Nice article explaining a geometrical derivation of Kepler First law (planets move in ellipses around the sun)

November 8, 2019 zimblesLeave a comment

Click to access Hodograph.pdf

Nice image about derivation

March 18, 2018 zimblesLeave a comment

Understanding the dirac equation

January 13, 2017January 16, 2017 zimblesLeave a comment

$(i \gamma^{\mu} \delta_{\mu} - m) \psi = 0$

i is the lateral number (complex number). It is the solution to equation: $x^2+1=0$
m is the ‘mass’ of the particle
$\psi$ is the solution of the wave equation: ie a linear combination of the equation: $\psi = e^{i(kx-\omega t)}$
The ‘delta’ is na ‘operator’. It is defined as: $\delta_{\mu} =( \frac {\partial}{\partial t} , \frac {\partial}{\partial x}, \frac {\partial}{\partial y}, \frac {\partial}{\partial z})$
The ‘gamma’ are 4 matrices. They are defined as:

There are two new entities in the formula: the Delta-operator and the Gamma-matrices:

Delta-operator:

Just like a function has a number as input and another number as output, a ‘operator’ takes a function as input and another function as output. The delta differentiates a 4-vector vs their coordinates. In other ways it eats a function that assigns a number to all coordinates in 4-space and gives the slope at every coordinatie in 4-space. It looks like the nabla operator, or the regular differntial operator. $\delta_{\mu} \approx \nabla \approx \frac {\partial}{\partial x}$ .

Gamma-matrices:

Where the gamma factor is defined to be non-abelian, or non-commutative, i.e.:

$\gamma_0 = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix}$

$\gamma_1 = \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{bmatrix}$

$\gamma_2 = \begin{bmatrix} 0 & 0 & 0 & -i \\ 0 & 0 & i & 0 \\ 0 & i & 0 & 0 \\ -i & 0 & 0 & 0 \end{bmatrix}$

$\gamma_3 = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{bmatrix}$

$(i \gamma^{\mu} \delta_{\mu} - m) \psi = 0$

The gamma-matrices are related to Pauli spin matrices

$\sigma_x = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$

$\sigma_y = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}$

$\sigma_z = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$

The relation between these Pauli Spin matrices and the Gamma Matrices is according to:

$\gamma_0 = \begin{bmatrix} 0 & I \\ I & 0 \end{bmatrix}$

$\gamma_i = \begin{bmatrix} 0 & \sigma_i \\ \sigma_i & 0 \end{bmatrix}$

The gamma-matrices are the central thing that Dirac added. The reason because the algebra with gamma-matrices solves the problem people had in finding a partial differential equation first order in time and space. Schrodinger equation is first order in time and second order in space, and Klein-Gordon equation is second order in both time and space.

The complete written out Dirac Equation:

So the actual meaning of this equation:

$(i \gamma^{\mu} \delta_{\mu} - m) \psi = 0$

Is the following collection of equations:

$i\frac{\partial \psi^1 }{\partial t} - m \psi^1 + i \frac{\partial \psi^3}{\partial z} + i \frac{\partial \psi^4}{\partial x} +\frac{\partial \psi^4}{\partial y} = 0$

$i\frac{\partial \psi^2 }{\partial t} - m \psi^2 - i \frac{\partial \psi^4}{\partial z} + i \frac{\partial \psi^1}{\partial x} - \frac{\partial \psi^3}{\partial y} = 0$

$-i\frac{\partial \psi^3 }{\partial t} - m \psi^3 - i \frac{\partial \psi^1}{\partial z} -i \frac{\partial \psi^2}{\partial x} - \frac{\partial \psi^2}{\partial y} = 0$

$-i\frac{\partial \psi^4 }{\partial t} - m \psi^4 + i \frac{\partial \psi^2}{\partial z} -i \frac{\partial \psi^3}{\partial x} +\frac{\partial \psi^1}{\partial y} = 0$

Written the Dirac equation like this, we get:

$i \gamma^{\mu} \delta_{\mu} \psi = m \psi$

the above set of equations are:

$i\frac{\partial \psi^1 }{\partial t} + i \frac{\partial \psi^4}{\partial x} +\frac{\partial \psi^4}{\partial y} + i \frac{\partial \psi^3}{\partial z} = m \psi^1$

$i\frac{\partial \psi^2 }{\partial t} - i \frac{\partial \psi^1}{\partial x} - \frac{\partial \psi^3}{\partial y} + i \frac{\partial \psi^4}{\partial z} = m \psi^2$

$-i\frac{\partial \psi^3 }{\partial t} - i \frac{\partial \psi^2}{\partial x} - \frac{\partial \psi^2}{\partial y} - i \frac{\partial \psi^1}{\partial z} = m \psi^3$

$-i\frac{\partial \psi^4 }{\partial t} -i \frac{\partial \psi^3}{\partial x} +\frac{\partial \psi^1}{\partial y} + i \frac{\partial \psi^2}{\partial z} = m \psi^4$

The relation between the Dirac Equation and the Klein-Gordon Equation:

$i \gamma^{\mu} \delta_{\mu} \psi = m \psi$

Let’s start with the Dirac equation and ‘square’ it, or multiply it with itself:

$i \gamma^{\mu} \delta_{\mu} \psi * i \gamma^{\mu} \delta_{\mu} \psi = m \psi * m \psi$

$-\gamma^{\mu} * \gamma^{\mu} \delta_{\mu}^2 \psi = m^2 \psi^2$

Now we can see why the gamma-matrices are defined the way they are, since:

$\gamma^{\mu}*\gamma^{\mu} = 1$

Thus, rewriting:

$-\delta_{\mu}^2 \psi^2 = m^2 \psi^2$

$-( \frac{\partial \psi}{\partial t} - \nabla^2 \psi) = m^2 \psi^2$

$latex – \frac{\partial^2 \psi}{\partial t^2} + \nabla^2 \psi = m^2 \psi^2 &S=4$

Sources, Dirac giving lectures and:

Why derivations are hard

January 13, 2017 zimblesLeave a comment

Because of notation.

But proper notation can make derivations very quick and easy.

Often the underlying concepts and logic can be VERY easy, but notation can be VERY hard.

I never found a correct derivation of E=mc2

January 13, 2017January 13, 2017 zimblesLeave a comment

So there are lots of derivations of this equation floating around on the web. But so often they are wrong! They often derive E=delta mc^2, which is actually not that hard.

A really ugly derivation uses the Taylor expansion of the relativistic kinetic energy equation and then just says that the first term is e=mc^2. But you cant get it like that!

Examples of wrong derivations:

An example is:

Click to access 0308039.pdf

This derives the delta equation.

Another example: this is even a whole webpage devoted to ‘explaining’ E=mc2:

http://www.emc2-explained.info/Emc2/Derive.htm#.WHjnu1wpm7A

WRONG!!! Arg so frustrating!! Even de advanced page is wrong: http://www.emc2-explained.info/Emc2/Deriving.htm#.WHjn91wpm7A

This is derivation of E=delta mc^2 instead of E=mc^2. In other words, the equation derived tells us that an object with mass M and kinetic energy E has extra inertia equal to E/c^2. It doesn’t say why ‘rest mass’ is equal to energy. I want to know!

http://www.askamathematician.com/2011/03/q-why-does-emc2/

This guy seems to have a lot of knowledge, but still derives it wrong! This is with the infamous Taylor expansion.

Now one of the most sad things I know is that even in the great Feynman lectures it is derived wrongly with the Taylor expansion…

http://www.feynmanlectures.caltech.edu/I_15.html

Better derivations:

http://www.adamauton.com/warp/emc2.html

Although this derivation doesn’t really show how spacetime geometry shows the equivalence of mass and energy. It just shows that photons have some apparant inertia.

Another awkard derivation is just postulating the existence of 4-vectors, then defining relativistic momentum as mass times the four vector, and then calling one of the components (the ‘time’) component as energy. Then saying, see: E=mc^2. It is correct but really unsatisfactory and artificial.

I think the best derivaiton is with the Lorentz/Poinceare group and showing a symmetry in spacetime. But I don’t know that one yet. I believe that one can be even pretty simple if you understand the notation.

The laws of planetary motion.

May 29, 2016 zimbles1 Comment

Copernicus formulated his planetary laws in the 1500’s as follows:

The planetary orbit is a circle
The Sun is at the center of the orbit
The speed of the planet in the orbit is constant

These are only true for orbital paths with no eccentricity, i.e. circular orbital paths. They were wrong, but nevertheless revolutionary. Earth was not in the center of the universe anymore!

The laws of Copernicus are superseded by the laws of Kepler’s corrections in the around 1600:

The planetary orbit is not a circle, but an ellipse. The Sun is not at the center but at a focal point of the elliptical orbit.
Neither the linear speed nor the angular speed of the planet in the orbit is constant, but the area speed is constant.
(Extra) The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit

Angular momentum ‘protects’ the orbit.

The angular momentum ‘protects’ the planet form spiralling inward. Because if angular momentum was not conserved, then this would happen ….

What the angular momentum does is it creates a barrier that prevents the Earth from collapsing into the Sun as one may expect. The reason for this is because at smaller distances, the angular momentum term in the energy dominates over the potential energy term thus creating this barrier.

The orbit of every planet is an ellipse with the Sun at one of the two foci.

This is the simplest law, but it is notoriously hard and opaque to derive. Calculus is to be used. I don’t know of any geometric proof.

With the use of a ‘hodograph‘ is maybe the ‘easiest’. Or with the use of abstract algebra, which is suprising.

Carl Runge and Wilhelm Lenz much later identified a symmetry principle in the phase space of planetary motion (the orthogonal group O(4) acting) which accounts for the first and third laws in the case of Newtonian gravitation, as conservation of angular momentum does via rotational symmetry for the second law.^[17]

A line joining a planet and the Sun sweeps out equal areas during equal intervals of time

$\text{Area} = A = \frac {1}{2} \Delta x r$

$L = mvr = \text{constant}$

$L = m\frac {\Delta x}{\Delta t} r = m \frac {\Delta x r}{\Delta t} = m \frac {2A}{\Delta t}$

$\text{constant} = \frac {\Delta A}{\Delta t}$

It seems to me that this is independent of the path taken. So for any curve the area swept is equal time.

The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

$F_\text{centripetal} = \frac {mv^2}{r} = m \omega^2 r$

$\omega = \text{angular velocity}$

$F_\text{gravity} = G \frac{mM}{r^2}$

Combining:

$m \omega^2 r = G \frac{mM}{r^2} \Rightarrow \omega^2 r^3 = GM$

The square of the angular speed times the cube of the orbital radius is constant.

Simple derivation of Newtonian mechanics from principle of least action.

May 15, 2016May 19, 2016 zimblesLeave a comment

I’m still in the process of completing this introductory post to lagrangians.

Overview of the concepts:

The lagrangian is the difference between the Kinetic energy and the potential energy:

$\text{Lagrangian} = L = KE - PE$

The action of a pathline is the Lagrangian times the Time.

$\text{Action} = S = \text{ (Kinetic energy - Potential energy) x the Time}$

$dS = L dt$

In no potential:

$\Delta S = L \Delta t = KE * \Delta t$

$\Delta S = \frac {1}{2}m (\frac { \Delta x}{ \Delta t})^2 * \Delta t$

In a potential:

$\Delta S = L \Delta t = (KE - PE) * \Delta t$

$\Delta S = (\frac {1}{2}m (\frac { \Delta x}{ \Delta t})^2 - V )* \Delta t$

These are abstract concepts, because we don’t experience them in everyday life. The action and lagrangian initially make no sense. But the reason the laws of physics are the way they are, is because of these concepts. The familiar newton laws betray these deeper lying abstract concepts that literally rule this world.

I will apply these concepts hopefully in a trivial way and with actual calculations to make them familiar to you. The three hypothetical situations I apply them to are:

In rest
In multiple constant velocities
In acceleration

Since most of the motion are combinations of these three elements, this will take you a long way of truly understanding the Newton’s laws.

We will consider the mass to be 1 kilogram.

Relationships:

The action scales quadratically with the space, inversely with the time and negative linearly with the potential.

At rest

What is the action of an object at rest? Lets calculate it:

$\Delta x = 0 \text{ meters}$

$\Delta t = 2 \text{ seconds}$

Lets calculate the Lagrangian:

$L = KE - PE = KE$

$L = \frac {1}{2} m (\frac { 0 }{ \Delta t})^2 = 0$

Now we can calculate the action:

$\Delta S = L \Delta t = 0 \Delta t = 0$

Conclusion: An object at rest results in zero action. This makes intuitive sense.

At constant speed (First law)

What is the action of an object at constant speed going from 0 to 10 meters in 5 seconds?

The values of the variables in question are: dt = 5 seconds, dx = 10 meters.

$\Delta x = 10 \text{ meters}$

$\Delta t = 5 \text{ seconds}$

Lets calculate the Lagrangian:

$L = KE - PE = KE$

$L = \frac {1}{2} m (\frac { 10 }{ 5 })^2 = 2m$

Now we can calculate the action:

$\Delta S = L \Delta t = 2m * 5 = 10m$

What is the action of an object at constant speed going from 10 meters to 20 meters in 2 seconds?

The values of the variables in question are: dt = 2 second, dx = 10 meters.

$\Delta x = 20 - 10 = 10 \text{ meters}$

$\Delta t = 2 \text{ seconds}$

Lets calculate the Lagrangian:

$L = KE - PE = KE$

$L = \frac {1}{2} m (\frac { 10 }{ 2 })^2 = 12.5m$

Now we can calculate the action:

$\Delta S = L \Delta t = 12.5m* 2 = 25m$

What is the action of an object going at constant speed from 0 to 10 meters in 2 seconds and then going from 10 to 20 meters in 1 second.

$\Delta S = S_1 + S_2 = 10m+ 25m= 35m$

What is the action of an object going at constant speed from 0 to 20 meters in 7 seconds?

$\Delta t = 7$

$\Delta x = 20$

$L = \frac {1}{2} m (\frac { 20}{ 7 })^2 = 4.1m$

$\Delta S = L \Delta t = 4.1m * 7 = 28.6m$

Going with a constant velocity results in lower action than in speeding up and slowing down. This is the preferred path if there is no potential.

Overview:

At rest: $\Delta S = 0$

10 meters:

In 5 seconds: $\Delta S = 10m$

In 2 seconds: $\Delta S = 25m$

20 meters:

The first 10 in 5 seconds and the second 10 in 2 seconds:

$\Delta S = 10m + 25m = 35m$

The whole 20 meters in 7 seconds at constant speed.

$\Delta S = 28.6m$

I hope you can see what I am trying to show you. Changing speed ‘costs’ more Lagrangians and is therefore avoided by nature. This is expressed in the first law of Newton.

{Here images}

In acceleration (Second law):

So when does is pay off to change your speed? In the case there is a potential. Then it is profitable for nature to take a different path.

Let’s say that there is a potential at BC of 25m.

The kinetic energy at BC was 50m but the lagrangian and thus the action changes like so:

$L = KE - PE = 50m - 25m = 25m$

$S_{BC} = L* \Delta t = 25m * 1 = 25m$

$S =S_{AB} + S_{BC} = 25m + 25m = 50m$

This is identical to refracting light! The pathline of the particle ‘refracts’. Instead of a different refractive medium we have a different ‘potential’.

Short derivation of the Schrödinger equation

May 11, 2016May 23, 2016 zimblesLeave a comment

Click to access 0610121.pdf

https://plus.maths.org/content/schrodinger-1

https://plus.maths.org/content/schrodingers-equation-what-does-it-mean

https://plus.maths.org/content/schrodiQngers-equation-action

Schrödinger equation:

If you want to know how a soccer ball behaves when kicked, you use Newton’s equations, if you want to know how an electron behaves in the orbit of an atom, you use the Schrödinger equation.

It has input the time and position of a particle and as output the probability you will find it there.

Energy conservation equation:

$E = \text{KE}+ \text{PE}$

$E = 1/2 mv^2+ V$

Momentum definition:

$p = mv$

$E = \frac {p^2}{2m} + V(x)$

Wave equation:

$\psi = e^{i(kx-\omega t)}$

Second derivative with respect to distance.

$\psi_x = ik e^{i(kx-\omega t)} = ik \psi$

$\psi_{xx} = - k^2 e^{i(kx-\omega t)} = -k^2 \psi$

Second derivative with respect to time.

$\psi_t = -i\omega e^{i(kx-\omega t)} = -i\omega \psi$

$\psi_{tt} = \omega^2 e^{i(kx-\omega t)} = \omega^2 \psi$

Quantum mechanical postulate:

$p = \frac {h}{\lambda} \Rightarrow \lambda = \frac {h}{p}$

Wave number k:

$k = \frac {2\pi}{\lambda} = \frac {2\pi}{\frac {h}{p}}= \frac {2\pi p}{h}$

$k = \frac {p}{\hbar}$

$p = \hbar k$

Wave equation:

$\psi_{xx} = - (\frac {p}{\hbar})^2 \psi$

$- (\hbar)^2 \psi_{xx} = p^2 \psi$

Multiply wave equation with the Energy conservation equation:

$E\psi = \frac {p^2}{2m}\psi + V(x)\psi$

Plug in to get the Time-Independent wave equation:

$E\psi = \frac {- \hbar^2} {2m}\psi_{xx} + V(x)\psi$

Lets find a beter definition for E \psi:

$E = hf = \hbar \omega$

$E\psi = \hbar \omega \psi$

$\psi_t = -i\omega \psi \Rightarrow \frac {1}{-i\omega} \psi_t = \frac {i}{\omega} \psi_t=\psi$

$E\psi = \hbar \omega \psi = \hbar \omega \frac {i}{\omega} \psi_t= i \hbar \psi_t$

Substitute in the time-independent wave equation to get the time-dependent wave equation:

$E\psi = \frac {- \hbar^2} {2m}\psi_{xx} + V\psi$

$i \hbar \psi_t = \frac {- \hbar^2} {2m}\psi_{xx} + V\psi$

And in 3D with fancy notation:

$i\hbar \frac{\partial}{\partial t}\Psi(\mathbf{r},\,t)=-\frac{\hbar^2}{2m}\nabla^2\Psi(\mathbf{r},\,t) + V(\mathbf{r})\Psi(\mathbf{r},\,t)$

As you can see it is not equal to the wave equation. So officially it is a ‘diffusion’-equation. The Dirac-equation is a wave equation though. I am making a post concerning the relation between schrödinger equation and continuity equations.

Following up on de Broglie’s ideas, physicist Peter Debye made an offhand comment that if particles behaved as waves, they should satisfy some sort of wave equation. Inspired by Debye’s remark, Schrödinger decided to find a proper 3-dimensional wave equation for the electron. He was guided by William R. Hamilton’s analogy between mechanics and optics, encoded in the observation that the zero-wavelength limit of optics resembles a mechanical system — the trajectories of light rays become sharp tracks that obey Fermat’s principle, an analog of the principle of least action.

What is the stress tensor?

May 10, 2016May 19, 2016 zimbles1 Comment

Overview:

What is stress?
What is the stress tensor?
Why do we split the stress-tensor?
What is the divergence of the stress tensor?
The stress-strain relationship.

Stress

Stress is force per area. You are more familiar probably with pressure, which is also force per area. Pressure is a type of stress. There is another type called ‘shear stress´.

The two types of stress:

Compressive stress (pressure). The force is parallel to the normal vector. $P$
Shear stress. The force is perpendicular to the normal vector. An example is friction when riding a bike or something. $\tau$

Force is a vector and area is also a vector. Why is area a vector? It has a magnitude (how big the area is) and a direction (normal to that surface).

You can have two kinds of stress, both are familiar to you:

$P = {F_\text{normal} \over A} \qquad \text{and} \qquad \tau = {F_\text{parallel} \over A}$

However, bear in mind that these definitions are not always valid; (1) assumes that the deformations are infinitesimal and (2) assumes that the solid is elastic (obeys Hooke’s law) and isotropic.

The stress tensor: If you have a scalar conserved quantity, the current density of the charge is a vector. If you have a vector conserved quantity (like momentum), the current density of momentum is a tensor, called the stress tensor.

The stress tensor has input two vectors (force and area) and returns one vector (stress).

Notation:

Tij is nothing more or less than the flow of i-momentum across surfaces with normal j.

$P_{x}=\sigma_{xx} = {F_x \over A_x} \qquad \text{and} \qquad \sigma_{xy}= \tau_{xy} = {F_y \over A_x}$

The first subscript denotes the direction of the normal vector for the surface and the second subscript denotes the direction of the force on that surface.

So a subscript of ‘xx’ means that the surface normal vector points into the x-direction and the force also points into that x-direction. A subscript ‘xy’ denotes on that same x-surface an y-force.

$\vec{\vec {\bf \sigma}} = \begin{bmatrix} \sigma_{xx}& \sigma_{xy}&\sigma_{xz} \\ \sigma_{yx}& \sigma_{yy}&\sigma_{yz} \\ \sigma_{zx}& \sigma_{zy}&\sigma_{zz}\end{bmatrix}$

At a point in 3D-space there are three possible surfaces and three possible direction of forces. This would mean that you need nine number to fully specify the stress on a point. You need a tensor to fully specify the ‘stress’ at a point. A vector is magnitude and direction. A tensor is just paired vectors. When do you need paired vectors? For example when writing down the ‘stress’ in a material.

But because at one point:

$\tau_{ij}=\tau_{ji}$

$\vec{\vec {\bf \sigma}} = \begin{bmatrix} P_{x}& \tau_{xy}&\tau_{xz} \\ \tau_{xy}& P_{y}&\tau_{yz} \\ \tau_{xz}& \tau_{yz}&P_{z}\end{bmatrix}$

Split the tensor

We can ‘split’ the stress-tensor in a ‘pressure’-tensor and a ‘shear’-tensor.

Pressure is defined as such:

$p = \frac {1}{3} (\sigma_{xx} + \sigma_{yy} + \sigma_{zz})$

$\vec{\vec {\bf \sigma}} = - \begin{bmatrix} p&0&0 \\ 0& p&0\\ 0& 0&p\end{bmatrix} + \begin{bmatrix} \sigma_{xx} + p& \sigma_{xy}&\sigma_{xz} \\ \sigma_{yx}& \sigma_{yy} + p &\sigma_{yz} \\ \sigma_{zx}& \sigma_{zy}&\sigma_{zz} + p\end{bmatrix}$

$\vec{\vec {\sigma}} = -p \, {\bf I} + \vec{\vec {\tau}}$

The engine of the stress tensor is entropy or the second law. The probable drive to disorder.

Divergence of the stress tensor:

Imagine an object in a stress tensor field. At every point on the surface of the object is a stress-applied. How do we know what the resultant stress is? The divergence of the stress tensor is the ‘netto-stress’ at that point. We can understand that by applying Gauss’ law:

$\text{Resultant stress} = \int {\vec{\vec {\sigma}}} dS = \int \nabla \cdot {\vec{\vec {\sigma}}} dV$

$\nabla \cdot {\vec{\vec {\sigma}}} = \text{netto-stress}$

$\nabla \cdot {\vec{\vec {\sigma}}} = \nabla \cdot {(-P \, {\bf I} + \vec{\vec {\tau}})} = -\nabla P + \nabla \cdot \vec{\vec {\tau}}$

Explain what the divergence is.

Application of Gauss theorem.

Explain the relation between stress and strain.

$\epsilon_{ij}=\frac{1}{2}\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\right)$

$\sigma_{ij}=2\mu\epsilon_{ij}+\lambda\epsilon_{kk}\delta_{ij}$

test

Can you apply the law of laplace to orbits?

May 6, 2016May 7, 2016 zimblesLeave a comment

Curvature of a curve drawn out by planet has to be proportional to the force experienced. Does this automatically lead to an ellipsical path?

The second derivative of y w.r.t. x is proportional to the force. This means that the second derivative of y w.r.t. x is equal to the second derivative of the y w.r.t. time. A solution is a sine wave. So the y coordinate of makes a sine wave and the x coordinate has a similar argument, also makes a sine wave.

But clearly, the curve at the other side of the ellipse has a high curvature but a low force (and low accerlation?)

This gives the ellipse.

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A writing place.

Category: Derivations

Nice article explaining a geometrical derivation of Kepler First law (planets move in ellipses around the sun)

Nice image about derivation

Understanding the dirac equation

Why derivations are hard

I never found a correct derivation of E=mc2

The laws of planetary motion.

Simple derivation of Newtonian mechanics from principle of least action.

Short derivation of the Schrödinger equation

What is the stress tensor?

Can you apply the law of laplace to orbits?