On PhD and related realities of my life

scp command to transfer directory

2011-05-18T23:43:00.000-07:00

Suppose I want to transfer the 'book' directory from machine 'x' to machine 'y'. Be in the machine 'x', in the parent directory of the 'book' directory, then type the following command:

]$ scp -r book kbhaskar@IP_y:/home/bhaskar

Give password and enter.

2011-05-18T04:05:00.001-07:00

ifort lapacktest.f90 -L/opt/intel/Compiler/11.1/

072/mkl/lib/em64t -lmkl_intel_lp64 -lmkl_sequential -lmkl_core -lpthread

sftp

2011-03-23T00:29:00.000-07:00

I am in one machine, say 'x'. I want to ftp files from this machine to another machine, say 'y'.
1) Be in machine 'x' in the directory from which you want to copy the files.
2) Now type:
    sftp kbhaskar@ipy (ipy is IP address of 'y'.)
    It will ask for password for 'y'. Give it.
3) The following prompt comes up:
    sftp>
    Type in the pathname of the directory in machine 'y' where you want the files to be copied with the 'cd' command:
    sftp> cd 'pathname'
4) mput filename
    Above command will transfer the specified file from machine 'x' to machine 'y'.

Constructing an ellipse

2010-11-21T07:57:00.000-08:00

Cut a sheet of paper into the form of a circle. F2 is the centre of the circle. F1 is an arbitrary point. Now fold the circle so that some point on the circumference coincides with F1. The crease thus formed is a tangent to an ellipse with F1 and F2 as its foci. More such tangents, and you're left with a pretty ellipse!

2009-11-05T01:51:00.001-08:00

1) J4 and J1 values
2) Hole in orb. AF
3) 4-dim basis
4) Magnon softening
5) Jahn-Teller effect
6) Pseudo gap in onset of orb. ordering.

Some Linux commands (for fedora 11) and notes on my OS

2009-08-09T23:42:00.000-07:00

It's a 32-bit Fedora 11 installed on my system.

Below are some commands for searching and installing softwares on the system:

1) export http_proxy="http://username:password@vsnlproxy.iitk.ac.in:3128"
(the http:// after the '=' may not be required)

The above command connects you to the users outside of iitk.

2) yum search player

The above command gives a list of the players available for playing video files. (mplayer etc.) One can also try out dcpp in the same fashion. Works only for free software.

3) yum -y install kmplayer.i586

The above command installs kmplayer.i586 in my computer.

yum -y update This command gives the updates (to which softwares?)

**********************************************

The command for running matlab is:

/opt/matlab7.4/bin/matlab &

***********************************************

This is how to use scp:

kbhaskar@phypc2:~$] scp ./kbhaskar/CMP/orbital_ordering/onset/orb_fluc/further/piavg_.m kbhaskar@172.28.72.29:~/.

Explanation: I am in phypc2, root directory. I am copying the file piavg_.m from phypc2 into the computer with IP address 172.28.72.29. My login ID is kbhaskar on both computers.

This way I don't need to open a mozilla browser and email the file each time I want to get the file from that computer into my present computer.

*******************************************************************************

2008-12-21T20:25:00.000-08:00

An oriental/Arabic sounding scale:
e f g# a b c d# e (ascending)

Improvising in the scale starting from c: C7,Em (largely, though Am may also be used, and even Ebm)

Improvising around d#: B

The G and D chords also may be used, but judiciously, to release the tension. G sounds good with Em, and so does D, but don't mix all three chords together.

Try Dm also.

2008-10-29T11:14:00.000-07:00

check out www.deezer.com

polyrhythms

2008-10-12T11:27:00.000-07:00

i just got an idea for representing polyrhythms. (the latter is a sequence of a fixed number of notes repeated again and again, but beginning each time at a different note, such that the number of notes in the sequence does not match with the number of subdivisions a beat is divided in. an eg. is: c d e d e f e f g f g a, each note being a 16th note in 4/4 time)

take two concentric circles, each rotating at different speeds. the outer one represents the main beat...the inner one represents the notes of the sequence.

Music Planning

2008-09-16T04:57:00.000-07:00

Upcoming music projects:
1) Rondo alla turca
2) ajeeb dastan hai yeh
3) team up with swanand to explore the idea developed with bittu
4) look up the jazz pieces in the book given by ashwin to explore the possibilities in ajeeb dastan...
5) Ventures: "Tequila" and "Let's Go"
6) check out Wes Montgomery's numbers also.

i'm confused...where do i begin?
finish off rondo alla turca and record/mix it. then finish the backing track for ajeeb dastan, then look up ashwin's book, and only then come back to develop ADHY further. project number (3) comes later.

Will I be a great physicist? Well, que sera sera!

2008-06-14T21:20:00.000-07:00

Here's my experience with physics during the last six years or so I've been at IIT doing my PhD:

"In Physics you never understand anything; you just get used to them"

(I believe this statement was originally made by von Neumann in the context of quantum mechanics. But my IIT experience has proved it to be universal to all of Physics, and hence I've modified it accordingly.)

A further modification of that statement:

"In Physics you never start understanding, you only stop asking questions."

- a fact to which I can attest, on the basis of direct experience.

I like the following site:

http://www.ap.smu.ca/~smuaps/quotes.shtml

2008-05-18T10:39:00.000-07:00

Today I attended the second lecture on the photography workshop by Dr. Anupam Pal. I'm putting down some of the things I learnt for future reference.

Buying Guide:

1) Fix your budget
2) If going for DSLR make sure you're getting a lens with the body!
3) Keep some money for a UV filter
4) Keep some money for memory card and a battery charger too (if Eveready type pencils cells are used)
5) Keep some money aside for Memory card reader and camera bag too.

1) 5 mega pixels resolution is sufficient.
2) For point and shoot camera, go for the best OPTICAL zoom. Digital zoom is of no use.
3) Look for exposure control. (A, T, M etc.)
4) look for anti-shake technology.
5) Go for the largest aperture (i.e. smallest f-number)
6) Look for minimum lag time
7) Look for swivel and largest LCD. (swivel means one can change the orientation of the LCD if, for example, the sun is shining directly on to it.)
8) Multimedia capability (just an extra)
9) Go for camera manufacturing companies like Canon, Nikon, Olympus, Pentax, Sony, etc.)
Avoid electronics companies like HP, Panasonic, Epson etc.

Features

Lens types:

1) Prime lenses: have fixed focal length.
2) Lenses with variable focal lengths: zoom lens and varifocal lens.

Lens with focal length of 50 mm is a normal lens (I don't know if that is the technical name). The image looks nearly as you see it.

Lens with focal length < 50 mm is called wide angle lens. Allows more elements to be captured in the frame.
While buying a lens, go for the one which has a smaller lower limit on the focal length (i.e. more wide-angle). The upper limit probably does not matter so much (I'll have to confirm this part)

focal length > 50 mm: telephoto lens.

Macro lenses are used for close-ups. There are something called tilt-shift lenses.

Filters

UV filter for protection (to camera sensor or the eye?)
Circular polarizer (removes reflection and saturates color)
Graduated neutral density filter (can be used for sunsets when the sky is too bright and the earth is too dark)
Neutral density filter.

Flash

Built-in flash: good for fill flash (what's that?) but bad if used as a primary source of light. Creates red eye.

Accessory flash.

Using flash in the sun can be useful. For eg. , if the sun is overhead and illuminating the head or hat of some person, but leaving the face in darkness or the eyes as black spots because of the shadow, then flash can be used to reveal these features.

Using flash at slow shutter speeds can create interesting effects while capturing movement. There will be a general blur with a sharp image at the instant the flash was on.

Light metering: amount of light entering the camera.
1) Auto: camera decides everything.
2) Aperture priority: You set the aperture, camera sets the shutter speed- this is MOST USEFUL.
3) Shutter priority: other way round as above.

Gray scale: camera tends to make images gray by 18%. So a black object will try to be whiter and a white object will try to be darker. For this it may be necessary to do full manual light metering. for eg. it may be necessary to underexpose a dark object and over expose a bright object.

Calculation for mega pixels: if you need a 8" by 12" print, you need 8 * 12 * 300 * 300 number of pixels.

There are two minor problems with DSLRs:
1) small sensor size compared to 35 mm film. There's a 1.5 or 1.6 multiplication factor which makes wide angle lens not so wide but a tele photo lens super-tele.
Sensor in DSLR has to be small because large sensor size makes the image quality at the edges bad. (But why does this problem not exist in 35 mm film?)

2) Dust accumulation on sensor. But Nikon has come up with ways to deal with this.

2008-03-27T02:34:00.000-07:00

"A two-electron spin-triplet state is symmetric under exchange of the two spins."
-Frohlich, Ueltschi, arXiv:cond-mat/0404483 v2 10 Nov 2004

Correlations

2008-03-24T22:05:00.000-07:00

If the probability of finding electron 'a' at r_a is independent of the probability of finding electron 'b' at r_b, then the probability of finding a at r_a AND the probability of finding b at r_b is simply the product of the probabilities i.e.

( | psi_ab (r_a, r_b) |^2 ) (dr_a)(dr_b) = (|psi_a (r_a)|^2 dr_a) (|psi_b(r_b)|^2 dr_b)

But if we take the Coulomb repulsion U into account then the position of b will influence the position of a and vice versa because the electrons do not want to be close, and the lhs above cannot then be described by the simple product ecause it will not then reflect this dependence. This dependence of one function on another is called 'correlation', as written in wikipedia.

Hence, U gives rise to correlation effects in the Hubbard model as for eg., if the up-spin electron is absent in a state, then the down-spin electron present in the same state will have a lower energy and so there will be a spectral weight transfer of the down spin electrons to lower energies.

2008-03-19T00:06:00.000-07:00

If the unit cell of a solid has an odd number of electrons it means there's an unpaired electron and so electrons can hop among adjacent sites making it conductor. But NiO and CoO are insulators even though they have odd number of electrons in adjacent sites. These are called Mott insulators.

2008-03-14T05:22:00.001-07:00

Gamma - X - M - R - Gamma

the nomeclature for the Brillouin zone branches in the simple cubic lattice.
Gamma: (0,0,0)
X: (pi,0,0)
M: (pi,pi,0)
R: (pi,pi,pi)

2008-02-25T20:40:00.000-08:00

The correlation function gives the response function (comes from the fluctuation-dissipation theorem: check it out). When the response function diverges, one can have spontaneous ordering, as happens in the case of magnetization, where we have spontaneous ordering when the response function diverges. Why this happens is obvious from the following equation:

m = XB

where m is the magnetization and B is the magnetic field and X is the response function, known as susceptibility. So when X diverges, one can have finite m even when B is zero.

Tonight, I ...

2008-02-14T11:04:00.000-08:00

But of course that is not the theme of this post. Instead I want to concretize some results and put them in one place. Else they'll get lost in myriad pages of notebooks scattered in here and there. I'd have preferred writing with my hands and putting up a scanned image. But the keyboard will have to do tonite.

If you're one of those rare visitors to this blog, you'd be better off reading Herstein's Topics in Algebra than the rest of the post, which is more for my reference than for your edification, but I welcome you with arms wide open if you choose to proceed.

EUCLID'S ALGORITHM

Let Z denote the set of integers. Let a, b Є Z with b not equal to 0. Then there exist unique m, r Є Z such that

a = mb + r with 0 ≤ r < |b|

In more familiar parlance, m is the quotient and r is the remainder when a is divided by b.

Proof:

Consider the set of integers S = { (a-nb) | n Є Z }

Obviously all elements are unequal for different valuesof n.

If a-nb = 0 for some value of n, say n = m1, then a = m1*b and r = 0, and we are done. (The uniqueness follows from the fact that all elemetns in S are different for different values on n)

Now consider the case when a-nb ≠ 0 for all n Є Z. (1)

Let b > 0. (the case where b.lt.0 can be treated in a similar vein)

From (1), S consists only of non-zero integers. Let q be the largest -ve integer and let r be the smallest positive integer.

So,
q = (a - m2*b) .lt. 0 (2)

and
r = (a - m2*b + b) > 0 (3)

If we add b to equation (2), we see that the resulting equation is the same as

r .lt. b

And from (3) we see

0 .lt. r

Obviously, the next element in the series, r+b is greater than r. (Since b > 0)

From (3), we can write a in the required form

a = mb + r with 0 ≤ r < |b| with unique m and r.

In considering the set S we are repeatedly subtracting b from a until we are left with the least positive integer r, which is why we were told in our school days that division is repeated subtraction (and multiplication, of course, is repeated addition)

-QED

Notation: If a is a factor (or divisor) of b, we will write a|b. It means we can write b=ka.
(a,b,k Є Z)

Definition 1: The greatest common divisor (gcd) of two integers a and b, denoted by (a,b)
is defined by the following:

Note that (a, b) is taken to be positive by convention.
Obviously, (a, b) is unique because if c=(a, b) and d=(a, b) then bythe definition of gcd, c|d and d|c => c=d.

The next lemma demonstrates some surprising properties of the gcd which I found extremely fascinating.

Lemma 1:If a and b are integers, not both zero, then (a,b) exists; moreover, we can find integers m and n such that (a,b) = ma+nb

Proof : Consider the set S defined by S = { Ma+Nb | M, N Є Z }

Since a and b both are not zero, there exist some non-zero integers in S. And since, if x Є Z, then so does -x, S contains positive integers. Let c = ma+nb be the smallest positive integer. We will now demonstrate a surprising fact: c divides all members of the set S! For, let x Є S, whence

x = pa+qb for some integers p and q. (6)

By Euclid's algorithm, we can write

x = tc + r with 0 ≤ r < .lt. c (7)

or, since c = ma+nb and x = pa+qb, we can write (7) as:

r = (p-tm)a+(q-tn)b .lt. c

If r>0, then, contrary to our assumption, r is the smallest positive integer of S and not c. Thus we are forced to conclude that r=0. Which means, from (7), that x is a multiple of c. But x was an arbitrary element of S, whence we conclude that c|x for all x Є S.

Now S includes a and b also (as can be seen by taking M or N to be zero in the definition of S), from which we conclude that c|a and c|b. Thus c satisfies the first part of the definition of the gcd. As regards the second part, note that if k|a and k|b, then k|(Ma+Nb) for any integers M and N, which proves that k|c.

Hence, c = ma+nb is the gcd of a and b.

-QED

Lemma 2: If a, b, c Є Z such that a | bc and (a,b) = 1, then a | c.

Proof: We're given that a | bc. Obviously, a | ac. Since a divides both bc and ac, a has to divide (ac, bc). This follows from the definition of gcd given earlier. What is (ac, bc) ? Well, since it is given that (a,b) = 1, it follows that (ac,bc) = c. Hence a | c.

-QED

Lemma 3: If (a, b) = (a, c) = 1, then (a, bc) = 1.

Proof: Since (a, b) = (a,c) = 1, using lemma 1, we know that there exist integers p, q, r, s such that

pa + qb = 1 (9)

and

ra + sc = 1 (10)

Multiplying the two equations (9) and (10) we get

(pra + psc + qbr)a + (qs)bc = 1

Since the least positive integer of the set S defined by S = { mu + nv | m, n Є Z } is (u, v) (as can be seen from the proof of lemma 1), it follows that (a,bc) = 1

-QED

Lemma 4: We'll be needing one more result later on, which is being presented here. This lemma states that if (n, x) = 1, and x = kn + r, with n > r > 0 by the Euclid algorithm, then (n, r) = 1.

Proof: Suppose (n, r) > 1. Then n and r have a common factor > 1, which implies that kn + r and n have a common factor > 1, but since kn + r = x, it implies that x and n have a common factor > 1, which contradicts the fact that (n, x) = 1.

-QED

Where do we go with all these lemmas? Well, I'd started out originally with algebra, so that is where we must go. More specifically, to Group Theory. According to Problem 14, pg. 36 of Herstein, we have the following result:

Lemma 5: If a finite set G is closed under an associative product, and both cancellation laws hold in G, then G is a group.

Proof: I'll give a detailed proof later. It's quite simple, and the idea is to consider any element a belonging to G, and consider it raised to various powers, which are all contained in G due to the closure property, and which are moreover finite in number because G is a finite group.

- almost QED!

Now look at Problem 15(a), on the same page. Initially I had some problems in understanding what it really meant, so I had to refer to a book on Number Theory (Niven, Zuckerman, Montgomery) for more explanation. So with this hindsight, I'll rephrase the result in the following theorem:

Theorem 1: If p is a prime number, then all positive integers less than p form a group under multiplication mod p.

Proof: First of all, what is meant by a mod p ? It is simply the remainder that is left when a is divided by p. (By Euclid's algorithm, this is less than p) Now we can see what is meant by multiplication mod p. Let a and b Є S, where S is defined by S = { x Є Z | p > x > 0 }. The binary operation * is called 'multiplication mod p' if a*b = ab mod p i.e. a*b is the remainder when ab is divided by p.

Let us go about proving the theorem.

(1) Closure:

Let a and b Є S. Since p is a prime, (p,a) = (p,b) =1. By lemma 3, (p, ab) = 1. Using Euclid's algorithm, we can write:

ab = mp + r

with p > r ≥ 0. If r = 0, ab is a multiple of p, which contradicts (p, ab) = 1. Thus p > r > 0, which implies a*b (= r) Є S. Thus the set S is closed under the operation multiplicaton mod p.

(2) Associativity:

Consider a*(b*c), where a, b, c Є S. By the definition of *, it equals

a*(bc mod p) = [a (bc mod p)] mod p = [a (bc - mp)] mod p = abc mod p

where we let bc = mp + b*c.

Similarly, one can show (a*b)*c = abc mod p.

Hence, associativity of * is also proved.

(3) Cancellation laws:

Let a*b = a*c, with a, b, c Є S. Using the Euclid algorithm and the definition of *, this implies

ab = mp + r
ac = np + r

with p > r > 0. From the above, it's obvious that p | a(b-c). Since (p, a) = 1, we have from lemma 2 that p | (b-c). But |b-c| is less than p, and so the only way p can divide b-c is by having

b-c = 0, i.e.

b = c

Similarly we can show that if b*a = c*a, then b = c. Thus both cancellation laws hold in S. Thus by theorem 1, S is a group.

-QED

We can generalize the above theorem to the case when p can be any positive integer. The modified theorem reads:

Theorem 2: If n is a positive integer, then all positive integers less than n and relatively prime to n (i.e. the gcd of any one of those numbers and n is 1) form a group under multiplication mod n.

Proof:

Let S = { x Є Z | n > x > 0 and (n, x) = 1}

(1) Closure:

Let x1 and x2 Є Z.
As before, x1*x2 = x1 x2 mod n.
From lemma 3, (n, x1 x1) = 1.
From lemma 4, x1*x2 Є S. Hence Closure of S under * is proved.

The proofs for the associativity and cancellation go through exactly the same way as in the previous theorem.

-QED

Now I'll just state some theorems on groups without proof which will be required for proving a theorem by Fermat (not his last!) and another one by Euler. But first, some definitions:

Definition 2: The number of elements of a finite group G is called the order of G. It is denoted by o(G).

Definition 3: If a Є G, where G is a group, then the order of a is the smallest positive integer such that a raised to that integer is the identity element in G.

Let us now state the required theorems.

Theorem 3: If G is a finite group, and H is a subgroup of G, then o(H) | o(G)

Theorem 4: If G is a finite group and a Є G, then o(a) | o(G). (This is just a special case of theorem 3 and can be seen by considering the cyclic subgroup generated by a.)

Theorem 5: If G is a finite group and a Є G, then a**o(G) = e, where e is the identity element in G. (This is obvious from the definition of o(a) and Theorem 4)

Let us now apply Theorem 5 to the set G of positive integers less than and relatively prime to n > 0. Theorem 2 assures us that this set is a group under multiplication mod n. Let a Є G and let the order of G be denoted by Φ(n). Φ(n) is the number of positive integers less than n and relatively prime to n. The identity element in G under * is 1. Thus from Theorem 5 we have

a*a*a*a......Φ(n) times = 1 (11)

The quantity on the left hand side is the remainder when a**Φ(n) is divided by n. Associativity of the group operation ensures us that it does not matter in what order we take the product a*(b*c) or (a*b)*c, and a question in Herstein, which I confess I haven't yet solved, ensures that this can be extended to any number of elements by induction, i.e. the order of the brackets does not matter in a product; only the order of the elements does. That is why I haven't put brackets on the quantity on the LHS of (11). Assuming that the order of the brackets does not matter in a product, a slight extension of the associativity proof in theorem 1 shows that the LHS is a**Φ(n) mod n, whence the statement in italics. It can be rephrased in the following form:

1 = a**Φ(n) mod n (12)

(12) states that n is a factor of the integer a**Φ(n) - 1, where n > a and (n, a) = 1.

Now consider the case when a > n and (n, a) = 1. By Euclid's algorithm we can write:

a = kn + r with n > r > 0. (13)

Thus, r Є G. From (13), we can see that (a**m mod n) equals (r**m mod n), where m is any integer and (a mod b) as before denotes the remainder when a is divided by b. This can be seen from the fact that the only term in the expansion of a**m not containing n is r**m. All the rest are multiples of n. If we take m = Φ(n), from (12) we see that r**Φ(n) mod n = 1, and from the statement in red, we get

a**Φ(n) mod n = 1,

i.e. (12) holds even when a > n, all that is required is that (n, a) = 1. (The above equation is correct only when n > 1. This will be made clear shortly.) Thus we have the following theorem by EULER:

Theorem 6 (Euler): If n and a are positive integers relatively prime to each other, then

a**Φ(n) mod n ≡ 1 (14)

where Φ(n) is the number of integers less than and relatively prime to n.

We use the congruence relation ≡, because the remainder is not 1 when n=1, rather it is 0 when a**Φ(1) is divided by 1, because then Φ(1) = 0. However we can use the equality sign when n > 1. The complicatiom arises from the fact that 1 divides both itself and the number just before itself i.e. 0. Since G is a finite group with Φ(n) number of elements and Φ(1) = 0, the case n=1 has to be treated separately and can be included with the other cases in the more general statement (14) only if we use the congruence relation instead of the equality)

*********************

Now let n = p, where p is a prime. Obviously, Φ(p) = p-1. From Theorem 6, if (a, p) = 1,

a**(p-1) mod p = 1 (15)

The above equation implies that p | (a**(p-1) - 1). From this, obviously, p | (a**p - a) i.e.

a**p ≡ a mod p (16)

Suppose now that (a, p) ≠ 1. Since p is a prime, the only way this can happen is if p | a. This also implies that p | a**p. Thus p | (a**p - a) and (16) holds in this case also. This proves the following theorem by FERMAT:

Theorem 7 (Fermat): If p is a prime number and a is any integer, then

a**p ≡ a mod p

(Since p is prime, p > 1 and the complications mentioned in theorem 6 do not arise in equation (15) where we have used the equality to denote that 1 is the remainder when a**(p-1) is divided by p and (a, p) = 1. But we have to use the congruence relation in (16) because a may NOT be the remainder when a**p is divided by p, as will happen when a > p. To rephrase: while 1 is definitely the remainder when a**(p-1) is divided by p and (a, p) = 1, a may not be the remainder when a*p is divided by p. Hence we have to use the congruence relation. It is important to note that (16) holds for any a, but (15) holds only when (a, p) = 1)

************************

Proof by Contradiction

2008-02-12T00:26:00.000-08:00

Suppose A and B are two statements and we want to prove that if A is true, then B is also true.
i.e.
A => B

Proving the above is the same as proving

NOT(B) => NOT(A)

The above is same as reductio-ad-absurdum . (Proof by contradiction)

It can be seen as follows:
Assume A to be true.
Assume, contrary to what we want to prove, that B is not true. i.e. NOT(B) is true.
From the above, show that A is not true, i.e. NOT(A) is true.
But we started out with the assumption that A is true, i.e. we have reached a contradiction.
Hence our assumption that B is not true when A is true is a false assumption, and we are left with the only(?) option that B is true.

The question mark in the above statement refers to the belief that we have only two options: either B is true, or B is false (not true).
What if there is a third option?
If there are more than two options, can we always categorize/rephrase the statement B such that all these options can be included in either of these B and NOT(B) ?
If not, reductio-ad -absurdum will fail.

For instance, the length of a stick can be anything...there are an infinite number of possibilities.
But we can make the statement B:
B = the stick is 5 meters in length.
So either B will be true or not; no third option is left
OR
B = the stick is longer than 5 meters
OR
B = the stick is longer than or equal to 5 meters etc.

All the infinite possibilities for the length of the stick can then be described by either B or NOT(B).

Can we always do that?

2007-09-26T14:20:00.000-07:00

RESPECT THE BEAT !!!

"At the same time, however, they should never be played faster than they can be performed cleanly and free of mistakes. The key to playing is fast is practicing slowly, and building the tempo incrementally. This is a practice technique that is almost always neglected by overzealous students! Some other techniques I like to use when practicing these exercises include the following:

1) Start very slowly, deliberately, and staccato. This helps builds articulation.
2) Use a metronome. It will help build your sense of time.
3) When you master an exercise at a given speed, increase the tempo one notch on your metronome.
4) Keep your jands low profile and your fingers curved. (only for pianists)
5) Don't tense up. Monitor the tension in your entire body.
6) Push yourself, but stop if you are experiencing pain. Technique exercises won't help you if you injure yourself!"

-extract from Peter Deneff's book on jazz piano (Jazz Hanon)

A simple proof of Gauss's Divergence Theorem

2007-08-14T21:51:00.000-07:00

The proof is presented from a physicist's point of view. All reasonable assumptions about the continuity, differentiability etc. of A(x,y,z) are made. The proof is for the usual 3-dimensional space we are familiar with. Volume V is a simple volume, without any complications of exotic mathematical surfaces like the Klein bottle or Moebius strip etc. In other words, V can be taken to be a cube, or a sphere, or an ellipsoid, or any other toplogically equivalent volume.

The case of other volumes like a doughnut (topologically equivalent to a tea cup) etc. should be amenable to a similar treatment.

The case of more complicated dimensions like 4-dimensional Euclidean space-time will involve the metric tensor and contra- and co-variant tensors.

On Photography

2007-07-15T06:43:00.000-07:00

Given the shutter speed and type of film being used, the lighting conditions are determined.
Given the shutter speed and lighting conditions, the type of film is determined.

Faster film is more sensitive to light and hence it needs less light to develop. In other words, fast shutter speed, which permits less light to fall on the film OR poor lighting conditions, are conditions in which fast film must be used. If there had been an AND instead of the OR, even faster films must be used.

Since my camera's shutter speed is very fast (a statement which I have to confirm), I must use sensitive film (I'm using 400) even for daylight photography. And if you want to shoot pics in less bright surroundings, go in for even faster films.

What if I use 100 film? Maybe in that case only the sun would be visible in the photographs.

2007-06-26T04:38:00.000-07:00

You are now 25 and a half years old. You are not getting any younger and you do not even have a job, age is catching up with you. Elsewhere, your other friends are working, are financially independent, and some are even married. To make matters worse for you, you are stuck up in a weird city and hemmed in by the walls of the campus. You are like a prisoner in the Shawshank prison, and time is catching up with you. Like Andy, you need to break out of here where you've been imprisoned for no fault of yours.

Remember: time is catching with you. you've wasted the best part of your life here. Don't lose what can still be salvaged.

It is a race against time. I have such a lot to experience, there is such a lot to see, and every moment is precious, and I cannot afford to lose it.

I do not want to be 'institutionalized' like Brooks was. Brooks got 'institutionalized' and commited suicide after he was discharged from the prison, having spent nearly 40 years of his life there. He did not know any other life except that of the prison.

Get away from this place before it gets you.

It is a race against time.