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Angular and magnetic momentum 
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#37
Apr614, 03:20 AM

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So, is the total angular momentum of 1H ([itex]\hbar[/itex]+1/2+1/2) =2 [itex]\hbar[/itex] =h/[itex]\pi[/itex]? 


#38
Apr614, 05:46 AM

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No. You should be able to look that stuff up ;)
Note: electrons in an atom have orbital angular momentum as well as spin. 


#39
Apr714, 01:00 AM

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According to what you said I considered a vector at the nucleus : L_{o} = [itex]\hbar[/itex] + L_{p} (you said) =1/2 [itex]\hbar[/itex] =3/2 [itex]\hbar[/itex] Ant the vector of the spin at the electron L_{e} = 1/2 [itex]\hbar[/itex] So the total momentum of the ^{1}H atom should be 3/2+1/2 = 2 [itex]\hbar[/itex]= h/[itex]\pi[/itex] I have read that the the spin being 1/2 implies the electron behaving like on a Moebius strip, is that true? why so? if it where ,say, 32 couldn't it run on a Moebius strip? 


#40
Apr714, 02:00 AM

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Sounds like a reference to Dirac's Plate Trick http://en.wikipedia.org/wiki/Plate_trick Also see: http://scienceblogs.com/principles/2...fortoddlers/ ... since it is still bothering you. Hydrogen: http://en.wikipedia.org/wiki/Hydroge...gular_momentum ... when you are thinking of the angular momentum of an atom, you have to consider what state it is in. Usually you want to do this for the ground state. http://www.physicsforums.com/showthread.php?t=69992 Beware of articles which go out of their way to make QM sound weird and mysterious. That would include most popscience and about 99% of the material found on the Discovery Channel. 


#41
Apr1214, 08:20 AM

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What I meant is: 1/2 [itex]\hbar[/itex] it's just a numerical value, it could be 2π [itex]\hbar[/itex] or 32 or anything. Why does this particular value determine the fact that it rotates twice before pointing in the same direction? or I misinterpreted the article? how do we notice that it rotates by 720°?  I read that in ground state 1s there is no angular momentum, this being one of the main differences from Bohr's model, is this true?  if so, as there is no pseudovector L(o) and no plane of an orbit xy, how can we detect the spin component on z? if there is no xy where is z?  I do not think QM is weird, but it seems to have rules that cannot be verified and contradict the rules outside the atom. It's difficult to know where it is just theory. Thanks. 


#42
Apr1214, 09:54 PM

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The value of angular momentum is not arbitrary if that's what you mean.
It is determined empirically and there is a deeper mathematical model that makes sense of it. Right now you are exploring the results of that model. The value assigned to mean a particular symmetry is arbitrary though, and there are several schemes. So you could be asking "why use that particular scheme?" Consider: It would be natural and intuitive, for many people, to classify rotational symmetry by the angle A you have to rotate through for the object to map on to itself. It's not nice to use this method in practice (give it a go) because you really want more symmetry to mean a bigger number and you want to avoid fractions if you can help it. But to see how that may work we have to be careful about what we mean when we talk about angles. The angle is the pointy bit in a corner  the size of the angle would be the inverse of it's sharpness. The less pointy the corner, the bigger the angle. The size of the angle is most conveniently defined on the unit circle as follows  the length of the circumference of the unit circle that lies inside the angle is called "the size of the angle". But that begs the question of the definition of the unit circle. A unit circle may be defined with the radius, diameter, or circumference equal to 1. Pick one. The choice is arbitrary  so long as you are consistent. Having decided on how to measure angles, we can talk about rotational symmetry. In general, if the object must be rotated through angle A to map onto itself, then we can define a rotational symmetry value is: ##S=1/A## (circumference = 1) ##S=\pi/A## (diameter = 1, unused) ##S=2\pi/A## (radius =1: this is "radians") ##S=360/A## (using degrees: 1deg = 1/360th of the unit circumference.) ... this is a useful way to define a symmetry number, because it gets bigger the more symmetry you have and, for all common experience, it will always be bigger than one (no fractions). Notice that S, defined this way, will usually be an integer? (The S value ends up being the same as the number of times the object maps onto itself when you rotate it once. i.e. a pentagon would have S=5.) In a purely abstract way, though, you can ask yourself what happens for noninteger values of S. i.e. what is S=8/3? That's bigger than 1 and not an integer. What is S were irrational? But that's just pure maths  there was no reason, before QM experiments, to suppose that noninteger symmetries would have any meaning in Nature. Physics uses Maths as a language for describing Nature. The formal language involves going through this process of making definitions and proposing theorems. Maths is a powerful language capable of describing things that do not exist in Nature too. We use the principles of empirical inquiry to check. 


#43
Apr1314, 12:13 AM

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http://gabrielse.physics.harvard.edu...ticMoment.html . Can you give me a link or explain how L_{e} is concretely determined? How can you detect and determine mechanical angular momentum at distance, without manipulating a body? 


#44
Apr1314, 12:55 AM

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That is pretty much the definitive experiment: how much more "concrete" do you need?



#45
Apr1314, 01:54 AM

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L_{e} is = 1* h/2*2π (= [itex]\hbar/2[/itex]), whereas [itex]\mu[/itex]_{e} is 1.001156 * h/2*2 π, which value is right? 


#47
Apr1314, 03:39 AM

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Can we determine concretely the magnetic moment of a bigger object? Suppose we have a coil where current is flowing, we have a formula to find the value of B and [itex]\mu[/itex], but is it possible to verify the real value of [itex]\mu[/itex]? what apparatus do we need? a magnetic field of any strength would do or we need a critical value? any direction will do? do we need more than a measurement? Thanks again for your patience, Simon 


#48
Apr1314, 09:00 AM

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If you have a charge with an angular momentum you must also have a magnetic moment.
They go together. The other demonstration for angular momentum is a torque experiment  where a spinpolarized beam is incident on a target, the target starts to rotate. You seem to be going in circles now. I think you need to have more time to go through these materials. Intrinsic angular momentum is an established property in physics. 


#49
Apr1414, 01:11 AM

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I'd appreciate if you could answer my previous question (is this the usual method? :http://www.serviciencia.es/notapli/NAS01i.pdf) suppose we make that coil spin around the axis of [itex]\mu[/itex], then that coil will have a magnetic moment [itex]\mu[/itex] = k * [itex]\mu[/itex]_{B} and a mechanichal angular momentum L (=mvr) = j * [itex]\hbar[/itex], just like an electron in a ^{1}H atom, am I right so far? Now, if we measure [itex]\mu[/itex] when the coil is rotating, I assume we get a different value of [itex]\mu[/itex], surely greater, since the applied field must win the resistance to torque also offered by L.  is that right? naively I assume that, now, the value of [itex]\mu[/itex] must be k+j, or (for some obscure reason) k*j ? I'd appreciate very much if you could tell me if anything is wrong there. That would save me a lot of more stupid or circular questions. If my assumptions are right, it will be clear to you the fact that I do not understand why:  if intrinsic angular momentum of the e on the zaxis is [itex]\hbar[/itex]  and [itex]\mu[/itex] is 1.001156[itex]\hbar[/itex] when we put a ^{1}H atom in a SternGerlach machine or a single electron in a Penning trap we measure in the first case only L and in the second only [itex]\mu[/itex] (ì 1.1156 L) Why not always both, and why angular momentum in SternGerlach? Do you follow me, Simon? probably I should start a new thread on this. Thanks for your patience!!! 


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