Greatings! I'm trying to understand special relativity and have one question bugging me.

In almost every book or video about the subject there is a thought experiment with moving light clock. I hope I need not elaborate on the sutup and the outcome of the experiment.

So the question is this: When a light source on the MOVING light clock emits a photon of light the path of the photon is triangular.

How this can be?

I thought since light speed is constant in every reference frame the movment of the "emmiter" can't affect the movment of the light so light should shoot right up (from the point of stationary observer) and thus move backward and up from the point of moving observer.

So you see my logic is: light speed is invariant => you can't make it move UP & FORWARD only UP. i.e. you can't change it's direction once it's emited in UP direction. Why the hell light moves sideways then?

And I'm puzzeled. There is flaw in the logic but I don't see it.

Thanks for help.


There are actually more than one ways to look at this problem. I am going to explain why the light must not go straight upward in the ground frame if the light clock is moving horizontally wrt ground frame.

The fundamental postulates of relativity is:

Laws of physics are invariant in all inertial frames.

Suppose the in the ground frame, the mid-point of the lower plate of the light clock is at (0,0) at time $t=0$. The upper plate of the light clock is at ($l$,0) at the same time. The clock is moving with constant speed $v$ in wrt the ground frame. A photon leaves the lower plate at $t=0$ (in ground's clock) in such a way that it is going straight upwards in the light clock frame. Now suppose there is a switch of some bomb attached with the midpoint of the upper-plate which if hit by a photon blows the light clock. Since the photon is going straight upwards in the light clock frame it will surely hit the upper plate and blow the light clock itself. Now if the photon moves straight upward even in the ground frame then it can never hit the upper plate as the light clock is moving and the upper plate must have travelled forward and thus the light clock actually survives in the ground frame. Which just can't happen because otherwise it indicates that laws of physics are different in the light clock frame and in the ground frame which just can't be the case in accordance with the first postulate of special relativity. Further, in order to be consistent with the first postulate of relativity, light has to go precisely in the direction so that it can hit the upper plate. So rather than contradicting with the relativity, the different direction of light is a necessity for the principle of relativity to hold.

Another thing: It is not dictated by the maxwell equations that the direction of light must be the same for all the inertial frame. If it would be the case it would actually break the directional symmetry of space. Because it would define a particular direction with a preference - the direction in which the light goes. Maxwell's equations just dictate that speed of light is same in all frames where Maxwell's equations are valid. i.e. at least all inertial frames.

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An important thing to notice about relativity is the changing observers doesn't change the events. All observer must be able to agree on what events happened.

Which doesn't mean that they can agree on all the measurements (angles, times distances) between those events.

If an observer in one frame observe that the light reflected from the mirror, than observers in other frames also see that the light reflected from the mirror because the light reflecting from the mirror is a thing that happened.

With that knowledge in hand you can analyze the path of the light in an easy frame (say, the light clocks own rest frame) and use the events that happen to help you fill in the details of paths in another frame.

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I am going to add a 4th point of view, hopefully complementary to the various versions based on the given that "reflection events have to be observed on the mirrors in both frames". After all, if one observer emits a light pulse in the "y-direction", the other observer must see the pulse propagating at an angle to "y" regardless of the presence or absence of a mirror setup.

The idea here has to do with two observations:

1) In any given frame light propagates not only at light speed c, but also in a direction $\vec{k}$ perpendicular to the light front at any given moment.

2) Different frames perceive different orientations of a light front due to relativity of simultaneity.

Let's see how this works. The first statement is nothing new. It follows from the Maxwell equations and it is a safe assumption. As for the second one, consider (a patch of) a plane wave light front propagating in the $y'$ direction as seen in the rest frame O' of the clock. Say it was emitted at $y'=0$ at time $ct'=0$. At later times $ct'>0$ it will be located at $y'(ct') = ct'$. Let's imagine that we validate this propagation by monitoring the light front with at least two detectors $A$ and $B$ positioned at different locations in the $y'=h$ plane, say at $(x'_A = 0, y'_A = h)$ and $(x'_B = L, y'_B = h)$, respectively. In the clock frame, both detectors will record the light front passing at $ct'= h$.

The other frame, O, must attribute the recording events at $A$ and $B$ to the passage of the same light front. But since $A$ and $B$ occur simultaneously in O', they are no longer observed simultaneously in O. Instead O sees the interactions at $A$ and $B$ at coordinates $$ \left\{\begin{eqnarray}x_A &=& \gamma(x'_A + \beta ct'_A) = \gamma\beta h\\ y_A &=& h\\ ct_A &=& \gamma(ct'_A + \beta x'_A) = \gamma h \end{eqnarray}\right. $$ and $$ \left\{\begin{eqnarray}x_B &=& \gamma(x'_B + \beta ct'_B) = \gamma(L+\beta h)\\ y_B &=& h\\ ct_B &=& \gamma(ct'_B + \beta x'_B) = \gamma(h +\beta L) = ct_A + \gamma\beta L \end{eqnarray}\right. $$
where $\beta = \frac{v}{c}$, $\gamma = \frac{1}{\sqrt{1-\beta^2}}$, and $v$ is the relative velocity. In fact, all points of the light front that O' observes at his time $ct'$ are each observed at a different time in O, depending on their position along $x'$. Conversely, events O observes at one moment $ct$ of his time at different locations along $x$, necessarily correspond to different times $ct'$ in O'. But then, what exactly does O observe of the light front at any $ct$?

Take for instance $A$'s time in O, $ct_A = \gamma h$. We know that for $x_A = \gamma\beta h$ the light front is located at $y_A = h$. Now consider what happens at $x_B = \gamma(L+\beta h)$ at the same moment $ct_A$. The front is definitely not at $y_B = h$ because it will only arrive there at $ct_B = ct_A + \gamma\beta L > ct_A$. So it has to be at some other location $\bar{y}_B < h$. Since we know $x_B$ and $ct_A$, we can use the reverse Lorentz transformations to find the time in O' for any event $(x_B, y, ct_A)$, regardless of the exact $y$ coordinate (yes, all events occurring simultaneously in the plane $x=x_B$ at time $ct_A$ in O also occur simultaneously in O'). Once we have the corresponding $ct'$, we know the $y$ location of the front since $y' = ct'$ in O'. We obtain $$ \bar{y}_B = ct' = \gamma(ct_A - \beta x_B) = \gamma [\gamma h - \beta\gamma(L+\beta h)] = \gamma^2 \left(\frac{h}{\gamma^2} - \beta L\right) = h - \gamma^2\beta L $$ The two points $(x_A, y_A)$ and $(x_B, \bar{y}_B)$ now give us the orientation of the light front in O at $ct_A = \gamma h$ as
$$ \frac{y - y_A}{\bar{y}_B - y_A} = \frac{x - x_A}{x_B - x_A}\;\;\text{or}\;\; \frac{y - h}{h - \gamma^2\beta L - h} = \frac{x - \gamma\beta h}{\gamma L + \gamma\beta h - \gamma\beta h} $$ Simplify and we are left with $$ y = -\gamma\beta x + \gamma^2 h $$

In other words, as seen in O the light front is tilted in the direction of motion at a slope $-\gamma\beta$. Correspondingly, its direction of propagation in O (the normal to the front) is also tilted in the direction of motion, at a slope $\frac{1}{\gamma\beta}$.

All that is left now is to compare this direction of propagation with the one obtained from the light clock setup in the usual way. If the clock has rest length $H$, in O' a light front travels across it in time $cT'=H$. In O the corresponding travel time is time dilated to $cT = \gamma cT' = \gamma H$, during which the clock itself moves a distance $D = \beta cT = \beta \gamma H$. So the direction of the light front as seen in O is tilted in the clock's direction of motion at a slope $\frac{\Delta y = H}{\Delta x = D} = \frac{H}{\beta\gamma H} = \frac{1}{\beta\gamma}$. Done.

Note that the first derivation of the slope did not make any use of the light clock, but only of the Lorentz transformations. Detectors $A$ and $B$ are meant to single out two distinct locations on the light front at a given moment in O', but can otherwise be removed without consequences.

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  • $\begingroup$ Thanks, the notion of a light-front helped me to understand this. Having said that, isn't this argument cyclic, since it relies on consequences of the premise (namely, the Lorentz transformations) in order to derive the premise (from which the Lorentz transformations arise)? $\endgroup$ – Jivan Pal Oct 14 at 16:01

The components of the velocity vector can change, it's the magnitude c to be invariant.

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  • $\begingroup$ Why then we(moving sideways) can shoot light upward and affect velocity vector of the light so it moves UP and SIDEWAYS? I mean... why we are able to affect not the MAGNITUDE of vector C, but the direction of it? And I must say the wording "we affect the direction" is moot the problem is: I can't formulate it better yet. $\endgroup$ – coobit Nov 5 '15 at 13:22
  • $\begingroup$ Because it's the theory. It has been developed to match with experiments (take for example the one from Michelson and Morley!) $\endgroup$ – Turms Nov 5 '15 at 13:32
  • $\begingroup$ Isn't it that the light appears to move sideways because the moving observer is moving sideways? $\endgroup$ – Kyle Kanos Nov 5 '15 at 13:34
  • 1
    $\begingroup$ Yes, that is a question. Why for stationary observer light moves sideways? It is easily understood for a ping pong clock (a ball insted of photon). Emission machine moves sideways and it adds sideways components to perfectly vertical shoot of a ball in the clock. but a photon... adding sideways components to a photon which was emitted perfectly up?!!! This just means to ME that emission machine affected the light speed (not in the magnitude but in direction). How is this possible? $\endgroup$ – coobit Nov 5 '15 at 13:40

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