A related question has been asked multiple times (like here, here and here). But none of those answers are clearing my doubt. In fact they're furthering my confusion. My specific doubt is regarding the contrasting explanations in the two common cases described below.

Consider Alice and Bob. Alice is moving horizontally rightward at a uniform velocity of v with respect to Bob.

Case 1:

Alice shoots a light pulse ahead of her (rightward).

  • Alice observes velocity of light to be c moving rightward.
  • Bob also observes velocity of light to be c moving rightward (and not c + v, because speed of source doesn't get added).

Case 2:

Alice is still moving rightward with a velocity v holding a light clock with her this time. A light pulse is shot vertically upward (according to her frame of reference) from the bottom mirror as shown in the image below.

  • Alice observes the light pulse travel vertically upward at speed c.
  • Bob observes the light pulse travel in north-east direction.

My question is regarding this Bob's observation highlighted in Bold. Why is the direction of light pulse changing according to Bob? It seems like vector addition of v and c but only considering a unit vector corresponding to v. This observation also clearly indicates that the motion of source does affect velocity of light (albeit the speed remains constant).

Kindly note that my question is not about light clock and its impossibility. As far as my question is concerned the second case is just about Alice shooting a light pulse vertically upward.

enter image description here

  • 4
    $\begingroup$ The rules aren't that you can't affect the velocity of light, it is that all frames agree on the magnitude of that vector being c. $\endgroup$ – Triatticus Nov 30 '18 at 21:18
  • $\begingroup$ @Triatticus So the net direction of the light pulse has to be determined by vector addition of the velocity of source and the velocity of light pulse? i.e., if $\vec{c}$ is original velocity of light pulse and $\vec{v}$ is velocity of source, then $\vec{c}_{resultant} = c(\frac{\vec{v}+\vec{c}}{\lVert \vec{v}+\vec{c}\rVert})$. Here, $\frac{\vec{v}+\vec{c}}{\lVert \vec{v}+\vec{c}\rVert}$ represents unit vector of the resultant. Please confirm if this interpretation is correct. $\endgroup$ – yathish Dec 1 '18 at 2:59
  • $\begingroup$ You just have to use the relativistic velocity addition formulas as the Galilean ones won't work here $\endgroup$ – Triatticus Dec 2 '18 at 18:22

A related question has been asked multiple times (like here, here and here).

I have no time to carefully examine those posts and answers, so I will take the matter anew. My problem however is that I didn't understand what your doubt is. So I'll simply try to present the light clock operation more carefully, and hope this may be of some help to you.

A minor note on your wording

  • Alice observes the light pulse travel vertically upward at speed c.
  • Bob observes the light pulse travel in north-east direction.

The way you describe directions is not consistent: "upwards" and then "north-east". Maybe Alice sends the pulse northward?

Be that as it may, consider that light pulse is of a very small transverse size and that the mirror too is very small. This means that the pulse has to be very accurately aimed at the mirror, otherwise will miss it.

Another point. If Alice sees the pulse hitting the mirror, the same will happen to Bob. Pulse hitting mirror is an objective phenomenon. It's impossible it does happen for one experimenter and doesn't for the other.

I understand a possible objection could be raised. Bob could say: "I can measure that Alice's light source stayed "vertical", as both its ends are moving, in my frame, at the same speed $v$. I don't understand" - it's always Bob speaking - "how a vertically directed source (e.g. a laser) could emit a light pulse in an oblique direction. Yet I see so it is, but I would like to understand a physical reason for that, independent of light clock and SR".

Bob is absolutely right. I could make him satisfied, by showing that - because of Maxwell equations and of laser functioning - a moving laser really outputs oblique pulses. Or I could give the proof for other possible collimated sources. There is nothing paradoxical in it. But this answer already consumed a significant part of my evening, so I apologize if I refrain from exhibiting these proofs now.

  • $\begingroup$ Thanks for your time. I get the objective phenomenon part now. My question was whether the velocity of source effects the velocity of light (velocity being a vector). Speed of light, of course, remains constant. $\endgroup$ – yathish Dec 1 '18 at 3:03

I'm not sure exactly what level answer you're looking for, so let me try twice:

  1. Alice, traveling eastward with respect to Bob at with velocity $v$, emits a light beam that travels (according to Alice) with an eastward velocity component $x$. Bob measures the velocity of the light beam. Is or is not the result of his measurement affected by Alice's velocity $v$?

I think you are troubled by the fact that the answer seems to be no when Alice sends the lightbeam eastward and yes when she sends it northward, and these seem to be inconsistent.

But everything is consistent with the velocity addition law: Bob has to measure the light beam traveling eastward at speed $$w={v+x\over1+vx}$$

(I am choosing units so $c=1$.)

When Alice shoots her lightbeam due east, we have $x=1$, so Bob gets $w=1$. When she shoots her lightbeam due north, she measures its eastbound velocity component as $x=0$, so Bob gets $w=v$.

These are two extreme cases, but they're both special cases of the same general formula. Once you know that formula, the apparent contradiction goes away.

  1. You might object that this formula was somehow cooked up just to fit the apparently contradictory facts and therefore there's no reason to believe in it. But now we go back to first principles:

Alice is entitled to consider herself stationary (the first postulate of relativity!), so when she sends the beam northward, it must hit the mirror that's due north of her.

If Alice says it hits the mirror, Bob says it hits the mirror.

So the first postulate of relativity tells us that when Alice says the lightbeam moves northward, Bob must say that it moves obliquely. At the same time, the second postulate of relativity tells us that when Alice says the lightbeam moves rightward at speed $1$, Bob must agree.

So if the postulates are correct, then there has to be some correct general formula that matches both cases.

  • $\begingroup$ Thank you. You have understood my precise confusion although by now I have myself learned how to analyse the situation quantitatively. Your analysis now makes sense to me too :) $\endgroup$ – yathish Dec 8 '18 at 13:50

Thanks to @Triatticus for pointing me in the right direction. I'm pasting what I've learned as an answer here (for my own future reference as well). I've also posted this answer here.

The velocity of moving clock does get added to the velocity of light but the velocities don't add up in a Galilean (linear) fashion. So it would be incorrect to use the phrase "add" here. Instead we can say the velocity of moving clock changes velocity of light pulse. The speed of light pulse still remains $c$, but its velocity changes (i.e., only the direction changes).

Qualitative reasoning - According to the stationary observer, the light pulse has to travel at an oblique angle in order to catch up to the top half of the light clock. This explains the oblique angle seen by the stationary observer.

Quantitative reasoning -

The relativistic velocity addition formulas are:

$\mathbf u_\parallel = \frac{\mathbf u_\parallel' + \mathbf v}{1 + \frac{\mathbf v \cdot \mathbf u_\parallel'}{c^2}}, \quad \mathbf u_\perp = \frac{\sqrt{1-\frac{v^2}{c^2}}\mathbf u_\perp'}{1 + \frac{\mathbf v\cdot \mathbf u_\parallel'}{c^2}}$

Here, the primed frame of reference is that of the moving light clock and the unprimed frame of reference is that of stationary observer. So, if the moving light clock is moving at a velocity $v$, then for the light pulse we have $u_\parallel'=0, u_\perp'=c$. Plugging in these values in above two formulas we get,

$\mathbf u_\parallel = v,\quad \mathbf u_\perp = \sqrt{1-\frac{v^2}{c^2}}\mathbf c$

These are the components of velocity of the light pulse as seen by the stationary observer. One can also calculate the oblique angle by using $\theta=\arctan{\frac{u_\perp}{u_\parallel}}$. Also, the magnitude of this velocity i.e., $\lVert \mathbf u\rVert = \sqrt{\mathbf u_\parallel ^2 + \mathbf u_\perp ^2} = c$. Thus it can be seen that the speed of light is consistent!


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