Does this have to do with the orientation?

In summary, the conversation is about integrating in the lower half plane and the orientation of the contour. The final answer should be s + kappa - g^2 / (s + gamma + gamma_n), but there is an extra factor of -1 if the contour is closed clockwise instead of counterclockwise. To integrate in the bottom region, one should go in a positive direction along the x-axis and then circle around the singularity in the negative half plane before returning to the x-axis, resulting in a clockwise motion.
  • #1
Nusc
760
2

Homework Statement


The final answer should be
[tex]
\begin{equation}
+ g^{2} \int_{-\infty}^{\infty} d \Delta'\; {\cal \rho}(\Delta')\, \frac{1}{s+\gamma+i\Delta'} = + g^{2} \frac{1}{s+\gamma+\gamma_{n}}
\end{equation}
[/tex]

but I get a - in front of g

[tex]
\begin{subequations}
\begin{eqnarray}
\int_{-\infty}^{\infty} d \Delta'\; {\cal \rho}(\Delta')\, \frac{1}{s+\gamma+i\Delta'}
&=&
\frac{1}{\pi} \int_{-\infty}^{\infty} d \Delta'\; \frac{\gamma_{n}}{(\Delta'^{2}+\gamma_{n}^{2})}\, \frac{1}{(s+\gamma+i\Delta')}
\\
\nonumber
&=&
\frac{1}{\pi} \int_{-\infty}^{\infty} d \Delta'\; \frac{\gamma_{n}}{(\Delta'+i\gamma_{n})(\Delta'-i\gamma_{n})}\, \frac{1}{(s+\gamma+i\Delta')}
\end{eqnarray}
\end{subequations}
[/tex]
in which case the poles are at $\pm i\gamma_{n}, +i(s+\gamma)$.

[tex]
\begin{subequations}
\begin{eqnarray}
\lim \limits_{\Delta' \to i\gamma_{n}} \frac{\gamma_{n}}{(\Delta'+i\gamma_{n})}\, \frac{1}{(s+\gamma+i\Delta')} &=& \frac{1}{2i}\frac{1}{s+\gamma-\gamma_{n}}
\\
\lim \limits_{\Delta' \to -i\gamma_{n}} \frac{\gamma_{n}}{(\Delta'-i\gamma_{n})}\, \frac{1}{(s+\gamma+i\Delta')} &=& \frac{-1}{2i}\frac{1}{s+\gamma+\gamma_{n}}
\\
\lim \limits_{\Delta' \to i(s+\gamma)} \frac{\gamma_{n}}{(\Delta'+i\gamma_{n})(\Delta'-i\gamma_{n})}\frac{1}{(s+\gamma+i\Delta')} &=& \lim \limits_{\Delta' \to i(s+\gamma)} \frac{\gamma_{n}}{(\Delta'+i\gamma_{n})(\Delta'-i\gamma_{n})}\frac{-i}{(-i(s+\gamma)+\Delta')}
\\
\end{eqnarray}
\end{subequations}
[/tex]
It is simpler integrate in down region where we have only one pole [tex] $ -i\gamma_{n}$.[/tex]
While we have two poles in the region above the x:
[tex]$\Delta'= i\gamma_n$,[/tex]
and
[tex] $\Delta'= i(s+\gamma)$.[/tex]
[tex]
\begin{subequations}
\begin{eqnarray}
\oint_{\gamma_{R}(t)} f(z) \, dz &=& 2\pi i[\frac{1}{2i}\frac{-1}{s+\gamma+\gamma_{n}} ]
%\frac{-1}{2i}\frac{1}{s+\gamma-\gamma_{n}} + \frac{\gamma_{n}}{(-s-\gamma+\gamma_{n})(s+\gamma+\gamma_{n})}]
\\
%&=&2\pi i[-\frac{(1+i)\gamma_{n}}{(s+\gamma-\gamma_{n})(s+\gamma+\gamma_{n})}]
\end{eqnarray}
\end{subequations}
[/tex]
[tex]
\begin{subequations}
\begin{eqnarray}
\int_{-\infty}^{\infty} d \Delta'\; {\cal \rho}(\Delta')\, \frac{1}{s+\gamma+i\Delta'}
&=&
\frac{-\pi}{s+\gamma+\gamma_{n}}
\\
\frac{1}{\pi}\int_{-\infty}^{\infty} d \Delta'\; {\cal \rho}(\Delta')\, \frac{1}{s+\gamma+i\Delta'}
&=&
\frac{-1}{s+\gamma+\gamma_{n}}
\end{eqnarray}
\end{subequations}
[/tex]
I get:
[tex]
\begin{equation}
+ g^{2} \int_{-\infty}^{\infty} d \Delta'\; {\cal \rho}(\Delta')\, \frac{1}{s+\gamma+i\Delta'} = s + \kappa - g^{2} \frac{1}{s+\gamma+\gamma_{n}}
\end{equation}
[/tex]

Homework Equations



The Attempt at a Solution

 
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  • #2


Does this have to do with the orientation?
 
  • #3


I didn't check the whole thing, but if you are integrating in the lower half plane, the contour does close clockwise instead of counterclockwise, so there is an extra factor of -1.
 
  • #4


If I'm integrating in the bottom region, how do I know if I'm going cw or ccw? Sorry it's been along time since I've done this.
 
  • #5


S'ok. You have go in a positive direction (since that's the direction of the real integral) along the x-axis and then circle around the singularity in the negative half plane and then return to the x-axis. You are sort of stuck moving clockwise.
 

1. What is orientation in scientific terms?

Orientation refers to the spatial arrangement or alignment of an object or system relative to a reference point or axis. It can also refer to the position of an object in relation to its surroundings or to other objects.

2. How is orientation determined in scientific experiments?

Orientation in scientific experiments is typically determined through the use of specialized equipment such as compasses, gyroscopes, or motion sensors. Researchers may also use mathematical calculations or rely on visual observations to determine orientation.

3. Why is orientation important in scientific research?

Orientation is important in scientific research because it can affect the results of experiments and studies. For example, the orientation of a microscope or telescope can impact the accuracy of observations, and the orientation of a molecule can affect its chemical properties.

4. Can orientation change over time?

Yes, orientation can change over time due to various factors such as motion, external forces, or internal processes. For example, the orientation of planets in our solar system changes as they orbit around the sun, and the orientation of a building can change due to erosion or seismic activity.

5. Are there any practical applications of studying orientation?

Yes, there are many practical applications of studying orientation in various fields such as engineering, biology, and geology. For example, engineers may need to consider the orientation of a structure to ensure its stability, and biologists may study the orientation of animals to understand their behavior and movement patterns.

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