B Measuring Michelson–Morley light beams

[P J Strydom]

The phrase "travel in a direction", or indeed just the word "travel" implies that there is something else involved, because there must be something that you're traveling relative to. An observer at rest relative to that something will indeed see both pulses leave at the same time and arrive at the same time. But that observer will see those pulses move further distances and therefore take more time.

Note that we are comparing what's observed by two different observers. We are not comparing the travel times of two different pulses measured by one of the observers.

To either observer both beams take the same amount of time. The observers disagree on the amount of that time.

Damn, let me try to understand this part first.
First. If we...
are in a spaceship in space.
And we do not know if we are traveling, or stationary...
and we pulse a light...
Einstein says that the light will travel at c independently from it's source.
we know the light will expand in a 3D shell from it's source.
Now, if the spaceship is stationary, We on this ship will observe the light traveling to the ceiling, tail and nose, arriving at its destination at the same time.
however, if this ship is moving in some direction, let's say forward in relation to the nose of the ship for this argument,
we on this ship should observe the light arriving at the tail, before the nose.
However, this is what length contraction stipulates...
The ship will contract in length and the 3D shell of light, will then reach the nose and the tail at the same time.

But, someone observing the ship from a distance. because this is what a different time frame is in my understanding,
he will see the ship moving forward, and the 3D shell of light will move to the back of this ship we are on.

is this correct?

 

[Ibix]

however, if this ship is moving in some direction, let's say forward in relation to the nose of the ship for this argument,
we on this ship should observe the light arriving at the tail, before the nose.

How are you going to observe it? You have to wait for the light to get back to the middle of the ship to see when the reflection happened, and we already know the light will return at the same time from both ends regardless of the state of motion of the ship. If you think you can get round this by using clocks at the end of the ships, think about how you would synchronise the clocks.

You are correct that, if you regard the ship as moving, the reflections do not occur simultaneously. But there is no way to observe this - any observations you make can be consistently explained either in terms of a stationary ship (identical forward and backward travel times leading to a simultaneous return) or a moving ship (longer forward and shorter backward travel times leading to a simultaneous return).

 

[Herman Trivilino]

And we do not know if we are traveling, or stationary...

There's no difference between those two things, so of course there's no way to know.

Now, if the spaceship is stationary, We on this ship will observe the light traveling to the ceiling, tail and nose, arriving at its destination at the same time.

Then that's what happens if the ship moves in a straight line at a steady speed.

however, if this ship is moving in some direction, let's say forward in relation to the nose of the ship for this argument,
we on this ship should observe the light arriving at the tail, before the nose.

No. Why would you think that?

However, this is what length contraction stipulates...
The ship will contract in length and the 3D shell of light, will then reach the nose and the tail at the same time.

There is no contraction according an observer resting in the ship.