## The escape velocity for the Schwarzschild metric.

The Schwarzschild metric is given by

$\displaystyle ds^2 = \psi c^2 dt^2 - \frac{1}{\psi} dr^2 - r^2 \left( d\theta^2 + \sin^2(\theta) d\phi^2 \right)$

where $\displaystyle \psi = 1 - \frac{r_s}{r}$ and $\displaystyle r_s = \frac{2 GM}{c^2}$.

The geodetic equation is given by

$\displaystyle \frac{d^2x^\kappa}{ds^2} + \Gamma^\kappa_{\mu\nu} \frac{dx^\mu}{ds} \frac{dx^\nu}{ds} = 0$

The path for escaping is given by $\theta = \textit{constant}$ and $\phi = \textit{constant}$, so we only need

$\left[\begin{array}{rcl} \Gamma^r_{rr} &=& \displaystyle \frac{1}{2} \frac{\psi'}{\psi} \\\\ \Gamma^t_{rt} = \Gamma^t_{tr} &=& \displaystyle \frac{1}{2} \frac{\psi'}{\psi} \\\\ \Gamma^r_{tt} &=& \displaystyle \frac{1}{2} \psi \psi' c^2 \end{array}\right.$

And we obtain the equations

$\left[\begin{array}{rcl} \displaystyle \frac{d^2r}{ds^2} + \Gamma^r_{rr} \left(\frac{dr}{ds}\right)^2 + \Gamma^r_{tt} \left(\frac{dt}{ds}\right)^2 &=& 0\\\\ \displaystyle \frac{d^2t}{ds^2} + 2 \Gamma^t_{tr} \frac{dr}{ds} \frac{dt}{ds} &=& 0 \end{array}\right.$

So

$\left[\begin{array}{rcl} \displaystyle \frac{d^2r}{ds^2} + \frac{1}{2} \frac{\psi'}{\psi} \left(\frac{dr}{ds}\right)^2 + \frac{1}{2} \psi \psi' c^2 \left(\frac{dt}{ds}\right)^2 &=& 0\\\\ \displaystyle \frac{d^2t}{ds^2} + \frac{\psi'}{\psi} \frac{dr}{ds} \frac{dt}{ds} &=& 0 \end{array}\right.$

Now it is clear that $\displaystyle \frac{d^2r}{ds^2} = \frac{d}{ds} \left( \dot{r} \frac{dt}{ds} \right) = \ddot{r} \left(\frac{dt}{ds}\right)^2 + \dot{r} \frac{d^2t}{ds^2}$. So we obtain the equation

$\displaystyle \left( \ddot{r} - \frac{1}{2} \frac{\psi'}{\psi} \dot{r}^2 + \frac{1}{2} \psi \psi' c^2 \right) \left( \frac{dt}{ds} \right)^2 = 0$

Then

$\displaystyle \ddot{r} - \frac{1}{2} \frac{\psi'}{\psi} \dot{r}^2 + \frac{1}{2} \psi \psi' c^2 = 0$

Note that for $c \rightarrow \infty$ we obtain $\ddot{r} + \phi' = 0$ – which is the Newton-limit, where we use $\displaystyle \psi = 1 - \frac{2 \phi}{c^2}$.

Now

$\displaystyle \frac{d}{dt} \left( \frac{1}{2} \frac{\dot{r}^2}{\psi} + \frac{1}{2} \psi c^2 \right) = \frac{\dot{r}}{\psi} \left( \ddot{r} - \frac{1}{2} \frac{\psi'}{\psi} \dot{r}^2 + \frac{1}{2} \psi \psi' c^2\right) = 0$

So

$\displaystyle \frac{1}{2} \frac{\dot{r}^2}{\psi} + \frac{1}{2} \psi c^2 = \textit{constant}$

Escaping means that there is some limit $r \rightarrow \infty$ and $\dot{r} \rightarrow 0$. Thus we obtain

$\displaystyle \frac{1}{2} \frac{\dot{r}^2}{\psi} + \frac{1}{2} \psi c^2 = \frac{1}{2} c^2$

Then we obtain

$\displaystyle \left( \frac{\dot{r}}{c} \right)^2 = \psi - \psi^2$

Let the escape velocity be denoted as $u$ then we can write

$u = c \sqrt{\psi - \psi^2}$

As $\displaystyle \psi = 1 - \frac{r_s}{r} = 1 - \frac{2\phi}{c^2}$ we obtain

$u = c \sqrt{ \displaystyle \frac{r_s}{r} - \left( \frac{r_s}{r} \right)^2 } = c \sqrt{ \displaystyle \frac{2\phi}{c^2} - \left(\frac{2\phi}{c^2}\right)^2}$

The only solutions are given by $\displaystyle \frac{r_s}{r} \ge 0$ and $\displaystyle \frac{r_s}{r} \le 1$ – thus $r \ge r_s$. Or $\displaystyle \frac{2\phi}{c^2} \le 1$ thus $\displaystyle \frac{2GM}{rc^2} \le 1$ so $\displaystyle r \ge \frac{2GM}{c^2}$

The escape velocity for the Schwarzschild metric is given by

$u = c \sqrt{ \displaystyle \frac{2\phi}{c^2} - \left(\frac{2\phi}{c^2}\right)^2}$

Hope you liked this post!

Posted in General Relativity, Mathematics, Physics, Relativity | 11 Comments

## The universe as a sphere.

Consider a uniform sphere that has mass $M$ and radius $R$. The gravitational energy of an isotropic sphere is given by $\displaystyle -\frac{16}{15} \pi^2 G \sigma^2 R^5$ so using $\displaystyle \sigma = \frac{3M}{4 \pi R^3}$ we obtain

$\displaystyle U = -\frac{3 G M^2}{5 R}$

The energy is given by $\displaystyle \int \frac{dm c^2}{\sqrt{1 - v^2/c^2}}$ and defining $\displaystyle v = \frac{r}{R} \dot{R}$ we obtain

$\displaystyle T = \int \frac{dm c^2}{\sqrt{ \displaystyle 1 - \dot\zeta^2 \left( \frac{r}{R} \right)^2}}$

where $\displaystyle \dot\zeta = \frac{\dot{R}}{c}$. So

$\displaystyle T = 4 \pi \sigma R^3 c^2 \int_0^R d(r/R) \frac{(r/R)^2}{\sqrt{1 - \dot\zeta^2 (r/R)^2}} = 4 \pi \sigma R^3 c^2 \int_0^1 dx \frac{x^2}{\sqrt{1 - \dot\zeta^2 x^2}}$

Using $\displaystyle \sigma = \frac{3M}{4 \pi R^3}$ we can write

$\displaystyle T = 3Mc^2 \frac{\sin^{-1}(\dot\zeta) - \dot\zeta \sqrt{1 - \dot\zeta^2}}{2 \dot\zeta^3} = \frac{3Mc^2}{f(\dot\zeta)}$.

Where $\displaystyle f(z) = \frac{2z^2}{\sin^{-1}(z) - z \sqrt{1-z^2}}$ and we focus on the function $f(z)$, the plot is given by

The function $f(z)$ looks like a circle. Note that $f(0)=3$ and $f(1) = 4/\pi$ – so we consider $\displaystyle \frac{f(z) - 4/\pi}{3-4/\pi}$ – the plot is given by

We therefore propose $\displaystyle \frac{f(z)-4/\pi}{3-4/\pi} \simeq \sqrt{1-z^2}$.

The total energy is given by

$\displaystyle E = T + V = 3Mc^2 \left\{ \frac{1}{f(\dot\zeta)} - \frac{GM}{5Rc^2}\right\}$

We assume that $E=0$ and that $c$ is constant in time, then we obtain

$\displaystyle f(\dot\zeta) = \kappa\zeta$

where $\displaystyle \kappa = \frac{5c^3}{GM}$ So

$\displaystyle \frac{f(\dot\zeta)-4/\pi}{3-4/\pi} = \frac{\kappa\zeta-4/\pi}{3-4/\pi} \simeq \sqrt{1-\dot\zeta^2}$

The solution is given by

$\displaystyle \zeta = \frac{GM}{c^3} \frac{4}{5\pi} \left\{ 1 + \left(3\pi/4-1\right) \sin\left(\frac{c^3}{GM} \frac{5}{3-4/\pi} t\right) \right\}$

Or using $R = c\zeta$ and $\displaystyle R_\textit{max} = \frac{3}{5} \frac{GM}{c^2}$ we can write

$\displaystyle R = R_\textit{max} \left\{ 4/3\pi + \left(1 - 4/3\pi\right) \cos\left(\frac{1}{1-4/3\pi} \frac{ct}{R_\textit{max}}\right) \right\}$

Then $R_\textit{max} = R(0)$ and $R(T_\textit{max}/2)=0$. So we obtain

$\left[\begin{array}{rcl} R_\textit{max} &=& \displaystyle \frac{3}{5} \frac{GM}{c^2}\\\\ T_\textit{max} &=& \displaystyle \tau \frac{R_\textit{max}}{c}\\\\ \tau &=& 2(1-4/3\pi) \cos^{-1} \left(1/(3\pi/4-1)\right) \simeq 2.7627547843822696042358013453991... \end{array}\right.$

A plot of the radius of the universe in time:

And a plot of the speed of the radius of the universe in time:

The radius is accelerating at the beginning. Is it possible that we are just at the beginning of the universe so that the expansion is still accelerating?

Here we have just used special relativity. The total energy of the universe in this model simply is zero (kinetic energy is positive – but the gravitational energy is negative) – so before the ‘big bang’ there was simply nothing – not energy – no matter and no time.

Posted in Cosmology, Mathematics, Physics, Relativity, Universe | 1 Comment

## Spontaneous creation of matter and antimatter.

In this post I will write something about the spontaneous creation of matter and antimatter. Let us consider matter as an elementairy particle. This elemenrairty particle has some mass and several quantum-numbers. We assume that charge itself is also a quantum number. Let the mass be denoted as $m$ and the quantum-numbers be denoted as $q_k$. The particle can be denoted as $\{m_0,q_k\}$ and the antiparticle can be denoted as $\{\overline{m}_0,\overline{q}_k\}$. Consider a time-line such that at $t=0$ the particle and the anti-particle are spontaneous created. Before $t=0$ there is no energy, no mass and no quantum-numbers; so after $t=0$ the total mass-energy must be zero as well as the total quantum-numbers. We assume that the mass of both the particle and the antiparticle are the same – but that the quantum-numbers are the opposite. Thus we have $\{m_0,q_k\}$ for the particle and $\{m_0,-q_k\}$ for the anti-particle. For simplicity the point of creation is given by $(x,y,z)=(0,0,0)$ and the particle moves in the positive $x$ direction and the anti-particle moves in the negative $x$ direction. The distance between the particle and the origin is latex $r$ and that is also the distance between the anti-particle and the origin. The mass-energy of each particle is given by $\displaystyle \frac{m_0 c^2}{\sqrt{1 - v^2/c^2}}$ – and the static energy is given by $\displaystyle -\frac{\phi}{r}$ – where $\displaystyle \phi = G m^2 + \sum_k \kappa_k q_k^2$. For example if we consider only mass and charge then $\displaystyle \phi = Gm^2 + \frac{q^2}{4\pi\varepsilon_0}$. Since the total energy is zero before $t=0$ we obtain the equation

$\displaystyle \frac{2 m_0 c^2}{\sqrt{1 - v^2/c^2}} - \frac{\phi}{r} = 0$

We ‘solve’ this equation by proposing the solution of the form $\displaystyle r = \frac{c}{\omega} \sin(\omega t)$ – then it is clear that $v = c \cos(\omega t)$ – so we obtain

$\displaystyle \frac{2 m_0 c^2}{\sin(\omega t)} - \frac{\phi\omega}{c\sin(\omega t)} = 0$

Thus $\displaystyle \omega = \frac{2m_0 c^3}{\phi}$. So

$\displaystyle r = \frac{\phi}{2m_0 c^2} \sin\left( \frac{2 m_0 c^3}{\phi} t\right)$

The maximum distance between the particle and the antiparticle is given by $\displaystyle r_\textit{max} = \frac{\phi}{m_0 c^2}$ and the life-time is given $\displaystyle \frac{\pi \phi}{m_0 c^3} = \frac{\pi r_\textit{max}}{c}$. So we can also write

$\displaystyle r_\textit{max} \sin \left( \frac{ct}{r_\textit{max}} \right)$

In case of gravity alone we have $\phi = G m_0^2$ – so the maximum distance is given by $\displaystyle \frac{G m_0}{c^2}$ which is half the Schwarzschild radius $r_s$ so for elementairy particles this is very small – the life time is even smaller as it is half the Schwarzschild radius divided by the speed of light. The case considered is based on creation in an empty space and based on Special Relativity. Special Relativity does allow creation of matter and antimatter – but it holds for only a very small time.

Posted in Mathematics, Physics, Relativity, Special Relativity | 9 Comments

## The gravitational energy of an isotropic sphere.

In this post we find an expression for the gravitational energy of an isotropic sphere. An isotropic sphere is a sphere such that the mass density can be written as $\sigma(r)$.

The gravitational energy of two point masses $m_p$ and $m_q$ is given by

$\displaystyle U_{pq} = - \frac{G m_p m_q}{|\vec{r}_p-\vec{r}_q|}$

The gravitational potential of a point mass $m_p$ is given by

$\displaystyle \Phi_p(\vec{r}) = -\frac{Gm_p}{|\vec{r}-\vec{r}_p|}$

Thus we can write

$\displaystyle U_{pq} = m_p \left( - \frac{G m_q}{|\vec{r}_p-\vec{r}_q|}\right) = m_p \Phi(\vec{r}_p)$.

When we use a mass density $\sigma(\vec{r})$ we can write

$\displaystyle U = -\frac{1}{2} G \int dm' \int dm \frac{1}{|\vec{r}_{dm'} - \vec{r}_{dm}|}$

where $dm = \sigma dx dy dz$. Let $\ell = |\vec{r}_{dm'} - \vec{r}_{dm}|$ then it is clear that

$\ell^2 = \left(r_{dm}\right)^2 + \left(r_{dm'}\right)^2 - 2 r_{dm} r_{dm'} \cos(\theta)$

Therefore

$\ell d\ell = r_{dm} r_{dm'} \sin(\theta)$

where $r=r_{dm}$ and $r'=r_{dm'}$ Now

$\displaystyle \int dm \frac{1}{\ell} = \int_0^R dr \int_0^\pi d\theta \int_0^{2\pi} d\phi \frac{\sigma(r) r^2 \sin(\theta)}{\ell}$

So

$\displaystyle \int dm \frac{1}{\ell} = 2 \pi \int_0^R dr \int_{|r-r'|}^{|r+r'|} d\ell \frac{r \sigma(r)}{r'}$

Thus

$\displaystyle \int dm \frac{1}{\ell} = 2 \pi \int_0^R dr \frac{r \sigma(r)}{r'} \left(|r+r'| - |r-r'|\right)$

Then it is clear that

$\displaystyle \int dm \frac{1}{\ell} = 2 \pi \left\{ \int_0^{r'} dr \frac{r \sigma(r)}{r'} \left(|r+r'| - |r-r'|\right) + \int_{r'}^R dr \frac{r \sigma(r)}{r'} \left(|r+r'| - |r-r'|\right) \right\}$

So

$\displaystyle \int dm \frac{1}{\ell} = 4 \pi \left\{ \frac{1}{r'} \int_0^{r'} dr r^2 \sigma(r) + \int_{r'}^R dr r \sigma(r) \right\}$

For the total energy we have

$\displaystyle U = -8 \pi^2 G \int_0^R dr' r'^2 \sigma(r') \left\{ \frac{1}{r'} \int_0^{r'} dr r^2 \sigma(r) + \int_{r'}^R dr r \sigma(r) \right\}$

In case $\sigma$ is constant we obtain

$\displaystyle U = -8 \pi^2 G \sigma^2 \int_0^R dr' r'^2 \left\{ \frac{1}{r'} \int_0^{r'} dr r^2 + \int_{r'}^R dr r \right\}$

Then

$\displaystyle U = -8 \pi^2 G \sigma^2 \int_0^R dr' r'^2 \left\{ \frac{1}{2} R^2 - \frac{1}{6} r'^2 \right\} = -8 \pi^2 G \sigma^2 \left\{ \frac{1}{6} - \frac{1}{30}\right\} R^4 = -\frac{16}{15} \pi^2 G \sigma^2 R^5$

as expected and well known.

Let us define $\displaystyle M(r) = \int dr r^2 \sigma(r)$ and $\displaystyle Q(r) = \int dr r \sigma(r)$. Then we can write

$\displaystyle U = -8 \pi^2 G \left\{ \int_0^R dr' r' \sigma(r') M(r') - \int_0^R dr' r' \sigma(r') M(0) + \int_0^R dr' r'^2 \sigma(r') Q(R) - \int_0^R dr' r'^2 \sigma(r')Q(r') \right\}$

So

$\displaystyle U = -8 \pi^2 G \left\{ 2\int_0^R dr' r' \sigma(r') M(r') - Q(R) M(0) - Q(0) M(0) + M(R) Q(R) - M(0) Q(R) - M(R)Q(R) + M(0)Q(0) \right\}$

Thus

$\displaystyle U = -16 \pi^2 G \left\{ \int_0^R dr' r' \sigma(r') M(r') - Q(R) M(0) \right\}$

But $M(0)=0$ so

$\displaystyle U = -16 \pi^2 G \int_0^R dr' r' \sigma(r') M(r')$

and using the definition of $M(r)$ we can write

$\displaystyle U = -16 \pi^2 G \int_0^R dr' r' \sigma(r') \int_0^{r'} dr r^2 \sigma(r)$

Again the case of a constant $\sigma$ then we get

$\displaystyle U = -16 \pi^2 G \sigma^2 \int_0^R dr' r' \int_0^{r'} dr r^2 = -16 \pi^2 G \sigma^2 \int_0^R dr' \frac{1}{3} r'^4 = -\frac{16}{15}\pi^2 G \sigma^2 R^5$

The gravitational potential of an isotropic sphere is given by

$\displaystyle U = -16 \pi^2 G \int_0^R dr' r' \sigma(r') \int_0^{r'} dr r^2 \sigma(r)$

where $\sigma(r)$ is the radial mass density.

## Sum of sinus.

In this post we ask the question: what is $\displaystyle \sum_{k=0}^n \sin(kx)$? It is clear that

$\displaystyle \sum_{k=0}^n \exp(kx) = \frac{\exp([n+1]x)-1}{\exp(x)-1}$

Then

$\displaystyle \sum_{k=0}^n \sin(kx) = \frac{1}{2i} \left\{ \sum_{k=0}^n \exp(ikx) - \sum_{k=0}^n \exp(-ikx) \right\}$

So

$\displaystyle \sum_{k=0}^n \sin(kx) = \frac{1}{2i} \left\{ \frac{\exp(i[n+1]x)-1}{\exp(ix)-1} - \frac{\exp(-i[n+1]x)-1}{\exp(-ix)-1} \right\}$

Thus

$\displaystyle \sum_{k=0}^n \sin(kx) = \frac{\sin(nx) - \sin([n+1]x) + \sin(x)}{2-2\cos(x)}$

Therefore

$\displaystyle \sum_{k=0}^n \sin(kx) = \frac{1}{2} \sin(nx) + \frac{1}{2} \sin(x) \frac{1-\cos(nx)}{1-\cos(x)}$

A simple test:

$\displaystyle \displaystyle \sum_{k=0}^{n+1} \sin(kx) - \displaystyle \sum_{k=0}^n \sin(kx) = \sin([n+1]x)$.

Now

$\displaystyle \frac{\sin([n+1]x) - \sin([n+2]x) + \sin(x)}{2-2\cos(x)} - \frac{\sin(nx) - \sin([n+1]x) + \sin(x)}{2-2\cos(x)}$

can be written as

$\displaystyle \frac{2\sin([n+1]x) - \sin([n+2]x) - \sin(nx)}{2-2\cos(x)}$

thus we obtain

$\displaystyle \frac{\sin(nx)\left[2\cos(x)-\cos(2x)-1\right] + \cos(nx)\left[2\sin(x)-\sin(2x)\right]}{2-2\cos(x)}$

which can be written as

$\sin(nx)\cos(x) + \cos(nx)\sin(x) = \sin([n+1]x)$

Therefore it is clear that IF this expression $\displaystyle \sum_{k=0}^n \sin(kx) = \frac{1}{2} \sin(nx) + \frac{1}{2} \sin(x) \frac{1-\cos(nx)}{1-\cos(x)}$ is TRUE for $n$ THEN it is also TRUE for $n+1$. As the expression is TRUE for $n=1$ it is TRUE for all positive integers $n$. So we obtain

$\displaystyle \sum_{k=0}^n \sin(kx) = \frac{1}{2} \sin(nx) + \frac{1}{2} \sin(x) \frac{1-\cos(nx)}{1-\cos(x)}$

Now we may wonder about $\displaystyle \int_0^z dx \sin(x)$. It is clear that

$\displaystyle \int_0^z dx \sin(x) = \lim_{n \rightarrow \infty} z/n \left( \frac{1}{2} \sin(z) + \frac{1}{2} \sin(z/n) \frac{1-\cos(z)}{1-\cos(z/n)} \right)$

thus

$\displaystyle \int_0^z dx \sin(x) = 1-\cos(z)$

As

$\displaystyle \lim_{n \rightarrow \infty} \frac{z/n \sin(z/n)}{1-\cos(z/n)} = \lim_{\xi \downarrow 0} \frac{\xi \sin(\xi)}{1-\cos(\xi)} = 1$

Hope you liked the post!

## Sum of powers.

What is $\sum_{j=1}^n j^k$? – this is a particular questions in mathematics. In this post, I would like to work this question out.

First I define some function $P_k(z)$, defined as

$\displaystyle P_k(z) = \prod_{j=1}^k (z+j)$

Note that $P_k(0)=k!$ and $P_k(-1)=0$.

It is clear that

$\displaystyle P_{k+1}(z) - P_{k+1}(z-1) = \prod_{j=1}^{k+1} (z+j) - \prod_{j=0}^k (z+j) = (k+1) P_k(z)$

It is also clear that

$\displaystyle \sum_{z=0}^n \left\{ P_{k+1}(z) - P_{k+1}(z-1) \right\} = P_{k+1}(n)$

Then we obtain that

$\displaystyle \sum_{z=0}^n P_k(z) = \frac{1}{k+1} P_{k+1}(n)$.

For example

$\displaystyle \sum_{z=0}^n (z+1)(z+2) = \frac{1}{3} (n+1)(n+2)(n+3)$.

Then

$\displaystyle \sum_{z=0}^n \left\{ z^2 + 3(z+1) - 1 \right\} = \frac{1}{3} (n+1)(n+2)(n+3)$.

So

$\displaystyle \sum_{z=0}^n z^2 = \frac{1}{3} (n+1)(n+2)(n+3) - \frac{3}{2} (n+1)(n+2) + (n+1)$.

Therefore

$\displaystyle \sum_{z=0}^n z^2 = \frac{1}{6} n (n+1) (2n+1)$

Hope you enjoyed this post!

## Fundamental definitions in Mechanics

In this post I will write something about fundamental definitions in Physics with regard to (Classical) Mechanics. Classical Mechanics is based on the concepts absolute space and absolute time. A position in the absolute space is denoted as $\vec{s}$ and time is denoted as $t$. Any particular mass $m$ has a path that is described simply as a mapping $\mu$ from the absolute time to the absolute space, given by

(1) $\mu:{\texttt{I\hspace{-0.2em}R}} \ni t \mapsto \vec{s}(t) \in {\texttt{I\hspace{-0.2em}R}}^3$,

assuming that the absolute space has dimension 3. The interval-velocity is simply defined as a change in position during a time-interval and by definition it is given by:

(2a) $\displaystyle \langle \vec{v} \rangle = \frac{\vec{s}_2 - \vec{s}_1}{t_2 - t_1}$,

where $\vec{s}_\jmath$ is defined as $\vec{s}(t_\jmath)$. A special interval-velocity is given when the time-interval tends to zero. Then there is only one time involved and we simply speak about THE velocity (at a given time), following the definition we have

(2b) $\displaystyle \vec{v} = \lim_{\delta t \rightarrow 0} \frac{\vec{s}(t + \delta t) - \vec{s}(t)}{\delta t}$,

which is normally denoted as $\displaystyle \vec{v} = \frac{d\vec{s}}{dt}$. similar we have a definition of the interval-acceleration, it is simply defined as a change in velocity during a time-interval and by definition it is given by:

(3a) $\displaystyle \langle \vec{a} \rangle = \frac{\vec{v}_2 - \vec{v}_1}{t_2 - t_1}$,

where $\vec{v}_\jmath$ is defined as $\vec{v}(t_\jmath)$. A special interval-acceleration is given when the time-interval tends to zero. Then there is only one time involved and we simply speak about THE acceleration (at a given time), following the definition we have

(3b) $\displaystyle \vec{a} = \lim_{\delta t \rightarrow 0} \frac{\vec{v}(t + \delta t) - \vec{v}(t)}{\delta t}$,

which is normally denoted as $\displaystyle \vec{a} = \frac{d\vec{v}}{dt}$. Note that we might also write $\displaystyle \vec{v} = \left(\frac{d}{dt}\right) \vec{s}$ and $\displaystyle \vec{a} = \left(\frac{d}{dt}\right) \vec{v}$. Combination gives $\displaystyle \vec{a} = \left( \frac{d}{dt} \right)^2 \vec{s}$. More there is a convention that $\displaystyle \left( \frac{d}{dt} \right)^n = \frac{d^n}{dt^n}$, so we obtain by definition and convention the following relations:

(4) $\left[ \begin{array}{rcl} \vec{v} &=& \displaystyle \frac{d\vec{s}}{dt} \\\\ \vec{a} &=& \displaystyle \frac{d^2\vec{s}}{dt^2} \end{array} \right.$

We call $\displaystyle \frac{d}{dt}$ the first order and $\displaystyle \frac{d^2}{dt^2}$ the second order. Classical Mechanics only used the first and the second order. However modern Mechanics uses higher orders as well. The following table gives an idea of the used orders and the used symbols:

(5) $\begin{array}{r|ccc} & \textrm{Order} & \textrm{Classical} & \textrm{Modern} \\ \hline \textrm{position} & & \vec{s} & \vec{r} \\\\ \textrm{velocity} & \textrm{first} & \vec{v} & \vec{v} \\\\ \textrm{acceleration} & \textrm{second} & \vec{a} & \vec{a} \\\\ \textrm{jerk} & \textrm{third} & - & \vec{j} \\\\ \textrm{snap} & \textrm{fourth} & - & \vec{s} \\\\ \textrm{cracle} & \textrm{fifth} & - & \vec{c} \\\\ \textrm{pop} & \textrm{sixth} & - & \vec{p} \end{array}$

Important is that the symbols might conflict as $\vec{p}$ might be momentum as well as the pop. However when we restrict to the symbols of Classical Mechanics then there is no conflict in the symbols.

The general idea is to speak of an average-foo during a time-interval that is defined as the change in bar during a particular time-interval. There are certain relations between the foo and the bar, given by:

(6) $\begin{array}{c|ccl} \textit{foo} & \textit{bar} && \textrm{remark} \\ \hline \\ \textrm{velocity} & \textrm{position} && \displaystyle \vec{v} = \frac{d\vec{s}}{dt} \\\\ \textrm{acceleration} & \textrm{velocity} && \displaystyle \vec{a} = \frac{d\vec{v}}{dt} \\\\ \textrm{force} & \textrm{momentum} && \displaystyle \vec{F} = \frac{d\vec{p}}{dt} \\\\ \textrm{force} \times \textrm{velocity} & \textrm{energy} && \displaystyle \vec{F} \cdot \vec{v} = \frac{dE}{dt} \Leftrightarrow E = \int dt \vec{F} \cdot \vec{v} \Leftrightarrow E = \int d\vec{s} \cdot \vec{F} \end{array}$

Note that momentum is defined as $\vec{p} = m \vec{v}$. Then it is clear that the force by definition is given by $\displaystyle \vec{F} = \frac{d\vec{p}}{dt}$. Therefore we obtain in general $\displaystyle \vec{F} = m\vec{a} + \left( \frac{dm}{dt} \right) \vec{v}$. So only IF $\displaystyle \frac{dm}{dt} = 0$ we MAY write $\vec{F} = m \vec{a}$. Assuming that the mass only depends on time, position and velocity, we may write in general

(7) $\displaystyle \vec{F} = m \vec{a} + \left( \frac{\partial m}{\partial t} + \left\{\vec\nabla_{\displaystyle \vec s} m \right\}\cdot \vec{v} + \left\{ \vec{\nabla}_{\displaystyle \vec v} m \right\} \cdot \vec{a} \right) \vec{v}$.

It is interesting to note that $\displaystyle \frac{dm}{dt} = \frac{\partial m}{\partial t} + \left\{\vec\nabla_{\displaystyle \vec s} m \right\}\cdot \vec{v} + \left\{ \vec{\nabla}_{\displaystyle \vec v} m \right\} \cdot \vec{a}$ IF the mass only depends on time, position and velocity. In this case it is clear that whenever we obtain

(8) $\displaystyle \frac{\partial m}{\partial t} + \left\{\vec\nabla_{\displaystyle \vec s} m \right\}\cdot \vec{v} + \left\{ \vec{\nabla}_{\displaystyle \vec v} m \right\} \cdot \vec{a} = 0$

then we might actually write $\vec{F} = m \vec{a}$.

Interesting to note is the relation $\displaystyle \vec{F} \cdot \vec{v} = \frac{dE}{dt}$. As $\displaystyle \vec{F} = \frac{d\vec{p}}{dt}$ and $\vec{p} = m \vec{v}$, we obtain

(9) $\displaystyle \frac{d\vec{p}}{dt} \cdot \vec{p} = m \frac{dE}{dt}$,

thus

(10) $\displaystyle \frac{d}{dt} \left( \frac{1}{2} p^2 \right) = \frac{d}{dt} \left( m E \right) - \left( \frac{dm}{dt} \right) E$,

where $p^2 = \vec{p} \cdot \vec{p}$. Therefore we obtain

(11) $\displaystyle \frac{d}{dt} \left( m E - \frac{1}{2} p^2 \right) = \left( \frac{dm}{dt} \right) E$.

The case $\displaystyle \frac{dm}{dt} = 0$ gives that $\displaystyle m E = k + \frac{1}{2} p^2$, which gives for $k=0$ the relation $\displaystyle E = \frac{1}{2} m v^2$, where $v^2 = \vec{v} \cdot \vec{v}$, also known as the Classical Kinetic Energy. Sofar some definitions involving Classical Mechanics.

## Fractional calculus

In calculus we consider the idea $\displaystyle f'(x) = \lim_{\Delta x \rightarrow 0} \frac{f(x+\Delta x) - f(x)}{\Delta x}$ where it is assumed that for certain $f(x)$ such a limit is defined. An alternative notation is $\displaystyle f'(x) = \frac{df(x)}{dx}$ or $\displaystyle f'(x) = \left(\frac{d}{dx}\right) f(x)$. We may consider $\displaystyle \frac{d}{dx}$ as an operator, let $\displaystyle \mathfrak{D}_x = \frac{d}{dx}$. Then $f'(x) = \mathfrak{D}_x f(x)$. We can also say that $f''(x) = \mathfrak{D}_x f'(x) = \mathfrak{D}_x ^2 f(x)$.

As simple example, say that $f(x) = x^3$ then we obtain that

(1.1) $\displaystyle \mathfrak{D}_x x^3 = \lim_{\Delta x \rightarrow 0} \frac{(x+\Delta x)^3 - x^3}{\Delta x} = \lim_{\Delta x \rightarrow 0} \frac{x^3 + 3 x^2 \Delta x + 3 x \Delta^2 x + \Delta^3 x - x^3}{\Delta x}$

therefore

(1.2) $\displaystyle \mathfrak{D}_x x^3 = \lim_{\Delta x \rightarrow 0} \left( 3 x^2 + 3 x \Delta x + \Delta^3 x\right) = 3 x^2$

Thus $\mathfrak{D}_x x^3 = 3 x^2$ as is well-known. Now we may wonder, what is $\mathfrak{D}_x^k x^y$ where $k$ is a positive integer and $n$ is any integer, it is clear that

(2) $\mathfrak{D}_x^k x^n = \mathfrak{D}_x^{k-1} n x^{n-1} = \mathfrak{D}_x^{k-2} n (n-1) x^{n-2} = \cdots = n (n-1) \cdots (n-k+1) x^{n-k}$

So we may write

(3) $\displaystyle \mathfrak{D}_x^k x^n = \frac{n!}{(n-k)!} x^{n-k}$

Now we use the definition of $\Gamma$ and we know that $\Gamma(n+1) = n!$ so we can extend the relation as

(4) $\displaystyle \mathfrak{D}_x^z x^y = \frac{\Gamma{(y+1)}}{\Gamma(y-z+1)!} x^{z-y}$

We now have a definition for the fractional operator $\mathfrak{D}^z$, as acting on $x^y$. Another example, consider $\mathfrak{D}_x^{1/2} x^4$ then we obtain

(5.1) $\displaystyle \mathfrak{D}_x^{1/2} x^4 = \frac{\Gamma(5)}{\Gamma(9/2)} x^{7/2}$

Now we continue and we have

(5.2) $\displaystyle \mathfrak{D}_x^{1/2} \frac{\Gamma(5)}{\Gamma(9/2)} x^{7/2} = \frac{\Gamma(5)}{\Gamma(9/2)} \frac{\Gamma(9/2)}{\Gamma(4)} x^3 = 4 x^3$

Therefore

(5.3) $\mathfrak{D}_x^{1/2} \mathfrak{D}_x^{1/2} x^4 = 4 x ^3$

as expected as $\mathfrak{D}_x^{1/2} \mathfrak{D}_x^{1/2} x^4 = \mathfrak{D}_x x^4 = 4 x^4$. similar we find that

(6) $\mathfrak{D}_x^z \exp(ax) = a^z \exp(ax)$

We now can consider something like $\mathfrak{D}_x^z \cos(ax)$ and we find that

(7.1) $\mathfrak{D}_x^z \cos(ax) = \mathfrak{D}_x^z \frac{1}{2} \left( \exp(\textbf{i}ax) + \exp(-\textbf{i}ax) \right)$ so

(7.2) $\mathfrak{D}_x^z \cos(ax) = \frac{1}{2} \left\{ (\textbf{i}a)^z \exp(\textbf{i}ax) + (-\textbf{i}a)^z \exp(-\textbf{i}ax) \right\}$ then

(7.3) $\begin{array}{rcl} \mathfrak{D}_x^z \cos(ax) &=& \frac{1}{2} a^z \left\{ \left( \cos(z\pi) + \textbf{i} \sin(z\pi) \right) \left(\cos(ax) + \textbf{i} \sin(ax)\right) \right. \\&& \hspace{2em} + \left. \left(\cos(z\pi) - \textbf{i} \sin(z\pi) \right) \left(\cos(ax) - \textbf{i} \sin(ax)\right) \right\} \end{array}$ therefore

(7.4) $\mathfrak{D}_x^z \cos(ax) = \cos(z\pi) \cos(ax) - \sin(z\pi) \sin(ax) = \cos(ax - z\pi)$

Similar we have $\mathfrak{D}_x^z \sin(ax) = \sin(ax - z\pi)$, for fractional calculus we have

(8.1) $\displaystyle \mathfrak{D}_x^z x^y = \frac{\Gamma{(y+1)}}{\Gamma(y-z+1)!} x^{z-y}$
(8.2) $\mathfrak{D}_x^z \exp(ax) = a^z \exp(ax)$
(8.3) $\mathfrak{D}_x^z \cos(ax) = \cos(ax - z\pi)$
(8.4) $\mathfrak{D}_x^z \sin(ax) = \sin(ax - z\pi)$.

Now using the operators $\mathfrak{D}_x^z$, we may also consider some operator like

(9) $\displaystyle \mathfrak{D}_{x;f(z)} = \int_0^\infty dz f(z) \mathfrak{D}_x^z$.

The $\exp(ax)$ is then fun part, as we have

(10.1) $\displaystyle \mathfrak{D}_{x;f(z)} \exp{ax} = \int_0^\infty dz f(z) \mathfrak{D}_x^z \exp{ax}$ thus

(10.2) $\displaystyle \mathfrak{D}_{x;f(z)} \exp{ax} = \left( \int_0^\infty dz f(z) a^z \right) \exp{ax}$

Note the form $\mathfrak{D}_{x;f(z)} \mathfrak{e} = \mathfrak{v} \mathfrak{e}$ – thus $\mathfrak{e}$ are eigenvectors and $\mathfrak{v}$ are eigenvalues.

Fractional calculus is an extension for normal calculus. The purpose of this post is simply to give some idea about fractional calculus. Have fun!

## What amount of energy is required to maintain constant acceleration?

Given is a body $\mathcal{B}$ that has a constant acceleration. Let the acceleration be $a$. Let us assume that the body starts at rest. If so then the travelled distance is given by $s(t) = \frac{1}{2} a t^2$ and the velocity is given by $v(t) = at$. So there is a critical time $t_c$ for what the velocity becomes the speed of light. This critical time is given by $\displaystyle t_c = \frac{c}{a}$. However, to maintain the constant acceleration $a$, energy is required. By definition, the amount of energy is given by

(1) $\displaystyle W = \int_{s_0}^{s_t} d\vec{s} \vec{F}$

The force is given by $\displaystyle \vec{F} = \frac{d\vec{p}}{dt}$, valid both for Classical Mechanics and Special Relativity. The difference is however, that the momentum is different for both Classical Mechanics and Special Relativity. For Classical Mechanics, the momentum is given by $\vec{p} = m \vec{v}$ and for Special Relativity, the momentum is given by $\displaystyle \vec{p} = \frac{m_0 \vec{v}}{\sqrt{1 - v^2/c^2}}$. Let us assume that the force is acting in the direction of the velocity, then we obtain

(2.1) $\vec{F}_\textrm{C} = m \vec{a}$ and

(2.2) $\displaystyle \vec{F}_\textrm{R} = \frac{m_0 \vec{a}}{\sqrt{1 - v^2/c^2}^3}$

Then it is clear that

(3.1) $\displaystyle W_\textrm{C} = \int_{s_0}^{s_t} ds m a = \left[ \frac{1}{2} m v^2 \right]_{v_0}^{v_t}$ and

(3.2) $\displaystyle W_\textrm{R} = \int_{s_0}^{s_t} ds \frac{m_0 a}{\sqrt{1 - v^2/c^2}^3} = \left[ \frac{m_0 c^2}{\sqrt{1 - v^2/c^2}} \right]_{v_0}^{v_t}$.

Using that $v_0=0$ and $v_t = a t$ we obtain

(4.1) $\displaystyle W_\textrm{C} = \frac{1}{2} m a^2 t^2$ and

(4.2) $\displaystyle W_\textrm{R} = \frac{m_0 c^2}{\sqrt{1 - \left( \displaystyle \frac{at}{c} \right)^2}} -1$

Then we obtain a special time for Special Relativity – where the amount of energy goes to infinity. It is clear that this time is given by $\displaystyle t_{W_\infty} = \frac{c}{a}$. Then we see that $t_c = t_{W_\infty}$, meaning that the critical time is reached when an infinite amount of energy is required to maintain the constant acceleration of the body $\mathcal{B}$. Therefore in Special Relativity, a body $\mathcal{B}$ can have a constant acceleration $a$ during some time-interval that is given by $[0,c/a]$; assuming that the body $\mathcal{B}$ starts at rest. Since an infinite amount of energy is already required to maintain this interval, exceeding the time $c/a$ is not possible. In Special Relativity a body can have a constant acceleration, but not for ever.

Consider a body $\mathcal{B}$ that has a constant acceleration. Does body $\mathcal{B}$ exceeds the speed of light? Let the constant acceleration be $a$. Assume that body $\mathcal{B}$ starts at rest, then the travelled distance is given by $\displaystyle s(t) = \frac{1}{2} a t^2$ and $v(t) = a t$. So the critical time is given by $\displaystyle t_c = \frac{c}{a}$. So it looks like as if such a body would exceed the speed of light.
But HOW can body $\mathcal{B}$ have a constant acceleration? The force acting on a body is given by $\displaystyle \vec{F} = \frac{d\vec{p}}{dt}$, and $\displaystyle \vec{p} = \frac{m_0 \vec{v}}{\sqrt{1 - v^2/c^2}}$. If the force is acting in the direction of the velocity, then we have
(1) $\displaystyle F = \frac{m_0 a}{\sqrt{1 - v^2/c^2}^3}$