Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

145 submissions , 122 unreviewed
3,928 questions , 1,396 unanswered
4,846 answers , 20,597 comments
1,470 users with positive rep
501 active unimported users
More ...

Doubt while computing the second functional derivative of the Euclidean action of a point particle

+ 4 like - 0 dislike
200 views

I am reading the chapter on instantons on Coleman's Aspects of Symmetry 8http://www.amazon.com/Aspects-Symmetry-Selected-Erice-Lectures/dp/0521318270).

Consider the Eucidean action of a point particle in 1+1 dimensions, under a potential

$$S[x]=\int{}dt\,(\frac{1}{2}\dot{x}^2+V)$$

I wanna get the second functional derivative of the action

$$\frac{\delta{}^2S}{\delta{}x(t)\delta{}x(t)}$$

I have been able to compute the first one

$$\frac{\delta{}S}{\delta{}x(t)}=-\ddot{x}+V'(x)$$

but I am clueless about how to proceed. Any hint will be much apreciated.

asked Jan 26, 2016 in Theoretical Physics by Dmitry hand me the Kalashnikov (720 points) [ revision history ]
edited Jan 26, 2016 by Dmitry hand me the Kalashnikov

2 Answers

+ 6 like - 0 dislike

If you take the functional derivative of first order, i.e $\frac{\delta S[x(t)]}{\delta x(t)}=- \ddot{x}(t)+V'(x)$.

Then express these with dirac deltas i.e:

$\int \ddot{x}(s)\delta(s-t')ds= \int x(s)\ddot{\delta}(s-t')ds$ where I have used to integration by parts and boundary conditions obviously vanish.

So if you take another functional derivative you get:

$$\int \frac{\delta x(s)}{\delta x(t)} \ddot{ \delta}(s-t') ds = \int \delta(t-s) \ddot{\delta}(s-t') ds = \ddot{\delta}(t-t')$$

And for the other part also:

$$V'(x(t')) = \int V'(x(s)) \delta(s-t')ds$$

So now we have:

$$\int \frac{\delta V'(x(s))}{\delta x(t)} \delta(s-t')ds = \int V''(x(s)) \delta(s-t)\delta(s-t') ds = V''(x(t))\delta(t-t')$$

So if we gather all the terms we get:

$$-\ddot{\delta}(t-t')+V''(x(t))\delta(t-t')$$

Which is a rigorous treatment of the maple calculation.

answered Jan 31, 2016 by MathematicalPhysicist (120 points) [ no revision ]

Very nice computation, congratulations and many thanks. 

thanks

+ 4 like - 1 dislike

Using Maple I am obtaining the obvious result:

$$\left( D^{ \left( 2 \right) } \right)  \left( V \right)  \left( x \left( \tau \right)  \right) {\it Dirac} \left( \tau-t \right) -{\it
Dirac} \left( 2,\tau-t \right)$$

which in a more standard notation reads

$$\frac{\delta{}^2S}{\delta{}x(\tau)\delta{}x(t)}= V''(x)\delta(\tau-t)-\ddot{\delta}(\tau-t)$$

answered Jan 27, 2016 by juancho (860 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\varnothing$ysicsOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...