enter image description here

I always see this picture for proof of the time relation between frames that moves relatively constant speed in the special theory of relativity.

The time for the observer who is in the box which travel at constant speed $V$.

$$ t'=\frac{h}{c} \tag 1$$

The time for observer who is outside of the box.

$$ t=\frac{\sqrt{x^2+h^2}}{c} \tag 2$$

$$ t=\frac{x}{V} \tag 3$$ and the time relation between observers is $$ t'=t \sqrt {1-\frac{V^2}{c^2}} \tag 4$$ Everything is Ok till here.

According to outside observer, the person who adjusts the laser gun (yellow drawn in picture) angle in the box must change the angle ($\alpha$) as shown in right side of the picture if the light is not affected from the speed of the box as the special theory of relativity claims. However, According to inside observer who adjust the laser angle, the angle should be exactly $90^0$ because the inside observer will not understand that the box moves.

How does the special theory of relativity explain this dilemma?

(Please assume that the sensor on top is very small and the laser has very focused slim light beam and the box is quite speedy otherwise the angle always would be very near $90^0$ and the angle cannot be detected for low speeds. )

  • 1
    There isn't an adjustment to be made or an angle to be changed, in the box you shoot it at 90° degrees, the observer outside the box sees a different angle, that's an effect called aberration of light. Nothing is in contrast with relativity, the speed of light is not affected by the motion of the box. You may say that its components change, that's true, but its speed is always c. – Run like hell Feb 20 '17 at 9:43
  • @Runlikehell Will the outside observer see that the laser equipment angle changed to $\alpha$ although the inside observer claims that no need to change from $90^0$? Is not it absurd? Because when the box stopped the events will not be not much . If the laser equipment angle is also $90^0$ for outside observer , the light should be clever to know the direction of the box. It means there should be an ether as the same we observe this kind of actions in the river. I wonder how we eliminate this dilemma. – Mathlover Feb 20 '17 at 10:59
  • Yes I think the observers see different angles. Because of Lorentz contraction you see an object in motion relative to you rotated, that's an apparent rotation known as Terrell rotation. But the guy inside doesn't make any adjustment, if he did change the angle the light beam wouldn't hit the detector in his own reference frame. – Run like hell Feb 20 '17 at 13:45
  • Observer who is at rest will see photon coming at oblique angle. In this case the source, the photon and the sensor on the top always have the same x coordinate. Terrell rotation is purely optical illusion. If the sphere is semi - transparent and a measuring rod is placed inside the sphere, on a picture we will see that sphere is rotated and measuring rod is contracted. – Albert Feb 20 '17 at 14:00
  • Of course it is only an apparent rotation, plus he would see the rod tilted with its head pointing in the direction opposite to the one of the motion. But the OP asked in comments if the two observers saw different angles, that's why I advocated the rotation, even if it is an apparent rotation, the observer at rest sees another angle. – Run like hell Feb 20 '17 at 17:17

The laser gun is not rotated for the outside observer. It is still at 90 degrees. Just imagine that some laser light leaks out of the back of the laser gun, then you would have two different angles that the laser gun would have to rotate in ....

The laser gun is still at 90 degrees relative to the box but it sends out light under an angle because of non-simultaneity. The phase over the surface of its output is not constant in the x-direction due to non-simultaneity. Such a varying phase is also how phased array radars can send out a beam under an angle. See the image below (source: https://en.wikipedia.org/wiki/Phased_array)

enter image description here

It sends out the beam under an angle and the wavefront of the laser beam also rotates (The wave front should always be in the direction of the propagation). See the middle image below for the Lorentz transformed frame.

The image at the left depicts a photon bouncing up and down in the box for the observer inside the box. The middle image shows the photon bouncing up and down as seen by the observer outside the moving box.

enter image description here

Without non-simultaneity, as in the "Mansoury-Sexl" transform in the image on the right, the wavefront would not rotate.

The image below explains the effect. The term $~t' =\gamma(t-\beta x)~$ shows how $t'$ depends on x. Looking at the image below: At the left side time has progressed further and thus the phase of the wave front has progressed further. At the right side it's the opposite. The net effect is that the wavefront rotates.

enter image description here

For more see the relevant chapter of my book here: http://www.physics-quest.org/Book_Chapter_Non_Simultaneity.pdf

  • Do you claim that the laser gun angle is $90^0$ for outside observer too? If so, you think that the light has the same effect as a river affects an object for outside observer. Sorry but I did not understand with your explanations how the light knows the direction of the box and the speed of the box and goes to sensor's direction on top if the angle of laser gun is $90^0$? – Mathlover Feb 20 '17 at 13:23
  • Yes the laser gun angle is 90 degrees for the outside observer as well off course. Only imagine that some laser light leaks out of the back of the laser gun. then you would have two different angles that the laser gun would have to rotate in. Non simultaneity has nothing to do with a 'River' Maybe you could first try to understand the 'phased array' example first and then understand why non-simultaneity causes the phase to vary over the surface of the laser gun output. – Hans de Vries Feb 20 '17 at 13:34
  • Great explanation. Just one comment: don't lasers work more like the "bouncing photon clock" than like phased array radars? – Peter Shor Feb 22 '17 at 12:55

From the point of view of the outside observer the laser gun should not be tilted. It will be directed straight up too. The source and photon will move in positive x direction with the same velocity and will always have the same x coordinate. Please look at this video:


In the first episode moving source (a lamp) launches a photon. Both photon and the lamp always have the same X coordinate. Then a small tube appears on the top of the lamp. This tube imitates laser pointer, which always directed at right angle to direction of motion.

I think I understand what confuses you. You think that for moving observer velocity of light in different directions is the same. Not quite. Imagine two observers A and B, who move relatively to each other. Let A conduct observations from his reference frame K. A admits, that velocity of light is c in his frame K. But, if observer B, who moves in this frame K flashes a lamp, observer A will see, that light recedes from B with different velocities in different directions.

However, if B wants to conduct measurements, he introduces his own frame K’ and admits that velocity of light in different directions is c. But now B will see, that if A flashes a lamp, light recedes from A with different velocities.

In Special Relativity observer NEVER admits state of his own motion. Observer is always at rest in his own rest frame.

Thus, in Special Relativity every observer covers “mutual for all observers” space by his own reference frame. This reference frame implies that there is an Einstein – synchronized clock in each point of space. Special Relativity is the change of frames. Observers never use one mutual one, but each has his own.

In the same frame moving observer sees, that clock of observer at rest runs faster, not slower. If he wants to measure time dilation, he has to change reference frame and introduce his own one.

Please compare how many ticks moving clock does and how many any synchronized one. Moving clock makes less ticks that one at rest. That means, from the point of view of a moving clock time in reference frame tick faster. You will see, that once moving clock turns into one at “rest” so as to see time dilation.


  • This action that you defined is the same effect as we throw a rock to upside in a train. The rock which we throw to upside always goes as you defined in your answer. If the light is not affected by the speed of box and the angle of gun is $90^0$ for outside observer, how the light can go in the same speed with box in x direction . The principle you defined brakes the special theory ."... light is always propagated in empty space with a definite velocity [speed] c which is independent of the state of motion of the emitting body" Best Regards – Mathlover Feb 22 '17 at 8:26
  • Light propagates with velocity c in observer's frame. But "outside observer" will see, that beam of light recedes from source with velocity lesser than c. For outside observer beam moves on the hypotenuse, but it's components by triangle legs less than c. For example, please look at mathpages.com/rr/s2-05/2-05.htm, starting from " Another interesting aspect of aberration is illustrated by considering two separate light sources S1 and S2, and two momentarily coincident observers A and B as shown below " – Albert Feb 22 '17 at 10:26
  • I have added some my reflections into the answer – Albert Feb 22 '17 at 12:31
  • I understand your point of view ,The gift in end explains very well. I know If the laser gun angle is $90^0$ for outside observer and If we wants that the laser light hits sensor, X direction speed must be $V$ in your approach. The problem is in this idea: To have a constant the speed of light need to decrease the speed that goes to upside way ( $V_y^2=c^2-V^2$). How does the light know the speed of the box and decrease upside speed ($V_y$) to adjust and to get the total speed is $c$ if special relativity claims that the light is independent of the state of the motion of the emitting body. – Mathlover Feb 22 '17 at 13:24
  • I think that the particle approach of photon has problem in here if the special relativity is true. – Mathlover Feb 22 '17 at 13:26

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