Minkowski spacetime and 组合数学 (Fall 2017)/Problem Set 1: Difference between pages

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[[File:World line.svg|thumb|Example of a light cone.]]
== Problem 1 ==
#有<math>k</math>种不同的明信片,每种明信片有无限多张,寄给<math>n</math>个人,每人一张,有多少种方法?
#有<math>k</math>种不同的明信片,每种明信片有无限多张,寄给<math>n</math>个人,每人一张,每个人必须收到不同种类的明信片,有多少种方法?
#有<math>k</math>种不同的明信片,每种明信片有无限多张,寄给<math>n</math>个人,每人收到<math>r</math>张不同的明信片(但不同的人可以收到相同的明信片),有多少种方法?
#只有一种明信片,共有<math>m</math>张,寄给<math>n</math>个人,全部寄完,每个人可以收多张明信片或者不收明信片,有多少种方法?
#有<math>k</math>种不同的明信片,其中第<math>i</math>种明信片有<math>m_i</math>张,寄给<math>n</math>个人,全部寄完,每个人可以收多张明信片或者不收明信片,有多少种方法?


In [[special relativity]], the '''Minkowski spacetime''' is a four-dimensional [[manifold]], created by [[Hermann Minkowski]]. It has four dimensions: three dimensions of space (x, y, z) and one dimension of time. Minkowski spacetime has a metric signature of (-+++) and is always flat. The convention in this article is to call Minkowski spacetime simply [[spacetime]].  
== Problem 2 ==
Find the number of ways to select <math>2n</math> balls from <math>n</math> identical blue balls, <math>n</math> identical red balls and <math>n</math> identical green balls.
* Give a combinatorial proof for the problem.
* Give an algebraic proof for the problem.


However, Minkowski spacetime only applies in special relativity. [[General relativity]] used the notion of curved [[spacetime]] to describe the effects of gravity and accelerated motion.
== Problem 3 ==
*一个长度为<math>n</math>的“山峦”是如下由<math>n</math>个"/"和<math>n</math>个"\"组成的,从坐标<math>(0,0)</math>到<math>(0,2n)</math>的折线,但任何时候都不允许低于<math>x</math>轴。例如下图:


==Definition(s)==
    /\
===Mathematical===
  /  \/\/\    /\/\
Spacetime can be thought of as a four-dimensional coordinate system in which the axes are given by
  /        \/\/    \/\/\
  ----------------------
:长度为<math>n</math>的“山峦”有多少?


<math>(x, y, z, ct)</math>
*一个长度为<math>n</math>的“地貌”是由<math>n</math>个"/"和<math>n</math>个"\"组成的,从坐标<math>(0,0)</math>到<math>(0,2n)</math>的折线,允许低于<math>x</math>轴。长度为<math>n</math>的“地貌”有多少?


They can also be denoted by
== Problem 4==
李雷和韩梅梅竞选学生会主席,韩梅梅获得选票 <math>p</math> 张,李雷获得选票 <math>q</math> 张,<math>p>q</math>。我们将总共的 <math>p+q</math> 张选票一张一张的点数,有多少种选票的排序方式使得在整个点票过程中,韩梅梅的票数一直高于李雷的票数?等价地,假设选票均匀分布的随机排列,以多大概率在整个点票过程中,韩梅梅的票数一直高于李雷的票数。


<math>(x_1, x_2, x_3, x_4)</math>
==Problem 5==
A <math>2\times n</math> rectangle is to be paved with <math>1\times 2</math> identical blocks and <math>2\times 2</math> identical blocks. Let <math>f(n)</math> denote the number of ways that can be done. Find a recurrence relation for <math>f(n)</math>, solve the recurrence relation.


Where <math>x_4</math> represents <math>ct</math>. The reason for measuring time in units of the speed of light times the time coordinate is so that the units for time are the same as the units for space. Spacetime has the differential for arc length given by
== Problem 6 ==
Let <math>a_n</math> be a sequence of numbers satisfying the recurrence relation:
:<math>p a_n+q a_{n-1}+r a_{n-2}=0</math>
with initial condition <math>a_0=s</math> and <math>a_1=t</math>, where <math>p,q,r,s,t</math> are constants such that <math>{p}+q+r=0</math>, <math>p\neq 0</math> and <math>s\neq t</math>. Solve the recurrence relation.


<math>ds^2=-c^2dt^2+dx^2+dy^2+dz^2</math>
== Problem 7 ==
* 令<math>s_n</math>表示长度为<math>n</math>,没有2个连续的1的二进制串的数量,即
*:<math>s_n=|\{x\in\{0,1\}^n\mid \forall 1\le i\le n-1, x_ix_{i+1}\neq 11\}|</math>。
:求 <math>s_n</math>


This implies that spacetime has a metric tensor given by
*令<math>t_n</math>表示长度为<math>n</math>,没有3个连续的1的二进制串的数量,即
*:<math>t_n=|\{x\in\{0,1\}^n\mid \forall 1\le i\le n-2, x_ix_{i+1}x_{i+2}\neq 111\}|</math>。
*#给出计算<math>t_n</math>的递归式,并给出足够的初始值。
*#计算<math>t_n</math>的生成函数<math>T(x)=\sum_{n\ge 0}t_n x^n</math>,给出生成函数<math>T(x)</math>的闭合形式。


<math>g_{uv}=\begin{bmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix}</math>
注意:只需解生成函数的闭合形式,无需展开。
 
As before stated, spacetime is flat everywhere; to some extent, it can be thought of as a plane.
 
===Simple===
Spacetime can be thought of as the "arena" in which all of the events in the universe take place. All that one needs to specify a point in spacetime is a certain time and a typical spatial orientation. It is hard (virtually impossible) to visualize four dimensions, but some analogy can be made, using the method below.
 
==Spacetime diagrams==
[[File:Minkowski diagram - asymmetric.svg|thumb|In the theory of relativity both observers assign the event at A to different times.]]
 
Hermann Minkowski introduced a certain method for graphing coordinate systems in Minkowski spacetime. As seen to the right, different coordinate systems will disagree on an object's spatial orientation and/or position in time. As you can see from the diagram, there is only one spatial axis (the x-axis) and one time axis (the ct-axis). If need be, one can introduce an extra spatial dimension, (the y-axis); unfortunately, this is the limit to the number of dimensions: graphing in four dimensions is impossible. The rule for graphing in Minkowski spacetime goes as follows:
 
1) The angle between the x-axis and the x'-axis is given by <math>tan \theta=\frac{v}{c}</math> where v is the velocity of the object
 
2) The speed of light through spacetime ''always'' makes an angle of 45 degrees with either axis.
 
==Spacetime in general relativity==
{{Main|Spacetime}}
 
In the general theory of relativity, [[Einstein]] used the equation
 
<math>R_{uv}-\frac{1}{2}g_{uv}R=8 \pi  T_{uv}</math>
 
To allow for spacetime to actually curve; the resulting effects are those of gravity.
 
==Related pages==
*[[Spacetime]]
*[[Special relativity]]
*[[General relativity]]
 
[[Category:Relativity]]

Revision as of 13:09, 17 September 2017

Problem 1

  1. [math]\displaystyle{ k }[/math]种不同的明信片,每种明信片有无限多张,寄给[math]\displaystyle{ n }[/math]个人,每人一张,有多少种方法?
  2. [math]\displaystyle{ k }[/math]种不同的明信片,每种明信片有无限多张,寄给[math]\displaystyle{ n }[/math]个人,每人一张,每个人必须收到不同种类的明信片,有多少种方法?
  3. [math]\displaystyle{ k }[/math]种不同的明信片,每种明信片有无限多张,寄给[math]\displaystyle{ n }[/math]个人,每人收到[math]\displaystyle{ r }[/math]张不同的明信片(但不同的人可以收到相同的明信片),有多少种方法?
  4. 只有一种明信片,共有[math]\displaystyle{ m }[/math]张,寄给[math]\displaystyle{ n }[/math]个人,全部寄完,每个人可以收多张明信片或者不收明信片,有多少种方法?
  5. [math]\displaystyle{ k }[/math]种不同的明信片,其中第[math]\displaystyle{ i }[/math]种明信片有[math]\displaystyle{ m_i }[/math]张,寄给[math]\displaystyle{ n }[/math]个人,全部寄完,每个人可以收多张明信片或者不收明信片,有多少种方法?

Problem 2

Find the number of ways to select [math]\displaystyle{ 2n }[/math] balls from [math]\displaystyle{ n }[/math] identical blue balls, [math]\displaystyle{ n }[/math] identical red balls and [math]\displaystyle{ n }[/math] identical green balls.

  • Give a combinatorial proof for the problem.
  • Give an algebraic proof for the problem.

Problem 3

  • 一个长度为[math]\displaystyle{ n }[/math]的“山峦”是如下由[math]\displaystyle{ n }[/math]个"/"和[math]\displaystyle{ n }[/math]个"\"组成的,从坐标[math]\displaystyle{ (0,0) }[/math][math]\displaystyle{ (0,2n) }[/math]的折线,但任何时候都不允许低于[math]\displaystyle{ x }[/math]轴。例如下图:
   /\
  /  \/\/\    /\/\
 /        \/\/    \/\/\
 ----------------------
长度为[math]\displaystyle{ n }[/math]的“山峦”有多少?
  • 一个长度为[math]\displaystyle{ n }[/math]的“地貌”是由[math]\displaystyle{ n }[/math]个"/"和[math]\displaystyle{ n }[/math]个"\"组成的,从坐标[math]\displaystyle{ (0,0) }[/math][math]\displaystyle{ (0,2n) }[/math]的折线,允许低于[math]\displaystyle{ x }[/math]轴。长度为[math]\displaystyle{ n }[/math]的“地貌”有多少?

Problem 4

李雷和韩梅梅竞选学生会主席,韩梅梅获得选票 [math]\displaystyle{ p }[/math] 张,李雷获得选票 [math]\displaystyle{ q }[/math] 张,[math]\displaystyle{ p\gt q }[/math]。我们将总共的 [math]\displaystyle{ p+q }[/math] 张选票一张一张的点数,有多少种选票的排序方式使得在整个点票过程中,韩梅梅的票数一直高于李雷的票数?等价地,假设选票均匀分布的随机排列,以多大概率在整个点票过程中,韩梅梅的票数一直高于李雷的票数。

Problem 5

A [math]\displaystyle{ 2\times n }[/math] rectangle is to be paved with [math]\displaystyle{ 1\times 2 }[/math] identical blocks and [math]\displaystyle{ 2\times 2 }[/math] identical blocks. Let [math]\displaystyle{ f(n) }[/math] denote the number of ways that can be done. Find a recurrence relation for [math]\displaystyle{ f(n) }[/math], solve the recurrence relation.

Problem 6

Let [math]\displaystyle{ a_n }[/math] be a sequence of numbers satisfying the recurrence relation:

[math]\displaystyle{ p a_n+q a_{n-1}+r a_{n-2}=0 }[/math]

with initial condition [math]\displaystyle{ a_0=s }[/math] and [math]\displaystyle{ a_1=t }[/math], where [math]\displaystyle{ p,q,r,s,t }[/math] are constants such that [math]\displaystyle{ {p}+q+r=0 }[/math], [math]\displaystyle{ p\neq 0 }[/math] and [math]\displaystyle{ s\neq t }[/math]. Solve the recurrence relation.

Problem 7

  • [math]\displaystyle{ s_n }[/math]表示长度为[math]\displaystyle{ n }[/math],没有2个连续的1的二进制串的数量,即
    [math]\displaystyle{ s_n=|\{x\in\{0,1\}^n\mid \forall 1\le i\le n-1, x_ix_{i+1}\neq 11\}| }[/math]
[math]\displaystyle{ s_n }[/math]
  • [math]\displaystyle{ t_n }[/math]表示长度为[math]\displaystyle{ n }[/math],没有3个连续的1的二进制串的数量,即
    [math]\displaystyle{ t_n=|\{x\in\{0,1\}^n\mid \forall 1\le i\le n-2, x_ix_{i+1}x_{i+2}\neq 111\}| }[/math]
    1. 给出计算[math]\displaystyle{ t_n }[/math]的递归式,并给出足够的初始值。
    2. 计算[math]\displaystyle{ t_n }[/math]的生成函数[math]\displaystyle{ T(x)=\sum_{n\ge 0}t_n x^n }[/math],给出生成函数[math]\displaystyle{ T(x) }[/math]的闭合形式。

注意:只需解生成函数的闭合形式,无需展开。