Evaluating the non-relativistic Action

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Homework Help Overview

The discussion revolves around evaluating the classical action for a non-relativistic point particle, expressed as (m/2) ∫ dt ̇{x}², where m is the mass and ̇{x} = dx/dt. The problem involves determining the action along a path defined by specific initial and final positions and times.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss transforming the time-bounded integral into a position-bounded integral and question the validity of this transformation. There is an exploration of the implications of using the chain rule in this context.

Discussion Status

Some participants have provided insights into the transformation of the integral and the reasoning behind it, while others seek further clarification on the underlying principles. The discussion remains open with multiple perspectives being considered.

Contextual Notes

Participants note that the bounds of the integral must be either spatial or temporal, indicating a constraint in the analysis of the action integral.

wam_mi
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Homework Statement



Consider the classical action [tex](m/2) \int dt \dot{x}^{2}[/tex] where m is the mass of the non-relativistic point particle, and [tex]\dot{x} = dx/dt[/tex] .

This classical path starts at [tex](x_{1}, t_{1})[/tex] and ends at [tex](x_{2}, t_{2})[/tex]. Could someone help me evaluate this classical action explicitly please?

Homework Equations


The Attempt at a Solution



Thanks a lot!
 
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wam_mi said:

Homework Statement



Consider the classical action [tex](m/2) \int dt \dot{x}^{2}[/tex] where m is the mass of the non-relativistic point particle, and [tex]\dot{x} = dx/dt[/tex] .

This classical path starts at [tex](x_{1}, t_{1})[/tex] and ends at [tex](x_{2}, t_{2})[/tex]. Could someone help me evaluate this classical action explicitly please?


Homework Equations





The Attempt at a Solution



Thanks a lot!


Simply add the time bounds of the path to the integral to get

[tex](m/2) \int^{t_2}_{t_1} \dot{x}^{2}dt[/tex].

Now turn this into a position-bounded integral as follows:

[tex](m/2) \int^{t_2}_{t_1} \dot{x}^{2}dt=(m/2) \int^{x_2}_{x_1} \dot{x}dx[/tex].

Finally take the time derivative out of integrand and solve the integral!

AB
 
Altabeh said:
Simply add the time bounds of the path to the integral to get

[tex](m/2) \int^{t_2}_{t_1} \dot{x}^{2}dt[/tex].

Now turn this into a position-bounded integral as follows:

[tex](m/2) \int^{t_2}_{t_1} \dot{x}^{2}dt=(m/2) \int^{x_2}_{x_1} \dot{x}dx[/tex].

Finally take the time derivative out of integrand and solve the integral!

AB

Hi there,

Could you kindly explain why one can turn the time-bounded integral into a position-bounded integral?

Thanks a lot!
 
wam_mi said:
Hi there,

Could you kindly explain why one can turn the time-bounded integral into a position-bounded integral?

Thanks a lot!

As in the classical theory of action integrals, the bounds must be either spatial or temporal, so we cannot do the analysis over the dynamics of particles along a bounded path with both elements of time and space simultaneously. So in the integral in question, you can use the chain rule to kill two of time differentials dt by expanding one of [tex]\dot{x}[/tex]'s and cancelling out dt's to only have a dx left behind. The remaining is straightforward to be treated. Since now integral element is dx, so the bound must be spatial and then you can easily factor the operator d/dt out of the integral to get

[tex]\frac{m}{2}\frac{d}{dt}\int^{x_2}_{x_1} x dx[/tex].

Is it all clear now?

AB
 

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