# SR Issue

Discussion in 'Alternative Theories' started by chinglu, Jun 11, 2014.

1. ### chingluValued Senior Member

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This is a thought experiment with strange results.

Assume M and M' are the origins of 2 frames and in the M' frame, there is an observer C' located at $(\frac{-vd'}{c},0,0)$ with $d'>0$.

When M and M' are co-located, lightning strikes their command location.

Here is the question, when C' and M are co-located, where is the lightning along the positive x-axis for both frame coordinate systems?

First, we have to know the time on the clocks at M and C' when they are co-located.

M frame calculations.
1) M clock - Apply LT $x'=(x-vt)\gamma$ with $x'=\frac{-vd'}{c}$ and $x=0$. Then, solve for t so $t=\frac{d'}{c\gamma}$
2) C' clock. Apply LT $t'=(t-vx/c^2)\gamma$ with $t=\frac{d'}{c\gamma}$ and $x=0$. Then $t'=\frac{d'}{c}$

M' frame calculations
1) C' clock - Apply LT $x=(x'+vt')\gamma$ with $x'=\frac{-vd'}{c}$ and $x=0$ and solve for t'. Then, $t'=\frac{d'}{c}$.
2) M clock - apply LT $t=(t'+vx'/c^2)\gamma$ with $t'=\frac{d'}{c}$ and $x'=\frac{-vd'}{c}$. Then, $t=\frac{d'}{c\gamma}$.

So far so good. SR agrees in the calculations of both frames the times on the clocks at C' and M when the two are co-located.

Now we ask the question where is the lightning along the positive x-axis when C' and M are co-located?

M frame calculations for the space-time coordinates of the lightning along the positive x-axis when C' and M are co-located.

1) M coordinate system location. Since the time on the M clock is $t=\frac{d'}{c\gamma}$, apply the light postulate $x=ct$. So, $x=d'/\gamma$.
Thus, the space-time coordinate of the lightning in the M coordinate system is $(d'/\gamma,0,0,\frac{d'}{c\gamma})$
2) M' coordinate system location. Apply LT $x'=(x-vt)\gamma$ and $t'=(t-vx/c^2)\gamma$ with $x=d'/\gamma$ and $t=\frac{d'}{c\gamma}$. Then, $x'=d'(1-v/c)$ and $x'=d'(1-v/c)/c$.
Thus, the space-time coordinate of the lightning in the M' coordinate system is $(d'(1-v/c),0,0,d'(1-v/c)/c)$

M' frame calculations for the space-time coordinates of the lightning along the positive x-axis when C' and M are co-located.
1) M' coordinate system location. The time on the clock at C' is $t'=\frac{d'}{c}$. The lightning struck at M', so apply the light postulate from M', which is also the origin of the primed frame $x'=ct'$, with $t'=\frac{d'}{c}$. Then, $x'=d'$.
Thus, the space-time coordinate of the lightning in the M' coordinate system is $(d',0,0,d'/c)$
2) M coordinate system location. Apply LT $x=(x'+vt')\gamma$ and $t=(t'+vx'/c^2)\gamma$ with $x'=d'$ and $t'=\frac{d'}{c}$. Then, $x=d'\gamma(1+v/c)$ and $t=d'\gamma(1+v/c)/c$.
Thus, the space-time coordinate of the lightning in the M coordinate system is $(d'\gamma(1+v/c),0,0,d'\gamma(1+v/c)/c)$.

Conclusions:

When C' and M are co-located, SR claims the lightning is located at M frame space-time coordinates of $(d'/\gamma,0,0,\frac{d'}{c\gamma})$ and $(d'\gamma(1+v/c),0,0,d'\gamma(1+v/c)/c)$.

When C' and M are co-located, SR claims the lightning is located at M' frame space-time coordinates of $(d'(1-v/c),0,0,d'(1-v/c)/c)$ and $(d',0,0,d'/c)$.

Therefore, if these calculations are correct, then SR claims when M and C' are co-located, one lightning strike is located at 2 different positions along the positive x-axis in both coordinate systems, which of course is inconsistent with nature.

So, where is the error in the calculations?

3. ### rpennerFully WiredValued Senior Member

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By definition a thought experiment is a rigorous exploration of the predictions of a physical theory. You lack rigor and so there is not enough thought in your thought experiment for it to merit that description.
You have ignored relativity of simultaneity and therefore ignored basic characteristic of special relativity, dating back to Einstein's original 1905 paper.

Let f,g,h be time-like inertial world lines. Let j, k be space-like straight lines. Let ℓ be a light-like straight line. Then we have in both coordinate system the following descriptions of these lines:
$\begin{array}{c|c|c} \textrm{Line} & \textrm{Frame} \; \Sigma & \textrm{Frame} \; \Sigma' \\ \hline \\ f & x = 0 & x' = -vt' \\ g & x = vt & x' = 0 \\ h & x = vt - \frac{d'}{c} v \sqrt{1 - \frac{v^2}{c^2}} & x' = - \frac{d'}{c} v \\ j & t = \frac{d'}{c} \sqrt{1 - \frac{v^2}{c^2}} & x' = - \frac{c^2}{v} \left( t' - \frac{d'}{c} \left(1 - \frac{v^2}{c^2} \right) \right) \\ k & x = \frac{c^2}{v} \left( t - \frac{d'}{c} \sqrt{1 - \frac{v^2}{c^2} } \right) & t' = \frac{d'}{c} \\ \ell & x = ct & x' = ct' \end{\array}$ $\begin{array}{c|c|c} \textrm{Event} & \textrm{Frame} \; \Sigma & \textrm{Frame} \; \Sigma' \\ \hline \\ O=f \cap g \cap \ell & \left( x=0, \; t=0\right) & \left(x'=0, \; t'=0 \right) \\ P = f \cap h \cap j \cap k & \left( x = 0, \; t = \frac{d'}{c} \sqrt{1 - \frac{v^2}{c^2}} \right) & \left( x' = - \frac{d'}{c} v, \; t' = \frac{d'}{c} \right) \\ Q = j \cap \ell & \left( x = d' \, \sqrt{1 - \frac{v^2}{c^2}}, \; t = \frac{d'}{c} \sqrt{1 - \frac{v^2}{c^2}} \right) & \left( x' = d' \, \left(1 - \frac{v}{c} \right) , \; t' = \frac{d'}{c} \left(1 - \frac{v}{c} \right) \right) \\ R = k \cap \ell & \left( x = \frac{c d' \sqrt{1 - \frac{v^2}{c^2}}}{c - v}, \; t = \frac{d' \sqrt{1 - \frac{v^2}{c^2}}}{c - v} \right) &\left( x' = d' , \; t' = \frac{d'}{c} \right) \end{\array}$
So by ignoring relativity of simultaneity, you improperly confuse lines j and k and therefore confuse events Q and R.

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Nice Mr rpenner.

I was going to answer him [not on the mathematics] but on the sheer audacity of his propostions.

What was the movie, with that little girl shouting out, "He's baaack!"

7. ### rpennerFully WiredValued Senior Member

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More on straight lines

In addition to writing lines as $y = mx + b$, one may describe the same line as $mx - y = b$ or more generically as $mx + n y = b$ with the advantage than nothing goes wonky when the line goes vertical. This applies to our lines in the x-t plane and we can rewrite the lines using $\beta \equiv \frac{v}{c}, \; \gamma = \frac{1}{\sqrt{1 - \beta^2}}$ as:

$f: \quad x + 0 c t = 0 \\ g: \quad x - \beta c t = 0 \\ h : \quad x - \beta c t = - \beta \frac{d'}{\gamma} \\ j : \quad 0 x + c t = \frac{d'}{\gamma} \\ k : \quad \beta x - c t = - \frac{d'}{\gamma} \\ \ell : \quad x - ct = 0$

Substituting in both right sides of $x = \gamma ( x' + \beta c t' ), \; t = \gamma ( t' + \beta c^{\tiny -1} x' )$, we immediately get:

$f: \quad \gamma ( x' + \beta c t' ) + 0 c \gamma ( t' + \beta c^{\tiny -1} x' ) = \gamma x' + \beta \gamma c t' = 0 \\ g: \quad \gamma ( x' + \beta c t' ) - \beta c \gamma ( t' + \beta c^{\tiny -1} x' ) = \left( 1 - \beta^2 \right) \gamma x' + 0 c t' = 0 \\ h : \quad \gamma ( x' + \beta c t' ) - \beta c \gamma ( t' + \beta c^{\tiny -1} x' ) = \left( 1 - \beta^2 \right) \gamma x' + 0 c t' = - \beta \frac{d'}{\gamma} \\ j : \quad 0 \gamma ( x' + \beta c t' ) + c \gamma ( t' + \beta c^{\tiny -1} x' ) = \beta \gamma x' + \gamma c t' = \frac{d'}{\gamma} \\ k : \quad \beta \gamma ( x' + \beta c t' ) - c \gamma ( t' + \beta c^{\tiny -1} x' ) = 0 x' - \left(1 - \beta^2 \right) \gamma c t' = - \frac{d'}{\gamma} \\ \ell : \quad \gamma ( x' + \beta c t' ) - c \gamma ( t' + \beta c^{\tiny -1} x' ) = \left( 1 - \beta \right) \gamma x' - \left( 1 - \beta \right) \gamma c t' = 0$

Removing common factors and exploiting $\left( 1 - \beta^2 \right) \gamma^2 = 1$ we have:

$f: \quad x' + \beta c t' = 0 \\ g: \quad x' + 0 c t' = 0 \\ h : \quad x' + 0 c t' = - \beta d' \\ j : \quad \beta x' + c t' = \frac{d'}{\gamma^2} = d' \, \left( 1 - \beta^2 \right) \\ k : \quad 0 x' + c t' = d' \\ \ell : \quad x' - c t' = 0$
in agreement with post #2.

So the take-away is that the Lorentz transform transforms straight lines in the x-t plane to straight lines in the x'-t' plane. It transforms parallel lines in the x-t plane to parallel lines in the x'-t' plane. It is an example of a linear transformation, or affine transformation. As such, it is totally impossible for it transform two intersecting lines into a pair of intersecting lines that meet at place that is not the image of the intersection of the first pair. Thus it is impossible for you to demonstrate that special relativity is internally inconsistent by just transforming straight lines and intersections.

Where you got confused is by trying to assert absolute time, even in the face of two different inertial frames and events which don't happen in the same place.

In frame $\Sigma$ events P and Q happen in different places at the same time. In frame $\Sigma'$ events P and R happen in different places at the same time. In neither frame do events Q and R happen at the same time. That can't happen because line j is not the same as line k when $0 \lt \beta \lt 1$.

8. ### chingluValued Senior Member

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1,637

1) You said the thought experiment is wrong or something. So, perhaps we can make it simple. Are the calculations for the times on the clocks at C' and M correct when they are co-located.

2) You claimed I ignored ROS. I used LP and LT. ROS is built into LT. LP is an axiom. Further, I could not follow your events. Perhaps you can change it a bit. For example, where is the lightning along the positive x-axis located in M and M' when C' and M are co-located? Just put down an answer or answers on its location. You see in nature, it is somewhere in the coordinates of the M and M' systems. So, where is it?

9. ### chingluValued Senior Member

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1,637
Can you show me specifically in the calculations where I asserted absolute time? If I came up with the same time somewhere for both frames, that means I assert absolute time. All times above for events have different times just as one would expect with ROS.

So, please show the specific calculation(s) where I used absolute time.

Thanks.

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Of course chinglu if you are ultimately proposing another alternative unreviewed theory, you are in the wrong forum.Let's see how this pans out anyway.

11. ### chingluValued Senior Member

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I am not proposing anything except SR calculations.

Do you see any errors?

12. ### chingluValued Senior Member

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Maybe You can help me see an error.

I calculated that the event C' and M co-locating will be simultaneous with some lightning event in M. M' will disagree these events are simultaneous.

Then the calculations showed the event C' and M co-locating will be simultaneous with some lightning event in M'. M will disagree these events are simultaneous.

Is this ROS yes or no? Or, is there some other error?

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Your errors have been noted by rpenner. :shrug:

14. ### rpennerFully WiredValued Senior Member

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The essentials are here:

No. I said it lacked rigor (数学的严谨), therefore it was not a thought experiment in the same way a mathless fantasy of FTL isn't a physical hypothesis.

Wrong way to go about fixing your problem. Your problem is you make sloppy assumptions without checking each assumption that it is compatible with theory.

That doesn't matter when the opposite of ROS is built into your assumptions.
It's not "lightning" (闪电)-- it's a flash of light (点样光的颗粒).

It's not in the calculations, it's in the assumptions made before the calculations.
The danger happens when you say something happens "at the same time ... somewhere else".

Nothing is consistently numbered or labeled until post #2.

Here is your post #1 marked up with corrections and correspondence with post #2:
Therefore, M corresponds directly with line f. Likewise, M' and C' correspond with lines g and h, respectively. How can that be confusing?

This is event O which is part of line f, part of line g, and also part of line ℓ which makes sense because line ℓ is the flash of light which travels in the +X direction from the lightning strike.
Not unsurprisingly, event O has coordinates $( x = 0, \; t = 0, \; x' = 0, \; t' = 0 )$

That is asking for what event is common to line f and line h. This I call event P, and has coordinates $\left( x = 0, \; t = \frac{d'}{c} \sqrt{1 - \frac{v^2}{c^2}}, \; x' = - \frac{d' \, v}{c} , \; t' = \frac{d'}{c} \right)$

This is not surprising.

Here you ignore relativity of simultaneity because there are two, physically and geometrically distinct definitions of when C' and M are co-located. Line j are all the events where $y=z=y'=z'=0$ and $t = \frac{d'}{c} \sqrt{1 - \frac{v^2}{c^2}}$ while line k are all the events where $y=z=y'=z'=0$ and $t' = \frac{d'}{c}$. Since $d' > 0$ and $0 < v < c$ it is impossible for j and k to be the same lines.

We can even solve in frame Σ where line j and line k intersect. That's the solution to simultaneous equations:
$\begin{eqnarray} 0 x & + & c t & = & \frac{d'}{\gamma} \\ \beta x & - & c t & = & - \frac{d'}{\gamma} \end{eqnarray}$
Adding these two equations term by term we compute $\beta x = 0$ so the intersection has $x = 0, t = \frac{d'}{c \gamma}$ which is event P again.
Therefore event P is where lines f, h, j and k all meet.

If at any event which is not P you talk about the time when C' and M are co-located, then you have made an assumption about time. If you don't explicitly talk about line j or line k, then you have ignored relativity of simultaneity. If you ignore that both j and k are physically distinct concepts of when C' and M are co-located then you have made the assumption of absolute time.

This is the intersection of lines j and ℓ which I have named Q. The coordinates of event Q are $\left( x = d' \, \sqrt{1 - \frac{v^2}{c^2}}, \; t = \frac{d'}{c} \sqrt{1 - \frac{v^2}{c^2}} , \; x' = d' \, \left(1 - \frac{v}{c} \right) , \; t' = \frac{d'}{c} \left(1 - \frac{v}{c} \right) \right)$.

Our calculations agree. But I never said the problem was your calculations, but rather your assumptions.
This is the intersection of lines k and ℓ which I have named R. The coordinates of event R are $\left( x = \frac{c d' \sqrt{1 - \frac{v^2}{c^2}}}{c - v}, \; t = \frac{d' \sqrt{1 - \frac{v^2}{c^2}}}{c - v} , \; x' = d' , \; t' = \frac{d'}{c} \right)$.

Our calculations agree.

Since your assumption that line j was identical to line k turned out to not be consistent with Special Relativity, it is only natural that event Q and event R are physically distinct in both position and time. As there is no frame which says Q and R are simultaneous, it is not a contradiction with nature that Q and R happen in different places. Indeed every coordinate frame says Q and R are light-like separated, which does directly agree with nature.

Last edited: Jun 13, 2014
15. ### chingluValued Senior Member

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Good, you put the concepts in your own way, but agree with everything up to the conclusions. We are at that point.

Assume P is the logical predicate defined as follows. P = C' and M are co-located.

Ok, now we can talk about things like if the predicate P is true, then this and that.

When P is true, where does SR claim the lightning flash is located in the coordinates of the M frame?

When P is true, where does SR claim the lightning flash is located in the coordinates of the M' frame?

You see, SR is a theory and must observe logic. For brevity define STCL as the space time coordinate of the lightning flash.

Let us define the predicate Q as $STCL=(d'/\gamma,0,0,\frac{d'}{c\gamma})$ and R as $STCL=(d',0,0,d'/c)$

Now, since we both agree when P is true, in M frame coordinates STCL is located at $(d'/\gamma,0,0,\frac{d'}{c\gamma})$ and $(d'\gamma(1+v/c),0,0,d'\gamma(1+v/c)/c)$

and also, when P is true, in M' frame coordinates STCL is located at $(d'(1-v/c),0,0,d'(1-v/c)/c)$ and $(d',0,0,d'/c)$.

Therefore, we find $p\to (Q\wedge\neg Q)$

Also, $p\to (R \wedge\neg R)$

So, when P is true, are these the correct conclusions?

16. ### arfa branecall me arfValued Senior Member

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Your predicate makes no sense; it says when P is true that an observer at M sees the lightning strike in two places, likewise an observer at M' sees two strikes.

Your statement is true only if $(d'/\gamma,0,0,\frac{d'}{c\gamma}) \,=\,(d'\gamma(1+v/c),0,0,d'\gamma(1+v/c)/c)$, and $(d'(1-v/c),0,0,d'(1-v/c)/c)\,=\, (d',0,0,d'/c)$.

If you can show this your logic might have legs, but I wouldn't try putting a kilt on it.

17. ### rpennerFully WiredValued Senior Member

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Line f is the trajectory through space-time of chinglu's object M. Line g is the trajectory throught space times of chinglu's object M'. Line h is the trajectory through space-time of chinglu's object C'. Lines f and g intersect in event O. Lines f and h intersect in event P. Lines j and k are lines of simultaneity for coordinate frames Σ and Σ', respectively. Because no physical phenomena depends on simultaneity of events not located in the same location, lines j and k are inventions of man, not nature. So confusing lines j and k does not advance understanding of nature. Line ℓ is light-like and can be said to originate at event O. Line ℓ intersects lines j and k in events Q and R, respectively.

Chinglu makes a big point about there being two events, Q and R, but this follows because we are talking about two distinct imaginary coordinate systems with different notions of coordinate time and therefore different definitions of simultaneity. Chinglu says he respects that relativity of simultaneity is part of special relativity but seems incapable of acknowledging that coordinate time is a human choice, not a law of nature. Two human-defined coordinate frames in relative motion leads to two different definitions of what events are simultaneous with event P. Lines j and k encapsulate those definitions, and lead to intersecting with two different events on line ℓ.

It's not nature's fault that chinglu doesn't understand lines j and k. It's not special relativity's fault that chinglu doesn't understand lines j and k.
Corrected Conclusions from [post=3198939]Post #15[/post]:
When C' and M are co-located, observers at rest in frame Σ claim the flash is located at Σ frame space-time coordinates of $(d'/\gamma,0,0,\frac{d'}{c\gamma})$.

When C' and M are co-located, observers at rest in frame Σ' claim the flash is located at Σ' frame space-time coordinates of $(d',0,0,d'/c)$. But this is not the same event as the event described by the the observers at rest in frame Σ, because observers in relative motion disagree on what space-like separated events are to be considered simultaneous.

Therefore, SR describes how moving observers or coordinate systems disagree on what events are to be considered simultaneous with each other. As they disagree on what events (Q or R) are simultaneous with when C' and M are co-located , event P, they also, naturally, disagree on the location of the event (Q or R) which is simultaneous, for events Q and R are light-like separated and cannot be at the same position unless they also happen at the same time, which is not the case in any frame.

Working in coordinate system Σ we can compute the space-time interval, (c Δt)² − (Δx)² − (Δy)² − (Δz)², between Q and R, where Q is $\left( x = d' \, \sqrt{1 - \frac{v^2}{c^2}}, \; y = 0, \; z = 0 , \; t = \frac{d'}{c} \sqrt{1 - \frac{v^2}{c^2}} \right)$ and R is $\left( x = \frac{c d' \sqrt{1 - \frac{v^2}{c^2}}}{c - v}, \; y = 0, \; z = 0 , \; t = \frac{d' \sqrt{1 - \frac{v^2}{c^2}}}{c - v} \right)$ and so we compute :
$\left( c t_R - c t_Q \right)^2 - \left( x_R - x_Q \right)^2 - \left( y_R - y_Q \right)^2 - \left( z_R - z_Q \right)^2 = \left( c \frac{d' \sqrt{1 - \frac{v^2}{c^2}}}{c - v} - c \frac{d'}{c} \sqrt{1 - \frac{v^2}{c^2}} \right)^2 - \left( \frac{c d' \sqrt{1 - \frac{v^2}{c^2}}}{c - v} - d' \, \sqrt{1 - \frac{v^2}{c^2}} \right)^2 - \left( 0 -0 \right)^2 - \left( 0 - 0 \right)^2 = 0$.
So Q and R are two different events, but they are both separated by a light-like space-time interval, so nothing in physical law says a flash of light can't at one time be at the location described by event Q and at a later time be at the location described by event R. That's what light does, it moves.

Line ℓ is slanted in every inertial coordinate system because it's always a light-like line describing how light moves at speed c. Events Q and R a two different parts of line ℓ so they two describe the movement of light between two different concepts of "now". That's central to the concept of relativity of simultaneity.

LIAR! We disagree on many things. Here is an incomplete list of the disagreements that have been evidenced in this thread:
• We disagree that you understand relativity of simultaneity.
• We disagree that you have any great insight into special relativity.
• We disagree that your math skills even have the potential to show that special relativity is either internally inconsistent or at odds with natures
• To show special relativity is internally inconsistent you must show that Lorentz transformations are not diffeomorphisms of flat space time, but they are trivial diffeomorphisms and so you have not enough math skills to know you are wasting everyone's time
• To show special relativity is inconsistent with nature, you need data from physical experiments. Special relativity is already shown to be consistent with the summary of all physical experiments in the field of electromagnetism up through 1865 and turns out to explain pair-production which was confirmed in the 1930's.
• We disagree that you should be allowed to post in the Physics and Math section of this website.
• We disagree that "lightning" moves at the speed of light.
• We disagree that M and M' are frames when you described them as spatial origins of coordinate systems. The systems themselves need new names and I chose Σ and Σ'.
• We disagree that P, Q, and R are events when you try to re-label them as predicates.
• We disagree that you know enough logic to use the term predicate.

The laws of nature include that
• inertial motion in flat space-time are represented in inertial Cartesian coordinates straight time-like lines,
• that light in vacuum moves in straight light-like lines in in inertial Cartesian coordinates,
• events considered simultaneous by an inertial observer all fall on straight space-like lines in inertial Cartesian coordinates, and
• all inertial Cartesian coordinate systems agree on the value of the space-time interval, (c Δt)² − (Δx)² − (Δy)² − (Δz)², between the coordinates of two events
but they don't include that one inertial observer's description of simultaneity is universally applicable. Thus lines j and k are different lines and thus events Q and R are different events, separated in both space and time by a light-like space-time interval. Nature doesn't have a problem with light being at one position at one time and at another position at a later time so long as (c Δt)² − (Δx)² − (Δy)² − (Δz)² = 0.

Last edited: Jun 14, 2014
18. ### chingluValued Senior Member

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1,637
No, that is not what I said.

When P is true, SR in the context of the M frame claims the lightning flash is located at M coordinate $(d'/\gamma,0,0,\frac{d'}{c\gamma})$. When P is true, SR in the context of the M' frame claims the lightning flash is at $(d'\gamma(1+v/c),0,0,d'\gamma(1+v/c)/c)$

Therefore, when P is true, SR claims the lightning is at 2 different places in M frame coordinates.

Now, either the calculations are wrong, which even RPenner has not claimed, or when P is true SR claims the lightning is at 2 different M frame coordinates along the positive x-axis.

19. ### chingluValued Senior Member

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I am not sure why you need to be so nasty especially when you can't refute any of my calculations. And, yes your conclusions above are correct. I never disputed that at all.

However, what you wrote above has nothing to do with my conclusions.

Here they are again.

When C' and M are co-located, SR claims the lightning is located at M frame space-time coordinates of $(d'/\gamma,0,0,\frac{d'}{c\gamma})$ and $(d'\gamma(1+v/c),0,0,d'\gamma(1+v/c)/c)$.

When C' and M are co-located, SR claims the lightning is located at M' frame space-time coordinates of $(d'(1-v/c),0,0,d'(1-v/c)/c)$ and $(d',0,0,d'/c)$.

Now, you have already agreed to the following in your text above:
When C' and M are co-located, SR claims the lightning is located at M frame space-time coordinates of $(d'/\gamma,0,0,\frac{d'}{c\gamma})$.

When C' and M are co-located, SR claims the lightning is located at M' frame space-time coordinates of $(d',0,0,d'/c)$.

What you failed to do however is address the following:
When C' and M are co-located, using M' frame calculations, SR claims the lightning is located at M frame space-time coordinates of $(d'\gamma(1+v/c),0,0,d'\gamma(1+v/c)/c)$.

When C' and M are co-located, using M frame calculations, SR claims the lightning is located at M' frame space-time coordinates of $(d'(1-v/c),0,0,d'(1-v/c)/c)$.

These two calculations above are derived by applying LT to the two space-time coordinates you already agree are true.

So, is it your belief when C' and M are co-located in the M frame, they are not co-located in the M' frame? Or, are you saying the above two calculations are wrong. That means LT is wrong? Or, is it something else?

20. ### rpennerFully WiredValued Senior Member

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• P, Q, and R are not a logical predicates, but space-time events as described in post #2.
• You conflate two meanings of "when" so readers don't know if you mean "simultaneous with" or "contingent upon"
• If you wish to start to write predicates in formal logic, you are required to use a form of modal logic which is compatible with special relativity if you wish to talk about simultaneity.
• You can't talk about anything being at two different positions at the same time if they are only at different positions at different times. Looking at events Q and R, they are clearly distinct events but both are part of line ℓ so if Q and R are at different positions (as they are in every coordinate system) then they must also happen at different times (which is also true in every coordinate system).

21. ### rpennerFully WiredValued Senior Member

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Because you should have learned how to do special relativity 1200 posts ago.
These are the frame Σ coordinates of event Q.
There are the frame Σ' coordinates of event R.
Event Q and event R are both part of line ℓ and therefore represent that light can be at one position at one time and then a different position at some later time, without violating any law of nature. This I have enhanced post #18 to explain.

These are the frame Σ coordinates of event R which you can compare with the frame Σ coordinates of event Q to see that they happen in different places, but also at different times and both fall on the same space-time line: x = ct.

These are the frame Σ' coordinates of event Q which you can compare with the frame Σ coordinates of event R to see that they happen in different places, but also at different times and both fall on the same space-time line: x' = ct'.

More importantly, I already gave these events and their coordinates in post #2 and described exactly which lines they were part of. Post #2 was very short. Have you understood it yet?

(Using "when" as a marker of contingency in relativistically-informed modal logic

YES. Because that's the definition of relativity of simultaneity. Specifically I believe for an inertial observer, not located at M and at rest in frame Σ, the concept of "when C' and M are co-located" differs from the concept of "when C' and M are co-located" by a different inertial observer, not located at C' and at rest in frame Σ'. I have already calculated those concepts as straight space-like line in post #2.

Line j is composed of all events that observers at rest in frame Σ agree are simultaneous with event P (and have y= z= 0). Line k is composed of all events that observers at rest in frame Σ' agree are simultaneous with event P (and have y'= z'= 0). These lines are straight in any inertial coordinate system but the only place they intersect is at event P. So unless you are exactly at the position of event P, lines j and k correspond to different human-chosen conventions of what is simultaneous with event P. Therefore events Q and R happen at different times.

Everyone at rest in cordinate frame Σ agrees that P and Q happen at the same time, but that R happens later than Q. Everyone at rest in coordinate frame Σ' agrees that P and R happen at the same time, but that R happens later than Q. Both coordinate frames agree that light moves from O to Q to R, being at different places at different times.

Therefore you have not demonstrated any contradiction with nature.

Last edited: Jun 14, 2014
22. ### chingluValued Senior Member

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1,637
You are completely wrong and here is specifically why. I brought you to the point of making this statement below:
I will use your terminology, though it is not necessary. So, you agree, when C' and M are co-located, observers at rest in frame Σ claim the flash is located at Σ frame space-time coordinates of $(d'/\gamma,0,0,\frac{d'}{c\gamma})$. OK that is fine. Now, as any high school person would know, to determine the corresponding space-time coordinate in the Σ' frame, one applies LT. This is where you are having difficulty. Don't forget to apply LT.

Now, when you apply LT, you get the Σ' light flash space-time coordinate of$(d'(1-v/c),0,0,d'(1-v/c)/c)$ when M and C' are co-located.

From the Σ' frame, you agree the flash is at $(d',0,0,d'/c)$ when C' and M are co-located. Again, apply LT and you get the Σ frame light flash coordinate of $(d'\gamma(1+v/c),0,0,d'\gamma(1+v/c)/c)$ when M and C' are co-located.

As we can clearly see, when C' and M are co-located, SR claims 2 different light flash positions in the Σ frame and also SR claims 2 different light flash positions in the Σ' frame.

23. ### chingluValued Senior Member

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1,637
Ummm, for you to have such an opinion, exactly where is anything I posted wrong?

Science is not about personal opinions that cannot be supported with any facts or calculations.