Understanding Light Behavior in a Moving Train: Explained

In summary: The light on the train doesn't know that it is "moving to the right". As far as the person on the train, the torch, the light and the mirror are concerned, they are at rest and it's the person on the platform that is moving to the left.Also, the train is on the Earth and the Earth is moving round the sun, so souldn't the light to go off in some entirely different direction depending on how fast the train is moving round the sun?I understand that the light would behave differently if it was on a moving spacecraft
  • #36
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).
 
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  • #37
Ibix said:
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?
 
  • #38
Raymond Potvin said:
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.

jbriggs444 said:
How exactly is your hypothetical emitter designed?

jbriggs444 said:
How does the aiming on your hypothetical flashlight work, exactly?

jbriggs444 said:
A starting point would be to answer the question I've asked three times now: What is the design of your aiming device?
 
  • #39
Raymond Potvin said:
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.
 
  • #40
I made a drawing to explain what I mean:
aberra10.png

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?
 
  • #41
Raymond Potvin said:
if its direction is independent from the motion of the source
You keep using that phrase. Why?
 
  • #42
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.
 
  • #43
Raymond Potvin said:
I made a drawing to explain what I mean:
aberra10.png

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.
 
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  • #44
PeroK said:
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... :wink: Agree the rest, though.
 
  • #45
Raymond Potvin said:
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.
 
  • #46
Raymond Potvin said:
a photon is sent in the direction of observer X
According to which frame?
 
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  • #47
Raymond Potvin said:
I made a drawing to explain what I mean:
aberra10.png

[..]
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.
 

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