## What do we do in physics?

Physics is a discipline of science that has become extremely extended. Physics involves both ‘dropping a rock from a tower’ as well as ‘accelerating protons and bring them in collision’. Public has formed a particular vision about physics. I am not going to preach about how public should think about physics. We all have our own opinion about anything and we are free to have an opinion about anything, including an opinion about physics.

As I wrote above: physics is a discipline of science. But what is science? And what is physics? I am not going to promote the ‘official version’ – but rather I am just giving my own vision about it:

# Where does the word ‘science’ come from?

The word “science” comes from the Latin word “scientia” – meaning “knowing”. The word “science” in the German language is given by the word “Wissenschaft”. The word “know” in the German language is given by the word “wissen”. Can you see the relation between the words “wissen” and the word “Wissenschaft”? The word “science” in the Dutch language is given by the word “wetenschap”. The word “know” in the Dutch language is given by the word “weten”. Can you see the relation between the words “weten” and the word “wetenschap”? It may appear that ‘science’ is related to ‘knowing’. But is it?

The word “science” in the Danish language is given by the word “videnskab”. The Danish word “viden” means “knowledge”. But the first part “viden” looks more to words like “video” and “vision” than it does to the word “know”. And words like “video” and “vision” are related to “see”.

When we look to the phrase “I know” in several languages we find the following:

1. Catalonian: “Sé”
2. Italian: “so”
3. Portuguese: “sei”
4. Spanish: “sé”

Why do all these phrases look like “see”? Is there a relation between “know” and “see”?

I do believe that there is a strong relation between the words “know” and “see”. I also believe that due to the “evolution of languages” certain meanings of words have changed. I believe that “science” is more about “see” than it is about “knowing”. And when we make the extension from “see” to “observe”, I would say that “science” is more about “observing” in the first place than it is about “knowing”.

What more do we know from the language? Let us look to The Proto-Indo-European language.

## The Proto-Indo-European words chi and ved.

The Proto-Indo-European word chi means see. Words that are derived from the Proto-Indo-European word chi are the words: see (English), sehen (German), zien (Dutch), se (Danish), all meaning “see”. But also the words scientia (Latin), sapere (Italian), saber (Portuguese and Spanish), are derived from the Proto-Indo-European word chi, but all meaning “know”.

The Proto-Indo-European word ved means know. Words that are derived from the Proto-Indo-European word ved are the words: wissen (German), weten (Dutch), vide (Danish), all meaning “know”. But also the words videre (Latin), vedere (Italian), ver (Portuguese and Spanish), all meaning “see”.

The following table gives a small summary:

 Proto-Indo-European chi ved “see” “know” “know” “see” English see – – – German sehen – wissen – Dutch zien – weten – Danish se – viden – Latin – scientia – videre Italian – sapere – vedere Portugues – saber – ver Spanish – saber – ver

The English word “know” is related to similar words in other languages: kennen (German), kennen (Dutch), kender (Danish).

It appears that the words “know” and “see” are strong related. I therefore believe that science is more related to observe (what I “see”) than it is related to knowledge. So what is science?

Science is about what I have seen.

And that is just my vision about where science is all about. But I also claim that there is a strong relation between “know” and “seen” (read “observed”):

The only thing I know is what I have seen.

And this emphases the strong relation between the words “know” and “seen”.

• I see (reading a book or a web-page) that “Washington is the capital of the United States of America” and therefore I know (that I have seen) that “Washington is the capital of the United States of America”.
• I see (looking at the sky) that “the stars are moving through the sky” and therefore I know (that I have seen) that “the stars are moving through the sky”.
• I see (looking at a stone dropped from a tower) that “stones fall down towards the Earth” and therefore I know (that I have seen) that “that stones fall down towards the Earth”.

What “I know” is what “I have seen” and what “I have seen” is what “I know”. The words “know” and “seen” are synonyms. The word “seen” has to be taken in the broadest meaning of the word, it is not only restricted to “what I have seen with my eyes” – but it also applies to “what I have seen with my brain”, i.e. that “what I have imagined”.

# “Understanding the brain” is part of “understanding science”.

I wrote: “Science is about what I see (observe)” and “The only thing I know is what I see (observe)”. But what does our brain do with this? Our brain is not simply restricted to what we see. Our brain starts a process that classifies what we have seen. When it comes to observations, our brain also classifies certain observations as “not important” and such observations are ignored. This occurs in our daily life: “Did you see that?” – “No I didn’t”. However, “I did see it” – but my brain has not “classified” (read “memorized”) it, because my brain has simple rejected that observation, i.e. I cannot remember it and I would say “I did not see it”. Just go to a movie in the cinema. The next time you see the SAME movie – you will “pay attention” to different elements of the movie. The SAME movie – the SAME observation – yet a different classification is performed by the brain and therefore a different “experience” of the observation is obtained by the brain.

For this reason I would like to propose to call “primary knowledge” as “what we observe” and “secondary knowledge” as “what we do with that what we have observed”. Let us consider an example. The following list refers to “primary knowledge”:

• I have observed a stone falling down towards the Earth.
• I have observed an another stone falling down towards the Earth.
• I have observed yet another stone falling down towards the Earth.
• I have observed again another stone falling down towards the Earth.

However “secondary knowledge” is given by:

• All stones are falling down towards the Earth.

We can refer to “secondary knowledge” as to the “conclusions” we draw from the “primary knowledge”. But can we rely on the ‘conclusion’ drawn by our brain?

## Facts and opinions – true and false – possible and impossible.

What is a fact? A fact is related to “primary knowledge”. When I see “that door is red” – then “that door is red” is a fact. However, it is a fact for me. Meaning that the fact is subjective as it applies to an individual. There can be however no doubt that the fact is true. The words fact and true are synonyms. It is ridiculous to say: “the fact is that door is green while I have seen that door is red. But as the words fact and true are synonyms, the word true applies exclusively to “primary knowledge”.

Facts are true.

What is an opinion? An opinion is related to “secondary knowledge”. When I see “that door is red”; “another door is red”; “yet another door is red” – then “all doors are red” is an opinion. However, it is an opinion for me. Meaning that the opinion is subjective as it applies to an individual. But if the word true applies exclusively to “primary knowledge” – then what can we say about an opinion? We can at best say that an opinion is possible or impossible. Logic is therefore restricted to possible or impossible.

Opinion is about possible and impossible.

But what about the word false? Is false not related to logic? But what is logic? Logic is the process of drawing conclusions and is part of “secondary knowledge” as it involves drawing conclusions from both “primary knowledge” and “secondary knowledge”. But if logic is the process of drawing conclusions that is part of “secondary knowledge” – then only the words possible or impossible apply to logic. There simply is no false!

Logic should be about possible and impossible.

Much confusion in our daily life (including discussion) is due to confusion of facts with opinions. Just remember facts are true while opinions are possible or impossible and there is no such thing as false!

 Classical Fact Opinion True False True False Epistemological Primary knowledge Secondary knowledge True Possible Impossible

Using the previous examples, the following list are facts:

• I have observed a stone falling down towards the Earth.
• I have observed an another stone falling down towards the Earth.
• I have observed yet another stone falling down towards the Earth.
• I have observed again another stone falling down towards the Earth.

All these facts are true. However opinion is given by:

• All stones are falling down towards the Earth.

And such an opinion can be possible. In general (we may assume that) an opinion is possible unless a fact shows otherwise.

Facts and opinion are personal in the first place. However, opinion can be regarded as public as well, in the case that several persons agree about opinion. Such opinions are shared possible by a large group of persons.

Facts are personal – opinions can be public.

That we should be carefull with drawing conclusions is clear. An example:

Person A performs an experiment by dropping stones from a tower. Person A describes this experiment in a document. For person A the fact is “I have observed stones falling down towards the Earth”. Person B reads that document. However, for person B “(I have observed) stones falling down towards the Earth” is NOT a fact. For person B the fact is “I have read a document about dropping stones from a tower”. For person B the “observed stones falling down towards the Earth” is an opinion and person B can say possible or impossible. In case several persons have read the document and all agree with it, the opinion “stones fall down towards the Earth” can become a public opinion. But such a public opinion “stones fall down towards the Earth” is at best possible and is never true!

So what does that mean for science?

Science is about possible and impossible.

Science is about primary knowledge and secondary knowledge, about facts and opinions. However, there are much more opinions than facts in science. And science is not about true, but science is about possible and impossible. Every person can write a document about science, but any reader of such a document can judge possible and impossible and is entitled to do so.

But what about physics? Physics is a discipline of science, so physics is about possible and impossible. But what do we do in physics?

More about that question in another post… Hope you liked this post!!!

## Apostasimetry

In this post I will write about geometry. However, I speak about apostasimetry (Greek: apostasi – “distance”, -metron “measurement”) as I focus on distance.

The concept distance is related to a number that we assign to the gab between two objects, it tells something about “how far apart the two objects are”.

Imagine two rocks on the beach. We measure the distance by putting small pearls between the rocks such that all pearls tough each other. Counting the number of pearls gives a number that we can assign the concept distance. The two rocks are 781 pearls apart. That chain of pearls “connecting” the two rock has a particular shape. We may do the same using another shape and find that the two rocks are 781 pearls apart. This method therefore does not yield a “unique” number and has no practical use. But the question “what is the minimum numbers of pearls required to form a chain between the rocks?” gives a unique number. And THIS is where the concept distance is related to: the minimum number of unit to form a chain between the two objects.

Instead of objects we use ‘points’ denoted by $\mathcal P_\jmath$. The concept distance between the two points $\mathcal{P}_\jmath$ and $\mathcal{P}_k$ is denoted by $D(\mathcal{P}_\jmath,\mathcal{P}_k)$ and represents a number. Shortly we can write as $r_{\jmath k}$ for the distance between the two points $\mathcal{P}_\jmath$ and $\mathcal{P}_k$.

But because of the definition of the concept distance, refering to a ‘minimum’ we obtain a particular relation when we consider three points $\mathcal{P}_\jmath$, $\mathcal{P}_k$ and $\mathcal{P}_\ell$. That relation is given by:

Distance inequality: $r_{\jmath k} + r_{k \ell} \ge r_{\ell \jmath}$

The reason is simple, IF $r_{\jmath k} + r_{k \ell} < r_{\ell \jmath}$, then $r_{\ell \jmath}$ does not represent a distance, for it is not the minimum. Some recognize this inequalicty as the triangular inequality – but it is a fundamental property of the concept distance that follows from definition. I shall refer to it as the “distance inequality”. I shal however write the distance inequality as

Distance inequality: $r_{\jmath k} + r_{k \ell} - r_{\ell \jmath} \ge 0$

What properties can we derive from the distance inequality?

## Distance is allways positive.

The distance inequality is valid for any combination of points, therefore we obtain:

$\begin{array}{rcll} r_{\jmath k} + r_{k \ell} - r_{\ell \jmath} & \ge & 0\\ r_{\ell \jmath} + r_{\jmath k} - r_{k \ell} & \ge & 0\\ \hline &&& + \\ 2 r_{\jmath k} & \ge & 0 \end{array}$

So the distance is allways positive. Ok, we may also argue: “as distance is related to counting – it can only be positive!”

## Distance is bounded.

We can write

$\begin{array}{rcl} r_{\jmath k} + r_{k \ell} - r_{\ell \jmath} & \ge & 0 \\ r_{k \ell} + r_{\ell \jmath} - r_{\jmath k} & \ge & 0 \end{array}$

and therefore

$\begin{array}{rcll} r_{\ell \jmath} - r_{k \ell} & \le & r_{\jmath k}\\ r_{\jmath k} & \le & r_{\ell \jmath} + r_{k \ell} \end{array}$

so

$r_{\ell \jmath} - r_{k \ell} \le r_{\jmath k} \le r_{\ell \jmath} + r_{k \ell}$.

This may not seem like a usefull inequality – however, when $r_{k \ell} = 0$ we obtain

$r_{\ell \jmath} \le r_{\jmath k} \le r_{\ell \jmath}$

and therefore

$r_{\ell \jmath} = r_{\jmath k}$

When all points are equal – we can only write $r_{k k} = \zeta$ and therefore

$r_{k \jmath} - \zeta \le r_{\jmath k} \le r_{k \jmath} + \zeta$.

So the distance is bounded, based on the equality of points.

We also see that IF $\zeta = 0$ then

$r_{k \jmath} \le r_{\jmath k} \le r_{k \jmath}$

thus

$r_{k \jmath} = r_{\jmath k}$

so distances are symmetrical.

## Metrical distances are symmetric.

A metrical distance is defined as a distance such that $r_{k k} = 0$ and therefore

$r_{k \jmath} = r_{\jmath k}$

so metrical distances are symmetrical.

## The trinodal identity.

A trinodal is defined as three points.

Let us consider three points and denote the distances as $x$, $y$ and $z$. These values are all positive. We obtain the three inequalities:

$\begin{array}{rcl} x + y + z & \ge & 0\\ x + y - z & \ge & 0 \\ z + x - y & \ge & 0 \\ y + z - x & \ge & 0 \end{array}$

and therefore

$\begin{array}{rcl} \Big( x + y + z \Big) \Big( x + y - z \Big) \Big( z + x - y \Big) \Big( y + z - x \Big) & \ge & 0 \end{array}$

However, is the “reverse” also valid?

When we look to the signs we have several possibilities:

$\begin{array}{rcl} \Big( + \Big) \Big( + \Big) \Big( + \Big) \Big( + \Big) & \ge & 0\\\\ \Big( + \Big) \Big( + \Big) \Big( - \Big) \Big( - \Big) & \ge & 0\\\\ \Big( + \Big) \Big( - \Big) \Big( + \Big) \Big( - \Big) & \ge & 0\\\\ \Big( + \Big) \Big( - \Big) \Big( - \Big) \Big( + \Big) & \ge & 0 \end{array}$

But due to the symmetry we can permute the symbols such that

$x \ge y \ge z$

thus

$\begin{array}{rcl} y & = & z + p\\ x & = & z + p + q \end{array}$

where both $p$ and $q$ are positive. Therefore we obtain

$\begin{array}{rcl} \Big( 3z + p + q \Big) \Big( 2p + q \Big) \Big( z + q \Big) \Big( y + z - x \Big) & \ge & 0 \end{array}$

Two of the three inequalities are always satisfied. Therefore we can write

$\Big( x + y + z \Big) \Big( x + y - z \Big) \Big( z + x - y \Big) \Big( y + z - x \Big) \ge 0 \Longleftrightarrow \left[\begin{array}{rcl} x + y + z & \ge & 0\\ x + y - z & \ge & 0 \\ z + x - y & \ge & 0 \\ y + z - x & \ge & 0 \end{array}\right.$

Let us therefore define

$H(x,y,z) = \sqrt{ \Big( x + y + z \Big) \Big( x + y - z \Big) \Big( z + x - y \Big) \Big( y + z - x \Big) }$

It is clear that

$\begin{array}{rcl} H(x,y,z) & = & \sqrt{ \Big( x + y + z \Big) \Big( x + y - z \Big) \Big( z + x - y \Big) \Big( y + z - x \Big) }\\ & = & \sqrt{ \Big( [x + y] + z \Big) \Big( [x + y] - z \Big) \Big( z + [x - y] \Big) \Big( z - [x - y] \Big) }\\ & = & \sqrt{ \Big( [x + y]^2 - z^2 \Big) \Big( z^2 - [x - y]^2 \Big) }\\ & = & \sqrt{ \Big( 2 x y + \left[ x^2 + y^2 - z^2 \right] \Big) \Big( 2 x y - \left[ x^2 + y^2 - z^2 \right] \Big) }\\ & = & \sqrt{ \Big( 4 x^2 y^2 - \left[ x^2 + y^2 - z^2 \right]^2 \Big) }\\ & = & \sqrt{ \Big( 2 x^2 y^2 + 2 y^2 z^2 + 2 z^2 x^2 - \left[ x^4 + y^4 + z^4 \right] \Big) }\\ & = & \sqrt{ \Big( \left[ x^2 + y^2 + z^2 \right]^2 - 2 \left[ x^4 + y^4 + z^4 \right] \Big) }\\ \end{array}$

The function $H(x,y,z)$ is named after Hero of Alexandria – as the Heron’s ‘surface’ of a triangle is given by $\frac{1}{4} H(x,y,z)$. Note that we have not defined a concept like ‘surface’ – but that ‘form’ of ‘surface’ is embedded in the function $H(x,y,z)$ and is due to properties of distance.

Every trinodal satisfies

$H(x,y,z) \ge 0$

that I shall call the trinodal inequality.

## The linear trinodal

As we deal with inequalities in the form $\cdots \ge 0$ there is a special case associate with the equality $\cdots = 0$.

I shall call a trinodal linear IF $H(x,y,z) = 0$. A linear trinodal has the property that $H(x,y,z) = 0$. Note that this is just a definition.

Going back to the definition of $H(x,y,z)$ we find for a linear trinodal:

$H(x,y,z) = 0 \Longrightarrow x = y + z \vee y = z + x \vee z = x + y \vee x = y = z = 0$

And this is what we basically have for a line as we use it normal – but now we have it defined from the distance inequality.

The link with normal geometry is that the surface of a line (or a point) is zero and therefore $H(x,y,z) = 0$.

So-far this post – I hope you liked it!

## Screen updating in Excel

The property ScreenUpdating of the object Application tells if the screen is being updated or not. Normaly you have a kind of source-code like

Sub Foo()
Application.ScreenUpdating = False
'   do something here ...
Application.ScreenUpdating = True
End Sub

Now this in general might work. However, when we enter this sub when ScreenUpdating is False – after this sub ScreenUpdating is True! We may solve this using the following code:

Sub Foo()
Dim SaveScreenUpdating
SaveScreenUpdating = Application.ScreenUpdating
Application.ScreenUpdating = False
'   do something here ...
Application.ScreenUpdating = SaveScreenUpdatingEnd Sub

We need 4 extra lines to control the property ScreenUpdating… We may also consider to use the following class

Class ScreenUpdating

Dim TheScreenUpdating As Boolean

Sub Enable()
Let Application.ScreenUpdating = True
End Sub

Sub Disable()
Let Application.ScreenUpdating = False
End Sub

Private Sub Class_Initialize()
Let TheScreenUpdating = Application.ScreenUpdating
End Sub

Private Sub Class_Terminate()
Let Application.ScreenUpdating = TheScreenUpdating
End Sub


and we use the Class ScreenUpdating to write a sub like

Sub Foo()
Dim ScreenUpdating as New ScreenUpdating
ScreenUpdating.Disable
'   do something here ...
End Sub

We only need 2 lines and we don’t need to worry to restore the original value of ScreenUpdating; the usage of the Class ScreenUpdating ensures that ScreenUpdating is restored!

We may improve the sub using a catch-the-error, like

Sub Foo()
On Error Goto Error
Dim ScreenUpdating as New ScreenUpdating
ScreenUpdating.Disable
'   do something here ...
Error:
End Sub

Once this sub is terminated, the original value ScreenUpdating of is restored!

Hope you like this post!

Posted in Excel, VBA | 2 Comments

One can derive the Lorentz transformations without specific using the second postulate of relativity. You can find it in this document: Transformations between inertial systems of reference. Ok it is still a draft:-) Hope you enjoy it!

## Raising the power of a 2×2 matrix

The main question here is:

given a matrix $\left( \begin{array}{cc} a & b \\ c & d \end{array} \right)$ then what is $\left( \begin{array}{cc} a & b \\ c & d \end{array} \right)^\zeta$?

Given is the matrix

(1) . . . $\textbf{A} = \left( \begin{array}{cc} a & b \\ c & d \end{array} \right)$.

It is clear that

$\begin{array}{rcl} \textbf{A}^2 &=& \left( \begin{array}{cc} a^2 + bc & ab+db \\ ac+dc & d^2 + bc \end{array} \right)\\\\ &=& \left( \begin{array}{cc} a^2 + ad - (ad-bc) & ab+db \\ ac+dc & d^2 + ad - (ad-bc) \end{array} \right)\\\\ &=& \left( \begin{array}{cc} (a+d)a & (a+d)b \\ (a+d)c & (a+d)d \end{array} \right) - \left( \begin{array}{cc} ad-bc & 0 \\ 0 & ad-bc \end{array} \right) \\\\ &=& (a+d) \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) - (ad-bc) \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right) \end{array}$

This can be written as $\textbf{A}^2 = (a+d) \textbf{A} - (ad-bc) \textbf{1}$. Note that $a+d = \textrm{Tr}(\textbf{A})$ and $ad-bc = \textrm{Det}(\textbf{A})$. We shall write $\chi = a+d$ and $\Delta = ad-bc$. Then we obtain

(2) . . . $\textbf{A}^2 = \chi \textbf{A} - \Delta \textbf{1}$.

The matrix $\textbf{A}$ has eigen-values $\lambda$, defined by $\textrm{Det} \left( \textbf{A} - \lambda \textbf{1}\right) = 0$. This gives the equation

(3) . . . $(a-\lambda)(d-\lambda) - bc = 0$.

Then we obtain $\lambda^2 - \chi \lambda + \Delta = 0$, so

(4) . . . $\lambda_\pm = \chi/2 \pm \sqrt{\chi^2/4 - \Delta}$.

It is clear that

(5a) . . . $\lambda_\pm + \lambda_\mp = \left( \chi/2 \pm \sqrt{\chi^2/4 - \Delta} \right) + \left( \chi/2 \mp \sqrt{\chi^2/4 - \Delta } \right) = \chi$

and

(5b) . . . $\lambda_\pm \lambda_\mp = \left( \chi/2 \pm \sqrt{\chi^2/4 - \Delta} \right) \left( \chi/2 \mp \sqrt{\chi^2/4 - \Delta } \right) = \chi^2/4 - \chi^2/4 + \Delta = \Delta$.

Therefore we can write equation (2) as

(6) . . . $\textbf{A}^2 = \left( \lambda_\pm + \lambda_\mp \right) \textbf{A} - \lambda_\pm \lambda_\mp \textbf{1}$

We now define

(7) . . . $\textbf{B}_\pm = \mp \left( \textbf{A} - \lambda_\pm \textbf{1} \right)$.

It is clear that

$\lambda_\mp \textbf{B}_\pm + \lambda_\pm \textbf{B}_\mp = \left( \pm \lambda_\pm \mp \lambda_\mp \right) \textbf{A}$,

therefore

$\textbf{A} = \displaystyle \frac{ \lambda_\mp \textbf{B}_\pm + \lambda_\pm \textbf{B}_\mp }{\pm \lambda_\pm \mp \lambda_\mp}$,

so we can write

(8) . . . $\textbf{A} = \displaystyle \frac{\lambda_-}{\lambda_+ - \lambda_-} \textbf{B}_+ + \frac{\lambda_+}{\lambda_+ - \lambda_-} \textbf{B}_-$

It is also clear that

$\begin{array}{rcl} \textbf{B}_\pm \textbf{B}_\pm &=& \left\{ \pm \left( \textbf{A} - \lambda_\pm \textbf{1} \right) \right\} \left\{ \pm \left( \textbf{A} - \lambda_\pm \textbf{1} \right) \right\} \\\\ &=& \textbf{A}^2 - 2 \lambda_\pm \textbf{A} + \lambda_\pm^2 \textbf{1} \textrm{ ... using equation (6) for }\textbf{A}^2\textrm{ ...} \\\\ &=& \left( \lambda_\pm + \lambda_\mp \right) \textbf{A} - \lambda_\pm \lambda_\mp \textbf{1} - 2 \lambda_\pm \textbf{A} + \lambda_\pm^2 \textbf{1}\\\\ &=& \left( \lambda_\mp - \lambda_\pm \right) \textbf{A} - \left( \lambda_\pm \lambda_\mp - \lambda_\pm^2 \right) \textbf{1}\\\\ &=& \left( \lambda_\mp - \lambda_\pm \right) \left( \textbf{A} - \lambda_\pm \textbf{1} \right)\\\\ &=& \left( \lambda_+ - \lambda_- \right) \textbf{B}_\pm \end{array}$

and

$\begin{array}{rcl} \textbf{B}_\pm \textbf{B}_\mp &=& \left\{ \pm \left( \textbf{A} - \lambda_\pm \textbf{1} \right) \right\} \left\{ \mp \left( \textbf{A} - \lambda_\mp \textbf{1} \right) \right\} \\\\ &=& - \left\{ \textbf{A}^2 + \left( \lambda_\pm - \lambda_\mp \right) \textbf{A} + \lambda_\pm \lambda_\mp \textbf{1} \right\} \textrm{ ... using equation (6) for }\textbf{A}^2 \textrm{ ...} \\\\ &=& - \left\{ \left( \lambda_\pm + \lambda_\mp \right) \textbf{A} - \lambda_\pm \lambda_\mp \textbf{1} - \left( \lambda_\pm + \lambda_\mp \right) \textbf{A} + \lambda_\pm \lambda_\mp \textbf{1} \right\} = 0 \end{array}$.

Therefore we obtain

(9a) . . . $\textbf{B}_\pm \textbf{B}_\pm = \left( \lambda_+ - \lambda_- \right) \textbf{B}_\pm$

and

(9b) . . . $\textbf{B}_\pm \textbf{B}_\mp = 0$.

As $\textbf{B}_\pm \textbf{B}_\pm = \left( \lambda_+ - \lambda_- \right) \textbf{B}_\pm$ it is clear that

(10) . . . ${\textbf{B}_\pm}^\zeta = \left( \lambda_+ - \lambda_- \right)^{\zeta-1} \textbf{B}_\pm$

It is als clear that IF

$\textbf{A} = \displaystyle \frac{\lambda_-}{\lambda_+ - \lambda_-} {\textbf{B}_+} + \frac{\lambda_+}{\lambda_+ - \lambda_-} {\textbf{B}_-}$

and

$\textbf{B}_\pm \textbf{B}_\mp = 0$

then

$\textbf{A}^\zeta = \displaystyle \left( \frac{\lambda_-}{\lambda_+ - \lambda_-} \right)^\zeta {\textbf{B}_+}^\zeta + \left( \frac{\lambda_+}{\lambda_+ - \lambda_-} \right)^\zeta {\textbf{B}_-}^\zeta$

and as ${\textbf{B}_\pm}^\zeta = \left( \lambda_+ - \lambda_- \right)^{\zeta-1} \textbf{B}_\pm$ we obtain

$\textbf{A}^\zeta = \displaystyle \frac{{\lambda_-}^\zeta}{\lambda_+ - \lambda_-} \textbf{B}_+ + \frac{{\lambda_+}^\zeta}{\lambda_+ - \lambda_-} \textbf{B}_-$.

As $\textbf{B}_+ = - \textbf{A} + \lambda_+ \textbf{1}$ and $\textbf{B}_- = \textbf{A} - \lambda_- \textbf{1}$ we obtain

$\begin{array}{rcl} \textbf{A}^\zeta &=& \displaystyle \frac{{\lambda_-}^\zeta}{\lambda_+ - \lambda_-} \left( - \textbf{A} + \lambda_+ \textbf{1} \right) + \frac{{\lambda_+}^\zeta}{\lambda_+ - \lambda_-} \left( \textbf{A} - \lambda_- \textbf{1} \right)\\\\ &=& \displaystyle \frac{{\lambda_+}^\zeta - {\lambda_-}^\zeta}{\lambda_+ - \lambda_-} \textbf{A} - \lambda_+ \lambda_- \frac{{\lambda_+}^{\zeta-1} - {\lambda_-}^{\zeta-1}}{\lambda_+ - \lambda_-} \textbf{1} \end{array}$

So the general result is

(11) . . . $\displaystyle \textbf{A}^\zeta = \frac{{\lambda_+}^\zeta - {\lambda_-}^\zeta}{\lambda_+ - \lambda_-} \textbf{A} - \lambda_+ \lambda_- \frac{{\lambda_+}^{\zeta-1} - {\lambda_-}^{\zeta-1}}{\lambda_+ - \lambda_-} \textbf{1}$

From this follows

$\begin{array}{rcl} \textbf{A}^0 &=& \textbf{1}\\\\ \textbf{A}^1 &=& \textbf{A}\\\\ \textbf{A}^2 &=& \left( \lambda_+ + \lambda_- \right) \textbf{A} - \lambda_+ \lambda_- \textbf{1} \textrm{ ... equation (6) ...}\\\\ \textbf{A}^3 &=& \left( {\lambda_+}^2 + \lambda_+ \lambda_- + {\lambda_-}^2 \right) \textbf{A} - \lambda_+ \lambda_- \left( \lambda_+ + \lambda_- \right) \textbf{1} \textrm{ ... from equation (6) follows ...}\\\\ &=& \left( \lambda_+ + \lambda_- \right) \textbf{A}^2 - \lambda_+ \lambda_- \textbf{A} \\\\ &=& \left( \lambda_+ + \lambda_- \right) \left\{ \left( \lambda_+ + \lambda_- \right) \textbf{A} - \lambda_+ \lambda_- \textbf{1} \right\} - \lambda_+ \lambda_- \textbf{A} \\\\ &=& \left\{ \left( \lambda_+ + \lambda_- \right)^2 - \lambda_+ \lambda_- \right\} \textbf{A} - \lambda_+ \lambda_- \left( \lambda_+ + \lambda_- \right) \textbf{1}\\\\ &=& \left( {\lambda_+}^2 + \lambda_+ \lambda_- + {\lambda_-}^2 \right) \textbf{A} - \lambda_+ \lambda_- \left( \lambda_+ + \lambda_- \right) \textbf{1} \end{array}$

Given any 2 × 2 matrix $\textbf{A}$, where

$\textbf{A} = \left( \begin{array}{cc} a & b \\ c & d \end{array} \right)$,

then

$\displaystyle \textbf{A}^\zeta = \frac{{\lambda_+}^\zeta - {\lambda_-}^\zeta}{\lambda_+ - \lambda_-} \textbf{A} - \lambda_+ \lambda_- \frac{{\lambda_+}^{\zeta-1} - {\lambda_-}^{\zeta-1}}{\lambda_+ - \lambda_-} \textbf{1}$,

where

$\lambda_\pm = \chi/2 \pm \sqrt{\chi^2/4 - \Delta}$,

where

$\chi = a+d$

and

$\Delta = ad-bc$.

We have found that $\Delta = \lambda_\pm \lambda_\mp$, therefore we can write $\lambda_\pm = \sqrt{\Delta} \exp(\pm\psi)$. Then we find that $\lambda_+ + \lambda_- = 2\sqrt{\Delta} \cosh(\psi)$ and $\lambda_+ - \lambda_- = 2\sqrt{\Delta} \sinh(\psi)$. We obtain

(12) . . . $\displaystyle \textbf{A}^\zeta = \sqrt{\Delta}^{\zeta-1} \left\{ \frac{\sinh(\zeta\psi)}{\sinh(\psi)} \textbf{A} - \sqrt{\Delta} \frac{\sinh(\zeta\psi - \psi)}{\sinh(\psi)} \textbf{I} \right\}$.

As $\lambda_+ + \lambda_- = a + d$ we obtain $a + d = 2\sqrt{\Delta} \cosh(\psi)$, therefore we can write $a = \sqrt{\Delta} (\cosh(\psi) + \mu)$ and $d = \sqrt{\Delta} (\cosh(\psi) - \mu)$. Then we obtain $bc = \Delta (\sinh^2(\psi) - \mu^2 )$, therefore we can write $b = \sqrt{\Delta} \exp(\sigma) (\sinh(\psi) + \mu)$ and $c = \sqrt{\Delta} \exp(-\sigma) (\sinh(\psi) - \mu)$. Then

$\textbf{A} = \sqrt{\Delta} \left(\begin{array}{cc} \cosh(\psi) + \mu & \exp(\sigma) (\sinh(\psi) + \mu) \\ \exp(-\sigma) (\sinh(\psi) - \mu) & \cosh(\psi) - \mu \end{array}\right)$

And when we put this in equation (12) we get

(13) . . . $\displaystyle \textbf{A}^\zeta = \sqrt{\Delta}^\zeta \left(\begin{array}{cc} \displaystyle \cosh(\zeta\psi) + \mu \frac{\sinh(\zeta\psi)}{\sinh(\psi)} & \displaystyle \exp(\sigma) \sinh(\zeta\psi) \left( 1 + \frac{\mu}{\sinh(\psi)} \right) \\ \displaystyle \exp(-\sigma) \sinh(\zeta\psi) \left( 1 - \frac{\mu}{\sinh(\psi)} \right) & \displaystyle \cosh(\zeta\psi) - \mu \frac{\sinh(\zeta\psi)}{\sinh(\psi)} \end{array}\right)$

as the final result.

It is also clear that the one-parameter group of 2 × 2 matrices takes the form

$\textbf{A}(\psi) = \sqrt{\Delta} \left(\begin{array}{cc} \cosh(\psi) + \mu & \exp(\sigma) (\sinh(\psi) + \mu) \\ \exp(-\sigma) (\sinh(\psi) - \mu) & \cosh(\psi) - \mu \end{array}\right)$

where $\psi$ is the group-parameter.

Hope you liked this post!

## Neutrinos faster than light.

Recently experiment has shown that neutrinos move faster then light. Some claim that “neutrinos might break one of the most fundamental laws of physics”. The question would be “what law of physics would be broken?” In general this all is associated with relativity. But there is no law within relativity that tell us that nothing can move faster then light. What relativity tells us is that the speed of light is a special velocity.

Let us consider a body. We can change the velocity of the body by using of a force. The change in energy is by definition given by $d E = \vec{F} d\vec{r}$. Newton tells us that $\displaystyle \vec{F} = \frac{d\vec{p}}{dt}$. Newton tells us also that $\vec{p} = m \vec{v}$, while Einstein tells us that $\vec{p} = m_o \vec{v} / \sqrt{1 - v^2 / c^2}$. So when we follow Newton for the force $\vec{F}$ and Einstein for the momentum $\vec{p}$, we get

(1) $\displaystyle dE = m_o \vec{v} d \left( \frac{\vec{v}}{\sqrt{1 - \vec{v} \cdot \vec{v} / c^2}} \right) = d\left( \frac{m_o c^2}{\sqrt{1 - \vec{v} \cdot \vec{v}/c^2}} \right)$

So we clearly obtain $\displaystyle E(v) = E_o + \frac{m_o c^2}{\sqrt{1 - v^2/c^2}}$, where we have used $\vec{v}\cdot\vec{v} = v^2$. Since this equation depends on speed, direction is not of importance. Within the Special Theory of Relativity the convention is used that $E_o = 0$, thus we obtain

(2) $\displaystyle E(v) = \frac{m_o c^2}{\sqrt{1 - v^2/c^2}}$

and also the famous equation $E(0) = m_o c^2$. Let $E(v)$ be the energy of a body moving at the speed $v$. Let $\Delta E(v_1,v_2)$ be the energy required to change the velocity of a body from $v_1$ to $v_2$. It is clear that $\Delta E(v_1,v_2) + \Delta E(v_2,v_3) = \Delta E(v_1,v_3)$ as $\Delta E(v_1,v_2) = E(v_2) - E(v_1)$. Let $v_1 < c$ and $v_2 > c$. It is clear that $\displaystyle E(v_1) = \frac{m_o c^2}{\sqrt{1-v_1^2/c^2}}$ and $\displaystyle E(v_2) = \frac{- i m_o c^2}{\sqrt{v_2^2/c^2-1}}$. Therefore

(3) $\displaystyle E(v_1,v_2) = - m_o c^2 \left( \frac{1}{\sqrt{1 - v_1^2/c^2}} + \frac{i}{\sqrt{v_2^2/c^2 - 1}} \right)$.

The amount of energy required to change the speed of a body from a speed slower than light into a speed faster than light is complex. This is not allowed and therefore we cannot change the speed of a body from a speed slower than light into a speed faster than light is complex – but the reverse is also true due to the symmetry; the amount of energy required to change the speed of a body from a speed faster than light into a speed slower than light is complex and is not allowed and therefore we cannot change the speed of a body from a speed faster than light into a speed slower than light is complex. In other words: either bodies always move with a speed slower than the speed of light or bodies always move with a speed faster then the speed of light. A body that always moves faster than the speed of light does not have an observable rest-mass $m_o$ and such bodies a theoretically a complex rest-mass.

When neutrinos move faster than the speed of light then neutrinos have a complex rest mass and that does not “break one of the most fundamental laws of physics”.

Posted in Cosmology, Physics, Relativity, Special Relativity, Universe | 3 Comments

## Measuring the speed of light and the speed of the medium.

Let us consider a medium and an observer that is moving through that medium. Say that the magnitude of the velocity of light with respect to the medium is $c$ and say that the velocity of the observer with respect to the medium is $\vec{v}$. Can we construct a device that allows us to measure the velocity of the observer with respect to the medium?

We consider a simple device constructed by a measuring rod $L$, a pulsing light source and a sensor at one end of the measuring rod $L$ and a mirror on the other end of the measuring rod $L$. If this measuring rod $L$ is moving with respect to the medium such that $\phi$ is the angle between the measuring rod and the velocity of the measuring rod – then we find

$\displaystyle \frac{2L}{\Delta T} = c \frac{1 - \beta^2}{\sqrt{ 1 - \beta^2 \sin^2(\phi) }}$

This can be written as

$\displaystyle \left( \frac{\Delta T}{2L} \right)^2_\phi = \frac{1}{c^2} \frac{1 - \frac{1}{2}\beta^2}{1 - \beta^2} + \frac{1}{c^2} \frac{ \frac{1}{2} \beta^2 \cos(2\phi)}{1 - \beta^2}$

Say the we have three measuring rods – the main measuring rod and other measuring rods at angle $\pm 45^\circ$ then we have

$\displaystyle \left( \frac{\Delta T}{2L} \right)^2_{\phi \pm \frac{1}{4} \pi} = \frac{1}{c^2} \frac{1 - \frac{1}{2}\beta^2}{1 - \beta^2} \mp \frac{1}{c^2} \frac{ \frac{1}{2} \beta^2 \sin(2\phi)}{1 - \beta^2}$

Now it is clear that

$\displaystyle 2 \left( \frac{\Delta T}{2L} \right)^2_{\phi} - \left( \frac{\Delta T}{2L} \right)^2_{\phi + \frac{1}{4} \pi} - \left( \frac{\Delta T}{2L} \right)^2_{\phi - \frac{1}{4} \pi} = \frac{1}{c^2} \frac{\beta^2}{1-\beta^2} \cos(2\phi)$

and

$\displaystyle \left( \frac{\Delta T}{2L} \right)^2_{\phi - \frac{1}{4} \pi} - \left( \frac{\Delta T}{2L} \right)^2_{\phi + \frac{1}{4} \pi} = \frac{1}{c^2} \frac{\beta^2}{1-\beta^2} \sin(2\phi)$

Which gives

$\displaystyle \tan(2\phi) = \frac{\displaystyle \left( \frac{\Delta T}{2L} \right)^2_{\phi - \frac{1}{4} \pi} - \left( \frac{\Delta T}{2L} \right)^2_{\phi + \frac{1}{4} \pi}}{\displaystyle 2 \left( \frac{\Delta T}{2L} \right)^2_{\phi} - \left( \frac{\Delta T}{2L} \right)^2_{\phi + \frac{1}{4} \pi} - \left( \frac{\Delta T}{2L} \right)^2_{\phi - \frac{1}{4} \pi}}$

and

$\displaystyle \left(\frac{1}{c^2} \frac{\beta^2}{1-\beta^2}\right)^2 = \left\{ 2 \left( \frac{\Delta T}{2L} \right)^2_{\phi} - \left( \frac{\Delta T}{2L} \right)^2_{\phi + \frac{1}{4} \pi} - \left( \frac{\Delta T}{2L} \right)^2_{\phi - \frac{1}{4} \pi} \right\}^2 + \left\{ \left( \frac{\Delta T}{2L} \right)^2_{\phi - \frac{1}{4} \pi} - \left( \frac{\Delta T}{2L} \right)^2_{\phi + \frac{1}{4} \pi} \right\}^2$

but also

$\displaystyle \left(\frac{1}{c^2} \frac{2-\beta^2}{1-\beta^2}\right)^2 = \left\{ \left( \frac{\Delta T}{2L} \right)^2_{\phi - \frac{1}{4} \pi} + \left( \frac{\Delta T}{2L} \right)^2_{\phi + \frac{1}{4} \pi} \right\}^2$

thus

$\displaystyle \left( \frac{\beta^2}{2-\beta^2} \right)^2 = \left\{1 - \frac{\displaystyle 2 \left( \frac{\Delta T}{2L} \right)^2_{\phi}}{\displaystyle \left( \frac{\Delta T}{2L} \right)^2_{\phi - \frac{1}{4} \pi} + \left( \frac{\Delta T}{2L} \right)^2_{\phi + \frac{1}{4} \pi}} \right\}^2 + \left\{ \frac{\displaystyle \left( \frac{\Delta T}{2L} \right)^2_{\phi - \frac{1}{4} \pi} - \left( \frac{\Delta T}{2L} \right)^2_{\phi + \frac{1}{4} \pi}}{\displaystyle \left( \frac{\Delta T}{2L} \right)^2_{\phi - \frac{1}{4} \pi} + \left( \frac{\Delta T}{2L} \right)^2_{\phi + \frac{1}{4} \pi}} \right\}^2$

which allows to solve $\beta$, and

$\displaystyle \begin{array}{rcl} \displaystyle \frac{2}{c} &=& \displaystyle \sqrt{ \left\{ \left( \frac{\Delta T}{2L} \right)^2_{\phi - \frac{1}{4} \pi} + \left( \frac{\Delta T}{2L} \right)^2_{\phi + \frac{1}{4} \pi} \right\}^2 }\\\\&&\displaystyle -\sqrt{ \left\{ 2 \left( \frac{\Delta T}{2L} \right)^2_{\phi} - \left( \frac{\Delta T}{2L} \right)^2_{\phi + \frac{1}{4} \pi} - \left( \frac{\Delta T}{2L} \right)^2_{\phi - \frac{1}{4} \pi} \right\}^2 + \left\{ \left( \frac{\Delta T}{2L} \right)^2_{\phi - \frac{1}{4} \pi} - \left( \frac{\Delta T}{2L} \right)^2_{\phi + \frac{1}{4} \pi} \right\}^2 }\end{array}$

which allows to solve $c$. So we can find the speed of light with only one clock – assuming that there is some medium for light propagation. The question is if such an experiment can be performed as $\Delta T$ is very small. Once we can find the speed of light by using only one clock then we are able to synchronize two remote clocks. We may wonder what that does with the principles of Relativity?

Hope you like the post!

Posted in Mathematics, Physics, Relativity, Special Relativity | 1 Comment

## The escape velocity for the Schwarzschild metric (2).

In the other post The escape velocity for the Schwarzschild metric. we have found the escape velocity

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

however – this was based on the limit $r \rightarrow \infty$ and $\dot{r} \rightarrow 0$.

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

where $\displaystyle \psi = 1 - \frac{r_s}{r}$. At $r \rightarrow \infty$ the metric becomes Minkowski, thus we may assume that $\dot{r} = \beta c$, where $-1 \le \beta \le +1$. It is clear that $\psi \rightarrow 1$ for $r \rightarrow \infty$, then we obtain

$\displaystyle \displaystyle \frac{1}{2} \frac{\dot{r}^2}{\psi} + \frac{1}{2} \psi c^2 = \frac{1}{2} \left( 1 + \beta^2 \right) c^2$,

so the general escape velocity becomes

$\displaystyle u = c \sqrt{\beta^2 + \left(1-\beta^2\right) \frac{2\phi}{c^2} - \left(\frac{2\phi}{c^2}\right)^2}$.

For light we have $\beta=1$ thus

$\displaystyle u_\textit{light} = c \sqrt{1 - \left(\frac{2\phi}{c^2}\right)^2}$.

For matter we have the critical case $\beta=0$ thus

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

Here are some plots…

For $\beta=0$ (matter) we have

For $\beta=\frac{1}{4}$ (matter) we have

For $\beta=\frac{1}{2}$ (matter) we have

For $\beta=1$ (light) we have

Sofar part 2…

## Kerr metric and dragging.

The Kerr-metric is given by

$\begin{array}{rcl} ds^2 &=& \displaystyle \left( 1 - \frac{ r_s r }{ r^2 + \jmath^2 \cos^2(\theta) } \right) c^2 dt^2 - \frac{ r^2 + \jmath^2 \cos^2(\theta) }{ r^2 - r_s r + \jmath^2 } dr^2 - \left( r^2 + \jmath^2 \cos^2(\theta) \right) d\theta^2\\ && \displaystyle - \left( r^2 + \jmath^2 + \frac{ r_s r \jmath^2 \sin^2(\theta) }{ r^2 + \jmath^2 \cos^2(\theta) } \right) \sin^2(\theta) d\phi^2 + \frac{ 2 r_s r \jmath \sin^2(\theta) }{ r^2 + \jmath^2 \cos^2(\theta) } c dt d\phi \end{array}$,

where $\displaystyle r_s = \frac{2GM}{c^2}$ and $\displaystyle \jmath=\frac{J}{Mc}$. We solve the geodesic equation using

$\displaystyle \frac{d}{ds} \left( g_{\alpha\mu} \frac{dx^\mu}{ds} \right) = \frac{1}{2} \left( \frac{\partial g_{\kappa\lambda}}{\partial x^\mu} \right) \frac{dx^\kappa}{ds} \frac{dx^\lambda}{ds}$

It is clear that $g_{\kappa\lambda} = g_{\kappa\lambda}(\sin^2(\theta))$, then for $\theta$ we obtain the equation

$\displaystyle \frac{d}{ds} \left( g_{\theta\theta} \frac{d\theta}{ds} \right) = \sin(\theta) \cos(\theta) \left( \frac{ \partial g_{\kappa\lambda} }{ \partial \sin^2(\theta) } \right) \frac{dx^\kappa}{ds} \frac{dx^\lambda}{ds}$

So there is a solution $\theta = \pi/2$. Next we look to the equation for $r$ and we have

$\displaystyle \frac{d}{ds} \left( g_{rr} \frac{dr}{ds} \right) = \frac{1}{2} \left( \partial_r g_{tt} \left( \frac{dt}{ds} \right)^2 + \partial_r g_{rr} \left( \frac{dr}{ds} \right)^2 + \partial_r g_{\phi\phi} \left( \frac{d\phi}{ds} \right)^2 + 2 \partial_r g_{t\phi} \frac{dt}{ds} \frac{d\phi}{ds} \right)$

An orbit with a constant $r$ gives the equation

$\displaystyle \left( \partial_r g_{tt} + \partial_r g_{\phi\phi} \dot\phi^2 + 2 \partial_r g_{t\phi} \dot\phi \right) \left( \frac{dt}{ds} \right)^2 = 0$

Thus

$\displaystyle \partial_r g_{\phi\phi} \dot\phi^2 + 2 \partial_r g_{t\phi} \dot\phi = - \partial_r g_{tt}$

So

$\displaystyle \dot\phi_\pm = - \frac{ \partial_r g_{t\phi} }{ \partial_r g_{\phi\phi} } \pm \sqrt{ \left( \frac{ \partial_r g_{t\phi} }{ \partial_r g_{\phi\phi} } \right)^2 - \frac{ \partial_r g_{tt} }{ \partial_r g_{\phi\phi} } }$

Now we can write

$\left[\begin{array}{rcl} g_{tt} &=& \left( 1 - \psi(r) \right) c^2\\ g_{\phi\phi} &=& -\left( r^2 + \jmath^2 + \jmath^2 \psi(r) \right)\\ g_{t\phi} &=& \jmath \psi(r) c \end{array}\right.$

where $\displaystyle \psi(r) = \frac{r_s r}{r^2+\jmath^2}$. Thus

$\left[\begin{array}{rcl} \partial_r g_{tt} &=& - \psi'(r) c^2\\ \partial_r g_{\phi\phi} &=& -2r - \jmath^2 \psi'(r)\\ \partial_r g_{t\phi} &=& \jmath \psi'(r) c \end{array}\right.$

Then we can write

$\displaystyle \dot\phi_\pm = \frac{\jmath \psi'(r) c}{2r + \jmath^2 \psi'(r)} \pm \sqrt{ \left( -\frac{\jmath \psi'(r) c}{2r + \jmath^2 \psi'(r)} \right)^2 - \frac{ \psi'(r) c^2 }{2r + \jmath^2 \psi'(r)} }$

which can be written as

$\displaystyle \dot\phi_\pm = \frac{ (\jmath/c) \Psi(r) \pm \sqrt{-\Psi(r)} }{1 + (\jmath/c)^2 \Psi(r)}$,

where $\displaystyle \Psi(r) = \frac{\psi'(r) c^2}{2r}$ and $\displaystyle \frac{\psi'(r) c^2}{2r} = -\frac{GM}{r^3} \frac{1 - (\jmath/r)^2}{\left( 1 + (\jmath/r)^2\right)^2}$. The case $\jmath=0$ gives

$\displaystyle \dot\phi_\pm = \pm \sqrt{\frac{GM}{r^3}}$,

which reduces to the Third law of Kepler, known as $\displaystyle \dot\phi^2 = \frac{GM}{r^3}$.

The first order of $\jmath$ gives

$\displaystyle \dot\phi_\pm \simeq -\frac{\jmath}{c} \frac{GM}{r^3} \pm \sqrt{\frac{GM}{r^3}}$

This can we written as

$\displaystyle T_\pm \simeq \pm \sqrt{ \frac{4 \pi^2 r^3}{GM} } - \frac{\jmath}{c}$,

this effect is very small and therefore difficult to test by experiment. For the Sun we have $\displaystyle \frac{\jmath}{c} \simeq 6.3 \times 10^{-6} \texttt{s}$, for the Earth we have $\displaystyle \frac{\jmath}{c} \simeq 1.3 \times 10^{-6} \texttt{s}$

## Solving the geodesic equation.

In this post, I will write about the geodesic equation. I suggest a simplified form that is simpler to solve.

The geodesic equation is given by

$\displaystyle \frac{d^2 x^\mu}{ds^2} + \Gamma^\mu_{\kappa\lambda} \frac{x^\kappa}{ds} \frac{x^\lambda}{ds} = 0$,

where $\displaystyle \Gamma^\mu_{\kappa\lambda} = \frac{1}{2} g^{\mu\alpha} \left\{ \partial_\kappa g_{\lambda\alpha} + \partial_\lambda g_{\alpha\kappa} - \partial_\alpha g_{\kappa\lambda} \right\}$. We have used the convention that $\displaystyle \partial_\kappa g_{\lambda\alpha} = \frac{\partial g_\lambda\alpha}{\partial x^\kappa}$. Then we may write

$\displaystyle \frac{d}{ds} \left( g_{\alpha\mu} \frac{x^\mu}{ds} \right) = \frac{1}{2} \left( \partial_\alpha g_{\kappa\lambda} \right) \frac{x^\kappa}{ds} \frac{x^\lambda}{ds}$

Then it is clear that IF the metric $g_{\mu\nu}$ does not depend on $x_\alpha$ we obtain the equation

$\displaystyle \frac{d}{ds} \left(g_{\alpha\mu} \frac{dx^\mu}{ds} \right) = 0$,

hus $\displaystyle g_{\alpha\mu} \frac{dx^\mu}{ds} = \textit{constant}$.

Consider for example the Schwarzschild metric, given by

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

where $\displaystyle \psi(r) = 1 - \frac{2GM}{rc^2}$.

Then for $\theta$ we obtain the equation

$\displaystyle \frac{d}{ds} \left( g_{\theta\theta} \frac{d\theta}{ds} \right) = - r^2 \sin(\theta) \cos(\theta) \left(\frac{d\phi}{ds}\right)^2$,

where $\theta = \pi/2$ is a solution. Next we consider the equation

$\displaystyle \frac{d}{ds} \left( g_{rr} \frac{dr}{ds} \right) = \left( \psi'(r) + \frac{\psi'(r)}{\psi^2(r)} \dot{r}^2 - 2r \dot{\phi}^2 \right) \left(\frac{dt}{ds} \right)^2$.

Say we consider circular orbits – thus $dr=0$ then we obtain

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

thus we obtain

$\displaystyle \frac{GM}{r^2} - r \dot{\phi}^2 = 0$,

so

$\displaystyle \dot{\phi}^2 = \frac{GM}{r^3}$,

which is know as the Third Law of Kepler.

We have here derived the Third law of Kepler for the Schwarzschild metric, without calculating the affine connection $\Gamma^\mu_{\kappa\lambda}$.

Have fun!