QM Weinberg Probl 9.1: Solving for Eq of Motion with Defined Lagrangian

[Kyueong-Hwang]

Homework Statement

specific lagrangian is defined.
Have to get a equation of motion

Homework Equations

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Lagrangian is defined as

The Attempt at a Solution

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eq of motion that i drive like

i guess term of f(x) have to vanish or form a shape of curl.
but it didn't be clear.
May i did something wrong but i can't catch that...

 

[BvU]

Hello Kyueong-Hwang,

I find it hard to read what you write after | )
You want ##\partial {\mathcal L}\over \partial x_i## so I expect to see a ##\partial x_i## in the denominator, not ##\partial x_i \; f(\vec x)## ?

Time to learn some ##\LaTeX## (item 7 here) !

 

[Kyueong-Hwang]

$$ \frac {\partial \mathfrak {L}} {\partial x_i} =\frac {\partial \vec x} {\partial t} \cdot \partial x_i \vec f \left(\vec x \right) - \partial x_i V \left(\vec x \right)$$
$$ \frac {\partial} {\partial t} \left(\frac {\partial \mathfrak {L}} {\frac {\partial x_i} {\partial t}} \right)= m \ddot x_i + \vec \nabla f_i \left(\vec x \right) \cdot \frac {\partial \vec x} {\partial t}$$

I thought ##\LaTeX## is hard to do.
But its quite fun XD

so. with these eqs we can make euler-lagrangian, eq of motion.
I guess changing index i in ##x_i## to ##\vec x## yield that vanish or assemble terms of ##f \left(\vec x \right)##

 

[Kyueong-Hwang]

if ##x_i## is a axis of 3-dim cartesian coordinate two ##f \left( \vec x \right)## terms yield ##\vec \nabla \times \vec f \left(\vec x \right)##
But there is no such statement assure that ##x_i## is 3-dim cartesian coordinate.

And someone in this forum site says in higher than 3 dimension curl is hard to define.
also on wiki only 3, 7 dimension define curl

 

[BvU]

$$\frac {\partial \mathfrak {L}} {\partial x_i} =\dot{ \vec x_i} {\partial f_i \left(\vec x \right)\over \partial x_i } - {\partial V \left(\vec x \right)\over \partial x_i }$$
don't write$$
\frac {\partial} {\partial t} \left(\frac {\partial \mathfrak {L}} {\frac {\partial x_i} {\partial t}} \right)$$for two reasons: 1. it is ##d\over dt##, not ##\partial\over \partial t##, and 2. this way you will mix up ##x_i## and ##\dot x_i## -- they should be treated as independent variables !

So $$
\frac {\partial \mathfrak {L}} {\partial \dot x_i} = m\dot x_i + \vec f_i(\vec x)$$

 

[BvU]

if ##x_i## is a axis of 3-dim cartesian coordinate two ##f \left( \vec x \right)## terms yield ##\vec \nabla \times \vec f \left(\vec x \right)##

Curl doesn't appear here.

 

[Kyueong-Hwang]

alright thanks
1, at latex q&a page i miss d notation so i use partial
2. \dot \vec x isn't work so i think x_i does too

 

[Kyueong-Hwang]

ah
your right. my fault.

is there any way to make this eq more clear?
next question is to make hamiltonian and find Schrodinger eq.
It may should be more clear.

 

[BvU]

is there any way to make this eq more clear?

Well, if ##\vec f(\vec x)## is time independent, you get ##m\ddot x = ... ##

Re Schr eq:
Please post a clear and complete problem description. For a different exercise, start a different thread.