Recently I encountered this (quite interesting) question:
Rain is falling vertically. A man running on the road keeps his umbrella tilted but a man standing on the street keeps his umbrella vertical to protect himself from the rain. But both of them keep their umbrella vertical to avoid the vertical sun-rays. Explain.
The answer that I thought is as follows:
The speed of man is comparable to that of the rain and hence his act of walking cause the relative velocity vector to change in direction and magnitude. Whereas his speed is nowhere comparable to that of light (about $3 \times 10^8\ \mathrm{m\ s^{-1}}$) and hence there is no significant deviation of the relative velocity vector of light.
But an obvious question that occurred to me was:
- What if the speed of man was quite comparable to that of light (say $0.9c$, just to give some figures!) then would there be a change in direction and hence the need of tilting the umbrella?
I think there might not be any deviation at all as the speed of light is invariant in all inertial frame of references (though this conclusion might not be correct as my understanding of SR is quite superficial.)
Best Answer
This is a simple but important point. While the speed of light is invariant among all inertial observers, the direction of light is not. Let's show this explicitly. Let's say that light is propagating in the negative$-y$ direction w.r.t. observer $\mathcal{O}_1$ who is standing still. Now, consider an observer $\mathcal{O}_2$ running at a uniform speed $v$ in the positive$-x$ direction w.r.t. $\mathcal{O}_1$. The relativistic relation between the velocity $\vec{u}$ of an object as observed by $\mathcal{O}_1$ and the velocity $\vec{u'}$ of the same object as observed by $\mathcal{O}_2$ is the following
$$u'_x=\frac{u_x-v}{\sqrt{1-\frac{v^2}{c^2}}}$$
$$u'_y=\frac{u_y\sqrt{1-\frac{v^2}{c^2}}}{{1-\frac{vu_x}{c^2}}}$$
In our case, we are concerned with the transformation of the velocity of light which is $c$ in the negative$-y$ direction as observed by $\mathcal{O}_1$. Thus, $u_x=0,u_y=-c$. Putting in these values in the above formulae would give you
$$u'_x=\frac{-v}{\sqrt{1-\frac{v^2}{c^2}}}$$ $$u'_y=-c\sqrt{1-\frac{v^2}{c^2}}$$
Thus, in the reference frame of $\mathcal{O}_2$, light makes an angle of $\arctan\frac{u_x}{u_y}=\arctan\big(\frac{v/c}{{1-{v^2}/{c^2}}}\big)$ with the vertical. So, indeed, you are correct. If $v/c$ is not ridiculously small, the person would have to tilt their umbrella noticeably.
You can easily verify that the speed of the light is still $c$ in this reference frame.