I Understanding Light Behavior in a Moving Train: Explained

[PeroK]

At the right, a ball sent in the direction of the Earth would not hit the earth, it would have to be sent at an angle to the direction of the Earth to account for the motion of the star if it had to hit the earth.

Another way to look at this is to imagine a pipe leading vertically down from the star. A photon that is fired down the pipe stays in the pipe.

In the left animation, the photon strikes the Earth as the Earth moves to the end of the pipe as the photon emerges.

And, in the right animation the photon also stays in the pipe - it must as its the same scenario. In this animation, the photon hits the Earth because the source and pipe are moving towards the Earth and the end of pipe reaches the Earth just as the photon emerges.

In both cases, the source, pipe and photon all have the same motion in the horizontal direction. On the left, in the source's frame, this is zero motion. On the right, in the Earth's frame this is the relative speed of the star and the Earth.

The same is true whatever the source fires down the pipe. If it stays in the pipe and hits the Earth in one frame, it must stay in the pipe and hit the Earth in the other frame.

 

[PeroK]

An idea to show that light and particles must behave in the same way.

Imagine a source emits a particle (ball) at near light speed and shortly afterwards a photon in the same direction. Imagine this in the rest frame of the source.

The two emissions move in the same direction and after a time, the photon catches the particle and collides with it.

This motion and collision must be observed in all other reference frames. If another reference frame moves perpendicular to the motion of the particle and photon, then it must measure the two moving in the same straight line and colliding.

Any "sideways" motion of the particle caused by the relative motion of the reference frame must be exhibited by the photon as well.

If the photon refused to move sideways in the second reference frame and went in a different direction, it would miss the particle and give an inconsistent observation.

Therefore, photons and particles must behave in the same way in this respect.

 

[Raymond Potvin]

Please show your math that leads you to believe this. This is such an absurdly wrong claim that I cannot believe that you have ever even attempted to work it out.

It is not a mathematical problem, it's a factual one. If we aim a photon from a moving vehicle and at 90 degree to its direction, it should follow a 90 degree path whatever the speed of the vehicle if the direction of light is independent from the motion of the source, but if we throw a ball in the same direction from the same vehicle, it will not follow a 90 degree path because a ball suffers the motion of its source.

 

[A.T.]

if the direction of light is independent from the motion of the source,

But it isn't.

 

[Ibix]

if the direction of light is independent from the motion of the source

You keep noticing that this is contradictory. I have said, @PeroK has said, @harrylin has said, @Dale has said, @jbriggs444 has said, and now @A.T. has said, that the direction of light is not independent of the motion of the source. When are you going to stop worrying about the physical consequences of something that everyone agrees doesn't happen...?

 

[vanhees71]

It's quite easy to explain. Take for simplicity a plane wave (any em. wave can be built by plane waves in the sense of a Fourier integral anyway). Then the electromagnetic field is of the form
$$(\vec{E},\vec{B})(t,\vec{x})=\text{Re}[(\vec{E}_0,\vec{B}_0) \exp(-\mathrm{i} x \cdot k)].$$
This means the propagation direction is $\vec{k}$, and you have $k^0=|\vec{k}|$ from Maxwell's equations, i.e., the phase velocity is $c=1$ (natural units).

So to know the direction of propagation in another frame you just have to use the Lorentz-transformation properties. The phase under the exponential is manfestly covariant and thus it's very easy to determine $k'$; it's simply
$$k'=\Lambda k.$$
Suppose you have a boost in $x$ direction. Then, with $\gamma=(1-v^2)^{-1/2}$,
$$\Lambda =\begin{pmatrix} \gamma & -v \gamma & 0 & 0\\ -v \gamma & \gamma & 0 & 0 \\ 0& 0 & 1 & 0 \\ 0&0 &0 &1 \end{pmatrix}.$$
So you get
$$k'=\Lambda \begin{pmatrix} |\vec{k}| \\ \vec{k} \end{pmatrix}=\begin{pmatrix} \gamma (|\vec{k}|-v k^1) \\ \gamma (k^1-v |\vec{k}|) \\ k^2 \\ k^3 \end{pmatrix}.$$
As you see, the frequency (0-component) is not the same, which is the relativistic Doppler shift, and also the direction is not the same in the new frame (aberration effect).

Of course, all the effects together conspire such that the physics doesn't change at all due to Lorentz boosts, as it must be due to the principle of special relativity (no change of physics due to change of the inertial frame).

 

[Raymond Potvin]

You keep noticing that this is contradictory. I have said, @PeroK has said, @harrylin has said, @Dale has said, @jbriggs444 has said, and now @A.T. has said, that the direction of light is not independent of the motion of the source. When are you going to stop worrying about the physical consequences of something that everyone agrees doesn't happen...?

Is there an experiment that shows that the direction of light depends on the motion of its source?

 

[jbriggs444]

Is there an experiment that shows that the direction of light depends on the motion of its source?

You really need to stop and answer the following question

Edit: Sorry, didn't notice that it was in a different thread.

How exactly is your hypothetical emitter designed?

How does the aiming on your hypothetical flashlight work, exactly?

A starting point would be to answer the question I've asked three times now: What is the design of your aiming device?

 

[Ibix]

Is there an experiment that shows that the direction of light depends on the motion of its source?

jbriggs444 and I have both linked you to Wikipedia on stellar aberration and I pointed out the FAQ at the top of this forum. PeroK has given you any number of clear reasons why your analysis of that is wrong.

You need to correct your understanding instead of simply repeating questions that you have already had answered. I strongly advise you to read PeroK's posts #31 & #32 and think about them carefully.

 

[Raymond Potvin]

I made a drawing to explain what I mean:

A star and an observer X are traveling in the same direction and at the same speed towards the left. When they align with the direction of the earth, a photon is sent in the direction of observer X. If the photon travels like a ball, it will hit the observer X at the left, but if its direction is independent from the motion of the source, it will hit the earth. Obviously, if a photon leaves a star in the direction of the earth, it should hit the earth, no? When we look at the stars, we see them where they were when they sent their light, no?

 

[jbriggs444]

if its direction is independent from the motion of the source

You keep using that phrase. Why?

 

[Ibix]

If your diagram is drawn in the rest frame of the Earth and the ball and the light pulse are launched at the angle you have shown then both will hit Earth; neither will hit X because both follow your right hand line and X will have moved out of the way by the time they get there.

In the rest frame of the star this is interpreted as leading a moving target. You don't aim at where the Earth is now (behind X), you aim where it will be when your ball/light pulse gets there - to the right of X.

 

[PeroK]

I made a drawing to explain what I mean:

A star and an observer X are traveling in the same direction and at the same speed towards the left. When they align with the direction of the earth, a photon is sent in the direction of observer X. If the photon travels like a ball, it will hit the observer X at the left, but if its direction is independent from the motion of the source, it will hit the earth. Obviously, if a photon leaves a star in the direction of the earth, it should hit the earth, no? When we look at the stars, we see them where they were when they sent their light, no?

What about the point I made in post #32? If a photon and a ball follow the same path in the rest frame of the source, then they must follow the same path in all reference frames.

In your diagram above both a ball and a photon will hit x but miss the earth. That is clear from looking at the problem from the rest frame of the star and x. The ball and photon take the same path in that frame and hence in all frames.

 

[Ibix]

In your diagram above both a ball and a photon will hit x but miss the earth

That depends on your interpretation of the diagram, I think... Agree the rest, though.

 

[Dale]

It is not a mathematical problem, it's a factual one. If we aim a photon from a moving vehicle and at 90 degree to its direction, it should follow a 90 degree path whatever the speed of the vehicle if the direction of light is independent from the motion of the source,

It is a mathematical problem, specifically a geometry problem. The geometry in relativity is given, mathematically, by the Lorentz transform. You cannot separate the "factual problem" from the mathematical problem, your post is simply a transparent evasion of the issue.

 

[A.T.]

a photon is sent in the direction of observer X

According to which frame?

 

[harrylin]

I made a drawing to explain what I mean:

[..]
Obviously, if a photon leaves a star in the direction of the earth, it should hit the earth, no? When we look at the stars, we see them where they were when they sent their light, no?

No, we don't generally see them were they were when they sent their light; your question suggests to me that you did not understand the animation that you posted in post #19; t may be useful to have another look at it.
And maybe, just maybe, the issue that is puzzling you (or part of it) is different from what you bring up. A star (at least, an ordinary star) is not like a laser pointer; it sends light in all directions. The starlight that we see is light that was sent in a different direction.