Time dilation, length contraction, but velocity invariant

[Kairos]

If a frame is moving at constant velocity relative to an observer, this observer perceives a time dilation and a length contraction. But in this case how the velocity (length/time) can appear constant ? It is expected to be contracted.. Thank you in advance for the explanation

 

[harrylin]

If a frame is moving at constant velocity relative to an observer, this observer perceives a time dilation and a length contraction. But in this case how the velocity (length/time) can appear constant ? It is expected to be contracted.. Thank you in advance for the explanation

Almost certainly you overlooked (or don't even know) about relativity of simultaneity.
If so, please search this forum for relativity of simultaneity and read §1 and 2 of http://fourmilab.ch/etexts/einstein/specrel/www/
Next, in particular this animation by Janus is very useful: https://www.physicsforums.com/threads/twin-paradox-and-asymmetry.816455/#post-5126008

 

[Kairos]

Thank you, but I don't see precisely the answer in these refs (except that Enstein confirms that V=D/t)

What is wrong in my question: D contracted?, t dilated? or D/t constant?

 

[PeroK]

Thank you, but I don't see precisely the answer in these refs (except that Enstein confirms that V=D/t)

What is wrong in my question: D contracted?, t dilated? or D/t constant?

Both length contraction and time dilation have the same factor. If we have an object moving with speed $v$ with respect to you, from point A to point B, then:

You measure the distance from A to B as $d$ and the time as $t$, so calculate the speed as $v = d/t$.

The moving object calculates the distance from A to B as $d/\gamma$ and the time as $t/\gamma$, so it calculates the speed as $v = d/t$

So, you both agree on the speed, but not on the distance traveled or elapsed time.

 

[Kairos]

The moving object calculates the distance from A to B as d/γd/\gamma and the time as t/γt/\gamma, so it calculates the speed as v=d/tv = d/t

that would be nice but d/gamma looks a length dilation?

 

[PeroK]

that would be nice but d/gamma looks a length dilation?

To the moving object, the distance from A to B is contracted by a factor of $\gamma = \frac{1}{\sqrt{1- v^2/c^2}} > 1$.

And to you (at rest with respect to A and B), the moving clock is dilated by a factor of $\gamma$.

So, it is nice!

 

[Kairos]

OK I did not remember the meaning of gamma ( I thought it was the inverse)

In this case, d/gamma is indeed a length contraction, but t/gamma is now a time contraction..

d and t should be both either dilated or contracted to maintain the ratio constant. Correct?

 

[PeroK]

OK I did not remember the meaning of gamma ( I thought it was the inverse)

In this case, d/gamma is indeed a length contraction, but t/gamma is now a time contraction..

d and t should be both either dilated or contracted to maintain the ratio constant. Correct?

The moving clock runs slow from your reference frame. That means time in that reference frame is "stretched" or "dilated". The maths and the transformation are the important things. "Contraction" and "dilation" are just words and need a context to know what they mean.

Time in your frame: 0...1...2...3
Time in moving frame: 0...1...2...3

 

[Kairos]

These words seem to be used inversely for lengths and time. This is very misleading.

Do you confirm that viewed from my reference frame (arbitrarily considered immobile) : (1) the lengths in the moving frame appear contracted and (2) the time marked by the moving clock appears contracted (runs faster)?

 

[PeroK]

These words seem to be used inversely for lengths and time. This is very misleading.

Do you confirm that viewed from my reference frame (arbitrarily considered immobile) : (1) the lengths in the moving frame appear contracted and (2) the time marked by the moving clock appears contracted (runs faster)?

No:

(1) the lengths in the moving frame appear contracted and (2) the time marked by the moving clock appears dilated (runs s-l-o-w-e-r).

But, here it's your lengths as viewed in the moving frame that are important. The moving object measures your distances as shorter than you do.

 

[Kairos]

Thanks a lot for your patience but I confess I am lost.
All these effects are supposed to be reciprocal because the inertial frames can of course be permuted

With respect to your post #4, $t/\gamma$ gives an apparent time shorter than $t$. To me shorter means contraction

 

[PeroK]

Thanks a lot for your patience but I confess I am lost.
All these effects are supposed to be reciprocal because the inertial frames can of course be permuted

With respect to your post #4, $t/\gamma$ gives an apparent time shorter than $t$. To me shorter means contraction

Words are not important. Less elapsed time can mean contraction or dilation, whatever, according to taste. This is why maths trumps words in the end. The maths is unambiguous.

But, there is something deep here and, in fact, understanding this little problem fully will take you a long way to understanding these two effects of SR.

Let's start again:

You set up two points A and B, a distance $d$ apart. You observe an object travel from A to B at speed $v$ and measure time $t = d/v$. That's just standard kinematics. You also notice that the object's clock has only measured $t' = t/\gamma$ while it traveled from A to B.

Now, what does the moving object observe? It must measure $t' = t/\gamma$ for its journey from A to B (although from its perspective A and B are moving towards it).

There are then two ways to look at this. If we agree that the relative velocity must be the same for both frames (I won't go into the argument here, but let's accept this). Then, it must conclude that the distance from A to B is only $d' = d/\gamma$

Alternatively, if it measures the distance from A to B and finds this to be $d' = d/\gamma$, then it agrees that its speed (or the speed of A and B) is $v$.

That's the basic symmetry of the situation. It's worth batting your head against this for a bit. Just keep going until it's clear. It may take some time but it's important.

 

[Kairos]

your sentence :

You also notice that the object's clock has only measured t′= t/\gamma while it traveled from A to B

seems contradictory with :

the time marked by the moving clock appears dilated (runs s-l-o-w-e-r).

In fact, my main question can be summarized: what means t' < t ?
according to me, if seconds' are shorter than seconds, time t' runs faster than time t, not slower

 

[PeroK]

your sentence :
seems contradictory with :

In fact, my main question can be summarized: what means t' < t ?
according to me, if seconds' are shorter than seconds, time t' runs faster than time t, not slower

If you have two clocks and one is running slower than the other. Let's say clock C' is running slower than clock C. And, if you start them both a 0, then: $t' < t$

For example, if C' is running at half the speed of C, then $t = 2t'$.

 

[Kairos]

OK, I understand the misunderstanding! I reasoned in "time units"
many thanks

 

[harrylin]

Thank you, but I don't see precisely the answer in these refs (except that Enstein confirms that V=D/t)

What is wrong in my question: D contracted?, t dilated? or D/t constant?

There are different ways to measure velocity, and it wasn't very clear from your question what velocity you meant. As you used the word "invariant", I supposed that you were thinking of measuring from the point of view of each frame as "rest frame", the change of position in that frame of a moving point that has a fixed position in the other frame (e.g. the motions of the origins of each frame as measured in the other frame).

If so, then you compare the time as read on one clock, with the time as read on another clock at a distance. When you write D/t you probably mean D/(t₂-t₁): t₂ is read on one clock, and t₁ on another clock; that depends on clock synchronization as you saw. And as Einstein explained there, the different systems synchronize their clocks differently.

If the "moving" system with contracted rulers and slow clocks used the same synchronization as the "rest" system, they would measure as speed a longer distance / less time. Observers of the "stationary" system interpret the fact that observers of the "moving" system measure the same speed as themselves as due to the wrong clock synchronization of the "moving" system.

 

[ghwellsjr]

If a frame is moving at constant velocity relative to an observer, this observer perceives a time dilation and a length contraction. But in this case how the velocity (length/time) can appear constant ? It is expected to be contracted.. Thank you in advance for the explanation

Good question. I think some spacetime diagrams may help to answer your question.

Let's take a scenario where an observer sends out a light pulse to a reflector 6 feet away and measures with his clock how long it takes for the reflection to get back to him. Since light travels at 1 foot per nanosecond, it will take 12 nsecs for him to see the reflection and he will validate that the light is traveling at 1 foot per nsec for the 12 feet of the round trip that the light takes. If he is following the precepts of Special Relativity, he will define the time at which the reflection took place at 6 nsecs according to his clock and I have made that dot black:

In this diagram, the observer is shown as the thick blue line with dots marking off one-nanosecond increments of time according to his clock. Since he is at rest in this Inertial Reference Frame (IRF), the horizontal grid lines align with his clock and the vertical grid lines align with his measurement of distance.

The reflector is shown as the thick red line with similar dots. The thin blue line represents the pulse of light that the observer emits and the thin red line represents the reflection. Hopefully, this all makes perfect sense to you.

Now let's transform the coordinates of all the dots in the above diagram to a new IRF moving at 0.6c with respect to the original IRF:

At 0.6c, gamma is 1.25 and you will note that the dots are placed 1.25 nsecs apart as defined by the gird lines. Also, the distance to the reflector is contracted to 4.8 feet (6/1.25) as defined by the grid lines. You will also note that the light pulse and its reflection continue to travel at 1 foot per nanosecond but it takes less time in this IRF (3 nsecs) for the light to reach the reflector and more time (12 nsecs) for the reflection to get back to the observer. The total Coordinate Time is 15 nsecs and the total Coordinate Distance the light traveled is 15 feet so the velocity is maintained at 1 foot per nanosecond.

Just for the fun of it, let's see what would happen if we applied the method of your question to establish the speed of the light. Here is a diagram that depicts the parameters that would apply if we just took the Length Contraction and Time Dilation factors into account:

Note that we would be using the length at the start of the scenario and the time as defined by the observer as when the reflection occurred to establish where the light reached at the time of the reflection which is way off base and would result in a contracted velocity as shown by the extra thin blue line. Instead, we need to use the Coordinate Distance of where the reflector has moved to when the light pulse reaches it and the Coordinate Time at which this occurred, not the time that the observer defines the reflection to occur according to his clock.

Does this all make perfect sense to you? Any questions?

 

[Stephanus]

The moving clock runs slow from your reference frame. That means time in that reference frame is "stretched" or "dilated". The maths and the transformation are the important things. "Contraction" and "dilation" are just words and need a context to know what they mean.

Time in your frame: 0...1...2...3
Time in moving frame: 0...1...2...3

Good! Very clear!
You should use "Courier New"

 

[Stephanus]

Now let's transform the coordinates of all the dots in the above diagram to a new IRF moving at 0.6c with respect to the original IRF[..]
Does this all make perfect sense to you? Any questions?

First,...$\gamma = \frac{1}{\sqrt{1-0.36}} = 1.25$
Let's say BR Frame is the moving blue/red frame at 0.6c
B is Blue
R is Red
W is the rest observer/watcher
Okay...

1) Why vertical line = 15?
Because $\sqrt{15^2-(-9)^2} = 12$, the time the event takes place. 12 seconds wrt BR. Hyptenuse in 4D, Is this right? 15 seconds wrt Rest, is this right?

2) Why -9? from (-9)²
When the light bounces from RED, it has to catch BLUE moving 0.6c 4.8ls away.
So it takes $ct = 4.8+vt; t = 4.8 + 0.6t; t = 12$ seconds
So $3-12 = -9$, so it will catches BLUE at -9 wrt rest, is this right?

3a) What is 3? from $3-12 = -9$
Because this is where and when wrt W, the light from (0,0) reaches Red WL and bounce back. Is this right?
3b) Why 3?
Because $3 = \frac{4.8}{1+0.6}$, is this right?

4)Why 4.8 not 6?
It comes from the Lorentz contraction, length transformation, which is 200% proven, irefutable, undeniable. $6*\sqrt{1-0.6^2}$ I think this is right.

5)Why the origin of the red line (6,0) wrt W, before now seems in (7.5,4.8)? Okay, let's say I don't know if it's in 7.5, 4.8, just say at (7.?,4.?). I think we should calculate when and where RED started.
I can't draw, can I just type?
Supposed, all this events from BR frames takes 12 seconds.
There are 3 events.
There are 3 object
W, R, and B
E0: Clocks are synchronized.
E1: Blue emit light to Red, and BR moves.
E2: The light touches Red and bounces back and R also sends a digital signal to W the clock wrt R this happen and the location wrt O this happen.
R can send the location wrt W because R is entering W light cone. Can R do that?
E3: The light touches Blue again.

Okay...
E1, starts
E2: Will W know when E2 wrt R happens?. R, record its clock and W's clock from W light cone sends to W back.
R clock wrt R -> 6
E2 location wrt W -> 3 from the data that R collect from W light cones.
Because R is entering W light cone, R also sends its speed to W

Is this how W translate the data?
R receive light signal when its clock is 6 wrt R.
R is in 3 lys away from where W start move wrt W
R speed is 0.6c based on R calculation, and doppler and Lorentz transformation. Can R calculate its speed?
If it moves at 0.6c so gamma is 1.25.
But I'm lost here. How can R clock read 6? Does R clock ticks twice? Does R clock start earlier?

 

[Kairos]

thank you all for your support.
In fact, I better understand equations that diagrams supposedly didactic...

 

[ghwellsjr]

First,...$\gamma = \frac{1}{\sqrt{1-0.36}} = 1.25$
Let's say BR Frame is the moving blue/red frame at 0.6c
B is Blue
R is Red
W is the rest observer/watcher
Okay...

View attachment 84896
1) Why vertical line = 15?
Because $\sqrt{15^2-(-9)^2} = 12$, the time the event takes place. 12 seconds wrt BR. Hyptenuse in 4D, Is this right? 15 seconds wrt Rest, is this right?

2) Why -9? from (-9)²
When the light bounces from RED, it has to catch BLUE moving 0.6c 4.8ls away.
So it takes $ct = 4.8+vt; t = 4.8 + 0.6t; t = 12$ seconds
So $3-12 = -9$, so it will catches BLUE at -9 wrt rest, is this right?

3a) What is 3? from $3-12 = -9$
Because this is where and when wrt W, the light from (0,0) reaches Red WL and bounce back. Is this right?
3b) Why 3?
Because $3 = \frac{4.8}{1+0.6}$, is this right?

4)Why 4.8 not 6?
It comes from the Lorentz contraction, length transformation, which is 200% proven, irefutable, undeniable. $6*\sqrt{1-0.6^2}$ I think this is right.

5)Why the origin of the red line (6,0) wrt W, before now seems in (7.5,4.8)? Okay, let's say I don't know if it's in 7.5, 4.8, just say at (7.?,4.?). I think we should calculate when and where RED started.
I can't draw, can I just type?
Supposed, all this events from BR frames takes 12 seconds.
There are 3 events.
There are 3 object
W, R, and B
E0: Clocks are synchronized.
E1: Blue emit light to Red, and BR moves.
E2: The light touches Red and bounces back and R also sends a digital signal to W the clock wrt R this happen and the location wrt O this happen.
R can send the location wrt W because R is entering W light cone. Can R do that?
E3: The light touches Blue again.

Okay...
E1, starts
E2: Will W know when E2 wrt R happens?. R, record its clock and W's clock from W light cone sends to W back.
R clock wrt R -> 6
E2 location wrt W -> 3 from the data that R collect from W light cones.
Because R is entering W light cone, R also sends its speed to W

Is this how W translate the data?
R receive light signal when its clock is 6 wrt R.
R is in 3 lys away from where W start move wrt W
R speed is 0.6c based on R calculation, and doppler and Lorentz transformation. Can R calculate its speed?
If it moves at 0.6c so gamma is 1.25.
But I'm lost here. How can R clock read 6? Does R clock ticks twice? Does R clock start earlier?

I made this diagram to illustrate why you don't want to take shortcuts in transforming IRF's, you should always apply the Lorentz Transform. The reason why I made the tick mark black when the R clock reaches 6 was to show that it doesn't have any significance to R, it is only a calculation that R makes after he receives the reflection at his time 12. He simply divides the time in half and defines that as the time of the reflection

All your calculations look correct but I'm wondering why you are doing this. Could you have done it if I had not already supplied the above diagram?

Instead of following your reasoning, if you start with the first diagram which defines the scenario, then all you have to do is apply the Lorentz Transformation process at 0.6c to the coordinates of each event and plot them on a new chart. You don't even have to think about what you are doing.

Do you know how to use the Lorentz Transform?

 

[Stephanus]

Continued...
From the above diagram:
RB distance: 6 light seconds
V: 0.6c
$\gamma = 1.25$
R: Red/Moves 0.6c
B: Blue/Moves 0.6c
BR: R and B frame
W: Watchers/observer/Rest
Event:
E0: Clock syncrhonization
E1: BR start moves
E2: R meets light from B
Here, R sends a digital signal to W containing:
R's clock time
W's clock time, according to W light cone
E3: Signals from R reaches W containing messages from R. R's clock time and W's clock time according to W's light cone
E4: Light meets B again

R's clock time at E2: 6, according to RB distance
W's clock time:
Because RB moves at 0.6c entering W light cone, then RW distance wrt R is 4.8 ls.
$E2 = \frac{4.8}{1+0.6} = 3$
So when E2 happens it's 3 seconds wrt W. and at x = 3ls wrt W
E2:
R's clock time: 6 (wrt is not relevant here. It's the number that R recorded)
W's clock time: 3 (wrt is not relevant here. It's the number that R recorded)
Is this right?

E3: Reply from R reaches W
W's clock time: 6
Reply from R
R's clock time: 6
W's clock time: 3What can W reconcile from this?
W has messages R's: 6 seconds at his 3 seconds.
Is this true, that if R light cone has reached W world line, W will be able to detect R speed?
If this is true then W will reason...
$6 = \sqrt{x^2-(0.6x)^2}$
$36 = x^2 * (1-0.36)$
$36 = 0.64x^2$
$x = 4.8$
Sorry, it's not 7.5 but 7.8.
So R must have started from 7.8
Is this right?
And from WHEN?
So, v = 0.6 D = 4.8, so time is 8 for R

Sorry, stuck here. To be continued...

 

[Stephanus]

Thanks for your graph ghwellsjr.

I made this diagram to illustrate why you don't want to take shortcuts in transforming IRF's, you should always apply the Lorentz Transform. The reason why I made the tick mark black when the R clock reaches 6 was to show that it doesn't have any significance to R, it is only a calculation that R makes after he receives the reflection at his time 12. He simply divides the time in half and defines that as the time of the reflection

Good point. The thick black mark in B you mean? I've suspected there's some detective story here. B has to deduct where R is Based on the data that B receives. Because R is away from B.

All your calculations look correct but I'm wondering why you are doing this. Could you have done it if I had not already supplied the above diagram?

Why am I doing this? I was alone in the office at night, waiting for Jakarta Traffice Jam. That's why I'm doing this.
Without your graph? No, of course not. I can't do this without your graph. But I think I don't know why. I make all the calculation based on your graph. And in some calculations I just match the calculation with your graph. If my calculation didn't match with your graph, I change the calculation so it will match your graph.
And second. At this stage, I have to calculate from provided/someone else graph. I can't make the graph myself. But I'm sure, later I can make all the calculation by hands withouth graph.

 

[Stephanus]

Both length contraction and time dilation have the same factor. If we have an object moving with speed $v$ with respect to you, from point A to point B, then:

You measure the distance from A to B as $d$ and the time as $t$, so calculate the speed as $v = d/t$.

The moving object calculates the distance from A to B as $d/\gamma$ and the time as $t/\gamma$, so it calculates the speed as $v = d/t$

So, you both agree on the speed, but not on the distance traveled or elapsed time.

Good, speed is the same!

 

[harrylin]

thank you all for your support.
In fact, I better understand equations that diagrams supposedly didactic...

OK. As I answered with equations, perhaps my answer #16 was not elaborate enough...

I supposed that you were thinking that measuring a distance with a contracted ruler gives you a greater distance, and measuring a duration with a slowed down clock gives you less time, so that distance/time -> greater speed.

However, the t that you wrote down refers in fact not to a single clock reading but to the difference (t2 - t1) of clock readings on different clocks. The answer depends therefore on clock synchronization. If you work it out, you will find that different clock synchronization exactly compensates for it.
A straightforward way to work it out is by means of the Lorentz transformations, see my post here:
https://www.physicsforums.com/threads/lorentz-transformation.818223/#post-5137123

As a matter of fact, the first Lorentz transformation was approximate as the last transformation equation in that post, which only accounted for relativity of simultaneity. Length contraction and time dilation were later added to obtain the perfect symmetry that relativity demands.

 

[Kairos]

understood , many thanks

 

[ghwellsjr]

Continued...
From the above diagram:

I presume you mean this diagram which I have redrawn with the units you are using (seconds and light-seconds rather than nanoseconds and feet). I'll explain later why I have extended the Red and Blue wordlines:

RB distance: 6 light seconds

But now we have a problem because in this diagram, the RB distance (if by that you mean the distance between Red and Blue) is 4.8 light seconds, not 6 light seconds.

V: 0.6c
$\gamma = 1.25$
R: Red/Moves 0.6c
B: Blue/Moves 0.6c

And now we have another problem because V is the speed that this frame has been constantly moving at relative to the original frame which makes the velocities of Red and Blue be -0.6c, not 0.6c.

BR: R and B frame
W: Watchers/observer/Rest

Two more problems:
# 1) BR is the frame in which B and R are at rest, which is not represented by the above diagram.
# 2) You never defined who W was. What frame is he at rest in?

Just for the record, here is a diagram for the BR frame, that is, the frame in which Blue and Red are constantly at rest:

Event:
E0: Clock syncrhonization

Clock synchronization does not take place in one event. It's a process that takes place over a significant amount of time and a significant amount of space. I think what you really intend by this event is nothing more than the origins of the two frames, that is, the event where the coordinates are all zero in both frames.

E1: BR start moves

I see two more problems here:
1) I think you probably are thinking in terms of the BR frame as being stationary for negative times and starting to move at time zero and continuing to move for positive times. That's the wrong way to think about these frames. Instead, you should think of the original frame, the one I would call the BR frame (the second one above), as being stationary all the time and the first frame as constantly moving all the time with respect to the original frame.

2) In any case, I don't understand what would be different about E0 and E1. I think you mean them to have the same coordinates (all zeroes) in both frames.

E2: R meets light from B
Here, R sends a digital signal to W containing:
R's clock time
W's clock time, according to W light cone
E3: Signals from R reaches W containing messages from R. R's clock time and W's clock time according to W's light cone

I cannot figure out where W is supposed to be. None of the above makes any sense to me. It would make sense if you didn't have W but merely stated that R sends a digital signal to B containing R's clock time.

E4: Light meets B again

R's clock time at E2: 6, according to RB distance
W's clock time:
Because RB moves at 0.6c entering W light cone, then RW distance wrt R is 4.8 ls.
$E2 = \frac{4.8}{1+0.6} = 3$
So when E2 happens it's 3 seconds wrt W. and at x = 3ls wrt W
E2:
R's clock time: 6 (wrt is not relevant here. It's the number that R recorded)
W's clock time: 3 (wrt is not relevant here. It's the number that R recorded)
Is this right?

I don't know since I don't understand where W is.

E3: Reply from R reaches W
W's clock time: 6
Reply from R
R's clock time: 6
W's clock time: 3What can W reconcile from this?
W has messages R's: 6 seconds at his 3 seconds.
Is this true, that if R light cone has reached W world line, W will be able to detect R speed

I don't know since I don't understand where W is.

If this is true then W will reason...
$6 = \sqrt{x^2-(0.6x)^2}$
$36 = x^2 * (1-0.36)$
$36 = 0.64x^2$
$x = 4.8$
Sorry, it's not 7.5 but 7.8.
So R must have started from 7.8
Is this right?
And from WHEN?
So, v = 0.6 D = 4.8, so time is 8 for R

Sorry, stuck here. To be continued...

You are trying to do something I had never intended in my diagrams so I'm at a complete loss to answer your questions and I don't think it belongs in this thread, since the velocity of an observer or object is not invariant. Only the velocity of light is invariant and that is what I was showing in my diagrams.

 

[ghwellsjr]

Thanks for your graph ghwellsjr.

I made this diagram to illustrate why you don't want to take shortcuts in transforming IRF's, you should always apply the Lorentz Transform. The reason why I made the tick mark black when the R clock reaches 6 was to show that it doesn't have any significance to R, it is only a calculation that R makes after he receives the reflection at his time 12. He simply divides the time in half and defines that as the time of the reflection

Good point. The thick black mark in B you mean? I've suspected there's some detective story here. B has to deduct where R is Based on the data that B receives. Because R is away from B.

That's almost correct. B does not deduce when the reflection occurs by this method, he defines when it occurs. Do you see the difference? Simultaneity is not something that nature provides for us that we have to figure out, instead, it is up to us to figure out a convenient and satisfying but totally arbitrary way to establish the concept of simultaneity of remote events. In other words, we are putting the meaning into nature, we are not drawing the meaning out of nature.

All your calculations look correct but I'm wondering why you are doing this. Could you have done it if I had not already supplied the above diagram?

Why am I doing this? I was alone in the office at night, waiting for Jakarta Traffice Jam. That's why I'm doing this.
Without your graph? No, of course not. I can't do this without your graph. But I think I don't know why. I make all the calculation based on your graph. And in some calculations I just match the calculation with your graph. If my calculation didn't match with your graph, I change the calculation so it will match your graph.
And second. At this stage, I have to calculate from provided/someone else graph. I can't make the graph myself. But I'm sure, later I can make all the calculation by hands withouth graph.

If you were given an assignment to draw a graph showing a blue observer with a red mirror six feet away who flashes a light pulse which travels at 1 foot per nanosecond toward the mirror and reflects back to him, could you not even make this first graph I provided in post #17 without having seen it before hand?

If you do have trouble with that assignment, could you please explain where you are having a problem?

Also, you didn't answer my previous very important question:

Do you know how to use the Lorentz Transform?

 

[Stephanus]

Ahh, it's good to be back again!
$v = -0.6c$
$\gamma = 1.25$
$D = 6 \text { } D' = 4.8$

I presume you mean this diagram which I have redrawn with the units you are using (seconds and light-seconds rather than nanoseconds and feet). I'll explain later why I have extended the Red and Blue wordlines:

I'm sorry, I should have read really slow. Okayy, it's seconds rather than nanoseconds. My mistake

But now we have a problem because in this diagram, the RB distance (if by that you mean the distance between Red and Blue) is 4.8 light seconds, not 6 light seconds.

MANY THANKS! Otherwise, I'll make a mistake in my calculation. So, how to draw space time diagram is this... beside the angle should be rotated to $\frac{1}{atan(v)}$, the x distance should be contracted according to \gamma.

But how can I pin point the coordinate for the rest frame?
For blue line (the point from the origin):
Let t₀ is the original time from B worldline before it moves.
Let t₁ is the time from B worldline after it moves.
In B frame...
$t_0=\sqrt{t_1^2-(vt_1)^2}$ Supposed from your diagram. The light from B bounce back to B for 12 nanoseconds. Speed 0.6c
$t_0=\sqrt{t_1^2*(1-v^2)}$
$t_1=\frac{t_0}{\sqrt{1-v^2}}$
So $t_1=t_0 * \gamma$ Is this right?
If BR moves -0.6c so 12 nanoseconds in rest frame is15 nanoseconds in moving frame, is this right?
If $t_1=t_0 * \gamma$, and $x_1 = vt_1$, so
$x_1=vt_0\gamma$, okay...

How can I pin point coordinate in Red Line?
I can't do that directly, All I can do is this.
How can I pin point Red (6,6) in rest frame to moving frame?
In R rest frame.$x=vt+6 \implies x = 6$
If R is moving $x=vt+4.8 \implies x = -0.6t + 4.8$
Let D = 4.8, the contracted length
Coordinate Red Rest(6,6) is in the blue light cone Rest (0,0)...
So the equation for light word line... $x = t$
combined
$x = t$
$x = vt+.4.8$, eliminated
$1.6 t = 4.8 \implies t=3, x =3$. This is how I calculated point in R moving frame?

How can I pin point Red Rest(6,0) to Red moving (x,t)?
The origin of light cone to reach (6,0) from Blue line is in (0,-6)
Blue Rest (0,-6) is...
$t_1 = t_0 * \gamma = -6 * 1.25 = -7.5$
$x_1 = v t_1 = 4.5$
What is the formula for line crossing (4.5,-7.5) at 45⁰ angle?
$x_2 = t_2 + 12$. How can I get 12?
$12 = x_1 - t_1 = vt_1-t_1 = t_1(1-v)$
$x_2 = t_2 + t_1(1-v)$
$x_2 = t_2 + t_0*\gamma*(1-v)$ and if $\gamma = \frac{1}{\sqrt{1-v^2}} = \frac{1}{\sqrt{(1+v)(1-v)}}$, so
$x_2 = t_2 + t_0*\gamma*(1-v) = t_2 + t_0*\frac{1}{\sqrt{(1+v)(1-v)}}*(1-v)$
$x_2 = t_2 + t_0*\sqrt{\frac{1-v}{1+v}}$

Red Rest(6,0) it will cross Blue in
$x = 0$
$x = t + 6$
$t_0 = -6$
$t_1 = -7.5$
So what is x_2,t_2 from 4.5,-7.5 that meets Red?
$x_2 = t_2 + t_0*\sqrt{\frac{1-v}{1+v}} = t_2 + 12$, light from blue
$x_2 = vt_2 + D'$ D' is 4.8 is the original length contracted by Lorentz factor.
$t_2 + 12 = -0.6t_2 + 4.8$
$t_2 = -4.5 \implies x_2 = 7.5$

So for every location in Red Line in rest frame, we can transform it in moving frame, by.
$t_2 = \frac{(t_0-D)*\sqrt{\frac{1-v}{1+v}} - D*\sqrt{1-v^2}}{1+v}$
Perhaps this is wrong?

And now we have another problem because V is the speed that this frame has been constantly moving at relative to the original frame which makes the velocities of Red and Blue be -0.6c, not 0.6c.

I would have dropped an airplane if I were an engineer with my endless careless calculation!.

Two more problems:
# 1) BR is the frame in which B and R are at rest, which is not represented by the above diagram.
# 2) You never defined who W was. What frame is he at rest in?

BR is the frame where Blue line and Red line lies.
W is the stationary observer watching the event unfold.
If B and R move, this is what W sees.

This is what B and R sees.
This is what BR sees.

I see two more problems here:

1) I think you probably are thinking in terms of the BR frame as being stationary for negative times and starting to move at time zero and continuing to move for positive times. That's the wrong way to think about these frames. Instead, you should think of the original frame, the one I would call the BR frame (the second one above), as being stationary all the time and the first frame as constantly moving all the time with respect to the original frame.
2) In any case, I don't understand what would be different about E0 and E1. I think you mean them to have the same coordinates (all zeroes) in both frames.
I cannot figure out where W is supposed to be. None of the above makes any sense to me. It would make sense if you didn't have W but merely stated that R sends a digital signal to B containing R's clock time.

So, if it's the wrong way to think, then, perhaps those last qestion I ask are wrong.
Thanks for your answer.

Btw, seeing your diargram. Does Red clock start first?

 

[ghwellsjr]

Ahh, it's good to be back again!
$v = -0.6c$

You misunderstood. When I said in post #27 that the velocities were negative, I was talking about the velocities of Red and Blue. The term "v" is the speed of the moving frame relative to the stationary frame. It's the value that we plug into the Lorentz Transformation equations which I will show you later since you apparently do not know how to use them.

$\gamma = 1.25$
$D = 6 \text { } D' = 4.8$

I'm sorry, I should have read really slow. Okayy, it's seconds rather than nanoseconds. My mistake

Since you switched back to nanoseconds in the rest of your post, let's stick with nanoseconds and feet, OK?

MANY THANKS! Otherwise, I'll make a mistake in my calculation. So, how to draw space time diagram is this... beside the angle should be rotated to $\frac{1}{atan(v)}$, the x distance should be contracted according to \gamma.

But how can I pin point the coordinate for the rest frame?

Here is the diagram for the stationary frame:

View attachment 84880​

This is the frame in which we define the scenario. As I said in post #28 we want to:

...draw a graph showing a blue observer with a red mirror six feet away who flashes a light pulse which travels at 1 foot per nanosecond toward the mirror and reflects back to him...

Does the above diagram make sense to you based on the specification of the scenario?

For blue line (the point from the origin):
Let t₀ is the original time from B worldline before it moves.
Let t₁ is the time from B worldline after it moves.

There's no before and after B moves or before and after its worldline moves. B is at rest in our defining stationary frame so its worldline is shown as a vertical line because its x-coordinate is a constant 0. Later on, when we transform to the frame moving at 0.6c with respect to the stationary frame, B will be moving at -0.6c all the time. B does not change its motion. We are merely describing its state of motion, either constantly at rest or constantly moving, in two different frames but never changing its state of motion in either frame.

In B frame...

$t_0=\sqrt{t_1^2-(vt_1)^2}$ Supposed from your diagram. The light from B bounce back to B for 12 nanoseconds. Speed 0.6c
$t_0=\sqrt{t_1^2*(1-v^2)}$
$t_1=\frac{t_0}{\sqrt{1-v^2}}$
So $t_1=t_0 * \gamma$ Is this right?
If BR moves -0.6c so 12 nanoseconds in rest frame is15 nanoseconds in moving frame, is this right?

You got the right answer but I don't know how you did it.

It's time for you to see how the Lorentz Transformation works. It's real simple and you don't have to think about what you are doing. You can take the (x,t) coordinates of any event (dot) from the stationary frame and transform to the (x',t') coordinates of the moving frame. You repeat this process for all the events (dots) or at least some strategic ones so that you can see how the new worldlines appear.

Here at the three equations which apply for units where c=1 as in our case:

γ = 1/√(1-v²)
x' = γ(x-vt)
t' = γ(t-vx)

You've already determined gamma to be 1.25 but I'll go through the motions here for anyone else that might be looking on. Remember, v=0.6:

γ = 1/√(1-v²) = 1/√(1-0.6²) = 1/√(1-0.36) = 1/√(0.64) = 1/0.8 = 1.25

Now we calculate the new time coordinate, t', for the new frame from the old coordinates, x=12 and t=0:

t' = γ(t-vx) = 1.25(12-0.6(0)) = 1.25(12) = 15

As you can see, we got the same answer you got but with a lot less effort, or at least, a lot less thinking.

Now it's just as important to obtain the x-coordinate which you did not do:

x' = γ(x-vt) = 1.25(0-0.6(12)) = 1.25(-7.2) = -9

So now the uppermost event (dot) on B's blue worldline has been transformed from (12,0) in the stationary frame to (15,-9) in the moving frame as shown in this diagram:

View attachment 84881​

If $t_1=t_0 * \gamma$, and $x_1 = vt_1$, so
$x_1=vt_0\gamma$, okay...

How can I pin point coordinate in Red Line?
I can't do that directly, All I can do is this.
How can I pin point Red (6,6) in rest frame to moving frame?
In R rest frame.$x=vt+6 \implies x = 6$
If R is moving $x=vt+4.8 \implies x = -0.6t + 4.8$
Let D = 4.8, the contracted length
Coordinate Red Rest(6,6) is in the blue light cone Rest (0,0)...
So the equation for light word line... $x = t$
combined
$x = t$
$x = vt+.4.8$, eliminated
$1.6 t = 4.8 \implies t=3, x =3$. This is how I calculated point in R moving frame?

Well, you got the right answer again but I don't know how. Better to use the Lorentz Transformation on (6,6), the event of the reflection:

x' = γ(x-vt) = 1.25(6-0.6(6)) = 1.25(6-3.6) = 1.25(2.4) = 3.0
t' = γ(t-vx) = 1.25(6-0.6(6)) = 1.25(6-3.6) = 1.25(2.4) = 3.0

How can I pin point Red Rest(6,0) to Red moving (x,t)?
The origin of light cone to reach (6,0) from Blue line is in (0,-6)
Blue Rest (0,-6) is...
$t_1 = t_0 * \gamma = -6 * 1.25 = -7.5$
$x_1 = v t_1 = 4.5$
What is the formula for line crossing (4.5,-7.5) at 45⁰ angle?
$x_2 = t_2 + 12$. How can I get 12?
$12 = x_1 - t_1 = vt_1-t_1 = t_1(1-v)$
$x_2 = t_2 + t_1(1-v)$
$x_2 = t_2 + t_0*\gamma*(1-v)$ and if $\gamma = \frac{1}{\sqrt{1-v^2}} = \frac{1}{\sqrt{(1+v)(1-v)}}$, so
$x_2 = t_2 + t_0*\gamma*(1-v) = t_2 + t_0*\frac{1}{\sqrt{(1+v)(1-v)}}*(1-v)$
$x_2 = t_2 + t_0*\sqrt{\frac{1-v}{1+v}}$

Red Rest(6,0) it will cross Blue in
$x = 0$
$x = t + 6$
$t_0 = -6$
$t_1 = -7.5$
So what is x_2,t_2 from 4.5,-7.5 that meets Red?
$x_2 = t_2 + t_0*\sqrt{\frac{1-v}{1+v}} = t_2 + 12$, light from blue
$x_2 = vt_2 + D'$ D' is 4.8 is the original length contracted by Lorentz factor.
$t_2 + 12 = -0.6t_2 + 4.8$
$t_2 = -4.5 \implies x_2 = 7.5$

For sure, I don't know what you are doing here. Part way down, it looked like you got the wrong answer but these last ones are correct for transforming (6,0):

x' = γ(x-vt) = 1.25(6-0.6(0)) = 1.25(6) = 7.5
t' = γ(t-vx)= 1.25(0-0.6(6)) = 1.25(-3.6) = -4.5

So for every location in Red Line in rest frame, we can transform it in moving frame, by.
$t_2 = \frac{(t_0-D)*\sqrt{\frac{1-v}{1+v}} - D*\sqrt{1-v^2}}{1+v}$
Perhaps this is wrong?

Perhaps...

Can you please start using the Lorentz Transformation?

I would have dropped an airplane if I were an engineer with my endless careless calculation!.

BR is the frame where Blue line and Red line lies.

This doesn't make any sense. The Blue and Red lines lie in all frames.

W is the stationary observer watching the event unfold.
If B and R move, this is what W sees.

I've asked you several times, where is W? is he stationary in one of the frames? If so, which one and what are his coordinates and state of motion? Why have you introduced him anyway?

View attachment 85108

This is what B and R sees.
This is what BR sees.View attachment 85109

So, if it's the wrong way to think, then, perhaps those last qestion I ask are wrong.
Thanks for your answer.

Whatever B and R see in one frame, they see in all other frames. For example, B sees the light pulse being emitted when his clock is at zero in both frames. B sees the reflected light pulse when his clock is at 12 in both frames. R sees the reflection occur when his clock reads 6 in both frames. Different frames do not change anything that any observer sees or measures.

Btw, seeing your diargram. Does Red clock start first?

The Red clock starts first in the moving frame. Both clocks start at the same time in the stationary frame.

 

[Stephanus]

Does the above diagram make sense to you based on the specification of the scenario?

Absolutely, 200% makes sense. B >>------------------------>> R and B <<---------------------------------<< R
Just like that! Because we are ALWAYS moving in spaceTIME, so the light makes 45⁰ angle from B to R instead of horizontal lines.

There's no before and after B moves or before and after its worldline moves. B is at rest in our defining stationary frame so its worldline is shown as a vertical line because its x-coordinate is a constant 0. Later on, when we transform to the frame moving at 0.6c with respect to the stationary frame, B will be moving at -0.6c all the time. B does not change its motion. We are merely describing its state of motion, either constantly at rest or constantly moving, in two different frames but never changing its state of motion in either frame.

I always consider B and R are always moving. But this does not register in my unconscious mind. Next time, I should have to picture it instinctively rather than mathematically.

You got the right answer but I don't know how you did it.

It's time for you to see how the Lorentz Transformation works. It's real simple and you don't have to think about what you are doing. You can take the (x,t) coordinates of any event (dot) from the stationary frame and transform to the (x',t') coordinates of the moving frame. You repeat this process for all the events (dots) or at least some strategic ones so that you can see how the new worldlines appear.

Here at the three equations which apply for units where c=1 as in our case:

γ = 1/√(1-v2)
x' = γ(x-vt)
t' = γ(t-vx)

I think
t' = γ(t-vx) is simpler than
$t' = \frac{(t-x)*\sqrt{\frac{1-v}{1+v}} - x*\sqrt{1-v^2}}{1+v}$

You've already determined gamma to be 1.25 but I'll go through the motions here for anyone else that might be looking on. Remember, v=0.6:

γ = 1/√(1-v²) = 1/√(1-0.6²) = 1/√(1-0.36) = 1/√(0.64) = 1/0.8 = 1.25

Now we calculate the new time coordinate, t', for the new frame from the old coordinates, x=12 and t=0:

t' = γ(t-vx) = 1.25(12-0.6(0)) = 1.25(12) = 15

As you can see, we got the same answer you got but with a lot less effort, or at least, a lot less thinking.

Now it's just as important to obtain the x-coordinate which you did not do:

x' = γ(x-vt) = 1.25(0-0.6(12)) = 1.25(-7.2) = -9

Forgot that. I would have calculated x from my graphic equation $x = vt = -0.6 * 15 = -9$ But later, I'll do it with Lorentz boost. in x direction.

Can you please start using the Lorentz Transformation?

I wouldn't be in Physics Forum if I had known Lorentz before. Thanks, I'll use them later. But I think this is important for me. Because it is geomatrically correct. I'm doing it in cartesian system. Because I have to be sure, how to calculate any points based on simpler rule.
1. Hypotenuse in 4D $\sqrt{(ct)^2-x^2-y^2-z^2}$
2. Lorentz factor $\gamma = \frac{1}{\sqrt{1-v^2}}$

This doesn't make any sense. The Blue and Red lines lie in all frames.

This, I don't understand. Perhaps later.

I've asked you several times, where is W? is he stationary in one of the frames? If so, which one and what are his coordinates and state of motion? Why have you introduced him anyway?

W is at rest while B and R are moving 0.6c to the "West"
W I think is in 0,0. W is the green line.

Whatever B and R see in one frame, they see in all other frames. For example, B sees the light pulse being emitted when his clock is at zero in both frames. B sees the reflected light pulse when his clock is at 12 in both frames. R sees the reflection occur when his clock reads 6 in both frames. Different frames do not change anything that any observer sees or measures.The Red clock starts first in the moving frame. Both clocks start at the same time in the stationary frame.

I'll contemplate it.
Thanks gwhellsjr for your invaluable effort and insight.

 

[Stephanus]

Dear PF Forum,
Can someone explain to me?

Let's take a scenario where an observer sends out a light pulse to a reflector 6 feet away and measures with his clock how long it takes for the reflection to get back to him. Since light travels at 1 foot per nanosecond, it will take 12 nsecs for him to see the reflection and he will validate that the light is traveling at 1 foot per nsec for the 12 feet of the round trip that the light takes. If he is following the precepts of Special Relativity, he will define the time at which the reflection took place at 6 nsecs according to his clock and I have made that dot black:

In this diagram, the observer is shown as the thick blue line with dots marking off one-nanosecond increments of time according to his clock. Since he is at rest in this Inertial Reference Frame (IRF), the horizontal grid lines align with his clock and the vertical grid lines align with his measurement of distance.

The reflector is shown as the thick red line with similar dots. The thin blue line represents the pulse of light that the observer emits and the thin red line represents the reflection.

Now let's transform the coordinates of all the dots in the above diagram to a new IRF moving at 0.6c with respect to the original IRF:

At 0.6c, gamma is 1.25 and you will note that the dots are placed 1.25 nsecs apart as defined by the gird lines. Also, the distance to the reflector is contracted to 4.8 feet (6/1.25) as defined by the grid lines. You will also note that the light pulse and its reflection continue to travel at 1 foot per nanosecond but it takes less time in this IRF (3 nsecs) for the light to reach the reflector and more time (12 nsecs) for the reflection to get back to the observer. The total Coordinate Time is 15 nsecs and the total Coordinate Distance the light traveled is 15 feet so the velocity is maintained at 1 foot per nanosecond.

Here is a diagram that depicts the parameters that would apply if we just took the Length Contraction and Time Dilation factors into account:

Note that we would be using the length at the start of the scenario and the time as defined by the observer as when the reflection occurred to establish where the light reached at the time of the reflection which is way off base and would result in a contracted velocity as shown by the extra thin blue line. Instead, we need to use the Coordinate Distance of where the reflector has moved to when the light pulse reaches it and the Coordinate Time at which this occurred, not the time that the observer defines the reflection to occur according to his clock.

BR is the frame where Blue line and Red line lies.

This doesn't make any sense. The Blue and Red lines lie in all frames.

Why it doesn't make sense?
What does "The Blue and Red lines lie in all frames" mean?
Thanks.

 

[Ibix]

Each of the diagrams is a representation of a frame of reference. The red and blue lines are manifestly in both. You could transform to another frame and draw a third diagram, and the red and blue lines would be there too.

The point is that a frame of reference is just a different choice of point of view. It's closely analagous to turning a map upside down because you happen to be facing southwards. Things on the map don't appear and disappear as you do it, they just change place.

I'm not sure of the context of your comment, but you probably meant something like "BR is the frame where the red and blue lines are vertical". That's the rest frame of the red and blue observers.

 

[Stephanus]

Each of the diagrams is a representation of a frame of reference. The red and blue lines are manifestly in both. You could transform to another frame and draw a third diagram, and the red and blue lines would be there too.

The point is that a frame of reference is just a different choice of point of view. It's closely analagous to turning a map upside down because you happen to be facing southwards. Things on the map don't appear and disappear as you do it, they just change place.

I'm not sure of the context of your comment, but you probably meant something like "BR is the frame where the red and blue lines are vertical". That's the rest frame of the red and blue observers.

You mean BR is in TD,LC,ConstantC1.PNG, TD,LC,ConstantC2.PNG and TD,LC,ConstantC3.PNG. Yes they are.
It's my English. I tought ghwellsjr meant, the frame exists in space time diagram, Up and down, left - right, far left - far right, everywhere in the space diagram. Okay,.. it's an English problem. Thanks for making it clear for me to understand ghwellsjr answer.

 

[Ibix]

Langauge is very important in relativity, mainly because it's a complex topic and quite subtle errors of language can mean something completely different. Having less-than-perfect English isn't a problem, but you will need to use the language precisely.

A frame of reference is just a choice of point of view. Once you've chosen a frame, there are a natural set of coordinates to use (at least in SR), and you can draw a space-time diagram from that perspective. You can use the Lorentz transforms to switch to another frame's natural set of coordinates, and draw the diagram from a that perspective.

For example, when drawing maps we typically choose the frame of reference in which North appears as straight up. But there's nothing stopping us picking a frame where West is straight up, or where 037° is straight up. Here's a map of mainland Britain drawn from each of those three reference frames:

I marked London with a red dot; it has different coordinates in each map - something like (1.5,-4.25), (-4.25,-1.25) and (4.9,-1.5). I could write down a mathematical relationship between the coordinates of any point in one map and its coordinates in another; that would be the way to transform coordinates between those two frames.

Similarly, ghwellsjr has picked two different frames of reference and drawn a map of the x-t part of space-time as seen from each one. He could write down a mathematical relationship between the coordinates of any event in one diagram and its coordinates in another - and that would be the Lorentz transforms (in fact, he knows the transforms and used them to generate the maps, but working from the maps to deduce the transforms is - conceptually, at least - what Einstein did).

NB: Map is CC attribution/share-alike licensed - please see here for the original, with details of licencing and attribution.