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Why does light continue on forever if it was created from some source whose radiation dwindles at a rate of the inverse square of distance. Clearly light can be viewed as an interdependent phenomena, the E field pulling the B field along with it, but if all light must come from a source, and that source creates a field dependent on distance, why doesn't the light die off?

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You are thinking about static fields. EM waves are described by time dependent fields. Light does "die off" as you move further from the source - why do you think stars don't appear as bright as the sun? –  Will Jul 8 '13 at 4:51
    
I thought that was due to dispersion of the rays. –  Anthony Jul 8 '13 at 4:53
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You are treating your waves as plane waves. Do you think this makes sense? The best way to think of this is as having a spherical wavefront. The energy density of an EM wave is proportional to the amplitude of the electric field. Conservation of energy then dictates that the amplitude must decrease with distance (this is effectively your "dispersion" concept). –  Will Jul 8 '13 at 5:05
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Maybe you're after this very short answer: Individual photons don't die, but if you imagine a source sending out photons in all directions, the number of phonons you'd find in a fixed area (such as $1m^2$) decreases. –  Lagerbaer Jul 8 '13 at 5:39
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@Tony rays are a geometrical concept imaging the direction of energy propagation but have no attributes other than that, geometry. Very useful for optics, but not for field arguments. Classically the E and B fields describing the EM wave decrease like 1/r, as Will shows, and quantum mechanically the photons do disperse, within the Heisenberg uncertainty principle . –  anna v Jul 8 '13 at 7:14

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up vote 8 down vote accepted

You seem to be trying to compare static electric fields with EM waves. To do a fair comparison you should set up to two scenarios as close as possible.

You are talking about the static field dropping with distance proportional to $\frac{1}{r^2}$, which tells me you are thinking about a point charge or some equivalent, spherically symmetric configuration (a sphere of charge perhaps). To do a fair comparison with an EM wave, you should also consider a point (or spherical) EM source.


Note: a(n infinite) plane wave is produced by an infinite plane source and is usually described by a sinusoidal EM wave propagating in a particular direction (perpendicular to the plane source), and thus has the form (for example): $$\vec{E} = E_0 \sin(k(x-ct))~\hat{z}$$ This obviously doesn't decay with distance ($x$). But to make a fair comparison, one should really compare this to an infinite plane of charge. Freshman electrostatics tells us that the associated electric field is $$\vec{E} = E_0~\hat{z}$$ ($E_0$ is dependent on the charge density) a constant as well! But these are not very physical scenarios as the require sources on infinite planes.


Now, back to the spherical source for an EM wave:

We are talking about EM waves, so the problem can be explored mathematically, by analyzing the Maxwell equations in vacuum: $$\left[\vec{\nabla}^2-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right]\vec{E} = 0~~~~~~~~~~~~~~~~~\mbox{(and the same equation for $\vec{B}$)}$$ For a spherically symmetric source, it makes sense to use spherical polar coordinates $(r,\theta,\phi)$, under which $$\vec{\nabla}^2\vec{E} = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial \vec{E}}{\partial r}\right)$$ leaving our equation as $$\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial \vec{E}}{\partial r}\right)-\frac{1}{c^2}\frac{\partial^2 \vec{E}}{\partial t^2} = 0$$ The general solution of which is $$\vec{E}(r) = \frac{\vec{E}_{out}(r-ct)}{r} + \frac{\vec{E}_{in}(r+ct)}{r}$$ where $\vec{E}_{out}(r)$ and $\vec{E}_{in}(r)$ are vector valued functions, specified by the boundary conditions (similar results for $\vec{B}$).


Application to your problem:

Let's assume we only have an outgoing wave, that is, $\vec{E}_{in}(r) = 0$ (this is the situation you are thinking of). Now $\vec{E}_{out}(r)$ specifies the profile of the wave - you can assume is sinusoidal if you like - this shape will propagate out. We see the dying off of the amplitude with distance is $\frac{1}{r}$, thus the energy density per second (which is $\propto |\vec{E}|^2$) of the wave dies of at $\frac{1}{r^2}$ and so we do see a "dying off" of light with distance.


Comparison to the photon approach

It's nice to see that the result using photons gives the same result. If we assume that a point source is (isotropically) emitting $N$ photons per second, each of the same energy $E$. Then at a distance $r$ from the source the density of photons each second is $\frac{N}{4\pi r^2}$ and so the energy density per second is $\frac{E~N}{4\pi r^2}$, which gives the same $\frac{1}{r^2}$ "dying off" of light with distance.

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So something like a laser-that has a linear output, doesn't it? And also, I was referencing static charges, but what about the light gives it any 'strength' besides the field strength given by the moving charges that produced it? Is it the displacement current in maxwell's equations? –  Anthony Jul 8 '13 at 7:36
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@Tony Actually, unfortunately a laser also ultimately has a $\frac{1}{r^2}$ "dying off". It's just that instead of filling the whole $4\pi$ steradians of solid angle, it is filling a cone with a very small vertex angle. In ray terms it's like rays coming from a point a long way off: they're almost parallel but not quite, so that the dependence is actually like $\frac{1}{\left(r + r_0\right)^2}$, where $r_0$ is the distace to the virtual source and some very big value. So, for small distances $r \ll r_0$ there isn't much dying off. But for $r \gg r_0$ the $\frac{1}{r^2}$ dependence resumes. –  WetSavannaAnimal aka Rod Vance Jul 8 '13 at 9:08

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