Are the laser guns inside the spaceship? Can you attach a rough sketch of this situation? It will be easier to comprehend.

The Lorentz transformation contains square root of 1-v^2/c^2. You have not it shown. Why?

I have not related X1 and X1' . Those terms will come when we relate these two frames and that is my question how can we relate.

I am just applying laws of motion assuming length and time may vary for different frame. 

Actually i was trying to derive Lorentz transformation on my own. I think i have confused myself. 

But the main question is two light waves propagating in opposite direction then how those light waves will be seen by observer at rest ?

Yes both the laser guns are inside the spaceship

Light waves remain unaffected by relativity laws; c+ c = c and not 2c. So, for an observer at rest, the two light beams will have velocity +c and -c respectively, where the signs represent the relative direction of propagation.

Ok i understand that concept about speed of light. What i meant to ask is how we can write equations relativity length and time

Dear Sarvesh,

Forget Lorentz transformations, beside use galilean transformations by adding light aberration and retardation of fields.

Sincerely and happy new year.

@ Sarvesh

Introduce Lorentz factor "gamma" with your length and time terms. You can find details in any standard book of special theory of relativity, or even wikipedia.

A crucial difficulty arises from the concept ``simultaneously'': if two events are simultaneous in one reference frame, then, due to Lorentz transformation, they ar not so in a frame that is moving. In the case you present, if I assume that the two laser are fired *simultaneously* as viewed from the spaceship, then the first step should be to calculate, using the Lorentz transformation, what is the time interval separating the firing of the first laser gun and that of the second, in the reference frame of the observer at rest. This is what usually creates conceptual difficulties: simultaneous events do not remain simultaneous in a moving reference frame.

Thanks Leyvraz. 

I think i am getting something. 

One more question i want to ask. 

Suppose the spaceship is moving to the right with velocity c/2 and a man inside that spaceship is with velocity c/2 to the left. Now the man moving left is in rest with respect to a stationary observer on the ground. Is the time and space for observer at rest is same for the man moving left ? 

And if the answer is yes for above question is it possible that ( with respect to the scenario in main question) light travelling in right and light travelling in left feels different space and time ?

``Is the time and space for observer at rest is same for the man moving left ? '' Yes: both aer at rest with respect to each other. To your last question: it is not really possible to define a reference frame moving at the speed of light, since the Lorentz transform becomes singular for v=c.

Thanks again Leyvraz 

Just one more last question

If the spaceship is moving to the right, two similar particles say muons are fired with same speed. One in the left direction and other in the right direction with same speed and speed is near the speed of light  ( speed w.r.t observer in spaceship) . Now will the average lifetime of particles moving in opposite direction will be same or different w.r.t. observer in the spaceship ?

The two muons will be moving at different speeds, so the apparent decay times will be different.

This is a variation of Einstein's train thought experiment, except that the flashes are simultaneous in the train (spaceship) frame rather than the observer frame.  See video below.  As Leyvraz already pointed out, simultaneity will not be preserved in both frames.  

This means that just as you have two distances x1 and x2 in your setup, you also need two times t1 and t2.  

You have to specify where the observer's are.  Your problem statement seems to be confused on this as you ask does x1=x2?  Just specify that the observer is in the center so this is true, so t1=t2=x1/c.  You have made the spaceship your un-primed frame.  If these are the distances measured within the spaceship then they are in the rest frame of the space ship, and the outside observer is moving at -v.  

If two things happen at the same time and place, they are simultaneous for everyone.  So all see the laser pulses arrive at the center of the train simultaneously. In this respect it is backwards from the train experiment, in which it is the  platform observer at which they arrive simultaneously.  If this happens when the two observers are adjacent, then all frames see the arrival as simultaneous, and since distances are equal in the spaceship, then the spaceship observer sees the pulse initiations as simultaneous.  The platform observer sees them at different past times: t1’=(t1-vx1/c²)γ  where γ is the usual Lorentz factor (see 2nd link).  You can let the x coordinate be zero wherever you want, so for simplicity make it zero at the position of the observers.  

For one of the times you'll add vx and the other you'll subtract.  It is easy to make a mistake.  I suggest using my heuristic "Leading clocks lag" (see 3rd link).  From the point of view of the outside observer, the clock at the front of the space ship will be behind, lagging, making that pulse start later.  This is consistent with the fact that in his reference frame the pulse does not have to go as far because the spaceship is catching up with it.  If t1 is the lead clock, then t1'

Dear all Einsteinists,

Following: time(ground) x velocity(ground) = time(move) x velocity(move) you all use a Minkowski space-time with v=v as an equivalence to an euclidian space-time with v=v. It is totally wrong !

Thanks Robert .

I think i should give more time on my thoughts and questions. I will get back to you in case i still have doubts

Leyvraz one final question.

If two muons are moving with same speed in opposite direction w.r.t. the observer inside the spaceship .

Is it moving with same or different speed w.r.t. the stationary observer ? How will be the average lifetimes as seen by the two observers ? 

If the muons have the same speed, but opposite velocities, in the spaceship, then, in the rest frame, their velocities will arise via relativistic velocity addition of the muon's velocity with the spaceship. Since the two muons' velocities are different (their signs are opposite) it follows that their speeds in the rest frame will be different, and thus, so will their decay times.

Please read the paper, may be helpful for you  .A new approach to the correction of

Galilean transformation.

Light cones in the rest system: (±x,ct) = (±1,1)ct.

Lorentz transform to the moving (primed) system.

chw     shw

shw     chw

Light cones in the moving system:

 x’ = (chw + shw)ct = exp(w)ct

ct’ = (shw + chw)ct = exp(w)ct

################

x’’  = (-chw + shw)ct =                             - exp(-w)ct

ct’’ = (-shw + chw)ct = - (-chw + shw)ct = exp(-w)ct

ct’’< ct’: 

equal time condition in the moving system:

ct’’ = ct’ : ct -> ct* =  exp(2w)ct: -x’’ = x’

Pay attention to the following two points, can understand your problem,

1 in any inertial frame of reference to measure speed of electromagnetic radiation are c.Invariance is the speed of light.

2 the frame of reference velocity v and the speed of light c cannot add (subtract).Because the light is not inertia carrier.

Lorentz transformation hidden logic contradiction, cannot describe the electromagnetic radiati。