Analytic Number Theory - 1

For those who have been following my blogs earlier, you would know that I am an amateur, self-made mathematician, and I like to write as I learn new concepts or explore old ones.

My old blog, which is now abandoned to start this new and more organized blog, can be accessed from here: Infinity & Beyond

Analytic number theory is taught as a part of higher mathematics in post-graduate Pure Science (Mathematics Hons.) courses and under-graduate engineering Courses in India.

While going through the internet, we find various sources for learning about this subject. Therefore, I will break this blog up into various pieces to cover this subject in significant details.

Coming to this, we will keep Tom M Apostol’s Introduction to Number Theory as the foundational text for this blog.

In this blog, we will cover the contents of Chapter 1 and Chapter 2 of Apostol.

Therefore, Elementary Number Theory, Fundamental Theorem of Arithmetic, Arithmetical Functions and Dirichlet Multiplication will be covered, amognst various other concepts.

It is again to be noted that, readers may take this is a disclaimer before we start, that I am writing this blog as a way to teach myself this subject and would therefore run this blog in a “note making/understanding” tone, rather than as a tutorial.

Fundamental Theorem of Arithmetic:

Many proofs of Analytic number theory makes use of the concepts that we learn in this section.

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The principle of Induction:

If $Q$ is set of integers such that,
(a) $1 \in Q$
(b) $n \in Q \text{ implies } n+1 \in Q$
then,
(c ) $\text{all integers }\geq1\text{ belongs to } Q$

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Classical Mechanics - 2: Lorentz Transformation

Lorentz Transformations are a set of linear transformations from one co-ordinate in space-time to another frame that moves at a constant velocity (the parameter) relative to the former Read more.

Doing justice to the title of the blog, I will straightaway dive into the mathematics of the concept,

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Imagine a basic cartesian co-ordinate system in three dimensions, we this this $\text{S- Frame}$, refer Figure-1 below,

Now, the image below, denoted as Figure-2 shows another frame $\text{S'-Frame}$.

Note that the origins of both frames, i.e., $O$ and $O'$ coincide at $t=0$.

Now imagine a ray of light being emitted from the origins of both the origins.

As per three-dimensional co-ordinate system, the distance covered by light will the distance of the point where light currently is at that particular instant of time.

So, if light is a point $P(a,b,c)$ at $t=t_1$, then the distance from origin is,
$$d = \sqrt{a^2+b^2+c^2}$$
And, since light is moving at speed $\mathrm{c}$ and the time is $t_1$, we can write above equation as,
$$\mathrm{c}t_1=\sqrt{a^2+b^2+c^2}$$

Now that the basic is clear, let us formulate the Progress of light equations for fram $S$ and $S'$ when light is emitted from both its origins,

Progress of light described in relation to frame $S$ is,
$$ct = \sqrt{x^2+y^2+z^2} \\ c^2t^2 =x^2+y^2+z^2 \\ x^2+y^2+z^2-c^2t^2 =0\tag*{(1)}$$

Similarly, progress of light described in relation to frame $S$ is,
$$ct' = \sqrt{x'^2+y'^2+z'^2} \\ c^2t'^2 =x'^2+y'^2+z'^2 \\ x'^2+y'^2+z'^2-c^2t'^2 =0 \tag*{(2)}$$

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Now, it is important that you pay attention to Figure-2, there are 4 cases we consider to formulate the next equations leading us to Lorentz Transformations,

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Case 1:
You (the observer) is standing at $\text{S-Frame}$ and loking at $O'$.
You will see that $O'$ is moving at velocity$v$ (not using $c$ for generalization’s sake) along $x$-axis.
[time = $t$, distance = $x$ and velocity = $v$]
$$v = \frac{x}{t} \\ x - vt = 0 \tag*{(3)}$$

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Case 2:
You (the observer) is standing at $\text{S'-Frame}$ and loking at $O'$.
You will see that $O'$ is not moving with respect to $\text{S'-Frame}$.
So, $$x'=0$$,
which can be written as,
$$x' = a( x - vt) \tag*{(4)}$$
Since, $x-vt=0$.
$a$ is a constant that is used to show that both the equations are in different frames of references.

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Case 3:
You (the observer) is standing at $\text{S'-Frame}$ and loking at $O$.
PAY ATTENTION
Note that $v$ is the velocity and hence is dependent of direction (a vector quantity).
You will see that $O$ is moving at velocity$-v$ (not using $c$ for generalization’s sake) along $x$-axis.
$$-v = \frac{x'}{t'}\\ x'+vt'=0 \tag*{(5)}$$

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Case 4:
You (the observer) is standing at $\text{S-Frame}$ and loking at $O$.
You will see that $O$ is not moving with respect to $\text{S-Frame}$.
So, $$x=0$$,
which can be written as,
$$x = a'( x' + vt') \tag*{(6)}$$
Since, $x'+vt'=0$.
$a'$ is a constant that is used to show that both the equations are in different frames of references.

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Please note that the velocity is same in all frames, which is in line with the special theory of relativity.

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See, eqn $(4)$,
$$x' = a( x - vt)$$
See eqn $(6)$,
$$x = a'( x' + vt')$$
Now, use $(4)$ in $(6)$, to find $t'$
$$x = a'x'+a'vt' \\ \implies x = a' (a(x-vt))+a'vt' \\ \implies x = a'ax - a'avt + a'vt' \\ \implies -a'vt' = a'ax - a'avt-x \\ \implies a'vt' = x + a'avt - a'ax\\ \implies t' = \frac{1}{a'}\frac{1}{v}x + at - a \frac{x}{v} \\ \implies \boxed{t' = a [ t + \frac{x}{v}[\frac{1}{aa'}-1]]} \tag*{(7)}$$

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We will have to now solve for $a$ and $a'$, which will help us in expressing $x'$ and $t'$ in terms of $x$, $v$ and $c$.

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In the $\text{S'-Frame}$, there is no motion along the $y'$ and $z'$ axis, thus,
(We do this step to bring motion in $S'$ frame in terms of parameters of $S$ frame)
$$y'=y \\ z' = z$$
And from $(4)$,
$$x' = a(x-vt)$$
Also, from $(7)$,
$$t' = a [ t + \frac{x}{v}[\frac{1}{aa'}-1]]$$
Substitute the above values in eqn $(2)$, we get,
$$[a(x-vt)]^2+y^2+z^2 - c^2a^2[t + \frac{x}{v}[\frac{1}{aa'}-1]]^2=0$$
After some algebraic rigor (which I have gone through, here is the proof :p), you will arrive at,
$$[a^2 + \frac{c^2a^2}{v^2}(\frac{1}{aa'}-1)^2]x^2+y^2+z^2-t^2(-a^2v^2+c^2a^2)\\-2xt[a^2v+\frac{c^2a^2}{v}(\frac{1}{aa'}-1)]=0 \tag*{(8)}$$
Comparing coefficients of eqn $(8)$ with that of eqn $(1)$,

Comparing $x^2$,
$$a^2 + \frac{c^2a^2}{v^2}(\frac{1}{aa'}-1)^2 = 1 \tag*{(9)}$$


Comparing $y^2$ and $z^2$ will yield no significant information as the coefficients are same.


Comparing $t^2$
$$-a^2v^2+c^2a^2=c^2 \tag*{(10)}$$


Comparing $-2xt$,
$$a^2v+\frac{c^2a^2}{v}(\frac{1}{aa'}-1)=0 \tag*{(11)}$$

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Solve $(10)$ for $a$,
$$a^2 = \frac{c^2}{c^2-v^2}\\ \boxed{a = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}} \tag*{(12)}$$


Solve $(11)$ for $a'$,
$$a^2v+\frac{c^2a^2}{v}(\frac{1}{aa'}-1)=0 \\ v+\frac{c^2}{v}(\frac{1}{aa'}-1)=0 \\ \frac{c^2}{v}(\frac{1}{aa'}-1)=-v \\ (\frac{1}{aa'}-1)=-\frac{v^2}{c^2} \\ a' = \frac{1}{(1-\frac{v^2}{c^2}).a} \\ a' = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \tag*{(13)}$$

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$$\therefore \boxed{a = a' = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}} \tag*{(14)}$$
$$\text{This is important.}$$

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Now, use derived $a$ and $a'$ in equations describing $x'$ and $t'$ to find the space and time Lorentz Transformations,

Eqn $(4)$,
$$x' = a(x-vt) \\ \implies \boxed{x' = \frac{x-vt}{\sqrt{1-\frac{v^2}{c^2}}}} \tag*{(15)}$$


Eqn $(7)$
$$t' = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[t + \frac{x}{v}[\frac{1}{\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}.\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}}-1]] \\ \implies t'=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[t+\frac{x}{v}[1-\frac{v^2}{c^2}-1]\\ \implies \boxed{t' = \frac{t-\frac{xv}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}} \tag*{(16)}$$
Equation $(15)$ and $(16)$ are the Lorentz Space and Time Transformation.


Cheers and a very Happy Diwali 2019 $\smile$ !!