B Relativity of Measures: A vs B Frame & Light Source S

[jbriggs444]

When I make reference to an inertial frame by its label alone (E, S, A, B etc), I’m refering to its spatial origin x0, y0, z0.

Three numbers is enough to identify the origin of a three dimensional coordinate system. It is not enough to identify the origin of a four dimensional coordinate system.

The position of the origin is not enough to specify a coordinate system.

What about the state of motion of the system? What about the angles of its reference axes?

Note that being able to specify $x_0, y_0$ and $z_0$ assumes that you already have another coordinate system already established. If you want a new coordinate system what you want is a transformation rule. Possibly expressed with the use of a coordinate transformation matrix.

 

[PeroK]

My point was unlike balls and cars, light does not move with speed c+v, or c-v its speed is always c.

I'm sure it's already been pointed out that this is wrong. Also, critically, we are talking about two dimensional motion. In which case we have the transformation rule for velocity components.

Note that the same rule of velocity transformation applies to a ball as to light. The difference is that the speed of light is an invariant of the transformation.

But as a scientist you will understand I cannot believe what I don’t understand.

My point is simply that we all go down a rabbit-hole now and again. It's important to recognise that, stop digging and get yourself out.

 

[Dale]

When I make reference to an inertial frame by its label alone (E, S, A, B etc), I’m refering to its spatial origin x0, y0, z0.

Please don’t. You will be consistently misunderstood. The label alone should refer to the reference frame itself. If you mean “the spatial origin of $E$” then say “the spatial origin of $E$”. If writing that is too cumbersome then use a new symbol. For example you could define $e^{\mu’}=(t’,0,0,0)$ for the spatial origin of $E$.

5. When the light pulse is emitted from E not S, does this change the aberration observed in the rest frame of S or E from that agreed in #1?

I still don’t understand the use of the word change here. What does “change the aberration” mean?

If a device at rest wrt $S$ emits light along the worldline $r^\mu=(t,0,t,0)$ then the same worldline has coordinates $r^{\mu'}=(t',-t' v, t' \sqrt{1-v^2},0)$ in $E$.

Now “when the light pulse is emitted from $E$ not $S$” is ambiguous. You could mean that a device at rest wrt $E$ emits light along the same worldline as above $r^{\mu'}=(t',-t' v, t' \sqrt{1-v^2},0)$ which is still the same worldline as $r^{\mu}=(t,0,t,0)$. Alternatively you could mean that a device at rest wrt $E$ emits light along a different worldline $\rho^{\mu’}=(t’,0,t’,0)$ which is the same worldline as $\rho^{\mu}=(t,t v, t \sqrt{1-v^2},0)$

So I cannot link my statements to your question about “change the abberation”, but hopefully you can.

 

[Ibix]

I've made an animation of what I think the OP is talking about. The $x$ and $y$ directions are horizontal and vertical as usual. There is a green-painted laser lying parallel to the $y$ axis and at rest in the frame depicted (this is what the OP calls frame $S$) and a blue-painted laser also lying parallel to the $y$ axis but moving left-to-right at 0.5c (this is at rest in the frame the OP calls $E$). As the two lasers pass they emit a short pulse of green or blue light (respectively) , and each says that the pulse is emitted perpendicular to the $x$ axis.

(The above is an animated gif - depending on your browser I think you may need to click on it and/or open the image in its own tab to see the animation.) I've additionally marked concentric circles centered on the emission point in this frame, and vertical lines upwards from the lasers. You can see that the laser pulses always remain directly above their emitting lasers, and that both laser pulses are doing the same speed because they always lie on the same circles. It may be helpful to see all of the frames of the animation above superimposed:

Here you can see that both pulses move in straight lines, but at an angle to one another.

The question, I think, is what does this look like in the rest frame of the blue laser, frame $E$. The answer is that it looks exactly the same, except left-right reversed and with the colours swapped.

 

[PeroK]

I've made an animation of what I think the OP is talking about. The $x$ and $y$ directions are horizontal and vertical as usual. There is a green-painted laser lying parallel to the $y$ axis and at rest in the frame depicted (this is what the OP calls frame $S$) and a blue-painted laser also lying parallel to the $y$ axis but moving left-to-right at 0.5c (this is at rest in the frame the OP calls $E$). As the two lasers pass they emit a short pulse of green or blue light (respectively) , and each says that the pulse is emitted perpendicular to the $x$ axis.
View attachment 322484
(The above is an animated gif - depending on your browser I think you may need to click on it and/or open the image in its own tab to see the animation.) I've additionally marked concentric circles centered on the emission point in this frame, and vertical lines upwards from the lasers. You can see that the laser pulses always remain directly above their emitting lasers, and that both laser pulses are doing the same speed because they always lie on the same circles. It may be helpful to see all of the frames of the animation above superimposed:
View attachment 322485
Here you can see that both pulses move in straight lines, but at an angle to one another.

The question, I think, is what does this look like in the rest frame of the blue laser, frame $E$. The answer is that it looks exactly the same, except left-right reversed and with the colours swapped.

I believe that @AlMetis thinks your animation contradicts the second postulate.

 

[Ibix]

I think we all see that, but @AlMetis thinks your animation contradicts the second postulate.

I'm completely lost about what @AlMetis thinks because he (I guess) seems to agree with everything that we say but still thinks something's wrong somewhere. I'm curious to see if I get any commentary on the animation.

 

[jbriggs444]

I'm completely lost about what @AlMetis thinks because he (I guess) seems to agree with everything that we say but still thinks something's wrong somewhere. I'm curious to see if I get any commentary on the animation.

Not sure, but perhaps the disconnect is with the idea of a launcher which is unambiguously vertical and a resulting trajectory which is at an angle that depends on the state of motion of the launching mechanism.

One can explore different techniques for collimation to see how these translate to a trajectory. The fun one is the direction of the wave form being emitted by the flat top surface of a laser. The easy one is a light rifle.

 

[AlMetis]

I'm completely lost about what @AlMetis thinks because he (I guess) seems to agree with everything that we say but still thinks something's wrong somewhere. I'm curious to see if I get any commentary on the animation.

First I want to thank you Dale for your patients and help. My language, and lack of knowledge has made it far more difficult for you than it should be, so for that I am very grateful.
I don’t want to derail our conversation, and I will come back to if I still don’t get it, but I think Ibix’s animation will help clear up my problem quicker than my ability to explain it.@Ibix, thank you. I think your animation should solve my problem. (I was in the middle of making an animation myself.)
You have set the viewer at rest with the green source. The emission point is illustrated by its constant position relative to the concentric circles and its constant position relative to the viewer, i.e. the rest frame of the green emitter.
As you mentioned, this “event” observed at rest with the blue frame, is identical except mirrored across y with colors reversed.

How does the center of the concentric circles remain at rest with both green and blue when the pulse is only emitted by green?
I understand (at least I think I do) the reciprocal nature of all the observations of this “one” event resulting from the relativity of motion between the two observers. I don’t understand how the center of the concentric rings can follow/remain at rest with, both frames when it represents a single event.

 

[Ibix]

How does the center of the concentric circles remain at rest with both green and blue when the pulse is only emitted by green?

The concentric circles are drawn at rest with respect to green. Blue would see them moving and length contracted into ellipses. A similar set of circles could be drawn for the blue emitter; they would remain centered on the blue laser and would appear elliptical in the animation.

I don’t understand how the center of the concentric rings can follow/remain at rest with, both frames when it represents a single event.

The center of the circles is not a single event. It is a place in space, extended in time. The two frames agree that the pulses were emitted when the two lasers passed one another, but their ideas of "where the emission was in space" differ after that.

 

[AlMetis]

The concentric circles are drawn at rest with respect to green. Blue would see them moving and length contracted into ellipses. A similar set of circles could be drawn for the blue emitter; they would remain centered on the blue laser and would appear elliptical in the animation.

If Blue sees one ring (for simplicity) moving away, its width (on axis of motion) contracted, this is because the light was emitted by Green.
Green sees the same ring, same event of emission, remain at rest (increasing diameter at c) and it remains circular.
If Blue had emitted the light, Green would see the (Blue) ellipse moving away, and Blue would see it remain at rest and remain circular.

Green and Blue disagree after emission on were in space the emission took place, i.e. where their positions coincided, because that place is relative to their frame of reference and their motion means they are in different frames of reference.
Have I got that correct?

 

[PeroK]

If Blue sees one ring (for simplicity) moving away, its width (on axis of motion) contracted, this is because the light was emitted by Green.
Green sees the same ring, same event of emission, remain at rest (increasing diameter at c) and it remains circular.
If Blue had emitted the light, Green would see the (Blue) ellipse moving away, and Blue would see it remain at rest and remain circular.

The speed of light is independent of the motion of the source. If we are now talking about a spherical wave of light emitted in all directions, then both light spheres are spheres in both frame.

Green and Blue disagree after emission on were in space the emission took place, i.e. where their positions coincided, because that place is relative to their frame of reference and their motion means they are in different frames of reference.

I don't understand what you mean.

Have I got that correct?

Sadly not.

 

[Ibix]

If Blue sees one ring (for simplicity) moving away, its width (on axis of motion) contracted, this is because the light was emitted by Green.

No, the rings are (or could be) material objects tied to the lasers. If blue sees a ring moving away, it's because green fitted one around their laser. Blue does not see the laser pulses reaching green's ring at the same time, but does see them reaching their own ring at the same time. Green does not see the pulses reaching blue's rings at the same time but does see them reaching their own ring at the same time.

If we have an all-directions light flash instead of a laser emission, both frames see the flash as circular. Both say their own (material) ring is illuminated all at once. Both see the other's ring moving, and thus off-center except at the moment of emission, length contracted, and will say the light did not reach all parts of it at the same time.

 

[robphy]

No, the rings are (or could be) material objects tied to the lasers. If blue sees a ring moving away, it's because green fitted one around their laser. Blue does not see the laser pulses reaching green's ring at the same time, but does see them reaching their own ring at the same time. Green does not see the pulses reaching blue's rings at the same time but does see them reaching their own ring at the same time.

If we have an all-directions light flash instead of a laser emission, both frames see the flash as circular. Both say their own (material) ring is illuminated all at once. Both see the other's ring moving, and thus off-center except at the moment of emission, length contracted, and will say the light did not reach all parts of it at the same time.

Elaborating on this explanation by @Ibix ...

The rings could be a circular arrangement of mirrors facing the source.
Each observer has its own set. It's a "circular light clock".

At the event O when the worldlines meet,
a flash of light begins to trace out the future light cone of event O.
It doesn't matter which worldline (if any) is "the source" of the light.
It is the flash from event O.From my Circular Light Clock visualization: https://www.geogebra.org/m/pr63mk3j ,
(T_MOV is tunable).

The future light cone of O (cyan) is intersected
by the worldtube of the arrangement of mirrors of each inertial observer.

As @Ibix says, each sees the other arrangement as length-contracted...

• Blue observes the Blue-mirror-intersection-events are simultaneous in Blue-time and circular in Blue-space,
  but observes the Green-mirror worldtube as elliptical in Blue-space (contracted in the direction of relative-motion)
  and observes the Green-mirror-intersections-events are not-simultaneousin Blue-time and are not-circular in Blue-space.
• In fact, Blue observes the Green-mirror-intersections-events trace out points over Blue-time an ellipse in Blue-space where the meeting event is a focus of that ellipse.
• The other focus of that ellipse is the Blue-space position of where those Green-reflected signals refocus at a common event.
  (This is due to the reflective properties of an ellipse.)
• The elapsed Blue-time of the Green-refocused signal is longer than
  the elapsed Blue-time of the Blue-refocused signal. The ratio is time-dilation factor.
• The spacetime-volume enclosed by these "causal diamonds" is invariant, proportional to the square-interval between the emission and refocusing events.

Of course, the situation is symmetric.
Green observes the same things about Blue.... in accord with the Relativity Principle.Here's the overhead profile (the xy-plane):

Here's the side profile (the tx-plane),
where one can see the (1+1)-dim causal diamonds
used in my "rotated graph paper" method.

 

[Ibix]

I think it's the last graph above that encapsulates everything.

The same light cone (pale blue) interacts very differently with a ring that is stationary in this frame and a ring that is moving in this frame. The latter is length contracted, which can be seen from the narrow horizontal cross section of the brown region, but its interaction with the light is spread out in space and time (green diagonal). If you include reflections from the rings (pale red) they return to their respective centers and do so at the same proper time, which can be seen from the black dotted hyperbola.

 

[AlMetis]

The rings could be a circular arrangement of mirrors facing the source.
Each observer has its own set. It's a "circular light clock".

Thank you Rob, this is an excellent visual technique and very helpful in communicating what I have failed to do for days.
I have taken the liberty of using your design to demonstrate the mechanics I don’t understand.
The origin of the Green inertial frame is represented by the green dot.
The origin of the Blue inertial frame is represented by the blue dot.
Rob’s mirror rings are fixed around the origin of each frame, hence the reflected (top) light cones.
The light green cone that continues beyond the reflection shows the future without the mirror rings.

Diag. 1 is similar to Rob’s with the colours changed to match our Green and Blue examples.
It is assumed in these diagrams that the viewer takes the symmetry of the cone across the time axis and across the reflection axis (ring of mirrors) to indicate you, the viewer are at rest with the world line of the event 0. When/if the light cone of an inertial frame holds this same symmetry across both axis, it is at rest with the world line of the event 0.
As Rob said:
”At the event O when the world-lines meet, a flash of light begins to trace out the future light cone of event 0. It doesn't matter which world-line (if any) is "the source" of the light.
It is the flash from event O.” (my emphasis)

In diagram 2.0 Green is at rest with the world line of the event.
In diagram 3.0 Green is moving to the right of the world line of the event.
In diagram 4.0 Green and Blue are moving to the right of the world line event, with Blue moving faster than Green.
If you look at the difference between diagram 1.0 and 4.0, the symmetry of observations (the light cones) predicted after the event based on the symmetry of motion before the event, is determined by whether the motion of each frame is known/measured relative to the world line of the event, or relative to the motion each frame.Diagram 5.0 shows the predicted observations based soley on the symmetry of motion between Red, Green and Blue. But we can see from diagram 4.0 the same motion (one side only shown in 4.0) is not predicted if we know the motion of Green relative to the world line of event 0.

 

[robphy]

Thank you Rob, this is an excellent visual technique and very helpful in communicating what I have failed to do for days.
I have taken the liberty of using your design to demonstrate the mechanics I don’t understand.
[snip]

I haven't been following the conservation.
What is the specific question? (I couldn't find it in the previous post.)

 

[Ibix]

the world line of the event,

This is nonsense. Events don't have worldlines. Thinking that they do may be part of your problem.

 

[PeroK]

Thank you Rob, this is an excellent visual technique and very helpful in communicating what I have failed to do for days.

I have a question for you.

We have an electron gun that fires electrons at $0.8c$. Now, we have a reference frame where the gun is moving at $0.5c$ in the direction the gun is pointing. In this reference frame, the speed of the electrons is $0.5c + 0.8c = 1.3c$?

Why not?

 

[Ibix]

But we can see from diagram 4.0 the same motion (one side only shown in 4.0) is not predicted if we know the motion of Green relative to the world line of event 0.

Reading through this more closely, I really think your problem is "the worldline of the event". You seem to be using it to mean "the time axis of the arbitrary frame you chose to draw the diagram in". Of course you can't draw the diagram without choosing a frame, but that makes no difference to the physics and "worldline of the event" is meaningless.

The other thing you seem to do is draw different rings of mirrors in different states of motion and then observe that the results are different. That shouldn't really be surprising.

 

[AlMetis]

This is nonsense. Events don't have worldlines. Thinking that they do may be part of your problem.

I’m sorry, I am talking about the light cone of the flash form the event. Similar to world lines in Minkowski space, rest is a chosen frame, I as did Rob, chose the light cone of the event.

 

[AlMetis]

Why not?

Because you just said the electrons are fired at 0.8c.
If I can measure where the electron left the gun and time of its travel over a fixed distance, I will find it to be 0.8c.

 

[AlMetis]

I haven't been following the conservation.
What is the specific question? (I couldn't find it in the previous post.)

 

[PeterDonis]

@AlMetis in your post #122, there is no asymmetry. The motion of Red and Blue is equal and opposite relative to Green in every frame. You have just chosen to draw your diagram 2 in a frame other than Green's rest frame, which makes things look "asymmetrical" even though they are not. The physical fact is that the motion is symmetrical relative to Green, and so the best frame to view the motion is Green's rest frame, as in diagram 1. But that doesn't mean the symmetry goes away in some other frame; it just means you've picked a bad frame to view it in.

 

[PeroK]

Because you just said the electrons are fired at 0.8c.
If I can measure where the electron left the gun and time of its travel over a fixed distance, I will find it to be 0.8c.

I was hoping for a serious answer.

 

[Ibix]

I’m sorry, I am talking about the light cone of the flash form the event. Similar to world lines in Minkowski space, rest is a chosen frame, I as did Rob, chose the light cone of the event.

This still doesn't make sense. The light cone doesn't form a standard of rest. It's always the same cone in all frames. And it isn't a worldline either.

I still don't understand what your problem is. You seem to be able to draw accurate diagrams, you just don't seem to believe them, or you see some contradiction between them.

 

[Dale]

View attachment 322741View attachment 322742

What is the SPECIFIC symmetry and the asymmetry you are asking about here?

I have asked a long time ago for you to identify the specific symmetry, and you still have not done so. You seemed to be making progress in that direction with your points 1-5. But after 5 you jumped to these diagrams. That is fine, but I still don’t know what is the symmetry in question:

What symmetry? You must be misunderstanding something because all of the predicted symmetries of SR have been experimentally confirmed. So either you think some symmetry is predicted and it isn’t, or you are not actually looking at the symmetry that you think you are.

From my perspective we have not made it past my post 72, now more than 50 posts ago.

(For the future, please don’t post images of text. The figures are great, but the images of text are unquotable)

By the way, in your second figure the velocities are wrong. You need to use relativistic velocity addition.

 

[AlMetis]

But that doesn't mean the symmetry goes away in some other frame; it just means you've picked a bad frame to view it in.

Thank you PeterDonis

I think you have just made it clear to me why everyone here keeps asking me the same question. I thought we were on the same page and looking for where I’ve misunderstood SR. But it now appears we are reading different books.
All of which makes me that much more grateful for everyone’s help and patients.

I agree, the relative motion of Red, Green and Blue remains constant, that is crucial to my question.
If I am free to choose any rest frame (inertial frame) from which to define the laws, and I choose the rest frame of Green which sets the motion of Red and Blue identical relative to Green, how do I explain the change in the time of light between Red and Green, and Blue and Green which is clearly shown in the light cone diagrams?

 

[Dale]

how do I explain the change in the time of light between Red and Green, and Blue and Green which is clearly shown in the light cone diagrams?

It is not clearly shown to me. What do you mean by "the change in the time of light"?

 

[AlMetis]

This still doesn't make sense. The light cone doesn't form a standard of rest. It's always the same cone in all frames. And it isn't a worldline either.

As I understand it, a light cone is a causal (light speed) limit representing the spatial separation on the horizontal axis x and time on the vertical axis y. If I diagram the spatial separation as constant on x, over increasing y, I have set the viewer at rest with the event with respect to the causal horizon of the event.

 

[Ibix]

how do I explain the change in the time of light between Red and Green, and Blue and Green which is clearly shown in the light cone diagrams?

You mean, in the rest frame of green, why does it take longer for the reflections to reach red and blue? Because red and blue are moving and it takes more time for the light to reach them because it has further to go.

In the rest frame of red, it takes longer for the light to reach green and even longer to reach blue, and for the same reason.

 

[Ibix]

As I understand it, a light cone is a causal (light speed) limit representing the spatial separation on the horizontal axis x and time on the vertical axis y. If I diagram the spatial separation as constant on x, over increasing y, I have set the viewer at rest with the event with respect to the causal horizon of the event.

None of that makes sense. A light cone is a causal boundary - events in its interior or on it can be affected by the event at its tip, events outside cannot. You can't be at rest with respect to one.

 

[PeroK]

By the way, in your second figure the velocities are wrong. You need to use relativistic velocity addition.

Look at the above posts:

I have a question for you.

We have an electron gun that fires electrons at $0.8c$. Now, we have a reference frame where the gun is moving at $0.5c$ in the direction the gun is pointing. In this reference frame, the speed of the electrons is $0.5c + 0.8c = 1.3c$?

Why not?

Because you just said the electrons are fired at 0.8c.
If I can measure where the electron left the gun and time of its travel over a fixed distance, I will find it to be 0.8c.

This suggests to me that the OP needs to learn more about the basics of SR. We are now 130+ posts down a rabbit hole, with little or no prospect of ever emerging into daylight.

 

[AlMetis]

It is not clearly shown to me. What do you mean by "the change in the time of light"?

The diagrams show the time of light from Green to Red is longer (higher on the time axis) than the time of light from Green to Blue.

 

[AlMetis]

You mean, in the rest frame of green, why does it take longer for the reflections to reach red and blue? Because red and blue are moving and it takes more time for the light to reach them because it has further to go.

In the rest frame of Green it takes longer for the unreflected light from Green to reach Red than it does to reach Blue.

 

[AlMetis]

You can't be at rest with respect to one.

How should I refer to a frame that remains at a constant distance from the boundary?

 

[DrGreg]

How should I refer to a frame that remains at a constant distance from the boundary?

First of all, a frame is everywhere. I presume you mean "the spatial origin of a frame".

The boundary is expanding at the speed of light, so no origin of a frame can be a constant distance from something moving at the speed of light.

 

[Dale]

The diagrams show the time of light from Green to Red is longer (higher on the time axis) than the time of light from Green to Blue.

OK, so on your diagram I have labeled four events $E_1=(t(E_1),x(E_1))$, etc.

If I understand you correctly the asymmetry that you are worried about is that $t(E_2)-t(E_1)<t(E_4)-t(E_3)$. Is that a correct understanding of the asymmetry you are looking at?

If not then please be explicit what it is. Preferably in less than 50 posts more.

 

[robphy]

In the last diagram of @AlMetis, the green event should not be collinear with the blue and red events (what @Dale labels as E2 and E4),
if that green event is the same event in the first spacetime-diagram of that series of diagrams where the three events lie on a common hyperbola... which means that those time-intervals from the meeting event are equal in magnitude).

 

[PeterDonis]

a frame that remains at a constant distance from the boundary?

If by "the boundary" you mean the light cone, there is no such frame, because there is no such thing as "constant distance from the light cone".

 

[PeterDonis]

As I understand it, a light cone is a causal (light speed) limit

No, it's the set of curves through a given point that are null, i.e., that are possible worldlines for light rays.

representing the spatial separation on the horizontal axis x and time on the vertical axis y.

That's not a light cone, that's a spacetime diagram.

If I diagram the spatial separation as constant on x

What spatial separation?

I have set the viewer at rest with the event with respect to the causal horizon of the event.

This is word salad. It makes no sense.

 

[AlMetis]

If I understand you correctly the asymmetry that you are worried about is that t(E2)−t(E1)<t(E4)−t(E3). Is that a correct understanding of the asymmetry you are looking at?

Yes.
With one correction, that does not change your question or my answer.
As robphy points out I have the green too far up the timeline, it should be below the blue.

 

[Dale]

Yes.
With one correction, that does not change your question or my answer.
As robphy points out I have the green too far up the timeline, it should be below the blue.

OK, then the answer is

... you think some symmetry is predicted and it isn’t, ...

Relativity does not predict that $t(E_2)-t(E_1)=t(E_4)-t(E_3)$ in all frames

 

[AlMetis]

Relativity does not predict that t(E2)−t(E1)=t(E4)−t(E3) in all frames

I am not saying it does.

I am saying that what relativity does predict will be observed when at rest with Green, t ( E 2 ) − t ( E 1 ) = t ( E 4 ) − t ( E 3 ) in diagram 1. changes when the information not available in diagram 1 is revealed in diagram 2. when t ( E 2 ) − t ( E 1 ) ≠ t ( E 4 ) − t ( E 3 ).

We are still at rest with Green and the symmetry of the motion between Red, Green and Blue still exists, but the time of the light pulse, i.e. the distance travelled observed at rest with Green, is not what is predicted from the information available in diagram 1.

I thought about this last night and realized there is a much more familiar way to pose my question.
I’ll post it shortly.

 

[Dale]

I am not saying it does.

Yes, you did:

But we don’t regain the symmetry.
The symmetry is not exclusive to Galilean relativity, it is also predicted by Einsteinian relativity on the left and lost on the right.

Why did it disappear?

(emphasis added)

I don’t need you to defend your words, but that was my reason to be involved in the thread.

the information not available in diagram 1 is revealed in diagram 2

The amount of information is the same in both diagrams. The Lorentz transform is a lossless transformation in terms of information. The representation is different, but the amount of information is the same. There is no unavailable information that is revealed.

I thought about this last night and realized there is a much more familiar way to pose my question.
I’ll post it shortly.

I think not. This is useless. We are nearly 150 posts in and you are still trying to formulate your question. Nobody coming to this thread later will know to look on page 5 for the question.

I am going to close this thread. Please do not post your new question shortly. There are only a few days left in the month. Take that time to formulate the question well and do not post a new thread until the beginning of next month.

When you do, please label all important events, worldlines, measures, and frames. Do not refer to any frame variant quantity (speed, distance, time) without explicitly mentioning the frame. And use LaTeX to write symbols.