For the last couple of months David had been teaching me a crash course on general relativity. He’d gone through the material but decided to give me one more lecture - scheduled for last Tuesday afternoon. Unfortunately, he passed away unexpectedly that morning.

I sat in that empty lecture room last week, reminiscing and trying to get my head around this awful thing.

The ESA reports on this spectacular lensed quasar image obtained by the Hubble space telescope. This is the first time a 5 times lensed object has been discovered.

Gravitational lensing results in an odd number of images - but some images are very weakly magnified and difficult to see, especially if the image is close to the lensing object. In this case the geometry is quite good and the fifth image is discernable from the lensing object. The five images of the distant quasar are marked by blue circles.

Also shown are three lensed images of a background galaxy, marked by red circles, and a distant supernova marked by the yellow circle.

Full story:European Space Agency

Some papers on gravitational lensing from the arXiv and elsewhere

• Lectures on Gravitational Lensing
• A Newtonian pre-introduction to gravitational lenses
• The Basics of Lensing
• Quasar Lensing
• Gravitational Lensing - A brief review
• Gravitational Lenses

The blurb claims these are some of the most detailed maps of Mars ever made. I don’t know if the maps cover the whole planet, as admittedly my Martian geography is pretty poor.

Just for the fun of it I tried moon.google.com and it resolves. Go take a look and make sure to zoom in

The rest of the planets don’t resolve. You can aslo get to Google Mars via mars.google.com.

Let [tex]\cal{O}[/tex] be an object moving along a path. The position of the object [tex]\cal{O}[/tex] is a function of the time [tex]t[/tex]. We denote the position vector [tex]\bold{x}(t)[/tex].

The time derivative of the position vector [tex]\bold{x}(t)[/tex] is the velocity vector [tex]\bold{v}(t)[/tex] of the object [tex]\cal{O}[/tex].

[tex]\bold{v}(t) \equiv \frac{d\bold{x}(t)}{dt} = \bold{\dot{x}}(t)[/tex].

The time derivative of the velocity vector [tex]\bold{v}(t)[/tex] is called the acceleration vector [tex]\bold{a}(t)[/tex] of the object [tex]\cal{O}[/tex] .

[tex]\bold{a}(t) \equiv \frac{d\bold{v}(t)}{dt} = \frac{d^2\bold{x}(t)}{dt^2} = \bold{\ddot{x}}(t) [/tex].

Integrate this last equation with respect to [tex]t[/tex],

[tex]\int \bold{\ddot{x}}(t) \ dt=\int \bold{a}(t) \ dt \implies \bold{\dot{x}}(t) =\bold{a}( t)t + \bold{C}[/tex],

where [tex]\bold{C}[/tex] is a constant vector to be determined by the initial conditions. Now, [tex]\bold{\dot{x}}(t) =\bold{v}(t)[/tex] so that

[tex]\bold{v}(t) =\bold{a}(t) t + \bold{C}[/tex]

We assume that at time [tex]t =0[/tex] the object [tex]\cal{O}[/tex] is moving with initial velocity [tex]\bold{v}(0) =\bold{v_0}[/tex]

[tex] \bold{v}(0) = \bold{v_0} =\bold{a}(0) \ 0 + C \implies C = \bold{v_0}[/tex]

and substitution gives the familiar formula for the velocity of an object

[tex]\bold{v}(t) = \bold{v_0} + \bold{a}(t) t[/tex].

We can integrate the equation

[tex]\bold{\dot{x}}(t) =\bold{v_0} + \bold{a}(t) t[/tex] with respect to [tex]t[/tex] a second time

[tex]\int \bold{\dot{x}}(t) \ dt = \int \bold{v_0} + \bold{a}(t) t \ dt[/tex] to give

[tex] \bold{x}(t) = \bold{v_0} t+ \frac{1}{2}\bold{a}(t) t^2 + \bold{K} [/tex], where [tex]\bold{K}[/tex] is a constant vector to be determined from the initial conditions.

We assume that at [tex]t =0[/tex] the object [tex]\cal{O}[/tex] has position vector [tex]\bold{x}(0)=\bold{x_0}[/tex]

[tex] \bold{x}(0) = \bold{x_0} = \bold{v_0} t+ \frac{1}{2}\bold{a}(0) 0^2 + \bold{K} \implies \bold{K}=\bold{x_0}[/tex]

and we have the familiar formula for the displacement (position) of the object at time [tex]t[/tex]

[tex] \bold{x}(t) = \bold{x_0} + \bold{v_0} t+ \frac{1}{2}\bold{a}(t) t^2 [/tex].

In a first treatment we often simplify things a bit. Firstly, to simplify notation we don’t write in the explicit time dependence for the position [tex]\bold{x}(t)[/tex] , so that it becomes [tex]\bold{x}[/tex] and similarly for [tex]\bold{v}[/tex] and [tex]\bold{a}[/tex]. Secondly we assume that the object starts from the origin rather than from some arbitrary position, thus we take [tex]\bold{x_0}=0[/tex], (which is really just a change of co-ordinates). This gives the familiar form of the equations of motion from high school text books:

[tex] \bold{x} = \bold{v_0} t+ \frac{1}{2}\bold{a} t^2 [/tex]
[tex]\bold{v} = \bold{v_0} + \bold{a} t[/tex].

These equations are commonly written as

[tex] s = u t+ \frac{1}{2} a t^2 [/tex]
[tex] v = u+ a t[/tex],

where we have used the alternate notation [tex]s \equiv x[/tex] for the position and [tex]u \equiv v_0[/tex] for the intial velocity. Note that we drop the bold face symbols [tex]\bold{v} \rightarrow v[/tex] which are used to denote vectors, since in a first treatment we consider only the one dimensional or scalar case.

Squaring the last equation gives
[tex]v^2 = (u+ a t)^2 = u^2 + 2 u a t + a^2t^2 = u^2 + 2a (ut+\frac{1}{2}at^2) [/tex]
[tex]\phantom{v^2}= u^2 + 2as[/tex]

another commonly used equation in high school physics texts.

The average velocity over the time interval [tex] (t_1,t_2) \equiv (t_i,t_f)[/tex] is given by

[tex]v_{av} = \frac{s_2 -s_1}{t_2-t_1} = \frac{s_f-s_i}{t_f-t_i} = \frac{\Delta(s)}{\Delta(t)}[/tex]

where, [tex] s_1 \equiv s_i[/tex] is the intial displacement and [tex]s_2 \equiv s_f[/tex] is the final displacement. Finally, in words: it is the change in position [tex]\Delta(s)[/tex] divided by the change in time [tex]\Delta(t)[/tex].

We begin our considerations with 2 by 2 matrices [tex]A[/tex] and [tex]B[/tex] over a field [tex]\mathbb{F}[/tex], where
[tex] A= \left ( \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right ) \quad \textrm{and} \quad B= \left ( \begin{array}{cc} b_{11} & b_{12} \\\ b_{21} & b_{22} \end{array} \right )[/tex]

The tensor (or Kronecker) product of [tex]A[/tex] and [tex]B[/tex] is defined as

[tex] A \otimes B = \left ( \begin{array}{cc} a_{11} B & a _{12} B \\\ a_{21} B & a_{22} B \end{array} \right ) [/tex]

[tex] = \left ( \begin{array}{cccc}a_{11} b_{11} & a_{11} b _{12} & a _{12} b_{11} & a _{12}b _{12} \\ a_{11} b_{21} & a_{11} b_{22} & a _{12} b_{21} & a _{12}b_{22} \\ a_{21} b_{11} & a_{21} b _{12} & a_{22} b_{11} & a_{22}b_{12} \\ a_{21} b_{21} & a_{21} b_{22} & a_{22} b_{21} & a_{22} b_{22} \end{array} \right )[/tex].

The tensor product of [tex]B[/tex] with [tex]A [/tex] is
[tex] B \otimes A = \left ( \begin{array}{cc} b_{11} A & b _{12}A \\ b_{21} A & b_{22}A \end{array} \right )[/tex]

[tex] = \left ( \begin{array}{cccc} b_{11} a_{11} & b_{11} a _{12} & b _{12} a_{11} & b _{12} a _{12} \\ b_{11} a_{21} & b_{11} a_{22} & b _{12} a_{21} & b _{12} a_{22} \\ b_{21} a_{11} & b_{21}a _{12} & b_{22} a_{11} & b_{22}a _{12} \\ b_{21}a_{21} & b_{21}a_{22} & b_{22} a_{21} & b_{22}a_{22} \end{array} \right )[/tex] .

Clearly, [tex]A \otimes B \ne B \otimes A[/tex] , so that the tensor product is non-commutative ^[1] (it is however associative).

We now consider the effect of multiplication by a scalar [tex] \nu \in \mathbb F[/tex] . We set

[tex] \nu A = \nu \left ( \begin{array}{cc} a_{11} & a _{12} \\\ a_{21} & a_{22} \end{array}\right ) = \left ( \begin{array}{cc} \nu a_{11} & \nu a _{12} \\\ \nu a_{21} & \nu a_{22} \end{array} \right) [/tex]

now
[tex] \nu A \otimes B = \left ( \begin{array}{cc} \nu a_{11} & \nu a _{12} \\\ \nu a_{21} & \nu a_{22} \end{array}\right) \otimes \left ( \begin{array}{cc} b_{11} & b _{12} \\\ b_{21} & b_{22} \end{array}\right) [/tex]
[tex] = \left ( \begin{array}{cc} \nu a_{11} B & \nu a _{12} B \\\ \nu a_{21} B & \nu a_{22} B \end{array} \right ) [/tex]

[tex] = \left ( \begin{array}{cccc} \nu a_{11} b_{11} & \nu a_{11} b _{12} & \nu a _{12} b_{11} & \nu a _{12}b _{12} \\ \nu a_{11} b_{21} & \nu a_{11} b_{22} & \nu a _{12} b_{21} & \nu a _{12}b_{22} \\ \nu a_{21} b_{11} & \nu a_{21} b _{12} & \nu a_{22} b_{11} & \nu a_{22}b_{12} \\ \nu a_{21} b_{21} & \nu a_{21} b_{22} & \nu a_{22} b_{21} & \nu a_{22} b_{22} \end{array} \right )[/tex]

[tex] = \nu \left ( \begin{array}{cccc} a_{11} b_{11} & a_{11} b _{12} & a _{12} b_{11} & a _{12}b _{12} \\ a_{11} b_{21} & a_{11} b_{22} & a _{12} b_{21} & a _{12}b_{22} \\ a_{21} b_{11} & a_{21} b _{12} & a_{22} b_{11} & a_{22}b_{12} \\ a_{21} b_{21} & a_{21} b_{22} & a_{22} b_{21} & a_{22} b_{22} \end{array} \right )[/tex]

[tex] = \nu (A \otimes B) = A \otimes \nu B [/tex] .

For arbitrary matrices [tex]A = a_{ij}, B = b_{kl}[/tex] the tensor product takes the form
[tex]A \otimes B = \left( \begin{array}{ccc} a_{11} B & \cdots & a_{1j} B \\ \vdots & \ddots & \vdots \\ a_{i1}B & \cdots & a_{ij}B\end{array} \right)[/tex]

For column vectors, [tex]u=\left( \begin{array}{c}u_1 \\\ \vdots \\\ u_j\end{array} \right), v= \left( \begin{array}{c} v_1 \\\ \vdots \\\ v_k \end{array} \right)[/tex] the tensor product is of the same general form.

A column vector is simply a [tex]k \times 1[/tex] matrix and the tensor product of [tex]u[/tex] with [tex]v[/tex] is

[tex]u \otimes v =\left( \begin{array}{c}u_1 v\\\ \vdots \\\ u_j v\end{array} \right)[/tex].

We can begin abstracting a bit now. Let [tex]V [/tex] and [tex]W[/tex] be two finite dimensional vector spaces. We define the tensor product [tex]V \otimes W[/tex] as the vector space spanned by all formal products [tex]v \otimes w[/tex] , [tex]v \in V[/tex] , [tex]w \in W [/tex] where [tex]\otimes [/tex] satisfies

(i) [tex] ( v_1 + v_2 ) \otimes w = v_1 \otimes w + v_2 \otimes w[/tex]

(ii) [tex] v \otimes ( w_1 + w_2 ) = v \otimes w_1 + v \otimes w_2 [/tex]

(iii) [tex] \nu ( v \otimes w ) = ( \nu v ) \otimes w = v \otimes ( \nu w ) [/tex],

for all [tex]v, v_1, v_2 \in V, w,w_1,w_2 \in W, \nu \in \mathbb{F}[/tex].

The first two parts of the defintion are straightforward, we want [tex]\otimes[/tex] to be linear in both the first and the second spaces (i.e. bilinear). The third axiom is the most confusing one. As you can see from the results derived from the Kronecker product of matrices, the third axiom makes sense if we are going to model tensor products algebraically.

Note that a general element of [tex]V \otimes W[/tex] is of the form [tex]\sum_{i,j} \alpha_{ij} v_i \otimes w_j[/tex] and cannot in general be expressed in the simpler form [tex]v \otimes w [/tex], for some [tex]v,w \in V[/tex].

Example, let [tex]V[/tex] be a 2-dimensionl vector space with basis [tex]\{e_1, e_2\}[/tex], then consider the space [tex]V \otimes V[/tex]. A typical element of [tex]V \otimes V[/tex] is
[tex]e_1 \otimes e_2 + e_2 \otimes e_1[/tex].
It cannot be expressed as a tensor product of two vectors, for suppose that there are vectors [tex]v, v’ \in V[/tex] such that
[tex] v \otimes v’ = e_1 \otimes e_2 + e_2 \otimes e_1 [/tex].
Now since [tex]\{e_1,e_2\}[/tex] is a basis of [tex]V[/tex] we can express [tex]v,v’[/tex] in terms of the basis vectors as
[tex]v = \alpha_1 e_1 + \alpha_2 e_2,\quad v’ =\beta_1 e_1 + \beta_2 e_2[/tex]
where [tex]\alpha_1,\alpha_2, \beta_1, \beta_2 \in \mathhbb{F}[/tex], so that

[tex]v \otimes v’ = (\alpha_1 e_1 + \alpha_2 e_2) \otimes (\beta_1 e_1 + \beta_2 e_2)[/tex]
[tex] =\alpha_1 \beta_1 \ e_1 \otimes e_1 + \alpha_1 \beta_2 \ e_1 \otimes e_2 + \alpha_2 \beta_1 \ e_2 \otimes e_1 + \alpha_2 \beta_2 \ e_2 \otimes e_2 [/tex]
[tex]= e_1 \otimes e_2 + e_2 \otimes e_1[/tex].

Now by equating coeffecients we must have [tex]\alpha_1 \beta_1=0[/tex], which implies that either [tex]\alpha_1 \beta_2 =0[/tex] or [tex]\alpha_2 \beta_1 =0[/tex], so that there is no solution, hence there are no vectors [tex]v,v’[/tex] such that [tex] v \otimes v’ = e_1 \otimes e_2 + e_2 \otimes e_1 [/tex].

The tensor product comes up a lot in quantum mechanics, for example in the addition of angular momenta.
It is also the basis of entanglement in quantum mechanics, a state such as
[tex] | \uparrow \ \rangle \otimes |\downarrow \ \rangle + |\downarrow \ \rangle\otimes |\uparrow\ \rangle [/tex]
is called an entangled state because it has no decomposition, just replace [tex]e_1[/tex] with [tex]| \uparrow \ \rangle[/tex] (spin up) and [tex]e_2[/tex] with [tex]| \downarrow \ \rangle[/tex] (spin down) in the previous argument.

A state such as
[tex] |\uparrow\ \rangle \otimes |\uparrow \ \rangle + |\downarrow\ \rangle\otimes |\uparrow \ \rangle [/tex] is not entangled because it can be expressed as [tex] (1|\uparrow \ \rangle + 1|\downarrow \ \rangle) \otimes ( 1|\uparrow \ \rangle + 0|\downarrow \ \rangle )[/tex].

[1]. To see how [tex]A \otimes B[/tex] and [tex]B \otimes A[/tex] are related, we introduce the permutation operator [tex]T[/tex] where,
[tex] T = \left ( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right )[/tex].

Note that [tex]T^2 =I[/tex], so that [tex]T = T^{-1}[/tex]. A quick calculation shows that
[tex] A \otimes B = T \cdot (B \otimes A ) \cdot T^{-1} [/tex].

ATHENS (AFP) - Greece has received copies of letters by Albert Einstein which suggest that the work of an unheralded Greek mathematician helped shape some of his theories, Greek and Israeli officials said.

Israel’s ambassador to Athens, Ram Aviram, presented the Greek foreign ministry with copies of 10 letters between Einstein and Greek mathematician Constantine Karatheodoris, part of a long correspondence which lasted from 1916 to 1930.

According to experts at the National Archives of Israel — custodians of the original letters — the mathematical side of Einstein’s physics theory was partly substantiated through the work of Karatheodoris, Aviram told AFP.

“The correspondence between the two mathematicians is intensive and quite close,” Aviram said. “At a certain moment, they called themselves in private names.”

The son of a Greek-born diplomat who served as the Ottoman Empire’s ambassador to Berlin, Karatheodoris who was born in 1873 and died in 1950 taught mathematics at four German universities — including those of Munich and Goettingen — and also worked on physics and archaeological engineering.

His scientific papers are in the collection of Goettingen University, and have never been translated into Greek, though a number of American universities have copies of his theories, said deputy foreign minister Evripidis Stylianidis.

The Greek authorities intend to create a museum honouring Karatheodoris in Komotini, a major town of the northeastern Greek region where his family came from.

Interesting that they call Einstein a mathematician, as I’ve always thought of him as a theoretical physicist, rather than a mathematical physicist. The only thing mathematical I associate with Uncle Albert is the Einstein summation convention.

Professor Rodney Baxter, from the Mathematical Sciences Institute at ANU, received both the 2006 Onsager Prize of the American Physical Society (APS) and the separate Onsager Lectureship and Medal for 2006 from the Norwegian University of Science and Technology.

Both prizes are named for widely respected theoretical physicist and Nobel Laureate Lars Onsager, who exactly calculated the order parameter of the Ising model in 1949. This was the first such calculation for a statistical mechanical model of magnetism. He received the 1968 Nobel Prize in Chemistry for his earlier work on irreversible thermodynamics.

“These awards are particularly pleasing for me as it is recognition of work on the order parameters of the chiral Potts model, which is research in the Lars Onsager tradition,” Professor Baxter said.

In his research, Professor Baxter showed by careful mathematical analysis that numerical predictions about the order parameters of the chiral Potts model were exactly right, something which had been elusive for mathematicians in the 15 years before his proof.

Simply, the complicated chiral Potts model is a prototype of theoretical descriptions of the interaction and behaviour of materials at the molecular level. It includes the Ising model as a special case.

The “exact solution” of the chiral Potts model achieved by Professor Baxter has important implications in the physical sciences. It greatly increases confidence in theoretical models, particularly in materials science, where physicists around the world, and at ANU, are building next generation electronic devices using two-dimensional layers in ‘chips’. These specialised ‘chips’ may eventually be used in computing, audiovisual technologies and advanced telecommunications.

The American Physical Society prize is awarded to recognise outstanding research in theoretical statistical physics. It awarded Professor Baxter the Prize for “his original and groundbreaking contributions to the field of exactly solved models in statistical mechanics, which continue to inspire profound developments in statistical physics and related fields”.

Full Story from ANU

A University of Queensland research team led by senior mathematics lecturer Dr Yao-Zhong Zhang has successfully solved a major long-standing problem in mathematical physics.

Dr Zhang and his postdoctoral fellows Wen-Li Yang and Shao-You Zhao, from the North West University in China, have discovered the determinant representation of correlation functions of the supersymmetric t-J model.

Dr Zhang said the theoretical problem had been around for many years.

“The analytic computation of correlation functions was arguably one of the most challenging and notoriously difficult problems in mathematical physics and its solution will have important implications.”

“It opens doors to further research in the theory of exactly soluble models as well as in pure mathematics, statistical mechanics and condensed matter physics.”

Dr Zhang said the work had attracted a great deal of academic interest and had been described by world-leading authorities in the field as a “major breakthrough” and a “great discovery”.

A paper on the solution will appear in the January 2006 issue of the Journal of Mathematical Physics, a leading international scientific journal published by the American Institute of Physics.

A second paper has also been submitted to the prestigious mathematical physics journal Communications in Mathematical Physics.

Dr Zhang and his research team have also been approached by the International Journal of Modern Physics B to write a review article on their research.

Source: University of Queensland

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