Re: What does Planck's constant really mean?

franklinhu@xxxxxxxxx wrote: 

Bjoern Feuerbacher wrote:

franklinhu@xxxxxxxxx wrote:

Planck's constant is said to be the smallest amount of heat which can
be radiated at the vibration frequency v. (See.
http://www.hypertextbook.com/physics/modern/planck/)


Can't you read? What the page *actually* says is that hv, i.e.
Planck's constant *times the frequency*, is that "smallest amount of
heat". (BTW, I would prefer using "energy" here instead of "heat").



This is in
accordance with the formula Energy = (any integer value representing
number of photons) X (Planck's constant 6.63 X 10^-34 J*s) X (frequency
in cycles/second).

And where exactly did you get this formula from? 

General Chemistry 3rd Edition Ebbing, pg.244. The other term besides hv
(n) is just a multiplier of the energy of a single photon which would
be hv.


Looking into a chemistry book for concepts of physics? Uh oh. The book
doesn't state that this formula is only true for states with definite
energy (and photon number), right?



My question is whether this also means that this is also the smallest
amount of energy difference between any frequencices.


What is "energy difference between frequencies" supposed to mean?


You can calculate the energy of a single photon = hv. If you take 2
different frequencies, you should be able to calculate the difference
in energy between the frequencies.

Oh, so you meant the energy difference between *photons* of electromagnetic waves with different frequencies.


In other words,
the amount of energy has to be some whole number multiple of 6.63 X
10-34.


If you didn't notice: in physics, quantities have *units*. h isn't
simply 6.63 x 10^-34. It is 6.63 x 10^-34 Js (Joule times seconds)!

Since energy is measured in J, not in Js, the answer to your question
should be obvious...


So you have to multiply Planck's constant by a frequency which gives
you only the energy term.


Indeed.


The reason why 1 hz is special is that this
makes the value of Planck's constant expressible directly in Joules.


Err, so what??? That's simply an artefact of our unit system! Nothing
special about that! No physics behind that!

If we had defined the second in another way, you would get another
value for hv if you chose v = 1 Hz = 1/s!

I had a similar discussion with Porat about a year ago. I demonstrated
to him that a different choice of units leads to different results. He ignored my arguments totally (as usual). Will you, too?


[snip]


This would also
imply that an infinity of energy amounts would also be allowed. But
this would seem to be in contradition with the whole idea that energy
only come in quantized packets.


No, that is in no way in contradiction with that idea. Why do you
think so? Perhaps you should educate yourself a bit on what is meant
when one says that energy comes in "quantized packets"...


Perhaps there is no contradition, that is the heart of my question. I
had difficulty seeing how Planck had to invoke quantatization of energy
in solving the ultraviolet catastrophe. I still don't see how quanta
solved the problem. 

Well, this isn't an easy problem.

Do you know the "equipartition theorem"? It's a theorem in (classical) statistical physics which says that each "degree of freedom" of a physisal system in a thermal equilibrium, at temperature T, should contain in the mean an energy of 0.5 k T (where k is Boltzmann's constant). But an electromagnetic field has infinitely many degrees of freedom; according to the equipartition theorem, every mode of the field (and there are infinitely many of them!) should contain in the mean an energy of 0.5 k T. So in total, there is an infinite amount of energy in the field stored - classically!

Planck solved this by postulating that the energy in every mode can not be any arbitrary value, but has to be a multiple of h v, where v is the frequency of the mode. So as long as 0.5 k T is smaller than hv, a mode with frequency v simply contains *no* energy. Since there is only a finite amount of modes with frequencies so that 0.5 k T > hv, this implies that the total energy in the field is finite.

Does this help in any way? I'm aware that I still used some terms with which you are probably not familiar ("mode", "degree of freedom" etc.); but nevertheless I hope you got a rough idea...


It just appeared that the probability of a black
body emitting a high frequency photon was reduced since it had to
absorb this same amount of energy in order to radiate.


No, this has little to do with Planck's solution.


[snip]


According to the web site I had cited previously:

"Moreover, it is necessary to interpret UN [the total energy of a
blackbody radiator] not as a continuous, infinitely divisible quantity,
but as a discrete quantity composed of an integral number of finite
equal parts."

The part in brackets has been inserted in the translation; it isn't there in the original German text on the left side. And I think it's wrong. UN is *not* the total energy of the radiator; it's merely the
total energy *in one single mode*!


The contradition I am questioning is the infinity of energy states
versus the statement that the energy is a descrete quantity.

The energy *in one single mode* is discrete. But since there are infinitely
many modes, each with a different energy quantum hv, the total energy can still
take *any* value.



[snip]


Now the question is, can I modify the frequency to produce any
arbitrary amount of energy like 6.64 X 10-34 J,


YES!



or would I have to
modify the frequency enough to produce another Planck constant worth of
energy or 13.38 X 10-34 Joules?


NO!

Why on earth do you even call 6.63 x 10^-34 J "a Planck constant worth
of energy"???


Ok, thank you for the very clear response. Now please explain how this
infinity of energies fits with Planck's explanation of the spectrum as
integral number of finite equal parts.


That's not what Planck actually said, see above.


[snip]


If frequency can only come in particular quanta, some rough
calculations show that the frequency would have to differ somewhere in
the magnitude of 1 X 10-21 meters for such a small jump to occur.


Huh? What has the units meters to do with frequency? You apparently
mean the wavelength?


Sorry, yes I mean wavelength


Could you show that calculation, please? 

I can only estimate because I'd need a calculator with nearly 36
signficant digits.

Huh? Why? None of the numbers you start with has more than, say, 8 significant digits!


So I start with a specific example.

Energy of single photon at 672nm =2.95x10-19 J
The question is, what is the change in wavelength such that the energy
= 2.95x10-19 J + 6.63X10-34J.

Err, the result will depend *strongly* on the wavelength you start with, if you didn't notice!!! If you use a wavelenght of optical light, your result above (around 10^-21 m) will be right (probably; I didn't check your calculation), but for radio or gamma radiation, you will get a *totally* different result!


[snip]


This
would also imply that this would correspond to the maximum frequency
possible (based on the smallest possible weavelength of 1X10-21m) or
about 3 X 10^29 cycles/second. It might also imply that this is the
smallest possible unit of distance in the universe, since nothing can
seem to act at smaller distances that this. Going even further into
speculation, this could mean that space itself is quantized in 1 X
10-21 m cubes and explains why we see quantized amounts of energy &
frequency and why we see Planck's constant. (Lots of speculation, but
an interesting idea, I think).


Sorry, I don't see how your space quantization idea leads to the known
formula E=hv.



You'd have to work your way back from the assumption that 6.63X10-34 J
is the absolute minimum quanta of energy. Then E=nhv is trivial  since E
is always going to be some multiple of 6.63X1-34 J which happens to be
Planck's constant.

But this is contradictory. The formula E=nhv gives e.g. for n=1, v=0.1 Hz that E=6.63 x 10^-35 J. Such an energy should not be possible according to your claim that energy is always an integer multiple of
6.63 x 10^-34 J.


Of course, none of this makes sense if E is allowed
to take on arbitrary values


It is.


There is some experimental evidence from the dropping of neutrons in a
gravitational field that would suggest that the neutrons cannot take
arbitrary positions in space, but must drop in specific quanta,
indicating that space is quantized.
(See:http://www.users.csbsju.edu/~frioux/neutron/neutron.htm)


That has *nothing* to do with space quantization!!! These are simply
the energy eigenstates of the neutrons in this bound state!!!



There is a lot of technical information to weed through, so perhaps
this abstract makes it clearer:
http://www.nature.com/cgi-taf/DynaPage.taf?file=/nature/journal/v415/n6869/abs/415297a_fs.html&dynoptions=doi1074722794

"the falling neutrons do not move continuously along the vertical
direction, but rather jump from one height to another"

While this is meant to show that neutrons can only take up fixed
positions in space, like electrons can only take up fixed positions
around a nucleus, the raw data shows stair step like quantization in
the movement of the neutrons. Now it could be that this disappears if
you remove the neutron from the gravitational field, but maybe it
doesn't.

Err, if you didn't notice: such a behaviour was *predicted* theoretically for the neutrons, based on the fact that they move in the gravitational field!


All we know is that it does appear to take discrete steps on
it's way down.


No, that's definitely *not* all we know.

[snip]



Bye,
Bjoern
.

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