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  1. A photon hits an atom perpendicularly to its speed v and it is absorbed as is known immediately. So it can not act after the time when it is not perpendicular (e.g. the force is 0 after that). Certainly all its energy goes to the atom when it is perpendicular to v.

  2. But the impact (absorption) applies a force on the atom and it is postulated that a perpendicular force can not do work. So work has not been done on the atom. Consequently its energy can not change.

Isn’t this a contradiction?

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    $\begingroup$ you say " it is postulated that a perpendicular force can not do work." consider two cars crashing with perpendicular velocity on a crossing, No damage? $\endgroup$
    – trula
    Commented Apr 24, 2021 at 20:28
  • $\begingroup$ thinking in terms of classical mechanics with quantum mechanical entities (work is a classical mechanics variable) as are photons and atoms leads to confusions. Quantum mechanics and the four vectors of special relativity had to be discovered in order to explain atomic behavior.. The photon with energy E and three momentum p transfers its dp/dt to the atom and the four vector algebra defines the kinematics $\endgroup$
    – anna v
    Commented Apr 25, 2021 at 5:54
  • $\begingroup$ @annav Does your remark mean that work in Cl.Ph. is not correctly defined from point of view of QP? As far as I know the classical quantities are mean values in time of the 'real' quantum quantities. But one can use a coherent radiation (e.g. many photons) and get a close to the quantum quantity (e.g. there will be work in a classical sense). The 'force' powering this 'work' will be perpendicular to the body (to be more close to a perpendicular surface one can use a quality crystal grid). $\endgroup$
    – Mercury
    Commented Apr 25, 2021 at 17:14
  • $\begingroup$ @trula The answer to this is that at the very first instant of the crash (in infinitesimal time) the force (however big) changes only the direction of the velocity. After that the velocity is not perp. and the force can increase it. That's why I am asking about a photon and its 0 time of action and 0 force after the action. (example is the Moon and gravity). $\endgroup$
    – Mercury
    Commented Apr 25, 2021 at 17:23
  • $\begingroup$ It is not averages that lead to the classical fields from quantum fields , but complicated field theoretical equations are needed to show how quantum probabilities lead to classical well defined fields. See motls.blogspot.com/2011/11/… $\endgroup$
    – anna v
    Commented Apr 25, 2021 at 18:03

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it is postulated that a perpendicular force can not do work

A force that is perpendicular to the velocity of an object’s center of mass cannot do work on a point particle. However a perpendicular force certainly can do work on non point particle objects, including atoms. When such a force does work the energy increases some other degree of freedom besides the KE of the center of mass.

Note: if an object is treated as a collection of point masses then the point mass where the force is applied does not generally share the velocity of the center of mass. Thus work can be done by the force.

Note 2: Also, if a force is treated as an instantaneous impulse then the instantaneous force is infinite and it can do work even when oriented perpendicular to the velocity. Consider, for example, a object moving along the x axis and an impulsive force in the y direction. The rate of work is $\vec F \cdot \vec v = (0,\infty,0)\cdot (v,0,0)=(0,\infty 0, 0)$. Notice that this includes the mathematically indeterminate form $\infty 0$ which cannot be evaluated directly and must be determined through some other means. Usually we use conservation of momentum and integrate over the impulsive force.

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    $\begingroup$ 1. Can you elaborate more on this: "a perpendicular force certainly can do work on non point particle objects". Can not every object be regarded as made of a set of point objects (at least in classical physics) and the photon reacts in a dot (nevertheless its wave properties). 2. Also on "When such a force does work the energy increases some other degree of freedom besides the KE of the center of mass." --- The energy can and will go to raise the level of an orbit electron (say from E0 to E1) but the impulse of the photon should be transmitted to the nuclei - hence force on the atom ->body. $\endgroup$
    – Mercury
    Commented Apr 25, 2021 at 16:58
  • $\begingroup$ I added a note with the additional information $\endgroup$
    – Dale
    Commented Apr 25, 2021 at 19:39
  • $\begingroup$ 1. Well you are surely right that generally the point masses can have a slightly different velocity direction. But I consider a particular case. So only this particular case has a meaning. 2. Why should the force be infinite? F=p.S; p is finite and S too. Always the transfer of p is instantaneous for elementary particles. $\endgroup$
    – Mercury
    Commented Apr 27, 2021 at 10:40
  • $\begingroup$ 1 the general comment applies to your specific case. 2 what is S? F=dp/dt as dt->0 F->infinity. Also, it is a little overly-classical to assert that it is instantaneous, there are wavefunctions and uncertainty principles involved $\endgroup$
    – Dale
    Commented Apr 27, 2021 at 13:40
  • $\begingroup$ Well a bad mistake of mine about using F=p.S because of the same symbol p for pressure. Sorry. Nevertheless it is not following at all that an infinite force can do things that an arbitrary large force can not do. And as you insert dt must be final and F can not be infinite in the end. $\endgroup$
    – Mercury
    Commented Apr 28, 2021 at 12:44

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