Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

enter image description here

Light is the yellow arrow. Observer is the black arrow. Observer is moving at a constant speed of v, w.r.t to a Galilean frame of reference.

Now from the point of view of the observer (O), how will the motion of the light ray look like? Will it bend away from him?

Looking for a good explanation. Thanks!

share|improve this question

2 Answers 2

up vote 4 down vote accepted

The way to do problems like this is always to use the Lorentz transformations. Choose some sensible spacetime points in the rest frame $S$ and use the transformations to see what those points look like in the moving frame $S'$. In this case this is what the points look like in $S$:

Image1

The spacetime points are labelled as $(t, x, y)$ - we'll ignore $z$ since we only need two dimensions for this problem. We choose our coordinates in the rest frame the light starts at $(0, 0, 0$), and after a time $t$ it's reached a distance $y = ct$ so this point is $(t, 0, ct)$.

Now our moving frame, $S'$, is moving along the $x$ axis at velocity $v$. As usual we'll choose the coordinates so the two frames coincide at time zero, so the point $(0, 0, 0)$ is the same in both frames. We just need to find where the other point is in $S'$. The Lorentz transformations tell us:

$$ t' = \gamma \left( t - \frac{vx}{c^2} \right ) $$

$$ x' = \gamma \left( x - vt \right) $$

$$ y' = y $$

So we substitute $t = t$, $x = 0$ and $y = ct$ and we get:

$$ t' = \gamma t $$

$$ x' = -\gamma vt $$

$$ y' = ct $$

So in the moving frame $S'$ the light ray looks like:

Image2

So in the moving frame the light is moving at an angle to the vertical, and the angle is given by:

$$ \theta = \tan^{-1} \frac{\gamma vt}{ct} $$

There are a few followup calculations that might be interesting to make. For example what happens to the angle $\theta$ as $v$ approaches $c$. Our equation above gives:

$$\begin{align} \theta &= \tan^{-1} \frac{\gamma ct}{ct} \\ &= \tan^{-1} \infty \\ &= \frac{\pi}{2} \end{align}$$

So go fast enough and the light ray appears to be moving directly away from you.

The other check is that the ray is still moving at the speed of light in the frame $S'$. In the $S'$ frame the velocity is:

$$ v' = \frac{\sqrt{x'^2 + y'^2}}{t'} $$

Putting in the values we've calculated for $t'$ etc we get (I'll square everything to simplify the algebra):

$$ \begin{align} v'^2 &= \frac{\gamma^2 v^2 t^2 + c^2 t^2}{\gamma^2 t^2} \\ &= v^2 + \frac{c^2}{\gamma^2} \\ &= v^2 + c^2 \left( 1 - \frac{v^2}{c^2} \right) \\ &= v^2 + c^2 - v^2 \\ &= c^2 \end{align}$$

So the speed of light still equals $c$, as it should.

share|improve this answer

Light isn't a single vector. It propagates out in all directions, because it is a spherical wave.

edit: I stand corrected, but I was trying to imply being able to a light pulse would require some sort of spherical wave reflecting off of particles in air. If in vacuum then you would not see the light ray.

In this picture you wouldn't see anything because the light is highly collimated. It would have to reflect off of particles.

Anyway, light wouldn't appear to "bend" to observer. It would merely de doppler shifted, since light is at velocity c for all observers (in vacuum) regardless of observers position or velocity.

And for a Galilean reference frame the light (even though you tagged special relativity) would appear to "bend" away from observer. Because it is going to be closest when to the left of observer and when passing observer (making a perpendicular with observers view and lights path) then increasing distance after.

share|improve this answer
    
Why should the light be a spherical wave? A light wavepacket can be quite different. –  Urgje Feb 15 at 10:29

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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