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[tex]p=\frac{h}{\lambda}=\frac{hf}{c}=\frac {E}{c}[/tex]

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?

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- #1

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[tex]p=\frac{h}{\lambda}=\frac{hf}{c}=\frac {E}{c}[/tex]

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?

- #2

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

- #3

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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?

Vanadium 50, where are you?

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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.

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

has me doubting myself.

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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.

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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

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

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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.

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

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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.

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

(energy flux) = (speed)

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)

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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.

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

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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.)

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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?

- #16

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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).

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).

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- #17

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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?

- #18

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Here's a simpler derivation for one dimensional transverse waves on a string based on Lagrangian mechanics. Let [tex] \phi(x) [/tex] be the displacement of the string at x. The kinetic energy of the string (in appropriate units) is [itex] \int (1/2)(\partial{\phi}/\partial{t})^2 [/itex], and the potential energy of the string is [itex] \int (c^2/2)(\partial{\phi}/\partial{x})^2 [/itex]. For a right moving wave of the form [itex] \phi(x) = F(x - ct) [/itex], the energy is T + U = [itex] E = \int c^2 (dF/dx)^2 dx [/itex]. 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 [itex] \phi(x) \longrightarrow \phi(x) - (\partial{\phi}/\partial{x}) \epsilon [/itex]. The charge of this symmetry is

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

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

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

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

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I have three questions:

1. what is [tex] \Pi_{ \phi }(x)[/tex]?

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.

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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.

- #23

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1. what is [tex] \Pi_{ \phi }(x)[/tex]?

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

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

[tex]v_1=v\frac{m_1-m_2}{m_1+m_2}[/tex]

[tex]v_2 = v \frac{2m_1}{m_1+m_2}[/tex]

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

[tex] A_1 = A \frac {k_1 - k_2}{k_1 + k_2} [/tex]

[tex] A_2 = A \frac {2 k_1}{k_1 + k_2} [/tex]

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

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

[tex] \frac{A^2}{v_1} = \frac{A_1^2}{v_2} + \frac{A_2^2}{v_2} [/tex]

This is just the momentum conservation for waves since [itex] A^2 [/itex] is the energy and [itex] v [/itex] is the speed of propogation. From P = E/c, we see that [itex]A^2/v_1[/itex] 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.

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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.

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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.

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Other than that, it is all just semantics.

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

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V50:

Yup ;-)

Yup ;-)

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

For constant tension [tex]T[/tex], constant frequency [tex]f[/tex], [tex] \lambda = \frac {v}{f} [/tex] and [tex] v^2 = \frac {T}{ \rho} [/tex]

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

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

which suggests that

[tex] \frac{A^2}{v_1} = \frac{A_1^2}{v_1} + \frac{A_2^2}{v_2} [/tex]

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.

For constant tension [tex]T[/tex], constant frequency [tex]f[/tex], [tex] \lambda = \frac {v}{f} [/tex] and [tex] v^2 = \frac {T}{ \rho} [/tex]

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

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

which suggests that

[tex] \frac{A^2}{v_1} = \frac{A_1^2}{v_1} + \frac{A_2^2}{v_2} [/tex]

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.

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- #29

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- #30

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

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

except this means that the wave's momentum is

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

and conservation of wave momentum gives us

[tex] \frac {A^2}{v_1 ^2}= \frac {A_1 ^2}{v_1 ^2} + \frac {A_2 ^2}{v_2 ^2} [/tex].

However, substituting A

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

[tex] \frac {A^2}{v_1 ^2}= \frac {A_1 ^2}{v_1 ^2} + \frac {A_2 ^2}{v_2 ^2} [/tex] only if A

so I've made a mistake somewhere..

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- #33

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Minus signs? But [tex] \vec p [/tex] and [tex] \vec v [/tex] are in the same direction..

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I was thinking reflected and incident momenta?

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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.

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.

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