Physics-Based Simulation

Mass Matrix

Recall that the mass matrix can be calculated as $$M_{ab}=\sum _{e\in \mathcal{T}}{\int }_{{\Omega }_e^0}R(\mathbf{X},0)N_a(\mathbf{X})N_b(\mathbf{X})d\mathbf{X},$$ where ${\Omega }_e^0$ represents the material space of tetrahedron $e$. Changing the integration variable from $\mathbf{X}$ to $(\beta ,\gamma ,\tau )$ results in $$\begin{array}{rl}& {\int }_{{\Omega }_e^0}R(\mathbf{X},0)N_a(\mathbf{X})N_b(\mathbf{X})d\mathbf{X}\\ =& {\int }_0^1{\int }_0^{1-\tau }{\int }_0^{1-\beta -\tau }R(\beta ,\gamma ,\tau ,0)N_a(\beta ,\gamma ,\tau )N_b(\beta ,\gamma ,\tau )\left|\det \left(\frac{\partial \mathbf{X}}{\partial (\beta ,\gamma ,\tau )}\right)\right|d\gamma d\beta d\tau .\end{array}$$

For element $e$ with vertices ${\mathbf{X}}_1$, ${\mathbf{X}}_2$, ${\mathbf{X}}_3$, and ${\mathbf{X}}_4$, $$\left|\det \left(\frac{\partial \mathbf{X}}{\partial (\beta ,\gamma ,\tau )}\right)\right|=\left|\det ([{\mathbf{X}}_2-{\mathbf{X}}_1,{\mathbf{X}}_3-{\mathbf{X}}_1,{\mathbf{X}}_4-{\mathbf{X}}_1])\right|=6V_e,$$ where $V_e$ is the volume of tetrahedron $e$.

Here, we will omit the detailed derivations of each entry in the consistent mass matrix. Assuming uniform density $R$, for the lumped mass matrix, $$M_{aa}^{\text{lump}}=\sum _{e\in \mathcal{T}(a)}\frac{1}{4}R{V}_e\text{and}{M}_{ab}^{\text{lump}}=0(a\ne b),$$ where $\mathcal{T}(a)$ denotes the set of tetrahedra incident to node $a$. In other words, the mass of each tetrahedron is evenly distributed among its 4 nodes, which is intuitively analogous to the 2D case.