Length contraction and direction of tavel

[Stephanus]

I now know why length is contracted from both front and back. But the problem is how I would measure same speed of light.
What I know is because C must be same, there is time dilation and length contraction. I don't get how for distant moving clocks with same velocity there is a change in speed of time flow but what I got from you Harald is he will set those clocks to run for different speeds to measure same C according to them.

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

Glad to know that I'm not the only one

 

[Stephanus]

Simultaneity of events!

[PLAIN]http://home.earthlink.net/~jparvey/sitebuildercontent/sitebuilderpictures/trainsimul2.gif[/QUOTE]

Good, very Good, Janus.
I remember once, some one sent me a clip about simultaneity of events using clock synchronizations.
Using two clock and one observer. Stays and moves.
Was that you?

Now I understand that there's another thing to study to understand relativity: Simultaneity of events

 

[PWiz]

@OP There is a lot of confusion over here. To understand length contraction, you should first understand the two postulates of SR, and then understand the derivation behind the lorentz transformation (or boost, as its called in a trivial application). Time dilation and length contraction follow from lorentz transformation. Trying to find the speed of light from time dilation and other formulae would be circular, because the formulae themselves are derived on the basis that the speed of light in all inertial frames of reference in a vacuum is invariant.

I suggest that you look up the derivation of the Lorentz transformation first to clear up your confusion, and then ask any subsequent questions that come to your mind. Most people get confused in relativity when they start learning it (that included me) because they don't tackle the concepts in the correct order (and many textbooks don't help either, such as introducing spacetime intervals first without any proof and then deriving the lorentz transformation from it, which is actually backwards). Always start with the most fundamental postulate which agrees with empirical evidence and then climb up the ladder.

 

[Stephanus]

@OP There is a lot of confusion over here. To understand length contraction, you should first understand the two postulates of SR, and then understand the derivation behind the lorentz transformation (or boost, as its called in a trivial application). Time dilation and length contraction follow from lorentz transformation. Trying to find the speed of light from time dilation and other formulae would be circular, because the formulae themselves are derived on the basis that the speed of light in all inertial frames of reference in a vacuum is invariant.

I suggest that you look up the derivation of the Lorentz transformation first to clear up your confusion, and then ask any subsequent questions that come to your mind.

Janus' train is a very good example.
Is the Lorentz transformation derived from this example? The railway somehow has to be contracted at the picture below. If not, then the train will think he's chasing light.
There's also a simultaneity of events as the front train hits the trigger first, compared to the to picture. The front and the back of the train hit the trigger at the SAME time.

 

[PWiz]

Janus' train is a very good example.
Is the Lorentz transformation derived from this example? The railway somehow has to be contracted at the picture below. If not, then the train will think he's chasing light.
There's also a simultaneity of events as the front train hits the trigger first, compared to the to picture. The front and the back of the train hit the trigger at the SAME time.

I suppose one could "informally" derive the lorentz transformation using that example, but it's much better to use math for a general situation and then find examples which can validate your results.
Wikipedia has a good derivation, so I recommend that you have a look at it if it helps you: http://en.m.wikipedia.org/wiki/Derivations_of_the_Lorentz_transformations

 

[Stephanus]

I suppose one could "informally" derive the lorentz transformation using that example, but it's much better to use math for a general situation and then find examples which can validate your results.
Wikipedia has a good derivation, so I recommend that you have a look at it if it helps you: http://en.m.wikipedia.org/wiki/Derivations_of_the_Lorentz_transformations

I didn't mean math 'derivatve'
I mean that Lorentz Transformation can be based from that train example.
Based, created, postulated, derived.

 

[PWiz]

I didn't mean math 'derivatve'
I mean that Lorentz Transformation can be based from that train example.
Based, created, postulated, derived.

The motivation behind Lorentz transformation and SR came from Maxwell's equations, which describe light as an electromagnetic wave which can propagate without a medium and has a constant value in vacuum $c=\frac{1}{\sqrt{{\mu}_0 ε_0 }}$ .
When one quotes the speed of a wave, it is always with reference to the medium in which it propagates. This cannot be applied to light traveling in a vacuum. Only two plausible explanations can them exist:
1) There is a medium spread throughout the universe in which light propagates (the aether), and light speed is relative to this medium.
2) Light travels at the same speed with respect to all inertial observers. There is no aether.

The Michelson-Morley experiment strongly suggested that the aether does not exist, and hence we get the first postulate of SR (the 2nd case).
Although Lorentz initally derived his equations in an effort to preserve the existence of the "aether" strikethrough](a stance which he gave up after the MM experiment)[\strikethrough], his equations work correctly in the framework of SR as well.

 

[harrylin]

[..] Wikipedia has a good derivation, so I recommend that you have a look at it if it helps you: http://en.m.wikipedia.org/wiki/Derivations_of_the_Lorentz_transformations

Ah yes that's nice! Perhaps that one is understandable for amr.

 

[harrylin]

[..] The Michelson-Morley experiment strongly suggested that the aether does not exist, and hence we get the first postulate of SR (the 2nd case).
Although Lorentz initally derived his equations in an effort to preserve the existence of the "aether" (a stance which he gave up after the MM experiment), his equations work correctly in the framework of SR as well.

MMX may suggest that to you, but not to Lorentz and he never changed his stance. Why did you think he did?

 

[Stephanus]

Ah yes that's nice! Perhaps that one is understandable for amr.

A very smart amateur

 

[PWiz]

MMX may suggest that to you, but not to Lorentz and he never changed his stance. Why did you think he did?

Hmm, you're right, I just noticed it. I must've come across some bogus science history site while briefly searching up the history of relativity

 

[PWiz]

Ah yes that's nice! Perhaps that one is understandable for amr.

I can't tell if that's sarcastic or not, lol.

 

[harrylin]

[..] I did some real calculations, and I got C to be about 275000 kilometers per second. I hope you will tell me where I made a mistake in the following calculations: the same train is moving with 0.9C and it has a proper length of 300000 kilometers. For an observer on the platform he will measure length of train as 13076696.83 meters and the train will have moved 619422481.5 meters in 2.500631499 seconds, so total distance traveled by light from perspective of a still person on the platform is 750189449.8 meters. By dividing I will get exact C because that's how I got that time, but when I use time dilation to get time should be measured by observers on the train" difference in time between the clock in front and the one in the back when light moves between them" I get it as 1.09 seconds instead of 1 second, so observers on train should measure C as 300000000/1.09 which gives about 275000 kilometers per second. I feel like the mistake is about simultaneity, but used clocks should be on time according to observers on the train.
Thanks for caring. [..]

OK, at first sight you are accounting for everything that is relevant (great!), but there is an error somewhere. Likely you got the time difference wrong, as you give much less details than is necessary for a correct calculation. Here's my calculation of the measurements in the train, as calculated from the perspective of the platform observer :

We put c= 300000 km/s. The train goes at 0.9c => γ = 2.294 -> L_train = 130767 km.
Total measured time for a signal from rear clock to front of train and back according to platform observer:
forward (130767 km) / (30000 km/s) = 4.3589 s. (Check: in 4.3589s the light has moved 1307670 km and the train 9/10th of that; difference is length of train).
backward (130767 km) / 570000 km/s) = 0.2294 s.
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.

 

[harrylin]

I can't tell if that's sarcastic or not, lol.

Seriously! At first sight it is easier to follow than the link to the "simple derivation" that I gave.

Nevertheless, probably a step-wise derivation by means of "patching up" the Galilean transformations (as historically first happened) is the easiest to follow, even though it is more elaborate.

 

[Amr Elsayed]

The Lorentz transformations tell us how to calculate x′0x'_0 and t′0t'_0 from x0x_0 and t0t_0 and how to calculate x′1x'_1 and t′1t'_1 from x1x_1 and t1t_1. Try it... You will get x′1=172105263x'_1=172105263, t′1=0.573684209t'_1=0.573684209, x′0=0x'_0=0 and t′0=0t'_0=0, and the speed of light calculated in the train frame will be c=3×108c=3\times{10}^8 just as it was in the platform frame.

Thanks for care, it's not about knowing the Lorentz transform that will give me the time and distance from perspective of train. I need to know why this transform will give me such distance in such time. Of course it's related to what I will measure if I'm on train and there are reasons for those measures.

In this case, measuring the speed of light is done by a pair of synchronized clocks in each frame

I don't get the difference between the 2 methods. There can be a difference if the clocks on train are not on time from my perspective if I am on train. If "synchronized clocks" means clocks on time from their perspective then isn't it same to measure time for light to pass the train?? and the guy on platform will use another 2 clocks on time for him so it's as he has measured it. I get why 2 synchronized clocks on train are not so for an observer on the platform, but just tell me which clocks each observer will use. Again, If each observer will use different clocks that are stationary to him, then what should non-simultaneity of clocks do ?

Ah yes that's nice! Perhaps that one is understandable for amr.

I hope so

forward (130767 km) / (30000 km/s) = 4.3589 s

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

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 than the rear clock was at the time of sending.

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

 

[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.

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