A Light ( or laser ) is shone out of the side window of a spaceship ( which is moving as close to the speed of light as possible ), WILL THE LIGHT STILL TRAVEL AT THE SPEED OF LIGHT IN THE B AXIS ON THE DIAGRAM BELOW ?

NOTE - No one seems to have clearly answered the question ( although they do indicate a diagonal trajectory )

enter image description here

NOTE - There is no need for the concept of the observer in this question, since the B axis is always the B axis .

  • Also, if the light does continue travelling at the speed of light, IN THE B AXIS, but it also travels 'SIDEWAYS' in the axis that the spaceship was travelling ( AXIS A ON THE DIAGRAM ABOVE ) , would that indicate that light / photons have mass ( since I often read that light / photons have no mass ) ?

  • Yes, surely I should think of this as the resulting DIAGONAL trajectory of the light, or should I, I have no qualifications in this area.

  • [ I Should ask this in a different question - How would physicists consider this light when it hits an object ( maybe an object travelling as close to the speed of light as possible, but in the exact opposite direction that the original spaceship was travelling in ), could the light / photons hit an object 'SIDEWAYS', or will it be DIAGONAL, and could the combined speed of the impact cause the light to hit the object at a speed greater than the speed of light ( I know the answer is always no ) .

  • Surely this has been answered in other questions, but it would be difficult to find .

NOTE - - I had deleted this question, due to an error I made in the question, I originally accidentally used the word 'PERPENDICULAR', when I intended to mean 'DIAGONAL'. This was a very significant error I made, and the results may still be present in one or more peoples answers . And, I have now also made a correction / clarification in the question title and initial description .

NOTE - I cannot post any comments, or reply to any comments, due to a technical problem which cannot be fixed.

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    $\begingroup$ The good thing about light is , it will behave the same for everyone no matter how fast they are travelling. So, you will get the same results from the spaceship experiment if you were to do that experiment in a moving car $\endgroup$
    – Jdeep
    Apr 7, 2021 at 7:51
  • 1
    $\begingroup$ If you made a mistake in your question, please correct it rather than vandalizing your post. If correcting it would invalidate existing questions, you can ask it as a a new question without vandalizing this one. $\endgroup$
    – Chris
    Apr 8, 2021 at 0:28
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    $\begingroup$ In the diagram your B direction appears to be at right angles (perpendicular to) the A direction. Is that not what you meant? In any case there's no one diagram that can work for all observers. From the point of view of an observer moving with the spaceship the light goes straight in the B direction and not at all in the A direction. From the point of view of an observer moving relative to the spaceship the light goes partly in the A direction (by the same amount the rocket does) and partly in the B direction. $\endgroup$
    – Eric Smith
    Apr 10, 2021 at 0:41
  • $\begingroup$ Why are you still asking this? Eric Smith already gave a complete and correct answer. He already answered your edits. He even gave you an explicit formula answering your question quantitatively. See his last formula. It directly explicitly and clearly answers your edits $\endgroup$
    – Dale
    Apr 12, 2021 at 3:37

4 Answers 4


The total speed of the light beam will be $c$ (the speed of light is always the same) but the direction of the light will, obviously, depend on the observer. If the light is shining perpendicular to the axis of the spaceship, then the "forward" motion of the light (the component in the direction of travel) will be the same as the forward velocity of the ship. You can see this because from the Earth's perspective in the time between the light leaving the laser and arriving at the window, the window has moved forward by a factor of $v$ (the forward velocity of the spaceship) times the distance from laser to window. Since the light is shining through the window, it also has to move forward that much. The perpendicular component of the light's velocity from Earth's frame is therefore $\sqrt{c^2 - v^2}$.

To sum up:

Spaceship view: velocity of Earth is $(-v, 0, 0)$, velocity of light beam is $(0, c, 0)$

Earth view: velocity of spaceship is $(v, 0, 0)$, velocity of light beam is $(v, \sqrt{c^2-v^2}, 0)$.


The speed of light is constant. Photons only travel one speed and one direction forward, so there is no sideways. You cannot add or subtract the speed of light from the speed of the space ship. The best way to think of it is to picture the energy being carried away from the light source as apposed to being emitted or launched. So if you have a light source such as a laser in a space ship moving close to the speed if light, then anyway you point the laser, the light will be carried away at the speed of light. Even when you point it backwards or all direction at once, it would be carried away at the speed of light.

  • $\begingroup$ if it's a narrow laser beam, for example, presumably the locus of the beam center forms a diagonal line even though the the photons individually travel perpendicular to the path of the ship? $\endgroup$
    – Roger Wood
    Apr 7, 2021 at 6:17
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    $\begingroup$ Photons can travel in any direction! The speed of light (magnitude of the velocity) is constant, but the components of that velocity are frame dependent. $\endgroup$
    – Eric Smith
    Apr 7, 2021 at 15:57
  • $\begingroup$ @RogerWood you cannot form a beam of light with photons on different paths. $\endgroup$ Apr 7, 2021 at 16:20
  • $\begingroup$ @Bill Alsept It's probably just semantics, but if I spin round and round holding my flashlight, won't I generate a spiral beam of light? $\endgroup$
    – Roger Wood
    Apr 7, 2021 at 19:48

The movement is relative. So the reasoning is the same as if we consider the spaceship at rest and shining a laser.

In this case it is the observer that is travelling close to the light speed in relation to the ship. And of course the beam will be diagonal to him.


The simplest way to approach this is to consider a photon (or a laser, with identical photons), and then Lorentz transform the 4-momenta.

But first, since you did not name the reference frames (which makes answering easier): The inertial frame of the rocket (observer) is henceforth $A$ ($O$). Moreover, "up" is $z$, and the beam is shone in the $x$ direction.

In the rocket frame, the photon momentum is:

$$p^{\mu}_A = (\omega, \omega,0 ,0) = \omega(1,1,0,0)$$

Note that the mass is:

$$ m^2 \equiv p^2_A = \omega^2(1^2-1^2) = 0 $$

Now boost this into the observer's frame:

$$p^{\mu}_O = (\gamma\omega, \omega, 0, \beta\gamma\omega) = \gamma\omega(1,\frac 1 {\gamma}, 0,\beta)$$

The mass is:

$$ m^2 \equiv p^2_O = \gamma^2\omega^2(1^2-\frac 1 {\gamma^2}-\beta^2) = 0 $$

So $p^0_O$ tells us the energy is higher by the Lorentz factor $\gamma$, $p^3_O$ shows that it tracks with the rocket, and along with $p^1_O$ gives the polar angle in the $O$ frame:

$$ \tan{\theta} = \frac{p^1_O}{p^3_O} = \frac 1 {\gamma\beta}$$

You could then boost this into any other frame to see what it looks like.


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