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Virial theorem
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== Illustrative special case == Consider {{math|1=''N'' = 2}} particles with equal mass {{mvar|m}}, acted upon by mutually attractive forces. Suppose the particles are at diametrically opposite points of a circular orbit with radius {{mvar|r}}. The velocities are {{math|'''v'''<sub>1</sub>(''t'')}} and {{math|1='''v'''<sub>2</sub>(''t'') = β'''v'''<sub>1</sub>(''t'')}}, which are normal to forces {{math|'''F'''<sub>1</sub>(''t'')}} and {{math|1='''F'''<sub>2</sub>(''t'') = β'''F'''<sub>1</sub>(''t'')}}. The respective magnitudes are fixed at {{mvar|v}} and {{mvar|F}}. The average kinetic energy of the system in an interval of time from {{math|''t''<sub>1</sub>}} to {{math|''t''<sub>2</sub>}} is <math display="block"> \langle T \rangle = \frac{1}{t_2 - t_1} \int_{t_1}^{t_2} \sum_{k=1}^N \frac12 m_k |\mathbf{v}_k(t)|^2 \,dt = \frac{1}{t_2 - t_1} \int_{t_1}^{t_2} \left( \frac12 m|\mathbf{v}_1(t)|^2 + \frac12 m|\mathbf{v}_2(t)|^2 \right) \,dt = mv^2. </math> Taking center of mass as the origin, the particles have positions {{math|'''r'''<sub>1</sub>(''t'')}} and {{math|1='''r'''<sub>2</sub>(''t'') = β'''r'''<sub>1</sub>(''t'')}} with fixed magnitude {{mvar|r}}. The attractive forces act in opposite directions as positions, so <!--<math>\mathbf F_1(t) \cdot \mathbf r_1(t) = \mathbf F_2(t) \mathbf r_2(t) = -Fr </math>-->{{math|1='''F'''<sub>1</sub>(''t'') β '''r'''<sub>1</sub>(''t'') = '''F'''<sub>2</sub>(''t'') β '''r'''<sub>2</sub>(''t'') = β''Fr''}}. Applying the [[centripetal force]] formula {{math|1=''F'' = ''mv''<sup>2</sup>/''r''}} results in <math display="block"> -\frac12 \sum_{k=1}^N \langle \mathbf{F}_k \cdot \mathbf{r}_k \rangle = -\frac12(-Fr - Fr) = Fr = \frac{mv^2}{r} \cdot r = mv^2 = \langle T \rangle, </math> as required. Note: If the origin is displaced, then we'd obtain the same result. This is because the dot product of the displacement with equal and opposite forces {{math|'''F'''<sub>1</sub>(''t'')}}, {{math|'''F'''<sub>2</sub>(''t'')}} results in net cancellation.
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