- #1
junt
- 18
- 1
Homework Statement
[/B]
I have the following expression:
$$S=T+V$$
$$T=\frac{m}{\tau_0+it}((x_1-x_0)^2+(x_2-x_1)^2)+\frac{m}{2(\tau_1-it)}(x_2-x_0)^2$$
$$V= \frac{(\tau_0+it)}{2}(\frac{k_0 x_0}{2}+\frac{k_0 x_2}{2}+k_0 x_1)+(\tau_1-it)(\frac{k_1 x_0}{2}+\frac{k_1 x_2}{2})$$
The main goal is to find a coordinate transformation such that I can write my S as being linear in t. t should not have none of the higher orders.
2. The attempt at a solution
I tried to scale my x's by $$\sqrt{\tau_0+it}\sqrt{\tau_1-it}$$ This allows me to separate component in T into those containing tau's and t's. For me the important component is the ones with t's. This scaling will lead to components like the following that involves t:
$$T= ... + it((x_1-x_0)^2+(x_2-x_1)^2-\frac{1}{2}(x_2-x_0))$$
Now this is bad because of the (-) sign in the third component. If I could get rid of (-) sign and get rid of the factor 1/2, I could simply do another coordinate transformation called normal mode (q-coordinate) transformation, and write this T part as $$..+i t(q_1^2+q_2^2)$$ I can then choose final variable q_0 such that it is contained in V part.
At the end I want to end up with something like:
$$S=g(\tau_0,\tau_1,x_0,x_1,x_2)+ it(f(\tau_0,\tau_1,x_0,x_1,x_2))$$
where g and f are real function of those variables.
Does anyone know of a better coordinate transformation to achieve this form? I would really appreciate it.
MJ
P.S: For those curious, this is an imaginary time action of a particle in linear potential.
Last edited: