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General relativity is a classical theory. The Heisenberg uncertainty principle does not apply to it. The research frontier in physics now exists in quantizing gravity and unifying it with the other three forces (strong , weak, electromagnetic). Once that is done the solution for the black hole will become a probability distribution and the Heisenberg ...

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If black holes have mass but no size, does that imply zero uncertainty in position? If so, what does that imply for uncertainty in momentum? I mean to say that the particles which were originally separate have theoretically come to occupy the same point in space. Does the uncertainty principle apply to this phenomenon? Zero size doesn't ...

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To answer your first point, as far as I know, a test beam is used to calibrate the position and accuracy of the system, before the main beam is sent, to avoid, as you point out, any damage to the equipment. An indication of how few actual collisions there are is given by the fact that in 1978, a scientist did get a high energy beam right in the face, not at ...

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Given two observables $A$ and $B$ such that $[A,B] = iC$, the most general form of the uncertainty principle is $$\Delta_\omega(A)\Delta_\omega(B)\geq\frac12|\omega(C)|,$$ where $\omega$ is any state of the algebra of observables. By the Riesz-Markov theorem, there is a regular probability measure such that $$\omega(f(A)) = \int_{\sigma(A)}f(\lambda)\ \text ... 1 When defining the uncertainty principle one has first of all to be very careful with the domain of definitions the operators have: in particular, if A, B are the observables whose uncertainties we want to measure together, what needs to be calculated is the commutator \left[A,B\right]. In order this commutator to be well defined we must have that the ... 1 First of all I think I should clarify the confusion of the terminology "uncertainty in x". What we mean by uncertainty is simply the standard deviation of the observable defined by:$$\Delta x= \sqrt{ \left< x^2 \right> - \left< x \right>^2}$$where <A> denotes the expectation value of the operator A is some state \psi. You should ... 1 Did you cover the uncertainty principle? In quantum mechanics there is and uncertainty between energy and time:$$ \Delta E \Delta t > \frac{h}{4\pi} this means that if you try to measure Energy with perfect accuracy you will have a great uncertainly in time (actually an infinity uncertainty). I guess this is what the professor was referring to, and ...

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