Refutation of Special Relativity for Dummies

[Reality Check]

...Yet it is obvious that the numerical value of c only depends on the units meter and second.

Maybe you missed/have forgotten the dimensional analysis part of your science education, wogoga.
For an equation to remain correct you have to change the units of every quantity. For an equation containing v and c.
* You have to change the units of c and
* You have to change the units of v.

Of course it is smarter to look at the unit-less quantity v/c and realize that when v << c, that number is small for any units. Square v/c as in the Lorentz factor and it is really, really small. That is why we do not see relativistic effects in everyday life and why Newtonian mechanics works for v << c.

 

[slyjoe]

...

Instead of joining the ranks of clueless pack followers, you could have used software for symbolic computation and tried yourself to check what corresponds to the mathematical reality.

Here the problem formulated with Mathematica-syntax:

γ = 1/Sqrt[1–v^2/c^2]
Series [γ (x - v t), {v, 0, n}]
Series [γ (t - v/c^2 x), {v, 0, n}]​

If n = 0, the result is:

x' = x . . . t' = t​

For the first-order series (with n = 1) we get:

x' = x - t v . . . t' = t - x/c² v​

The second-order result:

x - t v + x/2/c² v² . . . t - x/c² v + t/2/c² v²​

The only possibility to get the Galilean transformation is taking the first order Taylor-approximation for the x-coordinate and the zeroth order approximation for the t-coordinate.

Does anybody know who is responsible for spreading the myth that the Lorentz transformation reduces to the Galilean transformation? I assume and hope it was not Einstein.

Cheers, Wolfgang

Why would I do that? I understand limits.

 

[Perpetual Student]

Does anybody know who is responsible for spreading the myth that the Lorentz transformation reduces to the Galilean transformation? I assume and hope it was not Einstein.

Cheers, Wolfgang

Tough question.
It might have been Lorentz.
What do you think?

Amazingly, the myth lives on, can easily be demonstrated mathematically, and is consistent with all observations and experiments for over 110 years. But somehow it lives on!

 

[Darwin123]

You're trying too hard to win an argument and not trying hard enough to understand it.

In order to understand arithmetic, the OP would have to understand the rules of rounding off. These are also called the rules for significant figures.

If the OPe doesn’t understand how to round off, then he couldn’t follow what you say. Basically, the concept of limits is correlated with the rules of rounding off. The terms that are cut off when one truncates an infinite series hold the decimal places that are discarded when rounding off.


http://www.usca.edu/chemistry/genchem/sigfig.htm
‘Significant figures are critical when reporting scientific data because they give the reader an idea of how well you could actually measure/report your data. Before looking at a few examples, let's summarize the rules for significant figures.’

http://www.usca.edu/chemistry/genchem/sigfig2.htm
‘When adding or subtracting numbers, count the NUMBER OF DECIMAL PLACESto determine the number of significant figures. The answer cannot CONTAIN MORE PLACES AFTER THE DECIMAL POINT THAN THE SMALLEST NUMBER OF DECIMAL PLACES in the numbers being added or subtracted.
…
When multiplying or dividing numbers, count the NUMBER OF SIGNIFICANT FIGURES. The answer cannot CONTAIN MORE SIGNIFICANT FIGURES THAN THE NUMBER BEING MULTIPLIED OR DIVIDED with the LEAST NUMBER OF SIGNIFICANT FIGURES.’



https://www.physics.uoguelph.ca/tutorials/sig_fig/SIG_dig.htm
‘The number of significant digits in an answer to a calculation will depend on the number of significant digits in the given data, as discussed in the rules below. Approximate calculations (ORDER-of-magnitude estimates) always result in answers with only one or two significant digits.’



I’ll bet Einstein knew how to round off. That trickster probably knew about significant figures too!

Will the OP claim that Einstein invented rounding off? Only undilated time will tell

 

[Perpetual Student]

Here is a simpler approach (forget about setting c = 1), which is clearly distracting you. In fact I see an error in my earlier post where I set c = 1 but then proceeded to use c = 3*10^8 anyway, which exaggerated the resulting orders of magnitude (dumb mistake). So:

We have: x' = γ(x - vt)
and t' = γ(t - vx/c^2

Using a slow velocity like v = 10 m/s and setting x = 1m for simplicity of calculation demonstrates the case clearly. I am using only orders of magnitude:

γ = (1 - 10^(-15))^(-1/2), which we would (hopefully) agree is infinitesimally close to and indistinguishable from 1.

Now, x - vt = x - 10m, which is obviously the Galilean transformation. End of story.

And t' = t - vx/c^2 = t - 10^(-16)s which is infinitesimally close to t, to about the same order of magnitude that γ = 1. So t' = t, also as Galileo would have had it.

 

[Darwin123]

Here is a simpler approach (forget about setting c = 1), which is clearly distracting you. In fact I see an error in my earlier post where I set c = 1 but then proceeded to use c = 3*10^8 anyway, which exaggerated the resulting orders of magnitude (dumb mistake). So:

We have: x' = γ(x - vt)
and t' = γ(t - vx/c^2
...

I think that I know a way to get the 'asymmetry' that Wogaga wants. We require the twin on earth to determine the time at the position of the rocket in the 'stationary' frame. Therefore, we set x' = 0.

We have: x' = γ(x - vt)
and t' = γ(t - vx/c^2)

Let us set the proper acceleration of earth and rocket to zero. So (v/c) is constant in time.

Now, the earth twin wants to know how fast the clock is moving at a particular point on the rocket. Say he wants to know what the time, t', is at a particular place, x'=0.

x'=0
so
x = vt.

Now the earth twin wants to calculate t' under the conditions set above. So he substitutes for x in t'. So,

t' = γ(t - v[vt]/c^2)
So
t'= γ(t - tv^2/c^2)
t'= γ(t - t|v/c|^2)
t' = γt (1 - |v/c|^2)

Now
1/γ = sqrt(1 - |v/c|^2)

So:
t'=t/γ when x'=0.

So not this is consistent with
t = γ t' when x'=0.


The problem here is justifying the condition x' = 0. The dynamic asymmetry in the problem is hidden in this condition. Because the clock in the
rocket is not reset during the proper acceleration, we can't set x = 0.

That is, we made the problem asymmetric when we set x'=0 rather than x=0. If we set x=0, we would have the 'opposite' result.

The physical justification of x'=0 would require a little knowledge of how the clocks communicate. In other words, we would have to know what forces are involved in synchronization. In other words, I haven't really avoided proper acceleration. I just delayed its introduction a bit.


The x'=0 is only a special case in the sense that Wogoga is claiming. However, that is not important for the issue of whether the Galilean transformation is the low order approximation of the Lorentz transformation.

We note that v<c in all systems of units. Therefore, |v/c|<1.


We expand γ by a Taylor series. Or, if you wish, by binomial seres.

γ= 1-0.5|v/c|^2 + 0.75|v/c|^4 -...

The sign of the coefficient alternates between + and - for all orders. Therefore, we note that the infinite series above converges for all values of |v/c|<1. The proof of convergence requires calculus just a little above introductory. However, all of us here at least understand the concept of limits, right? So we can look up the conditions of convergence when we can't derive them.

We can drop all terms higher than first order, leaving only terms first order in |v/c|. However, we note that there are not high order terms in the Taylor expansion above.

So to first order in |v/c|, γ= 1.
Substituting this value into the formula for t:
t=t'.

Further, γ= 1 so,
x'=x-vt.

These are the

The important assumption here is not c=1. That doesn't matter because v and c are in the same units. The important assumption here is that |v|<c. Then, the series converges because the signs alternate.

This is about as mathematically rigorous as I know how. I have used no mathematics that Newton didn't know. There are important differences between relativity and Newtonian physics, but are not with the arithmetic.

Also note that the sign of v doesn't matter. One can substitute +v for -v and get similar results. Thus, the results are symmetric with respect to the sign of v.

I suggest that anyone who disagrees try the special case of v=0.1c. Choose t to be any value you like just as long as the units of x, t, v and c are consistent. Use the rules for round off significant figures.

Anyway, I thank Wogoga for asking the question. I love these complicated questions that masquerade as simple ones.

 

[Perpetual Student]

If you want, then you can call Einstein 'the bastard grandson of calculus'. After all, he later helped to develop tensor calculus and differential geometry. Never the less, these are just extensions of that old time differential calculus. Which is what I conjecture you are missing.
...Darwin123

This is not historically accurate. Einstein struggled with tensor calculus, which was developed by two Italian Mathematicians, Gregorio Ricci-Curbastro and his student Tullio Levi-Civita. LINK

Here is a fascinating history of the correspondence between Einstein and Levi-Civita (LINK):

Before developing his 1915 General Theory of Relativity, Einstein held the "Entwurf"theory. Tullio Levi-Civita from Padua, one of the founders of tensor calculus,objected to a major problematic element in this theory, which reflected its globalproblem: its field equations were restricted to an adapted coordinate system. Einsteinproved that his gravitational tensor was a covariant tensor for adapted coordinatesystems. In an exchange of letters and postcards that began in March 1915 and endedin May 1915, Levi-Civita presented his objections to Einstein's above proof. Einsteintried to find ways to save his proof, and found it hard to give it up. Finally Levi-Civitaconvinced Einstein about a fault in his arguments. However, only in spring 1916, longafter Einstein had abandoned the 1914 theory, did he finally understand the mainproblem with his 1914 gravitational tensor. In autumn 1915 the Göttingen brilliantmathematician David Hilbert found the central flaw in Einstein's 1914 derivation. OnMarch 30, 1916, Einstein sent to Hilbert a letter admitting, "The error you found inmy paper of 1914 has now become completely clear to me".

And we have the role of Marcel GrossmannLINK

It was Grossmann who emphasized the importance of a non-Euclidean geometry called Riemannian geometry (also elliptic geometry) to Einstein, which was a necessary step in the development of Einstein's general theory of relativity. Abraham Pais's book[1] on Einstein suggests that Grossmann mentored Einstein in tensor theory as well. Grossmann introduced Einstein to the absolute differential calculus, started by Christoffel[2] and fully developed by Ricci-Curbastro and Levi-Civita.[3] Grossmann facilitated Einstein's unique synthesis of mathematical and theoretical physics in what is still today considered the most elegant and powerful theory of gravity: the general theory of relativity. The collaboration of Einstein and Grossmann led to a ground-breaking paper: "Outline of a Generalized Theory of Relativity and of a Theory of Gravitation", which was published in 1913 and was one of the two fundamental papers which established Einstein's theory of gravity.[4]

You said that Einstein made a mistake. Einstein said that you can replace a curve with a polygon with an infinite number of line segments. You blamed scientists of mindlessly agreeing with Einstein on this matter.

This approximation was used a long time before Einstein was even born. It was in fact thought out a few decades before the trial of Galileo. The Catholic Church banned this approximation. However, it was well established by the time Newton wrote Principia. BTW: Newton uses this approximation many, many times in Principia.
...Darwin123

Actually, the method (called the method of exhaustion) was used even earlier by Archimedes. LINK

 

[jmckaskle]

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-v^2/c^2)
t' = (t-vx/c^2)/sqrt(1-v^2/c^2)

Taking the limit as C goes to infinity:

x' = (x-vt)/sqrt(1-0)
t' = (t-0)/sqrt(1-v^2/c^2)

simplifying,

x' = x-vt
t' = t

The Lorentz transformations are a generalization of the Galilean transformations, as you should know, and can be derived without too much difficulty from solving the set of linear equations:

x' = Ax + Bt
t' = Cx + Dt

for A, B, C, and D

Give it a shot. Working out the solution for yourself will provide you with the some insight into the issue.

 

[Darwin123]

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-v^2/c^2)
t' = (t-vx/c^2)/sqrt(1-v^2/c^2)

Taking the limit as C goes to infinity:

x' = (x-vt)/sqrt(1-0)
t' = (t-0)/sqrt(1-v^2/c^2)

simplifying,

x' = x-vt
t' = t

The Lorentz transformations are a generalization of the Galilean transformations, as you should know, and can be derived without too much difficulty from solving the set of linear equations:

x' = Ax + Bt
t' = Cx + Dt

for A, B, C, and D

Give it a shot. Working out the solution for yourself will provide you with the some insight into the issue.

Ulp! Limits? Infinite value !?

 

[jmckaskle]

Was it too mathy? I hope not.

 

[Darwin123]

Was it too mathy? I hope not.

Too mathy for the OP, not for me! And not for most of the participants!

Read my previous replies. I point outed that the OP is having a lot of problems with the concept of limit. My conjecture is that he is having more trouble with the concepts involving limit than with the fundamental physics of relativity.

The OP has read some of Einstein's articles, apparently. However, he doesn't seem to have read any physics that came before. He is constantly pointing out hypotheses that Einstein used which had been well established before Einstein was born.

The high point of this discussion was, for me, the post where the OP blames Einstein for introducing infinitesimals. The OP stated that Einstein was being stupid when he approximated a smooth curve as a polygon with an infinite number of sides.

The OPs statement about the order of the approximation is consistent with not understanding limits. He doesn't realize one can keep a first order term while dropping second order terms. The OP also seems fuzzy about expressions involving inequalities.

The OP starts to make ad hominem arguments about physicists, including Einstein, every time a reply mentions something about an inequality, or a limit, or infinity. He makes nostalgic statements about that Old Time Physics. However, he hasn't acknowledged that the old time physicists originally stated the hypotheses on inequality, limits and infinities.

I predict that he won't reply to your post. I say this despite the fact that you made a rational suggestion. It wasn't personal. You made what appears to be a reasonable challenge that the OP

Einstein did not introduce the mathematics of limits. Archimedes introduced the method of exhaustion about 22 centuries ago, as another poster pointed out. I may point out that Zeno tried to dismiss the idea of infinitesimals about 24 centuries ago, which is a long time before the OP did.

The OP has challenged the readers to find a 'concrete' place that he is wrong. He asks readers to objectively find logical errors in his posts. However, he ignores or insults posts that point out concrete errors in his posts. I conjecture that this is because these 'concrete errors' involve the concept of limits.

Maybe I don't understand what he means by 'concrete'. He may be speaking in the language of developmental psychologists like Piaget. In terms of developmental psychology, the concept of limits is formal not concrete. I may be giving hm more credit than he deserves in making this hypothesis.

However, I am more interested in what he calls 'concrete' than his views on relativity. I would like to know why he dismisses the rather detailed answers that he has received.

Your suggestion was right on the money, though. I am waiting for his answer!

 

[wogoga]

Some may even argue that v/c² x can be removed as a term from the first-order limit because the constant c is as high as 3*10⁸ m/s. Yet it is obvious that the numerical value of c only depends on the units meter and second. If we use light-year and second then the value of c becomes as low as 3.17*10^-8 light-year/sec.

An illustrative example of this argument:

Here is a simpler approach (forget about setting c = 1, which is clearly distracting you). In fact I see an error in my earlier post [#140] where I set c = 1 but then proceeded to use c = 3*10^8 anyway, which exaggerated the resulting orders of magnitude (dumb mistake). So:

We have: x' = γ(x - vt)
and t' = γ(t - vx/c^2

Using a slow velocity like v = 10 m/s and setting x = 1m for simplicity of calculation demonstrates the case clearly. I am using only orders of magnitude:

γ = (1 - 10^(-15))^(-1/2), which we would (hopefully) agree is infinitesimally close to and indistinguishable from 1.

Now, x - vt = x - 10m, which is obviously the Galilean transformation. End of story.

And t' = t - vx/c^2 = t - 10^(-16)s which is infinitesimally close to t, to about the same order of magnitude that γ = 1. So t' = t, also as Galileo would have had it.

No objection can be raised against your choice of v = 10 m/s, leading to a negligible Lorentz factor, but the choice of x = 1 m and t = 1 sec is in principle completely arbitrary, and depends on the units meter and second. With your choices we get:

x' = x - vt = 1 m - 10 m
t' = t - vx/c² = 1 s - 1.11*10^-16 s​

Keeping vt = 10 m and dropping vx/c² ≈ 10^-16 s seems justified.

Now let us do the same with the (somehow less arbitrary) units second and light-second, i.e. let us use x = 1 LS and t = 1 sec. The Lorentz factor of v = 10 m/s = 3.33*10^-8 LS/s obviously does not change and remains negligible. This time, we get:

x' = x - t v = 1 LS - 3.33*10^-8 LS
t' = t - xv/c² = 1 s - 3.33*10^-8 s​

Keeping vt = 3.33*10^-8 LS and dropping vx/c² = 3.33*10^-8 s seems unjustified.

The Taylor expansion of the Lorentz factor (1-v²/c²)^-1/2 with respect to v yields:

1 v⁰ + 0 v¹ + 1/(2c²) v² + 0 v³ + 3/(8c⁴) v⁴ + . . .​

If meter per second is used as unit for velocity, and both v and c = 2.998*10⁸ express unit-less numeric values, then also higher-order terms of the Lorentz-factor expansion turn analogously into seemingly negligible quantities:

1 + 5.56*10^-18 v² + 4.64*10^-35 v⁴ + . . .​

By the way, the zero-order Taylor-expansion term of the Lorentz factor, 1 v⁰ = 1, becomes after multiplication with -xv/c² the first-order term of the expansion of the t'-transformation. The term t - xv/c² itself is its own full Taylor expansion t v⁰ - x/c² v¹.

Cheers, Wolfgang
As long as we remain in the abstract, and do not apply formulas or propositions to concrete situations, contradictions remain undetected

 

[Darwin123]

second[/I] and light-second, i.e. let us use x = 1 LS and t = 1 sec. The Lorentz factor of v = 10 m/s = 3.33*10^-8 LS/s obviously does not change and remains negligible. This time, we get:

x' = x - t v = 1 LS - 3.33*10^-8 LS
t' = t - xv/c² = 1 s - 3.33*10^-8 s​

Wow! This error is as concrete as a mathematical statement can possible be.


c = 1 Ls/s.

v is not negligible to first order. v^2 is negligible to first order.

v = 3x10^-8 LS/s.

v^2=9x10-16 LS^2/s^2




I can parse your error in thre parts.

1) You decided v was negligible to first order when the problem explicitly stated otherwise.

1) You didn't calculate the size of the coefficients in the new units.

2) You then discarded the units of 'v' and 'c', claiming that velocity is now unitless.


I have found with the help of Perpetual Student two more concrete errors in your analysis. I would like an explanation of why you decided to change only the units without changing the numeric values.

If you want to keep the units consistent with your assumption, you have to change the distance to a unit of 'Ls' and both velocities to a unit of 'Ls/s'. The numerical values have to change consistent with the change in units.

When a scientist writes that he is using 'dimensionless units', he really means that he is not going to bother writing the units down since everyone knows the rules for unit cancellation.

'Dimensionless units' is not a mistake in physics. It is a writer's convention based on the scientists hypothesis concerning his audience. Please explain your version of unit cancellation for the benefit of an audience that knows how to convert units.

I myself despise the 'dimensionless units' convention, myself. I never use them when I write MY articles. I am a bit of a bigot in that way.

I am sure that there are a sizable fraction of the Forum that can convert units. Perpetual Student and I are part of this narrow minded elite. So please correct us.


BTW: I present an offer that has nothing to do with relativity. I have land that cost me 3 $ per acre. However, I am willing to sell it for the price of 1 unitless currency. If you don't like unitless currency, I will sell it for 1 K$ per milliacre. It is a bargain either way !

 

[Perpetual Student]

An illustrative example of this argument:



No objection can be raised against your choice of v = 10 m/s, leading to a negligible Lorentz factor, but the choice of x = 1 m and t = 1 sec is in principle completely arbitrary, and depends on the units meter and second. With your choices we get:

x' = x - vt = 1 m - 10 m
t' = t - vx/c² = 1 s - 1.11*10^-16 s​

Keeping vt = 10 m and dropping vx/c² ≈ 10^-16 s seems justified.

x' = x - vt = 1 m - 10 m and t' = t - vx/c2 = 1 s - 1.11*10-16 s?
First let's deal with this misconception. I'll respond to the rest of your post separately.
The conclusion and significance of x' = γ(x - vt) when setting v = 10m/s is that x' = x - 10m for all x and x'.

And t' = γ(t - vx/c^2, with x = 1m and v = 10m/s implies that t '= t for all t and t'.

If v = 10m/s, only when x becomes very large (hundreds of millions of light years) does x come into play in the time dilation equation.

 

[Perpetual Student]

Next:

Now let us do the same with the (somehow less arbitrary) units second and light-second, i.e. let us use x = 1 LS and t = 1 sec. The Lorentz factor of v = 10 m/s = 3.33*10^-8 LS/s obviously does not change and remains negligible. This time, we get:

x' = x - t v = 1 LS - 3.33*10^-8 LS
t' = t - xv/c² = 1 s - 3.33*10^-8 s​

Keeping vt = 3.33*10^-8 LS and dropping vx/c² = 3.33*10^-8 s seems unjustified.

x' = x - t v = 1 LS - 3.33*10-8 LS t' = t - xv/c2 = 1 s - 3.33*10-8 s
As already shown, this is not an appropriate conclusion.

In any case, you made a significant error:
x' = x - 3.33*10^(-8)LS, which, of course is still x - 10m for all x and x', as we had before.


Now, t' = t - 3.33*10^(-8)LS*3.33*10^(-7)LS/s, which is t' = t - 10^(-15)s which reduces to t' = t, also as we had before.

Note that 3.33*10^(-8)LS = 10m, whereas 3.33*10^(-8)LS times 3.33*10^(-7) equals 10^(-15)s.

You continue to confuse yourself by setting c = 1.

 

[Darwin123]

An illustrative example of this argument:



No objection can be raised against your choice of v = 10 m/s, leading to a negligible Lorentz factor, but the choice of x = 1 m and t = 1 sec is in principle completely arbitrary, and depends on the units meter and second. With your choices we get:

x' = x - vt = 1 m - 10 m
t' = t - vx/c² = 1 s - 1.11*10^-16 s​

Keeping vt = 10 m and dropping vx/c² ≈ 10^-16 s seems justified.

Now let us do the same with the (somehow less arbitrary) units second and light-second, i.e. let us use x = 1 LS and t = 1 sec. The Lorentz factor of v = 10 m/s = 3.33*10^-8 LS/s obviously does not change and remains negligible. This time, we get:

x' = x - t v = 1 LS - 3.33*10^-8 LS
t' = t - xv/c² = 1 s - 3.33*10^-8 s​

Keeping vt = 3.33*10^-8 LS and dropping vx/c² = 3.33*10^-8 s seems unjustified.

The Taylor expansion of the Lorentz factor (1-v²/c²)^-1/2 with respect to v yields:

1 v⁰ + 0 v¹ + 1/(2c²) v² + 0 v³ + 3/(8c⁴) v⁴ + . . .​

If meter per second is used as unit for velocity, and both v and c = 2.998*10⁸ express unit-less numeric values, then also higher-order terms of the Lorentz-factor expansion turn analogously into seemingly negligible quantities:

1 + 5.56*10^-18 v² + 4.64*10^-35 v⁴ + . . .​

By the way, the zero-order Taylor-expansion term of the Lorentz factor, 1 v⁰ = 1, becomes after multiplication with -xv/c² the first-order term of the expansion of the t'-transformation. The term t - xv/c² itself is its own full Taylor expansion t v⁰ - x/c² v¹.

Cheers, Wolfgang
As long as we remain in the abstract, and do not apply formulas or propositions to concrete situations, contradictions remain undetected

I didn't get your main problem before. Here it is:

x' = 0 because the man in the rocket can't measure his own displacement.

Therefore, x=vt.

You used x = vt when you calculated t.

By assuming x'=x, you made on concrete error. In reality, Galilean physics and Lorentzian physics, x' = 0.

Furthermore

t'=(t+xv/c^2)/sqrt(1-[v/c]^2)

This is a sign change that doesn't change much. So I won't count it.

 

[Darwin123]

An illustrative example of this argument:



No objection can be raised against your choice of v = 10 m/s, leading to a negligible Lorentz factor, but the choice of x = 1 m and t = 1 sec is in principle completely arbitrary, and depends on the units meter and second. With your choices we get:

x' = x - vt = 1 m - 10 m
t' = t - vx/c² = 1 s - 1.11*10^-16 s​

Keeping vt = 10 m and dropping vx/c² ≈ 10^-16 s seems justified.

Now let us do the same with the (somehow less arbitrary) units second and light-second, i.e. let us use x = 1 LS and t = 1 sec. The Lorentz factor of v = 10 m/s = 3.33*10^-8 LS/s obviously does not change and remains negligible. This time, we get:

x' = x - t v = 1 LS - 3.33*10^-8 LS
t' = t - xv/c² = 1 s - 3.33*10^-8 s​

Keeping vt = 3.33*10^-8 LS and dropping vx/c² = 3.33*10^-8 s seems unjustified.

The Taylor expansion of the Lorentz factor (1-v²/c²)^-1/2 with respect to v yields:

1 v⁰ + 0 v¹ + 1/(2c²) v² + 0 v³ + 3/(8c⁴) v⁴ + . . .​

If meter per second is used as unit for velocity, and both v and c = 2.998*10⁸ express unit-less numeric values, then also higher-order terms of the Lorentz-factor expansion turn analogously into seemingly negligible quantities:

1 + 5.56*10^-18 v² + 4.64*10^-35 v⁴ + . . .​

By the way, the zero-order Taylor-expansion term of the Lorentz factor, 1 v⁰ = 1, becomes after multiplication with -xv/c² the first-order term of the expansion of the t'-transformation. The term t - xv/c² itself is its own full Taylor expansion t v⁰ - x/c² v¹.

Cheers, Wolfgang
As long as we remain in the abstract, and do not apply formulas or propositions to concrete situations, contradictions remain undetected

I didn't get your main problem before. Here it is:

x' = 0 because the man in the rocket can't measure his own displacement.

Therefore, x=vt.

You used x = vt when you calculated t.

By assuming x'=x, you made on concrete error. In reality, Galilean physics and Lorentzian physics, x' = 0. So this is one concrete error.

Furthermore

t'=(t+xv/c^2)/sqrt(1-[v/c]^2)

This is a sign change that doesn't change much. So I won't count it.

 

[Perpetual Student]

The Taylor expansion of the Lorentz factor (1-v²/c²)^-1/2 with respect to v yields:

1 v⁰ + 0 v¹ + 1/(2c²) v² + 0 v³ + 3/(8c⁴) v⁴ + . . .​

If meter per second is used as unit for velocity, and both v and c = 2.998*10⁸ express unit-less numeric values, then also higher-order terms of the Lorentz-factor expansion turn analogously into seemingly negligible quantities:

1 + 5.56*10^-18 v² + 4.64*10^-35 v⁴ + . . .​

By the way, the zero-order Taylor-expansion term of the Lorentz factor, 1 v⁰ = 1, becomes after multiplication with -xv/c² the first-order term of the expansion of the t'-transformation. The term t - xv/c² itself is its own full Taylor expansion t v⁰ - x/c² v¹.

Cheers, Wolfgang
As long as we remain in the abstract, and do not apply formulas or propositions to concrete situations, contradictions remain undetected

So, is there a conclusion to this? What's the point?

 

[Darwin123]

...

x' = x - t v = 1 LS - 3.33*10^-8 LS
t' = t - xv/c² = 1 s - 3.33*10^-8 s​

...

You wrote the above. This is wrong. This is your 'concrete' error.

x'=0 LS = 0 m.

x' DOES NOT equal 1 LS.
x' DOES NOT equal 3.3x10^-8 LS
x' DOES NOT EQUAL 1 m.

Now t is the time measured on earth between the rocket and the earth.
t = x/v
t=1 m/ 10 m/s = 0.1 s
t=0.1 s

The man in the rocket can not determine his own displacement using local measurements. The man in the rocket is always at the origin of his own coordinate system. To put it another way, the man in the rocket travels 0 LS with respect to his own coordinate system.


The man on earth measures x where x>0. Indeed, you are assuming that:
x = 1 m.
This is equivalent to saying.
x =3.33*10^-9 LS

v in LS/s is simply v/c.

So,
x =3.33*10^-9 LS
t = 0.1 s
c = 1 LS/s
v=3.33*10^-8 LS/s

Now we already established that vt is not negligible because we set x'=0.


|vx/c^2|= |3.33*10^-8 LS/s||3.33*10^-9 LS|/|1 LS/s|^2

vx/c^2 = *10^-16 s

Obviously, 0.1 s is much larger than *10^-16 s.

Substitute the values of x, t, v and c into the expressions for x' and t.


x=1m
x'=0
Obviously, v is not negligible as far as displacement is concerned.

BTW: You made an error in sign when you wrote the expression for t'. This is not significant in the problem stated so that I won't count it as a concrete error. I simple will work with the correct sign, which is + not -.

t=0.1 s
t'=0.10000000000000001 s
Obviously, v was negligible so far as time is concerned.

This is a fundamental error. The 'moving' observer and the 'stationary' observer are both measuring the displacement of the rocket. However, the moving observer is IN the rocket at all times. The stationary observer sees the rocket move. So x' and x DON'T approximate each other to first order in v.

You assumed that x was measurable. Therefore, v can not be negligible so far as x' is concerned.

This would be as clear to Galileo as it would to Einstein and Newton.


I predict that you will object that x'=0. However, that is a logical consequence of the initial hypotheses. You violated this basic hypothesis when you assumed that x' DOES NOT equal 3.3x10^-8 LS.

It is rather embarrassing for me that I didn't notice this obvious error many posts ago. I don't think that Perpetual Student noticed it either. We were so busy figuring out what units that you were using that we didn't observer this fundamental error.

Both Galileo and Einstein hypothesized that an observer in an absolute space or an inertial frame can not determine his own nonzero motion by doing experiments in a closed container. The moving observer is hypothetically in a closed container. Therefore, the observer in the rocket is undergoing no motion according to his own measurement. The stationary observer can determine the nonzero motion of the rocket relative to himself.

Therefore, you violated both Lorentzian and Galilean invariance when you set
x' =3.33*10^-9 LS


I am not sure how many times to count this 'one' concrete error. To make that error, you had to violate more than two initial hypotheses. However, the different violations may be arriving from the same initial error.

Anyway, you promised to show how any errors that we present are really correct. Therefore, tell us why x' =3.33*10^-9 LS. Or alternatively, tell us why x' IS NOT equal to 0 m.

 

[Reality Check]

No objection can be raised against your choice of v = 10 m/s, leading to a negligible Lorentz factor, but the choice of x = 1 m and t = 1 sec is in principle completely arbitrary, and depends on the units meter and second. ...

Repeating ignorance of high school level science, wogoga, remains bad!
13 October 2015: Maybe you missed/have forgotten the dimensional analysis part of your science education, wogoga.
That is a almost a week for you to read a high school textbook to relearn the science, e.g. that when you divide two quantifies with the same units you get a number without any units.
The value of v/c is independent of the units used, wogoga. When v << c , v/c << 1 in any units of measurement, wogoga !

Relying on invalid math in examples is rather ignorant when there is general case that works for all units and all values of v/c: When v/c << 1 the Taylor series for the Lorentz factor can be truncated and we get Newtonian mechanics.