67

The right formula is $$\Delta X \Delta P \geq h/4\pi$$ where $P$ is the momentum which is approximatively $mv$ only for small velocities $v$ when compared with $c$. Otherwise you have to use the relativistic expression $$P = mv/ \sqrt{1-v^2/c^2}.$$ If $\Delta X$ is small, then $\Delta P$ is large but, according to the formula above, the speed remains of ...


24

What you've discovered is that "normal" Quantum Mechanics is incompatible with relativity. As Valter Moretti pointed out, using a relativistic expression for momentum solves this problem. There are, however, more problems that cannot be solved by simply using relativistic expression for energy and momentum. For example, The relativistic equation $...


10

To be more precise, it sounds like what you're describing is the path integral formulation of quantum mechanics. Within this framework, every possible path between $A$ and $B$ contributes a complex number to an infinite sum, which is used to calculate the "actual" path the particle takes. It is tempting at first to ask how a particle could take a ...


5

There are two problems with this setup. The first is here: Assume an electron which is moving very slowly If you know already that the electron is moving very slowly then you already have a small uncertainty in momentum. For example, if you know that the electron is moving at less than $1 \text{ m/s}$ then $\Delta v = 0.29 \text{ m/s}$ so we already have $...


4

So when you become a particle (or nuclear) physicist , one of the first things you need to memorize is that: $$ \hbar c \approx 200\,{\rm MeV\cdot fm}$$ where "fm" is a fermi ($10^{-15}\,$m), which is the scale of a nucleon. Thus, if your position uncertainty is 100 fm, you can immediately estimate a momentum uncertainty of 1 MeV/c. Since you've ...


3

Assume an electron which is moving very slowly and we observe it with a distance uncertainty of say Δx=1×10−13 m In QM, particles don't have velocities in the normal sense of the word. Velocity is an observable, and thus is represented by an operator applied to a quantum state. Talking of a particle's "velocity" implies that the particle has a ...


3

A particle cannot travel faster than the speed of light in the sense that you cannot measure the particle to be in position $A$ and then subsequently measure the particle to be in position $B$ unless enough time has passed (so that $\frac{|B-A|}{\Delta t}\leq c$). This is actually not true in normal quantum mechanics, but it is true in Quantum Field Theory. ...


3

Yes, it can, the limit is 299792458 m/s. For example in water the speed of light is reduced by the refraction index, and therefore a particle exceeding (c/n) does not violate the SR. Read more here: https://en.m.wikipedia.org/wiki/Cherenkov_radiation


3

Your first sentence is already not defined. If by "standpoint of a light ray" you mean "a hypothetical observer travelling at the speed of light", you are already in undefined territory. The Lorentz transformations, which mathematically tell us how to change between inertial reference frames contain a factor: $$\gamma=\frac{1}{\sqrt{1-v^2/...


2

The maximum speed of light is defined as the velocity of electromagnetic waves in a vacuum. Light is a range of electromagnetic wavelengths that can be detected by our eyes. Electromagnetic waves traveling through a non-vacuum region can be "slowed down" by interactions with matter/energy present in the non-vacuum region. As another commentator ...


2

Yes it is possible, for instance electrons regularly win the race against photons in water. And they emit radiation known as Cherenkov Radiation. It's properties are similar to those of the shockwaves produced when an object crosses the sound barrier: a classical conical shockwave, just with light in this case.


1

Yes, within the Feynman integral approach the photon (and any other particle treated with this machinery) may travel with a velocity grater than $c$. (This is the intuitive corresponding of a virtual particle one uses in quantum field theory.) However the path integral method cannot be taken too literally. Not everything one may imagine by looking at this ...


1

You don't really need the full Michelson-Morley experimental setup to point to that conclusion. It is enought to have one mirror moving at a velocity $v$ with respect to an observer and a photon moving at velocity $c$ in the opposite direction (one arm of the Michelson-Morley interferometer). If the momentum transfer from the photon to the mirror is ...


1

The warp drive in the form proposed by Alcubierre will violate causality globally, for the reason given in the clear and succinct answer here by Jerry Schirmer. It is always possible in physics to introduce something unphysical, such as negative mass, and then deduce something equally unphysical, such as an infinite energy source or an impossible propulsion ...


1

Hypnosifl's answer is good but I want to add some context. In any discussion of Alcubierre's warp drive, it's important to understand that there are no rules. You can take any smooth spacetime manifold you like, plug it into the GR field equation, and get a corresponding stress-energy tensor. You can then dress up the calculation with language about "...


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