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## Equi-partition of Energy and Classical Statistics, Physics tutorial

:IntroductionThe equal partition theorem on energy is defined and the theorem is employed to derive the Average-Translational Kinetic Energy of a particle in a gas and the utilization of standard equations from the statistical mechanics to derive the internal energy of the system.

The fundamental concept of classical mechanics similar to the basic of ideal gases and statistical distributions are highlighted.

:Equipartition TheoremThe equipartition theorem defines that the energy is shared uniformly amongst all energetically accessible degrees of freedom of a system. This is a system which will usually try to maximize its entropy (that is, how 'spread out' the energy is in the system) through distributing the available energy equally amongst all the accessible modes of motion.

The equipartition theorem can go further than simply expecting that the available energy will be shared uniformly amongst the accessible modes of motion, and can make quantitative predictions regarding how much energy will appear in each degree of freedom. Particularly, it defines that each and every quadratic degree of freedom will on average process an energy 1/2 KT. A quadratic degree of freedom is one of which the energy depends on the square of certain property. Let, consider the kinetic and potential energies related by translational, rotational and vibrational energy.

Translational degrees of freedom K = 1/2Mv

^{2 }Rotational degrees of freedom K = 1/2 I

^{2}Vibrational degrees of freedom K = 1/2Mr

^{2}V = 1/2 Kx

^{2}The above are the three kinds of degrees of freedom all encompass a quadratic dependence on the velocity (or angular velocity in case of rotation) and thus all follow the equipartition theorem.

It will be noted that whenever considering vibration in a harmonic oscillator potential (V, above) we consider both kinetic energy and potential that is, the P.E. counts as an extra degree of freedom. All the point regarding vibrations is that vibrational motion in molecules is highly quantized, and at room temperature most of the molecules are in their vibrational state and higher levels are not thermally accessible. As a consequence, equipartition contributions from vibrational degrees of freedom require only usually be considered at very high temperatures.

On the contrary, at room temperature numerous rotational and translational states are occupied, and they can be treated classically (that is, as if their energy levels were not quantized) to a very good approximation.

A simple derotation of the equipartition outcome for the translational motion.

We can employ the Maxwell Boltzmann distribution of molecular speeds to find out the average kinetic energy of partition in a gas, and exhibit that it agrees by the equipartition result.

The Maxwell Boltzmann distribution of molecule speeds is:

f(V) = 4π (m/2πKT)

^{3/2}V^{2}exp (-mv^{2}/2KT)The average kinetic energy of a particle in the gas is:

K = 1/2mv

^{2}=_{ 0}∫^{∞ }1/2mv^{2}f(v) dvReplacing for f(v) and taking the constant terms outside the integrals provides:

K = (1/2) m 4π (m/2πKT)

^{3/2}_{0}∫^{∞}V^{2}exp (-mv^{2}/2KT) dvWe can compute the integral by employing the general result that,

_{0}∫^{∞}X^{2S}exp (-ax^{2}) = [(2S - 1)/(2^{S+1}a^{S})] (π/a)^{1/2}Here n!! represents double factorial, n (n-2)(n-4)......etc. Recognizing X = Vanda = m/2 in our integral above provides,

_{0}∫^{∞}V^{4}exp (-mv^{2}/2KT) dv = [3!!/8(m/2KT)^{2}] (2πKT/m)^{1/2}= (3/2) (KT/m)^{2 }(2πKT/m)^{1/2}Replacing back into our expression for k provides,

K = 1/2 m 4π (m/2πKT)

^{3/2 }(3/2) (KT/m)^{2 }(2πKT/m)^{1/2 }K = 3/2 KTThe average translational K. Energy of the particle in a gas is thus (3/2) KT, or (1/2) KT per translational degree of freedom. In agreement by the equipartition theorem, a more common derivation of the Equipartition theorem needs statistical mechanics.

The partition function in statistical mechanics state us the number of quantum state of a system which are thermally accessible at a particular temperature.

It is stated as: q = Σ

_{i}exp (-E_{i}/KT)Here, E

_{i}are the energies of the quantum states i. Once we recognize the partition function, we can compute many of the macroscopic properties of our system by employing standard equations from the statistical mechanics.We will employ the 'P' function to compute the internal energy 'u' related by a single degree of freedom of the system and we require considering the difference between a quantum and a classical system.

If we are treating the particular motions classically, it does not make sense to deduce the partition function as a sum of discrete terms however classically, the position and moment of a particle can differ continuously and the energy levels are as well continuous. As an outcome, the classical partition function takes the form of the integral instead of a sum.

q = ∫exp [-E(X

_{1}X_{2}-------P_{1}P_{2}--------)/KT] dx_{1}dx_{2}----dP_{1}dP_{2}----Here the energy can be a function of the particle positions X

_{i}and momenta P_{i}. If we suppose that we can write the energy as a sum of contributions from each degree of freedom, then the exponential functional dependence on the energy signifies that we can separate the integral into the product of integrals over each and every degree of freedom that is,E(X

_{1}X_{2}-------P_{1}P_{2}--------) = E(X_{1}) + E(X_{2}) + ---- + E(P_{1}) + E(P_{2})----Therefore, on solving and taking integral we get:

= q(X

_{1}) q(X_{2}) ---- q(P_{1}) q(P_{2})The effect of this is that we here separated the partition into the product of partition functions for each and every degree of freedom. In common, we might write the P.f for a single degree of freedom in which the energy depends quadratically on the coordinate x (that is, ε(x)) = CX

^{2}having C as a constant) as:q(x) =

_{-∞}∫^{∞ }exp (-E(x)/KT)dx =_{-∞}∫^{∞}exp(-CX^{2}/KT) dx = (πKT/C)^{1/2}Here we have employed the standard integral:

_{-∞}∫^{∞}exp (ax^{2}) dx = (π/a)^{1/2}Once we recognize the P.f, we can compute the internal energy of the system according to the standard outcome from the statistical mechanics:

U = KT

^{2}d(ln q)/dTBy placing in a P.f, the internal energy related by one degree of freedom is thus:

U = (KT

^{2}/2) (c/πKT) (πK/C) (since dlny/dx = (1/y)(dy/dx) = (1/2) KTThe energy appreciated by each quadratic degree of freedom is thus (1/2) KT and we have proved the equipartition theorem.

:Classical StatisticsClassical Maxwell Boltzmann statistics is introduced to compute the occupancy of states.

This is derived on the basis of purely classical physics arguments. The fundamental concepts of classical statistics will be derived and the basics of ideal gases and statistical distributions are summarized as they are the basis of semi conductor statistics.

:Probability of Distribution FunctionAssume a large number 'N' of free classical particles like atoms, molecules or electrons that are kept at a constant temperature 'T', and which interact only weakly by one another. The energy of a single particle comprises of kinetic energy due to translatory motion and an internal energy for illustration due to rotations, vibrations, or orbital motions of the particle. In the following, we assume particles having only kinetic energy due to the translator motion.

The particles of the system can makes sure an energy 'E', where E can be either a discrete or a continuous variable. If N

_{i}particles out of 'N' particles have energy between ε_{i}and ε_{i}+ dε, the probability of any particle having any energy in the interval ε_{i}and ε_{i}+ dε, is represented by:f(ε

_{i})dε = N_{i}/NHere f(ε) the energy distribution function of a particle system in statistics is f(ε) is frequently termed as the probability density function. The total number of particles is represented by:

Σ

_{i }N_{i }= NHere the sum is over all the possible energy intervals. Therefore, the integral over the energy distribution function is:

_{0}∫^{∞}f(E)d(E) = Σ_{i}N_{i}/N = 1In another words, the probability of any particle having energy between zero and infinity is unity. The distribution functions that follow:

_{0}∫^{∞}f(E) dE = 1Are termed as normalized distribution functions.

The average energy or mean energy E‾ of a single particle is acquired by computing the total energy and dividing by the number of particles, that is:

E‾ = 1/N Σ

_{i}N_{i}E =_{0}∫^{∞ }E f(E) dEV

_{x, rms }= √v‾x^{2}= [_{-∞}∫^{∞}_{-∞}∫^{∞}_{-∞}∫^{∞ }Vx^{2}f (V_{x}, V_{y}, V_{z}) dV_{x}dV_{y}dV_{z}]^{1/2}In a closed system the mean velocities are zero, that is v‾

_{x}= v‾_{y}= v‾_{z}= 0. Though, the mean square velocities are, just as the energy not equal to zero.:Ideal gases of Atom and ElectronsThe fundamental of classical semi conductor statistics is ideal gas theory. It is thus essential to form a small excursion into this theory. The individual particles in such ideal gases are supposed to interact weakly, that is collisions between the atoms or molecules are a relatively rarely event. This is further supposed that there is no interaction among the particles of the gas (like electrostatics interaction), except the particles collide. The collisions are supposed to be (a) elastic (that is, total energy and momentum of the two particles comprised in a collision are preserved) and (b) of very short time period.

Ideal gases obey the universal gas equation:

Pv = RT

Here 'P' is the pressure, 'V' the volume of gas, 'T' its temperature and 'R' is the universal gas constant. This constant is independent of the species of the gas particles and consists of a value of R = 8.314 JK

^{-1}mol^{-1}:Maxwell velocity DistributionThe Maxwell velocity distribution illustrates the distribution of velocities of the parties of an ideal gas. This will be illustrated that the Maxwell velocity distribution is of the form.

fm (V) Fm (v) = Aexp [- (1/2)MV

^{2}/(KT)]Here, (1/2) mv

^{2}is the kinetic energy of the particles, if energy of the particles are purely kinetic.Whenever the energy of the particles is purely kinetic, the Maxwell distribution can be represented as fm (E) = Aexp (E/KT)

The proof of the Maxwell distribution of equation is suitably done in two steps. In the first step, the exponential factor is illustrated, that is, fm (E) = Aexp (-α E). In the second step it is represented that α = 1/(KT)

In the theory of ideal gas it is guaranteed that collisions between the particles are elastic. The total or net energy of two electrons before and after a collision remains similar, that is E

_{1}+ E_{2}= E_{1}^{1}+ E_{2}^{1}------Here E

_{1}and E_{2}are the electron energy before the collision and E_{1}^{1}and E_{2}^{1}are the energies after the collision. The probability of a collision of an election having energy E_{1}and of an election by energy E_{2}is proportional to the probability that there is an election of energy E_{1}and a second election by energy E_{2}. Whenever the probability of such a collision is 'P', thenP = B fm (E

_{1}) fm (E_{2})Here, 'B' is a constant. The similar consideration is valid for particles having energies E

_{1}and E_{2}. Therefore, the probability that the two electrons having energies E_{1}^{1}and E_{2}^{1}collide is represented by:P

_{1}= B fm (E_{1}^{1}) fm (E_{2}^{1})Whenever the change in energy before and after the collision is ΔE, then

ΔE = E

_{1}^{1}- E^{1}and ΔE = E_{2}- E_{2}^{1}Moreover, if the electron gas is in equilibrium, then P = P

_{1}and one getsfm (E

_{1}) fm (E_{2}) = fm (E_{1}+ ΔE) fm (E_{2}- ΔE)Only the exponential function satisfy the condition, that is

fm (E) = Aexp (-α E)

Here 'α' is a positive yet undetermined constant. The exponent is selected negative to make sure that the occupation probability reduces with higher energies. This will become obvious that 'α' is a universal constant and applies to all carrier systems like electron, heavy or light hole systems.

Subsequently, the constant 'α' will be determined it will be represented that α = 1/KT by using the results of the ideal gas theory. The energy of an election in an ideal gas is represented by:

E = 1/2 mv

^{2}= 1/2 m (V_{x}^{2}+ V_{y}^{2}+ V_{z}^{2})The Maxwell velocity distribution in the Cartesian co-ordinates is:

fm (V

_{x}, V_{y}, V_{z}) = (m/2πKT)^{3/2}exp [- {(1/2)m (V_{x}^{2}+ V_{y}^{2}+ V_{z}^{2})}/(KT)]Due to the spherical symmetry of the Maxwell velocity distribution, this is helpful to deduce the distribution in the spherical coordinates.

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