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When light is propagating in glass or other medium, it isn't really true, pure light. It is what (you'll learn about this later) we call a quantum superposition of excited matter states and pure photons, and the latter always move at the speed of light $c$. You can think, for a rough mind picture, of light propagating through a medium as somewhat like a ...

40

In a Newtonian/Galilean world, where $c$ is infinite, you could not escape Olbers' paradox with an infinite universe. Any line of sight would eventually intersect the surface of a star, and so the whole sky would be as bright as the Sun. This is true whenever two hypotheses are satisfied: The universe is spatially infinite (or rather, the distribution of ...

20

A photon will travel "at the speed of light" until obstructed. From the speed, and elapsed time, you can calculate how far the light will travel. Laser light consists of more than one photon "in phase", which has exactly the same property in this respect, as a solitary photon.

15

In vacuum $$\nabla \times \vec{B} = \frac{1}{c^2} \frac{\partial \vec{E}}{\partial t} = 0$$ so a changing E-field does not beget a changing B-field. Larmors formula for radiation from accelerating charges also has $c$ in the denominator. Therefore no (star)light at all ? [Or at least no electromagnetic waves].

12

Theoretically, the photon (or the beam of photons, there really isn't a difference) can go an infinite distance, traveling all the while at a speed $c$. Since photons contain energy, $E=h\nu$, then energy conservation requires the photon to only be destroyed via interaction (e.g., absorption in an atom). There is nothing that could make the photon simply ...

10

A classical explanation to supplement Rod's excellent quantum mechanical one: If you make a Huygens construction of wave propagation (I assume you know how to do that) then every point on the wave front is treated as the source of a new wave of the same frequency and phase. How that wave propagates depends on the medium it encounters. So the Huygens ...

8

Note that it is correct that a photon can travel an infinite distance in an infinite time, but it can not reach any desired point in the universe. This is caused by the expansion of the universe, which also leads to the fact that we can not receive information outside of the observable universe.

5

Changing c to infinite changes some important things. The actual effect depends on how you want to propose magnetic forces work (they're normally fictitious forces induced by relativity). If we assume the coupling constant (this constant doesn't appear in the equation as it's value is normally 1) goes to infinity as c goes to infinity so that magnetostatics ...

5

I reject the notion that $\epsilon_0=4\pi\times10^{-7}\,{\rm F/m}$ and $\mu_0\sim9\times10^{-12}\,{\rm H/m}$, they are precisely $\epsilon_0=1$ and $\mu_0=c^{-2}$. This obviously takes care of any issue with defining any unit with $\pi$ in it because, $$\sqrt{\frac{1}{\epsilon_0\mu_0}}=\sqrt{\frac{1}{1\cdot c^{-2}}}=\sqrt{c^2}=c$$ We use the MKSA system ...

4

One small addition to the other answers: While it is indeed true that the light will never stop if it doesn't hit anything, it will however get red shifted, and thus become less energetic, due to the expansion of the universe. For example, the cosmic microwave background consists of photons which were emitted back when the atoms formed. However, back then ...

3

The hand-wavy way to do it is to consider a wave solution like the one below, and apply Faraday's law to loop 1, and Ampere's law to loop 2: If you make the loops narrow enough, i.e., their widths are $dx$, then $$\oint_1\!\vec{E}\cdot \vec{ds} = -\frac{d\Phi_B}{dt} \to \frac{\partial E_y}{\partial x} = -\frac{\partial B_z}{dt}$$ $$\oint_2\!\vec{B}\cdot ... 3 Technically, \omega^2/1^2-k^2/c^2=0 is a degenerate hyperbola if that counts. But I don't think you can derive an equation of the form \omega^2/a^2 - k^2/b^2 = 1 for waves propagating in free space. You may however find something of the kind if you consider materials with fancier dispersion relations than \omega = ck, like e.g. plasmas. 3 For an object close to you, the speed of light is effectively infinite - i.e. the time taken for the light bulb 10m away from you to get to you is so close to zero that it can be considered immediate, and thus the speed of light is assumed to be infinite. With this in mind, this would mean that the sky would be brighter. In reality, the speed of light is a ... 2 A photon cannot be said to have its own inertial reference frame, because inertial reference are defined to be a family of coordinate systems that satisfy the two fundamental postulates of SR, one of which is that light moves at c in all frames. You could construct a coordinate system where the photon was at rest, but since this coordinate system wouldn't be ... 2 Of course you know from special relativity that no information can propagate faster than c, which includes your directing the Mars end of the stick to move through your Earth end push. But you don't need to think about SR at all. Simply think carefully about what would happen if you did give the end of this rod a sudden shove. It has a great deal of mass ... 2 The freespace dispersion equation is \omega^2 = k^2\,c^2 and this cannot change: this simply follows from considering plane wave components of propagating fields, which all fulfil the Helmholtz equation$$\nabla^2 A_j + \frac{\omega^2}{c^2} A_j = 0\tag{1} which is fulfilled by all Cartesian components of the moncrhomatic EM field vectors and, for a ...

2

The problem with FTL and causality has to do with two issues: 1) the relativity of simultaneity between inertial frames (not an issue in classical physics with sound waves, since in classical physics all inertial frames agree about simultaneity), which implies that a signal moving FTL but forward in time in one frame is moving backwards in time in other ...

2

Andrew the problem is that the speed of light is not an exact number. You give it as $2.99792458\times 10^{8}$ m/s; but suppose I instead tell you that it is really $1.8026175\times 10^{12}$ furlongs per fortnight. Neither of these numbers is any more correct than the other, but we have an accepted and defined system of units that we work in. Maxwell's ...

2

It sounds like that site is discussing the Standard Model neutrino. Neutrinos were presumed to be massless for a long time and the SM still models them as massless. We now know that this is not true, that (at least two of the three flavors of) neutrinos do have a small mass, although the SM is still a good approximation. So, like other massive particles ...

2

A ray of light or a laser beam will not stop until it reaches an obstruction. If there is no obstruction, light will NEVER stop. It has no end.

2

A "ray of light" must be respelled as "photon" because here we are talking physics. Between a single photon and a laser beam, in this case, there is no difference. Every photon will continue his travel until stopped, every single photon is "indistinguible" from others (in the sense that they are no different intrinsecally). The photons of a laser beam are ...

2

The differential and integral forms of Maxwell's equations are truly equivalent; they are essentially the same set of equations. One can convert between the two using two mathematical theorems: Divergence Theorem (Wikipeda - Divergence Theorem) Stokes' Theorem (Wikipedia - Stokes Theorem) The divergence theorem states that the flux over a closed surface ...

1

OK... I can't give a definitive answer to the problem. My intuition tells me that any massive particle or macroscopic mass, boosted high enough, has to look like a black hole. Why? Because it is very hard to see why/how gravity, if we believe in the equivalence principle, should be able to distinguish between kinetic energy and other forms of internal energy ...

1

Provided that there is nothing for the photon to interact with (i.e. we look at it in vacuum), the mean free path will be infinite; that is, it will continue travelling forever in a given direction. There's nothing which will stop the photon's path. Hence, it will go arbitrarily far. Whether you have a single photon or a laser, the answer won't change. The ...

1

As soon as the universe came out of its dark age if light speed was infinite then it would be able to keep up with the expansion of the universe. It would be very much brighter all around, perhaps intolerably to us. The universe would appear very active since event far far away would appear to us instantly. We might be blind as our light sensory organs might ...

1

At the Event Horizon of Black Holes, the acceleration an object experiences is infinite (Here, I am NOT talking about freely falling objects; Freely Falling objects are in inertial motion in General Relativity). As for you, you can't feel that on the Event Horizon of Black Hole (Stephen Hawking was wrong in A Brief History of Time) because you won't survive ...

1

As pointed out you can't travel at the speed of light but you can look at the limits we are tending towards as we approach it. So, if I were to travel in a spacecraft at the speed of light, would I freeze and stop moving? From the perspective of a stationary observer if your spacecraft was traveling at close to the speed of light, time on the ...

1

Stated simply, causality means that all causes should precede their effects, for all observers. The timings of the causes and effects aren't the times at which a human registers them, they are the times at which they occur in an observers reference frame - i.e. the time on the observer's watch at the moment they occur. If faster-than-light signals were ...

1

It is a matter of the standpoint of the observer. Because time comes to a standstill at the speed of light, to the photon, no time passes, whatever the distance traveled and its speed is therefore infinite.

1

The speed of the push would be roughly the speed of the sound in whatever medium the stick was made of. One thing is certain - it would not exceed the speed of light.

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