# Forces and the light

1. Do external forces can affect the light? Can any external force make the light accelerate? And if it can, will it accumulate mass? (according to the second Newton's law of motion $m = F/a$ )
2. We know that: $${E} = \frac{m_0c^2}{\sqrt{1 - \frac{v^2}{c^2}}} = mc^2$$ $$m=\frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$$

$m_0$ - rest mass

As you can see, as the velocity of a body is higher, the less mass it has. When we exert a tremendous force on an object, will it's mass will change along with the time?

1. According to the last equation, as slower the body, the more mass it has. If we could make the light slower, will it accumulate mass?
• The question is not very clear,but you cannot apply $F=ma$ to photon.This is only applicable In Newtonian mechanics. – Paul Jul 13 '15 at 14:30
• The actual Newton's first law is $\mathbf{F} = \frac{\mathrm{d}\mathbf{p}}{\mathrm{d}t}$. Thus any you further questions and ideas are completely meaningless. – m0nhawk Jul 13 '15 at 14:32
• If we could make light slow down locally, it would accumulate mass. However, we cannot make it slow down locally. This is because all massless objects travel at the speed of light. Objects not at light speed must have mass. But again, one cannot slow light down locally. One can only observe light at other speeds non-locally due to relativistic effects – Jim Jul 13 '15 at 14:49
• "Relativistic mass" is a misleading concept, see this Physics.SE post as well as this one. – Kyle Kanos Jul 13 '15 at 14:55
• I upvoted this question because it shows interest in the subject as long as (unsuccessful) attempts to deal with it. – Prof. Legolasov Jul 13 '15 at 15:20

Do external forces affect light?

Yes. See for example Faraday rotation:

GNUFDL image by Dr Bob, see Wikipedia

Can any external force make the light accelerate?

Yes. See for example Compton scattering. The photon doesn't change speed, but it's accelerated in the vector sense:

Image courtesy of Rod Nave, see hyperphysics

And if it can, will it accumulate mass?

In a way, yes. When you slow down a photon to less than c in say glass, some of its energy-momentum is exhibited as "effective mass". If you then slow the photon down to an effective speed of zero by trapping it in a mirror box, all of its energy-momentum is exhibited as mass. And it is indeed effective, because it increases the mass of that system. The box is harder to move when the photon is inside it. Then when you open the box, it's a radiating body that loses mass, just like Einstein's E=mc² paper.

(according to the second Newton's law of motion m=F/a)

That doesn't apply until you've trapped the photon in the mirror-box. Then you have a "body". The photon doesn't count as a "body".

As you can see, as the velocity of a body is higher, the less mass it has.

I'm sorry, that's not right. A faster-moving body comprises more energy, but when we talk about the mass of a body we typically mean rest mass, which doesn't change when you increase the speed.

When we exert a tremendous force on an object, will its mass change along with the time?

It's rest mass won't.

According to the last equation, as slower the body, the more mass it has.

It's true that a photon exhibits effective mass when you slow it down, but a body's rest mass doesn't increase when you slow it down.

If we could make the light slower, will it accumulate mass?

In a way, yes. More and more of the photon's energy-momentum will be exhibited as effective mass.

• +1 for answering all the questions in a clean and simple way. – Daniel Jul 13 '15 at 22:46
• Thanks @Daniel. IMHO it can be surprisingly difficult to give a clear answer. And putting one together helps one to organise one's own thoughts. It's an "explain it to your grandmother" thing. If I can't explain it to my grandmother, do I really understand it myself? – John Duffield Jul 14 '15 at 12:32

One has to think broader in order to answer those questions. Sure you can imagine a magical 'spring' being attached to a baseball, but there is no way to attach it to the light ray. There is simply not enough magic in this world.

Instead, let's focus on what forces could be actually applied to light. Currently, we are aware of four different types of interactions: electromagnetism, strong and weak nuclear forces, and finally, gravity.

Classical electromagnetism does not influence light because of the linearity of its equations of motion (the Maxwell equations). In classical electrodynamics, light is just an electromagnetic wave propagating in space-time. This wave is a solution of the linear homogeneous Maxwell equations, which is a specific realization of the general principle:

Any solution of a linear inhomogeneous differential equation can be expanded as a sum of some particular solution (vanishing at infinity and corresponding to the physical electromagnetic-interaction part of electromagnetism) and any solution of the corresponding homogeneous differential equation. Moreover, all solutions of the homogeneous Maxwell equations are plane waves (identified by momenta and polarization) which we experience as light.

Therefore, classical electromagnetic fields can not in any way influence the motion of light.

In quantum electrodynamics, which is a much more fundamental theory of nature, however, everything is much more complicated. Light is actually made of tiny quanta called photons which have electromagnetic charge zero (and therefore aren't influenced by E/M fields just like in classical theory). However, the overall equations of motion are no longer linear because of other charged quantum fields (for example, the electron/positron field).

This makes photons behave very much differently. They can spontaneously transform into pairs of electrons and positrons (this was actually experimentally observed millions of times and is called pair production). They can even interact with each-other by means of exchanging virtual electrons. Light can also be emitted or absorbed by electrons, which is how we experience its presence after all.

But these effects can only be observed with photons of much greater frequency than those visible to us, humans. Photons from visible spectrum would still travel through weak electromagnetic fields relatively untouched.

Now comes the most interesting part: gravity. The current widely-accepted theory of gravity is Einstein's General Relativity. One of the basic ideas of General Relativity is that gravity is not a physical field in the usual sense, but is described by the geometry of spacetime itself. This immediately implies that even the propagation of light is influenced by gravity (since light propagates in spacetime). So yes, light rays get bent by gravity. This was actually observed experimentally several times during the 20-th century by looking at how the light from stars is bent by the gravitational field of the Sun (these experiments are quite famous, because the only time one can observe the Sun and stars simultaneously is at the time of a total solar eclipse).

Note that the gravitational bending of the light can only be explained by General Relativity. Therefore you can not use special relativistic formulas (like you do in your question).

• and no gravitational lensing discussion is complete without a link to an image of "Einstein's cross", where the image of a single galaxy appears to be four different galaxies, because the light coming from that galaxy was bent by another galaxy in the foreground: observing.skyhound.com/archives/sep/ec.jpg – Jerry Schirmer Jul 13 '15 at 19:22

So the basic mass/energy/momentum relation in relativity can be phrased as$$\left(m_0 c^2\right)^2 = E^2 - c^2 p^2$$and the conventional explanation here is that $m = 0$ so $E = c p$, a formula known more or less since Maxwell showed that light is an electromagnetic wave.

It has also been known that light comes in packets with angular frequency $\omega = E /\hbar$ and wavenumber $k = p / \hbar$, leading to the familiar expression $$\omega(k) = c k,$$which is called a linear dispersion relation. [As for why it's called a "dispersion" relation, dispersion is about different wavelengths travelling at different speeds. Something which travels at velocity $v$ can be described with a combination of coordinates $x - v t$, while a pure wave is described with $k~x - \omega t$. If you have a wavelength-dependent velocity we can write it as $v(k) = v(2\pi/\lambda)$ and then make the analogy complete with $v(k) = \omega/k$ or $\omega(k) = k~v(k)$. So a "linear dispersion" corresponds to all wavelengths travelling at the same speed through the medium. We like to use this $\omega(k)$ function because a pure plane wave extending out to infinity travels with the "phase velocity" $\omega(k) / k$, but we can prove that a spatially-confined wave packet made of wavenumbers near $k$ actually travels at speed $\frac{d\omega}{dk}$.]

We sometimes euphemistically refer to, say, electrons which live in a conduction band which has linear dispersion as "relativistic". In those cases we are pretty clear about the fact that the electrons aren't literally moving as fast as the speed of light, and we can certainly inflict electromagnetic forces on them.

Can those "forces" be applied to light? Not really, not directly. Light only seems to interact with two things: charged particles, and gravity. The weirdness of the light/gravity interaction is seen in general relativity to be because we have a mistaken notion of what a "straight line" is: light always travels locally in a straight line, often the space of the universe itself is warped about energy concentrations, so that from far from, say, a star, it looks like the light is "bending" around the star even though locally the light is always travelling at speed $c$ in a straight line. The problem is that the coordinates themselves bend around the star. So our best model of how gravity "bends" light does not use forces at all to describe it!

As for charged particles: we can do a lot with them. We can bounce light off of mirrors, or even off of half-silvered mirrors which let some of the light through. We can slow light down to a fixed speed relative to a chunk of glass that the light is trapped in.

Can we describe those with "forces" and "masses"? Sometimes! The key is to go back to this dispersion relation. If we want the photons to have an effective "mass" $m_{\text{eff}}$, we need the analogous dispersion relation $$E \approx \text{const.} + \frac {p^2}{2~m_{\text{eff}}}$$from which we'd expect quadratic dispersion$$\omega(k) \approx \text{const.} + \frac{\hbar}{2~m_{\text{eff}}} ~ k^2.$$ So if you can find a suitable medium where this property holds, then you can meaningfully talk about light as if it had mass and lived in a potential energy.

If you want to search for these you'll probably need to work through a bunch of math related to things like the Abbe number to figure out what sort of glass this corresponds to. Many software packages give approximate forms that look like $$n^2 = 1 + A \lambda^{-2} + B \lambda^2 + \dots,$$so that $\omega(k)$ looks like $$\omega(k) = \frac {kc} {\sqrt{1 + a k^2 + b k^{-2} + \dots}}$$The bad news is, it's probably going to be really difficult to find a suitable glass. The good news is, as long as you're only dealing with a little bit of light scattering between a few different $k$, you can probably fit the equation to find an effective mass for any range where $\omega(k)$ is roughly a straight line, by fitting $\omega(k) = \alpha + \beta k^2$ to that linear region with $\beta = \frac 12 \frac{d\omega}{dk}$ and $\alpha = \omega_0 - \beta k_0^2.$ It's not great, but it's something.