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Notes on Ehresmann’s connections -Part 2, Frölicher–Nijenhuis bracket

Posted on 2010/12/272017/01/07 by arkajad

In this note I shall introduce the Frölicher-Nijenhuis bracket – a powerful tool, especially when dealing with connections on fibre bundles.
Let $(E,\pi,M)$ be a fibre bundle. In the previous note we have seen that the connection form $\omega$ is a one-form on $E$ with values in $TE$ (in fact, in values in $VE\subset TE.$ ) This leads us to a more general subject of natural operations on tangent valued-forms.

Let therefore, for now, $E$ be any differentiable manifold, and let us consider the space of tangent valued forms on $E.$ Let $\Omega (E)=\oplus_{i=0}^n \Omega^{i}(E)$ be the graded algebra of ordinary differential forms on $E.$ Then the (graded) space $\Omega (E,TE)$ of tangent-valued forms can be written as

$\Omega(E,TE)=\Omega(E)\otimes_E T(E).$

In the following we will use the capital indices $A,B,C\ldots$ to refer to local charts $x^A$ of $E.$

The elements of $\Omega^{0}(E,TE)$ are simply vector fields $\xi^A$ on $E.$ The elements of $\Omega^{1}(E,TE)$ are tangent-valued one-forms. If $\Phi$ is such a form, then $\Phi$ assigns to each tangent vector $\xi\in T_pE$ another tangent vector $\Phi(\xi)\in T_pE.$ In coordinates it is represented as $\Phi^A_B.$ In general, an element of $\Omega^{r}(E,TE)$ is represented by $\Phi^A_{B_1,\ldots B_r},$ antisymmetric in indices $B_1,\ldots B_r,$ or

$\Phi= \frac{1}{r!}\,\Phi^A_{B_1,\ldots B_r}\,dx^{B_1}\wedge\ldots\wedge dx^{B_r}\,\partial_A.$

In the following it will be convenient to introduce the following notation:

Notation

We will write

$\Phi=\sum \Phi^A_B\,d^B\otimes \partial_A,$

where $B=(1\leq B_1 < B_2 < \ldots < B_r\leq\mathrm{dim }\,E),$ $d^B=dx^{B_1}\wedge\ldots\wedge dx^{B^r}$ and $\partial_A=\frac{\partial}{\partial x^A}.$

Operations on differential forms

Frölicher-Nijenhuis (FN) bracket will assign to any two tangent valued forms $\Phi\in \Omega^{r}(E,TE),\,\Psi\in \Omega^{s}(E,TE)$ another form $[\Phi,\Psi]\in \Omega^{(r+s)}(E,TE).$

We will see (in the following posts) that the curvature of a connection can be simply expressed in terms of the FN bracket of the connection form with itself. Also Bianchi identities for the curvature will follow immediately from this definition and from the properties of the FN bracket. But first let us recall the three fundamental operations on differential forms $\Phi\in\Omega(E).$ These are: exterior derivative, Lie derivative, insertion. Exterior derivative $d:\omega\mapsto d\omega$ maps $\Omega^{r}(E)$ to $\Omega^{(r+1)}(E)$ according to the formula

$d\Phi(X_0,\ldots ,X_r)=\sum_{i=0}^r(-1)^iX_i(\Phi(X_0,\ldots,\hat{X_i},\ldots ,X_r)).$

Given a vector field $X\in\mathfrak{X} (E)$ the insertion operator $i_X$ maps $\Omega^{r}(E)$ to $\Omega^{(r-1)}(E)$ according to the formula

$(i_X\,\omega)(X_1,\ldots,X_{r-1})=\omega(X,X_1,\ldots,X_{r-1}).$

Lie derivative $\mathfrak{L}_X$ maps $\Omega^{r}(E)$ into itself, the definition being

$(\mathfrak{L}_X\omega)(X_1,\ldots,X_r)=X(\omega(X_1,\ldots,X_r))-\sum_{i=1}^r\omega(X_1,\ldots,[X,X_i],\ldots,X_r).$

These three operations on number-valued differential forms have the following important properties (note: the commutators are graded commutators):

$d(\omega\wedge\eta)=d\omega\wedge\eta+(-1)^{|\omega|}\omega\wedge d\eta$
$i_X(\omega\wedge\eta)=i_X\omega\wedge\eta+(-1)^{|\omega|}\omega\wedge i_X\eta$
$\mathfrak{L}_X(\omega\wedge\eta)=\mathfrak{L}_X\omega\wedge\eta+(-1)^{|\omega|}\omega\wedge \mathfrak{L}_X \eta$
$d^2=0$
$[i_X,i_Y]=i_Xi_Y+i_Yi_X=0$
$[\mathfrak{L}_X,d]=\mathfrak{L}_X\circ d+d\circ\mathfrak{L}_X=0$
$[i_X,d]=i_X\circ d+d\circ\ i_X=\mathfrak{L}_X$
$[\mathfrak{L}_X,\mathfrak{L}_Y]=\mathfrak{L}_X\circ\mathfrak{L}_Y-\mathfrak{L}_Y\circ\mathfrak{L}_X=\,\mathfrak{L}_{[X,Y]}$
$[\mathfrak{L}_X,i_Y]=\mathfrak{L}_X i_Y-i_Y\mathfrak{L}_X=i_{[X,Y]}$
If $f$ is a function on $E,$ then $\mathfrak{L}_{fX}\omega=f\mathfrak{L}_X\omega+df\wedge i_X\omega$

Frölicher–Nijenhuis bracket

Frölicher-Nijenhuis bracket associates to each pair of tangent-valued forms on $E$ the third tangent-valued form. Because of its bilinearity it is enough to define it on simple tensors. For $\alpha\in\Omega^{r}(E),\,\beta\in\Omega^{s}(E),$ and for $X,Y\in \mathfrak{X}(E)$ it is defined by the following formula:

$\begin{eqnarray*} [\alpha\otimes X,\,\beta\otimes Y]&=&(\alpha\wedge\beta)\otimes [X,Y]\\&+&(\alpha\wedge\mathfrak{L}_X\beta)\otimes Y\\&-&(\mathfrak{L}_Y\alpha\wedge\beta)\otimes X\\&+&(-1)^r(d\alpha\wedge i_X\beta)\otimes Y\\&+&(-1)^r(i_Y\alpha\wedge d\beta)\otimes X\end{eqnarray*}$

In coordinates the Frölicher-Nijenhuis bracket is given by the following explicit expression:

$\begin{eqnarray*} [\Phi,\Psi]&=&\left(\Phi^C_{B_1\ldots B_r}\partial_C\Psi^A_{B_{r+1}\ldots B_{r+s}}\right.\\ &-&(-1)^{rs}\Psi^C_{B_1\ldots B_s}\partial_C\Phi^A_{B_{s+1}\ldots B_{r+s}}\\ & -&r\Phi^A_{B_1\ldots B_{r-1} C}\partial_{B_r}\Psi^C_{B_{r+1}\ldots B_{r+s}}\\ &+&(-1)^{rs}s\Psi^A_{CB_{1}\ldots B_{s-1}}\partial_{B_{s}}\Phi^C_{B_{s+1}\ldots B_{r+s}}\left.\right)\,d^B\otimes\partial_A\end{eqnarray*}$

If we want to write the sam formula in a more explicit form, then antisymmetrizations and combinatorial factors enter:

$\begin{align*} &[\Phi,\Psi](X_1,\dots X_{r+s})=\\ &=\frac{1}{r!s!}\sum_\sigma (-1)^\sigma [\Phi(X_{\sigma 1}\ldots X_{\sigma r}),\Psi(X_{\sigma(r+1)}\ldots X_{\sigma(r+s)})] \\ &+(-1)^r\left( \frac{1}{r!(s-1)!}\sum_\sigma (-1)^\sigma\Psi([X_{\sigma1},\Phi(X_{\sigma2},\ldots ,X_{\sigma (r+1)})],X_{\sigma (r+2)},\ldots)\right.\\ &\left.-\frac{1}{(r-1)!(s-1)!2!}\sum_\sigma(-1)^\sigma\Psi(\Phi([X_{\sigma 1},X_{\sigma 2}],X_{\sigma 3},\ldots),X_{\sigma (r+2)},\ldots) \right)\\ &-(-1)^{rs+s}\left(\frac{1}{(r-1)!s!}\sum_\sigma (-1)^\sigma\Phi([X_{\sigma 1},\Psi(X_{\sigma 2},\ldots ,X_{\sigma (s+1)})],X_{\sigma (s+2)},\ldots)\right .\\ &-\left.\frac{1}{(r-1)!(s-1)!2!}\sum_\sigma(-1)^\sigma\Phi(\Psi([X_{\sigma 1},X_{\sigma 2}],X_{\sigma 3},\ldots ),,X_{\sigma (s+2)},\ldots )\right) \end{align*}$

As an exercise let us compute $[id_E,\beta\otimes Y],$ where $id_E=dx^A\otimes\partial_A$ is the identity tangent-valued form (the natural soldering form), with coordinates $\delta^A_B$ . Thus we set $r=1.$ The first term gives

$[dx^A\otimes\partial_A,\Psi]^A_{A_1\ldots A_{r+1}}=\delta^B_{A_1}\partial_B \Psi^A_{A_2\ldots A_{s+1}}=\partial^A_{A_1}\Psi_{A_2\ldots A_{s+1}}.$

The second and fourth terms vanishs because $\delta^A_B$ are constant.

The third is

$-\delta^A_B\partial_{A_1}\Psi^B_{A_2\ldots A_{s+1}}=-\partial_{A_1}\Psi^A_{A_2\ldots A_{s+1}}$

It follows that $[dx^A\otimes\partial_A,\Psi]=0$ – therefore $id_E$ commutes with every tangent valued form.

The most important properties of the FN bracket are the following ones

$[\Phi,\Psi]=-(-1)^{|\Phi|\,|\Psi|}\,[\Psi,\Phi]\quad$ – graded antisymmetry
$[\Phi_1,[\Phi_2,\Phi_3]]=[[\Phi_1,\Phi_2],\Phi_3]+(-1)^{|\Phi_1|\,\|\Phi_2|}[\Phi_2,[\Phi_1,\Phi_3]]$ – Jacobi identity

Suppose now $X$ is a vector field. It can be also considered as a tangent-valued 0-form: $X=1\otimes X.$ We then find that

$[X,\beta\otimes Y]=\beta\otimes [X,Y]+\mathfrak{L}_X\beta\otimes Y=\mathfrak{L}_X(\beta\otimes Y).$

Therefore for vector fields the FN bracket reduces to an ordinary Lie derivative.

Another insight into the nature of the FN bracket comes from considering tangent-valued forms as operators on diferential forms. First of all, for any $\Phi \in \Omega^{r+1}(E,TE)$ we can define a graded derivation $i_\Phi:\,\Omega^{s}E\rightarrow \Omega^{r+s} E$ by the formula:

$\begin{align*} (i_\Phi\, \omega)&(X_1,\ldots,X_{r+s})=\\ &\frac{1}{(r+1)!(s-1)!}\sum_{\sigma\in S_{r+s}}(-1)^\sigma\,\omega(\Phi(X_{\sigma 1},\ldots ,X_{\sigma (r+1)}),X_{\sigma (r+2)},\ldots) \end{align*}$

Now I have a problem. In references [1] and [2] we have the following local formulas:

In our notation this would be:

$i_\Phi\omega=\sum\,\Phi^A_{B_1\ldots B_{r+1}}\,\omega_{AB_{r+2}\ldots B_{r+s}}\,d^B$

But I am not able to reproduce this result. It seems to me that the factor $s$ is missing in front of the expression on the RHS. I will return to this enigma at the end of this note.

Remark. Notice that in these formulas we assume $\Phi \in \Omega^{r+1}(E,TE)$ and not in $\Omega^{r}(E,TE)$

Then we define the Lie derivative $\mathfrak{L}_\Phi$ of a differential form $\omega\in\Omega^s E$ along a tangent-valued form $\Phi\in \Omega^r(E,TE)$ by the formula

$\mathfrak{L}_\Phi\, \omega =i_\Phi\,d\omega-di_\Phi\,\omega$

Explicitly:

$\begin{align*} \mathfrak{L}_\Phi\,\omega=&\sum\,\left(\Phi^A_{B_1\ldots B_r}\partial_A\,\omega_{B_{r+1}\ldots B_{r+s}}\right.\\ &\left.+(-1)^r(\partial_{B_1}\Phi^A_{B_2\ldots B_{r+1}})\,\omega_{AB_{r+2}\ldots B_{r+s}}\right) d^B \end{align*}$

$\begin{align*} ( &\mathfrak{L}_{\Phi}\,\omega)(X_1,\ldots ,X_{r+s})=\\ &=\frac{1}{r!s!}\,\sum_\sigma\,(-1)^\sigma \mathfrak{L}_{\Phi (X_{\sigma 1},\ldots ,X_{\sigma r})}(\omega(X_{\sigma (r+1)},\ldots,X_{\sigma (r+s)}))\\ &+(-1)^r\left(\frac{1}{r!(s-1)!}\sum_\sigma (-1)^\sigma\,\omega([X_{\sigma 1},\Phi(X_{\sigma 2},\ldots,X_{\sigma (r+1)})],X_{\sigma (r+2)},\ldots) \right.\\ &\left.-\frac{1}{(r-1)!(s-1)!2!}\sum_\sigma\,(-1)^\sigma\,\omega(\Phi([X_{\sigma 1},X_{\sigma 2}],X_{\sigma 3},\ldots),X_{\sigma (r+2)},\ldots)\right) \end{align*}$

Again it is instructive to calculate the action of $i_{\Phi}$ and of this extended Lie derivative for $\Phi=id_E.$ Using the local formula from [1] or [2]

$i_\Phi\omega=\sum\,\Phi^A_{B_1\ldots B_r}\,\omega_{AB_{r+1}\ldots B_{r+s}}\,dx^B$

we obtain that for $\Phi=id_E$ , that is for $\Phi^A_B=\delta^A_B,$ we have

$i_\Phi\omega=\omega.$

But the identity map is not a graded derivative (of degree 0 in this case). On the other hand, from the formula involving vector fields I am getting

$i_\phi\omega = s\omega$

This looks good, because then

$i_\Phi(\omega_1\wedge\omega_2)=(s_1\omega_1)\wedge\omega_2+\omega_1\wedge (s_2\omega_1)=(s_1+s_2)\omega_1\wedge\omega_2$

as it should be.

At this point I have to pause until this point is clarified. Perhaps I am missing something evident? I don’t know.

And here are the conventions concerning differential forms used in [1] and [2]:

Differential Forms - Conventions in "Topics in Differential Geometry"

References

[1] Peter W. Michor, Topics in Differential Geometry (see hxxp://www.mat.uniroma1.it/people/manetti/GeoDiff0809/dgbook.pdf )
[2] Ivan Kolar, Peter W. Michor, Jan Slovak, Natural Operations in Differentila Geometry (see hxxp://147.251.48.205/pub/muni.cz/EMIS/monographs/KSM/kmsbookh.ps.gz )
[3] Andreas Kriegl, Peter W. Michor, The Convenient Setting for Global Analysis, Ch. 33.18

Notes on Ehresmann’s connections -Part 1

Posted on 2010/12/202017/01/06 by arkajad

Ehresmann’s connection

Consider a fibre bundle $\xi=(E,\pi,M)$ with a fibred chart $(x^\mu,v^a),$ where $x^\mu,\quad (\mu=1,\ldots m)$ are local coordinates on the base manifold $M,$ and $v^a\quad (a=1,\ldots n)$ are coordinates along the fibres. If we have two fibred charts, then on their intersection domain we have

$x^{\mu'}=x^{\mu'}(x^\mu),$
$v^{a'}=v^{a'}(x^\mu,v^a).$

The chart $(x^\mu,v^a)$ induces vector fields on the tangent space $TE:$

$\partial_\mu=\frac{\partial}{\partial x^\mu}$
$\partial_a=\frac{\partial}{\partial v^a}$

with the transformation laws:

$\frac{\partial f(x',v')}{\partial x^{\mu'}}=\frac{\partial f(x,v)}{\partial x^\mu}\frac{\partial x^\mu}{\partial x^{\mu'}}+\frac{\partial f(x,v)}{\partial v^a}\frac{\partial v^a}{\partial x^{\mu'}}$
$\frac{\partial f(x'v')}{\partial v^{a'}}=\frac{\partial f(x,v)}{\partial v^a}\frac{\partial v^a}{\partial v^{a'}}.$

Therefore

$\begin{eqnarray*}\partial_{\mu'}&=&\frac{\partial x^\mu}{\partial x^{\mu'}}\,\partial_\mu+\frac{\partial v^a}{\partial x^{\mu'}}\,\partial_a,\\ \partial_{a'}&=&\frac{\partial v^a}{\partial v^{a'}}\,\partial_a,\end{eqnarray*}$

and the inverse

$\begin{eqnarray*}\partial_{\mu}&=&\frac{\partial x^{\mu'}}{\partial x^{\mu}}\,\partial_{\mu'}+\frac{\partial v^{a'}}{\partial x^{\mu}}\,\partial_{a'},\\ \partial_{a}&=&\frac{\partial v^{a'}}{\partial v^{a}}\,\partial_{a'},\end{eqnarray*}$

Vector fields $\partial_a$ span what is called the vertical subspace in $TE.$ This subspace, consisting of vectors tangent to the fibers, is independent of the chart. There are no a priori distinguished `horizontal’ subspaces. The subspaces determined by vector fields $\partial_\mu$ change when the charts change. A preferred distribution of horizontal subspaces is determined by what is called an Ehresmann connection. Once such a connection is given, vector fields $\partial_\mu$ tangent to $M$ can be lifted to horizontal vector fields tangent to $E.$

Remark: In principle we should use a different notation for vector fields $\partial_\mu$ tangent to $M$ and vector fields induced on $E$ by a chart $(x,v).$ Using the same notation for both can lead to misunderstandings when is not paying attention to the context.

Let us denote by $\delta_\mu$ the horizontal lifts induced by an Ehresmann connection on $\xi,$ They can be written uniquely in the form:

$\delta_\mu=\partial_\mu+\Gamma^a_\mu(x,v)\,\partial_a.$

Remark: In Finsler geometry instead of $\Gamma^a_\mu$ one usually uses $N^a_\mu=-\Gamma^a_\mu,$ so that the formula for the horizontal lift reads $\delta_\mu=\frac{\partial}{\partial x^\mu}-N^a_\mu(x,v) \frac{\partial}{\partial v^a}$

The functions $\Gamma^a_\mu(x,v)$ may be called the coefficient functions of the Ehresmann connection with respect to the chart $(x,v).$

If $X=X^\mu\partial_\mu$ is a vector field on $M,$ we denote its horizontal lift by $\delta_X,$ thus

$\delta_X=X^\mu\,\delta_\mu.$

Suppose now we have two charts, $(x,v)$ and $(x'v').$ Corresponding to these two charts we will have two sets of horizontal lifts of the same connection $\delta_\mu$ and $\delta_{\mu'}.$ Since both are supposed to describe the same horizontal distribution, they must be related by an invertible matrix $A^{\mu'}_\mu$ :

$\delta_{\mu'}=A^\mu_{\mu'}\delta_\mu.$

Substituting now the definitions we have

$\partial_{\mu'}+\Gamma^{a'}_{\mu'}\,\partial_{a'}=A^\mu_{\mu'}\left(\partial_\mu+\Gamma^a_\mu\partial_a.\right)$

$\partial_{\mu'}+\Gamma^{a'}_{\mu'}\,\partial_{a'}=A^\mu_{\mu'}\frac{\partial x^{\mu'}}{\partial x^{\mu}}\,\partial_{\mu'}+A^\mu_{\mu'}\frac{\partial v^{a'}}{\partial x^{\mu}}\,\partial_{a'}+A^\mu_{\mu'}\Gamma^a_\mu\frac{\partial v^{a'}}{\partial v^{a}}\,\partial_{a'}$

Comparing the terms we find that

$A^\mu_{\mu'}= \frac{\partial x^\mu}{\partial x^{\mu'}}$

and therefore

$\Gamma^{a'}_{\mu'}=\frac{\partial x^\mu}{\partial x^{\mu'}}\left(\frac{\partial v^{a'}}{\partial v^a}\Gamma^a_\mu+\frac{\partial v^{a'}}{\partial x^\mu}\right).$

We can distinguish two special classes of transformations. First class consists of changes of coordinates on $M$ without reparametrizing of the fibers. For these transformations $\Gamma^a_\mu$ transforms like one–forms:

$\Gamma^a_{\mu'}=\frac{\partial x^\mu}{\partial x^{\mu'}}\Gamma^a_\mu.$

The second class consists of transformations of the form $x^{\mu'}=x^\mu,\, v^{a'}=v^a(x,v).$ For these transformations we have

$\Gamma^{a'}_{\mu}=\frac{\partial v^{a'}}{\partial v^a}\Gamma^a_\mu+\frac{\partial v^{a'}}{\partial x^\mu}.$

{\bf Remark}: If $\xi$ is a vector bundle, then the reparametrizations of the fibers compatible with the vector bundle structure are linear and of the form:

$v^{a'}={\Lambda(x)^{a'}}_a\,v^a.$

In this case the last formula reads:

$\Gamma^{a'}_\mu={\Lambda^{a'}}_a\Gamma^a_\mu+\partial_\mu {\Lambda^{a'}}_av^a$

Curvature

Given an Ehresmann connection represented by horizontal vector fields $\delta_\mu$ we can calculate their commutator. As it happens their commutator is a vertical vector field that can be written as

$[\delta_\mu,\delta_\nu]=R_{\mu\nu}^a\,\partial_a,$

where

$R_{\mu\nu}^a=\partial_\mu \Gamma^a_\nu-\partial_\nu \Gamma^a_\mu+\Gamma_\mu^b\,\partial_b\Gamma^a_\nu-\Gamma_\nu^b\,\partial_b\Gamma^a_\mu.$

More generally, for any two vector fields $X=X^\mu\partial_\mu,\,Y=Y^\mu\partial_\mu$ on $M$ one defines

$R_{X,Y}=[\delta_X,\delta_Y]-\delta_{[X,Y]}.$

If $f,g$ are functions on $M,$ then

$R_{fX,gY}(x,v)=f(x)g(x)\,R_{X,Y}(x,v).$

Under coordinate transformations $R_{\mu\nu}$ transforms like a tensor:

$R^a_{\mu'\nu'}=\frac{\partial x^\mu}{\partial x^{\mu'}}\frac{\partial x^\nu}{\partial x^{\nu'}}\,R^a_{\mu\nu}.$

Under fiber reparametrization we have:

$R_{\mu\nu}^{a'}=\frac{\partial v^{a'}}{\partial v^a}\,R^a_{\mu\nu}.$

Connection form

It is convenient to code the connection given by a horizontal distribution in one geometrical object. This can be done by introducing the curvature form – a one-form on $E$ with values in the vertical distribution. Given a vector tangent to $E$ at $(x,v)$ we decompose it into a horizontal and vertical part and define the connection form on this vector as the vertical part. It follows that the connection form is the identity map on vertical vectors. In coordinates we can uniquely write the connection form $\omega$ as
$\omega=(dv^a+N_\mu^a\,dx^\mu)\,\partial_a,$
The curvature form vanishes automatically on the horizontal vectors $\delta_\mu,$ therefore it easily follows that $N^a_\mu=-\Gamma^a_\mu.$

Einstein, prędkości, styczne

Posted on 2010/12/122017/01/09 by arkajad

W Relatywistycznym składaniu prędkości wyprowadziliśmy wzór na składanie prędkości w ramach szczególnej teorii względności. Oznaczmy przez $\beta$ bezwymiarową prędkość względną $v/c$ układu odniesienia S’ względem układu odniesienia S. Niech $\beta'$ będzie prędkością S” względem S’ – przy czym S” i S’ poruszają się w tym kierunku osi $x$ względem układu S i mają tak samo ustawione osie przestrzenne y,z jak układ S. Wtedy S” porusza się względem S z prędkością $\beta''$ daną przez formułę:

$\beta''=\frac{\beta+\beta'}{1+\beta\beta'}$

Jest to suma $\beta$ i $\beta',$ ale `zdeformowana’ poprzez obecność mianownika. Deformacja ta jest tym większa im bliższe są $\beta$ i $\beta'$ jedynki, czyli im bliższe są prędkości światła prędkości $v$ i $v'.$ Dla małych prędkości, dla $\beta$ i $\beta'$ bliskich zera, iloczyn $\beta\beta'$ jest mały w porównaniu z jedynką, można więc go zaniedbać. Jednak dla $\beta,\beta'$ bliskich jedności mianownik jest bliski dwójki. I to powoduje deformację zwykłego dodawania w liczniku.

Narysujmy wykres funkcji $f_0(\beta,\beta')=\beta+\beta'$ – czyli wykres zwykłego nierelatywistycznego dodawania prędkości. Wykres ten wygląda tak:

Dodawanie prędkości nierelatywistyczne — Zwykła suma

Jest to po prostu fragment płaszczyzny. Relatywistyczna formuła dodawania, w przedstawieniu graficznym ma taką postać

Relatywistyczne dodawanie prędkości — Suma zdeformowana

Widać wyraźnie deformację, widać ‘krzywiznę’. Ponieważ ‘krzywizna’ pojawi się dalej, gdy przejdziemy do ogólnej teorii względności, choć będzie w innym kontekscie, pojęciu krzywizny, jej dokładnej matematycznej definicji, warto poświęcić trochę czasu, warto weń ‘zainwestować’. Trzeba jednak zacząć od rzeczy prostszej, od stycznych.

Krzywizna krzywych płaskich.
1. Styczne

Od tego zaczniemy, bo dobrze zacząć od rzeczy prostych i potem stopniowo, przechodzić do tych bardziej złożonych. Tematowi temu poświęciłem kiedyś parę postów na moim blogu w Salon24, np. “Pędzi zawrót kolisty“. Tutaj jednak rzecz wprowadzę po kolei, w miarę systematycznie.

Krzywiznę odczuwamy intuicyjnie – znamy to z wesołego miasteczka:

Na zakrzywieniach albo nas coś wgniata, albo wręcz przeciwnie, stajemy się nieważcy. Czas więc to pojęcie sprecyzować, na razie w stosunku do krzywych płaskich.

Weźmy dla przykładu krzywą z tego obrazka:

Rollercoaster wielomianowy — Z górki na pazurki

(obrazek pożyczony z “Use Games to Motivate Your Calculus Students-Handout“)

Równanie tej krzywej ma postać:

$f(x)=0.5(x-5.8)(x-4)(x-0.7)(x+2.1)(x+6.19)$

Jest to wielomian piątego stopnia. Krzywa ma pięć przecięć z osią x, na obrazku widać jedynie trzy z tych przecięć. Pozostałe dwa są poza polem obrazka. Pierwszą rzeczą, z którą warto się zaznajomić, to pojęcie ‘stycznej’. Zamiast tłumaczyć co to jest ‘prosta styczna do krzywej w danym punkcie’ lepiej pokazać to na obrazku. W tym celu naniosłem najpierw wykres naszej krzywej, zrobiony przy pomocy programu plotującego, na obrazek w formacie jpg – tak będzie weselej. Oczywiście musiałem w tym celu nieco przeskalować wykres (w pionie, podzieliłem krzywą przez 120), by się ta krzywa z obrazka i ta krzywa ‘zmatematyzowana’ w miarę zgodziły. Wyszło nienajgorzej.