View source for Physics:Kelvin's minimum energy theorem

In [[Physics:Fluid mechanics|fluid mechanics]], '''Kelvin's minimum energy theorem''' (named after [[Biography:William Thomson, 1st Baron Kelvin|William Thomson, 1st Baron Kelvin]] who published it in 1849<ref>Thomson, W. (1849). Notes on hydrodynamics. V. On the vis-viva of a liquid in motion. Camb. Dubl. Math. J, 4, 90-94.</ref>) states that ''the steady irrotational motion of an [[Physics:Incompressible flow|incompressible fluid]] occupying a [[Simply connected space|simply connected region]] has less kinetic energy than any other motion with the same normal component of velocity at the boundary (and, if the domain extends to infinity, with zero value values there)''.<ref>Kelvin, W. T. B., & Tait, P. G. (1867). Treatise on natural philosophy (Vol. 1). Clarendon Press.</ref><ref>Lamb, H. (1932). Hydrodynamics. Cambridge university press.</ref><ref>Batchelor, G. K. (2000). An introduction to fluid dynamics. Cambridge university press.</ref><ref>Truesdell, C. (1954). The kinematics of vorticity (Vol. 954). Bloomington: Indiana University Press.</ref>
==Mathematical Proof==

Let <math>\mathbf{u}</math> be the velocity field of an incompressible irrotational fluid and <math>\mathbf{u_1}</math> be that of any other incompressible fluid motion with same normal component velocity <math>\mathbf{u}\cdot\mathbf{n}=\mathbf{u_1}\cdot\mathbf{n}</math> at the boundary of the domain, where <math>\mathbf{n}</math> is the unit vector of the bounding surface (and, if the domain extends to infinity, <math>\mathbf{u}\cdot\mathbf{n}=\mathbf{u_1}\cdot\mathbf{n}=0</math> there). Then the difference between the kinetic energy is given by

:<math>T_1-T = \frac{1}{2} \rho \int(\mathbf{u}_1^2-\mathbf{u}^2)\ dV</math>

can be rearranged to give 

:<math>T_1-T = \frac{1}{2} \rho \int(\mathbf{u}_1-\mathbf{u})^2\ dV + \rho \int (\mathbf{u}_1-\mathbf{u})\cdot\mathbf{u}\ dV.</math>

Since <math>\mathbf{u}</math> is irrotational and the domain is simply-connected, a single-valued [[Physics:Velocity potential|velocity potential]] exists, i.e., <math>\mathbf{u}=\nabla\phi</math>. Using this, the second integral in the above equation can be written as

:<math>\int (\mathbf{u}_1-\mathbf{u})\cdot\nabla\phi\ dV = \int \nabla\cdot[(\mathbf{u}_1-\mathbf{u})\phi]\ dV - \int\phi\nabla\cdot(\mathbf{u}_1-\mathbf{u})\ dV.</math>

The second integral is identically zero for steady incompressible fluid, i.e., <math>\nabla\cdot\mathbf{u}=\nabla\cdot\mathbf{u}_1=0</math>. Applying the Gauss theorem for the first integral we find

:<math>\int (\mathbf{u}_1-\mathbf{u})\cdot\nabla\phi\ dV = \int \phi(\mathbf{u}_1-\mathbf{u})\cdot\mathbf{n}\ dA</math>

where the surface integral is zero since normal component of velocities are equal there. Thus, one concludes

:<math>T_1-T=\frac{1}{2} \rho \int(\mathbf{u}_1-\mathbf{u})^2\ dV \geq 0</math>

or in other words, <math>T_1\geq T</math>, where the equality holds only if <math>\mathbf{u}_1=\mathbf{u}</math>, thereby proving the theorem.

==References==
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[[Category:Fluid dynamics]]

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