- #1

help1please

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## Homework Statement

L is the langrangian

[tex]\dot{q}[/tex] is the velocity with time derivative

## Homework Equations

[tex]\frac{\partial L}{\partial \dot{q}} = -M \frac{1}{2} \frac{1}{\sqrt{1- \dot{q} \cdot \dot{q}}} \frac{\partial}{\partial q_i} (-\dot{q} \cdot \dot{q})[/tex]

## The Attempt at a Solution

A few questions here. My work is telling me to calculate the right hand term

[tex]\frac{\partial}{\partial q_i} (-\dot{q} \cdot \dot{q})[/tex]

First of all, what is the dot in [tex]\dot{q} \cdot \dot{q}[/tex], is this just the dot product which would be the velocity squared?

Secondly, the work demonstrates how to calculate the right hand side as

[tex]\frac{\partial}{\partial q_i} (-\dot{q} \cdot \dot{q}) = -\frac{\partial}{\partial q_i}(\dot{q}_j \cdot \dot{q}_j) = -\delta_ij \dot{q}_j - \dot{q}_j \delta_ij[/tex]

This part has me totally confused. What exactly is happening here... ? How do you get, mathematically from

[tex]\frac{\partial}{\partial q_i} (-\dot{q} \cdot \dot{q})[/tex]

then subscripts of j appear and a negative sign shows up..

[tex]-\frac{\partial}{\partial q_i}(\dot{q}_j \cdot \dot{q}_j)[/tex]

Then what exactly is this

[tex]-\delta_ij \dot{q}_j - \dot{q}_j \delta_ij[/tex]

Can some one help me please, it has me stuck in my tracks... :(