In the OP, chinglu proposes that we consider objects M and C' moving inertially with non-zero relative velocity $$v = \beta c$$ and colliding at event P. Chinglu also asks us to consider a unidirectional flash of light originating at event O (in the past light cone of P) and asks for a description of the light "at the same time" as event P. Naturally, this does not give rise to a single description because "at the same time" is a function of which coordinate frame we use to determine when distant points are simultaneous. Specifically, if we use frame Σ, the frame where object M is at rest, then the definition of simultaneity in frame Σ finds only one event, Q, which is both in the future of the flash of light and "at the same time" as event P. And if we use the definition of simultaneity in frame Σ', the frame where object C' is at rest, then the definition of simultaneity in frame Σ finds only one event, R, which is both in the future of the flash of light and "at the same time" as event P. Because the direction of relative velocity between M and C' is not orthogonal to the direction of the propagation of the flash of light, it follows that $$Q \neq R$$ and necessarily, the time and space coordinates of Q will not match those of R in either frame.

There is no great mystery here unless one mistakenly assumes "at the same time" means something physical for events that don't happen in the same place.

For those that haven't seen any other post in this thread, object M travels on line f from event O to event P and is at rest in frame Σ. Object M' travels through event O and is at rest in frame Σ'. Object C' travels through event P and is at rest in frame Σ'. Line j is all events in the x-t plane of frame Σ which are considered simultaneous to P in that frame. Line k is all events in the x'-t' plane of frame Σ' which are considered simultaneous to P in that frame. Line ℓ is a light-like line which passes through O and intersects both lines j and k to the right of P. This intersections we call events Q and R, respectively.

You can't "agree" with anyone else in this thread, because both Special Relativity and myself have said that "when M and C' are co-located" has no

*physical* meaning except for objects whose world-lines pass through event P. Instead, I wrote in [post=3199686]post 65[/post]:

**you[r] only mistake is assuming the phrase "when M and C' are co-located" has some universal meaning independent of frames.** No law of nature is violated -- the only thing that is violated is *your assumption of the universality of simultaneity*. Lines j and k are different lines. Therefore $$Q \neq R$$, because no law of nature would make it possible for Q and R to be the same event.

Event P, the only place and time in the whole of space-time where M and C' are co-located, is far away from both Q and R, so the notion of which of Q or R happens at the same time as P is dependent on choice of coordinate systems. That's what *relativity of simultaneity* means.

The only way that the phrase "when M and C' are co-located" has any meaning when one's world line does not pass through event P is by the artificial establishment of an inertial coordinate frame. Thus line j is a line of simultaneity that applies to just frame Σ (or by the reckoning of people at rest in frame Σ). Likewise, line k is a line of simultaneity that applies to just frame Σ'.

I did not agree to any universal meaning independent of frames,

So you agree that lines j and k are distinct lines that only meet at event P? And you agree that any two points on line j have the same coordinate time in frame Σ? And you agree that any two points on line k have the same coordinate time in frame Σ'? Because understanding that about all of lines j and k except at event P is a general feature of special relativity.

I proved when M and C' are co-located, both frames agree on the time of their respective frames. This is simple SR.

It's very simple, I agree. But if you had any knowledge of linear algebra you wouldn't have to prove it for the coordinates of a single event P.

Given $$\Lambda$$ is a homomorphism between coordinates in frame Σ and frame Σ', then it follows that $$\Lambda^{-1}$$ exists and $$\Lambda^{-1} \Lambda = \Lambda \Lambda^{-1} = I$$ where $$I$$ is the identity transform. Thus this proves

$$\begin{pmatrix} x \\ y \\ z \\ t \end{pmatrix} = \Lambda^{-1} \Lambda \begin{pmatrix} x \\ y \\ z \\ t \end{pmatrix}, \quad \quad \quad \begin{pmatrix} x' \\ y' \\ z' \\ t' \end{pmatrix} = \Lambda \Lambda^{-1} \begin{pmatrix} x' \\ y' \\ z' \\ t' \end{pmatrix}$$

for the coordinates of all events in space-time.

Given $$x' = \gamma \left( x - \beta c t \right), t' = \gamma \left( t - \beta c^{-1} x \right)$$ it is easy to prove $$x = \gamma \left( x' - (-\beta) c t' \right), t = \gamma \left( t' - (-\beta) c^{-1} x' \right)$$ thus demonstrating that Lorentz transform has an inverse which is a different Lorentz transform. Thus you've labored to prove a trivial result while relying on a stronger theorem.

Are you saying this is false?

No. I guess you are admitting you didn't understand a single one of my posts, including the very short [post=3198606]post #2[/post]. To misunderstand my post to this extent looks like trolling.

Further I never said Q=R is required under SR, since it is not.

If it is not required that Q=R, then why are you arguing that Q and R can't be different?

Now, you again ran in terror from my post.

And you demonstrate that you are delusional in that you think I'm scared of your bad English and worse math and that you think you are entitled to immediate responses to misunderstandings that were addressed in post #2.

The only reason you would do this is because you know you are totally wrong.

Is it wrong to hope that one day you might finally understand trivial affine mappings between coordinate systems? Perhaps. But the only way you can demonstrate that I am totally wrong is by dying before learning better.

Are you terrified to answer these questions?

What questions? If you mean the sentences that end with the question mark, they appear to be predicated on false assumptions to the point where you are merely trolling.

Do not forget, you pretend to be an expert to all those reading this thread.

No. An expert has a subject matter that defines the boundaries of purported expertise. An expert's opinion in the area of expertise is revered. That I make pronouncements on your physics simply indicates I have walked the path that all physics experts walk, far enough to have a better idea where the line between layman and experts exists.

For example, I have read sections one and two of Einstein's 1905 paper that end with:

A. Einstein said:

So we see that we cannot attach any *absolute* signification to the concept of simultaneity, but that two events which, viewed from a system of co-ordinates, are simultaneous, can no longer be looked upon as simultaneous events when envisaged from a system which is in motion relatively to that system.

http://www.fourmilab.ch/etexts/einstein/specrel/www/

They will understand by your fear you have no idea what you are doing.

I really don't care who you think "they" are, but this purported thing you hope they will understand and my fear seem to be your baseless delusions.

Here it is again. Try to keep in mind all conclusions and calculations are based on M and C' being co-located.

M and C' are only co-located at one event in all of space and time: event P.

When C' and M are co-located, event Q is valid for the frame Σ. (This assumes LP is correct)

When C' and M are co-located, event R is valid for the frame Σ'. (This assumes LP is correct)

What do you mean "valid for the frame" ? I'll tell you what you mean. You mean that the t-value of event P is equal the the t-value of event Q and that Q is part of line ℓ. So a shorter and more mathematical way to say the same thing is that event Q is the intersection of lines j and ℓ. Or $$Q = j \cap \ell$$. Likewise $$R = k \cap \ell$$. See post #2.

Presumably LP is the speed of light postulate which says both O to Q and O to R are paths taken by something moving at the speed of light in every frame. But that was already true about all events on line ℓ. $$ O \in \ell, \quad Q \in \ell, \quad R \in \ell$$ are true independent of any choice of coordinate frame.

When C' and M are co-located, event LT(Q) is valid for the frame Σ'. (This assumes LT correctly translates)

When C' and M are co-located, event LT(R) is valid for the frame Σ. (This assumes LT correctly translates)

The Lorentz transform cannot change anything physically fundamental. It can't change inertial motion to non-inertial motion. It can't reverse cause and effect. It can't make something moving at the speed of light move at a different speed. So you are incorrect to assert that LT(Q) is a thing. The Lorentz transform changes

*coordinates*, not events. $$Q = j \cap \ell$$ and $$R = k \cap \ell$$ are statements about subsets of space-time, not coordinates. But since $$Q \in j, \quad R \not \in j$$ then of course R can never take the special role in frame Σ of being simultaneous with P that Q has. Likewise, $$Q \not \in k$$ so Q and P are not simultaneous in frame Σ'.

So it is untrue that the Σ' time coordinate of Q is the same as the Σ' time coordinate of P, even though the Σ time coordinate of Q equals the Σ time coordinate of P. That's as simple as noticing line j is different than line k.

However, LT(Q) != R and LT(R) != Q. (This show LP and LT do not agree when M and C' are co-located)

Again, you seem confused. Coordinate systems are just systematic ways of naming events. But the name Q or $$j \cap \ell$$ is just as good a name from the field of geometry. There is no reason to suspect a connection between the coordinates in one frame of R and the coordinates in another frame of Q. For one, in every frame R happens later than Q.

By this point you haven't done any logic. You have no definitions of what you mean, and you are just making empty assertions and confusing yourself.

You don't even know what universal property connects lines f and j ( or h and k).

Take any two events on line f, say P and O. Take any two events on line f, say Q and P. Then in every case, in every inertial coordinate system: $$c^2 \Delta t_f \Delta t_j - \Delta x_f \Delta x_j - \Delta y_f \Delta y_j - \Delta z_f \Delta z_j = 0$$

It works in frame Σ:

$$c^2 \Delta t_f \Delta t_j - \Delta x_f \Delta x_j - \Delta y_f \Delta y_j - \Delta z_f \Delta z_j =c^2 \left( \frac{d'}{\gamma c} - 0\right) \left( \frac{d'}{\gamma c} - \frac{d'}{\gamma c} \right) - \left( 0 - 0 \right) \left( \frac{d'}{\gamma} - 0 \right) - \left( 0 - 0 \right) \left( 0 - 0 \right) - \left( 0 - 0 \right) \left( 0 - 0 \right) = c^2 \times \frac{d'}{\gamma c} \times 0 - 0 \times \frac{d'}{\gamma} - 0 \times 0 - 0 \times 0 = 0$$

It works in frame Σ':

$$c^2 \Delta t'_f \Delta t'_j - \Delta x'_f \Delta x'_j - \Delta y'_f \Delta y'_j - \Delta z'_f \Delta z'_j =c^2 \left( \frac{d'}{c} - 0\right) \left( \frac{d}{c} \left( 1 - \frac{v}{c} \right) - \frac{d'}{c} \right) - \left( - \frac{v d'}{c} - 0 \right) \left( d' \left( 1 - \frac{v}{c} \right) - \left( - \frac{v d'}{c} \right) \right) - \left( 0 - 0 \right) \left( 0 - 0 \right) - \left( 0 - 0 \right) \left( 0 - 0 \right) = c^2 \times \frac{d'}{c} \times \left( - \frac{v d}{c^2} \right) - \left( -\frac{v d'}{c} \right) \times d' - 0 \times 0 - 0 \times 0 = 0$$

That's because this type of inner product is Lorentz invariant and represents a feature of the geometry of space-time, not merely a statement about coordinates.