Help with simple Einstein photon equation

[meemoe_uk]

I could do with some help understanding this.
That einstein photon particle energy equation

E=hf

Considering the mimimum energy ( Planck energy ) set E = h
That means f=1

What? Is the f dimension in 1/sec? It can't be right. Why would the minimum (planck) frequency happen to be 1 per sec? It looks like f must be non dimemsional. But if f is non dimensional, how does the equation relate to the wave aspect of the photon?

For photons c= f * wavelength

Somethings wrong with my understanding. I've looked at wiki and that didn't help.

Once I've got this sorted, I eventually want to undestand the limitations the Planck constant sets on the wave aspect of photons.

Thanks

 

[meemoe_uk]

Yep f is measured in Hz. My understanding was E had to be an integer multiple of h, since h is the smallest piece of energy possible.
Perhaps I'm confusing photon packet frequency with photon wave frequency? If these are independant then that explains it.

 

[pam]

In the photon effect, only one photon is absorbed.
The energy of an EM wave is E=nhf. In your case, the dinmensionless n has already been set equal to 1, so you use E=hf.

 

[Alfi]

Considering the mimimum energy ( Planck energy ) set E = h?? Minimum
1.95627185×10^9 J, Planck energy, the unit of energy in Planck units

That's a BIG minimum - wow

 

[peter0302]

The Planck Energy is neither the minimum unit of energy nor is it equal to 1h, although this is a very small unit of energy. There's no reason the frequency of a photon can't be less than one cycle per second (in theory).

The Planck Energy is the mass-energy of a black whole whose schwartzchild radius is equal to its Compton length divided by pi. It's the smallest mass that can be compressed into a black hole using known physics. It is on the order of 10^-8 kg, which is not the smallest unit of mass by a longshot.

The amount of energy 1h has no special significance AFAIK.

And I don't know where you get 1.956x10^9J from.

 

[jtbell]

Why would the minimum (planck) frequency happen to be 1 per sec?

It isn't. One cycle every two seconds gives f = 0.5 Hz. It's a very very low frequency for an electromagnetic wave, giving a very very tiny energy for a photon, but it's not forbidden in principle.

 

[Alfi]

"And I don't know where you get 1.956x10^9J from."

ummmm - here?

h**p://en.wikipedia.org/wiki/Planck_energy

 

[peter0302]

Ah, yes. But that is not equal to 1h. h is 6.626x10^-34 j*s.

 

[meemoe_uk]

Ok, I think I'm up to speed. Action is quite different to energy. Planck's constant is action. So an energy transfer can be smaller than h Joules as long as the energy transfer exists for a long time i.e. f<<1/sec. That's interesting, since we're usually talking about photons when we're considering energy transfer, and since photons travel at c, then f<1 means big wavelength. Or in other words Planck's constant says for photons, the smaller the energy transfer, the longer the distance it must occur over to create an attractive force. No wonder classical physics breaks down at atomic levels.

Now I'm savy about this I can answer some of the non understanding you lot are having trying to get me to understand.

That massive 1.956x10^9J is the Planck energy, ( from wiki ) the energy required to look at the Planck length scale.

 

[peter0302]

You're not making much sense. What does "Planck's constant is action" mean? Planck's constant is a conversion factor between frequency and energy, nothing more. I do not believe it is correct to say that the energy transfer takes longer for lower energy photons because the energy transfer is instantaneous.

Remember, the wavelength is NOT the _size_ of thre photon. It's the amount of space the photon travels in which it goes through one complete EM wave cycle. But when that photon strikes the electron, its entire energy is transferred immediately. Thus, for a photon of 1 Hz, it does NOT take 1 second for the electron to become energized.

You might look at it this way: the photon doesn't have energy because the EM field oscillates over spacetime; the EM field oscillates over spacetime BECAUSE the photon has energy. Once the photon hits the electron, ALL of that energy is transferred instantly and the EM oscillations cease. A better way to look at the equation E=hf would be f=E/h, ie. the frequency of the EM oscillations depends on the energy that the photon possesses intrinsically.

It is indeed true that when f < 1, we're talking about very large wavelengths. But that is of no moment. The photon still travels at the speed of light and still transfers the (little) energy it has instantaneously.

 

[meemoe_uk]

It means the dimension of Planck's constant is Joule Seconds. You wrote it yourself in your 2nd post here.

It's funny, many texts I have read gave the impression Planck's constant implies a discrete minimum quantum of energy, but when I try and work it out myself it doesn't figure. Like here, all I get is it's possible to have any energy, no matter how small, all that is needed is a wave of low enough frequency. Right or what?

 

[jtbell]

You get discrete energies only when you have a bound system, e.g. a hydrogen atom or the "particle in a box" that every QM textbook discusses as the first example of solving the Schrödinger equation. Or electromagnetic radiation confined in a cavity, which is what Planck was analyzing.

 

[meemoe_uk]

thanks jtbell, that was the missing link. But you didn't explain why. I will attempt an explanation.
It seems that because photon wavelength inside a bounded space is somewhat restricted, and so the frequency is somewhat restricted, then effectively frequency has a lower bound. Let this lower bound of f be F.
Then E=hF is the discrete energy of the given system. Fine.

It begs the question, does E and F vary with size of space? Atoms all have large discrete energy differences nearer the nucleus and that figures because of the small space. As larger concentric spheres of space around the atom are considered, the discrete energy differences become smaller.
And I know they approach zero. This is what I consider the edge of the atom, beyound which classical physics has some meaning.

Question : if this discrete energys behaviour diminishs to zero on the edge of the atom due to large enough space for F, how can F be restricting in macro objects such as a cavity? Cavities exhibit discrete energy behaviour, i.e. the black body radiation distribution.

My (guess) Answer : Differences between energy levels in atoms approach zero near the edge of the atom despite the size of the atom clearly not being big enough to have low F, because ermm...
(a) Somehow a low energy wave is allowed to be curved ( wraped around ) or constained inside the atom? Don't like this one
(b) Low energy, since low frequency waves from outside the atom can have low energy, it is the universe outside the atom that is equally* responsible for the low energy differences in the high electron energys. *

Question : If discrete energys is a result of bounded energy, then unbounded energy has a continuous distribution. How does the change from bounded-discrete to unbounded continuous energy take place?
My guess : Either the discreteness smoothly approaches zero as the size of the cavity increases, or there is a discontinuity at infinite size.

Hope your enjoying this as much as me. For any good student learning is an geometric process. The more that is learned just means more stuff to ask more questions about. And so on until mental exhaustion.

 

[pam]

You get discrete energies only when you have a bound system, e.g. a hydrogen atom or the "particle in a box" that every QM textbook discusses as the first example of solving the Schrödinger equation. Or electromagnetic radiation confined in a cavity, which is what Planck was analyzing.

An EM wave of a pure frequency is composed of a number, N, of photons, each of which has energy E=hf. The number N is often so large that the energy seems continuous.

 

[jtbell]

Sure. I was referring to the individual photons. If the radiation isn't confined, the energy and frequency of the individual photons isn't constrained. If the radiation is confined, the energy and frequency of each photon are constrained to specific discrete values, corresponding to the classical standing-wave modes inside the "container."

 

[jtbell]

Question : If discrete energys is a result of bounded energy, then unbounded energy has a continuous distribution. How does the change from bounded-discrete to unbounded continuous energy take place?
My guess : Either the discreteness smoothly approaches zero as the size of the cavity increases,

That's it! As the size of the cavity increases, the spacing between the discrete allowed energies decreases. As the size of the cavity goes to infinity, the spacing between the energies goes to zero.

 

[meemoe_uk]

Rock on.

But why did you ignore the other question? Couldn't you understand it?
I will rephrase it.
In terms of energy levels, discreteness changes to continuity at the edge of an atom, based on this, it would seem that anything bigger than an atom would also exhibit continuous ( classical ) energy levels. A macro body with cavity is such an object. Yet cavity radiation is subject to effects caused by discrete energys caused by boundedness.

Why does a macro object like a cavity exhibit behaviour associated with discrete energy levels, while at many scales smaller, space around an atom exhibits behaviour associated with unboundedness?

My guess answer : The shift to continuous energy levels around an atom is more an effect of the atom being in a practically infinite space than instrinsic properties of the atom. e.g. Indeed, if only the objects within a confined space around the atom are considered ( like other atoms ), then it is found that only discrete energys transfers are allowed. e.g.2 If the atom was put inside the forementioned BB cavity, it's energy levels would never attain continuity, but be restricted to discrete energys by of the BB cavity enviroment.
Right?

Going back to the question you answered. It's spawned another question but I think I've already got the answer.
If Increasing the size of the cavity to infinity means reducing the quantum effect to zero, doesn't this mean the classical derivation of BB would become significant in such a cavity? The rayleigh jeans distribution and the ultra violet catastrophe.
My Answer : Yes. But it's the part of the RJ distribution that agrees with Plancks distribution and reality - The low end of the graph dealing with low frequencies. So that's ok.

You know, I feel I've really made progress here. That was a lot of fun. Just like the good old days.