We can't have elastic scattering even theoretically!by hokhani Tags: elastic, scattering, theoretically 

#1
Nov2813, 12:19 PM

P: 241

Consider a photon which is scattered by a crystal elastically. In an elastic scattering we have [itex]
k_i=k_f and k_f  k_i = G [/itex] where [itex] G [/itex] is a reciprocal vector. But according to momentum conservation, the crystal must obtain a momentum [itex] \hbar G [/itex] and hence the kinetic energy, so according to energy conservation we no longer have [itex] k_i=k_f [/itex] because some of the energy of incident photon is transferred to the crystal and this is a contradiction! 



#2
Nov2813, 02:21 PM

P: 1,154

I am unsure of what you are trying to say. A change in momentum (being a vector) does not necessarily imply a change in energy (see for example a charged electron whose path is curved when entering a constant magnetic field). Additionally, in quantum mechanics momentum is related to the wavelength or the particle (or k) through
[tex]p=\hbar k [/tex] which relates to the SPATIAL behavior of the wave/particle where energy is related to the frequency (or ω) through [tex]E =\hbar \omega [/tex] which is to say it relates to the TEMPORAL behavior of the wave. Furthermore, you have to think about what G means spatially. It is just a bravais lattice vector and essentially is just encoding WHICH plane has interfered constructively with the incident surface ray in order to get a bragg peak. 



#3
Nov2813, 04:20 PM

P: 1,154

Actually, I've thought about it a bit more and I believe it's a little more subtle. Because you have perfect reflectance there is no net energy being transported to the sample, however, I believe you may have an entire net motion of the lattice (without any local distortion) along the direction of G. However, this motion carries no energy in the sense that it does not represent any heat.




#4
Nov2913, 12:22 PM

P: 241

We can't have elastic scattering even theoretically! 



#5
Nov2913, 12:26 PM

P: 1,154

Well of course it's impossible. No lattice is infinite, no lattice is a perfect reflector, no angle is perfect.




#6
Nov2913, 01:12 PM

P: 734

I'm not good at solid state physics but I think I can say this much.
At first,I remember that reciprocal space is not momentum space but pseudomomentum space because the symmetries of a crystal are not continuous.This may cause some changes in the reasoning! Second,the motion of the crystal as a whole,can be eliminated by a change of coordinate system so it shouldn't have physical meaning and so it shouldn't be taken into account in calculating the change of energy. Third [itex] k_i=k_f [/itex] is derived from assuming that there is no change of energy(The meaning of [itex] k_fk_i=G [/itex] being only a change of direction caused by scattering)and so the argument you're presenting becomes cyclic! 



#7
Nov3013, 06:10 AM

P: 241





#8
Nov3013, 07:10 AM

Mentor
P: 10,813

Not that it would be relevant... the total energy is conserved, if the photon loses a tiny fraction of its energy and the crystal gains a tiny bit (in the initial frame of the crystal), it is still called elastic scattering. 



#9
Nov3013, 06:43 PM

P: 638

A similar thing occurs in the Moessbauer effect. Here a photon is absorbed by a nucleus. The energy of the photon excites the nucleus to an excited level. The momentum of the photon is transferred onto the entire crystal lattice.
Because even a tiny crystal has a huge mass compared to the photon, the velocity gain of the crystal is extremely small, v = p/m. For the same reason the kinetic energy gained in this inelastic collision is very small, E=p^2/2m (assuming we are in the rest frame of the crystal before the absorption). In Moessbauer spectroscopy this has been verified experimentally. The relative line with of Moessbauer resonances is something like 10^15 (!!!), and the energy loss to the crystal is well below that. The same argument holds for neutron or xray diffraction. There is a real momentum transfer from the neutron or photon onto the crystal, but onto the crystal as a whole. Because of the mass of the crystal the associated energy transfer is so small that it can be safely neglected. In might be interesting to see if the energy shift is measurable for nanoparticles. I am afraid, however, that one would need free nanoparticles  if they are embedded or attached to something heavy we would be in the same case as above. If someone wants to do a backoftheenvelope calculation, I'd be interested in the result :) 



#10
Dec113, 04:55 AM

P: 241

All these reasoning don't convince the main problem. Though the crystal mass is very large that doesn't move significantly, but the energy transfered to the crystal may be at least comparable to the energy of incident particle so the energy of the scattered particle must be less than that of when impinging.




#11
Dec113, 05:21 AM

P: 638

momentum transfer = p = hbar (k_out  k_in) for a photon.
associated energy transfer = p^2/2m for a very large mass, the energy for a given momentum is simply very very small. In theory one often assumes infinite crystals, i.e. infinite mass. Then the energy transfer goes to zero. 



#12
Dec113, 07:47 AM

Mentor
P: 10,813





#13
Dec1913, 10:07 AM

P: 35

This is a very interesting question/problem on which I have been working for a long time. I believe, elastic scattering is impossible even theoritically. My analysis is based on the theory of elastic collisions in classical mechanics where the problem has its origin. The problem finds its way into every branch of modern physics  for example QM, Relativity, particle physics etc.
It is better and even simple to tackle the problem at its origin  in classical mechanics. Origin of the problem lies in the fact that in a binary elastic collision in the center of mass reference frame the velocity vectors of the particles change their directions through equal angles other than 180^{0} while keeping their respective magnitudes unchanged. That is, the relative velocity vector rotates through an angle not equal to 180^{0}! The relative velocity vector does not change its sign.This process came to be called elastic scattering. This essentially is Newton's result, in contrast to Huygen's result that demands that in a binary elastic collision process the relative velocity vecor changes sign as a result of elastic collision. The result of rotation (but not reversal) of relative velocity vector is a consequence of the model based on finite sized particles (in contrast to point masses) giving rise to the possibility of oblique collisions and there by to scatter. With point masses oblique collisions and consequential scatter are impossible even in principle! Thus, the model of finite sized masses in contrast to point sized masses leads to this paradox of elastic scattering. We may incidentally note that: Maxwell derived a result that showed transfer of kinetic energy (KE) from the mass of higher (KE) to the mass with lower KE during an elastic collision so that the difference in KEs after collision is necessarily less than the difference in KEs before collision. This is the origin of irreversibility being brought into thermodynamics through the model of ideal gases involving elastic collisions. 



#14
Dec1913, 05:53 PM

Mentor
P: 10,813

The velocity of objects relative to the center of mass changes by an angle other than 0°  it can be 180°, it does not have to be. The velocity of object 1 relative to object 2 changes by an angle other than 0°  it can be 180°, it does not have to be. There is no "paradox of elastic scattering". 



#15
Dec1913, 11:42 PM

P: 35

I think you are mixing two different "relative velocities" here, and I don't see where you see something special about 180°.
I am talking about the relative velocity of particle 1 with respect to velocity of particle 2. According to Huygens' analysis, this relative velocity changes sign as a result of elestic collision. In other words, reversal of relative velocity is a necessary consequence of a binary elastic collision. This makes elastic scattering impossible. However, according to Newton's analysis, restitution of relative speed is a necessary consequence of a binary elastic collision, but not reversal of relative velocity. This makes it possible for rotation of relative velocity vector through any non zero angle, including 180^{o}. This makes elastic scattering possible. Not if you include their spin ;). But arbitrary angles are possible with point masses (again apart from 0°, as this is a noninteraction and not called "collision"). In the case of point masses, the value of the 'b' parameter is zero and there is no possiility of oblique collision and no possibility of scattering. If you are not considering collisions we will not be discussing about one and the same issue. That is certainly wrong. The way Maxwell accounts for equivalization of temeratures of two gas samples at different temperatures goes somewhat like this: Suppose we mix in a rigid adiabatic container two ideal gas samples at different temperatures and leave them. They come to a common temperature  the equilibrium temperature. The mechanism of the process of equilibrization of temperatures is explained by Maxwell through the agency of binary elastic collisions between molecules wherein molecules with higher KE transfer energy to molecules with lower KE thereby bringing their KEs (temperatures) closer together. After a large number of such collisions the temerature difference of the two samples disappears and they assume the equilibrium temperature. This mechanism is untenable. Do you agre with that mechanism or think it is wrong mechanism? 



#16
Dec2013, 10:21 AM

Mentor
P: 10,813

Please use [quote]tags for quotes.
What do you mean with "restitution of relative speed"? Google finds exactly zero hits for this expression (in quotation marks). 



#17
Dec2013, 11:58 PM

P: 35

Source?
H. Erlichson, Am. J. Phys. 65 (2), 149154 (1997) No. Why should it? If we accept reversal of relative velocity as a necessary consequence of a binary elastic collision, then it follows that elastic scattering is impossible. What do you mean with "restitution of relative speed"? Google finds exactly zero hits for this expression (in quotation marks). I mean restoration of relative speed or coefficient of restitution is one. No. Point masses can have fields around them, like two charged pointparticles. Let us not bring fields into picture, they are not necessary to discuss elastic collision phenomena in classical mechanics. Even if you want to consider pointlike collisions (which have a probability of 0 to happen in real setups), this is still possible. With point masses, it is impossible to have oblique collisions and therefore scattering. We are interested what is possible in principle. Thus the probability arguments do not enter our discussion. Every collision is an interaction between fields and particles. If you smash two objects together, the electromagnetic interaction prevents them from going through each other. As I stated above, we shall restrict our discussion to classical mechanics for the sake of simplicity and shall not bring in fields. This is a statement about the average. Not for each collision. You may please this: J. C. Maxwell, The dynamical Theory of Gases in Selected readings in physics: Kinetic theory vol. 1, Stephen Brush, Pergamon Press, New York (1966), pp. 151152. 


Register to reply 
Related Discussions  
elastic scattering  Advanced Physics Homework  2  
elastic scattering  Atomic, Solid State, Comp. Physics  1  
Difference in quasi elastic scattering and non elastic scattering  Classical Physics  1  
Compton scattering elastic?  Quantum Physics  9  
elastic scattering  Atomic, Solid State, Comp. Physics  0 