My friend and I were working on a problem which is to find the speed of the laser spot on the wall when you flick the laser pointer [ I know it can go faster than the speed of light as said in many youtube videos, but I want to see it with math ].

These are the initial conditions (Using polar coordinates):

$\theta=0$ for $t<0$

$\theta=\omega t$ for $t\ge0$

The wall's equation is given by $r\cos\theta=d$ [ The wall is at a distance d from the origin ].

Now, we figured out that the equation of the curve, made by the photon jet coming out of the laser pointer [ Even though each photon individually follows a straight path, the photon jet follows a curved trajectory ].

That equation is $c(t-{ \theta\over \omega})=r\;;\;0\le\theta\le\omega t$ [ This is how the curve looks like when no wall is present at any time $t\ge0$].

So, now our goal reduces to finding the speed of the point of intersection of this curve and our wall.

$r\cos\theta=d\;\;\Rightarrow\;\;c(t-{ \theta\over \omega})\cos\theta=d$.

But how do we get the solution ($\theta = \theta(t)$ ; Here $\theta$ is the angle subtended by the line joining the origin and the point of contact of the curve on the wall, with the axis) for the above equation?

The idea is once we know this $\theta=\theta(t)$, we can also find $r=r(t)$ by back substituting. Then, $y=r\sin\theta$ is the position of the laser spot on the wall and differentiating with time, $\frac{dy}{dt}$ ,gives the speed of the laser spot!

EDIT : @Sanya, Suppose the particle is approaching you with a uniform velocity $v$ [ Remember that $\frac{dx}{dt}=-v$, because, $x$ is decreasing as $t$ increases ]. When the particle is at a distance $x$ from you at time $t$, the time light takes to arrive from the particle to you, is $\Delta t=x/c$. So, you will observe the particle to be at a distance $x$ at time $t+\Delta t$ ( This is exactly your clock's reading ). Therefore, the observed speed $\left |v_{obs} \right |=\left |\frac{dx}{d(t+\Delta t)}\right | = \frac{\left | \frac{dx}{dt} \right |} {\left | \frac{d(t + { x \over c }) }{dt} \right |} = {v \over (1-{v \over c})}={vc \over (c-v)} \ge v $

  • $\begingroup$ nothing we actually observe is going faster than the speed of light ;) $\endgroup$
    – Sanya
    Apr 24, 2017 at 21:30
  • $\begingroup$ You can observe ( perceive through your eyes ) a particle going faster than light, but it doesn't matter. What counts is the actual velocity of the particle. $\endgroup$
    – Ajay Mohan
    Apr 25, 2017 at 5:35
  • $\begingroup$ Your statement does not make sense to me :( Maybe you can edit your question to include a short, easy to comprehend example? $\endgroup$
    – Sanya
    Apr 25, 2017 at 19:28
  • $\begingroup$ Suppose there is an illuminating particle approaching head on towards you with a uniform speed v ( which is less than c ). The observed speed of that illuminating particle is (c*v)/(c-v) [ Try to derive it by yourself by considering the fact that there is a finite time taken for the light signal emitted by the particle to reach you. I'll attach the derivation in the question ]. If v=3c/4 ( less than c ), then the observed speed is 3c ( which is greater than c ). $\endgroup$
    – Ajay Mohan
    Apr 26, 2017 at 14:54
  • $\begingroup$ If an illuminating particle is approaching with $v=\frac{3}{4}c$, I think employing Galilean relativity is wrong - special relativity should be used in this case, would be my guess at least ... $\endgroup$
    – Sanya
    Apr 27, 2017 at 20:31

1 Answer 1


Nice calculations, but chosen approach isn't the easiest one. You've run yourself into transcendental equation. Probably, guys on math.stackexchange are cool enough to give you precise solution, but there is an easier way.

Here is the plan:

  • find moment $t_1$, when laser spot reaches point $A_1$, which is at angle $\theta$ from initial beam
  • find moment $t_2$, when laser spot reaches point $A_2$, which is at insignificantly larger angle $\theta + d\theta$ from initial beam
  • calc instantaneous velocity by finding $\lim_{d\theta \to 0} \frac{A_1A_2}{t_2 - t_1}$


Here $\angle{LOA_1} = \theta$, $\angle{A_1OA_2} = d\theta$, $OL = d$, $A_1H$ is auxiliary perpendicular to $OA_2$.

Well, time interval is composed of time-to-turn and time-for-light-to-cover-dist:

$$t_1 = \frac{\theta}{\omega} + \frac{d/\cos \theta}{c}$$ $$t_2 = \frac{\theta + d\theta}{\omega} + \frac{d/\cos (\theta + d\theta)}{c}$$

I won't give full solution, since homework/exercise tag policy explicitly (IMHO and quite unreasonably) restricts it, but main points are:

  • $\cos (\theta + d\theta) = \cos \theta \cos d\theta - \sin \theta \sin d\theta \approx \cos \theta - d\theta \sin \theta$ (removed second power of insignificant param)
  • $A_1A_2$ can be found from $\angle{HA_1A_2}$ and $A_1H$ which in turn can be found out of $OA_1$

My answer if calcs are right: $$v(\theta) = c \cdot \frac{1}{\frac{c}{d\omega} cos^2 \theta + \sin \theta}$$ Edge cases are fine:

  • $\theta = 0$ results into $v=\omega d$: only rotation matters since time to reach wall is about the same for close angles
  • $\theta \to \pi/2$ results into $v \to c$: "light stream" is almost parallel to wall

Since $\sin \theta$ is always less or equal to one, making the second item small enough guarantees speed faster than light. So big enough $\omega$ and you are done.

Fun fact: taking round wall with you in the center makes all calculations (both your approach ane mine) trivial, so dot speed will be $v = d\omega$. But who would make task trivial instead of having fun with math?


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