A simple harmonic oscillator whose Action is given by:

$S=\displaystyle\int dt\left(\frac{1}{2}\cdot m\left(\frac{dx}{dt}\right)^2-\frac{1}{2}m\omega x^2\right)$

here x is a function of time i.e $x(t)$ . Use the above equation and by doing integration by parts show that : $S=\displaystyle\int dt\left(-\frac{1}{2}\cdot m\cdot x\left(\frac{d^2x}{dt^2}\right)-\frac{1}{2}m\omega x^2\right)$ (here the upper limit is $t_1 = x_1$ and lower limit is $t_2 = x_2$)

I know what integration by parts is which is by defining a $u$ and $v$ functions according to simplicity then integrate according to it. Here I have to calculate everything with respect to Time how will I handle variable here which doesn't depends on time will I take them as constant ? and how did the second part$\left(-\frac{1}{2}m\omega x^2\right)$ remained the same? Please help me with the procedure.

My solution from @Y Tong hint ,

$=\int \frac{1}{2}\cdot m \frac{dx}{dt} dx-\int \frac {1}{2}m\omega x^2 dt$

taking , $\frac{1}{2}m \frac{dx}{dt}=u$ and $dx=v$

$=\int\frac{1}{2}m \frac{dx}{dt} x \biggr\rvert_{\mathbf{t_1}}-\int \frac{1}{2}m \frac{d}{dt}(\frac{dx}{dt})x $

Eventually I got ,

$=-\int \frac{1}{2}m \frac{d}{dt}(\frac{dx}{dt}) x -\int \frac{1}{2} m \omega x^2 dt$

And I am stuck here to generalize it into proven part . Can anyone help with the next steps ?

  • $\begingroup$ Yes, everything except $x$ is to be treated as a constant. $\endgroup$
    – Tavish
    Dec 25 '20 at 17:51
  • $\begingroup$ @Tavish can you tell me what does that t1=x1 mean in the limit part ? $\endgroup$
    – Hoppo
    Dec 25 '20 at 18:59
  • $\begingroup$ I’m not sure, but it could mean the upper limit of the integral, i.e. the integral is from $x_2$ to $x_1$. $\endgroup$
    – Tavish
    Dec 25 '20 at 19:29
  • $\begingroup$ The stated upper limits don't make sense given the context. I would instead expect $x_1=x(t_1)$ and $x_2=x(t_2)$, i.e., $x_1,x_2$ are the initial and final positions. $\endgroup$ Dec 25 '20 at 19:33
  • $\begingroup$ @Tavish i really have no idea how to use Integration by parts here! If i use integration by parts here, from (1/2 m x'') i should get two expression ... $\endgroup$
    – Hoppo
    Dec 25 '20 at 19:46

You want to use $\int udv=uv|_{t_1}^{t_2}-\int v du$ which is due to $(uv)'=u'v+uv'$.

Take $u=\frac12 m\frac{dx}{dt}$ and $dv=dx$, you have $\int_{t_1}^{t_2}\frac12 m\frac{dx}{dt} dx=\frac12 m\frac{dx}{dt} x|_{t_1}^{t_2}-\int_{t_1}^{t_2} \frac12 mx \frac{d^2x}{dt^2}dt.$ The variation should be there but for typical applications of the least action principle, you will fix the ends, so it doesn't matter. However, the potential term should have been $\frac12m\omega^2 x^2$ instead of $\frac12 m\omega x^2,$ or you won't even have matching units.

  • $\begingroup$ Do we have to touch the second part ? coz we are getting exactly what we want from the potential part ! $\endgroup$
    – Hoppo
    Dec 26 '20 at 8:19
  • $\begingroup$ Whoever designed the question just got some typos. The action has the unit of energylength (the unit of Planck's constant). You need masslength^2/time^2. So the first term is missing an m. That's also why I said the second term also has a typo. $\endgroup$
    – C Tong
    Dec 26 '20 at 12:36
  • $\begingroup$ I am sorry i made a correction where there is m ...:) $\endgroup$
    – Hoppo
    Dec 26 '20 at 16:48
  • $\begingroup$ Oh, I see what you were wondering about. I made a typo in my $vdu$ part. $v$ is just x$, and $du=\frac12 m \frac{d^2 x}{dt^2} dt$. So the integration by part was already complete. $\endgroup$
    – C Tong
    Dec 26 '20 at 18:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.