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Apparently all objects have wavelike properties, so, if we kick a football (soccer ball, if you must) through a pair of posts, does the ball in any sense diffract?

If this is ridiculous then let me know and I will delete the question.

Emilio Pisanty
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user27182
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1 Answers1

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The most straightforward thing to do is to simply estimate the diffraction. For a ball with momentum $p$, being kicked through a pair of posts separated by a distance $d$, the angular spacing of the diffraction pattern, in radians, would be $\theta=\lambda/d=h/pd$. Putting in p=1 kg.m/s and d=1 m, we get $\theta\sim10^{-33}$ radians. This diffraction angle is too small to be detectable.

The other issue you could worry about is whether it's valid to calculate the wavelength of a ball. The wavelength would be the wavelength corresponding to the ball's center of mass motion. If, for example, one part of the ball isn't coherent with another part of the ball, or there's no coherence between different points in space along the ball's motion, then this might not be valid. The largest objects I know of for which diffraction has been experimentally observed is a C60 molecule, in experiments by Zeilinger at al. in 1999. According to the analysis by Wu, decoherence is a strong effect in this experiment, and you don't get good agreement between theory and experiment unless you include decoherence. I'm not an expert on this kind of thing, but I think this suggests that in order to diffract much a larger object like a ball, you would need to prepare it in an extremely coherent state (which is probably not possible) and isolate it extremely well from its environment during the diffraction process (which is probably also not possible).

Zeilinger, http://www.nature.com/nature/journal/v401/n6754/abs/401680a0.html

Wu et al., http://arxiv.org/abs/1106.2307

  • Do we have to assume that the ball is a particle in order to do this? If we wanted to be accurate would we have to calculate the wavelengths of all of the particles of the ball and them superimpose them? – user27182 Mar 29 '13 at 21:31
  • @Hayeder: No, you don't need to assume that the ball is a structureless particle, and you don't need to take into account its internal structure (assuming no decoherence effects). The wavelength is simply the wavelength corresponding to the center of mass degree of freedom. This is pretty standard, for example, in nuclear physics, where the nuclei have quite a large number of particles in them. E.g., in nuclear fusion reactions, the nuclei have to tunnel through a potential barrier due to their electrostatic repulsion. It's a good approximation to treat this using the Schrodinger eqn this way. –  Mar 29 '13 at 21:35