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A question on a practice test I'm taking is as follows:

By shaking one end of a stretched string, a single pulse is generated. The traveling pulse carries:
A. mass
B. energy
C. momentum
D. energy and momentum
E. mass, energy and momentum

How would one describe the momentum of a wave?

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    $\begingroup$ As $p = \frac{E}{c}$. $\endgroup$ – ACuriousMind Apr 12 '15 at 22:17
  • $\begingroup$ Please note that "Explain how waves have momentum" is not an English question, so it should not have a question mark at the end. I edited the title to fix this. There are some good tips on writing question titles on this meta post. $\endgroup$ – DanielSank May 17 '15 at 17:15
  • $\begingroup$ It is possibly easier to grasp if you think about the momentum transfered by the coupling between elements of a wave medium. Of course that leaves of worrying over how it works in an electromagnetic wave. $\endgroup$ – dmckee May 17 '15 at 17:57
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This question is actually more complicated than it might seem.

To clarify the problem, let us consider a simplified model of the string: The string extends along the x-direction and is made up of masses connected by springs. For conceptual clarity, suppose these masses can only move up and down (along y; this could be enforced in a mechanical model by having the masses sliding up and down on little wires). In that case, the mechanical momentum is clearly only in the y-direction (all motion is along y), and there is no mechanical momentum in the direction of propagation of the wave (x). Depending on the shape of the wave packet, the total momentum in y-direction might also be zero, if some masses are moving up while others are moving down.

Nevertheless, there is indeed some kind of momentum associated with the motion of the wave along the x-direction, but it is a bit more abstract. To understand this, realize that the model is translationally symmetric along the x-direction, if we neglect the discrete spacing of the masses for a moment. This approximation will be fine in the continuum limit, i.e. for wave lengths much longer than the spacing between masses (and anyway I introduced the discrete model only to be very specific). According to Noether's theorem, (spatial) translational symmetry in a field theory implies conservation of momentum. Let us see how this works out here.

In the continuum limit, the equation of motion of our field theory is just the wave equation ${\partial^2u/\partial t^2} = v^2 {\partial^2u/\partial x^2}$, where $v$ is the velocity of the waves (which could be expressed via the properties of the springs and masses). The corresponding Lagrangian is $(\rho/2) \int (\partial u/\partial t)^2-v^2 (\partial u/\partial x)^2 dx$. Applying Noether's theorem to the translational symmetry of the Lagrangian reveals a conserved momentum density that is proportional to $-(\partial u/\partial t)*(\partial u/\partial x)$. If you integrate this expression over the extent of the wave packet, you get its total "wave momentum" (I am using this term to distinguish it from the simple microscopic mechanical momentum of the masses, mentioned above). And this momentum is indeed associated with the wave propagation along the x-direction (if we consider a 3d wave equation, we would get a momentum vector, pointing along the direction of motion of the wave packet). In a similar fashion, the momentum density of an electromagnetic wave would be derived, and it is proportional to the Poynting vector ${\vec E}\times {\vec B}$.

Whenever you have derived some momentum from the underlying translational symmetry of a field theory, it is then yet another, separate question whether/how that momentum can be converted into other forms of momentum. In the given mechanical example, we have already seen that the microscopic mechanical momentum only points along y, so if (for example) gas molecules collide with our masses, the momentum transferred onto the masses will also only be along the y-direction. However, there are situations where the conservation of the wave momentum is evidently useful. The one I can think of immediately is when the waves become nonlinear, such that wavepackets can scatter from each other (instead of just passing through each other). This could be brought about by springs that do no longer obey the simple linear Hooke's law if stretched too far. In that case, the dynamics of the waves during the scattering might be really complicated, but wave momentum conservation holds and puts a useful constraint on predictions about the end-result of the scattering process.

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  • $\begingroup$ It seems like the string should carry mechanical momentum. If we add gravity, breaking y-axis symmetry but not x-axis, and put a bead on the string, wouldn't it slide down the wave? The x-axis momentum it gains must come from the wave. $\endgroup$ – qbt937 Oct 29 '18 at 18:17
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Without going into wave equations, lets just say the segments of a string does not only move along the vertical direction. There are horizontal movements as well, although to a much smaller amplitude. A segment is being pulled towards the source horizontally when departing the equilibrium position and pulled back when heading back from maximum amplitude. For a segment in the middle of the wave, this makes no net contribution, but at the front of the wave a new segment is always being pulled towards the source, so the advancement of wave front should sustain momentum.

edit: updated sloppy language.

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According to quantum mechanics, $p=\frac{h}{2\pi}k$,where $k$ is wave vector and $h$ is Planck's constant. As we know $k={2\pi\over \lambda}$, where $\lambda$ is the wavelength of the wave. So momentum and wavelength are associated to each other. Moreover we can view as the motion of the every particles as well as wave and particle.This is known as duality of the quantum mechanics. The fact is that the wavelength of the macroscopic object is very large. so we use duality in the microscopic particles to describe the quantum mechanics.

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  • $\begingroup$ I am new in equation writing. Please if there is any mistake in symbols and any errors i highly appreciate for the correction. $\endgroup$ – rajuA1234 Apr 12 '15 at 22:46
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    $\begingroup$ Have a look at the MathJax tutorial. Also, note that the question is asking about classical waves, not quantum mechanics. $\endgroup$ – ACuriousMind Apr 12 '15 at 22:50

protected by Qmechanic Nov 1 '15 at 0:00

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