Think Science .

Super symmetric string theory does need 11 dimensions for exploiting the T and M symmetries but mathematically speaking, the number of dimensions cannot go to infinity. The reason string theory is gaining ground is because of the fact that it does not lead to infinities while considering fundamental particles as point particles does.

This is very similar to how Maxwell and Pauli both described the electrons behavior pretty accurately but one used a wave analogy while the other treated the electron as a particle. String theory is emerging as a good model for describing sub atomic interactions but that does not guarantee its physical reality. Science fiction writers have probably glorified the term “dimensions” and hence the common misconception that it is a fantastic concept. Mathematically speaking, constructing a higher dimensional set using matrices is kid stuff actually and any communication theorist (dealing with CDMA, Viterbi algorithms etc) does this on a daily basis.

I am probably explaining this for the billionth time.. well here goes with one another question.

 Q: suppose u r standing on new delhi station and u see a train moving toward kashmir and another train moving towards kanpur. so unless u find the trajectory of the train as a circle with new delhi station as is centre u can never say that they were at the same time in new delhi station . in the same way if u cant find the trajectory the end of the universe . and that to calculate it is a sphere how can u say that distant galaxies were at the same place .

The mathematical flaw in you approach is that you are regarding New Delhi to be a fixed reference point which is impossible to obtain. A more practical scenario is where you consider New Delhi itself to be moving towards/away from Kanpur/Kashmir. Thus this problem reduces to one of triangulation and it is possible to extrapolate the motion of all three bodies based on that. NOTHING is expanding INTO pre-existing space. Space is an outcome of the Big bang and it is space itself which is expanding. Space did not exist and if there is a boundary to the universe then there is no space beyond it.Also, we are limited to observing only a finite part of the universe and as such it is impossible to describe the geometry of the Universe. However, based on the mass density and certain sensible assumptions, it seems probable that the universe follows a Kaluza Klein geometry. The simplest example I can think of is probably the Mobious strip which is bounded in three dimensions but has an infinite edge. Quite possibly, as we are limited to observing only a few dimensions, we may see the universe as infinite but it may not be so. You can refer to the Brane theory for details.

True fundamental constants must be dimensionless. Lightspeed is fundamental given Lorentz invariance. A relativistic universe has four distinct distances: luminosity (inverse square), angular diameter, parallax, and proper motion. No two of them need agree to maintain consistency. Clocks can only be synchronized by being local. Then,

http://en.wikipedia.org/wiki/Natural_units

The Standard Model arrives massless. All primary particle masses must be empirically inserted plus more; eveything else is rationalized via the Higgs mechanism,

http://math.ucr.edu/home/baez/constants.html

Not done yet! Classical physics (e.g., gravitation) has fits over non-integral spins. Though forces are conveyed by bosons, all matter is fermionic.

It just seems that in this universe of simplicity our measurements and orders of magnitude are out of context. and nice round exponents would fit if the right measuring tools were used.

For some more information check out this page : http://physics.nist.gov/cuu/Units/current.html

During 2003-2006, there was trend in campus recruiting to ask questions from shakuntala Devi’s puzzle books. Just because Microsoft, google ask their employees puzzles and brainteaser before recruiting, in Silicon Valley and outside it every company started to ask such questions.

Shakuntala devi’s puzzle are indeed hair pulling and teaser ones. Yet they’re interesting as well. and reason they’re introduced in IT company screening process is to check how sharp a student thinks. that indeed a necessity as environment in IT company is like that.

So coming back to the point of shakuntala devi’s puzzles. She wrote more than 6 books on mathematical ability and puzzles. One book she wrote on homosexuals which is published but didn’t get much publicity for it. Some of her famous books are:

• Puzzles to Puzzle You
• More Puzzles to Puzzle You
• Book of Numbers
• Figuring: The Joy of Numbers
• In the Wonderland of Numbers
• Mathability: Awaken the Math Genius in Your Child

If what you want is the actual proof, I suggest you look at any good book on statistical mechanics. One of my favorite books for that subject is

Fundamentals of Statistical and Thermal Physics, by Frederick Reif
http://www.amazon.com/Fundamentals-Statistical-Thermal-Physics-McGraw-Hill/dp/0070518009
The law of equipartition of energy is valid for a system (classical or quantum) (a) whose hamiltonian is a quadratic function of the generalized coordinates and momenta describing the system in question, and (b) in thermal contact with a reservoir at constant temperature.

Because of (b), the probability of any microstate of the system is proportional to exp(-ßH), where ß = 1/(kT) and H is the hamiltonian of the system. Thus, the average value of any dynamical quantity F(q, p), depending on the generalized coordinates q and/or momenta p, is given by

<F> = ( ? F(q, p) exp[ -ß H(q, p) ] dq dp ) / ( ? exp[ -ß H(q, p) ] dq dp )

So far, this is perfectly general. Now, if (a) is true, then you can actually compute the average above quite easily. To make things easier here (because of orkut’s limitations), let’s say that the Hamiltonian is H = q^2, that is, it’s quadratic on the generalized coordinate q.

What’s the average value of the energy of the system, then? It’s simply

<H> = <q^2> = ( ? q^2 exp[ -ß q^2 ] dq ) / ( ? exp[ -ß q^2 ] dq )

Now consider the integral Z(ß) = ? exp[ -ß q^2 ] dq as a function of ß. It’s the integral appearing as the denominator of <H>. Taking the derivative with respect to ß, you find:

dZ(ß)/dß = - ? q^2 exp[ -ß q^2 ] dq

which, except for the minus sign, is exactly the numerator appearing in <H>. Thus,

<q^2> = - (1/Z) dZ/dß = - (d/dß) [ ln Z(ß) ]

So, all we need is to compute Z(ß). The limits of integration (which I haven’t shown in the above) are from negative to positive infinity, since the coordinate can take any of those values. So, with u = sqrt(ß) q,

Z(ß) = ? exp[ -ß q^2 ] dq = ? exp[ -u^2 ] du / sqrt(ß) = ß^(-1/2) x some constant

lnZ(ß) = -(1/2) lnß + some constant

and

<q^2> = - (d/dß) [ ln Z(ß) ] = (1/2) (1/ß) = kT/2

So, the average value of any quadratic generalized coordinate is kT/2. The same is true for the generalized momenta. Hence, for a system described by N generalized coordinates and N generalized momenta, all of which appearing quadratically in the hamiltonian, the average energy is going to be given by

<E> = N kT/2 + N kT/2 = N kT

That’s true also if the generalized coordinates and/or the generalized momenta appear in H as general quadratic functions, that is, with linear terms as well.