Why do you need a light clock?

[Atstill77]

TL;DR Summary

I came up with the thought experiment below, which makes

what is wrong with the thought experiment below?! If I consider a light beam going straight towards an observer in motion it makes it look the moving clock ticks faster, but if you use a light clock it makes it look like the clock turns slower. What am I missing here?

“Two observers convene in a rail station, each with a clock that ticks at the same rate when laid side-by-side. They synchronize their clocks so that both start at time zero. Observer 1 stays on the train platform and Observer 2 boards a high-speed train. The train reverses out of the station several hundred thousand miles and then accelerates forward and eventually holds a constant speed around that of half speed of light.

Now, Observer 2 is seated at the back of the railcar and to them the railcar appears at rest and the ground is moving at around half the speed of light, relative to them. A light beam emitter is stationed at the front of the railcar and pointed at the Observer 2.

The moment the railcar passes the train platform (where Observer 1 is standing) the emitter sends a beam of light towards Observer 2. Since the railcar is at rest with respect to Observer 2, the light is seen to propagate at the speed of light across the at-rest length of the railcar towards Observer 2.

HOWEVER this is not what Observer 1 sees. Since the speed of light is constant, Observer 1 also measures the light beam at the speed of light (“c”). But with respect to Observer 1, Observer 2 is in motion to the right. Once the light beam is emitted, Observer 1 sees Observer 2 move some distance towards the light beam before hitting it, and unusually sees that the railcar length is now shorter. Thus, Observer 1 measures a SHORTER length traversed by the light than the full at-rest length of the railcar.”

So if ^^ was true, Observer 1 would read less distance and less time versus the Observer 2 who reads more distance and more time. What’s the discrepancy here

 

[Dale]

Summary:: I came up with the thought experiment below, which makes

What’s the discrepancy here

The relativity of simultaneity.

It’s almost always the relativity of simultaneity

 

[Sagittarius A-Star]

A light beam emitter is stationed at the front of the railcar and pointed at the Observer 2.

A light-clock can only work with the 2-way-speed of light. Reason: A clock has only 1 worldline. So light needs to be sent by Oberserver 2 to the front of the railcar and reflected there back to Observer 2.

Time period for Observer_2:
$\tau = \frac{2L}{c}$

Time period for Observer_1:
$t = \frac{L'}{c-v}+\frac{L'}{c+v}= L' \frac{2c}{c^2-v^2} = \frac{2L'}{c}\gamma^2 = \frac{2L}{c}\gamma = \tau * \gamma$

 

[PeroK]

So if ^^ was true, Observer 1 would read less distance and less time versus the Observer 2 who reads more distance and more time. What’s the discrepancy here

When you take out all the extraneous material (of which there is a lot), you have:

1) A light source on a railcar moving relative to the platform.

2) An observer (A) on the platform.

3) An observer (B) moving with the light source, at the other end of the railcar.

Then you have two events:

Event 1: light source emits pulse of light

Event 2: light pulse is received by observer B.

If we assume the proper length of the railcar is $d$, then this is length contracted to $d/\gamma$ in the platform frame.

In the platform frame, the time between the events $\Delta t = \frac{d}{\gamma(v + c)}$, where $v$ is the speed of the railcar and $\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$ is the associated gamma factor.

And the distance between the events is $\Delta x = c\Delta t$.

In the railcar frame, the time between the events is $\Delta t' = \frac d c$ and the distance between the events is $\Delta x' = d$.

These time intervals and distances are different, as they must be. The thing to check is that the spacetime interval is null in both frames - as it must be as the events are separated by a light path. Let's check that:$$c^2(\Delta t)^2 - (\Delta x)^2 = (c^2 - c^2)(\Delta t)^2 = 0$$ And $$c^2(\Delta t')^2 - (\Delta x')^2 = d^2 - d^2 = 0$$ And we see that, as expected, both frames measure a null spacetime interval.

 

[Ibix]

So if ^^ was true, Observer 1 would read less distance and less time versus the Observer 2 who reads more distance and more time. What’s the discrepancy her

The problem is that you haven't described a clock here because there's no observer who can use it to measure a chunk of their proper time. That's what a clock does, and yours doesn't do this because no one can be at one end of the train when the light pulse leaves and also be at the other end when it arrives. So you aren't measuring time.

If you put in a mirror and reflect light back to its starting point then you measure a chunk of time for an observer at rest at the start point, and the maths is as @Sagittarius A-Star laid out above.

 

[PeroK]

What am I missing here?

There's a fundamental point here. You're assuming that, in some way, the time interval between two events must always be less in a "moving" frame. That cannot possibly be the case. If we take two events and the time interval between them as measured in two frames: $\Delta t$ and $\Delta t'$. We can take the unprimed frame to be the "stationary" frame, in which case you are expecting $\Delta t > \Delta t'$. And, we can also take the primed frame to be the "stationary" frame, in which case you are expecting $\Delta t < \Delta t'$.

But, you can't have both. This is not what time dilation means or implies.

 

[Janus]

Summary:: I came up with the thought experiment below, which makes

what is wrong with the thought experiment below?! If I consider a light beam going straight towards an observer in motion it makes it look the moving clock ticks faster, but if you use a light clock it makes it look like the clock turns slower. What am I missing here?

The first case includes Doppler shift which is a result of the distance between source and observer changing over time. With light clock examples, we factor that out because it is simply "layered over" the effect (time dilation) that we are interested in.

“Two observers convene in a rail station, each with a clock that ticks at the same rate when laid side-by-side. They synchronize their clocks so that both start at time zero. Observer 1 stays on the train platform and Observer 2 boards a high-speed train. The train reverses out of the station several hundred thousand miles and then accelerates forward and eventually holds a constant speed around that of half speed of light.

Now, Observer 2 is seated at the back of the railcar and to them the railcar appears at rest and the ground is moving at around half the speed of light, relative to them. A light beam emitter is stationed at the front of the railcar and pointed at the Observer 2.

The moment the railcar passes the train platform (where Observer 1 is standing) the emitter sends a beam of light towards Observer 2. Since the railcar is at rest with respect to Observer 2, the light is seen to propagate at the speed of light across the at-rest length of the railcar towards Observer 2.

HOWEVER this is not what Observer 1 sees. Since the speed of light is constant, Observer 1 also measures the light beam at the speed of light (“c”). But with respect to Observer 1, Observer 2 is in motion to the right. Once the light beam is emitted, Observer 1 sees Observer 2 move some distance towards the light beam before hitting it, and unusually sees that the railcar length is now shorter. Thus, Observer 1 measures a SHORTER length traversed by the light than the full at-rest length of the railcar.”

So if ^^ was true, Observer 1 would read less distance and less time versus the Observer 2 who reads more distance and more time. What’s the discrepancy here

Others here have pointed out the Relativity of Simultaneity issue. You can eliminate that by considering the "round trip" time of the light, The light leaves one end of the car, reflects off the other end and back to the first. You have basically set up a light clock "that is on its side".
Here is an illustration of a scenario where you have light clocks set up perpendicular to each other:
Here, the ( the expanding circle illustrates how all the light pulses initially emitted travel at c relative to rest frame of the animation. )

While the two legs of the round trip from the vertical pulse are equal in duration, the horizontal legs are not, but the summed duration of vertical vs horizontal pulses are the same.

 

[Atstill77]

The first case includes Doppler shift which is a result of the distance between source and observer changing over time. With light clock examples, we factor that out because it is simply "layered over" the effect (time dilation) that we are interested in.
Others here have pointed out the Relativity of Simultaneity issue. You can eliminate that by considering the "round trip" time of the light, The light leaves one end of the car, reflects off the other end and back to the first. You have basically set up a light clock "that is on its side".
Here is an illustration of a scenario where you have light clocks set up perpendicular to each other:
Here, the ( the expanding circle illustrates how all the light pulses initially emitted travel at c relative to rest frame of the animation. )View attachment 275687
While the two legs of the round trip from the vertical pulse are equal in duration, the horizontal legs are not, but the summed duration of vertical vs horizontal pulses are the same.

This is incredibly helpful. Thank you!