Time Dilation / Second Postulate

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Time Dilation / Second Postulate
« on: December 07, 2021, 03:46:53 AM »
Let M1 and M2 be two mirrors attached to the ceiling and floor of a spaceship respectively deep in the space. O1 is stationed inside the spaceship while O2 is at rest on an asteroid as shown in a tentative diagram.

Distance “d” = Ct for all stationary observers.

We get a time dilation triangle ABC as shown when a spaceship moves with velocity “V” relative to O2. The equation of timed dilation may be derived from the right-angled triangle of ABC as shown.

Relative to O2: 

Vt is the distance covered by the M1 or M2 in un-dilated time t in space.

Ct is the distance covered by a pulse in un-dilated time t in space

Base = Vt and Hypotenous = Ct, where t is un-dilated time

Relative O1

Perpendicular BC = Ct’ is the distance covered by a pulse in dilated time t’


Since the speed of light is the same for all observers irrespective of their motion, therefore, both space and time are not absolute in order to make “C” constant for all observers.

CRUX #1

Although O1 feels himself at rest yet space in the direction in which he moves is contracted relative to other axes. This means,


1-    Dimension of the “time and space” change in the direction in which O1 moves say x-axis ONLY. However

2-    Dimension of the “time and space” don’t change in the direction of the y-axis and z-axis.

So in order to make the speed of light constant for O1.

1-    Dilated time has to be used in the direction in which O1 moves (x-axis) as space is contracted for O1

2-    Un-dilated time in the rest of the axes (y-axis and z-axis) as space doesn’t contract for O1 in the foregoing axes.

Since a pulse moves back and forth in a y-axis, therefore, the perpendicular of the right-angled triangle used for the derivation of the equation of time dilation will be = Ct, not Ct’ where t is un-dilated time while t’ is dilated time.

CRUX #2:

Although, O2 is unable to see the contraction of space for O2 yet O2 still sees O1 at a point B despite knowing that the dimension of time and space are changed for O1. This is so sad because the actual position of O2 would be somewhere in between A and B say @ P due to Vt’ (the ACTUAL distance covered by M1 or M2 in dilated time t’).

M1, M2, and O2 would disappear for O2 with the passage of time as they are lagging behind for O2 in both un-contracted space un-dilated time “t”.

« Last Edit: December 07, 2021, 09:38:25 AM by E E K »

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Calen

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Re: Time Dilation / Second Postulate
« Reply #1 on: December 07, 2021, 05:07:56 AM »
Quote
1-    Dimension of the “time and space” change in the direction in which O1 moves say x-axis ONLY. However

2-    Dimension of the “time and space” don’t change in the direction of the y-axis and z-axis.

All movement in three dimensional space is three dimensional, so although the movement is along the x-axis, there is also a perpendicular movement with respect the y- and z-axes, and an associated relative change in spacetime contraction/dilation along those axes.
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Re: Time Dilation / Second Postulate
« Reply #2 on: December 07, 2021, 06:00:53 AM »
Quote
1-    Dimension of the “time and space” change in the direction in which O1 moves say x-axis ONLY. However

2-    Dimension of the “time and space” don’t change in the direction of the y-axis and z-axis.

All movement in three dimensional space is three dimensional, so although the movement is along the x-axis, there is also a perpendicular movement with respect the y- and z-axes, and an associated relative change in spacetime contraction/dilation along those axes.
Distance “d” = Ct for all stationary observers has been clarified at the start of the question.

According to you

Relative to O1 who feels himself at rest
The Distance “d”  will be then = Ct' on the y-axis but this must Ct as explained above.

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Calen

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Re: Time Dilation / Second Postulate
« Reply #3 on: December 07, 2021, 06:29:58 AM »
Quote
1-    Dimension of the “time and space” change in the direction in which O1 moves say x-axis ONLY. However

2-    Dimension of the “time and space” don’t change in the direction of the y-axis and z-axis.

All movement in three dimensional space is three dimensional, so although the movement is along the x-axis, there is also a perpendicular movement with respect the y- and z-axes, and an associated relative change in spacetime contraction/dilation along those axes.
Distance “d” = Ct for all stationary observers has been clarified at the start of the question.

According to you

Relative to O1 who feels himself at rest
The Distance “d”  will be then = Ct' on the y-axis but this must Ct as explained above.

There is no question in your post, just an assertion that something is 'sad'.

I can't say that I fully understand your post, as well, it's a mess, and confuses which observer sees time dilation.

« Last Edit: December 07, 2021, 06:47:18 AM by Calen »
We're all born naked, and the rest is drag.

Re: Time Dilation / Second Postulate
« Reply #4 on: December 07, 2021, 09:12:37 AM »
Quote
1-    Dimension of the “time and space” change in the direction in which O1 moves say x-axis ONLY. However

2-    Dimension of the “time and space” don’t change in the direction of the y-axis and z-axis.

All movement in three dimensional space is three dimensional, so although the movement is along the x-axis, there is also a perpendicular movement with respect the y- and z-axes, and an associated relative change in spacetime contraction/dilation along those axes.
Distance “d” = Ct for all stationary observers has been clarified at the start of the question.

According to you

Relative to O1 who feels himself at rest
The Distance “d”  will be then = Ct' on the y-axis but this must Ct as explained above.

There is no question in your post, just an assertion that something is 'sad'.

I can't say that I fully understand your post, as well, it's a mess, and confuses which observer sees time dilation.
I don’t think we need the equation.

For all stationary observer distance d=Ct, where t in un-dilated time. [s=3x10^8m, t=1 sec]

Let measure the speed of light in the x-axis direction in which O1 moves and the y-axis as well. 

Case #1: Measuring C along the x-axis

The relativistic effect happens only in the x-axis as both time and space change for O1. Although the dimensions of space and time are changed, the speed of light is still constant for O1.  For example, say O1 moves @ 0.8c relative to O2. If the contracted distance is 1.6 x10^8 (equivalent to 3x10^8 m for stationary observer O1 on an asteroid).  then its respective time would be 0.6 sec (equivalent to 1 sec for the stationary observer). As C=d/t therefore = (1.6x^10^8)/0.6 = 3x10^8 m/s.

Relativistic Doppler effect
Similarly, C= frequency x wavelength = 3x10^8 m/s=constant
The frequency of the wavelength of the light would be increased due to the decrease in the wavelength of light if light passes through the contracted space. Depending upon the high speed of the spaceship (close to C) relative to O2, O1 sees shifting in light towards blue.

Case #2: Measuring C along the y-axis
No change in space and time in this direction therefore here s=3x10^8m, t=1 sec just like for all stationary observers

All above shows, the perpendicular of the right-angled triangle of the time dilation diagram will be = Ct, not Ct'. I hope this may help. I may repost this question after rephrasing.
« Last Edit: December 07, 2021, 10:30:03 AM by E E K »

Re: Time Dilation / Second Postulate
« Reply #5 on: December 07, 2021, 10:47:44 AM »
Quote
1-    Dimension of the “time and space” change in the direction in which O1 moves say x-axis ONLY. However

2-    Dimension of the “time and space” don’t change in the direction of the y-axis and z-axis.

All movement in three dimensional space is three dimensional, so although the movement is along the x-axis, there is also a perpendicular movement with respect the y- and z-axes, and an associated relative change in spacetime contraction/dilation along those axes.
Distance “d” = Ct for all stationary observers has been clarified at the start of the question.

According to you

Relative to O1 who feels himself at rest
The Distance “d”  will be then = Ct' on the y-axis but this must Ct as explained above.

There is no question in your post, just an assertion that something is 'sad'.

I can't say that I fully understand your post, as well, it's a mess, and confuses which observer sees time dilation.

For simplicity, let you are moving at 0.8c relative to anybody who is at rest.
Both space and time only change in the direction in which you move say x-axis relative to anybody (at rest) but there is no change in the dimensions of time and space along the y-axis and the z-axis in which you don't move in your frame. So the perpendicular (on the y-axis) of the right-angled triangle of the time dilation diagram will be Ct, not Ct' where t is undilated time and t' is the dilated time. If this is true then the whole idea of Einstein's relativity is wrong. Right -  please refer to the time dilation triangle TY
« Last Edit: December 07, 2021, 10:53:37 AM by E E K »

Re: Time Dilation / Second Postulate
« Reply #6 on: December 07, 2021, 02:22:30 PM »
1-    Dimension of the “time and space” change in the direction in which O1 moves say x-axis ONLY. However

2-    Dimension of the “time and space” don’t change in the direction of the y-axis and z-axis.
Time and space are treated separately, as they are different dimensions of spacetime.
Time is not in any direction of space. If you want to treat it as a direction, it is a direction all on its own.

There is no "time along the x axis", there is just time.

If there is relative velocity, time does change. It is not separated into different directions. It is simply changes based upon motion of the observer.
The only one that doesn't "change" are directions of space perpendicular to the direction of motion.

This is why the formula for time dilation just uses v, while the formula for length contraction is applied separately for each direction with the velocity for each direction, often done by setting the motion to be directly along the x axis and setting vy and vz to be 0, and thus no length contraction for the y and z directions.

That means the length of that leg of the triangle will be ct', not ct.

So your triangle is constructed from 2 lines at right angles with lengths ct' and vt, and the hypotenuse of length ct.

Although, O2 is unable to see the contraction of space for O2 yet O2 still sees O1 at a point B despite knowing that the dimension of time and space are changed for O1. This is so sad because the actual position of O2 would be somewhere in between A and B say @ P due to Vt’ (the ACTUAL distance covered by M1 or M2 in dilated time t’).
Why wouldn't O2 be able to see the contraction of space for O1? Because they didn't have a reference for it?

But no, the position would be B, not P. Why would you use t' for an observation which is meant to be in time t?
There is no actual distance covered by M1 in dilated time.
In the reference frame using t', there is no motion of M1.

Stop just randomly switching between reference frames and using the wrong measurement for a reference frame.

Re: Time Dilation / Second Postulate
« Reply #7 on: December 08, 2021, 05:20:32 AM »
Sorry, it was my bad due to multiple interruptions.
 
Please refer to the Wikipedia article on “Time Dilation” for time dilation triangle
 
From the frame of reference of a moving observer traveling at the speed v relative to the rest frame of the clock (diagram at right), the light pulse traces out a longer, angled path - apothem.
 
Thus the base of the time dilation triangle is AA=s=vt’ where t’ is dilated time.
 
The velocity of the moving frame relative to the stationary frame is v in time t where t is not dilated.
 
Assume both mirrors of the light clock are permanently attached to the ceiling and floor respectively inside of the moving frame. This means mirrors also move at v in time t (where t is not dilated) relative to the stationary observer.
 
The spatial distance covered by the moving frame at any time t is s=vt (where t is not dilated) relative to the stationary observer. This means
 
The spatial distance covered by mirrors of the moving frame is also s=vt (where t is not dilated) relative to the stationary observer.
 
Since the aforementioned distances (s=vt’ and s=vt) are not equal therefore what would be the real position of mirrors or any stationary object inside the moving frame relative to the stationary observer?
 
I’ll come to the perpendicular of the triangle after.

In reference to the following diagram,

Event #1 is simultaneous for both O1 and O2 while Event #2 happens first for O1 and then O2 due to a change in dimension of spacetime for O1.
 
O1 moves at 0.8c relative to O2. There is a conflict in the timing of taking place of event #2 but not on Event #1.
 
How does O2 or O1 claim that Event #1 takes place at the same time for both O1 and O2 when spacetime changes  (even before event #1) for O1 while moving in front of O2 according to Einstein relativity?

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« Last Edit: December 08, 2021, 09:09:39 AM by E E K »

Re: Time Dilation / Second Postulate
« Reply #8 on: December 08, 2021, 12:49:48 PM »
Please refer to the Wikipedia article on “Time Dilation” for time dilation triangle
Other than switching around t and t' compared to you, it explains it quite well.
There is no problem.


Thus the base of the time dilation triangle is AA=s=vt’ where t’ is dilated time.
The velocity of the moving frame relative to the stationary frame is v in time t where t is not dilated.
You are mixing up frames of reference.
Wikipedia has t in the frame with the clock. It has t' in the frame outside in which the clock is observed to be moving.
So for wikipedia, AA=s=vt'.

But you switch the convention and have t for the frame the clock is moving in and t' for the frame the clock is stationary in.
This means AB=vt

It doesn't matter which convention you use, as long as you understand and stick to it.

Since the aforementioned distances (s=vt’ and s=vt) are not equal therefore what would be the real position of mirrors or any stationary object inside the moving frame relative to the stationary observer?
Again, this is because you are switching back and forth between different conventions.
In reality, that t' is that t, as one uses one convention and the other uses another.

That means it is actually the same.

Event #1 is simultaneous for both O1 and O2 while Event #2 happens first for O1 and then O2 due to a change in dimension of spacetime for O1.
This makes no sense.
You cannot discuss an event being simultaneous for 2 observers in their own reference frame.
Time is relative. You can set an arbitrary 0 point and then discuss how long after something occurs.
But if you want to discuss if 2 events are simultaneous you need to be using the same reference frame and discussing 2 different events for an observer in that reference frame.

Re: Time Dilation / Second Postulate
« Reply #9 on: December 09, 2021, 02:33:22 AM »
Quote
It doesn't matter which convention you use, as long as you understand and stick to it.
True, but what I’m trying to say the velocity "v" of the moving clock is always measured by the clock which is at rest.

O and O’ are two observers in two inertial frames F and F’ respectively. F=F’. A vertical light clock is installed in each frame. Initially, F and F’ are superimposed.  At t=t’=0, C = 2L/t.

At t=t’=0, we get the perpendicular of the time dilation triangle.

Let F’ move with v=s/t in the x-axis with respect to O in F which is at rest. Since movement is relative, O’ inside F’ feels that F’ is at rest while F moves with v=s/t in negative x-axis with respect F’.

For O’, spacetime changes in a direction in which O' moves w.r.t O

O’ measures C=2l/t’ in the x-axis (in ref to O)- Here space is contracted for O' but he doesn't feel.

For O’, spacetime doesn’t change in a direction in which he doesn’t move

O’ measures C=2L/t in the y-axis (vertical clock) and z-axis

So, O’ in moving F’ finds C=2L/t with the help of a vertcal light clock as spacetime doesn’t change for him in the y-axis or for the clock which is at rest.

O’ also measures the

Speed of F in time t as v=s/t, So base of the triangle = vt
Speed of light pulse in its longer path (Hypotenuse of the triangle)in time t = C =2L/t.

So if all the 3 sides of the triangle are measured in time t then what is the role of t’.


Similarly, if we don’t know the starting point of event #1 then how do we calculate Lorenz Factor?

For finding Lorenz Factor, a photon is fired from the origins of F and F’ in the x-axis direction at t=t’=0 or at the time of separation of F and F’.

Doesn’t acceleration involve before F’ attains v with respect to F and vice versa?

Re: Time Dilation / Second Postulate
« Reply #10 on: December 09, 2021, 03:22:15 AM »
Quote
It doesn't matter which convention you use, as long as you understand and stick to it.
True, but what I’m trying to say the velocity "v" of the moving clock is always measured by the clock which is at rest.
And the distance it moves is based upon that velocity and how long it is observed, in the same frame.
But what you are doing is switching back and forth between measuring time in the frame of the clock and measuring it in the frame of the stationary observer.
If you stop doing that, you don't have that problem.

For O’, spacetime changes in a direction in which O' moves w.r.t O
For O’, spacetime doesn’t change in a direction in which he doesn’t move
Again, there is no separate "spacetime" in different directions.
If you want to separate it into different directions, or more properly called dimensions, you have 3 spatial dimensions and 1 temporal dimension.
There is no "time in the x-axis".

As there is motion, time is relative.
This means it doesn't matter what direction the light goes in, they will be using t', not t.

Stop just ignoring key parts of relativity just because they are inconvenient for your argument as you need to ignore them to pretend there is a problem.

Re: Time Dilation / Second Postulate
« Reply #11 on: December 13, 2021, 02:50:49 AM »
Let you are in the F’ which is moving @ 0.8C  relative to me who am in F (stationary).

I measure the value of “C =  3x/10^8 m/sec“ in all directions. Here distance = 3x/10^8, time = 1 sec (C=distance/time)

I measure “C” for you in the direction in which you move (say x-axis)

As space is contracted for you [distance =1.8x10^8 while time = 0.6 sec] therefore
I measure C = 1.8x10^8 (m)/ 0.6 (sec) = 3x/10^8 m/sec 

Similarly, velocity of light = C = Frequency x Wavelength = 3x/10^8 m/sec - Wavelength decreases but frequency increases due to contraction in space.
Therefore, you also measure C =3x/10^8 m/sec in the x-axis direction.

As space doesn’t contract for you in a direction in which you don’t move, therefore

How do you measure “C” for yourself in the y-axis – vertical direction?
I believe it would be  C =  3x/10^8 m/sec as distance = 3x/10^8 and  time = 1 sec (C=distance/time)

How do I measure “C” for you in the y-axis – vertical direction?

I know it will be C =3x/10^8 m/sec but can you elaborate?

Here is my explanation.

Both F and F’ move forward in time but are always spotted on the surface on the base of the light cone, therefore, I always see a pulse on the surface of the base of a light cone. So, I measure “C” in between the two mirrors =3x10^8 m/s in the REAL space instead of tracing out a longer path of it.

I believe the speed of light (STANDARD "C") was measured on the surface of the base of a light cone.

A pulse (on space/base of the light cone) starts moving forward in time as soon as it fires from point "A" (M1/M2)  No one (including me) would be able to see this point “A” again because it becomes a past event within the light cone. We see the same space but at different time intervals/ time moments are not the same. Therefore

The reason one sees the longer path is that both time and space are counted/ added together right from the starting point say “A” from where a pulse was fired. There is no way one can go thereat "A" (in the past) and start drawing a line AB or BC purpose of measuring distances for triangle ABC.
« Last Edit: December 13, 2021, 03:19:10 AM by E E K »

Re: Time Dilation / Second Postulate
« Reply #12 on: December 13, 2021, 12:26:56 PM »
I measure “C” for you in the direction in which you move (say x-axis)
No, you don't.
You measure c for you, not for me.
You don't measure c for other reference frames. You can calculate it, taking note of length and space contraction and motion, but you can't just measure it.
And when you do calculate it, you find out it is still c.

Re: Time Dilation / Second Postulate
« Reply #13 on: December 17, 2021, 04:06:56 AM »
Yes, you are right. I calculated that not measured but you didn't say anything about the addition of Time and Space that was mentioned as well.

Measurement of “C”:



The vertical line represents Time which passes in a vertical direction
A horizontal line represents a stationary Space but moves vertically in time “t”.

A pulse back and forth in b/t M1 and M2 of AB horizontal light clock of length L w.r.t Oo. Both a clock and Oo are stationary w.r.t each other. 

A pulse that contains in space moves horizontally as well as vertically in time as space moves in time.

The actual path traced by a pulse is AB’A” in both Space and Time – Simply, AB’A” is the resultant of Time and Space.

Time and Space are separated from each other for the purpose of calculating C = distance/time as it’s impossible to measure AB’A”.

AA’ = t1 and A’A”=t2 are the projected time intervals on the Time-Axis through which space/pulse passed while
AB=A’B’=A”B”=L are the projected lengths of a horizontal light clock at to, t1, t2 respectively. Thus

A'B'=L is covered by a pulse in time t1
A"B"=L is covered by a pulse in time t2
As total time needed by a pulse to cover A'B' and A"B" = t1 + t2 = t
Total projected distance covered by a pulse on the surface of the base light cone in t = 2L (represented by orange color)

Therefore, velocity of light C=2L/t

I get the same orange color distance on the surface of the base of the light cone for the clock if moves say v=0.8c w.r.t (in front of) stationary observer Oa upon the application of the aforementioned conclusion.



We can project time t through which the clock traveled on Time Axis while distances (covered by a pulse as well as the clock) on the surface of the light cone.

By doing so all observers see identical “C” irrespective of frames.

Both Oo and Oa always see a pulse traces out its path L on the surface of the base of the cone.
« Last Edit: December 17, 2021, 04:27:04 AM by E E K »

Re: Time Dilation / Second Postulate
« Reply #14 on: December 17, 2021, 01:48:39 PM »
Yes, you are right. I calculated that not measured but you didn't say anything about the addition of Time and Space that was mentioned as well.
Because it was all based upon that faulty premise.
Everything following from a faulty premise is unsubstantiated and thus can be dismissed without further comment if that premise is shown to be wrong.

A horizontal line represents a stationary Space but moves vertically in time “t”.
No, in a space-time diagram for a particular reference frame, a horizontal line represents a space-like line. This is a line of constant time, with various points in space.
For someone in that reference frame, 2 points on a horizontal line occur at the same time.
This can also represent a space-like path, that is a path that only goes through space without any temporal component.

L is a measurement of a length rather than something that physically exists so while it may change as a function of time, it does not move through time.

In general a path is a path through spacetime, which can have a spatial component (which can also be broken down into different directions), and a temporal component.
Bu I would not say that space itself moves through time.

Time and Space are separated from each other for the purpose of calculating C = distance/time as it’s impossible to measure AB’A”.
It is possible to measure such a path, but that would be focusing on the proper time for that object, rather than the time the observer sees it take.
However, formally the y axis has dimensions of ct for scaling purposes, and the speed of anything is based upon its gradient in the diagram.
A path with a gradient of 1 or -1 has a speed of c, and is light like.

But yes, you can just project the paths separately onto the space axis and the time axis to determine the lengths and duration to calculate the speed.

I get the same orange color distance on the surface of the base of the light cone for the clock if moves say v=0.8c w.r.t (in front of) stationary observer Oa upon the application of the aforementioned conclusion.
Your diagram is not drawn correctly, as it has the path going only through A, not B, and is not to scale.
This is what you want:


Or, if you want to take length contraction into account, this:


These are for a light clock with a path length of 1 light second.

But all observers get the speed of the light pulse being the speed of light.

Re: Time Dilation / Second Postulate
« Reply #15 on: December 19, 2021, 11:41:02 PM »
By my reckoning (non-sense to most) my diagram is right. We don’t need transformation as all objects may be seen one another on the surface of the base of the cone therefore I purposely ignore Einstein relativity because the time axis on each point on the world line of either a pulse or a moving clock is parallel to the Time-Axis of any stationary object on their journey through space and time. 

Re: Time Dilation / Second Postulate
« Reply #16 on: December 20, 2021, 03:17:33 AM »
By my reckoning (non-sense to most) my diagram is right.
Considering you show the light as going from A, to A' to A'' rather than going to B at all (it should go to B'), it is wrong.

Re: Time Dilation / Second Postulate
« Reply #17 on: December 24, 2021, 12:32:22 AM »
Please bear with me. I believe I am not wrong. Let

Oa is an observer @ A, BC is a train car or Moving Clock, Ob is an observer @ B, Oc is an observer @ C, and BC moves w.r.t Oa say @ 0.8c

Fig #1 shows the frame of Oa - BC moves w.r.t Oa

Fig #2 shows the frame of Ob and Oc OR train car - Oa moves w.r.t frame of a train car

A pulse can be seen if it moves within the Time-Like/Light-Like region.

Fig #3 shows the frame of Ob and Oc OR train car

A pulse of light is fired @ B. Before reflection @ Point “R”, this pulse is seen by the Ob and Oc, but not by Oa as it moves within the Space-Like region for Oa.

After reflection from Point “R”, the pulse is seen by Oa as it re-enters into the region of Time-Like region for Oa (The 45o angle in Figure shows).  So, here a pulse is seen by all the 3 observers.

Fig #4 shows the frame of Oa

The 45o angle shows, a pulse is seen by all 3 observers as it lies within the Time-Like region for all of them before and after the reflection.
 
From Fig # 3 and 4

Oa sees two pulses, one from his own frame and the other from the moving frame when it is reflected back @ Point “R” for Ob and Oc as it moves within the Time-Like region for all the observers.

Similarly, both Oa and Oc also see two pulses, one from their own frame and one from Oa frame as both the pulses move within the Time-Like region for all the 3 observers.

A pulse is making one and only one path in the spacetime continuum. 

Fig #5

Even if Oa wants to create a coordinated system for moving frame, Oa would have to tilt the time axis of Ob/Oc and align it with their world lines. By doing so, surfaces of the base of light cones of Oa and Ob/Oc would be tilted with each other and Oa wouldn’t be able to see the moving frame at all and vice versa





« Last Edit: December 24, 2021, 03:48:35 AM by E E K »

Re: Time Dilation / Second Postulate
« Reply #18 on: December 24, 2021, 12:50:32 PM »
A pulse of light is fired @ B. Before reflection @ Point “R”, this pulse is seen by the Ob and Oc, but not by Oa as it moves within the Space-Like region for Oa.


After reflection from Point “R”, the pulse is seen by Oa as it re-enters into the region of Time-Like region for Oa (The 45o angle in Figure shows).  So, here a pulse is seen by all the 3 observers.
What you should be focusing on here is the past light cone. A time like path is simply a path that only goes through time, never space.
But in order to see an event it must lie in your past light cone.

If we replace the pulse of light with something which can actually be seen (but we keep it moving at the speed of light), and position them like in your diagram, with an initial separation of 1 light second, A will not see the pulse which starts at B for 1 second.
After detecting it and forecasting backwards in time they will be able to determine when it left.

Even if Oa wants to create a coordinated system for moving frame, Oa would have to tilt the time axis of Ob/Oc and align it with their world lines.
It isn't simply a case of tilting the time axis. The spatial axis also tilts.
Here is an example from wikipedia:


And this means what constitutes a space-like and time-like path are both distorted.
But light-like paths remain undistorted because the speed of light is the same for everyone.

To draw it you don't simply draw a 45 degree line from any arbitrary time axis. Instead you pick a set of axes (1 time and the other space) and draw a line half way between the 2. When your axes are orthogonal, such as in the case of an observer in the reference frame you are drawing, then it will be 45 degrees. In other cases, it wont be.

Additionally, there really is no surface at the base.
The light cones extended infinitely into the past and future.

Re: Time Dilation / Second Postulate
« Reply #19 on: December 26, 2021, 09:38:11 AM »
It can be a continuous beam of light, not just a pulse. Anyway, let’s try the other way

There are two spaceships A and B as shown in the figure. Both moves with the same speed say 0.95c in the same direction relative to stationary observer “Oo”. A concave mirror is attached to the rear end of B. A turns on its headlights toward the back end of B. After some time,

Relative to Oo
Oo sees light beams of small lengths moving just in front of moving A and therefore yet to reach the mirror.

Relative to Oa
Oa sees light beams have already reached the mirror and reflected back as well into the space behind B as shown.

Conclusion:

Oa sees one type of light-beams (reflected and non-reflected) while Oa sees two types of the light beams.

1-    Light beams of small lengths which emit directly from headlights of moving A. The lengths of the light-beams increase slowly with the passage of time just in front of moving A but have not yet reached the mirror
2-   As soon as light beams are reflected for Oa, it acts as if it were discharged for Oo by the rear mirror of B in the opposite direction of motion of A and B - Teleported lights

"How about the following contradiction"

A remotely controlled vertical light clock "A" as shown in the figure moves with very high-speed w.r.t the stationary observer Oo. As soon as a pulse is detected by M2 for Oa, A stops moving just in front of Oo.

[Aforementioned spaceships A and B can also be used as examples if they are remotely controlled]

Relative to Oa
A pulse of light covers a distance of L.  Oa sees Oo also stationary.

Relative to Oo
A pulse has not yet reached the M2 and sees both clocks along with Oa is still moving.


Let Oo has also a light clock in the same scenario.  Light clocks of Oo and Oa are synchronized before Oa moves w.r.t Oo at very high speed.

Relative to Oa:
As soon as a pulse is detected by M2, A comes to rest just in front of Oo.
Oa sees Oo also at rest but a pulse of his light is not yet reached M2. Here

Before stopping
Oa sees a longer path of a pulse of a light clock of Oo but has not yet reached the M2.

After stopping
Oa sees all clock are stationary. Oa sees a vertical path of a pulse (no longer long path) of the light clock of Oo but has not yet reached M2.

Relative to Oo
A pulse of the light clock of Oa has not yet reached M2 and sees a clock along with Oa is still moving.

When Oa stops, Oa finds Oo and himself at rest but Oo who is already at rest doesn't find Oa at rest - why?

« Last Edit: December 26, 2021, 10:12:33 AM by E E K »

Re: Time Dilation / Second Postulate
« Reply #20 on: December 26, 2021, 12:32:57 PM »
After some time
Time is relative. Who is this time for?

1-    Light beams of small lengths which emit directly from headlights of moving A. The lengths of the light-beams increase slowly with the passage of time just in front of moving A but have not yet reached the mirror
Or, to more accurately represent it, the apparent passage of time is different.
Oo observes these people having time pass much more slowly, with their actions appearing slower, including the light.

2-   As soon as light beams are reflected for Oa, it acts as if it were discharged for Oo by the rear mirror of B in the opposite direction of motion of A and B - Teleported lights
No. Oo will not see the reflection until the beam of light reaches the mirror. You are trying to act like the passage of time is the same for both. It is not.

A remotely controlled vertical light clock "A" as shown in the figure moves with very high-speed w.r.t the stationary observer Oo. As soon as a pulse is detected by M2 for Oa, A stops moving just in front of Oo.
You have brought this up before and had it explained to you before.

This time you are again pretending that the same amount of time has passed for both observes. It hasn't.
Again, time is relative.
And associated with this there is also the issue of order of events. Different reference frames see different orderings of events. If you pick 2 events, A & B which are simultaneous in 1 reference frame; then you can pick a different reference frame where A proceeds B, and a different reference frame where B proceeds A.

Ignoring that doesn't show a contradiction. It just shows you are willing to ignore reality to pretend there is a problem.

There are lots of paradoxes based upon this strange behaviour.

Re: Time Dilation / Second Postulate
« Reply #21 on: December 28, 2021, 12:35:20 AM »
Quote
Time is relative. Who is this time for?
It meant with the passage of time.

Einstein’s theory requires synchronization of clocks of both frames before one departs from another for the derivation time dilation/ Lorenz transformation.

How is it possible to draw a time dilation triangle if we don’t know the starting point (asked earlier) or time at which a pulse was fired or reflected from either of the mirrors relative to the moving observer (w.r.t stationary observer)?

Quote
You have brought this up before and had it explained to you before.

This time you are again pretending that the same amount of time has passed for both observes. It hasn't.
Again, time is relative.
I’m not sure if you have gone through it in detail but it was explained clearly in accordance with Einstien relativity when both observers have light clocks. Here again

Light clocks of Oo and Oa are synchronized before Oa moves w.r.t Oo at a very high speed.

Relative to Oa:
As soon as a pulse is detected by M2, A comes to rest just in front of Oo.
Oa sees Oo also at rest but a pulse of his light has not yet reached M2. Here

Before stopping
Oa sees a longer path of a pulse of a light clock of Oo but has not yet reached the M2.

After stopping
Oa sees all clocks are stationary. Oa sees a vertical path of a pulse (no longer long path) of the light clock of Oo and yet to reach M2.

Relative to Oo
A pulse of the light clock of Oa has not yet reached M2 and sees a clock along with Oa is still moving.

When Oa stops, Oa finds Oo and himself at rest but Oo who is already at rest doesn't find Oa at rest - why?

How about the occurrence of events when two bolts of light strike the longitudinal sides of the oncoming train relative to the stationary observer who stands in the middle of railway tracks. Both lights strike at the same time for both inside and outside observers.

Time dilation seems similar to Aristotle's wheel paradox in which the inner circle slips/skids. IMPOV the inner circle's tangential contact time is more than the tangential contact time of the outer circle on the line that is traced by both circles upon rolling.

Quote
There are lots of paradoxes based upon this strange behavior.
For Example?

« Last Edit: December 28, 2021, 01:33:15 AM by E E K »

Re: Time Dilation / Second Postulate
« Reply #22 on: December 28, 2021, 02:27:52 AM »
Einstein’s theory requires synchronization of clocks of both frames before one departs from another for the derivation time dilation/ Lorenz transformation.
And that synchronisation holds only at that moment. While there is relative motion those clocks will go out of sync.
As such when you describe an event happening at a particular time you need to say what time reference you are using.

But regardless, you can set any arbitrary time to be time 0, for any reference frame.
it was explained clearly in accordance with Einstien relativity when both observers have light clocks. Here again
No it wasn't, as you continue to pretend time is absolute, ignoring the relativity of time and simultaneity.

You also ignore the change in reference frame when Oa stops.

How about the occurrence of events when two bolts of light strike the longitudinal sides of the oncoming train relative to the stationary observer who stands in the middle of railway tracks. Both lights strike at the same time for both inside and outside observers.
No, they don't.
That is a key point you keep on ignoring.
The order of events is relative.

If you have 2 events, A & B, occur simultaneously for 1 particular reference frame, then there exists another reference frame where A occurs before B, and there is exists another reference frame where B occurs before A.
The order of events is relative.

Ingoing that wont change it.

Quote
There are lots of paradoxes based upon this strange behavior.
For Example?
Well one based upon ignoring the relative order of events is a train going through a tunnel.
Say there is a train with a proper length of 100 m, and a tunnel with a proper length of 90 m.
This tunnel also has a door on the entrance and exit.
If the train passes through the tunnel at 50% of the speed of light, is it possible for both doors to be closed at once?

Well according to the stationary observer outside, length contraction reduces the length of the train to ~87 m, meaning it will fit in the tunnel and both doors can be closed at once.
But according to the train, the tunnel has had its length contract to ~78 m, so it definitely can't fit.
So there is an apparent paradox because it appears that from one view it should fit and from the other it shouldn't.
This paradox is resolved by the relative ordering of events.

To the stationary observer, they see the train entirely in the tunnel, with the entry door closing and then the exit door opening a short time later. Notice the order of events, entry door closes then exit door opens.
To someone on the train, they see the exit of the tunnel open before the train is entirely in the tunnel. Only after they exit the tunnel does the entry door close behind them. Again notice the order of events, the exit door opens before the entry door closes.
The order of the events changes due to the relative motion of the reference frame.
If you had someone going backwards on the train at the appropriate speed then the exit door would open at the same time as the entry door closes.

As time, and the ordering of events, is relative there is no problem.
There only falsely appears to be a problem when you ignore that relativity and pretend the order of events should be the same in all reference frames.

This is also known as the ladder paradox, and uses a ladder in a garage.
An extension of it is to ask what happens if the doors remain closed.
To the observer at rest, the train fits in the tunnel, with both doors closed, so you should be able to stop it and keep the doors closed and keep the ladder inside.
But from the trains point of view, it was never able to fit.

This is resolved by considering how acceleration happens.
If you want the train to stop in an instant, then that doesn't just apply to all reference frames, again, because of the relativity of simultaneity.
If you have it stop instantly for the observer at rest, then in the inertial reference frame where the train is momentarily at rest the front must stop before the back, which causes the train to be crushed/compressed.

A similar paradox to that instead uses 2 accelerating spaceships.
An observer at rest observes 2 spaceships, which are currently at rest, separated by some long distance, with a string tied between them.
Then, in this frame, both spacecraft start accelerating at the same time. What happens?

A very naive approach is to just treat the 2 spacecraft and string as a single object, and say the length is contracted and nothing much happens.

But in reality, the distance between these 2 spacecraft was defined by the initial separation. The 2 spacecraft accelerating together will keep that distance between them. In order for that distance to shrink, the lead spacecraft would need to accelerate at a slower rate.
But the string is a physical object, and its length would contract. This should cause the string to snap.

But from a naive view of the reference frame of the spacecraft, both accelerate at the same time, and thus should remain the same distance apart, so the string should remain connected.
But this is also wrong.
This is now akin to gravity, as you have an accelerating reference frame. In this accelerating reference frame, time passes differently for the 2 ships.
The simplest way to analyse it is by using a inertial reference frame momentarily comoving with one of the ships.
In this reference frame, the lead ship will have started to accelerate first, and the second ship will accelerate some time after. This means the distance between the 2 ahs increased, and the string breaks.

Again, it only appears to be a problem because the relative order of events is ignored. When it is properly analysed, there is no problem.