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58

Short answer: nobody knows, but the Planck length is more numerology than physics at this point Long answer: Suppose you are a theoretical physicist. Your work doesn't involve units, just math--you never use the fact that $c = 3 \times 10^8 m/s$, but you probably have $c$ pop up in a few different places. Since you never work with actual physical ...


42

I've worked on a camera that has as one of its core features the ability to increase the FPS until you are counting single photons. Here is one of the pdfs about it. You will see from the figures that there is an intrinsic tradeoff between the noise/image-quality and the FPS, which is simply due to the statistics of photon counting noise as you get fewer ...


37

The answer to all questions is No. In fact, even the right reaction to the first sentence - that the Planck scale is a "discrete measure" - is No. The Planck length is a particular value of distance which is as important as $2\pi$ times the distance or any other multiple. The fact that we can speak about the Planck scale doesn't mean that the distance ...


34

As we cannot resolve arbitrarily small time intervals, what is ''really'' the case cannot be decided. But in classical and quantum mechanics (i.e., in most of physics), time is treated as continuous. Physics would become very awkward if expressed in terms of a discrete time. Edit: If time appear discrete (or continuous) at some level, it could still be ...


26

In addition to the good answer of MSalters it is worth pointing out that individual photons arrive on the detector of the camera at different times. The way that cameras work is that they convert photons arriving into a signal that can be read out. No matter how high the fps rate there will be some point in time where one frame is finished and the new one ...


24

Are you aware that light consists of quanta (Sophia might have been a bit premature in her comment to say QM is not involved). Each photon captured by the camera is captured in one frame, for practical purposes. So the problem as we increase the FPS is that each frame is now based on fewer and fewer photons. This isn't just theory. We've done this ...


23

Quarks do not violate quantization of charge, it's simply that $\frac{1}{3}e$ instead of the electron charge $e$ is the smallest unit of electric charge.


22

Yes, there are an uncountable infinity of possible wavelengths of light. In general the frequency spectrum for Electromagnetic (e.g light, radio, etc) is continuous and thus between any two frequencies there are an uncountable infinity of possible frequencies (just as there are an uncountable number of numbers between 1 and 2). Two things to consider in ...


22

There are a couple different meanings of the word that you should be aware of: In popular usage, "quantized" means that something only ever occurs in integer multiples of a certain unit, or a sum of integer multiples of a few units, usually because you have an integer number of objects each of which carries that unit. This is the sense in which charge is ...


21

None of the above. Though there are many speculations about the significance of the Planck length, none is proven in any currently accepted theory. It is expected, though, that quantum gravity effects become definitely non-neglegible at the energy/distance scale set by the Planck length, so it provides a heuristic scale at which we should not expect our ...


19

That's a really good question! There are three cases, the third of which is the most fundamental and most interesting. The first case is incomplete absorption, such as a gamma ray knocking loose a few electrons as it passes. In that case the differences are taken care of locally and fairly trivially by allocating energy, momentum and spin appropriately ...


17

Put into one sentence, Noether's first Theorem states that a continuous, global, off-shell symmetry of an action $S$ implies a local on-shell conservation law. By the words on-shell and off-shell are meant whether Euler-Lagrange equations of motion are satisfied or not. Now the question asks if continuous can be replace by discrete? It should immediately ...


17

Let me add two references to points already mentioned in this discussion: Today, there is no reason known why the electric charge has to be quantized. It is true that the quantization follows from the existence of magnetic monopoles and the consistency of the quantized electromagnetic field, which was shown first by Dirac, you'll find a very nice exposition ...


16

No, there aren't any holes like that in the EM spectrum. There are other ways of creating photons than by having electrons bound in atoms transition from one level to another. (For example, you can create pretty much any frequency of photon you want by accelerating a free electron.)


16

Formally there are an infinite number of different wavelenghts. However, any given physical system can only be found in a finite number of distinct physical states. To create a light source with a wavelength $\lambda$ that is well defined up to some resolution $\delta\lambda$, requires observing it within a system of size of the order of ...


14

Frequency is not quantized, and has a continuous spectrum. As such, a photon can have any energy, as $E=\hbar\omega$. However, quantum mechanically, if a particle is restricted by a potential, i.e. $$\hat{H}=-\frac{\hbar^2}{2m}\nabla^2 + \hat{V}$$ for $V\neq 0$, the energy spectrum is discrete. For example, in the case of the harmonic oscillator, ...


14

For static charges, the relationship is V (voltage) = Q (charge) / C (capacitance). Capacitance is a function of the shape, size and distance between objects, which are all continuous values. (Well, I suppose you could argue that shape and size are quantized to the atomic spacing of the object's material, but you can't say the same thing for distance.) So ...


14

I recently was writing about this on wikipedia. The most intuitive way to see why an operator like $S_z$ has discrete values is based on its relation to rotation operators: $R_{internal}(\hat{z},\phi) = \exp(-i\phi S_z / \hbar)$ where the left side means rotation of angle $\phi$ about the $z$-axis, but only rotating the "internal state" of particles not ...


13

I think it's important to note that quantum or quantized time is not equal to discrete time. For instance, we have "quantized" space. By this we mean that it receives quantum treatment. But the underlying coordinates still form a continuum. So even if you live on a finite circle and only consider wavefunctions so that you get a countable set of basis ...


13

The Planck-Time is a little higher than $t_p=5\cdot 10^{-44} \, \text{sec}$ so the maximum frame rate allowed by quantum mechanics is less than $1 \, \text{frame}/t_p = 2\cdot 10^{43} \, \text{frames/sec}$.


12

Sir Elderberry, Punk_Physicist and the Count Iblis have all given correct answers in principle. There are two phenomena (really thought experiment, rather than practical, devices) that one needs to heed. A finite measuring time $T$ can only resolve frequencies to within an uncertainty of the order of $1/T$. This is the reciprocal relationship between the ...


11

Charge comes from discrete symmetries and is countable and additive. Mass comes from continuous 4d space, is exchangeable with energy and, in quantum mechanical dimensions not linearly additive, thus not countable. Suppose you have an elementary quantum of mass, $m_q$. In the world we know two such quanta would not end up as $2m_q$. One would add the ...


10

Perhaps the simplest and most intuitive approach is to regularize the hard wall potential $$V_0(x)~=~\left\{ \begin{array}{rcl} 0 &\text{for}& x<0 \cr\cr \infty &\text{for}& x>0\end{array}\right. $$ as $$ \lim_{\varepsilon \to 0^+} V_{\varepsilon}(x) ~=~V_0(x).$$ For instance, one could choose the regularized potential as $$ ...


10

The number of atoms (or molecules) in a body is given by Avogadro's constant, or $6.022 \times 10^{23}$ per mole. A mole is the amount of material, in grams, equal to the atomic or molecular mass of the substance in question. For example, for water ($H_2O$), 1 mole equals 18 grams. To get this number, remember that hydrogen ($H$) has an atomic mass of ...


9

Zeno used his paradoxes to proof movement was impossible. But of course he knew movement existed! If you were going to punch him, he will not trust your fist would have to get infinite times half of the way before reaching him; he would try to avoid it. His philosophical motivation was to "stirr" the reason, show that by logical arguments we can fall into ...


9

The other answers are correct, but I think they miss one important point. The energy levels are not really discrete, because: Frequencies are not exactly defined. Due to the Heisenberg uncertainty principle, the location and frequency of a photon cannot both be exactly determined. Therefore, emission (and absorption) does not happen at an exact ...


9

Dear asmailer, the reason is simple and completely understood: the electric charge is the generator of a $U(1)$ symmetry which is compact and may be parameterized by an angle, $\phi$. So wave functions may only depend on the angle $\phi$ in a periodic way, $\exp(iQ\phi)$ where $Q$ is integer (or an integer multiple of $e/3$, if I look at the elementary ...


9

If I'm only allowed to use one single word to give an oversimplified intuitive reason for the discreteness in quantum mechanics, I would choose the word 'compactness'. Examples: The finite number of states in a compact region of phase space. See e.g. this Phys.SE post. The discrete spectrum for Lie algebra generators of a compact Lie group, e.g. angular ...


9

Here is an experimentalist's answer: You state: The probability of a photon having just the right amount of energy for an atomic transition is 0. You must be aware that the statement falls just by the existence of lasers, so your question should have a how is it possible to have lasers. 1) An individual photon cannot be labeled as continuous. It has ...



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