Currently self-studying QFT and The Standard Model by Schwartz and I'm stuck at equation 1.5 in Part 1 regarding black-body radiation

So basically the equation is basically a derivation of Planck's radiation law and I can't somehow find any resources as to how he derived it by adding a derivative inside. Planck says that each mode of frequency $$\omega_n$$ can be excited by $$j$$ times which gives the energy of the form $$jE_n = j(\hbar\omega_n)$$

This is equation 1.5:

$$\langle E_n \rangle = \frac{\sum_{j=0}^\infty (jE_n)e^{-jE_n\beta}}{\sum_{j=0}^\infty e^{-jE_n\beta}} = \frac{-\frac{d}{d\beta}\frac{1}{1-e^{-\hbar\omega_n\beta}}}{\frac{1}{1-e^{-\hbar\omega_n\beta}}} = \frac{\hbar\omega_n}{e^{\hbar\omega_n\beta}-1}\tag{1.5}$$

Where $$\beta = \frac{1}{k_BT}$$.

In short, I just don't get where the $$-\frac{d}{d\beta}$$ comes from. I also don't get how the summations disappear. Hope to get answers before I continue this book.

• Can you be more specific what do you not get here? $-\partial_\beta(\mathrm{e}^{-jE_n\beta}) = (j E_n)\mathrm{e}^{-jE_n\beta}$, so the equation is certainly true, right? May 23 at 11:11
• a free pdf copy of the book can be found here pdfdrive.com/… May 23 at 11:11
• replying to ACuriousMind, but if that's true then why does the original equation has $-\partial_\beta \frac{1}{(e^{-jE_n\beta})$ instead of what you wrote? May 23 at 11:29

It just a clever use of the geometrical series:

$$\frac{1}{1-q} = \sum_{j=0}^\infty q^j$$

which is valid for any real number $$q<1$$.

Here $$q = e^{-E_n\beta} \equiv e^{-\hbar\omega_n\beta}$$. As long as the energies $$E_n >0$$ which is certainly fulfilled, we can make use of the geometrical series. And only this. The parameter $$q=q(\beta)$$ is a function of $$\beta$$, so we can take the derivative of it:

$$-\frac{d}{d\beta} \frac{1}{1-q(\beta)} = -\frac{d}{d\beta} \sum_{j=0}^\infty q(\beta)^j=-\sum_{j=0}^\infty \frac{d}{d\beta}q(\beta)^j =-\sum_{j=0}^\infty q'(\beta) j q(\beta)^{j-1}$$

where in the last step the chain rule was used. We compute the derivative of $$q(\beta)$$: $$q'(\beta) = -E_n e^{-E_n\beta} = -E_n q(\beta)$$.

In the next step we plug this result into the sum:

$$-\sum_{j=0}^\infty q'(\beta) j q(\beta)^{j-1} = -\sum_{j=0}^\infty (-E_n) q(\beta)j q(\beta)^{j-1} = \sum_{j=0}^\infty jE_n q(\beta)^j = \sum_{j=0}^\infty jE_n e^{-jE_n\beta}$$

quod est demonstrandum. One only has to plug it now into the expression for $$\langle E_n\rangle$$.

• Ah that makes so much sense thanks a lot man May 23 at 11:37
• "As long as the energies En>0 which is certainly fulfilled..." This could be true in some cases (and can be forced to be true in many cases), but is not necessarily true. In fact, this is certainly not fulfilled by convention for many common systems. For example, the conventional zero of energy when studying the Hydrogen atom is far from the atom where the field is zero. This means the energy of all the bound states are less than zero. For example -13.6 eV for the ground state.
– hft
May 23 at 22:07
• I guess in this case, since $\omega_n$ is called a frequency it is probably positive. But in general the energy value is not always positive.
– hft
May 23 at 22:08
• @hft the frequencies of photons involved in the black-body radiation are certainly positive. May 24 at 8:47
• @FredericThomas Agreed.
– hft
May 24 at 18:26