Momentum of a Classical Wave: Explained

[snoopies622]

What is the momentum of a wave? I know that for a photon

p=\frac{h}{\lambda}=\frac{hf}{c}=\frac {E}{c}

but what about a classical, mechanical wave? Is it also equal to the wave's energy divided by its speed, or is it more complicated than that?

 

[Meir Achuz]

For a classical EM wave, the h/lambda doesn't apply, but p=E/c still does.

 

[snoopies622]

Yes, but what about mechanical waves, like sound waves or pulses on a string? There must be some formula for momentum in terms of density, elasticity, amplitude, etc.

Vanadium 50, where are you?

 

[ChrisHarvey]

Vibration isn't really my area, so treat my comments with caution.

Surely it's meaningless to talk about the momentum of a mechanical wave? You can talk about momentum for a photon because it is not actually a wave, but does display wave like properties (I'm sure something I just said there will upset 'real' physicists). Mechanical waves (flexural and longitudenal) are just vibrations - they are not 'things'. It's just a way to describe a type of movement. The waves themselves do not have a mass (a photon does... more gasps from physicists, I'm sure) so can't have momentum. If you look at the medium a mechanical wave is transmitted in, an element inside of it has a displacement (and the derivatives of displacement), a mass and forces acting on it, which vary with time (not the mass!). A single element therefore has time varying momentum. The whole structure does not.

 

[snoopies622]

That's what I thought, but this thread

https://www.physicsforums.com/showthread.php?t=281373

has me doubting myself.

 

[ChrisHarvey]

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! It's just the geometry and movement of the wave that allows that to happen. Different parts of the wave making the surfer move have different amounts of potential and kinetic energy (and therefore momentum) at different times. Theres a continuous transfer between the two, but here we're not talking about the momentum of the wave, but the momentum of water particles, which increases and decreases.

I don't even know how you would define a wave for the purpose of deriving momentum. I only consider it a type of motion.

As I said, I'm no expert in this field. I'm sure Vanadium 50 can correct me.

 

[ChrisHarvey]

I was wrong. Here's an explanation and derivation for wave momentum:

http://books.google.co.uk/books?id=...X&oi=book_result&resnum=5&ct=result#PPA45,M1"

I don't think it disagrees so much with what I said, but it takes it a lot further and I learned some new stuff too.

The reference defines a momentum density, which corresponds to what I said about each individual 'elements' inside the medium having their own momentum. It takes it further by defining the momentum carried by a wave as been the integral of momentum density with respect to distance along the wave and time. It also shows the relationship between momentum density and energy flow. That energy carried by a wave is more obvious to see I think.

I think it comes down to definitions. To say that a wave motion causes the medium it is into carry momentum is obvious I think (see my first post). To then say that the wave has momentum is a bit more confusing, since I considered the wave to be the motion only.

Enjoy the link :p
Chris

 

[Vanadium 50]

Vanadium 50, where are you?

By starting a new thread, you threw me off the scent.

Clem is right: p = E/c is valid for mechanical waves. Note that c is the wave's velocity, not necessarily the speed of light.

 

[cesiumfrog]

Clem is right: p = E/c is valid for mechanical waves.

Really? I understood that wave momentum was a much more contentious and complicated issue than that. (An example paper is J.Fluid Mech. v106 p331 1981.)

 

[Vanadium 50]

There is a complication with transverse waves, as one can show that setting up a transverse wave also sets up a longitudinal wave, and this carries the energy and momentum. One can make the situation more complicated still by introducing an non-uniform elasticity tensor. But the fact that there are ways of making things more complicated is not, in my mind, an excuse to toss out a model that is explanatory. (Just like I wouldn't stop people from understanding Newton's law of universal gravitation because it's not GR)

It's an experimental fact (although it was disputed earlier) that one can lose energy and/or momentum to a mechanical wave, and one can gain energy and momentum from mechanical waves. Momentum and energy are conserved if one ascribes to the wave itself the energy and momentum being transferred. Some people say "it's not 'really' in the wave; it's in the medium that's oscillating", which is in my mind a quibble that replaces a useful simple concept with one that's so complicated as to be useless. It would be like saying, "there's not really any such thing as an electric current; it's all just the motion of individual electrons".

For those who still don't believe that waves carry momentum, take a metal rod, and clamp it down at both ends and as many points in the middle as you'd like to insure it doesn't move. Place an object that's free to move at one end, touching the rod, and strike the other end of the rod with a hammer. The object takes off. Where did its momentum come from? And when the hammer stopped, where did its momentum go? Remember, the rod is stationary.

 

[snoopies622]

Ah, Google Books! That saves me a trip to Durham, at least for Elmore and Heald. Thanks Chris. And thank you Vanadium 50 for returning to this topic .

The relationship I was looking for is expressed in equation 1.11.12:

(energy flux) = (speed)² x (momentum density)

which can be rearranged to make (momentum)=(energy)/(speed), just like Vanadium 50 said and just like with electromagnetic waves.

I can't say I understand the derivation yet - it's pretty long and so far I've barely looked at it - but the text says that this relationship applies to all plane mechanical waves 'traveling in linear isotropic media', so perhaps it is the clue to the puzzle I posed in the 'a clue for de Broglie?' thread. I am also beginning to wonder if the two conditions behind the wave reflection/transmission equations (continuity and differentiabilty of waves at a boundary) do in fact imply conservation of momentum and kinetic energy, just in a subtle way. Hmm...(scratches head)

 

[ChrisHarvey]

take a metal rod, and clamp it down at both ends and as many points in the middle as you'd like to insure it doesn't move. Place an object that's free to move at one end, touching the rod, and strike the other end of the rod with a hammer. The object takes off. Where did its momentum come from? And when the hammer stopped, where did its momentum go?

I like this... You've convinced me with it. The concept of wave-momentum was a bit alien to me - I've only ever looked at forces at various support types due to waves, and for this you don't need to consider momentum.

There is a complication with transverse waves, as one can show that setting up a transverse wave also sets up a longitudinal wave, and this carries the energy and momentum.

I hadn't realized this was the case, but now you say it, I can see that it would be true.

 

[snoopies622]

Newton's cradle seems like a good example, too. Of course, these are both cases involving longitudinal waves. That a transverse wave of the same energy and speed carries just as much momentum is less obvious to me.

 

[skeleton]

A wave in a classical material (eg: water or air) has no overall momentum, but it does have instantaneous momentum. At any moment in time, there is momentum; at the onset of the wave a crest may be developing with a positive momentum. Later, as the wave passes, the momentum would reverse and portray a negative value. The summation of momentum over time is zero.

A sound wave conveys compression of the air, followed by rarefraction. A water wave has a crest followed by a trough. As the crest is developing, the water molecules are actually moving foreward; later they retreat into the trailing trough. A sample molecule transcribes a circular motion as the wave passes by. (I am ignoring a different phenomenon of braking waves.)

 

[Vanadium 50]

A wave in a classical material (eg: water or air) has no overall momentum, but it does have instantaneous momentum.

Would you like to explain why if a wave carries no "overall momentum" the rod-and-hammer apparatus performs as described?

 

[skeleton]

V-50:

Re the rod-and-hammer, let's consider a segment within an elastic material rod.

- That segment initially has zero momentum.
- As the wave approaches the segment, the elastic structure compresses. A positive momentum can be calculated.
- Later as the wave recedes, that same structure extends (rebounds). A negative momentum is calculated.
- Summed over time, the segment has zero momentum.

Quite simply, the rod does not translate (move) overall. With zero overall velocity, there is zero overall momentum within the rod. But what of the momentum introduced by the hammer? Well, it is carried off by the object on the opposite end of the rod (here I am considering, for example, the opposite end of the rod placed against a concrete wall that is being chipped away by the hammer and chisel mechanism).

In contrast, if the rod were made of plastic (deformable) material, then *some energy would be lost to* the rod itself. Where steel is used in the rod, and stressed beyond its yield stress, then an elasto-plastic behaviour becomes manifested. Some of the energy is absorbed by the plastic deformation of the steel, while the remainder is imparted by the opposing medium (concrete wall in my example).

 

[snoopies622]

But what of the momentum introduced by the hammer? Well, it is carried off by the object on the opposite end of the rod...

Doesn't this imply that momentum was transported from one end of the rod to the other? In other words, that the compression wave carried the momentum along with it?

 

[dx]

Here's a simpler derivation for one dimensional transverse waves on a string based on Lagrangian mechanics. Let \phi(x) be the displacement of the string at x. The kinetic energy of the string (in appropriate units) is \int (1/2)(\partial{\phi}/\partial{t})^2, and the potential energy of the string is \int (c^2/2)(\partial{\phi}/\partial{x})^2. For a right moving wave of the form \phi(x) = F(x - ct), the energy is T + U = E = \int c^2 (dF/dx)^2 dx. The linear momentum of the wave is the Noether charge of space translation symmetry (you can check that the Lagrangian T-U is invariant under space translations). An infinitesimal space translation has the form \phi(x) \longrightarrow \phi(x) - (\partial{\phi}/\partial{x}) \epsilon. The charge of this symmetry is

-\int \Pi_{\phi}(x) \phi(x) dx = -\int (\partial{\phi}/\partial{t})(\partial{\phi}/\partial{x}) dx = \int c(dF/dx)^2 dx

which is just E/c, so E = cP.

 

[Vanadium 50]

Of course the rod doesn't move - by construction. But how does the momentum get from one end of the rod to the other if it is not carried by a wave?

 

[snoopies622]

dx, I think you have just given us the golden key.

I have three questions:

1. what is \Pi_{ \phi }(x)?
2. this proof seems to imply that longitudinal displacement is not even necessary for wave momentum - is this correct?
3. do you have any thoughts about this:

https://www.physicsforums.com/showpost.php?p=2012460&postcount=1

?

Thank you very much, by the way.

 

[Vanadium 50]

You always get longitudinal forces as well as transverse forces - the string is under tension. Also, you can't stretch it transversely without also stretching it longitudinally.

 

[skeleton]

Of course the rod doesn't move - by construction. But how does the momentum get from one end of the rod to the other if it is not carried by a wave?

I agree that the hammer's momentum is carried through the rod via the rod's *momentary* momentum. Of course, that same momentum is then given to an impulse to a receiving medium (concrete wall) at the other end of the rod. And yes, the momentary momentum in the rod is *seen* translating along the wave.

 

[dx]

1. what is \Pi_{ \phi }(x)?

\Pi_{ \phi }(x) is the conjugate momentum of \phi(x), which is \partial{\mathscr{L}}/\partial{\dot{\phi}} in general and \dot{\phi} in our case. Also, I made a mistake above: -\int \Pi_{\phi}(x) \phi(x) dx should have been -\int \Pi_{\phi}(x) (\partial{\phi(x)}/\partial{x}) dx.

2. this proof seems to imply that longitudinal displacement is not even necessary for wave momentum - is this correct?

Yes, wave momentum and momentum in general for any system doesn't depend on the kind of degrees of freedom of the system. Whether it is longitudinal or transverse or both or some other kind of motion, if there is a space translational symmetry, there is a corresponding conserved quantity which we call linear momentum.

3. do you have any thoughts about this: https://www.physicsforums.com/showpost.php?p=2012460&postcount=1?

The equations

v_1=v\frac{m_1-m_2}{m_1+m_2}

v_2 = v \frac{2m_1}{m_1+m_2}

Can be derived from the conservation of momentum and energy: m_1v^2 = m_1{v_1}^2 + m_2{v_2}^2 and m_1v = m_1{v_1} + m_2{v_2}. The second pair of equations that you gave for reflection and transmission of waves

A_1 = A \frac {k_1 - k_2}{k_1 + k_2}

A_2 = A \frac {2 k_1}{k_1 + k_2}

can be obtained from the first set by the following transformation: v_i \longrightarrow A_i, v \longrightarrow A, m_i \longrightarrow (1/v_i).

Lets apply this transformation to the energy conservation law for the particle collision m_1v^2 = m_1{v_1}^2 + m_2{v_2}^2. We get

\frac{A^2}{v_1} = \frac{A_1^2}{v_2} + \frac{A_2^2}{v_2}

This is just the momentum conservation for waves since A^2 is the energy and v is the speed of propogation. From P = E/c, we see that A^2/v_1 is the momentum of the incoming wave, and similarly for the other terms. Applying the transformation to the momentum conservation equation of the particle collision should give something related to energy conservation in the wave case although I can't see it at the moment. I'll finish this tomorrow since it's getting late.

 

[Vanadium 50]

I agree that the hammer's momentum is carried through the rod via the rod's *momentary* momentum. Of course, that same momentum is then given to an impulse to a receiving medium (concrete wall) at the other end of the rod. And yes, the momentary momentum in the rod is *seen* translating along the wave.

Since this "momentary momentum" acts like momentum, has the same units as momentum, and is conserved with momentum: i.e. it is the sum of momentary momentum and ordinary momentum that is conserved, shouldn't we just call it "momentum"? Particularly since it's not even momentary: it lasts for a time L/c where L is the rod's length.

Why invent a new name for this? The old one seems to work well enough.

 

[skeleton]

V-50:

In using the term "momentary", I was trying to draw attention to the rod as a whole has *never* moved. Instead, only a segment of the molecules - at any moment in time - were advancing then later retreating.

On the question of duration, the wave persists for almost forever. In real structural materials, the wave would travel to the end of the rod in a finite interval of time. Most of the energy may be imparted upon the concrete wall, but a determinable portion will rebound back down the rod to its starting point, and repeat ad infinitum.

Other than that, it is all just semantics.

 

[Vanadium 50]

Other than that, it is all just semantics.

Does this mean you now agree that a wave carries momentum?

 

[skeleton]

V50:

Yup ;-)

 

[snoopies622]

Thanks again, dx -- you're really bringing this matter home for me. Here's my minor contribution:

For constant tension T, constant frequency f, \lambda = \frac {v}{f} and v^2 = \frac {T}{ \rho}
I get the energy of a wave on a string to be

E = 2 \pi ^2 f T (\frac {A^2}{v})

which suggests that

\frac{A^2}{v_1} = \frac{A_1^2}{v_1} + \frac{A_2^2}{v_2}

is the one that goes with wave energy (not momentum) conservation, which is nice since it came from substituting into the ball energy (not momentum) conservation equation. If this is correct, perhaps it will make the meaning of the other substitution more clear.

 

[dx]

Yes you're right. You might find this interesting: http://www.atm.damtp.cam.ac.uk/people/mem/papers/RECOIL/wave-momentum-myth-scanned.pdf

 

[snoopies622]

I get the energy of a wave on a string to be

E = 2 \pi ^2 f T (\frac {A^2}{v})

except this means that the wave's momentum is

p = E/v = 2 \pi ^2 f T (\frac {A^2}{v^2})

and conservation of wave momentum gives us

\frac {A^2}{v_1 ^2}= \frac {A_1 ^2}{v_1 ^2} + \frac {A_2 ^2}{v_2 ^2}.

However, substituting A_i for v_i and 1/v_j for m_j into m_1 v = m_1 v_1 + m_2 v_2

yields \frac {A}{v_1}= \frac {A_1}{v_1} + \frac {A_2}{v_2} which is consistent with

\frac {A^2}{v_1 ^2}= \frac {A_1 ^2}{v_1 ^2} + \frac {A_2 ^2}{v_2 ^2} only if A₁=0 or A₂=0 (total transmission or total reflection)

so I've made a mistake somewhere..

 

[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

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

while its momentum is

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

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 &gt; v_2 is also the opposite of what I would expect.