# Energies and frequencies

1. Jan 3, 2008

### arivero

If one reads the current version of the wikipedia entry on Photoelectric_effect

we see that historically the postulate E=h f predates the relativistic theory. So it is kind of circular to argue in terms of $E=h \nu$ to justify $E=h f$. The empirical process first postulated E=h f for the black body and then Einstein justifies the cut in the spectrum of the photoelectric effect by postulating a minimal non-zero required energy and that $E = h \nu$, being $\nu$ the frequency of the then-hypothetical quantum of electromagnetic radiation.

Later on, Bohr suspects that the fundamental object to quantize is not the energy but the angular momentum. This rule works both for the 3D harmonic oscillator and for the Coulomb potential, and in this case it allows Bohr to calculate some of the spectrum of the hydrogen atom.

But it is important to reminder that E=h f as it is, for the harmonic oscillator, does not depend of relativistic formulae. Put V(x)= k x^2, solve the non relativistic Schroedinger eq, and you get E(n) = n h f + E(0).

A interesting related point, suggested by Okun, is that f is measurable in terms of space and time, while E seems to be more indirectly measured.

Last edited: Jan 3, 2008
2. Jan 3, 2008

### blechman

where, precisely, does relativity come into play with the photoelectric effect? And $\nu$ is exactly the same quantity as f (two different symbols for exactly the same thing). So what are you trying to say??

3. Jan 3, 2008

### arivero

I was trying to set up a framework to clarify (or to obscure, it seems) the initial comments in the thread https://www.physicsforums.com/showthread.php?t=206933 , where relativity was invoked. In that thread, the first answers to the OP were to see the relationship between energy and frequency as a consequence of relativity when applied to massless particles. My hope is that the OP and the people who answered to the OP will read this thread, and your remark, and to discuss on their conceptions.

4. Jan 3, 2008

### olgranpappy

But what is *your* point? blechman notes that you have only changed notation from 'f' to 'nu'.

so...

5. Jan 3, 2008

### blechman

I see. I didn't realize that this comes in from another post.

6. Jan 4, 2008

### lightarrow

If you mean that

$\Large {E=h \nu}$

cannot be derived, I agree with you.