How to implement length contraction in special relativity

Question

I am trying to construct a special relativity animated simulation, and am unclear about some aspects, please.

The animation is as follows:

From the simulation coordinate system's point of view, Einstein is stationary. Curie and Newton is onboard a train moving 1 unit/frame to the right. While on the train, Curie throws out a light moving also at 1 unit/frame to the right.

Then, we animate the situation simultaneously using 3 different "frames of reference". The top frame is from Einstein's viewpoint. The second is from Curie's viewpoint. The third if from Newton's viewpoint.

Please help correct any wrong concepts I might be harboring from this point on:

The central tenet for special relativity in this situation is that, in the case above, the speed of light seen by Einstein in this case is 2 units/frame to the right. The speed of light seen by Curie is 1 unit/frame to the right. The speed of light seen by Newton in -1 unit/frame as it is moving towards him.

The central problem in the above picture is that we do not want to allow the speed of light from Einstein's frame to be any number other than 1. Is this correct? The speed of light is calculated here by having the previous position of light stored in the "notebook" of each participant, then compared to the new position of light. Eg, for Einstein, on previous frame light is located at [2,0], and at the next frame it is located at [4,0], so the speed is calculated as 2 units/frame to the right.

Fine, we might say. If we would like the speed of light seen by Einstein to be 1 in this case, we can "contract the universe by half" along the x-axis, so the same measurement by Einstein in the new contracted coordinate be [1,0] on the earlier frame and [2,0] on the next frame, which gives the calculation we desired of speed of light = 1 unit/frame to the right, as follows:

Is this the essential thought process of special relativity in this case?

I realize that to get to a rigorous definition of this we might have to be more careful and consider the different boundary cases etc. But the essential intuition is that, we contract or expand the coordinate system/axis by the different participants in order to guarantee the result of speed of light = 1 for all parties, Einstein, Curie and Newton in this case. Is this correct?

If not, please explain the numerical derivation for how the computer (which runs the simulation) might come up algorithmically with the answer "Speed of light seen by Einstein = 1 unit/frame" in this case, without resorting to our (the simulator's) divine intervention?

Answer 1

First, you need to describe things completely in one frame. The best way to do that is to use the parametric equation of a worldline. So, for example I will use coordinates $(t,x)$ in units where $c=1$. So the worldline for Einstein would be $r_{ein}=(t,0)$ while the worldline for Curie would be $r_{cur}=(t,vt)$ and the worldline for Newton would be $r_{new}=(t,vt+x_0)$ and the worldline for the light would be $r_{lig}=(t,ct)=(t,t)$

So now you have expressions for the location of every object in the scenario as a function of time in Einstein's frame. Now, notice that you can write all of those in the form $r=(t,\alpha t + \beta)$ where for Einstein $\alpha = 0$ and $\beta = 0$ but for Newton $\alpha = v$ and $\beta = x_0$, etc.

So all that you have to do is to transform $r$ into another frame (the primed frame) and then substitute in the appropriate $\alpha$ and $\beta$. You can simply substitute in the Lorentz transform for a velocity $u$ to obtain $$r'=(t',x')=\left(\frac{t}{\sqrt{1-u^2}}-\frac{u (\beta +\alpha t)}{\sqrt{1-u^2}},\frac{\beta +\alpha t}{\sqrt{1-u^2}}-\frac{t u}{\sqrt{1-u^2}}\right)$$

Now, notice that this form is still expressed in terms of $t$, the time coordinate in Einstein's frame. But we can take the first component in $r'$ and write $$t'=\frac{t}{\sqrt{1-u^2}}-\frac{u (\beta +\alpha t)}{\sqrt{1-u^2}}$$ and then solve for $t$. We can then substitute that expression back into the formula for $r'$ to get $$r'=\left(t',\frac{t' (u-\alpha )-\beta \sqrt{1-u^2}}{\alpha u-1}\right)$$

Thus, we now have an expression to transform any of these worldlines into any frame simply by setting the appropriate values for $u$, $\alpha$, and $\beta$ and we can do so using either $t$ or $t'$ to parameterize it.

Answer 2

The speed of light is the same to all observers- you are making a fundamental mistake if you assume that the speed of light from Einstein's perspective is double the speed from Curie's.

If Curie throws out the beam at the moment she passes Einstein, then the front of the beam moves down the carriage towards Newton at a fixed speed relative to all three observers.

The crucial point is that at any given moment Curie and Einstein agree on where the front of the beam is- what they disagree about is what time it is at that location.

For example, when the beam is halfway down the carriage, from Curie's perspective the distance the beam has travelled is half the length of the carriage. From Einstein's perspective the beam has travelled further- it has travelled half the length of the carriage plus the distance from him to the rear of the carriage. Given that the distance travelled by light differs for each observer, the time the light has taken to arrive at the centre of the carriage must differ for each observer. So a clock on the carriage adjacent to the front of the beam at the middle of the carriage will show one time, while a clock on the trackside adjacent to the middle of the carriage will show a later time.

I think the conceptual difficulty with what you are trying to do is that you are linking the passage of time with the updating of the picture frame. Since time varies for Einstein and Curie, you cannot associate both of their times with a single sequence of frames- you will need to have something else on the screen to shown time explicitly.