Length contraction and direction of tavel

[Dale]

I don't get the difference between the 2 methods.

OK, so let me show you how time dilation is derived.

First, we start with the Lorentz transform (for simplicity I will use units where c=1: $t'=\gamma(t-vx)$. Now, consider two events and calculate the differences in coordiates, eg $\Delta x =x_2-x_1$, then clearly $\Delta t'=\gamma(\Delta t-v\Delta x)$. Now, if $\Delta x=0$ then this simplifies to the time dilation formula $\Delta t'=\gamma \Delta t$.

So the critical difference is that in the case where the time dilation formula is valid you must have $\Delta x=0$, and in the case of light $\Delta x$ is never 0. Therefore, the formula to calculate $\Delta t'$ has an extra term related to $\Delta x$ which must be accounted for and failure to do so results in the discrepancies that you have posted.

 

[Amr Elsayed]

in the case of light Δx\Delta x is never 0

I think this helps, but in my case, delta X is not worth zero because of length contraction since we will not agree about length of train??

 

[harrylin]

[..] I guess it's 300000 km/s not 30000 km/s , this will give different time

Not so: c - 0.9c = 0.1c = 30000 km/s. See also my clarification next to it. For the calculation you must therefore use c-v and c+v, as I did, or get with trial and error to the right amount .It's also explained here:
https://en.wikipedia.org/wiki/Michelson–Morley_experiment#Light_path_analysis_and_consequences

I thought synchronized clocks are on time for their perspective " train" .

Exactly, that's what I described: the 2 train seconds are divided in half. Exactly as in §1 of http://www.fourmilab.ch/etexts/einstein/specrel/www/

If not, I then know they are not on time, why I use them then, or this difference between clocks is not done by me, and clocks on same train with same velocity were on time for first time ??

The train observer assumes that light moves at the same speed in both directions relative to the train (he does not have to do that, but he can do so).

 

[Dale]

I think this helps, but in my case, delta X is not worth zero because of length contraction since we will not agree about length of train??

$\Delta x$ is not zero in either frame because light is not at rest in either frame. Think about what the variable means physically. Since x is the position coordinate then $\Delta x =0$ means that the position does not change. A position which does not change is what defines being "at rest". So $\Delta x=0$ means at rest, light is not at rest, so for light $\Delta x \ne 0$

 

[Amr Elsayed]

because light is not at rest in either frame

I thought X was distance moved. So, to apply simplified formula for dilation we need an object that is under C speed limit as for it delta x is zero and time is passing for us ??
I will try to apply the non-simplified formula on my example

4.3589 + 0.2294 s = 4.588 s. That total time is measured by the train observer as 4.588 s / 2.294 = 2.000 s.
The train observer next sets the time of the front clock by sending a light signal from rear to front; he sets the front clock to read 1s more at reception than the rear clock was at the time of sending

do you mean it's time to travel from back to front and return ?? If so. this doesn't tell that C is constant from both directions. It tells that the average velocity is C. You could get velocity for each direction if you calculated the exact distance covered by C. instead of getting the velocity of light that I think the observer on train will measure.
another thing, since the observer who measures is on train. he will doesn't take difference in clocks into account since he sees them synchronized

 

[harrylin]

[..]
do you mean it's time to travel from back to front and return ?? If so. this doesn't tell that C is constant from both directions.

Instead - and it seemed as if you understood that! - the clock synchronization procedure makes the measured speed the same in both directions. By definition. As I put it:
The train observer next sets the time of the front clock by sending a light signal from rear to front; he sets the front clock to read 1s more at reception than the rear clock was at the time of sending. You correctly did so in post #21 - so now I'm puzzled at why you don't understand it next. Could it be that a concept that is clear to you from a signal in the middle, is unclear with a signal from one end? It's the same concept, only a slightly different procedure.
If so, you should perhaps try the calculation with a light signal from the middle, as you did in post #21...

As a reminder, you wrote:
"The 2 guys on the train agree to set their clocks on 5 when they receive the laser. For them, both clocks are on time " pointing to same time" since C must have same speed and it was same distance so it will arrive to both guys at same time from their perspective."

But you seem to have again completely forgotten that, as now you wrote:
" this doesn't tell that C is constant from both directions."
Indeed, length contraction and time dilation are irrelevant for that, as we told you many times - and as you even understood in your post 21. Length contraction and time dilation tell us what he will find for the total time for light to make a round trip. That should be 2 seconds and it is 2 seconds as you saw. Next, clock synchronization takes care of the one-way speed of light, just as you understood in post 21.

It tells that the average velocity is C. You could get velocity for each direction if you calculated the exact distance covered by C. instead of getting the velocity of light that I think the observer on train will measure another thing, since the observer who measures is on train.

Once more: if he divides, as he must do, 2 seconds by 2, he can only find 1 second for the time each way because he assumes for his clock synchronization that the speed of light is the same in both directions. He can then only find 300000 km /1 s for one way, and 300000 km / 1s for the other way. Because it is 6000000 km / 2s both ways and he sets the clock so as to make the speed of light the same in both directions..

he will doesn't take difference in clocks into account since he sees them synchronized

That is the synchronization according to him! The whole point is that he assumes that he is in rest, and so he uses the method of §1 of http://fourmilab.ch/etexts/einstein/specrel/www/
One last time: he simply follows the procedure of §1: "we establish by definition that the “time” required by light to travel from A to B equals the “time” it requires to travel from B to A."

 

[Amr Elsayed]

You correctly did so in post #21 - so now I'm puzzled at why you don't understand it next

Later I thought I was wrong about it, because beams of laser will not strike both guys at same time according to them. The events of lighting lasers are not simultaneous according to the train frame. The difference is : if both guys are at same place as i said in #21 each one will see the other's watch as he exactly does. and so for the 2 guys they will not agree about time from their perspective. The other process: guys are away from each other so it will take some time for the clock at the back to reach the guy in front

That is the synchronization according to him!

In deed, yes but you said the we would calculate the second that light needs to travel at... I think I didn't get what you meant. If I'm one guy in front and my friend's clock at back is 1 more second, then I shall observe no difference, and If I'm measuring difference in clocks to calculate C then I will not take this 1 more second into account since I don't observe it and I'm the one who is measuring.

 

[Amr Elsayed]

Did you say that the front watch will be 1 sec more according to the guy there to avoid the difference due to light movement ? which means I am as the guy on front will not see both clocks on time ??

 

[harrylin]

Later I thought I was wrong about it, because beams of laser will not strike both guys at same time according to them.

You were right, and for one instant you understood it...

The events of lighting lasers are not simultaneous according to the train frame.

For really the last time, this is how they synchronize their clocks:
""we establish by definition that the “time” required by light to travel from A to B equals the “time” it requires to travel from B to A."

Please tell me, how can they not find that the “time” required by light to travel from A to B equals the “time” it requires to travel from B to A, after they set their clocks such, that, as you put it, "the guys set their clocks on 5 when they receive the laser [from the middle of the train] ??

[..] I think I didn't get what you meant. If I'm one guy in front and my friend's clock at back is 1 more second, then I shall observe no difference, and If I'm measuring difference in clocks to calculate C then I will not take this 1 more second into account since I don't observe it and I'm the one who is measuring.

Sorry I cannot follow what you are saying. Are you applying the clock synchronization procedure of §1?

You replied: "Indeed, yes [That is the synchronization according to him!]"
If so, you will find the same as me.
It's really simple, secondary school stuff. for the people in the train who think that the train is in rest:
t1 + t2 = 2s
By definition (necessarily, because they think to be in rest): t1 = t2 = ½ (t1 + t2)
=> t1 = t2 = 1s

There is nothing more to it than that simple calculation.

 

[harrylin]

Did you say that the front watch will be 1 sec more according to the guy there to avoid the difference due to light movement ? which means I am as the guy on front will not see both clocks on time ??

I don't understand what you mean, but please look at what I said and asked you to do in post #20.
Perhaps you did not yet try to do it?

 

[Amr Elsayed]

I didn't actually. I didn't even read it and I don't know why or how, but If I see 1 sec difference between my watch and the other one at back" after I modified it with your step" then a guy on platform will not measure this difference ? what's next ? When I measure light speed I shall take this second into account, but again since it's same person measuring I then don't have to take it into account . I just want difference in time according to me.
I would be grateful If you tell me what exactly I will measure and how as a one guy in front of moving train. and light is striking me from outside not from train in sth like brief scenario

 

[harrylin]

I didn't actually. I didn't even read it and I don't know why or how, but If I see 1 sec difference between my watch and the other one at back" after I modified it with your step" then a guy on platform will not measure this difference ? what's next ? When I measure light speed I shall take this second into account, but again since it's same person measuring I then don't have to take it into account . I just want difference in time according to me.
I would be grateful If you tell me what exactly I will measure and how as a one guy in front of moving train. and light is striking me from outside not from train in sth like brief scenario

Sorry, your questioning exhausted me, and I already gave all calculations and explanations that you need. More useful if you re-read this thread and try to get more out of it.

I don't remember if I put it to your attention, but we simply repeated here what is written and explained at the end of §2 of http://fourmilab.ch/etexts/einstein/specrel/www/ (starting with "We imagine further")

The essential calculations about clock synchronization are primary school stuff; it's therefore not much to do with calculations, it's conceptual understanding. And for a brief moment in time you had it.

For low speeds you can approximate the Lorentz transformations as follows, and it can be useful to sketch the time transformation along the X and X' axes as I suggested in post 20:
x' ≈ x - vt
t' ≈ t - vx/c²

PS: most likely you need just a few days to let the concept of "relativity of simultaneity" sink in. I recall that this was the case for me, when I finally "got" it!

 

[Amr Elsayed]

Thank you for help and care. you have been explaining in 4 pages now. I will search again about simultaneity and synchronized clocks, and I hope it will sink in.
Thanks once again
regards

 

[Stephanus]

Thank you for help and care. you have been explaining in 4 pages now. I will search again about simultaneity and synchronized clocks, and I hope it will sink in.
Thanks once again
regards

I think you should start from the beginning Amr Elsayed. But I don't know either where the "beginning" is.
I have asked about
Twin Paradox,
Twin Paradox and asymmetry
Twin Paradox and symmetry
Lorentz and Doppler
Motion in Space
Universe frame of reference.
Still I haven't reach my destination, But I know, I'm on the right track based on their answers.
Janus clip have helped me much. Two clips actually he gave me.

1. About synchronizing clocks with and without the observer moving.
https://www.physicsforums.com/threads/twin-paradox-and-asymmetry.816455/
You can find his clips somewhere from that thread.
I think, if an amateur like me understand (or at least grasp), perhaps you can understand more.
From his answer, I think I can understand that simultaneity events are affected by relativity.

2. And in the train clip above, I think I know WHY there has to be length contraction. With out this, the signal, from the train point of view, can't reach the observer at the same time. About the HOW, well, still don't understand. Yet. Or, still haven't calculated it yet. I think it somehow has something to do with Lorentz Transformation.
And that's why I create a new thread about Lorentz Transformation before I make some calculations.
And length contraction leads to affected simultaneity events.
Just don't be discourage if you don't understand

 

[Mentz114]

@Amr Elsayed

I think what you are trying to do is like proving that a triangle has three sides. The speed of light is a constant in SR by definition just like a triangle has three sides by definition.
If your numerical calculations tell you otherwise it is because you made a mistake. More studying won't help you find the mistake. I won't help you find the mistake.

It is easy to show from a spacetime diagram that in SR inertial frames the distance covered by light is always equal to the time taken. Always. Ever and ever.

Why don't you accept this ? It seems to me you won't believe it and have some reason that it must be wrong.

 

[Janus]

But let me explain what exactly I mean. suppose there is a spacecraft moving at 0.9C and at an exact moment it is far 3*10^8 meters from a space station. craft is moving away from station .At that moment the station lights a beam into the space craft. The distance between the craft and the station measured by the moving craft is 1307669683 using length contraction. and since C must be 3*10^8, then time for light beam to reach the craft must be 0.4358898944 second from craft's perspective. This means It should take 1 second from perspective of station according to time dilation, but in deed because the craft is moving away by 0.9C it should take 10 seconds for light beam to reach craft from station's perspective not 1 second. I feel confused about it

Again, your problem here is assuming that "at the same time" means the same for both the spacecraft and station.

To illustrate we add clocks to both the spaceship and station and assume that the spacecraft passed the station and at that moment they both set their clocks to 0

Now when the ship is 1 light sec from the station ( as measured by the station), the station sends its signal. At this moment, the station clock reads 1.111 sec. According to the station, the spaceship clock now reads 0.4843 sec because of time dilation. The signal takes 10 sec by the station clock to reach the spaceship, during which time the spaceship clock advances ~4.359 sec, which when added to the 0.4843 sec, means that the ship clock reads 4.843 sec when the light arrives.

Here's how things happen according to the spaceship. When the spaceship is the 0.4359 light sec away from the station( as measured by the ship) you mentioned above, its clock reads 0.4843 sec, however the station clock, due to time dilation only reads ~0.2111 secs. In other words, it hasn't yet reached the time it needs to read when it sends the signal . The ship must wait until its clock reads 2.549 before the clock on the station reads 1.111 sec and sends the signal. At which time, it will be 2.294 light seconds from the station. it will take the light 2.294 sec to travel the distance between them, meaning the ship clock reads 4.843 second when the light arrives, Which is the same answer we got according to the station.

So in order for both the ship and station to agree that the light left the station when its clock read 1.111 sec and arrived at the ship when its clock reads 4.843 sec, both the ship and station have to measure the light as traveling at c relative to themselves.

 

[Dale]

I thought X was distance moved.

No. In the Lorentz transform x is position in some inertial frame. If $x_1$ is the position at the beginning and $x_2$ is the position at the end, then $\Delta x=x_2-x_1$ is the distance moved.

 

[harrylin]

[..]

1. About synchronizing clocks with and without the observer moving.
https://www.physicsforums.com/threads/twin-paradox-and-asymmetry.816455/
You can find his clips somewhere from that thread.
I think, if an amateur like me understand (or at least grasp), perhaps you can understand more.
From his answer, I think I can understand that simultaneity events are affected by relativity.
[..]

Ah yes, you mean this one: #9
That's a very clear animation by Janus!

 

[Amr Elsayed]

So in order for both the ship and station to agree that the light left the station when its clock read 1.111 sec and arrived at the ship when its clock reads 4.843 sec, both the ship and station have to measure the light as traveling at c relative to themselves.

That's good. I got it. you exactly understood my problem and had a good way to explain it. Thanks for help, but I still have some wonders
For a laser being shot from the other direction " toward the ship" assuming our clocks were on time then I started to accelerate to reach 0.9C should I see the laser being shot after still observers do or before. I am asking because I don't know what acceleration has to do with time flow of ship clock might affect it, and I had to in order to make sure that both clocks were on time.

I think, if an amateur like me understand (or at least grasp), perhaps you can understand more.

Just don't be discourage if you don't understand

I really want to thank you, you deserve respect, but If there is an amateur here it's the guy that just is done with his first year at high school "me"
I am not discourage and I learned to be more specific while asking to get maximum help ,because If someone else understands your problem and give you what you need, you simply get it

Why don't you accept this ? It seems to me you won't believe it and have some reason that it must be wrong.

Actually, this never was the purpose of the question. I'm sure light always goes at C , but I wanted to know what nature does to keep it and it was time dilation and length contraction and simultaneity. And It did sink into some extent :)

 

[Stephanus]

Sorry, can I ask a question here?
In one of the Lorentz Transformation formula
http://en.wikipedia.org/wiki/Lorentz_transformation#Boost_in_the_x-direction
Instead of using x and t, I use x_a and t_a and instead of using x' and t' I use x_b and t_b
I use only x-axis here.
$x_b = \gamma (x_a - vt_a)$

No. In the Lorentz transform x is position in some inertial frame. If $x_1$ is the position at the beginning and $x_2$ is the position at the end, then $\Delta x=x_2-x_1$ is the distance moved.

$\text{is }x_a = \Delta x$?

 

[Stephanus]

@Amr Elsayed

I think what you are trying to do is like proving that a triangle has three sides. The speed of light is a constant in SR by definition just like a triangle has three sides by definition.
If your numerical calculations tell you otherwise it is because you made a mistake. More studying won't help you find the mistake. I won't help you find the mistake.

It is easy to show from a spacetime diagram that in SR inertial frames the distance covered by light is always equal to the time taken. Always. Ever and ever.

Why don't you accept this ? It seems to me you won't believe it and have some reason that it must be wrong.

Yeah, and it is that "ever and ever" that causes everything else to come up. Time dilation, length contraction, simultaneity of events, time machine idea, twin paradox and many things.

 

[Dale]

$\text{is }x_a = \Delta x$?

No. As you said earlier your $x_a$ is everyone else's $x$

Instead of using x and t, I use xa and ta

 

[Stephanus]

No. As you said earlier your $x_a$ is everyone else's $x$

Is X_a a coordinate or distance?
Or is X_a = X₁ and X_b = X₂ in your example?
$X_b = \gamma(x_a - vt_a)$

 

[harrylin]

[..]
For a laser being shot from the other direction " toward the ship" assuming our clocks were on time then I started to accelerate to reach 0.9C should I see the laser being shot after still observers do or before. I am asking because I don't know what acceleration has to do with time flow of ship clock might affect it, and I had to in order to make sure that both clocks were on time.

[emphasis mine]
I 'm not sure to understand your question, but still I think that I can give an answer - see next!

I wanted to know what nature does to keep it and it was time dilation and length contraction and simultaneity. And It did sink into some extent :)

Glad to hear that it is starting to sink in.

However, as I told you before, nature takes care of time dilation and length contraction, but nature does not take care of simultaneity as defined by Einstein.

Apparent simultaneity is taken care of by us, by means of clock synchronization. And if you accelerate to a different state of motion without touching your clocks, then, if you have very good clocks, you can detect a difference between the speed of light in forward and backward directions. You will have to do a new synchronization if you want to use your differently moving system as a standard reference system, in which light appears to propagate at c in all directions.

 

[Dale]

Is X_a a coordinate or distance?
Or is X_a = X₁ and X_b = X₂ in your example?
$X_b = \gamma(x_a - vt_a)$

You are the one who defined X_a so you should know! You defined x_a as the same as what everyone else uses for x, and everyone else uses x as a coordinate not a distance. So x_a is defined by you as a coordinate, not a distance.

Most people use primes (e.g. x vs x') to denote different frames. Most people use subscripts to denote different events. So most people would use x_a to denote the x coordinate of event "a". You have defined it differently, which is your prerogative, but I don't understand why you keep asking other people what is meant by the variables that you have defined. You defined it, so you should know.

x_a is NOT the same as x₁ and x_b is NOT the same as x₂. I used subscripts to denote different events, you used subscripts to denote reference frames. You yourself defined them!

 

[Stephanus]

I really want to thank you, you deserve respect, but If there is an amateur here it's the guy that just is done with his first year at high school "me"
I am not discourage and I learned to be more specific while asking to get maximum help ,because If someone else understands your problem and give you what you need, you simply get it

Oh, I just remember something
"A good theory must be able to be explained to a six years old" - Albert 1879 - 1955.
If you can't understand, perhaps you're not six years old?

 

[harrylin]

Is X_a a coordinate or distance?
Or is X_a = X₁ and X_b = X₂ in your example?
$X_b = \gamma(x_a - vt_a)$

Maybe the following helps.

Apparently you use the index a for system S, and the index b for system S'. However that will quickly be confusing because many people designate points with a, b etc. So I will use modern standard notation instead: x, t etc. relate to "stationary" system S, and x', t' etc. relate to "moving" system S'.

$x' = \gamma(x - vt)$

[EDIT:] Therefore, for points a and b on the X-axis (they are also points on the X' axis):

$x'_b = \gamma(x_b - vt_b)$
$x'_a = \gamma(x_a - vt_a)$
---------------------------------- - (subtraction)
$x'_b - x'_a = \gamma(x_b - x_a - v(t_b - t_a))$

For measurements of those points at the same time according to system S, t_b - t_a = 0.

Then we obtain for t_a=t_b:
$\Delta x' = x'_b - x'_a = \gamma(x_b - x_a)$

BTW, it is enlightening to figure out what you find for t'_a=t'_b

 

[Stephanus]

You are the one who defined X_a so you should know! You defined x_a as the same as what everyone else uses for x, and everyone else uses x as a coordinate not a distance. So x_a is defined by you as a coordinate, not a distance.

Most people use primes (e.g. x vs x') to denote different frames. Most people use subscripts to denote different events. So most people would use x_a to denote the x coordinate of event "a". You have defined it differently, which is your prerogative, but I don't understand why you keep asking other people what is meant by the variables that you have defined. You defined it, so you should know.

X_a is NOT the same as X₁ and X_b is NOT the same as X₂. I used subscripts to denote different events, you used subscripts to denote reference frames. You yourself defined them!

Okay, perhaps you would tell me what X is in Lorentz equation.
http://en.wikipedia.org/wiki/Lorentz_transformation#Boost_in_the_x-direction
I copy it literally
$t' = \gamma(t-\frac{vx}{c^2})$
$x'=\gamma(x-vt)$
What is X in this context?
Position/Coordinate?
Length?
And what is $x'=\gamma(x-vt)$ in English?
'His' position in my frame would be x' if I'm in x times gamma?

 

[Stephanus]

Apparently you use the index a for system S, and the index b for system S'

Yes, yes, that's right.
What are you? A language teacher?

Therefore, for points a and b on the X_a-axis (they are also points on the X_b axis):

$x'_b = \gamma(x_b - vt_b)$
$x'_a = \gamma(x_a - vt_a)$
---------------------------------- - (subtraction)
$x'_b - x'_a = \gamma(x_b - x_a - v(t_b - t_a))$

Thank you, thank you

 

[Stephanus]

Therefore, for points a and b on the X_a-axis (they are also points on the X_b axis):

"Therefore, for points a and b on the X-axis (they are also points on the X' axis)" in "Modern standard notation" you mean?

 

[harrylin]

"Therefore, for points a and b on the X-axis (they are also points on the X' axis)" in "Modern standard notation" you mean?

Ah thanks for spotting that - I had started answering in your notation but changed my mind. Corrected now.

 

[PWiz]

Okay, perhaps you would tell me what X is in Lorentz equation.
http://en.wikipedia.org/wiki/Lorentz_transformation#Boost_in_the_x-direction
I copy it literally
$t' = \gamma(t-\frac{vx}{c^2})$
$x'=\gamma(x-vt)$
What is X in this context?
Position/Coordinate?
Length?
And what is $x'=\gamma(x-vt)$ in English?
'His' position in my frame would be x' if I'm in x times gamma?

x is the coordinate in S frame, whereas x' is the coordinate in S' frame (NOT distance). Your final equation enables you to find the position coordinate in frame S' (which is denoted by x') using the position coordinate and time from frame S (which are denoted by x and t). It roughly the answer to the question "if that ball is recorded to have position coordinate x at time t in frame S, what will be that ball's position coordinate in frame S' ?".

 

[Stephanus]

x is the coordinate in S frame, whereas x' is the coordinate in S' frame (NOT distance). Your final equation enables you to find the position coordinate in frame S' (which is denoted by x') using the position coordinate and time from frame S (which are denoted by x and t). It roughly the answer to the question "if that ball is recorded to have position coordinate x at time t in frame S, what will be that ball's position coordinate in frame S' ?".

Thank you, thank you.

 

[Amr Elsayed]

I 'm not sure to understand your question, but still I think that I can give an answer - see next!

I wanted to see how simultaneity works in the other case " light goes toward the moving ship" . Briefly I was asking if I shall also see the light emission after still observers do knowing that they were as distant as me when light was emitted . If our clocks were on time before and due to time dilation I then should see light emission after they do since light emission is related to a specific time at the still clock of station . I'm not sure If my synchronization to set clocks on time first is right and I'm not sure if should see light before still observers does or after

Glad to hear that it is starting to sink in.
However, as I told you before, nature takes care of time dilation and length contraction, but nature does not take care of simultaneity as defined by Einstein.

Thank you
I agree with you that we take care of simultaneity

Oh, I just remember something
"A good theory must be able to be explained to a six years old" - Albert 1879 - 1955.
If you can't understand, perhaps you're not six years old?

Yeah Good point

 

[Stephanus]

This is 'Janus' train
Let
A1: Front Mark
A2: Bow of the Train, Front
B1: Rear Mark
B2: Aft of the Train, Back
M: Mid point between A1 and B1
D: Distance between A1 and M, let say D = 1.6 Light second
E: Distance between A1 and B1 = 2 * D = 3.2 lt
V: Velocity of the railway = 0.6c
L: is the length of the train = ...??
Yes! The railway is moving. The train stops. I don't know if this makes sense in real world.

Everything is in Train Frame.
Okay...
WHEN A2 meets A1, the train bow flashes a signal
Let V_a = c-V = 0.4c
T_a = D/V_a = 4 sec
The signal will reach M in 4 sec

WHEN B2 meets B1, the train aft flashes a signal
Let V_b = c+V = 1.6c
T_b = D/V_b = 1 sec

So B2 shouldn't meet B1 at the same time A2 meets A1, right? There's simultaneity different here
In fact B2 should 'wait' for 3 seconds or the length of the train should be longer than A1-B1, there's length contraction here
Additional train length is 3 seconds * V
So
L = 3 seconds * V + E
$L = (T_a - T_b)V + E$

$L = (\frac{D}{c-V}-\frac{D}{c+V})*V+E$

$L = (\frac{Dc+DV-(Dc-DV)}{c^2-V^2})*V+E$

$L = (\frac{2DV}{c^2-V^2})*V+E$

$E = 2D$, so

$L = (\frac{EV^2}{c^2-V^2})+E$

If we substitute C with 1 and V is a factor of C, so

$L = (\frac{EV^2}{1-V^2})+E$

$\frac{L}{E} = (\frac{V^2}{1-V^2})+1$
$\frac{L}{E} = (\frac{1}{1-V^2})$
So, the railway length is 3.2 lt and the train length is 5 lt for V = 0.6c
That way, 5 lt - 3.2 lt =1.8 lt
If V = 0.6 then, it takes 3 seconds for the back of the train to reach 1.8 lt. And flashes a signal which hit M at the same time the bow signal hits M
It makes sense,
But...
Why
$\gamma = \frac{1}{\sqrt{1-V^2}}$?
not this one
$\text{ratio} = \frac{1}{1-V^2}$
Do I mistakenly make the equation?
But if Lorentz is right, WHICH I KNOW HE IS!, the length of the train will be 4 lt not 5 lt. Doesn't make sense in real world. Or it does?

 

[Stephanus]

Why
$\gamma = \frac{1}{\sqrt{1-V^2}}$?
not this one
$\text{ratio} = \frac{1}{1-V^2}$
Do I mistakenly make the equation?
But if Lorentz is right, WHICH I KNOW HE IS!, the length of the train will be 4 lt not 5 lt. Doesn't make sense in real world. Or it does?

Ahhh, it works both ways. The train and the railway.
The length of the train AT REST is 5 light seconds.
And the length of the railway AT REST is 4 ls.
The ratio doesn't have to be
$\frac{1}{1-V^2}$
But,
$\frac{1}{\sqrt{1-V^2}}$ is enough.
The length of the moving rail is not computed from the train, but from the railway AT REST.
So $4\text{ ls } * \frac{1}{\sqrt{1-v^2}} \text { is } 3.2 \text{ls}$
And for the moving train...
$5 \text { ls } * \frac{1}{\sqrt{1-v^2}} \text { is } 4 \text{ ls}$
I had this answer when I drove at highway watching road milestone, coming back from my father in law house at the country.
But, it's been 24 hours since I post this posting. Why nobody answers?

 

[harrylin]

Ahhh, it works both ways. The train and the railway.
[..]
I had this answer when I drove at highway watching road milestone, coming back from my father in law house at the country.
But, it's been 24 hours since I post this posting. Why nobody answers?

There is a limited amount of people here, with limited free time, and too many topics. I didn't understand your question, and also I don't understand your new insight, except that it looks as if you now for the first time understand that time dilation is mutual. I'm glad that you apparently got it now.

 

[Stephanus]

There is a limited amount of people here, with limited free time, and too many topics. I didn't understand your question, and also I don't understand your new insight, except that it looks as if you now for the first time understand that time dilation is mutual. I'm glad that you apparently got it now.

Sorry, I didn't mean to rush everybody with my (endless) questions .
It's just a rhetoric statement.
Actually I want to know about twins paradox and why there's twins paradox but the universe HAS NO frame of reference.
But to get there, I've been asking about
Motion in Space
Doppler
Lorentz
Relativity
Signal
etc...
But now I know I'm on the right track, but not there, yet.
I want to thank you personally HarryLin for your helps to me all this time.

 

[Ibix]

There is a twin paradox because elapsed time is the "length" of a path through space-time, and the twins take different paths. There's not really any more mystery to it than that.

 

[Stephanus]

There is a twin paradox because elapsed time is the "length" of a path through space-time, and the twins take different paths. There's not really any more mystery to it than that.

Thanks Ibix for your answer.
But I think, it's still far away from my destination.
Now, I just understand how Hendrik discovered this formula.
$\gamma = \frac{1}{\sqrt{1-v^2}}$
I stil want to study why
$t' = \gamma(t-\frac{vx}{c^2})$ and
$x' = \gamma(x-vt)$
through train/platform simulation, by myself just like
https://www.physicsforums.com/threads/length-contraction.817911/#post-5137268And after that I'll fire my (endless) questions (again).

 

[Ibix]

The reason for the differential aging of the twins is due to the difference in length of the space-time paths taken by the two twins. That is all the reason there is. It is your destination if you are attempting to understand the twin paradox scenario. Dismissing it as "far away from your destination" is just daft - why ask the question if you aren't going to listen to the answer?

There are basically two sets of transforms between inertial frames that are consistent with the notion that physics is the same in all of them: these are the Galilean transforms and the Lorentz transforms. The Lorentz transforms are consistent with observation while the Galilean ones are not. So the Lorentz transforms are the right ones (but we were fooled for centuries because we didn't have sensitive enough experiments to spot the errors in the Galilean transforms).

In a Euclidean space, the distance between points is given by Pythagoras' Theorem, and the answer does not dependent on the choice of coordinates. I might say that two points are $\Delta x=L$ meters apart in the x direction and $\Delta y=W$ meters apart in the y direction. You might say that they ae $\Delta x'=W$ meters apart in the x' direction and $\Delta y'=-L$ meters apart in the x direction. We are using coordinates rotated by 90° with respect to each other. However, we will both agree that the distance between the points is $\sqrt{L^2+W^2}$ meters.

The Lorentz transforms imply that space and time are one four-dimensional whole called space-time - but it does not follow Euclidean geometry. In space-time, the equivalent of Pythagoras' theorem is $c\Delta\tau = \sqrt{c\Delta t^2-\Delta x^2-\Delta y^2-\Delta z^2}$. This quantity does not depend on the choice of coordinates - you and I might disagree on $\Delta x$ (that would be length contraction) or $\Delta t$ (that would be time dilation), but we will always come up with the same $\Delta\tau$ for any given path.

It's easy to see that, in your rest frame, $\Delta\tau=\Delta t$, because in your rest frame you are not moving so for you $\Delta x=\Delta y=\Delta z=0$. Someone at rest in one frame is moving in another, though, so in general $\Delta\tau$ is the time experienced by someone moving at constant speed from point A to point B, separated in space by ($\Delta x,\Delta y,\Delta z$), in time $\Delta t$.

It's then easy to see that if I move from an event at (t,x,y,z)=(0,0,0,0) to (t,x,y,z)=(10,0,0,0) (i.e., stay put for ten years) while you travel from (0,0,0,0) to (5,3,0,0) then to (10,0,0,0) (i.e., take five years to get three light years away in the x direction, then turn round and come back) that the $\Delta\tau$s are different - $\sqrt{10^2-0^2}=10$ years for me, $\sqrt{5^2-3^2}+\sqrt{5^2-(-3)^2}=8$ years for you.

I think that's a complete explanation of the twin paradox from top to bottom. Certainly you should learn the maths and be able to derive the Lorentz transforms if you intend to study SR. It will show you that what I have written above is self-consistent. But it will not give you any further insight into the twin paradox - this is the whole of the "why" there is differential aging.

 

[Stephanus]

The reason for the differential aging of the twins is due to the difference in length of the space-time paths taken by the two twins. That is all the reason there is. It is your destination if you are attempting to understand the twin paradox scenario. Dismissing it as "far away from your destination" is just daft - why ask the question if you aren't going to listen to the answer?

There are basically two sets of transforms between inertial frames that are consistent with the notion that physics is the same in all of them: these are the Galilean transforms and the Lorentz transforms. The Lorentz transforms are consistent with observation while the Galilean ones are not. So the Lorentz transforms are the right ones (but we were fooled for centuries because we didn't have sensitive enough experiments to spot the errors in the Galilean transforms).

In a Euclidean space, the distance between points is given by Pythagoras' Theorem, and the answer does not dependent on the choice of coordinates. I might say that two points are $\Delta x=L$ meters apart in the x direction and $\Delta y=W$ meters apart in the y direction. You might say that they ae $\Delta x'=W$ meters apart in the x' direction and $\Delta y'=-L$ meters apart in the x direction. We are using coordinates rotated by 90° with respect to each other. However, we will both agree that the distance between the points is $\sqrt{L^2+W^2}$ meters.

The Lorentz transforms imply that space and time are one four-dimensional whole called space-time - but it does not follow Euclidean geometry. In space-time, the equivalent of Pythagoras' theorem is $c\Delta\tau = \sqrt{c\Delta t^2-\Delta x^2-\Delta y^2-\Delta z^2}$. This quantity does not depend on the choice of coordinates - you and I might disagree on $\Delta x$ (that would be length contraction) or $\Delta t$ (that would be time dilation), but we will always come up with the same $\Delta\tau$ for any given path.

It's easy to see that, in your rest frame, $\Delta\tau=\Delta t$, because in your rest frame you are not moving so for you $\Delta x=\Delta y=\Delta z=0$. Someone at rest in one frame is moving in another, though, so in general $\Delta\tau$ is the time experienced by someone moving at constant speed from point A to point B, separated in space by ($\Delta x,\Delta y,\Delta z$), in time $\Delta t$.

It's then easy to see that if I move from an event at (t,x,y,z)=(0,0,0,0) to (t,x,y,z)=(10,0,0,0) (i.e., stay put for ten years) while you travel from (0,0,0,0) to (5,3,0,0) then to (10,0,0,0) (i.e., take five years to get three light years away in the x direction, then turn round and come back) that the $\Delta\tau$s are different - $\sqrt{10^2-0^2}=10$ years for me, $\sqrt{5^2-3^2}+\sqrt{5^2-(-3)^2}=8$ years for you.

I think that's a complete explanation of the twin paradox from top to bottom. Certainly you should learn the maths and be able to derive the Lorentz transforms if you intend to study SR. It will show you that what I have written above is self-consistent. But it will not give you any further insight into the twin paradox - this is the whole of the "why" there is differential aging.

Thank you Ibix for your answer.
It's not that I don't want to listen to explanations. It's just that sometimes I can't fully understand.
It seems that understanding SR is very difficult, not just like in discovery channel where the host says "..., so the other twin ages slower..."
And since joining this forum, I've learned (or read) about
Worldine, simultaneity events, and now space time path.
It takes time to study all those things.
Yesterday I just understood how Hendrik Lorentz formulated this formula $\gamma = \frac{1}{\sqrt{1-v^2}}$ trough Janus' train simulation.
Perhaps you should know that I have little (if not at all) background in Physics and Math.
And thanks for this answer, too. I'll contemplate it.

 

[Stephanus]

You might say that they ae $\Delta x'=W$ meters apart in the x' direction and $\Delta y'=-L$ meters apart in the x direction. We are using coordinates rotated by 90° with respect to each other.

Did you mean in the y' direction?

 

[Ibix]

Did you mean in the y' direction?

Yes. Typo - sorry.

 

[harrylin]

Sorry, I didn't mean to rush everybody with my (endless) questions .
It's just a rhetoric statement.
Actually I want to know about twins paradox and why there's twins paradox but the universe HAS NO frame of reference.
But to get there, I've been asking about [..]

Hi Stephanus, I notice a glitch in where you want to get with all your questions!

Probably you mean that the universe has no "absolute" frame of reference. I don't know who told you that, but we can only say that WE have no "absolute" frame of reference; we are ignorant if the universe has one. There is no twins paradox if the universe does have an absolute frame of reference. There is also no twin paradox if the universe is what is called a "block universe", and many people like that model more. And there may be other explanations as well. After long discussions and even debates about such models of SR on this forum, discussions in which people argue in favour or against them have been banned*. But the old discussions give enough information, you can search the forum for "block universe".

* See item 11 in this forum's FAQ: https://www.physicsforums.com/threads/relativity-faq-list.807523/

 

[Stephanus]

Hi Stephanus, I notice a glitch in where you want to get with all your questions!

Probably you mean that the universe has no "absolute" frame of reference. I don't know who told you that, but we can only say that WE have no "absolute" frame of reference; we are ignorant if the universe has one. There is no twins paradox if the universe does have an absolute frame of reference. There is also no twin paradox if the universe is what is called a "block universe", and many people like that model more. And there may be other explanations as well. After long discussions and even debates about such models of SR on this forum, discussions in which people argue in favour or against them have been banned*. But the old discussions give enough information, you can search the forum for "block universe".

* See item 11 in this forum's FAQ: https://www.physicsforums.com/threads/relativity-faq-list.807523/

I mean this.
If we sit at the back seat of the plane which flies at 250 m/s and we call a flight attendance in front of us. Our sound will travel at 330m/s from OUR frame in the plane.
But an observer on Earth if somehow he/she can see the sound wave travels, will see that our sound wave is traveling at 580 m/s, right?
Or imagine a sport car running at 200 m/s. It has a device that in a certain time produces an ultrasonic sound (not that its engine is silence). The ultrasonic sound will travel at 330 m/s no matter how fast the car moves. And observer on the podium if somehow can see the sound wave, he/she will see the sound wave travels at 330 m/s. But the driver will see that the sound wave travels at 130 m/s from his point of view/frame. But if the driver is communicating with it's paddock crew, assuming he/she's driving a Ferrari Testarosa, not Ferrari F1, his/her sound will travel at 530 m/s from the ground point of view, but from his/her frame it's 330 m/s.
Light is wave, sound is wave.
Light is affected by doppler effect, so is sound.
But...
If it's about light then...
But an observer on Earth if somehow he/she can see the sound wave travels, will see that our sound wave is traveling at 580 330 m/s, right?

But the driver will see that the sound wave travels at 130 330 m/s from his point of view/frame.

But if the driver is communicating with it's paddock crew, assuming he/she's driving a Ferrari Testarosa, not Ferrari F1, his/her sound will travel at 530 330 m/s from the ground point of view, but from his/her frame it's 330 m/s.
It's always 330m/s, 330m/s, 330m/s ever.
Argghh, these "light" things will surely make me go crazy.
I have to contemplate it slowly, really slow.
It's not that your and every other mentor explanations are blur or unclear, I think this thing have to be self understood. Just like riding a bike. There's no trainer in the world that can teach you how to ride a bike. You have to try it yourself, the trainer can only teach you WHAT you have to do.
So far I know how Hendrik Lorentz formulated $\gamma \text { is } \frac{1}{\sqrt{1-v^2}}$, and why $\gamma \text { has to be } \frac{1}{\sqrt{1-v^2}}$.
The rest I should have absorb slowly.
Thanks.

 

[A.T.]

If it's about light then...

It's not about light vs. sound, but fast vs. slow. The speed of anything that travels at relativistic speed in one frame, will transform in a significantly non- Galilean way to other frames.

 

[Stephanus]

It's not about light vs. sound, but fast vs. slow. The speed of anything that travels at relativistic speed in one frame, will transform in a significantly non- Galilean way to other frames.

Really? Is it not about light only? Because the speed of sound varies, but not light.

 

[A.T.]

Is it not about light only?

All large speeds don't transform to other frames in the Galilean way you assume sound speed does. Even for sound that is just an approximation.

 

[Stephanus]

Speeds slightly below c also don't transform to other frames the way sound speed does.

Are you trying to say that in plane,
when we call the attendants from the back seat,
our sound wave travels at
$\frac{s + p}{1+(s * p)/c^2}$?
s = speed of sound
p = speed of plane
c = maximum speed in the universe, I refrain myself to use the world "light"
And incidently, light travel at the top speed of the universe?