I order to address your question I need to get something out of the way first.

As we know, special relativity is well corroborated. There are the well known confirmations. The muons created in the upper atmosphere that make it all the way to the Earth's surface. In particle accelerators we have that unstable particles have a longer half life than they would have when stationary with respect to the observer, in accordance with special relativity.

It is experimental evidence like that that justifies confidence in a theory. This confidence then justifies confidence in the assumptions that underly the theory.

In the case of special relativity there is that crucial underlying assumption that is commonly expressed as the light postulate: for all members of the equivalence class of inertial coordinate systems the speed of light is the same.

My point is: the purpose of the light clock demonstration is to start with assuming that the speed of light is the same for all members of the equivalence class, and proceed to work out the ramifications of that.

It is incorrect to present the light postulate as a conclusion.

To your question.

In my opinion it is always best to make the setup completely symmetrical, as the phenomenon itself is symmetrical. Keeping the symmetry in mind may help you catch an error in your thinking.

We give each spaceship its own light clock. On each spaceship the light remains inside that spaceship. Each observer creates two plots: one plot represents the motion of the light inside his own spaceship, the other plot the motion of the light inside the other spaceship.

The setup is aligned such that the motion of the light inside each spaceship is perpendicular to the direction of the relative velocity between the two spaceships.

From here on I will use 'perpendicular' for the motion direction perpendicular to the direction of the relative velocity between the two spaceships. (And the same thing for 'parallel', of course.)

Each observer will plot the motion of the light in the other spaceship as a zigzag motion, as he is plotting that motion in his own coordinate system. The faster the relative velocity of the two spaceships, the more elongated the plot of the zigzag.

In each plot the velocity of the light can be decomposed in two velocity components: a perpendicular component and a parallel component.

In each plot: as the light is approaching the detector that it is aimed at it does not have a velocity component in the direction parallel to the detector.

So it's unclear what your concern is. This bouncing scenario will work in analogous way if you perform it with actual particles that are set up for perfect elastic bounce.

That is, even with particles with a rest mass an analogous scenario is possible.

By contrast: in a scenario where one ship is emitting light and the other spaceship is receiving that light, that is a whole different ball game.

I order to address your question I need to get something out of the way first.

As we know, special relativity is well corroborated. There are the well known confirmations. The muons created in the upper atmosphere that make it all the way to the Earth's surface. In particle accelerators we have that unstable particles have a longer half life than they would have when stationary with respect to the observer, in accordance with special relativity.

It is experimental evidence like that that justifies confidence in a theory. This confidence then justifies confidence in the assumptions that underly the theory.

In the case of special relativity there is that crucial underlying assumption that is commonly expressed as the light postulate: for all members of the equivalence class of inertial coordinate systems the speed of light is the same.

My point is: the purpose of the light clock demonstration is to start with assuming that the speed of light is the same for all members of the equivalence class, and proceed to work out the ramifications of that.

It is incorrect to present the light postulate as a conclusion.

To your question.

In my opinion it is always best to make the setup completely symmetrical, as the phenomenon itself is symmetrical. Keeping the symmetry in mind may help you catch an error in your thinking.

We give each spaceship its own light clock. On each spaceship the light remains inside that spaceship. Each observer creates two plots: one plot represents the motion of the light inside his own spaceship, the other plot the motion of the light inside the other spaceship.

The setup is aligned such that the motion of the light inside each spaceship is perpendicular to the direction of the relative velocity between the two spaceships.

From here on I will use 'perpendicular' for the motion direction perpendicular to the direction of the relative velocity between the two spaceships. (And the same thing for 'parallel', of course.)

Each observer will plot the motion of the light in the other spaceship as a zigzag motion, as he is plotting that motion in his own coordinate system. The faster the relative velocity of the two spaceships, the more elongated the plot of the zigzag.

In each plot the velocity of the light can be decomposed in two velocity components: a perpendicular component and a parallel component.

In each plot: as the light is approaching the detector that it is aimed at it does not have a velocity component in the direction parallel to the detector.

So it's unclear what your concern is. This bouncing scenario will work in analogous way if you perform it with actual particles that are set up for perfect elastic bounce.

That is, even with particles with a rest mass an analogous scenario is possible.

By contrast: in a scenario where one ship is emitting light and the other spaceship is receiving that light, that is a whole different ball game.

The scenario that I discussed (light bounces inside each spaceship) is presented in this light clock interactive animation

It is part of a larger collection of interactive animations by Michael Fowler.

I order to address your question I need to get something out of the way first.

As we know, special relativity is well corroborated. There are the well known confirmations. The muons created in the upper atmosphere that make it all the way to the Earth's surface. In particle accelerators we have that unstable particles have a longer half life than they would have when stationary with respect to the observer, in accordance with special relativity.

It is experimental evidence like that that justifies confidence in a theory. This confidence then justifies confidence in the assumptions that underly the theory.

In the case of special relativity there is that crucial underlying assumption that is commonly expressed as the light postulate: for all members of the equivalence class of inertial coordinate systems the speed of light is the same.

My point is: the purpose of the light clock demonstration is to start with assuming that the speed of light is the same for all members of the equivalence class, and proceed to work out the ramifications of that.

It is incorrect to present the light postulate as a conclusion.

To your question.

In my opinion it is always best to make the setup completely symmetrical, as the phenomenon itself is symmetrical. Keeping the symmetry in mind may help you catch an error in your thinking.

We give each spaceship its own light clock. On each spaceship the light remains inside that spaceship. Each observer creates two plots: one plot represents the motion of the light inside his own spaceship, the other plot the motion of the light inside the other spaceship.

The setup is aligned such that the motion of the light inside each spaceship is perpendicular to the direction of the relative velocity between the two spaceships.

From here on I will use 'perpendicular' for the motion direction perpendicular to the direction of the relative velocity between the two spaceships. (And the same thing for 'parallel', of course.)

Each observer will plot the motion of the light in the other spaceship as a zigzag motion, as he is plotting that motion in his own coordinate system. The faster the relative velocity of the two spaceships, the more elongated the plot of the zigzag.

In each plot the velocity of the light can be decomposed in two velocity components: a perpendicular component and a parallel component.

In each plot: as the light is approaching the detector that it is aimed at it does not have a velocity component in the direction parallel to the detector.

So it's unclear what your concern is. This bouncing scenario will work in analogous way if you perform it with actual particles that are set up for perfect elastic bounce.

That is, even with particles with a rest mass an analogous scenario is possible.

I order to address your question I need to get something out of the way first.

As we know, special relativity is well corroborated. There are the well known confirmations. The muons created in the upper atmosphere that make it all the way to the Earth's surface. In particle accelerators we have that unstable particles have a longer half life than they would have when stationary with respect to the observer, in accordance with special relativity.

It is experimental evidence like that that justifies confidence in a theory. This confidence then justifies confidence in the assumptions that underly the theory.

In the case of special relativity there is that crucial underlying assumption that is commonly expressed as the light postulate: for all members of the equivalence class of inertial coordinate systems the speed of light is the same.

My point is: the purpose of the light clock demonstration is to start with assuming that the speed of light is the same for all members of the equivalence class, and proceed to work out the ramifications of that.

It is incorrect to present the light postulate as a conclusion.

To your question.

In my opinion it is always best to make the setup completely symmetrical, as the phenomenon itself is symmetrical. Keeping the symmetry in mind may help you catch an error in your thinking.

We give each spaceship its own light clock. On each spaceship the light remains inside that spaceship. Each observer creates two plots: one plot represents the motion of the light inside his own spaceship, the other plot the motion of the light inside the other spaceship.

The setup is aligned such that the motion of the light inside each spaceship is perpendicular to the direction of the relative velocity between the two spaceships.

From here on I will use 'perpendicular' for the motion direction perpendicular to the direction of the relative velocity between the two spaceships. (And the same thing for 'parallel', of course.)

Each observer will plot the motion of the light in the other spaceship as a zigzag motion, as he is plotting that motion in his own coordinate system. The faster the relative velocity of the two spaceships, the more elongated the plot of the zigzag.

In each plot the velocity of the light can be decomposed in two velocity components: a perpendicular component and a parallel component.

In each plot: as the light is approaching the detector that it is aimed at it does not have a velocity component in the direction parallel to the detector.

So it's unclear what your concern is. This bouncing scenario will work in analogous way if you perform it with actual particles that are set up for perfect elastic bounce.

That is, even with particles with a rest mass an analogous scenario is possible.

By contrast: in a scenario where one ship is emitting light and the other spaceship is receiving that light, that is a whole different ball game.