Big Numbers: How much can you count??

            Nowadays it is very obvious for us to write a large number , whether it is the distance of Earth from Sun in centimeters or the age of our universe, by simply setting enough number of zeros in the right side of some figure and most surprisingly by doing this we may easily write down the number which is greater than the total number of atoms of the universe that is identically 300,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 or in a shorter notation 3x10⁷⁴. But this method was not known to the people of the ancient time. In fact this it was invented less than two thousand years ago by some unknown Indian Mathematician. Now in the next few minutes, I am going to talk about how was the world of writing numbers before this great discovery.

            Now let me tell you some facts about the 'Khoikhoi'  (a group of Khoisan people native to southwestern Africa. The Dutch settlers called them Hottentots). They don't have in their vocabulary the names for numbers larger than three. They used to describe any number, larger than three,  as "many". If you ask them how many fingers they have they will say 'many'. Coming back to the process of writing large numbers, ancient Egyptians used many different symbols to write a number as an example if you wish to write 8732 in their way you may write it as

                       

whereas in Rome, it was accustomed to use alphabets to write numbers. 8732 can be written as

MMMMMMMMDCCXXXII , which is known to us as roman numbers and still have some use in some particular places. In the both above mentioned cases it is very tedious work to write a large number like one billion. Even if you wish to write one million in these way you will need to use more than two or three pages and  more than few hours.

            The great scientist Archimedes was the first person to show that actually it is possible to write big numbers and the method was quite similar to the way large numbers are written in modern science. He begins with the largest number that found in the ancient Greek arithmetic: 'a myriad' or ten thousands. Then he introduced  'a myriad myriad' that is ' an Octade' or a units of second class, which is equal to a hundred million then 'octade octades' or a unit of third class ( a ten million billion) etc. I know this is a little bit confusing but please keep patience.

To revive and make the discussion more interesting let me tell you one another story. we all know the game of chess and some of us also know that chess was invented in India by Sissa Ben Dahir. For inventing and presenting this game to King shirham of India, king asks Sissa Ben to claim something large as a gift. Clever Sissa Ben desire to have one wheat grain to put in the first box of the chessboard then 2 grains to put in the second box and 4 grains to put in the third one and  8 grains in the fourth box and so for the next each box. Listening to this the king was happy because he thought this gift would not cost him a lot.

            Now let's calculate how much grains actually need to cover all the 64 boxes in the way mentioned by Sissa Ben.  It will be a geometric progression (G.P) series like this: total no of grains  = 1+2+4+8+16+32+64+..........................+(64^th term )

= 1+2¹ +2²+2³+2⁴+2⁵+2⁶+ ..................................+2⁶³   = 18446744073709551615   (approximately ) which is equal to the total wheat production of the world for more than one thousand years. 

   

** In Archimedes's notation the calculation should be like this

Don't expect me to discuss the next part of the story. 

In the above discussion, we have faced some really large numbers like the amount of wheat claimed by Sissa Ben. Still, you can deal with this numbers if enough time permits. But there exist some numbers which are really infinite. you can't write them no matter how long and how hard you work.

I'll talk about infinite numbers latter in my post.



For this post I thoroughly followed One Two Three Infinity by Gamow George

Five Fold Rotational Symmetry of Crystal - A Different Approach of Discussion

                Symmetry is a very vast and important part for not only crystallography but also for many other parts of physics. we will discuss the symmetry of a crystal more specifically about rotation symmetry of the crystal.

When a crystal can be divided into two mirror images about a plane or about an axis or about a point that is called crystal symmetry.

why symmetry is so important for crystallography ??

                Among many importance of symmetry three basic importance are  (i) Identification of materials, (ii) Prediction of atomic structure and (iii) Relation to physical properties ( optical, mechanical, electrical and magnetic etc.)

In crystallography, symmetry is used to characterize crystals, identify repeating parts of molecules and simplify both data collection and nearly all calculations. Also, the symmetry of physical properties of a crystal such as thermal conductivity and optical activity must include the symmetry of the crystal. Thus, a thorough knowledge of symmetry is essential to a crystallographer. 

Symmetry can be various types like translation symmetry, rotational symmetry etc. Here I will talk about rotational symmetry only. Rotational symmetry means rotation of the crystal by a certain angle about an axis (rotational axis) makes no difference to the crystal i.e crystal will remain exactly same as before after that rotation.

If we can rotate the crystal n times by an angle α where n and α satisfy nα=360 degrees then this kind of rotational symmetry is called n-fold symmetry. That means for n-fold symmetry n times rotation by angle α  make a total of 360 degrees rotation of the crystal. Let take an example of a square of side a. If we rotate the square by 90 degrees about the axis perpendicular to its center it will remain same as before and we can rotate the square 4 times to make a total rotation of 360 degrees. So the square has 4-fold rotational symmetry. If we take an equilateral triangle it a 3-fold rotational symmetry.

Coming back to the crystal, now we are going to check how many possible folds of symmetry a crystal can possess.
 Crystal can have 1,2,3,4,6-fold symmetry but can not have 5 or 7 fold symmetry. Now I have reached to the main topic of this discussion that is '5-fold symmetry can not exist in crystal'.

This can be shown by many different methods but I am going to discuss three methods of showing that 5-fold symmetry does not exist for crystal.

1.First one is very easy and you may find it in any standard introductory textbook of solid sate physics*.

                                                                               

Let the lattice have n-fold rotational symmetry about the axis passing through the lattice point and perpendicular to the plane of the paper. By performing a rotation of angle α= 2π/n we can get back the same lattice. So in the figure A' and B' are also lattice points as A and B thus we can write A'B' = m*AB where m is an integer.

                                                                 A'B' = m*AB

                                                Hence,   A'X + XY+ YB' = ma

                                                                acosα +a+ acosα = ma

                                                                cosα = (m-1)/2 ...................(1)

As  -1 ≤ cosα ≤ 1  so m can not be any random integer. m must have some specific values shown in the below table.

               

m	cosα	Α        (degrees)	Symmetry
0	-0.5	120	3-fold
1	0	90	4-fold
2	0.5	60	6-fold
3	1	360	1-fold (identity)
-1	-1	180	2-fold

So from this table, we can see 1,2,3,4,6-fold symmetry exist for crystal but 5-fold symmetry does not exist.

2. The second approach of showing that 5-fold symmetry doesn't exist**.

This approach is much more physical than the first one. Here we only use some simple geometry to reach our goal. Before that, we need to know that in the arrangement of lattice the most important thing is periodicity. The basic structure should repeat again and again to form an infinite lattice. There should not be any gap between these repetition.

Area filled with a connected array of square.

so if we use either a square or an equilateral triangle or a hexagon to cover up some place by repeating them again and again, we can arrange them such a way that there will be no gap between them. That is why there are no problems with these kind of symmetries.

Area filled with a connected array of a uniform triangle.

Now if we take a uniform pentagon and try to cover some plane area by repeating the pentagon many times it is not possible to find an arrangement where there will be no gaps (try it yourself). So it is impossible to fill all the area of a plane with a connected array of pentagons. This is the problem with five-fold symmetry of a crystal.

3.Now the third method.

The most simple method where we consider 5-fold symmetry is existed and then prove that something is wrong in this assumption.

Let’s begin by assuming 5-fold symmetry is possible. Set the shortest lattice vector

along the x-axis: a₀ = a₀(1, 0). Then any other lattice vector can be derived from this as:

                                                                a_n = a₀ [cos(2πn/5), sin(2πn/5)]

we know that some of two lattice vector is also a lattice vector but no lattice vector can be shorter than a₀ . let sum a₁ and a₄

                                                T= a₁ + a₄ =a₀ [{cos(2π/5)+cos(8π/5)}, {sin(2π/5)+sin(8π/5)}]

                                                Hence, T=a₀(0.62,0)

Here we found that the magnitude of the lattice vector T is shorter than a₀  but as per our consideration a₀ is the shortest lattice vector. This is a contradictory situation so something is wrong in our assumption that 5-fold symmetry does exist.

*Reference for the first method: book-1, book-2

** Reference for the second method: Introduction To Solid State Physics 8 Edition -Charles Kittel  or Introduction to Solid State Physics (Paperback) by Charles Kittel

Some Important Things:

First I would like to say that the statement  'five-fold symmetry does not exist'  is not completely true. In the above discussion all the things I mentioned is correct for periodic crystals (in general, all the crystals are periodic that is why we can use the statement of nonexistence of five-fold symmetry for the crystal in general ). There exist a different type of crystal called quasicrystal, which has a structure that is ordered but not periodic. Quasicrystals have long-range order but they are not periodic. Quasiperiodicity leads to unconventional symmetries (5, 8, 10, 12 -fold).

Dan Shechtman,  the Philip Tobias Professor of Materials Science, is the man behind the discovery of Quasicrystal and received the Nobel prize in chemistry in 2011.

Simple Harmonic Motion.

                Before studying Simple Harmonic Motion (SHM) we need to know something called vibration and restoring force.  A particle will vibrate if any displacement from its position of rest calls into a force which tries to bring the particle back to its rest position. This type of force is known as restoring force, which is generally a function of displacement.

                Restoring force F(x) = a₀+a₁x+a₂x²+a₃x³ +..................................

when x=0  i.e  the position of rest, F(x) is also zero so a₀ must be equal to zero. For small displacement, we can restrict the series up to the first order term. Now it becomes F(x)= a₁x. The vibration of this particular type for which restoring force is directly proportional to the displacement is called Simple Harmonic Motion (SHM).

Resorting force and displacement

                For SHM F(x) = -sx (where s is constant and the negative sign is due to the fact that restoring force is directed opposite to the displacement ). A particle vibrating in SHM is known as Linear Harmonic Oscillator (LHO).

                From Newton law of motion we can write F= m x''       here double prime indicates double derivative with respect to time.  so now we have,

                                                                                mx" = -sx

                                                                  hence,   x''= -ω²x  where  ω²=  s/m

                                                                                      x'' +ω²x=0  ............................... (1)

we may write the solution of eq (1) in many ways, one possible solution is  x(t)=acos(ωt +α) ....... (2)  where a is the amplitude and α is the initial phase of the SHM. The period T (time taken for one complete oscillation ) is given by  2π/ω i.e  T = 2π sqrt(m/s).

[ verify yourself  that after time T, x(t) comes back to its initial position i.e x(t+T)=x(t) ]

particle undergoing SHM

Energy of SHM ::

                                velocity of the oscillator v=  -aωsin(ωt +α) .......... (3)

                                acceleration   f = -aω² cos(ωt +α) = -ω²x(t) ................(4)

acceleration is proportional to the displacement and directed opposite to the displacement.

                            Kinetic Energy  E_k = mv² = ma²ω²sin²(ωt +α)  (from eq. (3) )  .................. (5)

                          Potential Energy E_p = mω²a²cos²(ωt +α)  .................... (6)   [ using (1) & (2)]

                                Total energy    E = E_k+E_p = ma²ω²  =constant   .................. (7)

Problems::

1.Show that the projection of a particle moving in a uniform circular motion is a SHM .

Hint::      

                      

Ref..  Vibrations and Waves (The M.I.T. Introductory Physics Series) - A.P.French.  [  use this link.:: click here.]

2.A particle of mass 10g is placed in potential field given by V = (50x²+100) erg/g. what will be the frequency of oscillation?

solution ::  F= -  = -100x = -sx

                        so, s = 100 erg/cm

                                ω²= s/m = 10

                                ω= sqrt(10)  s^-1

Some Simple Examples of SHM ::

1.Simple pendulum:  A simple pendulum is consists of a massless string of length L attached at the top to a rigid support and at the bottom connected to a bob of mass M. Let ψ be the angle made by the string with vertical ( see the figure).

The tangential displacement of the bob = Lψ

The tangential velocity of the bob = Lψ'

The tangential acceleration  of the bob = Lψ''       
 [Here the single and double prime denotes first and second derivative with respect to time respectively ]

Now we can write according to the Newton's law of Motion

                                    Restoring force F =  MLψ''

                                    -Mgsinψ = MLψ'' ........................... (8)

for small values of ψ (in rad ) sinψ = ψ

                        Hence, Lψ'' = - g ψ ................. (9)    

Here also the acceleration is proportional and oppositely directed to the displacement. So according to eq. (4) it is oscillating in SHM.

                                    ψ'' = - (g/L)ψ

                                    ψ'' + (g/L)ψ = 0 ................. (10)

comparing eq.(10) with eq.(1) we can get the frequency of oscillation  ω = sqrt(g/L) hence Time period of oscillation is T = 2π/ ω  = 2π*sqrt(L/g)

                       

                                                                       

For more examples ::
 ref. link-1: Waves : Berkeley Physics Course 1st Edition (Volume 3) -Franks S. Crawford


Summary and Formulas::

1. Restoring Force is always a function of displacement and oppositely directed.
2.  Equation of motion for a particle undergoing SHM is    x'' +ω²x=0.
3. Acceleration of the oscillating particle is always negative and given by  f=-ω²x(t).
4. Angular frequency ω= 2π/T , where T is the time period of oscillation.
5. Total energy of SHM is given by E= ma²ω²  =constant   where a is the amplitude of oscillation and m is the mass of the particle.