Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Spin 1 silver atoms have a definite spin axis, e.g. up or down along an axis labeled X. This in turn means that they carry angular momentum in an overt, visible fashion.

However, spin 1 photons do not seem to display an experimentally meaningful version of a spin axis, unless I'm completely missing something (very possible!).

Instead, photons in the form of light display an effect we call "polarization," by which we mean the light has a definite vibrational orientation in space. This effect follows the same 90$^\circ$ rules as the basis states for massive spin 1 particles, but lacks the directionality of those states.

So for example, if X is horizontal and Y is vertical, vertically polarized light could be interpreted as exhibiting a rotation in YZ plane, which is the same plane of rotation that gives rise to -X and +X spins for particles with mass. But since photons have no mass and travel at $c$, there is no way to assign $\pm$ to this nominal rotation. Instead, the light appears to be composed of particles with no visible spin, or (perhaps) alternatively as superpositions of equal and cancelling quantities of -X and +X spin.

Circular polarization seems at first glance to provide a solution by providing distinct clockwise and counterclockwise directions. I've gone that route myself, but the more I look at it, the more sure I am that it's a bogus assignment. If nothing else, circular polarization can always be decomposed into two planar polarization that have a relative phase shift. Furthermore, the idea of a spin axis pointing along the infinitely compressed direction of propagation of a massless particle is problematic at very best. Circular polarization still seems to me the most promising path for finding real angular momentum in photons, but if it is the solution, I must admit that the more I look at it the less I see how.

So, at last, my question:

If there is no way to demonstrate experimentally that an arbitrarily small unit of light carries explicit, non-zero angular momentum, how then does a photon manage to convey the 1 unit of spin required needed for it to balance angular momentum in particle interactions?

share|improve this question
2  
"If nothing else, circular polarization can always be decomposed into two planar polarization that have a relative phase shift." You could just as well say that linear polarizations can be decomposed into sums of circular polarization so I'm not sure that this argument is going anywhere. Indeed if you start from the idea that the photon has spin then this may be more "natural", though I am always a little leery of argument from naturalness. –  dmckee Aug 14 '13 at 19:57
    
Try this en.wikipedia.org/wiki/Angular_momentum_of_light If it's not clear perhaps you could refine your question somewhat. –  user27777 Aug 14 '13 at 21:10
2  
Just for reference @John, Terry knows his physics and likes to ask the kind of needling questions that make you examine just how well you understand a topic. In this case I read the question as "How can we motivate the photon spin starting from the classical theory?" And I don't have an answer except "I'd rather go the other way because I think I understand it that way". –  dmckee Aug 14 '13 at 21:28
1  
Undeveloped brainstorm that I don't have time to follow through: what if we examine the maximum of (angular momentum density)/(energy density) in the classical theory and claim that this has to be related to the (angular momentum)/(energy) of the "particle of light" because that maximum must correspond to all the bits "lined up"? –  dmckee Aug 15 '13 at 2:04
1  
I think that for users of this forum it should be a must to read at least the blog entry of Lubos Motl on how classical fields emerge from the underlying quantum mechanical substratum motls.blogspot.gr/2011/11/… . It ain't simple as the emergence of thermodynamics from quantum statistical mechanics is not a simple algebraic method. The classical field emerges from the coherent effect of the individual photons' spin and energy, given their quantum mechanical solution as photons. –  anna v Aug 15 '13 at 16:32
show 7 more comments

3 Answers

up vote 6 down vote accepted

Here's my haltingly offered answer - I am not sure I am not overlooking a subtlety that you can see but I can't. I'm going to try to answer @dmckee 's take on your question: "How can we motivate the photon spin starting from the classical theory?"

I relate to your worries that slipping between linear and circular polarized base states might just be nonphysical co-ordinate transformations, but I believe that there are three ways that I can think of wherein Nature does show a preference for particular base states, two theoretical and one experimental; these are:

  1. The diagonalisation and decoupling of the Maxwell equations by the Riemann-Silberstein vectors;
  2. The classical angular momentum transferred to an absorbing medium calculation and
  3. The conversion of linearly to circularly polarized light by a quarter waveplate birefringent crystal and the measurable torque exerted on the crystal by that process.


Diagonalisation of Maxwell's Equations by the Riemann-Silberstein Vectors

The Maxwell curl equations in freespace:

$$c\,\left(\begin{array}{c}\nabla \wedge \sqrt{\epsilon_0} \mathbf{E}\\ \nabla \wedge \sqrt{\mu_0} \mathbf{H}\end{array}\right) = \left(\begin{array}{cc}\mathbf{0}_{3\times3}&-\mathbf{1}_{3\times3}\\\mathbf{1}_{3\times3}&\mathbf{0}_{3\times3}\end{array}\right)\left(\begin{array}{c}\partial_t \sqrt{\epsilon_0}\mathbf{E}\\\partial_t \sqrt{\mu_0}\mathbf{H}\end{array}\right)$$

are obviously mutually coupled and can be decoupled by forming the Riemann-Silberstein vectors (arising from the diagonalisation of the block $6\times6$ matrix of ($2\times2$ matrix of four $3\times3$ scalar matrices); we get:

$$i\, \partial_t \mathbf{F}_\pm = \pm c\,\nabla \wedge \mathbf{F}_\pm$$

where:

$$\mathbf{F}_\pm = \frac{1}{\sqrt{2}}\left( \sqrt{\epsilon_0} \mathbf{E} \pm i \sqrt{\mu_0} \mathbf{H}\right)$$

(I apologize for using SI units exclusively - much of my career has been building numerical software and the only way to get such beasts debugged is by making everyone stick to the same units - now I can't think in Planck or natural units at all any more). Now, if $\mathbf{E}$ and $\mathbf{H}$ are real-valued fields, we only need one complex Riemann-Silberstein vector to encode the whole of the Maxwell equations. Equivalent information is encoded into the positive frequency parts alone of the two Riemann-Silberstein vectors. What's really neat about the second approach is that if the light is right circularly polarized, only $\mathbf{F}_+$ is nonzero; if left, only $\mathbf{F}_-$ is nonzero. So the positive frequency parts of the fields are decoupled precisely by splitting them into left and right circularly polarized components, NOT linearly polarized components. This is the first big clue that Nature indeed does show a preference for circularly polarized base states. Also note that the positive frequency (i.e. positive energy) part is the meaningful one if one is to think of Maxwell's equations as the propagation equation for the first quantized photon.

Now, in momentum (Fourier) space, the decoupled Maxwell equations become (we do spatial, not time - Fourier transform of both sides):

$$\mathrm{d}_t \tilde{\mathbf{F}}_{\pm,\mathbf{k}} = \pm c\,\mathbf{k} \wedge \tilde{\mathbf{F}}_{\pm,\mathbf{k}}$$

or, in matrix notation $\mathrm{d}_t \tilde{\mathbf{F}}_{\pm,\mathbf{k}} = \pm c\,\mathbf{K}(\mathbf{k}) \tilde{\mathbf{F}}_{\pm,\mathbf{k}}$ where $\mathbf{K}(\mathbf{k})$ is the $3\times3$ skew-Hermitian matrix corresponding to $\mathbf{k} \wedge$, i.e. the "infinitesimal" rotation in the Lie algebra $\mathfrak{so}(3)$ and the basic solutions are $\tilde{\mathbf{F}}_{\pm,\mathbf{k}} = \exp(c \mathbf{K}(\mathbf{k}) t)\tilde{\mathbf{F}}_{\pm,\mathbf{k}}(0)$, ie vectors spinning at uniform angular speed $\omega = c k$. There is a wonderful aside to the Riemann-Silberstein notation which I come back to at the end of my answer.

If you google Iwo Bialynicki-Birula and his work on the photon wave function, he has heaps more to say about such things. His personal website is http://cft.edu.pl/~birula and all his publications are downloadable therefrom. The particular scaling of the Riemann-Silberstein vectors above is Bialynicki-Birula's, and it means that $|\mathbf{F}_+|^2 + |\mathbf{F}_-|^2$ is the electromagnetic energy density. He defines the pair $(\mathbf{F}_+, \mathbf{F}_-)$, normalised so that $|\mathbf{F}_+|^2 + |\mathbf{F}_-|^2$ becomes a probability density to absorb the photon at a particular point, to be a first quantized photon wave function (without a position observable). There is special, nonlocal inner product to define the Hilbert space and in such a formalism the general Hamiltonian observable is $\hbar\, c\, \mathrm{diag}\left(\nabla\wedge, -\nabla\wedge\right)$. Please also see Arnold Neumaier's pithy summary (here) of a key result in section 7 of Bialynicki-Birula's "Photon wave function" in Progress in Optics 36 V (1996), pp. 245-294 also downloadable from arXiv:quant-ph/0508202. The Hilbert space of Riemann Silberstein vector pairs that Bialynicki-Birula defines is acted on by an irreducible unitary representation, defined by Bialynicki-Birula's observables $\hat{H}$, $\hat{\mathbf{P}}$, $\hat{\mathbf{K}}$ and $\hat{\mathbf{J}}$, of the full Poincaré group presented in the paper. So the two subspaces containing wholly right ($\mathbf{F}_-=\mathbf{0}$) and wholly left-polarized ($\mathbf{F}_+=\mathbf{0}$) states are the "particles" of the theory: you don't get the same thing with other nontrivial linear combinations of light base states (which are not eigenfunctions of the angular momentum observable).


The Classical Angular Momentum Calculation

Now we look at the classical angular momentum. The Wikipedia page on angular momentum of light gives the classical angular momentum as:

$$\frac{\epsilon_0}{2i\omega}\int \left(\mathbf{E}^\ast\wedge\mathbf{E}\right)d^{3}\mathbf{r} +\frac{\epsilon_0}{2i\omega}\sum_{i=x,y,z}\int \left({E^i}^{\ast}\left(\mathbf{r}\wedge\mathbf{\nabla}\right)E^{i}\right)d^{3}\mathbf{r}$$

when the positive frequency part alone of the fields is kept (hence the complex conjugates). The first term is the spin angular momentum, and, rewritten in positive frequency Riemann-Silberstein vectors when everything is roughly paraxial (i.e. near to a plane wave) it reads:

$$\hat{\mathbf{z}}\frac{1}{\omega}\int \left(|\mathbf{F}_+|^2-|\mathbf{F}_-|^2\right)d^{3}\mathbf{r}$$

i.e. $\frac{1}{\omega}$ times the right polarized energy density less the left polarized energy density in the direction of propagation of the light. Orbital angular momentum vanishes in the paraxial limit and so the last equation is the total angular momentum in this case. It is important to recall how this equation is derived: one imagines an electromagnetic field crossing the boundary into a conductive medium and being absorbed there, then one calculates the angular impulse exerted on the medium, exactly analogously with method 3 of the momentum calculation in my answer http://physics.stackexchange.com/a/72688/26076 . The point is that the angular momentum density $\left(|\mathbf{F}_+|^2-|\mathbf{F}_-|^2\right)/\omega$ calculated from this most basic (in the sense of fundamental) Newtonian-Maxwell physics is the difference between the intensities of the circularly polarized base states, not the linear ones. So again Nature shows her preference. This calculation says that the right and left circularly polarized components transfer angular momentum $\pm E/\omega$ in the direction of light propagation, respectively, whenever energy $E$ is absorbed. So now we see that, if the photon has energy $h\nu$, then if a high number of them are to transfer the same angular momentum as classical physics reckons, the photon's angular momentum has to be $\pm h\nu/\omega$ or $\pm \hbar$ in the direction of its propagation for right and left circularly polarized photons, respectively. I'll get to other polarization states in a moment.

General Photon Emission States and Conservation of Angular Momentum by Fluorophores

You ask about one photon emission and whether it is always circular. Absolutely NOT. General one photon states are pure quantum superpositions of circularly polarized one-photon number states. Suppose we have a linearly polarized photon and we work out the results of imparting the number observables: let $a_\pm^\dagger$ be the creation operators for right and left handed polarized states. The linear polarized states are:

$$\psi_x = \frac{1}{\sqrt{2}} \left(\psi_++\psi_-\right)$$ $$\psi_y = \frac{-i}{\sqrt{2}} \left(\psi_+-\psi_-\right)$$

where $\psi_\pm$ are the pure one photon right and left handed polarization states. Then the right and left handed number operators return:

$$n_\pm(\psi_x) = n_\pm(\psi_y) = \left<\psi|a_\pm^\dagger a_\pm | \psi\right> =\frac{1}{2}$$

and the total photon number operator returns:

$$n = \left<\psi|a_+^\dagger a_+ + a_-^\dagger a_-| \psi\right> =1$$

As you likely know, we can always change our one photon Fock state basis by any unitary transformation and the creation and annihilation operators transform likewise. The linear photon creation operators, for example, are $a_x^\dagger=\frac{1}{\sqrt{2}}\left(a_+^\dagger + a_-^\dagger\right)$ and $a_y^\dagger=\frac{-i}{\sqrt{2}}\left(a_+^\dagger - a_-^\dagger\right)$ and imparting the number operator formed from these and their respective Hermitian conjugates would return the result "1 photon" when applied to the respective polarization states and "0" when applied to the orthogonal linear polarization states. The general pure one photon state is of the form:

$$\psi(\alpha, \phi) = \alpha e^{i \frac{\phi}{2}} \psi_+ + \sqrt{1-\alpha^2} e^{-i \frac{\phi}{2}} \psi_-$$

where $\alpha \in [0, 1]$ and $\phi \in [0, 2\,\pi)$ and, from our classical calculation above, we know that its angular momentum must be:

$$\hbar \left<\psi|a_+^\dagger a_+ - a_-^\dagger a_- | \psi\right> = (2\,\alpha^2-1)\,\hbar$$

One photon emissions are always the right pure quantum superpositions that make sure angular momentum is conserved. For example, if a fluorophore absorbs linearly polarized light, and if it does not exert torque on its surroundings before spontaneous emission, the emission must be linearly polarized. Likewise for any other general polarization state absorbed by fluorophores. See the answer http://physics.stackexchange.com/a/73439/26076 for more details, and you also might like to look at the work of Gregorio Weber in the 1950s on the polarization of fluorescence. See see G. Weber, “Rotational Brownian motion and polarization of the fluorescence of solutions,” Adv. Protein Chem. 8, 415–459 (1953) and other works.

My own work in this field is summarised in J. Opt. Soc. Am. B, Vol. 24, No. 6/June 2007 p1369.

The general criterion, in my $\psi(\alpha,\phi)$ notation, where the base states are circularly polarized ones along the direction of propagation, is that if $\alpha_{in},\phi_{in}$ characterize the photon absorbed by an isolated (not transferring angular impulse) fluorophore, and $\alpha_{out},\phi_{out}$ the fluoresced photon, then $\alpha_{in} = \alpha_{out}$ and the phase angles are unrelated. Likewise, when a particle and antiparticle with opposite spins annihilate each other and only two photons are emitted, each characterized by $\alpha_{1},\phi_{1}$ and $\alpha_{2},\phi_{2}$, then $\alpha_{2} = \sqrt{1- \alpha_{1}^2}$ and the phase angles are unrelated and I'm guessing that all values of $\alpha_{1}^2$, $\phi_1$ and $\phi_2$ are equally probable


Torque Exerted on Birefringent Crystals

Now we get to birefringent crystals. It is obvious, but I believe important for this answer, to take heed that birefringent crystals are a way wherein Nature very explicitly tells the difference between linear and circular polarization. It is the linear, NOT the circularly polarized states that are the "eigenmodes" of a birefringent crystal, i.e. linearly polarized fields aligned with the fast and slow axes of the crystal are simply phase delayed and undergo no mixing. Circular polarized states are NOT eigenmodes: they mix in such crystals. The mixing induced by a quarter wave plate - in particular that which happens when the input field is linearly polarized and aligned at 45 degrees to the fast and slow axes therefore imparts a torque on the crystal: it most certainly should be possible to measure that torque and check it against classical calculations: from our calculations above, in this situation there will be a torque $P/\omega$, when the light's power is $P$. A thought experiment: a 100W, 1mm diameter linearly polarized collimated beam passes through a quarter wave plate suspended in a fluid. With infrared light at $193\mathrm{THz}$ (you can get fibre lasers at $193\mathrm{THz}$ outputting hundreds of watts) the torque will be of the order of $10^{-13}\mathrm{Nm}$, if the crystal is a millimeter in diameter and 3 millimeters or so long with a density of $3000\mathrm{kg\,m^{-3}}$, its mass moment of inertia is of the order of $4\times10^{-13}\mathrm{kg \, m^2}$, so this will be an accurately measurable effect (indeed, as the crystal spins and misaligns itself from the 45 degree position, its angular position will fulfill $\mathrm{d}_t^2 \theta = - \frac{1}{2}\Omega^2 \sin(2\theta)$ and we will have a torsional pendulum oscillating at $\Omega /(2\pi) = 0.1\mathrm{Hz}$!).

If there are experiments to observe the transfer of $\hbar$ angular momentum by one photon, then it might be to do with laser tweezer experiments: circularly polarized beams are used to spin things under the microscope in the laser trap and I believe Prof Halina Rubinsztein-Dunlop (http://physics.uq.edu.au/people/halina) was interested some years ago in what happens with such things at very low light levels - I got the impression she had an interest in directly observing one $\hbar$ of angular momentum transfer. She may know of any experiment along these lines.


Aside: More on the Riemann Silberstein Notation

I think you will like this one, Terry. The Riemann-Silberstein vectors are actually the electromagnetic (Maxwell) tensor $F^{\mu\nu}$ in disguise. We can write Maxwell's equations in a quaternion form:

$$\left(c^{-1}\partial_t + \sigma_1 \partial_x + \sigma_2 \partial_y + \sigma_3 \partial_z\right) \,\mathbf{F}_+ = {\bf 0}$$

$$\left(c^{-1}\partial_t - \sigma_1 \partial_x - \sigma_2 \partial_y - \sigma_3 \partial_z\right) \,\mathbf{F}_- = {\bf 0}$$

where $\sigma_j$ are the Pauli spin matrices and the electromagnetic field components are:

$$\begin{array}{lcl}\frac{1}{\sqrt{\epsilon}}\mathbf{F}_\pm &=& \left(\begin{array}{cc}E_z & E_x - i E_y\\E_x + i E_y & -E_z\end{array}\right) \pm i \,c\,\left(\begin{array}{cc}B_z & B_x - i B_y\\B_x + i B_y & -B_z\end{array}\right)\\ & =& E_x \sigma_1 + E_y \sigma_2+E_z\sigma_3 + i\,c\,\left(B_x \sigma_1 + B_y \sigma_2+B_z\sigma_3\right)\end{array}$$

As you know, these the Pauli spin matrices are the imaginary quaternion units reordered. When inertial reference frames are shifted by a proper Lorentz transformation:

$$L = \exp\left(\frac{1}{2}W\right)$$

where:

$$W = \left(\eta^1 + i\theta \chi^1\right) \sigma_1 + \left(\eta^2 + i\theta \chi^2\right) \sigma_2 + \left(\eta^3 + i\theta \chi^3\right) \sigma_3$$

encodes the transformation's rotation angle $\theta$, the direction cosines of $\chi^j$ of its rotation axes and its rapidities $\eta^j$, the entities $\mathbf{F}_\pm$ undergo the spinor map:

$${\bf F} \mapsto L {\bf F} L^\dagger$$

Here, we're actually dealing with the double cover $PSL(2,\mathbb{C})$ of the identity-connected component of the Lorentz group $SO(3,1)$, so we have spinor maps representing Lorentz transformations, just as we must use spinor maps to make a quaternion impart its rotation on a vector. The electromagnetic field is thus a bivector in the Clifford algebra $C\ell_3(\mathbb{R})$, and one neat thing about this notation is that polarization and spin are manifestly Lorentz covariant given the circular polarization base state interpretation of $\mathbf{F}_\pm$ afforded by the discussion above (note that the quaternion units in the de Rham derivatives also transform by the same spinor maps), something that is not so clear in other forms of Maxwell equations. Also, you can convert Bialynicki-Birula's ten observables $\hat{H}$, $\hat{\mathbf{P}}$, $\hat{\mathbf{K}}$ and $\hat{\mathbf{J}}$ also have very clean quaternionic forms.

share|improve this answer
    
You might find some of the modern treatments of Clifford algebra in physics (or "geometric algebra") interesting, given that you are already so familiar with the basic ideas (as evidenced by this answer). In particular, let $e_t, e_x, e_y, e_z$ be basis vectors. Let $\nabla = -e_t c^{-1} \partial_t + e_x \partial_x + e_y \partial_y + e_z \partial_z$, and your quaternion equations take the form of $\nabla (e_t F_+) = 0$ and $e_t \nabla F_-= 0$. Electromagnetism using the spacetime algebra is stunningly clear. –  Muphrid Aug 16 '13 at 3:22
    
Thanks, @Muphrid This is indeed booked on my intellectual journey! I began three years ago writing an exposition on Lie theory going back to something like Lie's original conceptions without the need for a manifold (incidentally important to the solution of Hilbert's fifth problem) that builds everything from the notion of $C^1$-paths. You very naturally get to the modern idea of a manifold (although not quite general, as the Lie group has an Abelian fundamental group) since $C^1\Rightarrow C^\omega$ (the much harder $C^0\Rightarrow C^\omega$ is Hilbert's fifth problem) from these ideas .... –  WetSavannaAnimal aka Rod Vance Aug 16 '13 at 3:40
    
... Then one can generalize from the intuition for the manifold of a Lie group to general differential geometry and GR, then .... (that's where I am now!). Since I am 50 now, how far do you reckon I'm going to get before I am worm-food? BTW I am reading through Doran and Lasenby at the moment - do you recommend any others? –  WetSavannaAnimal aka Rod Vance Aug 16 '13 at 3:41
    
+1, very informative answer. This is related to one of my old questions: physics.stackexchange.com/questions/28616/… –  Jia Yiyang Aug 16 '13 at 4:15
    
Well, 3 years ago I was only just beginning to learn about geometric algebra, so a good bit can happen in the future. For GA books, I still consider Hestenes and Sobczyk (Clifford Algebra to Geometric Calculus) my personal holy grail to understand completely. There's stuff they have there on eigenblades of linear operators that I still want to grasp fully, as well as their development of vector manifolds. I also like Alan MacDonald's books to make more contact with nitty-gritty linear algebra stuff (like SVD) in the GA language. –  Muphrid Aug 16 '13 at 4:17
show 6 more comments

The suggested experiment was performed by R. Beth already in 1936. In the experiment linearly polarized light was converted to circularly polarized light by means of a doubly refracting plate. The (macroscopic) reaction torque was measured and shown to conform with the angular momentum theory of the photon.

share|improve this answer
    
Thanks David. It's nice to know the right history: I should have looked for experiments myself, but there are so many experimentalists out there with brains the size of Jupiter and indeed bigger that it was just utterly unbelievable to me that the experiment had not been done and properly tested to high precision! –  WetSavannaAnimal aka Rod Vance Aug 15 '13 at 21:55
    
David Bar Moshe, very cool, thanks! This really soothes my mind, as I'd always assumed this kind of experimentally verifiable conveyance of angular momentum was "of course" true, and managed to wrap myself into a philosophical loop about it. –  Terry Bollinger Aug 20 '13 at 0:46
    
Also, a comment: This whole issue reminds me, perhaps tangentially?, of the curious ambiguity of angular momentum in electromagnetic fields. What ambiguity? Well, it's not a big deal, and perhaps I'm just over-thinking it again, but it's just this: When a material object conveys angular momentum, there is no ambiguity about sign. So in space, catching a ball that is spinning clockwise will impart clockwise spin to you. But since a photon is pure electromagnetism, I suspect if the apparatus measuring reaction torque was made of antimatter, the angular momentum sign of the light would reverse. –  Terry Bollinger Aug 20 '13 at 0:55
add comment

You actually answered your own question. The photon carries no mass. Angular momentum in the quantum world is a bit of a misnomer. They found this property of quantum objects, and the math worked out just like the math for classical spinning objects. Guess what they decided to call it? Anyways, without mass you won't see the same kind of angular momentum you see in massive particles.

share|improve this answer
1  
Cassius, thanks, but alas, it's not quite that simple. Quantized angular momentum -- also called spin -- is still very much angular momentum. And even if it was not, it still is the same quantity found in massive particles such as electrons and silver atoms, since spin must balance out in particle equation involving both massive and massless particles. And don't forget: Spin is just as weird and non-classical for electrons -- maybe more so! -- than it is for photons, since you need both mass and size to create conventional angular momentum. –  Terry Bollinger Aug 15 '13 at 0:52
    
The angular momentum quantities that you are thinking of are the total angular momentum (J) and the orbital angular momentum (L). The spin is a purely quantum property. The mathematics of spin are exactly the same as classical angular momentum, but it is NOT the same physical quantity. Spin is intrinsic to the particle you are discussing and does not arise from any motion or momentum of the particle. –  cassius Aug 22 '13 at 11:49
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.