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Fugie Runner
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Do photons reflected off prisms exert directional force however small as they are reflected off the hypotenuse side of a right triangle shaped prism?
Reflection of Photons on Prisms: Directional Force?
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jtbell
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Light carries momentum, which is a vector quantity, that is, it has both magnitude and direction. Therefore, if a beam of light changes direction, something must recoil in order for the total momentum to be conserved. This is true in both the classical and quantum models of light.
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humanino
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I fully agree with jtbell answers. I would like to point out that the original question suggests that photons can somehow bounce off a surface. Technically however, photons being massless cannot be put at rest, so must be absorbed and re-emitted. Maybe just nitpicking 
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jimgraber
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humanino said:I fully agree with jtbell answers. I would like to point out that the original question suggests that photons can somehow bounce off a surface. Technically however, photons being massless cannot be put at rest, so must be absorbed and re-emitted. Maybe just nitpicking
Wow, I never thought a reflection or an elastic scatter involved a state of rest. However, I can accept that absorption and re-emission is mathematically equivalent to a reflection or a scatter. Is this really the way it is treated in QM or QFT? Also, is the momentum of the reflection absorbed by a single atom, or by the entire prism or mirror, similar to the Mossbauer effect?
Inquiring minds want to know. TIA.
Jim Graber
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$$ \frac{d}{dr}[\delta_{i+1}] = \frac{2\mu}{\hbar^2}\frac{V(r)}{k^2}\sin(kr + \delta_i)$$
##\delta_i## is the phase shift for triplet and singlet state, ##\mu## is the reduced mass for neutron-proton, ##k=\sqrt{2\mu E_{cm}/\hbar^2}## is the wave number and ##V(r)## is the potential of interaction like Yukawa, Wood-Saxon, Square well potential, etc. I first...
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark.
Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true.
But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy
\langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67}
The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form
f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...
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