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Question: If a particle has a defined position at every time, must it necessarily also have a defined velocity? Consider a particle moving along a line, so its position along the line at time t is x(t). Suppose we define x(t) as follows:
          { 1   if t > 0
  x(t) =  {
          { 0   if t <= 0

If I remember my calculus correctly, x'(0) is undefined, while x(t) = 0. Does it therefore follow that at time t=0, the particle has a position and is moving but has no velocity? Would it be physically possible (i.e. compatible with the laws of physics as we currently understand them) for a particle with that behaviour to actually exist? -- SJK

In classical (non-quantum) mechanics a particle with mass cannot make such an instantaneous jump in position. It implies infinite acceleration which implies infinite force. So this case is not physically possible in classical mechanics (assuming zero-mass particles are not physically possible). -- Eob

the derivative of the step function x(t) you wrote above is the Dirac delta function. -- :RAE

I am in browse mode tonight... but at some point mention will have to made of tangent spaces and tie the discussion back to differential geometry.

Eob: What about in quantum mechanics? IIRC, quantum mechanics predicts instantaneous jumps in position (consider e.g. the Bohr model of the atom). And it has a zero-mass particle, the photon... -- SJK

I was not sure about quantum mechanics which was why I explicitly restricted my comments to classical mechanics. But now that I consider it more I would hazard that the question that was posed is not meaningful in quantum mechanics because you can never know x(t) exactly. As for the The Bohr Model, it has been superceeded by a model of the atom surrounded by orbitals which are standing waves of the wave function, so I am not sure it is relevant. --Eob

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