# Physical meaning of thermal parameters dimension (Å²)

In single-crystal x-ray crystallography both isotropic and anisotropic displacement parameters $U_{ij}$ of thermal ellipsoids have dimension of square angstrom ($Å^2$) as follows from the definition of Debye–Waller factor ($T$ is dimensionless):

• For isotropic approximation (conditions of Bragg's law): $$T_\text{iso} = 8\pi^2U\left(\frac{\sin \Theta}{\lambda}\right)^2 \to [U] = Å^2 \tag{1}$$

• For anisotropic displacement parameters (ADPs): $$T_\text{ani} = 2\pi^2 \sum_{i=1\\j=1}^3{H[i]H[j]U_{ij}e^\star[i]e^\star[j]} \to [U] = Å^2 \tag{2}$$ where $H$ is reflex's index; $e^\star$ is length of the reciprocal lattice basis vector.

In both cases $U_1$, $U_2$ and $U_3$ ($U_{ij}$ in general) are basically showing contribution of oscillation along three principal (orthogonal) axes and are measured in $Å^2$. So, is there any physical meaning that can be attributed to square angstrom in this case? At first glance it seems rather confusing that a parameter that seems to represent a linear quantity is measured in square units of length.

• Take a square root and be done with it. – Ivan Neretin Aug 27 '17 at 5:48
• The link you provided for the definition of Debye-Waller factor gives a different equation for $T_{iso}$ then you have included here. In the link, the U does seem to be squared. Also, it should be that the exponential of what you wrote is $T_{iso}$ – Tyberius Aug 27 '17 at 15:50
• @Tyberius I'm not sure which equation you are pointing to, and I mostly provided the link for DWF ($B$) as a quick reference what this thing is. In general $B=8\pi^2U=8\pi^2\bar{u^2}$, if that is what confused you ($\bar{u}$ differs from $U$ as it is a root-mean-square amplitude of the atomic vibration). – andselisk Aug 27 '17 at 15:59
• I'm referring to the first equation in that link. "However, as can be seen from the basic expression for the isotropic Debye-Waller factor, $T=\exp(-8\pi^2\langle u^2\rangle \sin^2(\theta)/\lambda^2)$, the larger the second moment, the more rapidly the scattering from the atomic center in question falls off with increase in the scattering angle." So it seems as though the factor to consider is $\langle u^2\rangle$, the variance, which should have units of angstroms squared. – Tyberius Aug 27 '17 at 16:16
• As given by @Tyberius the $\langle u^2 \rangle$ term is the mean square displacement of the lattice points perpendicular to the scattering plane. As the scattering also depends on $\sin^2(\theta)/\lambda^2$ for a given set of displacements the diminution in intensity becomes more important for large scattering angles. – porphyrin Aug 28 '17 at 12:31