The fastest trajectory that a particle should follow through two different mediums

Hello, after 3 failed attempts to solve this problem, I decided to start a bounty for this question. Please, I need a complete answer with an interpreation of the final result. Thank you in advance.

Problem: A particle travels at speed $v_a$ in a medium $A$ and at speed $v_b$ in an medium $B$. The particle departs on time $t=0$ from the point $P_i$ and has to arrive in the minimal time to the point $P_f$, as in the picture. Determine the trajectory that the particle has to follow to arrive from to the point $P_f$ in the minimal time.

Attempt 1: I know that the fastest way to go from a point to another point is the segment of straight line that joins the two points, but I don't know how to consider the velocities $v_a$ and $v_b$ of the problem, or if they're just distractors to confuse me and there's no need to consider them. Am I right or not?

Attempt 2: Let $P_c$ be the point where the particle crosses the boundary between $A$ and $B$, and let $r$ be the distance from the left point of the boundary to $P_c$. We have

$t_{P_i P_c}=d_{P_i P_c} /v_a=\frac{\sqrt{h^2+r^2}}{v_a}$

$t_{P_c P_f}=d_{P_c P_f} /v_b = \frac{\sqrt{m^2+(s-r)^2}}{v_b}$

So,

$t_{P_i P_f}=t_{P_i P_c}+t_{P_c P_f}=\frac{\sqrt{h^2+r^2}}{v_a}+\frac{\sqrt{m^2+(s-r)^2}}{v_b}$.

Now, we have to consider the function:

$t(r)=\frac{\sqrt{h^2+r^2}}{v_a}+\frac{\sqrt{m^2+(s-r)^2}}{v_b}$,

differentiating with respect to $r$ and setting to zero, we get:

$t'(r)=\frac{r}{v_a \sqrt{h^2+r^2}}-\frac{s-r}{v_b \sqrt{m^2+(s-r)^2}}=0$

So,

$rv_b \sqrt{m^2+(s-r)^2}=(s-r)v_a \sqrt{h^2+r^2}$

and

$r^2v_b^2 (m^2+(s-r)^2)=(s-r)^2v_a^2 (h^2+r^2)$

Rearranging, we get the following 4th degree polynomial in $r$:

$(v_b^2-v_a^2)r^4+2s(v_a^2-v_b^2)r^3+(v_b^2m^2-v_a^2h^2+v_b^2s^2-s^2v_a^2)r^2+(2sv_a^2h^2)r-(s^2v_a^2h^2)=0.$

So, now do I have to solve this equation?

Attempt 3: Let $P_c$ be the point where the particle crosses the boundary between $A$ and $B$, let $r$ be the distance from the left point of the boundary to $P_c$, and let $\alpha$ and $\beta$ be the angles in the following picture:

We have

$t_{P_i P_c}=d_{P_i P_c} /v_a=\frac{r \sin \alpha}{v_a}$

$t_{P_c P_f}=d_{P_c P_f} /v_b = \frac{(s-r) \sin \beta}{v_b}$

So,

$t_{P_i P_f}=t_{P_i P_c}+t_{P_c P_f}= \frac{r \sin \alpha}{v_a} +\frac{(s-r) \sin \beta}{v_b}.$

Now, we have to consider the function:

$t(r)=\frac{r \sin \alpha}{v_a} +\frac{(s-r) \sin \beta}{v_b},$

differentiating with respect to $r$ and setting to zero, we get:

$t'(r)=\frac{\sin \alpha}{v_a}-\frac{r \sin \beta}{v_b}=0.$

So,

$\frac{\sin \alpha}{v_a}=\frac{r \sin \beta}{v_b},$

and then

$r=\frac{\sin \alpha}{\sin \beta}\frac{v_b}{v_a}=1,$ ??

by Snell's law.

• We do have to consider them. Note that light particles manage to figure it out: To get from $P_i$ to $P_f$, where one medium is air and the other glass, or water, they manage to take the right path. – André Nicolas Sep 28 '13 at 2:47
• Hint: what happens as $v_a$ goes to infinity? – Trevor Wilson Sep 28 '13 at 2:50
• The picture suggests that "the velocity in area A" and "the velocity in area B" are like streams that apply friction-like force to the particle crossing them, causing the particle to have aligned velocity component asymptotically approaching that of the area. – abiessu Sep 28 '13 at 2:57
• @abiessu Could you clarify what you mean by "streams" and "aligned velocity component"? (By the way, the picture might be misleading.) – Trevor Wilson Sep 28 '13 at 3:05
• @TrevorWilson: true, that's just what the picture made me think of, a "stream" of something like air in area A that pushes anything within it to go at the same speed, and a similar "stream" in area B. Of course, the normal-vector component of the particle's speed would not change under this model, so the crossing time would be independent of $v_a,v_b$, but the crossing points would be dependent. – abiessu Sep 28 '13 at 3:10

Hint: While the particle is in A or B it must travel in a straight line, as that is the shortest distance. Once you have that, you just need to pick the point where it crosses the boundary. Write the equation for the time of flight based on the position of that point, differentiate with respect to the position of crossing the boundary, set to zero, etc.

• Ok, thank you so much. I tried to do this in Attempt 2 above, but I got a horrible equation to solve. Can you check if my procedure is correct please? – Twink Sep 28 '13 at 15:33
• You are looking for Snell's law where the ratio of the sines of the angles is the ratio of the velocities. What you are doing looks fine, but it does seem it should be easier than that. – Ross Millikan Sep 28 '13 at 16:06
• In which part of Attempt 2 should I use Snell's law? Before starting or after solving the equation? Or would it be a totally different Attempt 3? – Twink Sep 28 '13 at 16:11
• @Twink You are trying to prove Snell's law, or a version of it. – Neal Sep 28 '13 at 16:12
• It is the result you are looking for-I don't think you can use it. – Ross Millikan Sep 28 '13 at 16:12

Minimize the flying time $\tau\left(x\right)$: $$\tau\left(x\right) \equiv {\sqrt{x^{2} + h^{2}} \over v_{a}} + {\sqrt{\left(s - x\right)^{2} + m^{2}} \over v_{b}}$$

$$\tau'\left(x\right) \equiv {1 \over v_{a}}\,\overbrace{{x \over \sqrt{x^{2} + h^{2}}}}^{\sin\left(\alpha\right)}\ -\ {1 \over v_{b}}\,\overbrace{{s - x \over \sqrt{\left(s - x\right)^{2} + m^{2}}}} ^{\sin\left(\beta\right)} = 0 \quad\Longrightarrow\quad {1 \over v_{a}}\,\sin\left(\alpha\right) = {1 \over v_{b}}\,\sin\left(\beta\right)$$

Now, you have ${\large \it three}$ equations for ${\large \it three}$ unknowns: $x$, $\alpha$ and $\beta$.