2
$\begingroup$

enter image description hereConsider the following experiment:

A stationary observer A is in a 2D frame of reference F that has coordinates (x,y).

There is a light source L moving at velocity v in the direction of the X coordinate. The light source L is aimed in the direction of the Y coordinate, hence a photon is emitted at right angles to its direction of travel. The photon travels at the speed of light, c.

At time t = 0 , observer A is at (0,0) and light source L is at (1,0).

When t=$t_1$, the light source L moving at velocity v has travelled v$t_1$ (ie distance = speed x time), hence:

At time t = $t_1$ , observer A is at (0,0) and light source L is at (1+v$t_1$,0)

enter image description here

At t=$t_1$, with reference to observer A, what is the new position of the photon that exited the light source when the light source was at (1,0)?

Is it:

Option 1 - (1, c$t_1$)
Option 2 - (1+v$t_1$,c$t_1$)
Option 3 - (1+v$t_1$, $t_1$ $\sqrt{c^2-v^2}$ )

In Option 3 we made the distance traveled in the Y direction to be $t_1 \sqrt {c^2−v^2}$ so that the diagonal distance traveled by the photon would be $ct_1$. This means the speed of the photon along the diagonal would be c (ie $ct_1$/$t_1$), so A observes the photon to travel at speed c. See the diagram below for an explanation of how y was calculated.

enter image description here

If Option 1 is correct, then:

  • A observes the photon to travel at c
  • The velocity of the source does not affect the velocity of the photon (ie it does not change the speed of the photon nor the direction of the photon) as the photon has no velocity vector in the X direction.

If Option 2 is correct then:

  • A observes the photon to travel at a velocity greater than c
  • The velocity of the source does affect the velocity of the photon as the photon has a velocity vector in the X direction.

If Option 3 is correct then:

  • A observes the photon to travel at speed c
  • The velocity of the light source does affect the velocity of the photon as the photon has a velocity vector in the X direction. While the photon's speed has not changed its direction has been affected by the velocity of the source.

If Option 3 is correct, does it mean the velocity of a light source affects the velocity of light? While the speed of light did not change for observer A, the direction the light traveled was affected by the velocity of the light source and since velocity is a vector specifying speed and direction then the velocity of the light has been changed by the velocity of the light source. Is that correct?

$\endgroup$
  • $\begingroup$ PLEASE DON'T USE ALL CAPS. $\endgroup$ – user4552 May 24 '18 at 18:00
  • $\begingroup$ No. For each observer photons move as if they were emitted by a source that was stationary in the frame of the onserver. $\endgroup$ – safesphere May 24 '18 at 19:11
  • 3
    $\begingroup$ Please refer to relativistic aberration. $\endgroup$ – Ng Chung Tak May 24 '18 at 19:41
  • $\begingroup$ If a laser pointer emits light at right angle, x-coordinate of the light pulse will always be the same as the laser ponter's. Some videos show emission / reception at different angles. youtube.com/watch?v=hnphFr2Iai4; youtube.com/watch?v=FQKp3FU8vR8 $\endgroup$ – Albert May 27 '18 at 20:23
1
$\begingroup$

Let us analyse the options one by one. We use the fundamental postulate of special relativity and Lorentz transformations.

If option 1 is correct then the observer sitting on the light source would measure the speed of light as greater than $c$.

If option 2 is obviously not correct due to the same reason for the observer at $A$.

Option 3 is interesting. We see that speed of light is $c$ for $A$. Now let us place an observer such that they are at rest with respect to the source ($S$). Now, for this observer, $A$ is moving with a velocity $v$ in the negative $x$ direction. When time elapsed is $t_1$ in the $A$ frame, if we denote time for the $S$ frame with $t'$ then, from distance travelled $L$ and $L'$ in frames $A$ and $S$ respectively,

$$\frac{L}{t_1}=\frac{L'}{t'_1}$$

or $$\frac{\sqrt{v^2t_1^2+t_1^2(c^2-v^2)}}{t_1}=\frac{\sqrt{t_1^2(c^2-v^2)}}{t'_1}$$

which gives us the time-dilation relation

$$t'_1=t_1\sqrt{1-\frac{v^2}{c^2}}$$

So mathematically, option 3 does not disappoint.

A source moving with a uniform velocity is the same as an observer moving with a uniform velocity with respect to the source. In the frame $S$, which is just a light source and a detector at rest with respect to each other, everything works out intuitively. A better statement would be that light seems to tilt, towards the direction of the source's motion (relative to the observer) due to the motion of the observer. Or, you can think this way, as Wikipedia says

It is as if light emitted by a moving object is concentrated conically, towards its direction of motion; an effect called relativistic beaming.

https://en.wikipedia.org/wiki/Relativistic_aberration

Though I feel that perhaps it is better to think that the motion of observers themselves affect the measurements recorded by them.

To answer the question, yes, velocity of the source or observer with respect to each other does affect the measurement of the velocity of light.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.