Theory of Heat - Problme set 3

Transcription

1 Theory of Heat - Problme set Fran Essenberger Problem We have a discrete random variable L x with three values. determine this probabilities. With the two given equations we can This leads to: p + p + p p + p + p + p p By insert in in eq. we nd: p 6. Because of the normalization condition p. Figure : Setch for the pl x and fl x. Problem a For one particle there is only inetic energy is E in p m. This energy is between the Interval E in [E, E + δe]. This leads to a condition for the momentum, which is: me p me + δe b p a This case is illustrated in g.. In all pictures the phase space accessible is colored in gray.

2 Figure : Phase space for one particle, which is caught in a box of length an by a limited amount of energy. b α Know there is a wea interaction. We represent this be a x x term. For the phase space only involving x and x and a xed p and p this leads to me + p α m + p m x x me + δe + p α m + p m b α me + δe + p m + x x p α m me + p m + a p m Figure : Phase space just involving the spacial coordinates. L should be grater then b, so the particle can be located somewhere. If with wea interacting just a energy exchange is meant the equations reduce to: me p + p me + δe.

3 Figure 4: Phase space for wealy interacting meaning just energy exchange and not space dependent interaction term. For the phase space just involving p and p we get a condition lie: me + mα x x p + p me + δe + mα x x, This is just the condition for a circle, where the radius must be between two limits. Figure : Phase space just involving the momentum coordinates. a me + δe + mα x x me + mα x x and b

4 Problem a b Figure 6: The plot of the probability. was set to. The probability to nd a particle at the position x is: px dx px, x e x πx 4 We calculate the integrals separately: dx x e x x dxx e x x x + xe x + x π This leads to: x dx x e x x px e dx e x x πx 4 dx x x x + x x e dxe x dx xe x x 4 e x dxe x πx x + x π π e x [ ] d dxx dx e x + πx x +. For the other particle everything goes the same way, because the integral is symmetric for the permutation of x and x x x px e x +. πx 4

5 Figure 7: Plot for px x. was again set. For a independent event the probability has to be: px x px, x px px in our case this leads to: px x px x! px px e x + πx x x 4 + e x exp [ πx x x ] x πx e x πx x x + + x 4 + x + x x. This is not the case, so the events are not independent or disjoint. c Now we calculate the conditional probability: px x : px, x px» x exp» x x exp» exp x x ππx x x + πx x x x x x x πx + [ exp + x ] x x x πx. x + Problem 4 The probabilities must depend on the waiting time t and the number of cars, which come in this time. For such problem the Poisson distribution is the right one. p t, t. This is becomes cler, when we depart the waiting time in N parts with the length dt. In every dt intervall we have the probability dr to have a succesive event. This leads to the binomial distribution

6 for the probability to get events. p N N dt dt N dtn t N t t N N N. Because the dt is small the conditions for a poisson distribution lie in problemset are fulllled and we get bey fust inseting the eq.. We now calculate the mean and variance by the momentum generating function in general, so we can just insert later. We start with the discrete random variable and t xed. M x e x t x M x x t xm x x t t t! t! +x t +x x +x t xo t e t } {{ } + t + t! t! t e }{{} + } {{ } t t t 4! t t + At the the rst term of the sum was left, because it is zero. If we now set the xed and the t variable the probability distribution is no longer normalized and we have to chec the normalization for the given value. Because of that we can not calculate this terms in general. All given results are in minutes. a Our time constant is min and our waiting time is t 6 min. So we get with eq. : b 6 For the bus everything is very simple. If every ve minutes a bus passes by in ten minutes two buses pass. For the cars we get again the Poisson distribution, but this time with a dierent time constant t min. For the mean we get: p bus δ and p car e. Bus lim N i N δ 6

7 p,4,7,7,8 4,9,4 6, 7, 8, Table : p min with the a waiting time of min For the variance we get for the buses: Bus Bus lim and for the cars with the eq. and 4 : N i N δ 4. car t car t car + t t t. If we just calculate a few probabilities for the cars with eq. we get: c Now the waiting time t is varied and the number of buses and cars passing by is x. We have a continues random variable, which we want to call just t. For the bus the probability of two successive buses is zero for waiting times smaller then minutes and one for waiting times greater then minutes. So we get p bus t, δ t t [, ]. This function is normalized. Now it is ased for the mean and the variance. For the buses we get: bus bus bus dtδ tt dtδ tt. For the cars we just modify eq.. We want the probability for two successive cars passing by in the interval t. The events A st car arrives B min no car comes and C the second car arrives are disjunct so we get pa B C A C papbpc pb. papc So we set pa B C A C pb, which is just and for our poisson distrubution. t p car t, N! e t Ne t. 6 The norm N has to be found by dtnpt,. For a integral of the type dxx n e αx we get by n! integrating n times by part α. so we nd for the norm: n+ dtnne t N! N 7

8 For the cars the integrals are a bit more complex to solve. With the normed probability we get: car car car car dt e t! t dt e t t! d For the bus the highest possible waiting time, if you arrive at a randomly chosen time, is minutes. In this time interval the probability to get a bus must be distributed equally, because the moment was chosen randomly. p bus t t [, ] Now we can calculate the mean and variance for the time to wait for a bus: bus bus bus dtt t, dtt 4 dtt 4 t 4 For the cars nothing changes in comparison to question c. Because the event, that a car will pass is independent from what happened before. The cars just come at random. You can wait endless for the next one. A good idea is to thin, that the randomly arriving observer arrives with a car. This is OK, because the cars come at random. Now it is clear, that the the probability distribution for the waiting time is the same for the time between two successive cars. The mean and variance are not changed lie discussed before: car car p car t, e t. 7 dt e t! t dt e t t! car car. For the buses there is a dierence between c and d.in d the waiting time is limited to minutes because of the distribution in found in c. So the average waiting time must be. minutes if the observer arrives at random. In 8