I'm trying to figure out when is [tex]\frac{\delta L}{\delta p}=\dot{q} [/tex].(adsbygoogle = window.adsbygoogle || []).push({});

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?

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Derivative of Lagrangian with respect to momentum

Loading...

Similar Threads - Derivative Lagrangian respect | Date |
---|---|

A Deriving matrix element from Lagrangian | Mar 15, 2017 |

A Deriving the Lagrangian from the Hamiltonian operator | Mar 7, 2017 |

I Lagrangian for electromagnetic field derivation | Jan 23, 2017 |

Derivation of standard model lagrangian | Feb 18, 2015 |

Why does field Lagrangian depend on four-derivative? | Jan 27, 2015 |

**Physics Forums - The Fusion of Science and Community**