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I have only just finished High school physics so my understanding is still fairly simple but I'm having trouble with this question.

Imagine you are in space traveling at a relativistic speed with a laser/light source. When you fire the laser do you see a straight line?

If you do, then when the light leaves the laser, what makes it continue to have the same velocity as you? If you sent out a pulse and aimed it exactly at a target on a side, then would it miss the target by continuing to travel in the same direction as you?

From a stationary observer looking down they would see a laser moving (lets say to the left) and at the instant it fires the laser is pointing in a straight line but the light will move off in a diagonal, moving to the left at the same speed as the laser.

If that laser stopped when it is fired then it would essentially look to be firing a laser beam diagonally!

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If you fire a laser directly at a target that is moving relative to you, then of course you will miss the target. I'm not sure what this has to do with relativity. –  WillO Feb 3 at 0:03
I'm just having trouble with some ideas so what seems obvious to some might not to others, I'm still wrapping my head around the concepts of light. –  Jordan Brown Feb 3 at 0:26

3 Answers 3

Yes, you will see the light beam traveling in a straight line. The principle of relativity states that all laws of physics are the same for observers in relative inertial motion. Therefore, you should see no difference in how light propagates whether you are standing still or moving rapidly, as long as you are moving rapidly with a constant velocity. However, in special relativity you cannot simply add velocities the way you are used to from high school physics. Otherwise there would be nothing preventing you from breaking the "cosmic speed limit" of the speed of light. Any time you walk while holding a flashlight you would be producing light moving at a speed $c +$ your walking pace! If you add colinear velocities $v$ and $u$, the result will be $$\frac{v+u}{{1+\frac{vu}{c^2}}}$$

You can see that if you consider velocities much lower than $c$, the denominator will approach $1$, and you recover the familiar formula $v+u$.

If you fire the laser off at an angle from the direction in which you are moving, then there is an additional consideration you must take into account. Angles will also change depending on your reference frame. This is behind a phenomenon known as relativistic beaming. If you were to take a light source that radiated in all directions, like a light bulb, and suddenly set it moving at high speed, there would be a higher intensity of light emitted along the direction it is moving as compared to other directions. The reason for this is that when a light ray is emitted along the angle $\alpha$ from the light bulb's perspective, it is emitted at the angle $$\alpha' = \cos^{-1}\left(\frac{\cos\alpha + V}{1+V\cos\alpha}\right)$$ from your perspective.

The last part of your question is a little difficult for me to decipher, but I hope this has been of some help.

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When you fire a laser, you don't see anything unless you fire it at yourself.

However, from your point of view in the fast-traveling spacecraft, your laser beam is straight. To you, you are stationary and everything else is moving.

If you point the laser at a fast-moving object (from your point of view), then the laser could miss the object. That is because you are seeing the object where it was a little time earlier, so you are aiming the laser in the wrong direction. In addition, the object will move some more after you fire the laser and before the beam can reach it, making the error even bigger.

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Light in vacuum travels at lightspeed for all observers. The paradox is taken up by relativistic Doppler shifting of the observed light's frequency. Light travels along null geodesic paths. Light defines a straight path through spacetime.

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