137

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 ...


49

Let's say space is really a lattice with spacing $\Delta x$. It turns out that this idea has more trouble with experiment than you might think, but we can plow ahead for the purposes of this question. You might propose replacing integrals in physics with discrete sums over individual lattice points, to take a concrete example let's think about the work ...


48

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 ...


43

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 ...


36

The book is almost surely referring to the ultraviolet catastrophe. Classical physics predicts that the spectral energy density $u (\nu,T)$ of a black body at thermal equilibrium follows the Rayleigh-Jeans law: $$u(\nu,T) \propto \nu^2 T$$ where $\nu$ is frequency and $T$ is temperature. This is clearly a problem, since $u$ diverges as $\nu \to \infty$ (...


32

They are "fuzzy" in position space because they have well defined energies (and therefore give rise to sharp distributions of photon energy when they change states). Foundation Quantum states (the class to which orbitals belong) are elements in a Hilbert Space (which physicist sometime call "infinite dimensional vector spaces" just to watch the ...


31

since energy is quantized You have a misunderstanding here on what quantization means. At present in our theoretical models of particle interactions all the variables are continuous, both space-time and energy momentum. This means they can take any value from the field of real numbers. It is the specific solution of quantum mechanical equations, with given ...


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 ...


25

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 ...


25

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.


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 ...


22

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.)


22

You may also wonder why we use the concept of density of a material despite that material is made of molecules, or atoms, or in general quantized entities. This is the basis of hydrodynamics and solid mechanics (we use differential equations and integrals even if there are atoms). Electrostatics is just another branch of the "physics of continuous media&...


21

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 $$ V_{...


19

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 $\lambda^2/\delta\...


18

I want to offer a different perspective from the already existing answers, which all seem to somehow refer to the Standard Model or other specific physical theories to say that mass is not an integral multiple of some fundamental mass unit, hence not discretized. The reason why mass is not like that - and can indeed conceivably have continuous values in a ...


18

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 ...


18

This depends on the energy: Cold nuclei, low energies (typical order like 1-10 MeV): Usually, nuclei are not much excited or not at all. We can probe the nucleus experimentally (excite it somewhat) and we can interpret an observation by various models. There is number of models, most of them consider nucleus as a system composed of standalone nucleons. To ...


17

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 frequency,...


17

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, $$E_n=\...


17

There is no general rule. However there is a class of bounded self-adjoint operators whose spectrum is made of a bounded set of isolated points (proper eigenvalues) -- except for $0$ at most -- and the eigenspaces associated to these eigenvalues are finite dimensional. They are the so-called compact operators (this class includes classes of operators ...


16

Quantisation does not imply discreteness. If a system has been quantised, we just mean we have taken the set of states, and replaced it by a vector space of states. In other words, one can add states in quantum mechanics, allowing a system to be in two states "at once". Observable quantities become certain operators acting on this vector space of states. As ...


16

There are plenty of exact integers in physics. Take a wavefunction, for example: $\psi(x) = \langle x | \psi \rangle$. You can count the stationary points and the zero crossings. 1,2,3, .... In the room where I am sitting, there are exactly 2 windows. I know you wish to say that a macroscopic object is some sort of approximation, but you can if you like ...


15

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 $1 \, \text{frame}/t_p = 2\cdot 10^{43} \, \text{frames/sec}$.


15

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 ...


15

This is a comment, as Andrew's answer is adequate for the problem. I want to point out , which is not clear in your question, the difference between mathematical modeling and the object modeled. When modeling an object mathematically one can use continuous variables by the function of mathematics. If the object modeled has discontinuities, the mathematics ...


14

There is a tiny bit more going on than the otherwise excellent answer by zeldrege suggests. Imagine that you wish to probe an unspecified object to examine its structure. If we use light to look at the structure of an object, we need to have its wavelength smaller than the size of the details we wish to look at. Probing an object that has a (linear) size ...


13

There is an argument known as Weyl's tile argument which is not physics but philosophy, involves some really easy math and accessible to laymen like myself. Still, I'm tempted to put this here since it answers your question even though this probably doesn't belong on a physics forum. In a discrete space, say a square/rectangular tiled space, (for ...


13

This is ultimately a very deep question. We do not have a very easy ability to answer it at this time. Let me give you the pieces that we presently have. The basic interactions. The world as we know it today consists of five fundamental things that happen; they are all called "fields" and the first four are called "forces" or "interactions". In rough order ...


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