is consistent in the context of the time dilation equation. I have no idea how he is getting 10^(-!6) m/s or s/m.

 

[fuelair]

Look my post #177. The units in vx/c^2 clearly reduce to simply s, which is consistent in the context of the time dilation equation. I have no idea how he is getting 10^(-!6) m/s or s/m.

Nor do others. It is exactly like magic. Except the form "magic" takes works because smoke and mirrors. With which this may be quite easily and accurately compared.

 

[Darwin123]

Nor do others. It is exactly like magic. Except the form "magic" takes works because smoke and mirrors. With which this may be quite easily and accurately compared.

That is true. The OP doesn't like it when anyone answers his challenge. So when faced with his 'concrete error', he Segways into another topic altogether by adding another concrete error. That way, the last reply in the thread doesn't show the original concrete error that started it.

However, I want the end of this thread to always end in a 'concrete mistake'. So let me repeat a major concrete error. This way, no one forgets. I will rephrase it a bit in case he doesn't understand.

A body becomes an observer when his position is set to zero (x'=0). This is because an observer making local measurements can never detect changes in the observers position. So if the position of the observer as measured by himself starts at time t=0 is x', then x'=0 for all values of t>0. The value of x' remains zero regardless of the units of length.

So in the scenario that he was describing, x'=0 m = 0 LS = 0 km = 0 miles. However, he dismissed the hypothesis that x'=0. Instead, he decided that to first order in v, x'=x-vt where x and t is arbitrary.

His concrete error was in assuming that x and t is arbitrary when the observer is computing his own position, x'. So he wrote that x'=10 m - 1 m. So he said that x'=9 m. However, x'=0 by his initial hypotheses.

This one concrete error can be expressed in several different ways. For example:
1) The OP hypothesized that x'><0 when really x'=0.
2) The OP calculated x'=9 m when really x'=0.
3) The OP decided both x and t are arbitrary when in fact x is proportional to t.
4) The OP decided that x'=x to first order when in fact they are no equal to any order for t>0 in either Galilean or Lorentzian coordinates.

He is obligated by his own challenge to rebut any of the four claims above, which are numerically equivalent. He can challenge claims 1-4 without being accused of presenting a non sequitur. Any rebuttal that doesn't address claims 1-4 will be recognized by most posters here as a non sequitur.

So the thread ends with this claim. All his claims that result in x' not being zero are logical contradictions on his part, not relativity. In the scenario that the OP has chosen to numerically analyze, x'=0 in all units of length.

 

[wogoga]

A quote from Two myths about special relativity by Ralph Baierlein:

Q. Does the Lorentz transformation reduce to the Galilean transformation when the ratio v/c is small?
A. No.​

One argument of the author:

Consider the usual pair of inertial reference frames, the primed frame moving with speed v along the x axis of the unprimed frame. To avoid any spurious dependence on the origins of coordinate systems, consider a pair of physical events. The Lorentz transformation for the time interval between the events takes the form

Δt' = 1/√(1-v²/c²) (Δt - v/c² Δx)​

Let the ratio v/c be as small as desired (but nonzero). Then it is always possible to find an event pair for which Δx is large enough that the term with Δx dominates over the term with Δt. This behavior is entirely different from what the Galilean transformation Δt' = Δt asserts.​

It seems to me that this argument is not complete. Let us create an artificial time-transformation by replacing v/c² of the Lorentz transformation with v²/c²:

Δt' = 1/√(1-v²/c²) (Δt - v²/c² Δx)​

Any reasonable program for symbolic computation can calculate the first-order Taylor expansion with respect to v of the artificial time transformation. The result:

Δt' = Δt​

Nevertheless, for every given Δt it is possible to find a Δx large enough that

|v²/c² Δx| > |Δt|​

Therefore, from Baierlein's argument as quoted above we could conclude that our artificial transformation does not reduce to Δt' = Δt. Why is the argument valid for the Lorentz time-transformation, but not for our artificial transformation?

In the Lorentz case, in order to continue to "dominate" a given Δt, any further reduction from speed v << c to w < v must be compensated by increasing Δx with factor v/w. Thus, v Δx → const, if v → 0.

Yet in the analogue situation with the artificial transformation, length interval Δx must increase by factor v²/w² in order to "dominate" a given time interval Δt. This necessity to square v/w, leading to v Δx → infinity if v → 0, is a clear hint that there is no first-order effect.

In general, when "reducing" formulas, the order-number of the reduction is essential. E.g. the first-order reduction of kinetic energy E_kin = m v²/2 is zero. Thus, when we claim that the classical kinetic energy can be derived as a reduction from the energy-equivalent of the relativistic mass-increase

E_kin = m c² ( (1-v²/c²)^-0.5 - 1 )​

we assume a second-order Taylor expansion.


A further argument from the same chapter LOW-SPEED BEHAVIOR OF THE LORENTZ TRANSFORMATION of Ralph Baierlein:

For a sophisticated justification, note that the composition (the successive use) of two Lorentz transformations is equivalent to another Lorentz transformation. This equivalence is the group property of the Lorentz transformation. Moreover, the Lorentz transformation is differentiable with respect to v/c, and the derivative is nonzero at v/c=0. Consequently, any Lorentz transformation with finite speed can be constructed by iterating a Lorentz transformation with a small (and ultimately infinitesimal) ratio v/c.

If the Lorentz transformation for infinitesimal v/c were to reduce to the Galilean transformation, then the iterative process could never generate a finite Lorentz transformation that is radically different from the Galilean transformation. But the finite transformations are indeed radically different, and so—however subtly—the infinitesimal Lorentz transformation must differ significantly from the Galilean transformation​

Very interesting! A first-order effect with respect to v of a formula f[v] can only disappear, if the first derivative df[v]/dv is zero at v = 0. However, if we actually calculate the first derivative of the Lorentz time-transformation t' at v = 0, we get:

dt'/dv = -x/c²​

This is a clear indication that the Lorentz time-transformation reduces to t' = t only at point x = 0.

Cheers, Wolfgang

The Lorentz-transformation is dead. Long live the Lorentz-factor!!!

 

[Perpetual Student]

A quote from Two myths about special relativity by Ralph Baierlein:

Q. Does the Lorentz transformation reduce to the Galilean transformation when the ratio v/c is small?
A. No.​

One argument of the author:

Consider the usual pair of inertial reference frames, the primed frame moving with speed v along the x axis of the unprimed frame. To avoid any spurious dependence on the origins of coordinate systems, consider a pair of physical events. The Lorentz transformation for the time interval between the events takes the form

Δt' = 1/√(1-v²/c²) (Δt - v/c² Δx)​

Let the ratio v/c be as small as desired (but nonzero). Then it is always possible to find an event pair for which Δx is large enough that the term with Δx dominates over the term with Δt. This behavior is entirely different from what the Galilean transformation Δt' = Δt asserts.​

<...>

Before considering the rest of your post think about this.

In the equation Δt' = 1/√(1-v2/c2) (Δt - v/c2 Δx), at low velocities (e.g.: 10m/s) for Δt' not to reduce to Δt, Δx must be huge. We are talking about enormous distances. At , say, 10^15 meters, where Δt' would be significantly different than Δt, the distance is on the order of that of the Andromeda galaxy. But it would take light over two million years to get there. So, yes Δx must be taken into account. Nevertheless, at distances within our solar system, we are dealing with a reduction to Newtonian physics, as has been established many times.
It seems you are grasping at straws to demonstrate a preconceived notion that is patently wrong. Why? What is the motive behind this strange behavior?

 

[Darwin123]

Before considering the rest of your post think about this.

In the equation Δt' = 1/√(1-v2/c2) (Δt - v/c2 Δx), at low velocities (e.g.: 10m/s) for Δt' not to reduce to Δt, Δx must be huge. We are talking about enormous distances.

In the specific examples shown in the OPs website, Δx=0. If both observers were looking at some third body, then Δx would be arbitrary. Under that condition, Δx could be enormous. Under that condition, the Lorentz transform would not approach the Galilean transform in the limit of v/c approaching zero.

If the OP acknowledges that a third body is being observed, then maybe he has 'corrected' me. Of course, under this condition there is yet a second velocity. There is still a velocity, v, corresponding to the relative velocity between the two observers, and a velocity, w, corresponding to the velocity of this third body. Under this condition, the relativistic addition between velocities may be necessary to resolve the 'paradox'.


The OP did not mention a second velocity, however. He specifies the time intervals as measured by the two observers. So each observer can only measure time intervals where the corresponding Δx is zero (Δx=0).


Hence, the stationary observer measures Δt' where Δx'=0. The moving observer measures Δt where Δx=0. Hence, the time intervals measured by the observer in his own inertial frame approaches the same value when v/c approaches 0.

The reference was talking about the time intervals as measured by a third observer in a third inertial frame. A third observer would be determining a Δx not equal to zero. The distance Δx'' would be much different from zero. So the time intervals derived by subtraction would be different.

One of the ways a third observer is taken into account is by the relativistic addition of velocities. However, the OP probably thinks 'relativistic addition' is a contradiction, too. He can't separate the case of two observers from the case of three observers.

As I predicted, the OP did not correct the concrete mistake that I pointed out. He did not provide a reason to believe that x' is not equal to zero. Instead, he quoted a reference which was talking about a different scenario. He refers to a scenario where there is a third observer rather than one that is restricted to two observers. He merely tried to throw sand in our eyes.

I repeat the concrete error that he made. In the scenario imagined by the OP, where there are only two observers, either x'=0. This is because the moving observer can not detect his own motion.

So I am still waiting for him to explain how the moving observer can observer an arbitrarily large x'.

 

[Perpetual Student]

It seems that what Mr. wogoga is missing here is that the Lorenz transformations never actually become the Gallilean transformations, which we all know and readily concede. The Lorenz transformations are reduced to the Gallilean transformations only for practical purposes, but not in absolute terms. If we had the technology and ability to keep track of the infinitesimal differences from Gallilean to Lorentz transformations at ordinary speeds and velocities, we would still have no motivation to do so. Only in extreme circumstances (like our GPS technology), where we actually take into account the Lorenz transformation, is the difference important.

 

[Reality Check]

A quote from Two myths about special relativity by Ralph Baierlein:

wogoga: Text in various colors does not make a PDF on the Internet valid mathematics or science !

Ralph Baierlein is addressing a misinterpretation of the question.
"Does the Lorentz transformation reduce to the Galilean transformation when the ratio v/c is small?" is not the question "Does the Lorentz transformation become the Galilean transformation when the ratio v/c is small?". The term reduce to in this context means "when higher terms in an expansion become negligible" or "in the limit as v goes to zero".
Baierlein misses that the reduction is not for a specific v/c value. It is for a limit. In that limit we will get to a v/c value where the Lorentz factor is ~1 and the delta x term is ~0. If the delta x term is not negligible then we can always select a smaller v/c to reduce the Lorentz transformation to the Galilean transformation.

...
Very interesting! A first-order effect with respect to v of a formula f[v] can only disappear, if the first derivative df[v]/dv is zero at v = 0

Very irrelevant and even ignorant, wogoga!
The Lorentz transformation becomes the Galilean transformation at v = 0 ! When v = 0 equation 1 (Lorentz transformation for a time interval) reduces to the Galilean transformation.

 

[Darwin123]

the Lorentz time-transformation reduces to t' = t only at point x = 0.

Hence, the Lorentz transform EQUALS the Galilean transform at x=0. 'Reduce to does not mean the same as 'equal to'. At all values where x is not equal to zero, the value of t' asymtotically approaches t in the limit of v/c approaching=0. The word 'reduce' means to asymptotically approach as a limit. to here means


I already pointed out that for the moving observer, x'=0 for all values of time, t'. The Lorentz transform of x and t determined by the stationary observer has to equal the x' and t' of the moving observer. Hence, x'=0 regardless of the value of v at the point x'=0 and only at x'=0.

You have never addressed that concrete error. If the moving observer starts at x'=0, then it has to always be at x'=0.

The Lorentz-transformation is dead. Long live the Lorentz-factor!!!

The Lorentz factor without the synchronization term leads to the false paradoxes of relativity. The synchronization term is subtracted out when determining clock rates and ruler lengths. However, the synchronization term is not is not subtracted out when determining clock setting and ruler positions.

The experimental determination of body lengths and time intervals requires one to fix the position of both ends as seen by any one observer. The Lorentz time dilation is determined by setting ∆x=0. The Lorentz length contraction is determined by setting ∆x=0.

If there is proper acceleration, then one can't precisely guarantee that either ∆x=0 or ∆t=0. To define the proper acceleration, one needs nonzero intervals that may be infinitesimal. I think this is part of your problem. You don't accept infinitesimals.

How about that 'Principia', huh! Man, Newton really knew how to toss around those infinitesimals! Einstein just 'plagiarized' Newton with regards to infinitesimals!

The idea of infinitesimals was bouncing around Europe log before Newton, of course. However, the Jesuits banned any variation of 'atomic' theory including infinitesimals. No God fearing person could divide matter up into either atoms or infinitesimals.

You got to love those Jesuits! Not!

 

[wogoga]

The Lorentz transformations reduce to the Galilean transformations by selecting an infinite value for C because under Galilean relativity, light travel is instantaneous:

x' = (x-vt)/sqrt(1-v²/c²)
t' = (t-vx/c²)/sqrt(1-v²/c²)

Taking the limit as C goes to infinity:

x' = (x-vt)/sqrt(1-0)
t' = (t-0)/sqrt(1-v²/c²)

simplifying,

x' = x-vt
t' = t

Ok. This is the only way to save the claim that the Lorentz transformation reduces to the Galilean with v << c. The problem that vx/c² does not disappear in a first-order expansion with respect to v is obviously solved by c → ∞.

At least in case of a normal reduction resp. series-expansion, this trick only works if we replace c by k, where k → ∞ if v → 0, and k → c if v → c. A concrete example is replacing c with k = c∙(c/v) . This modified "Lorentz" transformation actually reduces to the Galilean transformation, yet the "Lorentz" factor becomes (1 - v⁴/c⁴)^-1/2 instead of (1 - v²/c²)^-1/2.

The solution of the dilemma by c → ∞ has also been adopted by Wikipedia:

"Another important property is for relative speeds much less than the speed of light, the Lorentz transformations reduce to the Galilean transformation in accordance with the correspondence principle."​

The chapter Relativistic kinetic energy of Correspondence principle presents the example of relativistic kinetic energy reducing to classical kinetic energy. Here v << c means: E_kin = ½ mv² is a good approximation for the relativistic formula if e.g. v = 0.01 c, a bad approximation if v = 0.5 c, and fully wrong if v= 0.99 c.

"Mathematically, as v → 0, c → ∞. In words, as relative velocity approaches 0, the speed of light (seems to) approach infinity."​

Increasing a fundamental constant such as light-speed c is quite different from reducing velocity v. If v → 0 and c → ∞ were interchangeable in this respect then it should also be possible to derive classical kinetic energy from the relativistic E_kin = mc² ((1-v²/c²)^-0.5-1) in such a way. Yet c → ∞ instead of v → 0 only leads to E_kin = mc² (1-1) = 0.

The effect of c → ∞ on the Lorentz transformation is simple: xv/c² becomes zero, and the Lorentz factor becomes one. By increasing c to infinity we do not get a series expansion but only a limit value.

"Hence, it is sometimes said that nonrelativistic physics is a physics of 'instantaneous action at a distance'.^[10]"​

"Nonrelativistic" physics? At least with respect to "classical" physics, the following is valid: Gravitational, electric and magnetic interactions were considered instantaneous actions at a distance, where total energy and momentum of the interacting parts are conserved. Yet light was considered as something different: Waves (or particles) leave a source and propagate at a finite speed. Already before Newton laid with his Principia the "official" foundation of classical mechanics in 1687, the finiteness of c had been established. Therefore, one cannot invoke "instantaneous action at a distance" of classical physics in order to justify c → ∞.

Reference [10] of the Wikipedia quote leads to Einstein's Relativity: The Special and General Theory. On page 57 we read:

Let me add a final remark of a fundamental nature. The success of the Faraday-Maxwell interpretation of electromagnetic action at a distance resulted in physicists becoming convinced that there are no such things as instantaneous actions at a distance (not involving an intermediary medium) of the type of Newton's law of gravitation. According to the theory of relativity, action at a distance with the velocity of light always takes the place of instantaneous action at a distance or of action at a distance with an infinite velocity of transmission.​

If c is an upper speed limit, then what was considered infinite velocity in classical physics can only be c, and Einstein's claim that propagation at c takes the place of instantaneous actions at a distance is correct. Yet Einstein's statement does not justify c → ∞, because it does not imply that according to classical physics, instantaneous action at a distance takes the place of light propagating at c.

Cheers, Wolfgang
www.pandualism.com

 

[Giordano]

Ok. This is the only way to save the claim that the Lorentz transformation reduces to the Galilean with v << c. The problem that vx/c² does not disappear in a first-order expansion with respect to v is obviously solved by c → ∞.

At least in case of a normal reduction resp. series-expansion, this trick only works if we replace c by k, where k → ∞ if v → 0, and k → c if v → c. A concrete example is replacing c with k = c∙(c/v) . This modified "Lorentz" transformation actually reduces to the Galilean transformation, yet the "Lorentz" factor becomes (1 - v⁴/c⁴)^-1/2 instead of (1 - v²/c²)^-1/2.

The solution of the dilemma by c → ∞ has also been adopted by Wikipedia:

"Another important property is for relative speeds much less than the speed of light, the Lorentz transformations reduce to the Galilean transformation in accordance with the correspondence principle."​

The chapter Relativistic kinetic energy of Correspondence principle presents the example of relativistic kinetic energy reducing to classical kinetic energy. Here v << c means: E_kin = ½ mv² is a good approximation for the relativistic formula if e.g. v = 0.01 c, a bad approximation if v = 0.5 c, and fully wrong if v= 0.99 c.

"Mathematically, as v → 0, c → ∞. In words, as relative velocity approaches 0, the speed of light (seems to) approach infinity."​

Increasing a fundamental constant such as light-speed c is quite different from reducing velocity v. If v → 0 and c → ∞ were interchangeable in this respect then it should also be possible to derive classical kinetic energy from the relativistic E_kin = mc² ((1-v²/c²)^-0.5-1) in such a way. Yet c → ∞ instead of v → 0 only leads to E_kin = mc² (1-1) = 0.

The effect of c → ∞ on the Lorentz transformation is simple: xv/c² becomes zero, and the Lorentz factor becomes one. By increasing c to infinity we do not get a series expansion but only a limit value.

"Hence, it is sometimes said that nonrelativistic physics is a physics of 'instantaneous action at a distance'.^[10]"​

"Nonrelativistic" physics? At least with respect to "classical" physics, the following is valid: Gravitational, electric and magnetic interactions were considered instantaneous actions at a distance, where total energy and momentum of the interacting parts are conserved. Yet light was considered as something different: Waves (or particles) leave a source and propagate at a finite speed. Already before Newton laid with his Principia the "official" foundation of classical mechanics in 1687, the finiteness of c had been established. Therefore, one cannot invoke "instantaneous action at a distance" of classical physics in order to justify c → ∞.

Reference [10] of the Wikipedia quote leads to Einstein's Relativity: The Special and General Theory. On page 57 we read:

Let me add a final remark of a fundamental nature. The success of the Faraday-Maxwell interpretation of electromagnetic action at a distance resulted in physicists becoming convinced that there are no such things as instantaneous actions at a distance (not involving an intermediary medium) of the type of Newton's law of gravitation. According to the theory of relativity, action at a distance with the velocity of light always takes the place of instantaneous action at a distance or of action at a distance with an infinite velocity of transmission.​

If c is an upper speed limit, then what was considered infinite velocity in classical physics can only be c, and Einstein's claim that propagation at c takes the place of instantaneous actions at a distance is correct. Yet Einstein's statement does not justify c → ∞, because it does not imply that according to classical physics, instantaneous action at a distance takes the place of light propagating at c.

Cheers, Wolfgang
www.pandualism.com

What I am getting from your statement is that the predictions of Relativity are detectably different from classical physics only under the conditions that the math says that they should be detectably different, and that Relativity predicts results that appear to be the same as classical physics under the conditions where Relativity would predict that they would appear to be detectably the same. Further, the actual experimental tests confirm this and Relativity helps resolve experimental results that classical physics cannot.

Is this what you are saying?

 

[Reality Check]

This is the only way to save the claim that the Lorentz transformation reduces to the Galilean with v << c.

Wrong, wogoga.
It is basic mathematics that everyone here except you understands. The Lorentz transformation reduces to the Galilean with v << c.

This is the case of "a normal reduction resp. series-expansion"m We can neglect higher terms by either letting v tend to zero or c tend to ∞. No imaginary "k" need apply.

Writing total ignorance is bad, wogoga. In those limits the actual Lorentz factor (not your imaginary one ) reduces to 1 and the Lorentz transformation reduces to the Galilean transformation.

Lots of irrelevant nonsense snipped.

The effect of c → ∞ on the Lorentz transformation is simple: xv/c² tends to zero and the Lorentz factor (even in that series expansion that you claim does not exist ) tends to one. And we get the Galilean transformation.

There is a thing called nonrelativistic physics, wogoga ! It existed for over 3 centuries before a guy called Einstein discovered relativistic physics.

Photons are not forces, wogoga - Duh!
Forces in classical physics can act instantaneously over distances while photons, electrons, planets, stars, etc., move at finite speeds. Therefore, one can invoke the "instantaneous action at a distance" in classical physics in order to justify c → ∞.

If c is an upper speed limit, then what was considered infinite velocity in classical physics can only be c, ...

You still do not understand what is going on, wogoga.
Classical physics has no upper velocity. In classical physics you can accelerate any body to any velocity. It is relativity that states that a massive body cannot travel at c and a massless body always travels at c (hypothetically there can be particles travelling faster than c (tachyons)).
Thus letting c → ∞ in relativity will reduce relativity to classical physics. This is obvious from the postulates of Special Relativity.

 

[Darwin123]

If c is an upper speed limit, then what was considered infinite velocity in classical physics can only be c, and Einstein's claim that propagation at c takes the place of instantaneous actions at a distance is correct. Yet Einstein's statement does not justify c → ∞, because it does not imply that according to classical physics, instantaneous action at a distance takes the place of light propagating at c.

Nominated for the 'Paragraph Most Impossible to Understand'.

I realize that there is a lot of competition on this board for this award. Further, you may think you understand it on first reading. However, reread it carefully paying close attention to definitions of the individual words.

The paragraph gets extra points because it looks almost rational on first reading. However, this itself is a misunderstanding.

 

[wogoga]

The Lorentz transformation reduces to the Galilean with v << c... In those limits the actual Lorentz factor (not your imaginary one ) reduces to 1 and the Lorentz transformation reduces to the Galilean transformation.

Fizeau's experiment concerning relative speed of light in a moving medium is considered a consequence of relativistic velocity addition. This means: The Lorentz transformation with a small speed v of a transparent medium relative to a laboratory leads to a relativistic effect, capable of explaining partial drag of light according to the Fresnel drag coefficient f = 1 - 1/n². This obviously could not be possible if the Lorentz transformation turned for v << c into the Galilean, as generally claimed.
In case of water, refraction of n ≈ 1.33 leads to a drag of f ≈ 0.44. If water in a tube starts moving at 1 m/s in direction of light propagation, the speed of this light increases relative to the laboratory from w = c/n ≈ 2.25∙10⁸ m/s by 0.44 m/s (instead of 1 m/s). This only partial drag of light by the movement of the medium is explained by the Lorentz transformation with v = 1 m/s.

If the Lorentz transformation actually reduced to the Galilean in case of a velocity as low as v = 1 m/s then the application of the Lorentz transformation with v = 1 m/s would increase speed w = c/n by 1 m/s and not by 0.44 m/s. Or does anybody deny that the Galilean transformation with v = 1 m/s changes any speed w by 1 m/s?

Light-speed w' = c/n in moving water means that in a moving frame F' where the water is at rest, light travels during time-interval Δt' distance Δx' = w'∙Δt'. In order to get light-speed w with respect to the laboratory frame F, we apply the Lorentz transformation and calculate w = Δx /Δt.

Δx = γ (Δx' + v Δt') = γ (w' Δt' + v Δt') = γ Δt' (w' + v)
Δt = γ (Δt' + v/c² Δx') = γ (Δt' + v/c² w' Δt') = γ Δt' (1 + v w'/c²)​

The Lorentz factor γ is irrelevant as it cancels out, and we get:

w = Δx/Δt = (v + w') /(1 + v w'/c²) = (v + c/n)/(1 + v/(c∙n))​

Series expansion of w with respect to v leads to:

w = c/n + (1 - 1/n²) v + (1 - n²)/(c∙n³) v² + …​

Thus, relativity of simultaneity (or Lorentz's local time of 1892) is assumed to explain Fizeau's experiment, and all the insolvable paradoxes arising from relativity of simultaneity can be adapted to partial dragging of light by a medium.

Cheers, Wolfgang
Human history is full of empiric and experimental confirmations of wrong beliefs

 

[Reality Check]

Fizeau's experiment ...snipped irrelevant and inanely highlighted text...

This has nothing to do with your continued ignorance of basic mathematics, wogoga.
29 October 2015 wogoga: Repeating a delusion about the Lorentz transformation not reducing to the Galilean transformation in the limit of v << c (or the classic limit of c goes to infinity)

 

[Perpetual Student]

This has nothing to do with your continued ignorance of basic mathematics, wogoga.
29 October 2015 wogoga: Repeating a delusion about the Lorentz transformation not reducing to the Galilean transformation in the limit of v << c (or the classic limit of c goes to infinity)

Clearly there is no amount of evidence, mathematics nor demonstration that can dissuade this Mr. Wogoga from his preconceived and sadly uninformed notions.
Like all dogmatists he is permanently "fixed" in his opinions. He makes no attempt to understand those who respond to him and deflects all attempts made to correct his errors -- and he will continue to do so. It's time to give up!

 

[Reality Check]

It is certainly time to give up trying to explain this basic mathematics to wogoga and just point out to lurkers how bad his understanding is.

 

[Darwin123]

It is certainly time to give up trying to explain this basic mathematics to wogoga and just point out to lurkers how bad his understanding is.

Wogoga may have a few valid points in his ramblings, which is why I choose to argue with him. For instance, Wogoga has made a lot of mistakes when 'proving' that the Lorentz transform does not 'reduce' to the Galilean transformation. However, he is repeating an argument made Ralph Baierlein.

I think that Baierlein is making a valid point. So Wogoga in a partly right.

I would like to discuss this statement of Ralph Baierlein that Wogoga cited. I partly agree with Baierlein. However, he is still making a small error.

Quote from Wogoga from Baierlein. Baierlein (RB) says:

'A quote from Two myths about special relativity by RB:
Q. Does the Lorentz transformation reduce to the Galilean transformation when the ratio v/c is small?
A. No.
One argument of the author:
Consider the usual pair of inertial reference frames, the primed frame moving with speed v along the x axis of the unprimed frame. To avoid any spurious dependence on the origins of coordinate systems, consider a pair of physical events. The Lorentz transformation for the time interval between the events takes the form
Δt' = 1/√(1-v2/c2) (Δt - v/c2 Δx)
Let the ratio v/c be as small as desired (but nonzero). Then it is always possible to find an event pair for which Δx is large enough that the term with Δx dominates over the term with Δt. This behavior is entirely different from what the Galilean transformation Δt' = Δt asserts.'

I think that RB is making a slight mistake here. However, it is not nearly as fundamental as the mistake that Wogoga is claiming. RB is not claiming that relativity is wrong.

So I think RB should have said differential instead of infinitesimal. All differential quantities are infinitesimal quantities. Not every infinitesimal is a differential quantity.

In thermodynamics, internal energy can be expressed as a perfect differential. Work can not be expressed as a perfect differential, even when it is an infinitesimal. Heat can not be expressed as a perfect differential, even when it is an infinitesimal.

In the relativity, the proper time ds can be expressed as a perfect differential. Δx and Δt are never differentials, even though they may be infinitesimal. However, v/c can not be treated as a differential either. It does not accumulate.

So I think RB made a slight mistake when using mathematical jargon. He is wrong when he used the word infinitesimal. However, he would have been right had he used the word differential instead of infinitesimal.

Or maybe he shouldn't have used the word 'reduce'. Ralph may think 'reduce' means 'accumulative'. If he used the word accumulative instead of reduce, he would have been unambiguously right.


The formal meaning of the word 'reduce' is critical here. The word 'reduce' is slightly ambiguous, so its meaning must be extracted from the context that is being used. It is probable that RB is using the word reduce a bit differently from most of the posters who reply to Wogoga.


The meaning of the above citation varies with domain of event pairs that the Lorentz transform is applied. A transform takes a domain and maps it on a range. The domain does not have to include the universal set. The remaining issue is what domain we are talking about.

If one |Δx| has a finite upper bound in the domain, then the Lorentz transform reduces to the Galilean transform. Any finite domain. One can use the size of the current universe as an upper bound of |Δx|. Then, Δx' and Δt' can be expanded in Taylor series of Δt as long as |vΔx/c^2|<1.


Although v/c may be treated as an infinitesimal, it can not be treated as differentials unless they are bound in the system being analyzed. The word infinitesimal is slightly different from the word differential. An infinitesimal quantity is a quantity that is smallest than the smallest real number. A differential is an infinitesimal quantity that satisfies a constraint condition.

I would interpret Ralph Baierlein (RP) as follows. Contrary to popular belief, v/c can not be treated as a differential quantity. It does not accumulate. Only Δs, the proper time, can be treated as a differential.

However, this does not show that relativity is wrong. This does not even show that special relativity is wrong. This in no way is a paradox.

If 'reduce' means to be equal after truncation of a Taylor series, then most definitely the Lorentz transform can be reduced to a Galilean transform. That is the way most of the replies have implied. However, RP does not use the word 'reduce' this way. He seems to be using a topological definition of reduce having nothing to do with Taylor series.

I will give Wogoga the benefit of the doubt just to lay my mind to rest. To end this part of the argument, let us assume that Wogoga understood and accepted what RP was saying.

So I will concede Wogoga the following point. If 'reduce' means to use v/c as a differential, then the Lorentz transform does not reduce to the Galilean transform. Basically, one can't assume that v/c is a continuous function of s, the proper time. It is physically possible for the acceleration of an observer to approach infinity. The term, v/c, should never be used as a differential.

However, v/c can sometimes be used as an infinitesimal. This means that one can expand the terms of the Lorentz transform into a Taylor series using v/c. So everyone else on the thread is right, too. One expand the Lorentz transformation in terms of v/c using a Taylor series. If one truncates the series to first order in v/c, one will obtain the Galilean transformation.

RB did not say that this invalidated relativity. If he thought this invalidated relativity, then he would have said so. When Wogoga says that this invalidates relativity, he is wrong. Galilean and Lorentzian relativity are BOTH logically consistent. If RB claimed that relativity is wrong, then Wogoga should cite that section.

I thank Wogoga for quoting that passage from RB. I am reading a book about infinitesimals right now. However, I was a little weak on the difference between infinitesimal and differential. The quotation made me realize that an infinitesimal is different from a differential. So although Wogoga is completely wrong, his errors SOMETIMES provoke thought!

I still recommend that everyone including Wogoga read: 'Infinitesimals:...' by Alexander Amir.

 

[Perpetual Student]

Consider the usual pair of inertial reference frames, the primed frame moving with speed v along the x axis of the unprimed frame. To avoid any spurious dependence on the origins of coordinate systems, consider a pair of physical events. The Lorentz transformation for the time interval between the events takes the form
Δt' = 1/√(1-v2/c2) (Δt - v/c2 Δx)
Let the ratio v/c be as small as desired (but nonzero). Then it is always possible to find an event pair for which Δx is large enough that the term with Δx dominates over the term with Δt. This behavior is entirely different from what the Galilean transformation Δt' = Δt asserts.'

Yes, however for ordinary velocities, Δx would have to be millions of light years for the Galilean transformation not to be adequate. We are talking about astronomical distances.

 

[TheAdversary]

The speed of light is the ultimate invariant quantity. I suspect that the law of inertia has something to do with the fact that this ultimate invariant quantity
is a speed, a change over time. Change over Time is an acceptable philosophical concept; You can define it no further; They seem to be fundamental concepts.
This is my philosophical reason why I think Special Relativity can't be refuted.