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Hoof47
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I am interested in quantum mechanics, but my maths isn't up to understanding some of the concepts. I've tried to look at spin, but all explanations are mathematical. Is it possible for it to be explained in words?
cristo said:Spin is a type of "intrinsic" angular momentum-- i.e. a type of angular momentum that a particle has, regardless of its motion in space. marlon has posted in his blog an https://www.physicsforums.com/blogs/marlon-13790/what-is-spin-152/ to this, which you may wish to look at.
A common misunderstanding is for one to interpret a paricle's spin as the "rotation about its own axis." However, spin is a purely quantum mechanical property, and it therefore makes no sense to interpret it in any classical sense.
Particles with spin *do* have angular momentum and magnetic momentum.Cane_Toad said:Are all the terms associated with spin all analogous? E.g. "intrinsic angular momentum"; a particle has neither angular momentum or "spin", right? Are all these just naming conventions?
I'm assuming the term "spin", etc., is due to the Stern-Gerlach experiment, where a particle's "spin" is measured in a way similar to what could be done for classical spinning object. True?
Since particles do deflect in a magnetic field as if they had angular momentum, but they don't actually have any, what is going on?
lightarrow said:Particles with spin *do* have angular momentum and magnetic momentum.
But they don't "spin".
If you wonder how could this be, you're not alone.
jostpuur said:...But you can have internal angular momentum that is not result of any kind of rotation also in classical context.
Sojourner01 said:...
So the answer to the question is, spin is spin. It doesn't need to go any deeper than that.
Cane_Toad said:Please expound.
But it is angular momentum, and it contributes to the total macroscopic angular momentum of an object, as observed in the Einstein - de Haas effect.
the Einstein-de Haas effect demonstrates that spin angular momentum is indeed of the same nature as the angular momentum of rotating bodies as conceived in classical mechanics. This is remarkable, since electron spin, being quantized, cannot be described within the framework of classical mechanics.
jostpuur said:This is where I asked about the matter:
https://www.physicsforums.com/showthread.php?t=160778
My understanding of the Dirac's field has been rather "changing" lately, as result of discussions here in PF, and the OP of my thread probably isn't relevant for this matter anymore, but pay attention to the later comments on the classical electromagnetic field too.
Photons are massless and yet have linear momentum...Cane_Toad said:But, come on! It's bad enough that an electron has angular momentum with rotation, but now EM fields have angular momentum without mass!?
Cane_Toad said:But, come on! It's bad enough that an electron has angular momentum with rotation, but now EM fields have angular momentum without mass!?
But, come on! It's bad enough that an electron has angular momentum with rotation, but now EM fields have angular momentum without mass!?
jtbell said:Well, EM fields carry energy and linear momentum, so they pretty much have to have angular momentum too, based on the definition of angular momentum as [itex]\vec r \times \vec p[/itex].
Photons are traditionally said to be massless. This is a figure of speech that physicists use to describe something about how a photon's particle-like properties are described by the language of special relativity.
...
Because the energy of a particle just equals its relativistic mass times c2, physicists have learned to economise the language by only ever referring to a particle's energy. When they use the term "mass", they mean rest mass. This is purely a linguistic convention.
That is an experimentally derived upper limit, from observations. The theoretical mass of the photon is defined as zero. Period.Cane_Toad said:It does go on to tell about a theoretical upper limit for the rest mass of light between 7 × 10-17 eV and 3 × 10-27 eV. I'm not sure why they bother to state the rest mass at zero, and then go on to give the estimate.
Cane_Toad said:Ok, so you're talking about the internal angular momentum of an EM field, so at least there's not some wacky compound I didn't know about.
But, come on! It's bad enough that an electron has angular momentum with rotation, but now EM fields have angular momentum without mass!?
What the hell, I guess QM is re-defining all my CM terms according to its own fancy. Sigh, at least some of the QM fathers had the sense to use non-conflicting terms, like "flavor".
You can define angular momentum without the need to have a rotating mass: think about light, as jtbell wrote. What is striking in that case is that *statics* electric and magnetic fields can have angular momentum: you only need a circulation of the Poyinting vector.Cane_Toad said:But...it isn't *real* momemtum, is it (i.e. no rotation -> no velocity)? I'm wondering if it's just a model to account for deflection in a mag field?
(And you meant magnetic moment?)
Example: if you shoot a very strong light pulse in a specific direction, you receive a recoil in the opposite direction. If you want to hold true the law of momentum conservation (higly advisable!) you are forced to conclude that light too carries momentum. That is a more general definition than p = m*v (for non zero rest mass and not relativistically) or p = m*v*gamma (for non zero rest mass and relativistically).Cane_Toad said:Ok, so you're talking about the internal angular momentum of an EM field, so at least there's not some wacky compound I didn't know about.
But, come on! It's bad enough that an electron has angular momentum with rotation, but now EM fields have angular momentum without mass!?
What the hell, I guess QM is re-defining all my CM terms according to its own fancy. Sigh, at least some of the QM fathers had the sense to use non-conflicting terms, like "flavor".
Please explain this (that seems as a nonsense to me).jostpuur said:This is getting slightly off topic, but I could make a remark, that electric and magnetic fields can have rest mass too. The plane waves which travel at the speed of light are massless, but static electric fields around some charge distributions are not.
lightarrow said:Please explain this (that seems as a nonsense to me).
Cane_Toad said:It does go on to tell about a theoretical upper limit for the rest mass of light between 7 × 10-17 eV and 3 × 10-27 eV. I'm not sure why they bother to state the rest mass at zero, and then go on to give the estimate.
If someone will tell me definitively that there is no explanation for how a particle is deflected in the Stern-Gerlach thing other than via a set of functions which have familiar sounds like "spin" and "angular momemtum", then I can be content, or at least as content as I am with duality (not!).
But to evaluate the electric field's energy and compare it with the energy you give to the system to establish if the mass is in the field or in the particles, you should (also) know the electron's shape and dimensions. Do you know them?jostpuur said:Just [tex]E^2=p^2c^2+(mc^2)^2[/tex]. For a plane wave we have [tex]E=pc[/tex] and [tex]m=0[/tex], but for a static field instead [tex]p=0[/tex] and [tex]E=mc^2[/tex]. Suppose you have electron and proton close to each other, and you then drag them far away from each other. You need energy to do this, and after the procedure the mass of the system has increased by the energy you inserted there. The energy is in the electric field, and that is also where the mass increase is.
I think I heard somewhere, that this has not been verified experimentally with electric interactions yet, but only with nuclear interactions. You know, in nuclear physics the mass excess stuff is plain clear. Anyway, it should be true also for electric interactions.
lightarrow said:But to evaluate the electric field's energy and compare it with the energy you give to the system to establish if the mass is in the field or in the particles, you should (also) know the electron's shape and dimensions. Do you know them?
Ok, nice work; however I don't understand why the other term of the integration by parts vanishes. Apart the constant terms, you have to evaluatejostpuur said:...
makes the interaction energy to be
[tex]
\int\frac{d^3x\; q_1q_2}{16\pi^2\epsilon_0} \frac{x-x_1}{|x-x_1|^3}\cdot \nabla\frac{-1}{|x-x_2|}
[/tex]
The nabla can be moved onto the first factor using integration by parts, and this gives
[tex]
\int \frac{d^3x\;q_1q_2}{16\pi^2\epsilon_0}\Big( \nabla\cdot\frac{x-x_1}{|x-x_1|^3}\Big)\frac{1}{|x-x_2|}
[/tex]
Now using an delta function representation
[tex]
\nabla\cdot\frac{x-x_1}{|x-x_1|^3} = 4\pi\delta^3(x-x_1)
[/tex]
the interaction energy becomes
[tex]
\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{|x_1-x_2|}
[/tex]
which is just the Coulomb's potential. So when you speak about the potential energy given by the Coulomb's potential formula, you are speaking about the energy in the electric field. And when the mass of the system is changing as result of changing potential energy, it is the field that is carrying the changing mass.
jostpuur said:I was afraid you were going to ask something about the delta function, because I don't know it very well. After substituting the delta function representation the energy is
[tex]
\int \frac{d^3x\; q_1q_2}{4\pi \epsilon_0}\delta^3(x-x_1) \frac{1}{|x-x_2|}
[/tex]
and the delta function works so that the integration can be carried out by substituting [itex]x_1[/itex] to there where [itex]x[/itex] is elsewhere. You know this stuff:
[tex]
\int\limits_{-\infty}^{\infty} dx\; \delta(x-x') f(x) = f(x')?
[/tex]
I have seen other calculations, where similar techniques were used rigorously, and there's a change that I could maybe understand what happens here too, but you should first be able to understand the hand waving calculation so that you see what is supposed to happen.
jostpuur said:This is getting slightly off topic, but I could make a remark, that electric and magnetic fields can have rest mass too. ...
Hoof47 said:I am interested in quantum mechanics, but my maths isn't up to understanding some of the concepts. I've tried to look at spin, but all explanations are mathematical. Is it possible for it to be explained in words?
jostpuur said:We can think that the integration domain has outer boundary that is a surface of some large ball B(0,R), and after integration we are taking the limit [itex]R\to\infty[/itex].
The asymptotic behavior of the integrand is
[tex]
\Big|\frac{x-x_1}{|x-x_1|^3}\frac{1}{|x-x_2|}\Big|\approx \frac{1}{R^3}
[/tex]
on the surface, when the ball is so big that we can assume the points [itex]x_1[/itex] and [itex]x_2[/itex] to be close to the origo. Since the area of the surface of the ball approaches infinity like [itex]R^2[/itex], the surface integral (that we get when the volume integral is made into a surface integral according to the Gauss's theorem) is going to give something like [itex]1/R[/itex], and it will vanish on the limit [itex]R\to\infty[/itex].
lightarrow said:Suppose you don't have two point charges q1 and q2 but two spherical conductive charges of non-zero radius r.
The boundary is:
1.an infinitely far spherical surface
2. the suface of the first conductive sphere
3. the suface of the second conductive sphere
Now, make r go to zero...