In an old question: How would night sky look like if the speed of light was infinite? the best answer was voted down to negative credits. I cannot understand why. From Maxwell's equations, we derive the scalar wave equation (https://maxwells-equations.com/equations/wave.php): $$(d^2 E_x)/dt^2 = 1/μϵ * (d^2 E_x)/(dz^2),$$ where $1/μϵ = c^2,$ the electric field has only the x-component and the propagation is along the $z$-direction. Assuming c is infinite, the second time derivative of the electric field would also be infinite. Could we still have electromagnetic waves and light with this wave equation? If not, shouldn't we with the infinite c throw away also the Maxwell's equations? Could the university still exist?
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1$\begingroup$ Neither the universe, as we know it, nor the university could possibly exist in the absence of electrodynamics. $\endgroup$– Albertus MagnusCommented Jun 2 at 13:49
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1$\begingroup$ @AlbertusMagnus This is the problem with hypothetical (aka: non-physical) questions: they are fun to entertain, but any answer violates some principle, so you get down voted, and into silly arguments in the comments. Moreover, the people who ask them don't have the chops (yet) to understand the good answers. It's not worth the effort $\endgroup$– JEBCommented Jun 2 at 16:36
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3 Answers
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The wave equation for the $j$-component of the electric field is: $$ \frac{1}{c^2}\frac{d^2E_j}{dt^2} -\Delta E_j = 0 $$ Where $\Delta$ is the Laplacian. If $c$ were infinite, $1/c^2$ would vanish and we would be left with Laplace's equation: $$ \Delta E_j = 0 $$ Which is solved by harmonic functions, which are not wavelike.
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3$\begingroup$ Right. The whole concept of electrodynamics depends on finite speeds of propagation. Essentially, electrodynamics reduces to electrostatics if one allows for instantaneous "spooky action at a distance". $\endgroup$ Commented Jun 2 at 13:48
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1$\begingroup$ Your argument is good for propagation in a vacuum, but it is still conceivable (not in this world but only on paper) that in matter the speed might still be finite even though in vacuum it is not. Not in this world though... $\endgroup$ Commented Jun 2 at 14:38
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$\begingroup$ Well now you have a problem: what is the range of these harmonic function? How does light reception work with infinite $c$? Maybe you can still see? $\endgroup$– JEBCommented Jun 2 at 16:41
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$\begingroup$ @JEB arguably the problems start when you take $c$ to be infinite. I did not really consider interactions, because everything we know about optics and light relies on the fact that $c$ is finite. It is certainly possible to have light "propagate" in this theory, and the intensity of light emanating from a point-like source would decrease as $1/r$ instead of $1/r^2$. $\endgroup$– paulinaCommented Jun 2 at 16:56
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If all light travels at the same infinite speed, you get Laplace equation as from other answers.
However, one more interesting thing you can assume is that light travels with a speed that is finite but that depends on its frequency. For example, you can assume that the speed of propagation is proportional to the frequency (the inverse of the wavelength). Under this assumption, you get, well... The Schrödinger equation :) $$ \frac{d}{dt} \psi-i\Delta \psi=0. $$
Schrödinger has infinite speed of propagation in the sense that wave packets can be arbitrarily fast, depending on their frequency.
You can assume other kinds of behaviours. The thing that has to be specified to tell how fast light travels is the so-called "dispersion relation": $$ \tau=\omega(k), $$ which gives a law on how waves of a certain spacial frequency oscillate in time. Light has dispersion relation $\omega(k)=c|k|$, Schrödinger has dispersion relation $\omega(k)=|k|^2$. The gradient of $\omega$ with respect to $k$ gives the so-called "group velocity", which tells you precisely the velocity of wave packets with frequency $k$. The gradient of $\omega(k)=c|k|$ is $\nabla \omega(k)= ck/|k|$, which is bounded in norm by $c$ :) this is why you get finite speed of propagation.
Other models exist with different dispersion relations:
- deep water waves - $\omega(k)=\sqrt{gk}$
- shallow water waves - $\omega(k)=k^3$ (Airy equation)
- 2D shallow water waves with low surface tension - $\omega(k_1,k_2)=k_1^3-k_2^2/k_1$ (KP-II equation)
- relativistic quantum waves - $\omega(k)=\sqrt{k^2+m^2}$ (Klein-Gordon and Dirac equations)
- ...
All these correspond to certain linear partial differential equations that are called "dispersive equations". Their main properties is that they describe wave-like phenomena which conserve energy, but wave packets have different velocities depending on their frequencies, so that they tend to spread over time. The first three in the list have unbounded speed of propagation, like Schrödinger. Essentially, most of the dispersive equations one can come up with have unbounded speed of propagation.
Light waves are very special in this sense: the wave equation assigns the same velocity to all wave packets, the only difference which makes light disperse is the direction of propagation of different wave packets. In one dimension, light does not disperse :D It just obeys a transport equation.
So, I would argue that light that travels with infinite speed makes mathematically sense, although not in the sense one can think of.
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In the Dirac and quantum telegraph equations, a wave travels at a group velocity equal to the speed of light. An infinite velocity for an electromagnetic wave is possible if the wave diffuses in a medium only, which is governed by $\nabla^2\vec{E}=\mu_0\sigma \frac{\partial\vec{E}}{\partial t}$.
