## Questions abouut:Special Relativity, Time Dilation, Light Clock, Velocity of light.

I am a new fish to SR and I have some questions about light clock. Hope someone to clear my concepts.

Q1.
As we all know, the speed (scalar) of light is "c" and is independent to the motion of the light source, but what about the velocity (vector) of light (with magnitude "c" but what direction ? )?

Q2.
Suppose the direction of a light ray is prependicular to the velocity of the light source (like the one in light clock.), can the light ray get the horizontal speed from the light source?

Q3.
In the frame of a person who stand beside a light clock on a moving train which having speed "u" to right,
why the path of a photon that he observed is vertically up and down ? Is it because the photon got the horizontal speed ("u" to right) .If yes, then it become a classical veloctiy addition ?; If no, the photon got no horizontal speed from the source, then how can the person on the train see a vertical up and down path instead of diagonal path?

I hope you can understand what I am talking about.

Thank you!!
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 Quote by Canaan_C I am a new fish to SR and I have some questions about light clock. Hope someone to clear my concepts. Q1. As we all know, the speed (scalar) of light is "c" and is independent to the motion of the light source, but what about the velocity (vector) of light (with magnitude "c" but what direction ? )?
While the speed is observer independent, the direction of the light is not. Different observers will see the same beam of light traveling different in different directions. (This is called aberration.)

 Q2. Suppose the direction of a light ray is prependicular to the velocity of the light source (like the one in light clock.), can the light ray get the horizontal speed from the light source?
Yes. In the sense that someone observing the light clock pass by will see the light moving on a diagonal. The horizontal component of the light's speed will equal that of the light clock.

 Q3. In the frame of a person who stand beside a light clock on a moving train which having speed "u" to right, why the path of a photon that he observed is vertically up and down ? Is it because the photon got the horizontal speed ("u" to right) .If yes, then it become a classical veloctiy addition ?; If no, the photon got no horizontal speed from the source, then how can the person on the train see a vertical up and down path instead of diagonal path?
It's not classical velocity addition, but relativistic velocity addition. In the frame of the train, the light beam travels vertically. To find the velocity of the light with respect to the ground, one adds--relativistically--the horizontal speed of the train. The net result of that addition is that the speed stays the same (still c) but that the light moves on a diagonal.

Look up: relativistic addition of velocity; aberration
 Mentor (Everything below is in units such that c=1). 1. The direction of the velocity depends on the frame. This is obvious in the case of rotations, but less obvious in the case of a velocity boost. If the velocity is in the direction of the vector (1,1) in the xy-plane, the velocity in a frame that's moving with speed v in the positive x direction of the first frame is in the direction of $(1-\sqrt 2v,1)$. 2. Yes, but the velocities add up in a complicated way: $$\vec u\oplus\vec v=\frac{1}{1+\vec u\cdot\vec v}\bigg(\vec u+\vec v+\frac{\gamma_{\vec u}}{1+\gamma_{\vec u}}\vec u\times(\vec u\times\vec v)\bigg)$$ The definition of $\gamma_{\vec u}$ is $$\gamma_{\vec u}=\frac{1}{\sqrt{1-\vec u^2}}$$ The specific case you're asking about is $\vec u=u\vec e_0,\ \vec v=\vec e_1$. 3. Why wouldn't it be up and down? The observer and the light clock have the same velocity (if I understand your description correctly), so from their point of view, it's the ground that's moving. So the only way the direction could be anything else is if a) the motion of other objects in the vicinity drags the light with it, or b) if this observer's point of view is less valid than the point of view of someone standing on the ground outside. Edit: After reading Doc AI's answer I think I probably misunderstood your description. If the clock is resting on the ground outside the train, then the answer is what he said.

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## Questions abouut:Special Relativity, Time Dilation, Light Clock, Velocity of light.

 Quote by Canaan_C Q2. Suppose the direction of a light ray is prependicular to the velocity of the light source (like the one in light clock.), can the light ray get the horizontal speed from the light source?
As written, NO.
The light ray "does not get" the horizontal speed from the light source (since there is no force acting horizontally to change the direction of the light ray).

Rather,
the light ray "happens to have" the same horizontal speed as the light source.

I think of it this way.
The initial flash has light rays in all directions.
The particular light ray that has the same horizontal component of its velocity as the moving light clock will be the one that will be reflected by the distant mirror of the light clock... and remain with this inertially moving light clock. The other light rays are lost (if not reflected by other light clocks moving inertially with different velocities).
 Thank you all of you, I think I wil adopt what Doc Al said at the stage. The idea from robphy is what I guess originally, but I don't think it is a appropriate explanation. What I want to know is why the light ray "happens to have" the same horizontal speed as the light source?

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 Quote by Canaan_C What I want to know is why the light ray "happens to have" the same horizontal speed as the light source?
Because if it the horizontal speed has any other value, the observer on the train will disagree with the observer on the ground about which spot the light hits. Imagine that the ray of light is a laser that hits an ant and kills it. The two observers would disagree about whether the ant is alive.

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 Quote by Canaan_C Thank you all of you, I think I wil adopt what Doc Al said at the stage. The idea from robphy is what I guess originally, but I don't think it is a appropriate explanation. What I want to know is why the light ray "happens to have" the same horizontal speed as the light source?
As I described, it's the only one that gets reflected by that mirror.
Watch my video:
physics.syr.edu/courses/modules/LIGHTCONE/LightClock/VisualizingProperTime-y-pair-A-with-photons.avi

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 Quote by Canaan_C I am a new fish to SR …
SR needs more fish!

Welcome to PF!

Preserve the bowliverse!

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 Quote by Canaan_C I am a new fish to SR
 Quote by tiny-tim SR needs more fish! :wink
Welcome indeed!

But how can SR have fish? Wouldn't the water define a preferred frame of reference?

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 Quote by atyy Welcome indeed! But how can SR have fish? Wouldn't the water define a preferred frame of reference?
water? what's water?
 What I am asking is what "JustinTime " asked in his post : Light shone in a train bouncing off mirrors

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Hi Canaan_C!
 Quote by Canaan_C What I am asking is what "JustinTime " asked in his post : Light shone in a train bouncing off mirrors
ooh, that's a very long thread

let's go back to …
 Quote by Canaan_C Q2. Suppose the direction of a light ray is prependicular to the velocity of the light source (like the one in light clock.), can the light ray get the horizontal speed from the light source? Q3. In the frame of a person who stand beside a light clock on a moving train which having speed "u" to right, why the path of a photon that he observed is vertically up and down ? Is it because the photon got the horizontal speed ("u" to right) .If yes, then it become a classical veloctiy addition ?; If no, the photon got no horizontal speed from the source, then how can the person on the train see a vertical up and down path instead of diagonal path?
It depends what angle the light starts.

If the light goes through two pinholes in plates attached to the train which are arranged vertically (as viewed by someone in the train), then the light will go vertically up in that frame, hit the mirror on the ceiling, and be reflected back along its path (as viewed by someone in the train), but will be seen as zigzagging by someone on the track.

If the light goes through two pinholes in plates attached to the track which are arranged vertically (as viewed by someone on the track), then the light will go vertically up in that frame, hit the mirror on the ceiling, and be reflected back along its path (as viewed by someone on the track), but will be seen as zigzagging by someone in the train.

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 Quote by tiny-tim It depends what angle the light starts. If the light goes through two pinholes in plates attached to the train which are arranged vertically (as viewed by someone in the train), then the light will go vertically up in that frame, hit the mirror on the ceiling, and be reflected back along its path (as viewed by someone in the train), but will be seen as zigzagging by someone on the track. If the light goes through two pinholes in plates attached to the track which are arranged vertically (as viewed by someone on the track), then the light will go vertically up in that frame, hit the mirror on the ceiling, and be reflected back along its path (as viewed by someone on the track), but will be seen as zigzagging by someone in the train.
That is not the answer I usually get for this question, which I presumed was the standard SR answer. I'll explain myself.

In your configuration, tiny-tim, it seems light is allowed to spread in all directions, as it is its tendency. Thus it is quite understandable that, no matter the state of motion of the frame in question, if not one photon, another one (if not one ray, another one, if you want to put it in terms of the ray model), will find its way through the pinholes.

But the question becomes more challenging if you consider another configuration where light does not have that liberty. Think for example of light projected from a laser gun, which follows a considerably restricted path in parallel to the arm of the laser gun.

In this second scenario, Lorentzian relativity (based on the existence of the ether) would claim, if I’m not mistaken, the following: if the frame in question has residual acceleration (by this I mean the net effect of all its historical accelerations wrt the ether) zero or whatever but along the line joining the laser and the target, it would reach its destination, but if the frame’s velocity wrt the ether has a component perpendicular to that line, the light would not make it to the target, it would hit the top of the train somewhere else.

Instead, the standard SR answer is that the laser light will “never” miss its target in the slightest. Why? Because light does not take the speed (scalar component) but does acquire the direction of the frame from which it is projected, the direction of the source, in other words. Why so? Because of the principle of relativity, which states that all experiments render the same results in all frames, no matter their diverse states of motion. Otherwise, you would be able to classify frames according to this criterion: those where the laser light hits the target and thus conform to the general laws of physics and those where it doesn’t. Why should we adhere to that principle, formulated in such a rigid manner? Because experiments have always corroborated it, even if we have no “physical” explanation for it.

That is why in the other thread mentioned by Canaan_C I asked this question, which remained unanswered: We usually say that LR and SR are conceptually different, but predict the same practical results. However, this seems to be a difference, doesn´t it? I think we should simply accept there is such difference, whatever it means.
 Hello Saw. Perhaps this will help. Imagine a laser beam, as thin as you like, pointing at a spot immediately above it on the ceiling. If you considered the whole system, laser/ceiling stationary you woulr have no problem imagining the laser hitting the spot. Now the question is - How do you know whether the system is moving or not? The answer of course is that you cannot know. So as the frame the laser/ceiling is in is completely arbitrary, and will be moving relative to other frames, the laser will hit the spot in any case. Its path will of course look different to someone in the laser/ceiling frame than to someone not at rest in that frame. There is probably a better way of phrasing it but this was quick answer. Matheinste.
 Light always propagates spherically, it just don't always have the same frequency and/or amplitude on a particular part of this spherical "wave front" emitted. A laser is just a device designed, tuned, and optimized to produce these waves with a specific frequency along only one part of this spherical wave front (mirrors and other optical materials are good at this. So, next time you see a laser beam, remember you are only seeing one tiny thin fraction of the spherical "light wave" because the rest of it is outside (probably far, far outside) the visible. For pedagogical purposes let's consider the waves to propagate in some fluid-like medium. In this case you can think of a ball in water which suddenly expands and then contracts. A spherical region of the water around the ball will be alternatingly compressed/rarified, similar to how sound propagates, and as such will have a characteristic wavelength w and frequency f. The wavelength is just the distance between compressed (or rarified) regions and the frequency is just the length of a single compressed (or rarified) region. Empirically we know that the ball cannot know if it is moving in this fluid or not (and, by extension, if the fluid is even there). How come? Let's analyze the situation in some detail. When the ball moves in this fluid the waves are compressed/shortened in its direction of motion and lengthened behind it. This is the standard Doppler effect, which you should familiarize yourself with if you haven't already. The ball moves a little forward, emits a wave, then moves a little forward and emits another, making the waves "in front" of it closer together and the ones "behind" it further apart, shortened and widened exactly by the magnitude of the velocity. If the ball is moving at 0.5*c (where c is how fast the waves propagate with respect to the fluid) then the wavelength of the light in the direction of the ball's motion (forward wavelength wD') is: wD' = 1-(0.5/1) = 0.5*w and its wavelength in the direction away from its motion (wD'') is: wD'' = 1+(0.5/1) = 1.5*w. These relations come from understanding the standard Doppler effect. So it should be readily apparent from measurement that the wavelength of light emitted by a moving body is 50% smaller in one direction and 33% larger in the other, right? This is an asymmetric effect and should be indicative of an aether. Unfortunately, the ball itself expands/contracts more slowly when it moves through this fluid, because of the way it interacts with the fluid. The wavelength (forward and backward) increases (greater distance between similar regions). The wavelength emitted by the ball moving at 0.5*c in the forward direction (before applying the regular doppler effect) is: wG'=w/sqrt(1–(0.5/1)^ 2) = 1.1547 In the reverse direction it's the same because the ball's slower expansion/contraction applies in all directions: wG''=w/sqrt(1-(0.5/1)^2) = 1.1547 So, the forward and reverse wavelengths with respect to the fluid are just the products of the doppler broadening/contraction and the Lorentzian symmetric wavelength broadening: w' = wD'*wG' = 0.577 w'' = wD''*wG'' = 1.732 The forward and backward wavelengths are exact inverses, w'*w'' = 1. So, if the ball is moving away from us we will see exactly the same wavelength shift as in the case were we are moving away from it! Since the wavelength is shifted by an identical proportion in both directions, the ball's directional motion within the aether cannot be identified. We can only detect relative motion. If I'm holding the ball and I detect its wavelength is w and I push it away from me, I cannot tell whether it's moving away from me or I'm moving away from it. The wavelength shift is the same proportion either way. This symmetry, which makes it impossible to observe this fluid medium directly, led most physicists to discard the fluid aether.
 Blog Entries: 8 Recognitions: Gold Member Matheinste, I can perfecty visualize, even for the LASER beam, a vertical path as measured in the coordinate system of the source frame and a diagonal path as measured in the coordinate system of the observer frame. I just wanted to point out that, if the beam is not part of a spherical wave (it is a laser beam), in the conceptual framework of SR there is no explanation for that effect, other than that if it were otherwise we would be able to detect the relative motion of the source frame wrt others. altonhare, the second part of your post is a brilliant description of why SR and LR are said to render the same practical results..., as a general rule. However, in the first part of your post, you seem to give me a reason why the issue posed in this thread should not be an exception, but I do not see the reason. According to LR, if there is an ether, that "tiny thin fraction of the spherical "light wave" that is visible should follow a path that is dependent on the state of motion (or no motion or whatever) of the ether but it should not follow the direction of motion of the source and should not hit the target it was pointing at in the source frame...
 Hello saw. According to SR the speed of light is constant and the same for all frames. This means that the point of emission, or source, remains central to the expanding sphere of light. In the case of a laser beam or ray, the ray is just a radius of this sphere. In LR the point of emission, or source would move away from the centre of the expanding light sphere if the source were moving relative to the ether. What i am trying to say is that in SR the source remains cetral to the expanding light sphere but in LR it can move away from it depending on whether or not the source is in motion relative to the ether. Does that help. Matheinste.

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