Editing Debye model (section)

==== One-dimensional chain in 3D space ====
The same derivation could be done for a one-dimensional chain of atoms. The number of modes remains unchanged, because there are still three polarizations, so
:<math>\sum_{\rm modes}3=3 N.</math>

The rest of the derivation is analogous to the previous, so the left hand side is rewritten with respect to the Debye frequency:
:<math>\sum_{\rm modes}3=\frac {3 L}{2 \pi} \int_{-k_{\rm D}}^{k_{\rm D}}d k = \frac {3 L}{\pi v_{\rm s}} \int_{0}^{\omega_{\rm D}}d \omega.</math>

The last step is multiplied by two is because the integrand in the first integral is even and the bounds of integration are symmetric about the origin, so the integral can be rewritten as from 0 to <math>k_D</math> after scaling by a factor of 2. This is also equivalent to the statement that the volume of a one-dimensional ball is twice its radius. Applying a change a substitution of <math>k=\frac{\omega}{v_s}</math> , our bounds are now 0 to <math>\omega_D = k_Dv_s</math>, which gives us our rightmost integral. We continue;
:<math> \frac {3 L}{\pi v_{\rm s}} \int_{0}^{\omega_{\rm D}}d \omega = \frac {3 L}{\pi v_{\rm s}} \omega_{\rm D} = 3 N .</math>

Conclusion:
:<math> \omega_{\rm D} = \frac {\pi v_{\rm s} N}{L} .</math>