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K^2 is offline
Nov23-12, 05:08 AM
Sci Advisor
P: 2,470

Symmetry and conservation law

Like I said in the edit, it's the translation of coordinates that doesn't change the Lagrangian. I suppose, a concrete example would be better. Say, I have two masses, m1 and m2 at locations x1 and x2 connected by harmonic potential.

[tex]L(x_1,x_2,\dot{x}_1,\dot{x_2}) = \frac{1}{2}(m_1 \dot{x}_1^2+m_2 \dot{x}_2^2) - \frac{k}{2}(x_2 - x_1)^2[/tex]

Now, let's say I introduced new coordinates, x'1=x1+c, x'2=x2+c.

[tex]L(x'_1,x'_2,\dot{x}'_1,\dot{x}'_2) = \frac{1}{2}(m_1 \dot{x}\prime_1^2+m_2 \dot{x}\prime_2^2) - \frac{k}{2}(x'_2 - x'_1)^2 = \frac{1}{2}(m_1 \dot{x}_1^2+m_2 \dot{x}_2^2) - \frac{k}{2}(x_2 + c - x_1 - c)^2 = L(x_1,x_2,\dot{x}_1,\dot{x_2})[/tex]

So the Lagrangian is exactly the same after coordinate got shifted. Not just the total action, but the actual value of Lagrangian at every point.