B Einstein thought experiment confusion: “light clock in a moving frame”

[ESponge2000]

TL;DR Summary

An area of confusion I have in demonstrating the Lorentz formula 1 / (1-(v/c)^2) Through stationary observer’s perception of the moving frame’s light clock appearing zig-zagged instead of straight up-down makes perfect sense to me until the question of how this motion plays out when v approaches as close to exactly c as possible

So for example, we can say a stationary observer and a moving ship are in relative motion horizontally and at rest with each other vertically, right? This is basic Pythagorean math.

Stationary observer has own light clock that dribles up-down like a pendulum and it drops a beam 1 light-second down in 1 second, as does the ship has a beam like this as well that drops same vertical distance down and is at rest with the ship


We know The ship’s light beam will appear diagonal not vertical to the stationary observer and will drop a vertical distance that both frames agree to be 1 light-second for any light beam that covers that vertical length when it completes that length . Hence what the stationary observer sees to be a diagonal line (longer than the vertical) will incur 1 second passing on the ship’s clock will take longer in the stationary observer’s reference frame because the diagonal is a longer distance and the beam will travel at the same speed to reach the bottom, hence time dilation

So when Lorentz is 2 is when v is circa 86% of the speed of light
The horizontal distance here is sqrt(0.75) like we said … the velocity, Solving Pythagorean theorem 1^2 - (0.86..)^2 =0.5^2 We backsolve that the stationary observer’s light beam vertical to the stationary observer would not have had time to go top
To bottom if it were to have a vertical
Motion timed with only the vertical velocity achieved perceived in the perceived diagonal
Beam from the moving ship’s vertical light clock
… in fact the stationary observer would calculate they only cover half a second ticked (half the distance to the bottom) on the stationary light clock ….if their own clock were to be paced to the vertical scalar of the diagonal beam And hence the time dilation is a factor of 1/0.5 =2


We can show this too by animating the motion of the ship and drawing the beam that’s relative to the ship’s position being the one at rest.



BUT….What trips me up here though is from the stationary observer, where the moving ship is so so so so so so so close to c say it’s 99.999999999999………..% of c with a billion 9’s ….. we would still draw out that the horizontal is not infinitely long so how do we show this in a practical animation when we have to not be able to know how the angle of diagonal will need to be without forcing the diagonal to be 1 unit ? This doesn’t seem like something the moving videos capture well…..
demonstration doesnt make it immediately obvious that there should be essentially no dipping at all in the diagonal when v approaches c… unless we already solved for the Lorentz to determine the angle before doing this kind of video demonstration

Forget math calculations for a moment and just construct 2 light beams , one moves with the stationary observer and the other one moves with the ship…. I guess it makes sense but we would have to make sure to make it crystal clear that it’s the “speed” NOT the vertical scalar that MUST be matched the same for both light clocks in all frames and these videos tend to not focus on how much strain that places on the derived angle of the dip , such that when v=c you should have no opportunity for ANY dipping of the ship’s light beam through the perspective of the stationary observer (Of course this exercise ignores the way communication signals are also restricted to the velocity of c so we perceive but don’t observe this relationship directly, but let’s keep this exercise simple, we are focused on what we WOULD see if we could hypothetically view the entire X axis for each time t

 

[ESponge2000]

again the math makes perfect sense to me, the visual demonstration of the animation could capture the effect that the ship’s light beam has to obey 1) being at the same horizontal position of ship at all time points and 2) its length MUST be forced the same as the other light beam’s vertical path length relative to stationary observer ….

Maybe the video should show frequent pauses to keep showing the lengths of the 2 beams the same at all times amidst different directions… and then it is more clear that given v=c, the length of the ship’s light beam can only be same as the stationary one by 100% moving horizontal always … hence translating to no vertical dip ever to be perceived by stationary observer and hence no time ever passing on the ship’s clock…. I think I solved my own confusion

Basic framework :
1) Light beam for stationary clock: moves down 1 unit in 1 second of time
2) Light beam for moving ship clock, the vertical distance it moves is its passage of time and 1 unit is 1 second of time for it
3) the vertical lengths are scaled identically for both light beams
4) The ship is moving left to right relative to the stationary observer and the distance in units is v/c
5) For the stationary observer, the moving ship’s light beam is always to move the same total distance as its own vertical light beam , and always must be in the same horizontal position as the ship’s … Now whatever vertical drop on the ship’s beam produces that same total length as the other beam, represents the vertical coordinate of the ship’s beam …. When v=c this same length is only achieved if it coincides with where all of the ship’s beam having not dipped at all… Hence showing a passage of 0 vertical for the ship and a Lorentz factor infinity

 

[Dale]

how do we show this in a practical animation

Yes, that close to $c$ you won’t get good animations. You will run into numerical issues almost immediately, and your animation would have to run for a really long time.

I tend to use $v=0.6c$ for all of my figures and plots. So I would probably do the same for animations.

 

[Ibix]

As Dale says, you can't draw that kind of number on a useful scale. That's why diagrams are only an aid to understanding the maths.

A relevant observation is that every light pulse emitter from an event lies on a circle (or sphere in 3d) centered on that event. So drawing an expanding circle is one way to show where all the light must be.

 

[ESponge2000]

The expanding

As Dale says, you can't draw that kind of number on a useful scale. That's why diagrams are only an aid to understanding the maths.

A relevant observation is that every light pulse emitter from an event lies on a circle (or sphere in 3d) centered on that event. So drawing an expanding circle is one way to show where all the light must be.

the expanding circle is very helpful actually but still with the drawback that you need to be careful to redraw lengths and time to the resting frame being applied I think.

There might be a more constructive way to draw the Einstein thought experiment with the light clocks but where we keep making new ships that each depart the stationary observer’s location 5 seconds apart, each with instant acceleration to a constant velocity and each ship has a different velocity …. Then the demonstration shows the closer that velocity is to c, the less steep its downward slope, And for each motion demonstration , we do a pause at 1 second after departure ….so we show
0-1 second , pause , 1-5 seconds pause
New ships departs
5-6 seconds pause , resume 6-10 pause
New ship departs

And so on
And for the beginning version the ships go into a fog after 5 stationary observer’s seconds… but after we digest all this then we try to do with multiple ships in one diagram

 

[Herman Trivilino]

lthe question of how this motion plays out when v approaches as close to exactly c as possible

There is no such thing. No matter how close you get you can always get closer, but you can never reach a speed of $c$.