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See the image first: See the image

Why are light rays able to cross each other? Air isn't able to.

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    $\begingroup$ Water waves can cross each other. This is more or less what light waves crossing would look like if you magnified them enough. $\endgroup$ – Peter Shor Jan 18 '17 at 15:53
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    $\begingroup$ @PeterShor ...that's probably for small amplitude. As for large amplitudes water waves are non-linear and they do interact $\endgroup$ – mikuszefski Jan 18 '17 at 16:08
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    $\begingroup$ @mikuszefski: I'm probably being too pedantic here, but the same is true for light (although the amplitudes here probably need to be truly enormous). Consider sending a lot of photons from all directions towards one spot: they'll form a black hole. $\endgroup$ – Peter Shor Jan 18 '17 at 16:11
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    $\begingroup$ @Carl: it doesn't matter!!! Low amplitude water waves pass through each other just like light waves. And demonstrate interference just like light waves. $\endgroup$ – Peter Shor Jan 18 '17 at 19:33
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    $\begingroup$ Why questions like this are notoriously difficult for physics because science doesn't explain "why." When you ask a physicist why something happens, what we give you instead is an explanation of the phenomena using a more general theory, or we tie it to a mathematical model and studies showing that model's validity. For example, we could explain this behavior of light by explaining it in quantum mechanics terms, or we could say "light acts like a wave" and discuss the mathematical behavior of waves. Can you give us anything to help us figure out what you are looking for from a "why" question? $\endgroup$ – Cort Ammon Jan 18 '17 at 20:14
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Note: this answer was in response to the original question:

My question is that Why the light rays able to cross each other weather water waves and air could not cross each other

Other waves pass through each other just as with light. This is easy to test. Place four people at the corner of a large room. Have two of them, at adjacent corners talk to the person at a diagonal corner. Use a cone such as a cheerleader might use to somewhat channel the sound. You may be a bit distracted by the other voice but you will clearly hear the voice from the opposite corner.

Here's a standard demo in a high school science class. Have two students hold each end of a moderately stretched slinky resting on a smooth floor. Have each student give the slinky a sharp snap to their right. Since the students are facing each other, the pulses will be opposite one another as they travel toward opposite ends. When the two pulses meet in the middle the slinky will appear relatively straight but only for an instant. The two pulses will continue to travel past one another as if they never had met.

Waves of the same kind traveling through one another maintain their original identity after the encounter. This is a basic property of waves, you can read about it in any introductory Physics text.

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    $\begingroup$ Bad answer even though it appears simple. Light waves simply are not like any kind of physical (water, acoustic) waves. $\endgroup$ – Carl Witthoft Jan 18 '17 at 19:33
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    $\begingroup$ We need some moderators to fix this rating -- it's a case of too many nonphysicists upvoting a false answer. Reminds me, sadly, of the time when (some) math PhDs angrily denounced the correct answer to the Monty Hall problem. $\endgroup$ – Carl Witthoft Jan 18 '17 at 19:37
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    $\begingroup$ @CarlWitthoft. Rude and inappropriate. This was answered in context of the original question before it was edited. My intrepretation was the focus on why other waves do not appear to pass through one another like light. I taught Physics for 24 years, quite successfully by most measures. $\endgroup$ – bpedit Jan 18 '17 at 19:54
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    $\begingroup$ @CarlWitthoft I think this answer is good. Light waves may be made of different things than sound waves, but they pass through each other for the same reason: because they can be modeled as linear equations. $\endgroup$ – Owen Jan 18 '17 at 20:07
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    $\begingroup$ @CarlWitthoft For the record, moderators cannot influence the voting total on any post except by voting on it themselves like any other user. Also, could you elaborate on why you think the analogy to other waves is "false"? To me, it's simply a result of the wave equation (no matter for what type of wave) being linear, so two different solutions just superpose instead of altering each other. $\endgroup$ – ACuriousMind Jan 18 '17 at 20:29
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I believe it's because you are thinking of the light as particles (little solid balls) that it seems a bit odd that "they can pass through each other." I think thinking in terms of (quantum) fields gets rid of this. As someone mentioned, photons are bosons and so there is no Pauli exclusion principle that applies to them. That is to say, they can be at the same place at the same time....so the fact that they can "pass through" each other seems less odd, knowing that.

Said more simply, the information of the photons spreads out, and there is nothing to prohibit this information from allowing the photons to cross the same point in space at the same time. This is easier to see if you know math and look at the equations explaining photons.

*Note: We are neglecting any interactions between photons in this sort of explanation.

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You seem to be looking for a more mathematically oriented answer, so let's try that. You cannot prove that light rays are able to cross each other because that is false: it only works in the approximation of very dim light. If you were to take your source of light and increase the frequency or amplitude of the electric field, you'd observe a non-linear behaviour: rays would very clearly influence each other as the pass through each other.

The key aspect of the superposition principle is linearity, that is, the fact that Maxwell's equations are linear. If you consider electromagnetic radiation in a certain material, it is well-known that for intense enough radiation the polarisation becomes non-linear, and you enter the realm of nonlinear optics. Here, the polarisation $\boldsymbol P$ becomes a (non-linear) function of $\boldsymbol E$, and therefore the wave equation becomes $$ \left(\nabla\times\nabla\times+\frac{n^2}{c^2}\frac{\partial^2}{\partial t^2}\right)\boldsymbol E=\frac{1}{c^2}\frac{\partial^2}{\partial t^2}\boldsymbol P(\boldsymbol E) $$ which is a non-linear equation for $\boldsymbol E$. For example, a very strong source of light (travelling through the air) the Kerr effect or other non-linear effects kick in. This is due to quadratic (and higher) terms in the polarisation tensor $\boldsymbol P\sim\chi_1\boldsymbol E+\chi_2\boldsymbol E\otimes\boldsymbol E+\cdots$.

The polarisation of realistic materials is always non-linear, and therefore light rays always influence each other. Only in the limit $\chi_2E^2\ll \chi_1E$ the wave equation becomes linear; in other words, only in the limit of very weak radiation are light rays oblivious to other light rays.

Even in vacuum you can observe non-linear behaviour. The Maxwell Lagrangian, corrected by quantum-mechanical effects, becomes $$ \mathcal L=\frac12(\boldsymbol E^2-\boldsymbol B^2)+\frac{2\alpha}{45m^4}\left[(\boldsymbol E^2-\boldsymbol B^2)^2+7(\boldsymbol E\cdot\boldsymbol B)^2\right] $$ where $\alpha\sim 1/137$ is the fine-structure constant, and $m$ is the mass of the electron. As $\mathcal L$ is non-linear in $\boldsymbol E,\boldsymbol B$, the propagation of electromangetic waves is no longer linear, and the superposition principle ceases to hold. Only when you can neglect the non-linear terms the superposition principle becomes valid.

In a nutshell, your question is based on a false premise. Light rays seem to be able to cross each other and continue their path unaffected, but this is because you are not using sensitive enough instruments (e.g., your eyes). If you were to measure the effect of light rays on each other with a very good piece of equipment, you'd observe that they do affect each other, both in a material and in vacuum.

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Why are the light rays able to cross each other

The underlying level of nature is quantum mechanical. Light is an emergent phenomenon from the quantum mechanical level of photons, where an enormous number of photons of energy $h\nu$ build up the classical electromagnetic wave which is light.

Photon–photon interactions are very very rare at energies below twice the mass of an electron . The quantum mechanical feynman diagram of two photons interacting, from which the probability of interaction can be calculated:

two photon

has four electromagnetic vertices, i.e. (1/137)^1/2 for the amplitude, and when squared as it multiplies the integral for the probability, the number becomes miniscule, so photon photon interactions are very rare. As the answer by @AccidentalFourierTransform states, using classical electromagnetic waves, some interaction can happen, but it would need very good instrumentation to see it.

One can see interference between two light beams, but interference is not interaction, it comes from the superposition of two beams collective wavefunctions, which when detected show the interference pattern from the way the photons' wave functions build up the macroscopic light beam. Superposition is not interaction, so the beams can cross and continue on their way, if a detector is not introduced in the overlap.(Note that anyway, to see interference patterns one should have coherent monochromatic beams).

For high energy photons other channels open with higher probability, but that is another story.

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  • $\begingroup$ From your statement: "... photons of energy hν build up" can I take it that an individual photon has a frequency and therefore a 'colour'? If so how does this (do these) transform into the " ... classical electromagnetic wave which is light."? If you can point me in the right direction I'd be grateful. Thanks. $\endgroup$ – JohnP Jan 25 '17 at 11:11
  • $\begingroup$ @JohnP The photon when measured can only give information for spin (+/-1) and energy.BUT as a quantum mecanical entity it is described by a wavefunction which is a solution of the quantized maxwell equations, see this arxiv.org/ftp/quant-ph/papers/0604/0604169.pdf . The way the classical field is build from QED can be seen here motls.blogspot.com/2011/11/… . the image here en.wikipedia.org/wiki/Spin_angular_momentum_of_light may help $\endgroup$ – anna v Jan 25 '17 at 11:37
  • $\begingroup$ A small correction perhaps: one needs two vertices to get one factor of $\alpha$, right? So, doesn't four vertices give two factors of $\alpha$? $\endgroup$ – flippiefanus Jan 27 '17 at 4:30
  • $\begingroup$ @flippiefanus I was using the "fundamental forces" number hyperphysics.phy-astr.gsu.edu/hbase/Forces/couple.html and according to www-pnp.physics.ox.ac.uk/~barra/teaching/feynman.pdf this is already squared. I edited , thanks. $\endgroup$ – anna v Jan 27 '17 at 5:15
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There is no force which could act between two photons so they cannot* interact with each other to cause interference.

*At least for the most part, it is possible for them to cause interference via decay to other particles.

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    $\begingroup$ @wizzwizz4 No, the double-slit experiment, extended, has conclusively shown that each photon interferes with itself. It's just a simplified model of light as interfering wavefronts that we use to keep non-physicists' heads from exploding. $\endgroup$ – Carl Witthoft Jan 18 '17 at 19:34
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    $\begingroup$ @Carl Witthoft: two different photons can interfere with each other, as well. It's just that demonstrating this is a lot more complicated because you somehow have to synchronize their phases. $\endgroup$ – Peter Shor Jan 18 '17 at 19:37
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    $\begingroup$ @PeterShor no, photons do not interfere with each other, or to be more precise, the quantum mechanical probability (remember $<\psi|$ and $|\xi>$ stuff?) is down by a factor of$ c^2$ $\endgroup$ – Carl Witthoft Jan 18 '17 at 19:41
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    $\begingroup$ @dmckee This is, perhaps, a misunderstanding on my part. I had thought the photon must first produce another particle in order to interact and, to conserve relevant quantum numbers they must produce a second particle too. (I admit this explanation has got away from me a little so this may not make sense) But the number of photons cannot be constant at every stage so I assumed a decay must take place. A feynman diagram would be best but I'll have to settle for: $\gamma ->f \bar{f}$ is what I am imagining, with the fermions being the particles which interact with our second photon. $\endgroup$ – Lio Elbammalf Jan 18 '17 at 21:32
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    $\begingroup$ @Carl: Please google "two-photon interference", and explain to me why all these experimentalists with published papers are wrong. They don't interact with each other very strongly, but interaction and interference are different things. $\endgroup$ – Peter Shor Jan 19 '17 at 2:52
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Because the light consists of photons. And photons are Bosons (not like for example electrons, which are Fermions. Bosons and Fermions obey different laws (search for Fermi-Dirac statistics and Bose-Einstein statistic on Wikipedia for more info). In short and mundaine words: Fermions can not be in the same place at the same time, but Bosons can. This is why light rays go through each other.

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    $\begingroup$ This has nothing to do with Fermion or Boson. Bosons can easily interact via electric charge etc. $\endgroup$ – mikuszefski Jan 18 '17 at 16:10
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    $\begingroup$ @mikuszefski but it does have to do with Bosons having integral spin. $\endgroup$ – Carl Witthoft Jan 18 '17 at 19:36
  • $\begingroup$ All the waves or wavelike stuff can be interpreted as bosons, like phonons in a crystal... $\endgroup$ – Žarko Tomičić Jan 18 '17 at 21:08
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    $\begingroup$ Light beams passing through each other necessarily have different momenta so exclusion never comes up: they are not trying to be in the same state (because the state space is the Hamiltonian phase-space: both position and momentum matter). $\endgroup$ – dmckee Jan 19 '17 at 2:03
  • $\begingroup$ @CarlWitthoft well that's sort of the definition of a Boson, is it not? Nevertheless, that does not exclude fermions from crossing, as this means they move in different directions. The momenta are, hence, not equal such that they can be at the same position without violating Pauli's principle. $\endgroup$ – mikuszefski Jan 19 '17 at 7:11
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A small addition to other answers. Light is both a wave and a particle, called Wave-particle Duality. Basically, It acts like both a Wave and a Particle. Its very complicated to get into, and I cant properly describe when it acts like what in all situations, but there are a few that are relevant here. First off, Air is a particle, or rather, many many particles. As such, Like you described, they can not cross eachother. What we have found though, is that Light does not act like a particle in that aspect, It acts like a wave. Peter Shor's comment links to this image showing this in water, which shows how waves act when they cross over each other. This is called Interference. When 2 waves cross, they interfere with eachother, changing the total size of the wave. 2 waves can stack on each other, becoming twice as tall, or they meet each others low points, and flatten out. They could do both, and everything in between, but they will not stop moving because of this meeting. The Interference link above to the Wiki page has some good animations of it in action.

Light will act in the same way. and this is most easily shown by the famous Double Slit experiment. There are numerous youtube videos animating this, such as this one, which shows that light, when passing through 2 slits, will hit a wall with the exact same pattern as waves of water would in the same situation, and this pattern is called the Interference pattern. It also shows off the Wave-Particle Duality, because when you actually monitor the light to determine how it passes through the slits, it acts like a particle, but that is a whole other question.

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protected by Qmechanic Jan 18 '17 at 19:54

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