Quantum Physics

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Offline Mr Andrew

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« on: 17/07/2007 21:28:21 »
Why the need for quantum phyiscs?  I know it all started with Plank's constant and the quantization of energy for electromagnetic waves, but why did he come up with it?  Why would he, one day, decided that energy can only have values equal to integral multiples of 'h'?  If anybody knows the answer to this, please, feel free to explain...this whole subject is confusing me, not the ideas or principles it involves but why the ideas and principles are necessary.  I have been unable to find this anywhere else.  Hopefully someone here knows the answer.
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Offline DoctorBeaver

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« Reply #1 on: 17/07/2007 21:56:01 »
There was a severe problem in classical physics with black body radiation. Any object with a higher temperature than its surroundings will lose heat by radiation. The hotter the object, the more it will radiate. A black body absorbs all frequencies, so it should radiate equally at all frequencies. But that does not happen. Black bodies emit more radiation at some wavelengths than at others. Planck decided to tackle this problem. He solved it by proposing that heat radiation is emitted only in specific amounts, which he called quanta.
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Offline maff

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« Reply #2 on: 17/07/2007 22:29:05 »
There was a severe problem in classical physics with black body radiation. Any object with a higher temperature than its surroundings will lose heat by radiation. The hotter the object, the more it will radiate. A black body absorbs all frequencies, so it should radiate equally at all frequencies. But that does not happen. Black bodies emit more radiation at some wavelengths than at others. Planck decided to tackle this problem. He solved it by proposing that heat radiation is emitted only in specific amounts, which he called quanta.
You cannot possibly comment on how black bodies radiate until you know the complete picture of the mass contained within them ie the number of free Electrons.
..maff

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Offline DoctorBeaver

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« Reply #3 on: 17/07/2007 23:17:49 »
I suggest you take it up with Warwick university then as their website says the following:-

A black body is a hypothetical object that is able to emit and absorb electromagnetic radiation of all possible wavelengths.

The observed frequency spectrum (variation of intensity of radiation versus frequency) was different from that predicted by the theory.

The theory predicted that the spectrum should be the same for all black bodies.
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« Reply #4 on: 17/07/2007 23:29:40 »
Why the need for quantum phyiscs?

There are very practical devices that could not have been developed or explained using classical physics (SQUIDs and tunnelling electron microscopes are some of the most obviously associated with quantum physics, but it helps explain many other devices and observed phenomena that could not be explained by classical physics).

Could there be a different non-classical (or maybe neo-classical) model of physics that could explain these phenomena as well?  Possibly, but nobody has found it yet.

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Offline maff

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« Reply #5 on: 17/07/2007 23:46:59 »
I suggest you take it up with Warwick university then as their website says the following:-

A black body is a hypothetical object that is able to emit and absorb electromagnetic radiation of all possible wavelengths.

The observed frequency spectrum (variation of intensity of radiation versus frequency) was different from that predicted by the theory.

The theory predicted that the spectrum should be the same for all black bodies.

A black body is an unknown mass of unknown density. It's density's contain unknown quantities of Electrons and other Photon absorbing particles. To try and determine what it's electromagnetic responses are is an act of utter madness. Every black body can have a different response and probably does. To estimate the electromagnetic response of a black body is the same as guessing it's mass.
Warwick guesses a lot.
..maff

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Offline Soul Surfer

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« Reply #6 on: 17/07/2007 23:48:41 »
Mr Andrew   To answer your question in a little more detail.  

Around the turn of the twentieth century Maxwells laws describing how electromagnetic waves are radiated by moving electrical charges was known.  Some understanding also existed about the dynamic electromagnetic structure of matter.  Ie that atoms consisted of positive and negative charges that were in continuous motion.  The electromagnetic radiation energy curve of a black body (ideal radiator) at different frequencies and temperartures had been observed practically.  Rayliegh had got a good model for the low frequency end of this spectrum by considering the motions of the electrical charges in atoms but it had a big problem it went off to very large radiation energies the higher the frequency got.   It was clear that the model was wrong if the radiation energies of higher frequencies could be arbitrarily small.  Planck realised that if for some reason that high frequency energy could only be radiated in relatively large chunks (quanta)  The problem could be solved.
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Offline DoctorBeaver

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« Reply #7 on: 18/07/2007 01:25:39 »
I suggest you take it up with Warwick university then as their website says the following:-

A black body is a hypothetical object that is able to emit and absorb electromagnetic radiation of all possible wavelengths.

The observed frequency spectrum (variation of intensity of radiation versus frequency) was different from that predicted by the theory.

The theory predicted that the spectrum should be the same for all black bodies.

A black body is an unknown mass of unknown density. It's density's contain unknown quantities of Electrons and other Photon absorbing particles. To try and determine what it's electromagnetic responses are is an act of utter madness. Every black body can have a different response and probably does. To estimate the electromagnetic response of a black body is the same as guessing it's mass.
Warwick guesses a lot.
..maff

I'm not a physicist so I can't argue authoritatively, but every book I've read on the subject - and there have been a lot - has said the same thing. None of them has raised the point that you have. Can you suggest why that should be?
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Offline Mr Andrew

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« Reply #8 on: 18/07/2007 02:59:00 »
You say that the energy curve of black body radiation has been observed experimentally but if a black body is an ideal radiator and ideal models don't ever exist, how did physicists observe this curve?
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Offline Soul Surfer

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« Reply #9 on: 18/07/2007 11:50:02 »
In many cases it is possible to gain a significant understanding of physics using non ideal experimentation techniques.

It is however quite possible to make an object that behaves like a perfectly black body for a wide range of optical and infra red frequencies of electromagnetic radiation.  Consider a hollow spherical cavity made of an electrically conducting material with a rough blackened inside and a smallish hole in its side.  Any electromagnetic radiation falling on this hole will enter the cavity and be reflected round inside it and is very unlikely to come out before it is absorbed by the cavity.  This hole absorbs and therefore radiates electromagnetic radiation just like a perfect black body.  For many purposes rough blackened suirfaces are pretty close too.

experiments to demonstrate the planck energy curve and its behaviour with temperature are quite easily within the capabilities of an undergraduate physcs lab.
« Last Edit: 18/07/2007 12:09:05 by Soul Surfer »
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Offline DoctorBeaver

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« Reply #10 on: 18/07/2007 12:28:32 »
You say that the energy curve of black body radiation has been observed experimentally but if a black body is an ideal radiator and ideal models don't ever exist, how did physicists observe this curve?

I didn't say it. Warwick Uni's website did. I merely copied & pasted what was there.
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Offline Mr Andrew

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« Reply #11 on: 21/07/2007 17:43:51 »
I am confused about this simulated black body.  Is the radiation going into the cavity being absorbed by the 'black body?'  Also, if black bodies absorb all radiation how do they emit any radiation?  Isn't their namesake from the fact that they DON'T emit any radiation and are thus black?
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Offline lightarrow

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« Reply #12 on: 21/07/2007 19:47:21 »
I am confused about this simulated black body.  Is the radiation going into the cavity being absorbed by the 'black body?' 
The radiation entering the cavity is completely absorbed by it; for this reason you see the cavity as the most black body you could ever seen. You can construct it yourself.
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Also, if black bodies absorb all radiation how do they emit any radiation? 
The fact they absorb all the incident light doesn't imply they don't emit it.
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Isn't their namesake from the fact that they DON'T emit any radiation and are thus black?
They don't emit visible light at room temperature, but they emit infrared radiation. Increasing the temperature, they emit light in the same way a hot piece of carbon emits it.

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Offline Soul Surfer

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« Reply #13 on: 21/07/2007 23:48:48 »
In fact black bodies are the most efficient radiators of thermal radiation as well as the most efficient absorbers.  The least efficient absorbers and radiators are shiny bright objects that reflect all the energy that falls upon them rather than absorb it.  That is why the inside of glass vacuum flasks are silvered and if you want really efficient roof insulation you should include a layer of shiny metal foil.
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Offline Mr Andrew

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« Reply #14 on: 24/07/2007 21:48:46 »
Do any of you know of the mathematical representation of why black bodies are the most efficient emitters of radiation?
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Offline Soul Surfer

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« Reply #15 on: 25/07/2007 11:55:33 »
I do not believe that there is a precise matematical answer to yor question.

The performance of various surfaces as emitters and absorbers of thermal infra red radiation was a standard physics experiment when I was at school.

It is easiest to understand the absorbtion properties  if a surface is bright and shiny and conducting the radiation is reflected and clearly not absorbed if it is dull and dark most is is absorbed.  Although visual appearance is not always a good guide for infra red.  The emission properties of surfaces are similar to the absorbtion properties but clealy a bright shiny surface at a high temperature will always emit some radiation
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Offline Mr Andrew

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« Reply #16 on: 25/07/2007 17:13:18 »
I understand the absorbtion properties and their correlation with color (color is a direct result of absorbtion properties of objects).  But what I don't understand is the emission and how it relates to the color of objects (effectively how emission relates to absorbtion).  Emission is due to the inherent vibration of molecules which result in temperature.  Molecules consist of charges and vibrating (oscillating) charges generate electromagnetic waves.  The frequency of these waves dependes on the kinetic energy of the vibrating charge and thus the molecule.  Therefore temperature (average kinetic energy) directly affects the frequency of the electromagnetic waves emitted.  I understand all of this.  But how do the absorbtion properties of matter affect its emission properties?  And shouldn't the most efficient absorber of radiation absorb most of the waves emitted from anywhere but the surface of the substance?  Granted, something that reflects all radiation would reflect the radiation around inside the material.  But that radiation is still in existence while the ones absorbed in a black body are not.  So, shouldn't the radiation emitted from inside a bright, shiny object eventually escape, thus making the object a more efficient emitter of radiation than a black body?  Can someone please clear this up for me?
--Life is the greatest experiment that any person will ever conduct.  It should be treated with the same scientific method as any other experiment.

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Offline JP

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« Reply #17 on: 25/07/2007 19:15:54 »
If you're assuming you have a black body and a non-black body heated up to the same temperature, and you allow them to cool, I think it can be explained by the quantum statistics of the body.  A black body is described by Bose-Einstein statistics, which means that it is possible for all emitting particles in the body to simultaneously be in their lowest energy state (meaning they've emitted all their possible energy). 

If you want a mathematical equation, Bose-Einstein statistics are given by an equation of the form:
p(E)=1/[Exp(E/T)-1],
which describes the probability density of finding a particle of finding a particle of a Bose-Einstein system in energy state E of at system temperature T
(which you'll notice is the same form as Planck's law for black body radiation:
http://scienceworld.wolfram.com/physics/PlanckLaw.html).

Notice that if you allow this system to cool to 0 temperature, p(E≠0)=0, and p(0)=∞.  This means at at its coldest temperature, every particle making up a black body system is in the 0 energy state, making it the most efficient possible emitter of thermal energy.  Any non-black body couldn't possibly reach this limit, since it would have parts that couldn't achieve their zero-energy state, even at absolute zero (these parts are described by Fermi statistics).

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Offline lightarrow

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« Reply #18 on: 26/07/2007 12:44:35 »
Do any of you know of the mathematical representation of why black bodies are the most efficient emitters of radiation?

You put some different bodies inside an oven with void inside and let them reach termic equilibrium. Then, every body receives and give away heat in the form of EM radiation and because of thermic equil., the EM energy given away from each body in 1 second is exactly the same as the energy absorbed in 1 second.

Since the black body is the one which absorbs the max amount of energy per unit time and unit area, it's also the one which emits the max amount of radiation.

See Kirchoff's law:
http://en.wikipedia.org/wiki/Kirchhoff's_law_of_thermal_radiation
« Last Edit: 26/07/2007 13:02:15 by lightarrow »

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Offline lightarrow

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« Reply #19 on: 26/07/2007 12:57:57 »
I understand the absorbtion properties and their correlation with color (color is a direct result of absorbtion properties of objects).  But what I don't understand is the emission and how it relates to the color of objects (effectively how emission relates to absorbtion).  Emission is due to the inherent vibration of molecules which result in temperature.  Molecules consist of charges and vibrating (oscillating) charges generate electromagnetic waves.  The frequency of these waves dependes on the kinetic energy of the vibrating charge
No. If you change the mass of the charge and not the frequency of oscillation, the kinetic energy changes, but not the frequency of the EM radiation emitted. This depends only on the frequency of the charge oscillation.
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and thus the molecule.  Therefore temperature (average kinetic energy) directly affects the frequency of the electromagnetic waves emitted.  I understand all of this.  But how do the absorbtion properties of matter affect its emission properties?  And shouldn't the most efficient absorber of radiation absorb most of the waves emitted from anywhere but the surface of the substance? 
I cannot understand what you mean.
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Granted, something that reflects all radiation would reflect the radiation around inside the material.
Don't confuse the inside with the surface. Inside happen other things. The emission of radiation that happens at the surface depends only on the properties of the surface and on the value of the temperature.
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But that radiation is still in existence while the ones absorbed in a black body are not.  So, shouldn't the radiation emitted from inside a bright, shiny object eventually escape, thus making the object a more efficient emitter of radiation than a black body? 
Sorry, I don't understand what you mean.

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Offline Mr Andrew

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« Reply #20 on: 26/07/2007 15:20:50 »
Ah, Kirchoff's Law.  Ok, that makes sense.  So, if I put an object in a state in which no EM radiation can reach it, it would not emit any radiation?  That would mean that charges stop oscillating because isn't it the oscillations that cause radiation to be emitted?  If the charges stop oscillating then they cease to have a wave-particle duality because they are no longer waves.  Does that mean they are just particles or that they are non-existent?  Is the absorbtion of EM radiation necessary for existence?  WOW, I probably made a huge leap in logic here and missed something important!  This is ridiculous!



Oh, lightarrow, those passages you didn't understand: in the first, I thought that KE of a molecule affected the frequency of EM radiation emitted and since <KE> is temperature, I said that tempurature affected the frequency of EM radiation; in the second passage, I assumed that radiation that is absorbed ceases to exist and radiation that is reflected still exists.  If this were the case the radiation reflected around inside a non-black body would eventually escape but the radiation emitted inside a black body would be absorbed and couldn't escape because it wouldn't exist.  Of course, you totally bypassed these passages with what you mentioned in your post so the parts you didn't understand are irrelevant now.  Thanks for the link, it helped a lot.
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Offline Mr Andrew

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« Reply #21 on: 30/07/2007 17:17:13 »
Alright, I understand all of this now.  However, on a related topic, I'm trying to solve the Schrodinger equation in 3-D for the Hydrogen atom and I can't figure out how to do it...I know you have to use spherical coordinates to seperate the variables but the Laplacian isn't the same in spherical coordinates as in rectangular coordinates.  Can someone explain why they are different?
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Offline JP

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« Reply #22 on: 30/07/2007 18:14:01 »
Look at the gradient in spatial coordinates first:

x∂/∂x + y∂/∂y + z∂/∂z

All coordinates here are measuring position.   You're trying to change variables so that two of your coordinates measure angles and one measures a position.  In order to do so, you need to scale the angular derivatives by something that tells you how much changing that angle will change your position.  For example, a change in the azimuthal angle θ by a small amount dθ means a spatial change given by dr=rdθ (see figure).  Since the other spatial variable (the zenith angle) has an extra projection involved, dr=r sinθ dφ.

If you'll notice these scaling factors (r, r sin θ) appear in the gradient in spherical coordinates

r∂/dr + θ(1/r) ∂/∂θ + φ 1/(r sin θ) ∂/∂φ.

These scaling factors are the reason why the Laplacian looks different in different coordinates.

Calculating the Laplacian gets messy.  You have to dot the gradient above with another gradient, so you need to use the vector identity: Divergence[fv]=f Divergence[v]+Gradient[f]•v, and keep in mind that you can act on the unit vectors r, θ, and φ with derivatives as well.  Hope that makes some sense. 

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Offline DoctorBeaver

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« Reply #23 on: 30/07/2007 18:43:40 »
Look at the gradient in spatial coordinates first:

x∂/∂x + y∂/∂y + z∂/∂z

All coordinates here are measuring position.   You're trying to change variables so that two of your coordinates measure angles and one measures a position.  In order to do so, you need to scale the angular derivatives by something that tells you how much changing that angle will change your position.  For example, a change in the azimuthal angle θ by a small amount dθ means a spatial change given by dr=rdθ (see figure).  Since the other spatial variable (the zenith angle) has an extra projection involved, dr=r sinθ dφ.

If you'll notice these scaling factors (r, r sin θ) appear in the gradient in spherical coordinates

r∂/dr + θ(1/r) ∂/∂θ + φ 1/(r sin θ) ∂/∂φ.

These scaling factors are the reason why the Laplacian looks different in different coordinates.

Calculating the Laplacian gets messy.  You have to dot the gradient above with another gradient, so you need to use the vector identity: Divergence[fv]=f Divergence[v]+Gradient[f]•v, and keep in mind that you can act on the unit vectors r, θ, and φ with derivatives as well.  Hope that makes some sense. 


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« Reply #24 on: 30/07/2007 19:16:42 »
Heh.  I assumed since the question started off "I'm trying to solve the Schrodinger equation in 3-D for the Hydrogen atom..." that he wanted a technical answer.  ;)

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Offline DoctorBeaver

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« Reply #25 on: 30/07/2007 20:52:34 »
Heh.  I assumed since the question started off "I'm trying to solve the Schrodinger equation in 3-D for the Hydrogen atom..." that he wanted a technical answer.  ;)

I wasn't complaining. It was meant to convey the fact that you totally lost me. Your answer seems very comprehensive and I wish I could understand it.
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Offline Mr Andrew

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« Reply #26 on: 01/08/2007 02:34:14 »
Thanks for the help, and yes, I did want a technical answer.  I'm right beside the Doctor doing mental push-ups but I will work it out eventually.  I'm probably in way over my head with this but it's summer and I've nothing better to do.

I understand where you got the scaling factors from but how do they fit into the equation for the gradient?  It looks like you have the derivative with respect to r added to itself there.  Is it because the gradient only act on positions?
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Offline JP

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« Reply #27 on: 01/08/2007 15:19:26 »
The gradient is supposed to "live" in position space, since it tells you in what spatial direction the function is increasing fastest. 

I'm not sure I follow your question about adding the derivative with respect to r to itself, but I probably should define the notation I used.  Bold text was indicating unit direction vectors, and what the gradient will do is tell you how much a function changes with respect to one of its arguments, and then you have to multiply that by a unit vector so that it tells you in what direction it's changing.  Hence you get a term
r∂/∂r
which means "find out how much the function is changing as we change r (i.e. take the derivative), and then make this value point in the direction it's changing (i.e. multiply it by the r unit vector).  You don't need to scale it since it's already measuring how the function changes in space, since r is a spatial coordinate.

If you have a function in spherical coordinates, you'll want to do the same things in the θ and φ directions.  The problem is that these aren't measuring spatial coordinates.  You start off using the same form for your derivative terms θ∂/∂θ, where the boldface indicates a unit vector pointing either latitudinally or longitudinally from r.  However, you couldn't simply add these terms to r∂/∂r since they're measuring angles, not spatial coordinates [you can check units here: the unit vectors are unitless, ∂/∂r is 1/length, and the ∂/∂θ and ∂/∂φ are 1/angle (which is unitless)].  To get these angular derivatives to measure how a function is changing in spatial position, you need to throw those scaling factors in.

Here's a couple of links that explain things as well (they show where the scaling factors come from, though they're pretty math-intense):
http://mathworld.wolfram.com/Gradient.html
http://mathworld.wolfram.com/Laplacian.html
http://mathworld.wolfram.com/CurvilinearCoordinates.html

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Offline Mr Andrew

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« Reply #28 on: 01/08/2007 20:43:45 »
Yes, thank you.  I am well aquainted with wolfram already and the explainations on their site are helpful, sometimes, but rarely do they include the origin of things such as these scaling factors.  They explain what they are but not where they come from.

As for my question about adding the ∂/∂r to itself three times.  It is in reference to the fact that (1/r) ∂/∂θ ≡ ∂/∂r ≡ (1/r sinθ) ∂/∂φ.  Ah, hold on... I feel an epiphany coming!  Grad[v] = vd/dr = rd/dr + θd/dr + φd/dr and the only way to make any sense of the last two terms is to make the substitutions dr = rdθ and dr = r sinθ dφ so that the unit vectors and independent variables match!  Does that make sense or am I losing my mind?
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« Reply #29 on: 01/08/2007 22:02:22 »
I've dug up another way of explaining this that should be clearer.  I think you're almost there, but to be honest I'm getting confused on notation at this point.  However, what I'll give you now should be a pretty easy-to-follow explanation.  First, let s measure a spatial position.  (r is now reserved for a unit vector pointing radially, not a spatial position).

If you want to know how s varies when you change the spherical variables a bit, you get:

ds=rdr+θ r dθ+φ r sinθ dφ

Now take a function of the spatial coordinate: f(s)
df=dr(∂f/∂r)+dθ(∂f/∂θ)+dφ(∂f/∂φ)
is a measure of how much f changes as the spherical coordinates change.  This has to be equivalent to how much f changes as you move a little bit in space.  The dot product ds•Grad[f] (the directional derivative) measures how f changes in space as you move by a small amount ds, so:

df=ds•Grad[f].

You already know how df and ds relate to the spherical coordinates, so you can solve for how Grad[f] must relate to them.
Match up terms in dr and likewise for dθ and dφ (you can do this since you could set dr, dθ and dφ to zero independently of each other):

dr (∂f/∂r) = dr r•Grad[f]

so the r component of Grad[f] should just be (∂f/∂r).

Matching up the other two parts:

dθ (∂f/∂θ) = r dθ θ•Grad[f]

so the θ component of Grad[f] must equal (1/r)(∂f/∂θ),

dφ (∂f/∂φ) = r sinθ dφ φ•Grad[f]  →  1/(r sinθ) (∂f/∂φ) =  φ•Grad[f]


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Quantum Physics
« Reply #30 on: 01/08/2007 23:44:03 »
I agree that changing the co-ordinates can be very hard work because it was one of the tings we did during my physics degree course many years ago.  There are lots of steps and the equations got so big that each step tended to cover a whole page of a notebook.  confocal elliptical co-ordinates were the worst!
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Offline ramphysix

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Quantum Physics
« Reply #31 on: 11/08/2007 12:02:48 »
you can find lots of good sources about quantum on the following link..
newbielink:http://www.whatusearch.net/Science/Physics/Quantum_Mechanics/ [nonactive]