Andy Octavian's Blog

New prospect for particle physics: Plasma accelerator

Posted by: Octavian on: November 12, 2009

In: Particle Physics
1 Comment

This morning I got my copy of Symmetry magazine and found an interesting article about new, tremendous, and insightful methods to increase the energy of accelerated particles using a centimeters-length tube, instead of gigantic tracks-and-circles-shaped accelerators. Just check it out.

I wonder about the car racing at the next millennium.

Tags: accelerators, Particle Physics, particles, symmetry

Rotationally invariant gradient Kahler-Ricci solitons on the complex plane

Posted by: Octavian on: September 30, 2009

In: Differential Geometry
2 Comments

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.

Tags: Differential Geometry, expanding, gradient Kahler-Ricci soliton, holomorphic, Kahler manifolds, Kahler metric, Kahler-Ricci soliton, shrinking

Two views in classical mechanics: Newtonian theory and variational principle

Posted by: Octavian on: September 24, 2009

In: Classical Mechanics
5 Comments

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.

Tags: Classical Mechanics, motion, Newtonian, theory, variational principle

Basics of the Kahler manifolds

Posted by: Octavian on: September 22, 2009

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.

Tags: complex, curvature, Differential Geometry, form, Hermitian, Kahler, manifold, metric, Ricci, Riemann, structure, symmetry, tensor

My very unimportant calculation with the integers

Posted by: Octavian on: September 18, 2009

In: Math
2 Comments

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:

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

Tags: integers, Kronecker delta, series

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Andy Octavian's Blog

New prospect for particle physics: Plasma accelerator

Rotationally invariant gradient Kahler-Ricci solitons on the complex plane

Two views in classical mechanics: Newtonian theory and variational principle

Basics of the Kahler manifolds

My very unimportant calculation with the integers

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