- #1
geoduck
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I'm trying to figure out when is [tex]\frac{\delta L}{\delta p}=\dot{q} [/tex].
From [tex]L=p\frac{\delta H}{\delta p}-H [/tex]
I get that [tex]\frac{\delta L}{\delta p}=p\frac{\delta^2 H}{\delta p^2}=p \frac{d}{dp}\dot{q} [/tex].
For this to equal just [tex]\dot{q} [/tex], it must obey the dfe [tex]\frac{d}{d ln p}\dot{q}=\dot{q} [/tex]
which implies that [tex]\dot{q}=cp [/tex] for some constant c.
However, I have verified that for the relativistic Lagrangian [tex]L=\sqrt{p^2+m^2} [/tex] it is still true that [tex]\frac{\delta L}{\delta p}=\dot{q}[/tex]. However, the relation between p and [tex]\dot{q} [/tex] is not linear:
[tex]p=\frac{m \dot{q}}{1-\dot{q}^2} [/tex]
How can this be? What did I do wrong?
From [tex]L=p\frac{\delta H}{\delta p}-H [/tex]
I get that [tex]\frac{\delta L}{\delta p}=p\frac{\delta^2 H}{\delta p^2}=p \frac{d}{dp}\dot{q} [/tex].
For this to equal just [tex]\dot{q} [/tex], it must obey the dfe [tex]\frac{d}{d ln p}\dot{q}=\dot{q} [/tex]
which implies that [tex]\dot{q}=cp [/tex] for some constant c.
However, I have verified that for the relativistic Lagrangian [tex]L=\sqrt{p^2+m^2} [/tex] it is still true that [tex]\frac{\delta L}{\delta p}=\dot{q}[/tex]. However, the relation between p and [tex]\dot{q} [/tex] is not linear:
[tex]p=\frac{m \dot{q}}{1-\dot{q}^2} [/tex]
How can this be? What did I do wrong?