SR is dead.

We can see SR contradicts itself with a simple example.

http://vixra.org/pdf/1109.0003v1.pdf

 

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Makes me want to quit school and go to Bovine University instead.

 

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It is a simple example.

With your intellect, you can certainly prove is is false.

I have emails from prd@aps.org that none could refute the logic, but the flat earthers would not publish the article.

 

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Hi chinglu,
can you summarize what you learned fromyour previous threads?
[thread=105498]Time dilation[/thread]
[thread=106240]MMX vs Earths' rotational sagnac[/thread]
[thread=109190]Special Relativity is Refuted[/thread]

 

Danger, Will Robinson, danger!

I would love to see those.

My prima facie reaction (without having to read much further than the abstract) would be to reject this on style alone. But that's just me: I tend to demand that papers are actually readable.

 

The style, the subject matter, the choice of particular words and the general incoherent abrasiveness makes me think of the ex-poster known as Jack_. He was obsessed with common origins of light spheres and thought there was a contradiction.

Even if it isn't Jack_ whose the author the claims have been thoroughly refuted on this forum in the many threads Jack_ started, not least by myself. Chinglu, I suggest you search the Pseudoscience subforum for threads started by Jack_ and you'll find lengthy discussions on precisely the subject matter of that 'paper'. It is completely refuted, that's why people won't publish it.

 

Cesspool.

 

No, _Jack had the grammar of a native English speaker, while I think English is not chinglu's first language.

But yeah, similar subject matter. I suspect they 'learned' from each other, or from similar sources.

 

Look, I just want to stay on this subject. If you have a problem with the author, I suggest your take that up with him.

The light sphere has attained the unprimed coordinate (2,10,0).

The light postulate claims further propagation away from the unprimed origin will occur.

Now, in the view of the unprimed frame, that implies the light sphere will move to say (2+h,10,0). But, that means the light sphere moves closer to the primed origin.

On the other hand, the primed frame claims the light sphere moves left after further propagation.

That is a contradiction.

 

Preamble
Working in seconds for units of time and light-seconds for units of length, we find c = 1 light-second per second. So we may, without risk of confusion, work with c=1 and drop all units. Thus the entire exercise is one in algebraic geometry.

Event O -- "a light pulse is emitted when the origins of two coordinate systems are common"
\(t_O = 0, x_O = 0, y_O = 0, z_O = 0, t'_O = 0, x'_O = 0, y'_O = 0, z'_O = 0\)

The speed of the primed coordinates relative to the unprimed coordinates is (3/5)c in the direction of the x-axis, so the Lorentz transformation is:
\(\begin{pmatrix} c t' \\ x' \\ y' \\ z' \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \frac{-\frac{v}{c}}{\sqrt{1 - \frac{v^2}{c^2}}} & 0 & 0 \\ \frac{-\frac{v}{c}}{\sqrt{1 - \frac{v^2}{c^2}}} & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} c t \\ x \\ y \\ z \end{pmatrix} = \begin{pmatrix} \frac{5}{4} & -\frac{3}{4} & 0 & 0 \\ -\frac{3}{4} & \frac{5}{4} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} c t \\ x \\ y \\ z \end{pmatrix}\)

Locus of all light world-lines emitted from O, SLW, describes the behavior of light in bulk.
\(t_{SLW} \ge 0, t'_{SLW} \ge 0, x_{SLW}^2 + y_{SLW}^2 + z_{SLW}^2 = c^2t_{SLW}^2, x'_{SLW}^2 + y'_{SLW}^2 + z'_{SLW}^2 = c^2t'_{SLW}^2\)

World-line P, a "place" in unprimed coordinates is moving in primed coordinates.
\(x_P = 2, y_P = 10, z_P = 0, x'_P = \frac{8 - 3 t'}{5}, y'_P = 10, z'_P = 0\)

Note that \(t' > \frac{8}{3}\) means \(x'_P < 0\)

Event A, the SLW reaches "place" P
\(t_A = 2 \sqrt{26}, x_A = 2, y_A = 10, z_A = 0, t'_A = \frac{ -3 + 5 \sqrt{26}}{2}, x'_A = \frac{ 5 - 3 \sqrt{26}}{2}, y'_A = 10, z'_A = 0\)

Note that \(t'_A > \frac{8}{3}\) so \(x'_A < 0\)

World-line L, the unique ray of light that starts at O and passes through World-line P

\(x_L = \frac{\sqrt{26}}{26} t, y_L = \frac{5 \sqrt{26}}{26} t, z_L = 0, x'_L = \frac{-375 + 16\sqrt{26}}{641} t', y'_L = \frac{60 + 100\sqrt{26}}{641} t', z'_L = 0\)

Events O and A are both on World-line L and Locus SLW. But Locus SLW is not synomymous with World-line L.

Name of a dog, NO!

The "Place" of the origin of the unprimed coordinate system is World-line M
\(x_M = 0, y_M = 0, z_M = 0, x'_M = - \frac{3}{5} t', y'_M = 0, z'_M = 0\)

The "Place" of the origin of the primed coordinate is World-line N
\(x_N = \frac{3}{5} t, y_N = 0, z_N = 0, x'_N = 0, y'_N = 0, z'_N = 0\)

Thus the Locus B1 of points "between" M and N (restricted to after O) is expanding slower than the speed of light and thus SLW and B1 only have a possible common point at event O and O is only in B1 if "between" is inclusive of N and M.

The Locus of all "places" stationary in the unprimed coordinate system with y=10, z=0 is Locus Y, The restriction on z was necessary to make the time-like slices "lines" as Mr. Banks describes it and is consistent with later wording.

\(y_Y = 10, z_Y = 0, y'_Y = 10, z'_Y = 0\)


Thus we may discuss the locus D, which is the intersection of locus Y and locus SLW.

\(t_D \ge 10, x_D = \pm \sqrt{t_D^2 - 100}, y_D = 10, z_D = 0, t'_D \ge 10, x'_D = \pm \sqrt{t'_D^2 - 100}, y'_D = 10, z'_D = 0\)

This is the spot of light on a wire at y=10 and z=0 which is illuminated by a flash of light from event O. Although each ray of light moves at speed c, the spot of light moves in two directions at once and everywhere moves faster than c, since,
\(| \frac{d x_D}{c d t_D} | = \frac{t}{\sqrt{t^2 - 100}} > 1\)

Marbhfháisc ort!
According to Nature, angular momentum is fixed and barring interaction light moves in straight lines. Like World-Line L, where the intersection with Locus D is the single Event A.

We can assume L propagates further, but the y coordinate immediately increases past 10.

We can talk about the future of D but because we are talking about something non-physical and moving faster than the speed of light, the mapping between coordinate systems does not make sense in physics.

 

You have an error right here.

\(x_P = 2, y_P = 10, z_P = 0, x'_P = \frac{8 - 3 t'}{5}, y'_P = 10, z'_P = 0\)

This is not the way things work under SR.

x' = (x - vt)γ
y'=y
z'=z
t'=( t - vx/c² )γ

Note how you have \(x'_P = \frac{8 - 3 t'}{5}\). Can you see your t'? That is an indication you do not know what you are doing.

So, let's do it correctly and try to follow along.

x' = (x - vt)γ = ( 2 - (3/5)(√(2² + 10²)) )(5/4) < 0.

Now, try again and once you match these results, you will know you have the correct anaser.

 

Here is another error.

This is the spot of light on a wire at y=10 and z=0 which is illuminated by a flash of light from event O. Although each ray of light moves at speed c, the spot of light moves in two directions at once and everywhere moves faster than c, since,
\(| \frac{d x_D}{c d t_D} | = \frac{t}{\sqrt{t^2 - 100}} > 1\)

Marbhfháisc ort!
According to Nature, angular momentum is fixed and barring interaction light moves in straight lines. Like World-Line L, where the intersection with Locus D is the single Event A.

We can assume L propagates further, but the y coordinate immediately increases past 10.

We can talk about the future of D but because we are talking about something non-physical and moving faster than the speed of light, the mapping between coordinate systems does not make sense in physics.

The fact is that the intersection of the light sphere along the line y=10 is greater than c. But, that reality exists or it does not. It that reality does not exist, then the light sphere refuses to intersect the line y=10 with further propagation. Therefore, the light postulate is false in the unprimed frame.

If the light sphere does intersect the line y=10 faster than c, then SR is false and the light postulate is false in the primed frame.

Take your pick.

 

Wait, I thought you had emails from PRD rejecting this work. Are you, or are you not, the author?

 

Oh stop, you counterrevolutionary stooge!

The primed coordinates are moving relative to the unprimed coordinates. So every stationary world line in one frame must be a moving world line in the other frame. So if there is no t-dependence for the x,y,z in one frame, there must be t-dependence for at least one of x,y,z in the other.

Pick a t', any t'. Then solve for x, y, z and t, starting with the associated point on the world-line.

\(\begin{eqnarray} x' = \frac{8 - 3 t'}{5} & = & \frac{5}{4} x - \frac{3}{4} t \\ y' = 10 & = & y \\ z' = 0 & = & z \\ t' & = & \frac{5}{4} t - \frac{3}{4} x \end{eqnarray}\)
So \( x = 2, y = 10, z = 0, t = \frac{4 t' +6}{5}\).

Or you can pick any t, get expressions for x', y', z' and t', then solve for x' in terms of t'.

\(\begin{eqnarray} x & = & 2 \\ y & = & 10 \\ z & = & 0 \\ x' & = & (x - vt) \gamma = ( 2 - \frac{3}{5} t) \frac{5}{4} = \frac{5}{2} - \frac{3}{4} t \\ y' & = & y = 10 \\ z' & = & z = 0 \\ t' & = & ( t - vx/c^2 ) \gamma = ( t - \frac{3}{5} 2 / 1^2 ) \frac{5}{4} = \frac{5}{4} t - \frac{3}{2} \end{eqnarray}\)
\(t' = \frac{5}{4} t - \frac{3}{2}\) means \(t' + \frac{3}{2} = \frac{5}{4} t\) which means \(t = \frac{4}{5} t' + \frac{6}{5}\). And so \(x' = \frac{5}{2} - \frac{3}{4} t = \frac{5}{2} - \frac{3}{4} ( \frac{4}{5} t' + \frac{6}{5} ) = \frac{25}{10} - \frac{9}{10} - \frac{3}{5} t' = \frac{8}{5} - \frac{3}{5} t' = \frac{8 - 3 t'}{5}\)

So the Lorentz transformation with non-zero v transforms a stationary world-line into a world-line moving at constant velocity -v. How is that confusing to you?

Peas and rice!
I disagree with the pedagogy of this approach, because you are not transforming the world-line of P but just a single event on world line P when t = √104 = 2 √26. But this approach is consistent with the relation between x' and t' I described for the whole world-line.
\(\begin{eqnarray} x & = & 2 \\ y & = & 10 \\ z & = & 0 \\ t & = & \sqrt{104} \\ x' & = & (x - vt) \gamma = ( 2 - \frac{3}{5} 2 \sqrt{26}) \frac{5}{4} = \frac{5}{2} - \frac{3 \sqrt{26}}{2} \\ y' & = & y = 10 \\ z' & = & z = 0 \\ t' & = & ( t - vx/c^2 ) \gamma = ( 2 \sqrt{26} - \frac{3}{5} 2 / 1^2 ) \frac{5}{4} = \frac{5 \sqrt{26}}{2} - \frac{3}{2} \end{eqnarray}\)
So when I claim \(x' = \frac{8 - 3 t'}{5} \), you should have calculated: \(\frac{8 - 3 t'}{5} = \frac{8 - 3 \frac{5 \sqrt{26} - 3}{2}}{5} = \frac{16 - 15 \sqrt{26} + 9}{10} = \frac{25 - 15 \sqrt{26}}{10} = \frac{5 - 3 \sqrt{26}}{2}\)

So it certainly appears I had the correct answer all along.

Egg of a turtle!
I didn't say the intersection didn't exist in a mathematical sense. I said it didn't correspond to a physical body moving a rate constrained by the speed of light or the law of inertia.

An equivalent parameterization of Locus D is in terms of s (and s').

\(t_D = 10 \cosh s, x_D = 10 \sinh s, y_D = 10, z_D = 0, t'_D = 10 \cosh s', x'_D = 10 \sinh s', y'_D = 10, z'_D = 0\)
Using the Lorentz transform, we may compute:

\( 10 \cosh s' = t'_D = \frac{5}{4} t_D - \frac{3}{4} x_D = \frac{25}{2} \cosh s - \frac{15}{2} \sinh s \\ 10 \sinh s' = x'_D = \frac{5}{4} x_D - \frac{3}{4} t_D = \frac{25}{2} \sinh s - \frac{15}{2} \cosh s\)
Which has solution
\(s' = s - \tanh^{-1}\left( \frac{3}{5} \right)\)

Thus
\(\frac{d s}{d t} = \frac{\pm 1}{\sqrt{t^2 - 100}}\)
\(\frac{d x}{d s} = 10 \cosh s\)
\(\frac{d x}{d t} = \frac{\pm t}{\sqrt{t^2 - 100}}\)


What does this mean for Mr. Banks' Event A ?
\(s_A = \sinh^{-1}\left( \frac{1}{5} \right) = \ln \frac{1 + \sqrt{26}}{5} > 0\)
\(t_A = 10 \cosh \sinh^{-1}\left( \frac{1}{5} \right) = 2 \sqrt{26}\)
\(x_A = 10 \sinh \sinh^{-1}\left( \frac{1}{5} \right) = 2\)
\(\left. \frac{d s}{d t} \right| _A = \frac{1}{2}\)
\(\left. \frac{d x}{d s} \right| _A = 2 \sqrt{26}\)
\(\left. \frac{d x}{d t} \right| _A = \sqrt{26}\)
\(s'_A = \sinh^{-1}\left( \frac{1}{5} \right) - \tanh^{-1}\left( \frac{3}{5} \right) = \ln \frac{1 + \sqrt{26}}{10} < 0\)
\(t'_A = 10 \cosh \ln \frac{1 + \sqrt{26}}{10} = \frac{-3 + 5 \sqrt{26}}{2}\)
\(x'_A = 10 \sinh \ln \frac{1 + \sqrt{26}}{10} = \frac{5 - 3 \sqrt{26}}{2}\)
\(\left. \frac{d s'}{d t'} \right| _A = - \frac{10 + 6 \sqrt{26} }{209} \)
\(\left. \frac{d x'}{d s'} \right| _A = \frac{-3 + 5 \sqrt{26}}{2}\)
\(\left. \frac{d x'}{d t'} \right| _A = - \frac{375+16 \sqrt{26}}{209}\)

But this is not paradoxical by itself, since
\(\left. \frac{d x'}{d t} \right| _A = \frac{d x'}{d s'} \frac{d s'}{d s} \frac{d s}{d t} = \frac{-3 + 5 \sqrt{26}}{2} \times 1 \times \frac{1}{2} = \frac{-3 + 5 \sqrt{26}}{4}\)

The product \(\frac{d x'}{d t} \frac{d x}{d t} = \frac{d x'}{d s'} \frac{d s'}{d s} \frac{d s}{d t} \frac{d x}{d s} \frac{d s}{d t} = \frac{d x'}{d s'} \frac{d x}{d s} ( \frac{d s}{d t} )^2 = ( 10 \cosh s' ) ( 10 \cosh s ) ( \frac{1}{t^2 - 100} ) = \frac{100 \cosh \left( s - \ln(2) \right) \cosh s}{100 \cosh^2 s - 100} > \frac{1}{2}\)

Thus a positive dx corresponds to a positive dx' during any dt and the motion is not paradoxical. Because the spot is moving faster than light, the primed observer can indeed see it moving a different direction in time, which is an affront to intuition, not special relativity and not physics.

Special relativity does not say nothing can move faster than light -- it says nothing can carry a signal faster than light. Indeed, the signal is sent from event O to A, not from A along the wire at y=10, z = 0.

 

I am not confused you are. By listing x' in terms of t', you are masking or not understanding the issue.

This is a problem of recursion. One starts at some state and proceeds to another state. The mechanism for this change of state is the light postulate in the unprimed frame.

Therefore, by you listing x' in terms of t', completely misses the point of the paper which is clear to list the start or initial state.

So, I have that is now cleared up.

So, we start at (2,10,0) which is translated to (x'<0,10,0).

Now, the question is asked which you have avoided, if the light postulate is true, where will the unprimed frame find the "next" position of the light sphere along the line y=10? The answer is some h at (2+h,10,0).

Now, from the view of the primed frame, allow the light sphere to propagate further from (x'<0,10,0), where will this recursive logic place the light sphere?

The answer is (x'-k,10,0) with k > 0.

Now do you understand a recursive function in a consistent theory? That's right, after a state change, you have only one answer.

However, SR claims after the initial state with the application of the light postulate, the light sphere will be left of the initial space position and to the right of the initial space position. If you understand math well enough, I am sure you understand the result is not functional.

1) You wrote
(a) This is the spot of light on a wire at y=10 and z=0 which is illuminated by a flash of light from event O. Although each ray of light moves at speed c, the spot of light moves in two directions at once and everywhere moves faster than c
(b) I said it didn't correspond to a physical body moving a rate constrained by the speed of light or the law of inertia.

Note how you did not specify in (a) anything about a physical body, you mentioned a spot of light instead. So, you are in error.

Now, you also correctly proved dx' > 0 and dx > 0 for any dt. You sure waster a lot of space proving this simple fact. I can do it in a few lines.

But, here is where you lack the understanding. If x' < 0 and dx' > 0, then any change in the position of the light sphere forces x' to increase meaning since x' < 0, that implies the light sphere moves closer to the primed origin given a fixed y and z with dx' > 0.

Therefore, you are proven the author's case. The author claimed LT calculations force the light sphere to move closer to the primed origin after any further propagation. However, the primed frame light postulate claims along the line y=10, after further propagation, the light sphere moves further from the primed origin.

So, without even realizing it, and you are smart BTW, you proved LT contradicts the light postulate in the primed frame. You just missed the fact that dx'>0 and x'<0 on the specified interval for the problem in the paper.

 

The author let me see their answer.

The response was actually amusing.

 

chinglu is not the author, he writes much BIGGER idiocies than the author

 

Yea, only RPenner has the balls to attack the math.

You sit on the sidelines crying about the temperature of your cocoa.

So, either offer math or confess it is above your intelligence.

And, if you offer math against the paper, then I will prove it is above your intelligence.

So, let's go.

 

I only said that you post much BIGGER idiocies than the author of the paper. This is perfectly irrefutable, fully confirmed by experience.