Howdy, readers!

Because I am in a highly-focused preparation for some tests, this blog is gonna be down for a while. Surely I still may write some words or more, but in a less frequency than my natural one. Just wait, I'll come back ^^

Rotationally invariant gradient Kahler-Ricci solitons on the complex plane

2009 September 30
2 Comments
tags: , , , , , , ,
by Octavian

This is the review of a part of discussion in a paper of M. Feldman, T. Ilmanen, and D. Knopf, titled Rotationally Symmetric Shrinking and Expanding Gradient Kahler-Ricci Solitons, published in Journal of Differential Geometry, 65 (2003) 169-209.

Suppose {h} is {U(n)}-invariant Kahler metric on {{\mathbb C}^n \backslash \{0\}}. We may assume that {h_{\alpha \bar{\beta}} = \partial_\alpha \partial_{\bar{\beta}} P} holds on all of {{\mathbb C}^n \backslash \{0\}}. {P} is also {U(n)}-invariant. Hence we can write

\displaystyle P = P(r), \quad r \equiv \log{|z|^2} \ \ \ \ \ (1)

Define {\phi \equiv P_r}. Then by using the chain rule we will get

\displaystyle h_{\alpha \bar{\beta}} = \phi e^{-r} \delta_{\alpha \bar{\beta}} + (\phi_r - \phi) e^{-2r} \bar{z}_\alpha z_\beta \ \ \ \ \ (2)

Therefore, we also have

\displaystyle h^{\alpha \bar{\beta}} = \phi^{-1} e^r \delta^{\alpha \bar{\beta}} + (\phi_r^{-1} - \phi^{-1}) z^\alpha \bar{z}^\beta \ \ \ \ \ (3)

and

\displaystyle \det{(h_{\alpha \bar{\beta}})} = e^{-nr} \phi^{n-1} \phi_r \ \ \ \ \ (4)

{h} is positive definite if and only if {\phi > 0} and {\phi_r > 0}.

Suppose now {h} is {U(n)}-invariant gradient Kahler-Ricci solitons. Then {h} satisfies

\displaystyle -R_{\alpha \bar{\beta}} = 2 \nabla_\alpha \nabla_{\bar{\beta}} Q + 2 \lambda h_{\alpha \bar{\beta}} \ \ \ \ \ (5)

where

\displaystyle R_{\alpha \bar{\beta}} = -2 \partial_\alpha \partial_{\bar{\beta}} \log{\det{h}} \ \ \ \ \ (6)

and also we have

\displaystyle \nabla_{\bar{\alpha}} X^\beta = \nabla_{\bar{\alpha}} (2 h^{\beta \bar{\gamma}} \nabla_{\bar{\gamma}} Q) = 0 \ \ \ \ \ (7)

from which we can guarantee that {X} is holomorphic.

Hence,

\displaystyle \partial_\alpha \partial_{\bar{\beta}} (\log{\det{h}} - Q - \lambda P) = 0 \ \ \ \ \ (8)

Assume that {Q = \log{\det{h}} - \lambda P}. Therefore,

\displaystyle \partial_{\bar{\alpha}} (h^{\beta \bar{\gamma}} \partial_{\bar{\gamma}} Q) = 0 \ \ \ \ \ (9)

\displaystyle \partial_{\bar{\alpha}} (h^{\beta \bar{\gamma}} \partial_{\bar{\gamma}} (\log{\det{h}} - \lambda P)) = 0 \ \ \ \ \ (10)

We want to write another description for equation (10). Remember that

\displaystyle X^\beta = 2 h^{\beta \bar{\gamma}} \frac{\partial Q}{\partial \bar{z}^\gamma} \ \ \ \ \ (11)

and by using the simple straightforward calculation we will arive at

\displaystyle X^\beta = 2 \frac{Q_r}{P_{rr}} z^\beta \ \ \ \ \ (12)

So {X} is holomorphic if and only if there exists {\mu \in {\mathbb R}} such that

\displaystyle Q_r = \mu P_{rr} \ \ \ \ \ (13)

But

\displaystyle Q = \log{\det{h}} - \lambda P \ \ \ \ \ (14)

\displaystyle Q = -nr + (n-1) \log{\phi} + \log{\phi_r} - \lambda P \ \ \ \ \ (15)

Hence

\displaystyle Q_r = -n + (n-1) \frac{\phi_r}{\phi} + \frac{\phi_{rr}}{\phi_r} - (n + \lambda \phi) = \mu \phi_r \ \ \ \ \ (16)

Inserting this into equation (13), we will have

\displaystyle \frac{\phi_{rr}}{\phi_r} + (n-1) \frac{\phi_r}{\phi} - \mu \phi_r - (n + \lambda \phi) = 0 \ \ \ \ \ (17)

We can state equation (17) in terms of {P} and its derivatives by finding the expression for {\mu} and differentiating with respect to {r}, i.e.

\displaystyle \mu = \frac{\phi_{rr}}{\phi_r^2} + \frac{n-1}{\phi} - \frac{n + \lambda \phi}{\phi_r} \ \ \ \ \ (18)

\displaystyle 0 = \frac{d \mu}{dr} = \frac{\phi_r^2 \phi_{rrr} - 2 \phi_r \phi_{rr}^2}{\phi_r^4} - \frac{(n-1) \phi_r}{\phi^2} - \frac{\lambda \phi_r^2 - (n + \lambda \phi) \phi_{rr}}{\phi_r^2} \ \ \ \ \ (19)

Then, after a little bit rearrangement, we have

\displaystyle P_{rrrr} - 2 \frac{P_{rrr}^2}{P_{rr}} + n P_{rrr} - (n-1) \frac{P_{rr}^3}{P_r^2} + \lambda (P_{rrr} P_r - P_{rr}^2) = 0 \ \ \ \ \ (20)

We assume in equation (17) above that {\mu \neq 0}, because we assume that {h} is not trivial (Einstein metric). Since {\phi_r > 0}, we can write {r} as a function of {\phi}. Define {F(\phi) \equiv \phi_r}. Then, from (17), by remembering that {\phi_{rr} / \phi_r = d \phi_r / d \phi = F'(\phi)},

\displaystyle F'(\phi) + ( \frac{n-1}{\phi} - \mu ) F(\phi) - (n + \lambda \phi) = 0 \ \ \ \ \ (21)

Let {A(\phi) \equiv \frac{n-1}{\phi} - \mu}, and {B(\phi) \equiv n + \lambda \phi}. Hence,

\displaystyle F'(\phi) + A(\phi) F(\phi) = B(\phi) \ \ \ \ \ (22)

\displaystyle e^{\int A(\phi) d\phi} F(\phi) = \int B(\phi) e^{\int A(\phi) d\phi} d\phi \ \ \ \ \ (23)

But,

\displaystyle e^{\int A(\phi) d\phi} = \phi^{n-1} e^{-\mu \phi} \ \ \ \ \ (24)

It yields

\displaystyle F(\phi) = \phi^{1-n} e^{\mu \phi} \int (n + \lambda \phi) \phi^{n-1} e^{-\mu \phi} d\phi \ \ \ \ \ (25)

\displaystyle F(\phi) = \phi^{1-n} e^{\mu \phi} (\nu + \lambda I_n + n I_{n-1}) \ \ \ \ \ (26)

where {I_m = \int \phi^m e^{-\mu \phi} d\phi}.

We are now trying to find {I_m}.

\displaystyle I_m = \int \phi^m e^{-\mu \phi} d\phi \ \ \ \ \ (27)

\displaystyle I_m = -\frac{1}{\mu} \phi^m e^{-\mu \phi} + \frac{m}{\mu} I_{m-1} \ \ \ \ \ (28)

Set {m=n}, then

\displaystyle I_n = -\frac{1}{\mu} \phi^n e^{-\mu \phi} + \frac{n}{\mu} I_{n-1} \ \ \ \ \ (29)

This recursive equation can be solved easily to find the expression for {I_n}, by noting that

\displaystyle I_1 = \int \phi e^{-\mu \phi} d\phi = -\frac{1}{\mu} \phi e^{-\mu \phi} - \frac{1}{\mu^2} e^{-\mu \phi} \ \ \ \ \ (30)

such that we can write

\displaystyle I_n = - e^{-\mu \phi} \sum_{j=0}^n \frac{1}{\mu^{j+1}} \frac{n!}{(n-j)!} \phi^{n-j} \ \ \ \ \ (31)

If we change the order of summation, hence change the index of summation, then

\displaystyle I_n = - e^{-\mu \phi} \sum_{j=0}^n \frac{1}{\mu^{n-j+1}} \frac{n!}{j!} \phi^j \ \ \ \ \ (32)

Therefore, we will have

\displaystyle F(\phi) = \frac{\nu e^{\mu \phi}}{\phi^{n-1}} + \frac{\lambda e^{\mu \phi}}{\phi^{n-1}} I_n + \frac{n e^{\mu \phi}}{\phi^{n-1}} I_{n-1} \ \ \ \ \ (33)

\displaystyle \phi_r = F(\phi) = \frac{\nu e^{\mu \phi}}{\phi^{n-1}} - \frac{\lambda}{\mu} \phi - \frac{\lambda + \mu}{\mu^{1+n}} \sum_{j=0}^{n-1} \frac{n!}{j!} \mu^j \phi^{j+1-n} \ \ \ \ \ (34)

\displaystyle \frac{d\phi}{dr} = \frac{\nu e^{\mu \phi}}{\phi^{n-1}} - \frac{\lambda}{\mu} \phi - \frac{\lambda + \mu}{\mu^{1+n}} \sum_{j=0}^{n-1} \frac{n!}{j!} \mu^j \phi^{j+1-n} \ \ \ \ \ (35)

This is the first order ordinary differential equation of the Kahler potential of Kahler-Ricci soliton, that in low dimensional manifold can be solved explicitly.

from → Differential Geometry

Two views in classical mechanics: Newtonian theory and variational principle

2009 September 24
5 Comments
tags: , , , ,
by Octavian

In classical mechanics, there are two approaches that we can use to describe the details of particle’s motion (note: since we are talking about classical theory of nature, we should expect that this particle behaves like a macroscopic thing, in a low speed region). These two approaches are Newton’s law and variational principle. In Newtonian theory, every single details of particle’s motion can be described by the ultimate formula (called Newton’s law II), which relates the force of interaction and the acceleration of particle. If we write down the appropriate data of particle’s motion in a certain time t, for instance its position, velocity, or its normal acceleration, in principle we can obtain all we want to know before, and after time t. In other words, sufficient information in a single time dictates what will happen in the next time, and records what has happened in the previous one. This is the root of the idea which says that we can predict future by using present data. It is also the thing that makes Newton’s theory become overwhelming for several centuries. At least this theory can make anyone who is contra becomes popular all over the world (just call it Einstein).

On the other hand, the variational principle has a contrast difference with the former theory, although its goal is the same. This principle says that if a particle moves from one point to another, then the actual path of that particle (the path that is really observed in experiment) is the one that has an extremum value of a certain functional. From the last statement, it’s very clear that our first strategy to find the actual path (and also another information related to the particle’s motion) is to compare. By comparing the values of that functional (next we call it action) for all possibilities of paths the particle can undergo (which is infinitely many), and finding out which one of them that has the extremum value, it’s done. Of course, this is not an easy stuff to find. In a complex case the only thing we can work out is just finding the differential equation that this path satisfies, not the equation which describes the curve explicitly.

The big difference between Newtonian theory and variational principle is in our view to tackle the nature. In Newtonian view, we are the follower of particle; everywhere and everytime this particle moves, we are watching and we are following wherever it is. But in variational view, we are watching also, but from a high hill; we are watching the motion for all time at a glance, and picking the suitable path among infinitely many possible ones. Despite of this difference, what we will get is the same: all informations and details related to particle’s motion in space.

from → Classical Mechanics

Basics of the Kahler manifolds

2009 September 22
Leave a comment
tags: , , , , , , , , , , , ,
by Octavian

Suppose we have a complex manifold {M} endowed with the Riemannian metric {g}. We know that this metric is positive-definite simmetric bilinear tensor. {g} is called Hermitian metric if

\displaystyle \begin{array}{rcl} g(X,Y) = g(JX,JY) \end{array}

for every {X, Y \in \chi (M)}, and {J} is a complex structure of {M}. Componentwise, we can state the Hermitian metric {g} as

\displaystyle g = g_{\mu \bar{\nu}} dz^\mu \otimes d\bar{z}^\nu + g_{\bar{\mu} \nu} d\bar{z}^\mu \otimes dz^\nu \ \ \ \ \ (1)

Define a Kahler form of {g} as

\displaystyle \begin{array}{rcl} \omega(X,Y) = g(JX,Y) \end{array}

for every {X, Y \in \chi (M)}, and {J} is a complex structure of {M}. Hence we can write the componentwise version of Kahler form as

\displaystyle \omega = i g_{\mu \bar{\nu}} dz^\mu \wedge d\bar{z}^\nu \ \ \ \ \ (2)

where {g_{\mu \bar{\nu}}} in equation (2) above is Hermitian. The complex manifold endowed with a Hermitian metric {g} is called Hermitian manifold.

The Riemann curvature tensor of a Hermitian manifold is defined as

\displaystyle R(X,Y)Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]} Z \ \ \ \ \ (3)

for {X,Y,Z \in \chi(M)}. From Riemann curvature tensor we can get the Ricci form defined as

\displaystyle \mathcal{R} = i R_{\mu \bar{\nu}} dz^\mu \wedge d\bar{z}^\nu \ \ \ \ \ (4)

where {R_{\mu \bar{\nu}}} is the contraction of Riemann tensor:

\displaystyle R_{\mu \bar{\nu}} = {R^\kappa}_{\kappa \mu \bar{\nu}} = - \partial_\mu \partial_{\bar{\nu}} \log{G} \ \ \ \ \ (5)

where {G = \det{(g_{\mu \bar{\nu}})}}.

The metric {g} of a Hermitian manifold {M} is called Kahler metric if the Kahler form is a closed {(1,1)}-form, i.e.

\displaystyle d \omega = 0 \ \ \ \ \ (6)

And from equation (6) above we can get

\displaystyle \frac{\partial g_{\mu \bar{\nu}}}{\partial z^\lambda} = \frac{\partial g_{\lambda \bar{\nu}}}{\partial z^\mu} \ \ \ \ \ (7)

\displaystyle \frac{\partial g_{\mu \bar{\nu}}}{\partial \bar{z}^\lambda} = \frac{\partial g_{\mu \bar{\lambda}}}{\partial \bar{z}^\nu} \ \ \ \ \ (8)

which from the Kahler conditions (7) and (8) above we can find an explicit component of {g}, i.e.

\displaystyle g_{\mu \bar{\nu}} = \partial_\mu \partial_{\bar{\nu}} K \ \ \ \ \ (9)

{K} is called the Kahler potential. For if the metric {g} is Kahler, then we can get the expression for Kahler form as

\displaystyle \omega = i \partial \bar{\partial} K \ \ \ \ \ (10)

The Hermitian manifold endowed with a Kahler metric is called Kahler manifold.

Kahler manifold is torsion free. The Ricci form of Kahler manifold is defined as before,

\displaystyle Ric = -i \partial_\mu \partial_{\bar{\nu}} \log{G} dz^\mu \wedge d\bar{z}^\nu \ \ \ \ \ (11)

where we have used the notation {Ric} instead of {\mathcal{R}} since in Kahler manifold, the component of Ricci form, {R_{\mu \bar{\nu}}}, is same as Ricci curvature {Ric_{\mu \bar{\nu}} = {R^\kappa}_{\mu \kappa \bar{\nu}}}, due to the additional simmetry of the components of Riemann tensor results from the Kahler condition.

from → Differential Geometry

My very unimportant calculation with the integers

2009 September 18
2 Comments
tags: , ,
by Octavian

Suppose you have the Kronecker delta defined for {i,j \in \mathbb{Z}^+} as

\displaystyle \delta_{ij} = 1 \ \ \ \ \ (1)

if {i=j}, and

\displaystyle \delta_{ij} = 0 \ \ \ \ \ (2)

if {i \neq j}. Then define a new tensor that invert the duty of Kronecker,

\displaystyle g_{ij} = 0 \ \ \ \ \ (3)

if {i=j}, and

\displaystyle g_{ij} = 1 \ \ \ \ \ (4)

if {i \neq j}. Obviously, we have this relation

\displaystyle \delta_{ij} + g_{ij} = 1 \ \ \ \ \ (5)

for all {i,j \in \mathbb{Z}^+}, and also

\displaystyle \delta^n_{ij} + g^m_{ij} = 1 \ \ \ \ \ (6)

for all {i,j \in \mathbb{Z}^+}, and {n,m \in \mathbb{Z}^+}. Clearly the relation (5) is just the very special case of relation (6).

By taking {n=m} in equation (6), and by substituting {g_{ij}} from equation (5), we will have

\displaystyle (1 - \delta_{ij})^n = 1 - \delta^n_{ij} \ \ \ \ \ (7)

Then by expanding the LHS of equation (7), we will get this relation

\displaystyle (1 + (-1)^n) \delta^n_{ij} + \sum_{k=1}^{n-1} (-1)^k \binom{n}{k} \delta^{k}_{ij} = 0 \ \ \ \ \ (8)

And by setting {i=j} in the equation (8) above, we will get

\displaystyle (1 + (-1)^n) + \sum_{k=1}^{n-1} (-1)^k \binom{n}{k} = 0 \ \ \ \ \ (9)

If {n} is even, then

\displaystyle 2 = \sum_{k=1}^{n-1} (-1)^{k+1} \binom{n}{k} \ \ \ \ \ (10)

and if {n} is odd, then

\displaystyle 0 = \sum_{k=1}^{n-1} (-1)^{k+1} \binom{n}{k} \ \ \ \ \ (11)

Take {n} as an even positive number, we are clearly led to this relation

\displaystyle a = \frac{1}{2} \sum_{n=1}^a \sum_{k=1}^{2n-1} (-1)^{k+1} \binom{2n}{k} \ \ \ \ \ (12)

for {a \in \mathbb{Z}^+}.

Here are some intricate questions:

  1. By knowing that

    \displaystyle \begin{array}{rcl} a = \frac{1}{2} \sum_{n=1}^a \sum_{k=1}^{2n-1} (-1)^{k+1} \binom{2n}{k} \end{array}

    and

    \displaystyle \begin{array}{rcl} b = \frac{1}{2} \sum_{n=1}^b \sum_{k=1}^{2n-1} (-1)^{k+1} \binom{2n}{k} \end{array}

    is it right that

    \displaystyle \begin{array}{rcl} a + b = \frac{1}{2} \sum_{n=1}^{a+b} \sum_{k=1}^{2n-1} (-1)^{k+1} \binom{2n}{k} \end{array}

  2. By knowing that

    \displaystyle \begin{array}{rcl} a = \frac{1}{2} \sum_{n=1}^a \sum_{k=1}^{2n-1} (-1)^{k+1} \binom{2n}{k} \end{array}

    and

    \displaystyle \begin{array}{rcl} 1 = \frac{1}{2} \sum_{k=1}^{2n-1} (-1)^{k+1} \binom{2n}{k} \end{array}

    can you write another description of {a} by considering that {a = a \cdot 1}?

from → Math

The complex projective plane

2009 September 10
Leave a comment
tags: , , , ,
by Octavian

The complex projective plane, {{\mathbb C} \mathbb{P}^{n-1}}, is the set of lines originated at the origin of {{\mathbb C}^n}. Define the coordinate system {\{ z^\mu \}} on {{\mathbb C}^n}, where {0 \leq \mu \leq n-1}. Define the equivalence relation {\sim} on {{\mathbb C}^n - \{0\}} such that {w,z \in {\mathbb C}^n - \{0\}} belong to the same equivalence class if there exists a number {a \in {\mathbb R} - \{0\}} such that {w=az}. Then the class {[w]} is just merely a line originated at the origin of {{\mathbb C}^n}. Hence {({\mathbb C}^n - \{0\})/\sim} is just the set of {[w]}, for {w \in {\mathbb C}^n - \{0\}}.

Take a point {z \in [z] \in ({\mathbb C}^n - \{0\})/\sim}, where {z=(z^0,\ldots,z^\mu,\ldots,z^{n-1})}, {0 \leq \mu \leq n-1}. Define a chart {U_\mu \subset {\mathbb C}^n - \{0\}} such that {z^\mu \neq 0}. Define {\xi_\mu = (\frac{z^0}{z^\mu}, \ldots, \frac{z^{\mu - 1}}{z^\mu}, \frac{z^{\mu + 1}}{z^\mu}, \ldots, \frac{z^{n-1}}{z^\mu}) = \{ \xi_\mu^\nu \}}, where {\nu = \{0, \ldots, \mu - 1, \mu + 1, \ldots, n-1 \}}. {\xi_\mu} acts as the coordinate of point in {{\mathbb C} \mathbb{P}^{n-1}}, in the chart {U_\mu}. If we introduce another chart {U_\alpha}, then the coordinate in this new chart is {\xi_\alpha = (\frac{z^0}{z^\alpha}, \ldots, \frac{z^{\alpha - 1}}{z^\alpha}, \frac{z^{\alpha + 1}}{z^\alpha}, \ldots, \frac{z^{n-1}}{z^\alpha})}, such that in the intersection {U_\mu \cap U_\alpha} the transformation function between coordinates {\xi_\mu} and {\xi_\alpha} must be holomorphic.

The coordinate system {\{z^\mu\}} is called homogeneous coordinate, while {\{\xi_\mu\}} is called inhomogeneous coordinate.

Define a new notation {\zeta_\mu^{\nu + 1} = \xi_\mu^\nu} for {0 \leq \nu \leq \mu -1}, and {\zeta_\mu^\nu = \xi_\mu^\nu} for {\mu+1 \leq \nu \leq n-1}. It’s just the renaming program of the coordinate notation for {{\mathbb C} \mathbb{P}^{n-1}}.

We need to find a function of these {\zeta}’s that can represent the Kahler potential in {{\mathbb C} \mathbb{P}^{n-1}}. The existence of this function will be the first sign for us to ensure that {{\mathbb C} \mathbb{P}^{n-1}} is Kahler manifold. Let’s define the function on the chart {U_\alpha},

\displaystyle K_\alpha (p) = \sum_{\nu = 1}^{n-1} |\zeta_\alpha^\nu (p)|^2 + 1 = \sum_{\nu = 1}^{n-1} \Big|\frac{z^\nu}{z^\alpha}\Big|^2, \quad p \in U_\alpha \ \ \ \ \ (1)

then on another chart {U_\beta} we have another function {K_\beta} defined similarly with the former, and this relation must hold in the intersection of those two charts,

\displaystyle \begin{array}{rcl} K_\alpha (p) = \Big|\frac{z^\beta}{z^\alpha}\Big|^2 K_\beta (p), \quad p \in U_\alpha \cap U_\beta \end{array}

Then,

\displaystyle \begin{array}{rcl} \ln{K_\alpha} = \ln{K_\beta} + \ln{\frac{z^\beta}{z^\alpha}} + \ln{\frac{\bar{z}^\beta}{\bar{z}^\alpha}} \end{array}

Since {\frac{z^\beta}{z^\alpha}} is a holomorphic function, then {\bar{\partial} \ln{\frac{z^\beta}{z^\alpha}} = 0}. Also, {\partial \ln{\frac{\bar{z}^\beta}{\bar{z}^\alpha}} = 0}. Then,

\displaystyle \partial \bar{\partial} \ln{K_\alpha} = \partial \bar{\partial} \ln{K_\beta} \ \ \ \ \ (2)

Define a closed two-form {\Omega} locally by

\displaystyle \Omega = i \partial \bar{\partial} \ln{K} \ \ \ \ \ (3)

then we can show that it is the Kahler form. We must show that the metric corresponding to this form is Hermitian, and positive definite. The closeness behavior of {\Omega} will guarantee that the metric is Kahler.

Take {X, Y \in T_p {\mathbb C} \mathop{\mathbb P}^{n-1}}, and define {g: T_p {\mathbb C}\mathop{\mathbb P}^{n-1} \times T_p {\mathbb C}\mathop{\mathbb P}^{n-1} \rightarrow {\mathbb R}} by {g(X,Y) = \Omega(X,JY)}, where {p \in M}. The metric {g} is Hermitian. Indeed,

\displaystyle \begin{array}{rcl} g(JX,JY) &=& -\Omega (JX,Y) = \Omega (Y, JX) = g(Y,X) \\ &=& g(X,Y) \end{array}

On a chart {U_\alpha}, we obtain

\displaystyle \Omega = i \frac{\partial^2}{\partial \zeta^\mu \partial \bar{\zeta}^\nu} \ln{K} d\zeta^\mu \wedge d\bar{\zeta}^\nu \ \ \ \ \ (4)

and from equation (3) we will have

\displaystyle \begin{array}{rcl} \Omega &=& i \frac{\partial^2}{\partial \zeta^\mu \partial \bar{\zeta}^\nu} \ln{(\sum_{\omega = 1}^{n-1} \zeta^\omega \bar{\zeta}^\omega + 1 )} d\zeta^\mu \wedge d\bar{\zeta}^\nu \\ &=& i \frac{\delta_{\mu \nu} (\sum_{\omega = 1}^{n-1} \zeta^\omega \bar{\zeta}^\omega + 1) - \zeta^\nu \bar{\zeta}^\mu}{(\sum_{\omega = 1}^{n-1} \zeta^\omega \bar{\zeta}^\omega + 1)^2} d\zeta^\mu \wedge d\bar{\zeta}^\nu \end{array}

Define a real vector field {X = X^\mu \frac{\partial}{\partial \zeta^\mu} + \bar{X}^\mu \frac{\partial}{\partial \bar{\zeta}^\mu}}, such that {JX = i X^\mu \frac{\partial}{\partial \zeta^\mu} - i \bar{X}^\mu \frac{\partial}{\partial \bar{\zeta}^\mu}}. Hence,

\displaystyle \begin{array}{rcl} g(X,X) &=& \Omega(X,JX) = 2 \sum_{\mu, \nu} \frac{\delta_{\mu \nu} (\sum_{\omega = 1}^{n-1} \zeta^\omega \bar{\zeta}^\omega + 1) - \zeta^\nu \bar{\zeta}^\mu}{(\sum_{\omega = 1}^{n-1} \zeta^\omega \bar{\zeta}^\omega + 1)^2} X^\mu \bar{X}^\nu \\ &=& 2 ( \sum_{\mu = 1}^{n-1} |X^\mu|^2 (\sum_{\omega = 1}^{n-1} |\bar{\zeta}^\omega|^2 + 1 ) - (\sum_{\mu = 1}^{n-1} |X^\mu \bar{\zeta}^\mu|^2 ) ) \\ && (\sum_{\omega = 1}^{n-1} |\zeta^\omega|^2 + 1 )^{-2} \\ &\geq & 0 \end{array}

from → Differential Geometry

Just a mental exercise

2009 September 9
9 Comments
tags: , ,
by Octavian

Suppose you have a Mobius strip.

  1. Draw the line from an initial point on the Mobius strip down the middle of that strip, and return to the initial point to form a loop. By definition, if you walk along that loop you will end up on the initial point you start to walk, with your feet have touched the two sides of the strip. This behavior will be the ultimate difference between Mobius and ordinary strip; in ordinary strip you just walk on one side of strip, and cannot touch another side.
    If you cut the strip along the loop, what object you will have after your scissor returns to the initial point?
  2. Suppose you also have the second identical Mobius strip. Place those two strips in front of you with their sides touch each other along the edge-to-edge line, such that at a glance it seems like a number 8. Cut along that line but immediately glue the corresponding sides such that they still stick together, merging two Mobius strips into one larger object, and making that number 8 like a big O. Yes, at a glance it’s just a big O, but precisely what kind of object it is? Anticipate all possibilities.

Compare the answers of question 1 and 2.

from → Differential Geometry, Puzzles

Persamaan Differensial Orde-1: Part 1

2009 September 8
2 Comments
tags: , , , ,
by Octavian

Misal kita mempunyai persamaan differensial orde-1 berbentuk

\displaystyle P(x,y) dx + Q(x,y) dy = 0 \ \ \ \ \ (1)

di mana {P(x,y)} dan {Q(x,y)} adalah fungsi dua variabel {x} dan {y}.

— 1. Separable differential equations —

Definition 1 Persamaan differential (1) kita sebut separable jika kita dapat mengubahnya menjadi bentuk

\displaystyle f(x) dx + g(y) dy = 0 \ \ \ \ \ (2)

di mana {f(x)} dan {g(y)} adalah fungsi 1 variabel.

Jika kita dapat mengubah persamaan differential (1) menjadi (2), maka kita dapat mengintegralkan suku demi suku pada persamaan (2) untuk mendapatkan solusi dari persamaan differential (1). Contoh dari persamaan differensial yang separable adalah

\displaystyle \begin{array}{rcl} x \sqrt{1-y^2} dx - y \sqrt{1-x^2} dy = 0 \end{array}

sedangkan contoh untuk persamaan differensial yang tidak bisa dipisahkan variabelnya adalah

\displaystyle \begin{array}{rcl} (xy + \frac{1}{y^2}) dx + y dy = 0 \end{array}

Dengan mengintegralkan suku demi suku pada persamaan ini kita dapat dengan mudah mencari solusi dari persamaan differensial tersebut.

— 2. Homogeneous differential equations —

Definition 2 Suatu fungsi 2 variabel {f(x,y)} disebut homogen orde-n jika kita dapat menyatakan fungsi {f} tersebut dalam bentuk

\displaystyle f(x,y) = x^n h(u) \ \ \ \ \ (3)

di mana {u = y/x}, {h(u)} adalah fungsi dalam {u}, dan {n} adalah sebarang bilangan bulat. Atau dalam bentuk lain, {f} homogen (orde-n) jika kita dapat menyatakannya dalam bentuk

\displaystyle f(x,y) = y^n h(u) \ \ \ \ \ (4)

di mana {u = x/y}, {h(u)} adalah fungsi dalam {u}, dan {n} adalah sebarang bilangan bulat.

Secara ekivalen kita juga dapat mengatakan bahwa {f(x,y)} homogen jika

\displaystyle f(tx, ty) = t^n f(x,y) \ \ \ \ \ (5)

untuk sebarang bilangan bulat {n}.

Example 1 Fungsi {f(x,y) = x^2 + y^2 + \ln{\frac{y}{x}}} bukan fungsi homogen karena

\displaystyle \begin{array}{rcl} f(tx, ty) = t^2 x^2 + t^2 y^2 + \ln{\frac{ty}{tx}} \neq t^2 (x^2 + y^2 + \ln{\frac{y}{x}}) = t^2 f(x,y) \end{array}

Kita juga tahu bahwa fungsi {g(x,y) = x^2 + y^3 + \ln{\frac{y}{x}}} tidak homogen karena kita tidak bisa mencari {n} sehingga berlaku {g(tx,ty) = t^n g(x,y)}.

Example 2 Fungsi {f(x,y) = x^5 + 5 x^3 y^2 + 18 x^2 y^3 - 3 x y^4 + y^5} adalah fungsi homogen orde-5, karena

\displaystyle \begin{array}{rcl} f(tx,ty) = t^5 (x^5 + 5 x^3 y^2 + 18 x^2 y^3 - 3 x y^4 + y^5) = t^5 f(x,y) \end{array}

Definition 3 Persamaan differensial parsial (1) disebut persamaan differensial homogen jika koefisien {P(x,y)} dan {Q(x,y)} adalah fungsi homogen berorde sama.

Example 3 Persamaan differensial ini,

\displaystyle \begin{array}{rcl} (\sqrt{x^2 + y^2} + x) dx + x dy = 0 \end{array}

adalah homogen, karena koefisien dari {dx} dan {dy} adalah fungsi yang homogen dan sama-sama berorde 1.

Metode untuk menyelesaikan persamaan differensial homogen adalah dengan mensubtitusi

\displaystyle y = ux, \qquad dy = u dx + x du \ \ \ \ \ (6)

ke dalam persamaan differensial, sehingga nantinya kita akan mendapat persamaan differensial yang separable. Setelah itu, tinggal mencari solusi untuk persamaan differensial separable seperti biasa.

Example 4 Cari solusi dari

\displaystyle \begin{array}{rcl} (\sqrt{x^2 - y^2} + y) dx - x dy = 0 \end{array}

Karena persamaan differensial di atas adalah homogen (just check it), maka kita substitusi {y = ux}, dengan catatan bahwa {|u| \leq 1} agar akar pada persamaan di atas tetap terdefinisi. Sehingga,

\displaystyle \begin{array}{rcl} (\sqrt{x^2 - u^2 x^2} + ux) dx - x (u dx + x du) = 0 \end{array}

Jika kita mengeset {x \neq 0}, maka kita dapat membagi kedua suku dengan {x}, dan akan kita dapat

\displaystyle \begin{array}{rcl} \pm \sqrt{1-u^2} dx - x du = 0 \end{array}

Kemudian jika kita membatasi {u} agar {u \neq -1}, akan kita dapat

\displaystyle \begin{array}{rcl} \frac{dx}{x} = \pm \frac{du}{\sqrt{1-u^2}} \end{array}

Kita telah mendapat bentuk persamaan differensial yang separable, jadi untuk mencari solusinya adalah dengan mengintegral setiap sukunya. An easy task!

from → Mathematical Physics

Hukum Gauss: Part 1

2009 August 30
16 Comments
tags: , , , , , , , , , , , ,
by Octavian

Pada post sebelumnya, kita telah membahas salah satu metode untuk mencari nilai medan listrik dari sebuah sistem muatan, yaitu dengan memakai hukum Coulomb. Untuk muatan titik, kita tahu (you should, heh) bahwa medan listriknya berbentuk sebagai berikut,

\displaystyle \bf{E} = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2} \hat{\bf{r}} \ \ \ \ \ (1)

dan untuk muatan listrik yang bukan titik (misalnya bola, silinder, atau apapun bentuknya) dapat kita cari dengan membagi-bagi muatan tersebut menjadi muatan-muatan yang kecil, sehingga setiap serpihan memberikan kontribusi seperti pada persamaan (1). Untuk mencari total medan listrik dari semua serpihan muatan tersebut, kita menggunakan integral, untuk menjumlahkan semua kontribusi dari setiap serpihan kecil muatan, yaitu

\displaystyle \bf{E} = \frac{1}{4\pi \epsilon_0} \int \frac{dQ}{r^2} \hat{\bf{r}} \ \ \ \ \ (2)

Persamaan (2) pada dasarnya adalah persamaan dewa; apapun bentuk medan listrik dan sesusah apapun kontribusi muatannya, kita secara teori dapat menghitung nilai medan listrik yang dia hasilkan. Tapi pening yang ditimbulkan oleh persamaan ini juga tidak kalah besarnya. Yeah, you must pay a price for a good thing. Untuk sistem muatan yang sederhana sekalipun, persamaan (2) dapat memberikan komplikasi yang tidak bisa ditolerir. Misal, coba cari saja medan listrik yang ditimbulkan oleh suatu bola pejal bermuatan {Q} dan berjari-jari {R}, dengan menggunakan persamaan di atas. It’s a kind of chaos. :)

Surely kita butuh sebuah metode lain yang tidak menimbulkan kesulitan seperti ini. Metode itu disebut hukum Gauss.

1. Flux

Sebelum kita membahas tentang hukum Gauss, kita bahas apa itu flux terlebih dahulu.

Misal ada sebuah permukaan persegi {S} dalam sebuah daerah bermedan listrik. Flux adalah ukuran seberapa besar permukaan tersebut ditembus oleh medan listrik. Well, just see a picture.

Pada gambar {(a)}, kita tahu bahwa medan listrik tersebut menembus permukaan {S} (karena arrow medan listriknya tegak lurus dengan bidang permukaan itu), sedangkan untuk gambar {(b)}, medan listriknya tidak menembus permukaan {S} (karena setiap arrow medan listriknya bersinggungan dengan bidang). Secara matematik, flux medan listrik {\Phi} pada permukaan yang sebelah kiri didefinisikan dengan

\displaystyle \Phi = E a^2

sedangkan untuk permukaan sebelah kanan,

\displaystyle \Phi = 0

Konsep flux ini mirip dengan konsep proyeksi. Jika kita memiliki sebuah persegi {S}, dan misalkan kita seperti orang yang melihat {S} itu dari arah datangnya medan listrik, maka nilai flux dapat memberi tahu kita seberapa miring permukaan tersebut relatif terhadap arah pandang kita. Kita telah tahu luasan {S} adalah {a^2} dan kita dapat informasi bahwa fluksnya bernilai {Ea^2}, maka kita simpulkan permukaan itu tegak lurus terhadap arah pandang kita. Jika kita dapat informasi bahwa fluksnya bernilai kurang dari {Ea^2}, maka kita simpulkan bahwa permukaannya miring: vektor yang tegak lurus dengan bidang {S} membentuk sudut tertentu dengan arah pandang kita. Well, jika sudut kemiringan tersebut adalah {\theta}, maka nilai flux yang dihasilkan didefinisikan sebagai

\displaystyle \Phi = E a^2 \cos{\theta} \ \ \ \ \ (3)

atau jika kita mendefinisikan suatu vektor {\bf{n}} yang tegak lurus dengan permukaan {S}, maka nilai fluxnya adalah

\displaystyle \Phi = \bf{E} \cdot \bf{n} \ \ \ \ \ (4)

yang sebenarnya hanyalah generalisasi dari persamaan (3). Gambar di bawah mengilustrasikan apa yang telah terjadi di atas.

Itu jika permukaan {S} yang kita punya adalah persegi. Jika permukaannya berbentuk sebarang, dan mungkin bisa melengkung-lengkung, atau bisa jadi medan listriknya yang tidak uniform, maka nilai fluxnya didapat dengan menjumlahkan semua flux yang dihasilkan untuk setiap luasan-luasan kecil pada permukaan itu. Singkatnya, total flux yang dihasilkan adalah

\displaystyle \Phi = \int \bf{E} \cdot d\bf{n} \ \ \ \ \ (5)

Exercise 1 Sebuah kubus dengan panjang rusuk {L} berada dalam sebuah daerah bermedan listrik uniform {\bf{E}}. Cari flux listrik pada kubus tersebut.

Exercise 2 Cari perbandingan flux yang diterima oleh permukaan {S} akibat medan listrik dari {Q} dengan flux total yang dikeluarkan oleh muatan {Q} tersebut, pada masing-masing kasus berikut.

  • a. Muatan {Q} diletakkan di titik tengah sebuah kubus yang panjang sisinya {a}, dimana permukaan {S} adalah salah satu sisi kubus tersebut.
  • b. Muatan {Q} diletakkan di atas salah satu titik sudut persegi {S}, setinggi {a}, yang juga merupakan panjang sisi {S}.
  • c. Muatan {Q} diletakkan pada sebuah titik yang terletak di diagonal ruang sebuah kubus, yang jaraknya infinitesimal terhadap salah satu titik sudut permukaan {S}.

from → Electromagnetism

Conference in honour of Murray Gell-Mann’s 80th birthday

2009 August 25
Leave a comment
tags:
by Octavian

Conference in Honour of Murray Gell-Mann’s 80th Birthday
Conference in Honour of Murray Gell-Mann’s 80th Birthday

Quantum Mechanics, Elementary Particles, Quantum Cosmology and Complexity

This cool conference will be held in NTU (Nanyang Technological Centre), on February 24-26, 2010.

A comment about this conference? Great!

Please visit the conference’s site, and see the list of invited speakers before you getting speechless.

from → Next Conferences

International Conference on Algebra (ICA) 2010, in honour of the 70th birthday of Professor Shum Kar-Ping

2009 August 25
Leave a comment
tags:
by Octavian

Next year UGM (Universitas Gadjah Mada) will host another big conference on mathematics (now especially about algebra), besides this one that will be attended this year. Reading the title, if you are not an algebraist you will ask about who Shum Kar-Ping is. Who is that guy, and what is his contribution to algebra. Well, you can do googling, or just registering for this conference.

This is the list of important dates:

  • January 30, 2010
    second announcement with instructions
    on the manuscript preparation
  • June 15, 2010
    abstract submission deadline
  • July 15, 2010
    notification on abstract acceptance
  • August 15, 2010
    deadline for registration
  • September 7, 2010
    confirmation of attendance
  • September 30, 2010
    deadline for paper submission for review
  • October 7-10, 2010
    conference time
  • December 25, 2010
    notification on paper acceptance
  • January 30, 2011
    deadline for paper sending to the authors for correction
  • March 31, 2011
    deadline for final paper submission for proceedings

Just visit the official site of this conference for further informations.

from → Next Conferences

« Older Entries