[alancalverd]

True, but most black body radiation is not caused by quantum transitions within atoms, just continuum vibration. Quantum transitions are not "black".

You really do need to get a grip on dimensional analysis before wading into the waters of physics.

[jeffreyH]

Cyclic count of a wavelength?? When did I say that? It was frequency!

[timey (OP)]

Alan - yes indeed, but the energy input to radiation frequency ratio is quantised as a result of the blackbody experiment data.

Jeff - yes, you were referring to frequency.  So am I correct in my interpretation that the count of frequency is dimensionless?

[Colin2B]

Jeff - yes, you were referring to frequency.  So am I correct in my interpretation that the count of frequency is dimensionless?

The count is dimensionless (remember Sesame Street), frequency and wavelength are not.

I realise you think we're products of our misinformed education and totally unable to think outside the box (well, he keeps telling us so) but if any of your ideas ever come to being judged it is this lack of understanding of basic principles that will lead to you being mis-judged.
I only say this because I feel you and your ideas would benefit from spending some time looking at these basics, it's not just terminology, it helps to keep ideas clear.

[timey (OP)]

I most certainly do not think you are products of a misinformed education.  If you are then I am as well.  The information that you have been given is the same info as I have read.  I do think that the method of education is a conditioning factor... and many phycologists would concur.

I never watched sesame Street, and indeed did not experience TV in the home, apart from a brief 4 months before travelling to India, until 1983.

I do not know why you keep referring to me in relation to Thebox.  It's highly annoying.  The dude can't even comprehend that there is an 8 light minute delay between the sun and the Earth.  I don't want to be insulting Thebox or anything, I have respect not to cause upset, but really Colin?

Yes the count is dimensionless, but the duration of time the count is held relative to is not dimensionless.  True or false?

[Colin2B]

I do not know why you keep referring to me in relation to Thebox. 

Not a reference to you at all, but aimed at the box - perhaps unfairly. I have a great deal of respect for the box, he does try to think, but gets bogged down by misunderstanding what he is observing. I'll try not to comment on him.

Yes the count is dimensionless, but the duration of time the count is held relative to is not dimensionless.  True or false?

Yes, that's what I said. The count of cycles is dimensionless, the frequency (cycles per unit time) is not.
What confused me was your suggestion that wavelength is dimensionless and meaningless. I can see now that you were not talking about wavelength.

PS My education was also nonstandard, long story, but I left school with no qualifications - unless you count a leaving certificate which said little more than you've been here and left.

[alancalverd]

Alan - yes indeed, but the energy input to radiation frequency ratio is quantised as a result of the blackbody experiment data.

Nothing happens as a result of experimental data. It's the other way round: data is the result of what happens. E=hf is not a statement of quantisation as f can take any value in a continuum.

The confusion arises because Planck's Law is derived from the concept of electromagnetic radiation in a hypothetical conducting cubical box: this allows us to count the number of photons at any wavelength that has nodes at the walls of the box, and hence the energy density as a function of frequency within the hypothetical box. However to get from there to the emitted spectrum from an actual object, you have to integrate over all possible boxes in the object, which gives you a continuous function with no forbidden transitions. 

[timey (OP)]

OK Colin - Thanks.  Though I too like Thebox and I have no wish to hurt his feelings.

The frequency of wavelength doesn't really take on a meaningful dimension until velocities are attributed.  Am I correct?

Alan - I'm giving what you've said some in depth thought before replying, but it occurs that a sphere is the most efficient shape.  Would a blackbody conducted in a sphere give different data?

[Colin2B]

OK Colin - Thanks.  Though I too like Thebox and I have no wish to hurt his feelings.

By the way, he does understand that light takes the 8mins, he is confused by an interesting variation of Zeno's paradox.

The frequency of wavelength doesn't really take on a meaningful dimension until velocities are attributed.  Am I correct?

I'm having difficulty deciphering your sentence because you are still misusing wavelength, I think? What do you mean by frequency of wavelength? Frequency or cycles?
Frequency is very meaningful without considering velocity. Similarly wavelength.

Would a blackbody conducted in a sphere give different data?

No, the shape of the body has no effect on the radiation curve. You need to reread Alan's post which explains it all.

[alancalverd]

Experimental black body standards are usually spherical with a small hole to let the radiation in or out, because in practice it's much easier to integrate over a sphere than an infinite number of cubes and you don't get any dominant modes. Think flute (clean sinusoid) compared with a rectangular organ pipe.

[Colin2B]

Think flute (clean sinusoid) compared with a rectangular organ pipe.

I like Alan's eg.
If you want to use Chladni you might imagine photographing 1000s of plates at different frequencies, if you then superimposed these photos you would see sand continuously covering the whole area but thicker in some places (like a bump in the curve). Not a perfect illustration, but you get the idea.

[jeffreyH]

Going back to E = hf it shows that for each cycle there is an action with the value h. Since frequency is in cycles per second, if the number of cycles are changed then the total energy expressed over the time interval changes. With Doppler shift we have such a change. This need have nothing to do with gravity and simply be a consequence of the postulates of special relativity. That is only related to the velocity of the emitting object. In a gravitational field you have to take acceleration into account.

[timey (OP)]

Colin - I like Alan's analogy as well...  A wave cycle is 1 up and 1 down cycle.  Frequency is a count of those cycles.  A wavelength is the duration of 1 up and 1 down cycle.  The duration of a wavelength doesn't seem to have much meaning as a dimension until the frequency of a wavelength is associated with velocity.  Is that more clear?

Alan - From what I gleaned from diagrams, the blackbody looks cylindrical and rounded at the ends. (I was thrown by your cubic description, but interested in the resulting thought train.)
The light does not change frequency in the cavity after it has been emitted though, right?
It is as the transferal motion at the molecular level causing electronic motion is increased by the applied temperature, that the emitted photons have more energy, right?

Jeff - just saw your post... Interesting.

[alancalverd]

NO!!!!

Frequency is the count of cycles per unit time. Dimension is T^-1

Wavelength is the distance between peaks. Dimension is L 

Keep it simple and use the same language as everyone else, if you want to be taken seriously.

https://en.wikipedia.org/wiki/Black_body  is good, as is the Wikipedia entry for "thermal radiation".

If I was making a practical source of blackbody radiation it would probably be cylindrical rather than spherical so I could heat it evenly, and just accept the fact that it might be slightly biassed in the very long wave region.

Not sure what "The light does not change frequency in the cavity after it has been emitted" means. How can it, once it has been emitted?

Re: Jeff's post. Doppler shift is not a relativistic effect.

[timey (OP)]

Alan - the count of cycles becomes more complex if the time T that the count is being held relative to is variable...  But my point is, (if I am correct) that it is this count of cycles, ie: frequency, that is being used to attribute red shift velocities.

Yes - the photons do not change frequency after being emitted, so does this mean that the shape of the blackbody, and how those light waves travel around inside it, is irrelevant to the data curve?

[alancalverd]

You can measure red shift by energy, frequency or wavelength. It's all the same thing, just depends on what instrument you are using to measure it.  If you don't use a consistent measure of time you may as well give up doing physics.

As I said, the optimum black body would be an infinite sphere at an absolutely contant temperature,  with an infinitesimal hole, but a cylinder with hemispherical ends is generally good enough over a small spectral range. 

[timey (OP)]

But Hubble adds the aspect of velocity to the scenario.

And yes, that's right, because a sphere is the most efficient shape.  But the change in the frequency of the light is occurring within the changes in the molecular and atomic structure of the blackbody not the cavity.  Correct?

[alancalverd]

But Hubble adds the aspect of velocity to the scenario.

Obviously, if the source is moving wirh respect to the observer, you will see both Doppler and relativistic changes in the received frequency.

And yes, that's right, because a sphere is the most efficient shape.  But the change in the frequency of the light is occurring within the changes in the molecular and atomic structure of the blackbody not the cavity.  Correct?

What change? Blackbody radiation is a continuous spectrum. If you aise the temperature the peak and upper limit shift towards a higher energy and the total photon flux increases. Atomic structure is not changed by temperatures below several zillion degrees. There are no molecules in a metallic structure.

[timey (OP)]

But it's not a continuum throughout the whole process. The applying of the temperature, the temperature transferring to motion, the motion transferring to electronic motion accompanied by photons being emitted of certain energies...
If it were then the data would not be quantised.  So where in the process is the data becoming quantised?

The resulting data of the experiment suggests that it is the input that is delivered in packets.  But what if the input is a continuum and there is another process occurring?  If the addition of a continuum of thermal energy to a molecular structure can cause the atoms to emit photons of escalating energy, where E=hf, hwill be the value of that other process.

Where the data curved diverges in higher frequency region, (or lower frequency region) by attributing lesser (or extra) length of seconds to the process proportionally as energy is added (or decreased), a bit like wavelength being inversely proportional to frequency, except it would be a variable second, the length of which being inversely proportional to frequency, and using these variable seconds to calculate as the applied energy is increased, the value of h will have been transferred and the energy input 'can' be a continuum...
Wavelength can be calculated as hc/E, or c/f.  With E=hf, I can see that the logic might work, but it's the maths that are the defining factor.  Does what I suggest actually do what I suggest it does mathematically?

[Colin2B]

... by attributing lesser (or extra) length of seconds to the process proportionally as energy is added (or decreased), a bit like wavelength being inversely proportional to frequency, except it would be a variable second, the length of which being inversely proportional to frequency, and using these variable seconds to calculate as the applied energy is increased, ...... I can see that the logic might work, but it's the maths that are the defining factor.  Does what I suggest actually do what I suggest it does mathematically?

You can get the same effect in many different ways
12+3=15
3x5=15

If I drive 100 miles at 50mph it will take 2hrs. If I do the same distance in 1.5hrs do I suggest that time has changed or my speed? Effect is the same.
You need to show just cause, not just an effect.