Daily Archives: January 10, 2024

3.6 Analytic continuation for gamma function, 4

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A First Course in String Theory

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Explain why the above right-hand side is well defined for \displaystyle{ \Re (z) > - N - 1 }.

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\displaystyle{ \begin{aligned} \Gamma (z) &= \int_{0}^{1}t^{z-1}e^{-t}dt + \int_{1}^{\infty}t^{z-1}e^{-t}dt \\\\ \end{aligned} \\ }

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Prove that \displaystyle{\int_{1}^{\infty}t^{z-1}e^{-t}dt} converges.

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Prove that \displaystyle{\int_{0}^{1}t^{z-1}e^{-t}dt} converges.

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Under this condition,

\displaystyle{ \begin{aligned} &\int_{0}^{1}t^{z-1}e^{-t}dt \end{aligned} }
\displaystyle{ = \int_{0}^{1} t^{z-1} \left( \sum_{n=0}^N \frac{(-t)^n}{n!} +\sum_{n=N+1}^\infty \frac{(-t)^n}{n!} \right) dt, }
\displaystyle{ = \int_{0}^{1} t^{z-1} \sum_{n=0}^N \frac{(-t)^n}{n!} dt + \int_{0}^{1} t^{z-1} \sum_{n=N+1}^\infty \frac{(-t)^n}{n!} dt}

\displaystyle{ = \sum_{n=0}^N \frac{(-1)^n}{n!} \frac{1}{z+n} + \int_{0}^{1} t^{z-1} \sum_{n=N+1}^\infty \frac{(-t)^n}{n!} dt}

So for \displaystyle{\Re (z) > 0},

\displaystyle{ \begin{aligned} &\Gamma (z) \\ &= \int_{0}^{\infty}t^{z-1}e^{-t}dt \\ &= \int_{0}^{1}t^{z-1}e^{-t}dt + \int_{1}^{\infty}t^{z-1}e^{-t}dt \\ &= \sum_{n=0}^N \frac{(-1)^n}{n!} \frac{1}{z+n} + \int_{0}^{1} t^{z-1} \sum_{n=N+1}^\infty \frac{(-t)^n}{n!} dt + \int_{1}^{\infty}t^{z-1}e^{-t}dt \\ \end{aligned} \\ }

If we define this as the analytic continuation of Gamma function:

\displaystyle{ \begin{aligned} \Gamma_c (z) \end{aligned} \\ }

\displaystyle{ \begin{aligned} = \sum_{n=0}^N \frac{(-1)^n}{n!} \frac{1}{z+n} + \int_{0}^{1} t^{z-1} \sum_{n=N+1}^\infty \frac{(-t)^n}{n!} dt + \int_{1}^{\infty}t^{z-1}e^{-t}dt \\ \end{aligned} \\ },

its convergence depends on the convergence of the middle term:

\displaystyle{ \begin{aligned} \int_{0}^{1} t^{z-1} \sum_{n=N+1}^\infty \frac{(-t)^n}{n!} dt \\ \end{aligned} \\ }

By the argument above, a necessary condition for this term to converge is

\displaystyle{ \Re (z) + n_{\text{min}} > 0 }

So

\displaystyle{ \Re (z) + N + 1 > 0 }

\displaystyle{ \Re (z) > - N - 1 }

However, besides the integral convergence, we have to consider also the summation convergence.

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To check whether the order of integration and summation can be interchanged, we rewrite the term as

\displaystyle{ \begin{aligned} &\int_{0}^{1} t^{z-1} \sum_{n=N+1}^\infty \frac{(-t)^n}{n!} dt \\ &= \int_{0}^{1} t^{z-1} \left( e^{-t} - \sum_{n=0}^N \frac{(-t)^n}{n!} \right) dt \\ \end{aligned} \\ }

Since both \displaystyle{e^{-t}} and \displaystyle{\sum_{n=0}^N \frac{(-t)^n}{n!}} are finite, \displaystyle{ \begin{aligned} \sum_{n=N+1}^\infty \frac{(-t)^n}{n!} dt \end{aligned} \\ } is also finite.

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— Me@2023-09-07 07:38:17 PM

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2024.01.10 Wednesday (c) All rights reserved by ACHK

What is absolute is the change of shape

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@dialectphilosophy, 1.3.4

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The key point is that the observer within the car can see the separation changes among some objects, including the car itself, within the car.

Besides seeing the separation changes, the observer can also feel the acceleration directly within his body. The feeling of force is also due to the separation changes, but among points of his body.

How come a free falling frame is equivalent to an inertia frame? In other words, if acceleration is really absolute, how come a free falling observer cannot feel the acceleration?

Short answer: Acceleration is absolute in general, but not in all cases. What is absolute are the particles’ positions relative to each other in a physical system.

Long answer:

For two free falling objects, if they start to fall at the same time, they have equal initial velocities. Then, according to the equation

s = ut + \frac{1}{2}at^2,

the displacements of them are always the same. So their separation is always constant.

For the two objects, their displacements are

\begin{aligned} s_1 &= u_1 t_1 + \frac{1}{2} a_1 t_1^2 \\ s_2 &= u_2 t_2 + \frac{1}{2} a_2 t_2^2 \\ \end{aligned}

If u_1 = u_2, t_1 = t_2, and a_1 = a_2, then s_1 = s_2. As a result, their separation remains unchanged.

Note that the separation would remain unchanged only if

1. the acceleration is uniform (in space) so that a_1 = a_2 always; and

2. both initial velocities have an identical value, i.e. u_1 = u_2.

So what really is absolute is not acceleration, but the separation changes among points seen by the observer within the physical system. Let us label “the separations among points” as “the spatial configuration”, or as an even simpler term: “the shape of the physical system”.

— Me@2023-12-06 11:06:23 AM

— Me@2023-12-24 05:58:54 PM

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2024.01.10 Wednesday (c) All rights reserved by ACHK