Momentum of a Classical Wave: Explained

[snoopies622]

I haven't found my mistake yet. Is my equation for the energy of a wave wrong (see post 28)? I know it's not in the usual form -- all I did was make a few substitutions. The other assumptions are the reflection/transmission equations and p=E/v.

 

[atyy]

Quick guess, no idea if it's relevant. E is scalar and p is vector, so how about sticking minus signs somewhere in the p equation?

 

[snoopies622]

Minus signs? But $$\vec p$$ and $$\vec v$$ are in the same direction..

 

[atyy]

I was thinking reflected and incident momenta?

 

[snoopies622]

Oh, OK.

I think I found my mistake but no time to type now. Back in a few hours.

Edit: Having computer problems - will (hopefully) continue this matter over the weekend.

 

[snoopies622]

Well atyy, after having spent more time this weekend struggling with this puzzle than I care to admit, I now think you're right about the minus signs.

The energy of a wave should be

$$E = \mid 2 \pi ^2 f T A^2 v^{-1} \mid$$

while its momentum is

$$<br /> p = \frac { \mid 2 \pi ^2 f T A^2 v^ {-1} \mid } {v}<br />$$

where $$v$$ is either positive or negative depending on the wave's direction. This seems to make everything work out.

Part of my confusion came from substituting terms back and forth between the wave-at-a-boundary equations (where 1/v can be either positive or negative) and the colliding ball equations (where mass is always positive). I'm still not clear about how far one can take that sort of thing, but for now I believe that there is something left there for me to discover and I find it very alluring.

Anyway, thanks for your help.

 

[snoopies622]

This seems to make everything work out.

or does it?

Consider again a wave hitting head-on a boundary between two media. Let E be the energy of the incident wave, E₁ the energy of the reflected wave and E₂ that of the transmitted wave. The speeds of the waves are v₁, -v₁ and v₂ respectively.

Conservation of energy and momentum gives us


$$E = E_1 + E_2 \hspace{15 mm} \frac {E}{v_1} = \frac {E_1}{-v_1} + \frac {E_2}{v_2}$$

Solving these we get

$$E_1 = E ( \frac {v_1 - v_2}{v_1 + v_2} ) \hspace {15 mm} E_2 = E ( \frac {2 v_2}{v_1 + v_2} )$$

But what if v₂ > v₁? E₁ can't be negative (can it?) and ignoring the minus sign gives two values that no longer solve the system of equations.

 

[snoopies622]

Here's a specific example of the problem mentioned in entry #30 and reframed in entry #37, for emphasis.

Suppose we have incident wave speed $$v_1 = 4$$, transmitted wave speed $$v_2 = 1$$, and the incident wave's amplitude $$A = 10$$.

$$<br /> A_1 = A (\frac {v_2 - v_1}{v_2 + v_1}) = -6 \hspace {10 mm} A_2 = A ( \frac {2 v_2}{v_2 + v_1})=4$$

Since the energy of a wave $$E = | 2 \pi ^2 f T ( \frac{A^2}{v} )|$$

we can choose frequency $$f$$ and tension $$T$$ such that $$2 \pi ^2 f T =1$$ and we get

$$<br /> E = | \frac {A^2}{v_1} |= 25 \hspace {10 mm}<br /> E_1 = | \frac {( A_1 ) ^2}{ - v_1} | = 9 \hspace {10 mm}<br /> E_2 = | \frac {(A_2) ^2}{v_2} | = 16$$

So far so good. But with $$p= \frac {E}{v}$$ for momenta we have

$$p = \frac {E}{v_1} = \frac {25}{4} =6.25$$

$$p_1 = \frac {E_1}{- v_1} = \frac {9}{-4} = -2.25$$

$$p_2 = \frac {E_2}{v_2} = \frac {16}{1}=16$$


Not only is the momentum not conserved, but in this case the combined reflected and transmitted momentum is greater than that of the incident wave. Where did the extra momentum come from? This is very confusing. Does anyone know how to resolve this?

Aside: Although it's probably not relevant, the quantity $$(\frac {A}{v})$$ -- whatever that might represent -- is conserved, not just in this case

$$\frac {10}{4} = \frac {-6}{4} + \frac {4}{1}$$

but in general:

$$\frac {A_1}{v_1} + \frac {A_2}{v_2} = \frac {A}{v_1} ( \frac {v_2 - v_1}{v_2 + v_1} ) + \frac {A}{v_2} ( \frac {2 v_2}{v_2 + v_1} ) = \frac {A v_2 (v_2 + v_1)}{v_1 v_2 (v_2 + v_1)} = \frac {A}{v_1}$$

 

[Vanadium 50]

You don't get to choose the speed of the wave: the speed is $\sqrt{T/\mu}$. Both the tension and the mass density are the same for both waves.

 

[snoopies622]

By "both waves", I don't know if you mean both the incident and the reflected wave, or the incident and the transmitted wave. If the first case, that is accounted for by using v₁ for the incident wave and -v₁ for the reflected wave above (I just edited it for the energy computation part), if the second case then there is no boundary and no reflection at all, which is not what I'm describing here. For a string with the kind of boundary I'm describing here, the tension is the same on both sides but the mass density is different, so there are two different speeds.

 

[jambaugh]

From that thread, I'd argue that it's not exactly the wave making the surfer go. The surfer is 'continuously falling' down a crest. That's gravity making the surfer move! ...

Remember the surfer (in an idealized setting) is not moving higher or lower so its not quite right to say gravity is pushing him. Also note the gravitational force vector is straight down and so has no lateral component.

The water pushes him laterally and since both before and after the wave passes the water is stationary you can group all the momentum of the moving water into a "momentum of the wave".

 

[snoopies622]

I guess what is called "radiation pressure" when applied to electromagnetic waves also exists for mechanical waves, and this is what accounts for the apparent non-conservation of wave momentum at a boundary indicated by the above calculations. But I do find it surprising that this pressure could be negative. This means that as waves pass from one medium to another they might push on to the new medium or instead might instead "push off" against the old medium. That this pushing off occurs when $$v_1 > v_2$$ is also the opposite of what I would expect.