# Validity of a differing frame of reference than that used in Landau and Lifshitz's solution [closed]

For the following problem, first problem in chapter 2 (page 16) of Landau and Lifshitz's Classical Mechanics text:

I am trying to see whether the picture I drew when originally solving the problem before looking at the solution

is valid. The answer would then be $$\frac{\cos(\theta_1)}{\cos(\theta_2)}=$$ ...

Based on the solution, the proper drawing should be

Can I get any input on whether the different cases are equivalent?

Put the solution that L & L give us out of mind and just read the problem. There is no reason NOT to draw this setup, right?

If we agree on that, my question boils down to a trigonometric one. Why use $$\theta_2$$ and $$\phi_2$$ as opposed to $$\theta_1$$ and $$\phi_1$$?

• Sorry, isn't your second drawing wrong? L & L define the angles with respect to the normal. But your theta 1 above is drawn with respect to the plane of the surface, while your theta 2 is with respect to the normal. Or am I missing something? – CGS Jul 14 at 20:17
• @CGS you're absolutely right, thank you! Will edit now. Any thoughts on if the other orientation of the system based on the wording of the problem hold? – Lopey Tall Jul 14 at 20:20
• Yes, I think your drawing (the first one) is not correct. L&L state that the component of the velocity parallel to the surface boundary is conserved by the nature of the potential. Your drawing would show it reversed by 180 degrees. I think this is very similar to refraction of light through a boundary that produces Snell's law. L&L are just using a particle here. – CGS Jul 14 at 20:28
• can you see my comment on the answer below? @CGS – Lopey Tall Jul 15 at 21:25
• Post discussed in the hbar chat room here. – Qmechanic Jul 16 at 15:53

## Solution

The momentum component is to be constrained constant along the Plane$$(x)$$ and not the Normal$$(y)$$. This is because the potential energy is independent of $$x$$. $$U=\begin{cases}U_1&y<0\\U_2&y>0 \end{cases}$$ So, we have the equations \begin{align*} v_1\sin\theta_1&=v_2\sin\phi_1\tag{1}\\ \frac12mv_1^2+U_1&=\frac12mv_2^2+U_2\tag{2}\\ \end{align*} Putting the value of $$v_2$$ from equation $$(1)$$ into $$(2)$$, one gets \begin{align*} \frac12mv_1^2+U_1&=\frac12m\left(v_1\frac{\sin\theta_1}{\sin\phi_1}\right)^2+U_2 \\ (U_1-U_2)&=\frac12mv_1^2\left[\left(\frac{\sin\theta_1}{\sin\phi_1}\right)^2-1\right] \\ \boxed{\frac{\sin\theta_1}{\sin\phi_1}=\sqrt{1+\frac{2}{mv_1^2}(U_1-U_2)}}\tag{1} \end{align*} Thus, we get the relation between the angles.

Why use $$\theta_2$$ and $$\phi_2$$ as opposed to $$\theta_1$$ and $$\phi_1$$?

The answer could look different by the use of trigonometric identity $$\cos\left(\frac{\pi}2-\alpha\right)=\sin\alpha$$ in the numerator and/or denominator of equation $$(1)$$ but is exactly the same in all the following appearances physically because they are all equal to the same physical quantity. $$\frac{\cos\theta_2}{\cos\phi_2}=\frac{\cos\theta_2}{\sin\phi_1}=\frac{\sin\theta_1}{\cos\phi_2}=\frac{\sin\theta_1}{\sin\phi_1}=\sqrt{1+\frac{2}{mv_1^2}(U_1-U_2)}=\frac{\frac{v_1^{\text{along plane}}}{{v_1}}}{\frac{v_2^{\text{along plane}}}{{v_2}}}$$

Note that the solution of the book has $$\theta_1=\theta_1$$ and $$\theta_2=\phi_1$$ with no other angles used in this answer or your figure.

• This is definitely a trigonometry question at this point. @CGS Can you put L&L's solution out of your mind and just imagine being faced the problem on an exam or something? I've edited the question which a better geometric question. – Lopey Tall Jul 15 at 21:25
• @LopeyTall Have a look at the edit. Hope it helps! – Sameer Baheti Jul 16 at 12:26
• Thank you so much for this Sameer! incredibly helpful! :D – Lopey Tall Jul 17 at 13:59
• @LopeyTall I missed the bounty after so much effort. Sad! – Sameer Baheti Jul 17 at 14:00
• Seriously! @Qmechanic this is unfair. Sameer answered the question while a bounty was active.... and then you all closed it for some ridiculous reason. – Lopey Tall Jul 17 at 14:02

OK given the diagram of the problem you present in your fourth figure, the answer is the same if you use pairs of angles relative to the normal, or pairs of angles relative to the plane. [My notation is a little different. Subscripts 1 and 2 are regions 1 and 2 respectively. Angle theta is relative to the normal and angle phi is relative to the plane.] We know:

$$V_1sin(\theta_1) = V_2sin(\theta_2)$$

But from the geometry of the problem we also know that $$\theta_x = \frac{\pi}{2} - \phi_x$$

If we substitute this in the above, we get: $$V_1sin(\frac{\pi}{2} - \phi_1) = V_2sin(\frac{\pi}{2} - \phi_2)$$

And there is a trigonometric identity that says $$sin(\frac{\pi}{2} -\theta) = cos(\theta)$$

[See wikipedia here, section "Reflections, shifts and periodicity".]

And this gives, of course $$V_1cos(\phi_1) = V_2cos(\phi_2)$$ which is how you viewed the problem.

So the ratios of sines or cosines in the problem are the same, as long as you use the correct angles.

Sameer Baheti correctly notes this too.

• Brilliant! thank you! :) – Lopey Tall Jul 17 at 14:03