Review of statistics and probability I• consider first a discrete distribution of possible outcomes of some measurement: each measurement has one value hi

and can be placed in one bin, bins labeled by the index i• the total number of possible different bins is H• the number of measurements that fall into bin i is ni

• Example: measurements of human height, to nearest cm• the set of ni values is the discrete distribution

H

iinN

1

:• the total number of measurements is N:

• rescale ni by 1/N to get probability pi of hi:Nin

pi :

• pi distribution is said to be normalized: 11

H

iip

• will shortly convert this discrete stuff to a continuous distribution by replacing summations (over heights) by integrations

Review of statistics and probability II

• it tells us NOTHING about the “spread” of the distribution

• the deviationi of a measurement hi from the average := hi – <h>

• what is the average deviation? What does it tell us?

0 hhhhhhi

• and this is a paradigm for calculating ANY average!

• if we lined up all the people end to end we would obtain the total length L:

H

iii

H

iii hpNhnL

11

• the average height of a person is <h>:

H

iiiH

iin

H

iihinH

iii hphn

NN

Lh

11

1

1

1

• How narrow or broad is the distribution?

Review of statistics and probability III

• consider the squared deviationi:= (hi - <h>)2

• this is always ≥ 0• now consider the average of this quantity: the variance

• the square root of this quantity is the standard deviation, with the same units as the quantity in question• it will turn out to be the uncertainty in a measurable

22

2

1

2

1

1variance:deviation standard

hh

hhphhnN i

H

iii

H

ii

2222

1

2

11

2

2

1

2

1

2

2

1:variance

hhhhhh

hphhphp

hhphhnN

H

ii

H

iii

H

iii

i

H

iii

H

ii

Now convert to a continuous distribution• now the bin index is a continuous variable: i x• now the sums from i =1 to i = H become integrals over the domain of x (say, –∞ ≤ x ≤ ∞)• define the probability density (x) as follows: the probability that a measurement of x occurs between x and x + dx is (x) dx• the dimensions of (x) are the dimensions of 1/x

• probability that measurement x is between a and b: b

a

ab dxxP )(:

• (x) distribution is said to be normalized:

dxx)(1

• average value of x in that interval a < x < b: b

a

dxxxx )(

• average value of f(x) in that interval a < x < b: b

a

dxxxff )()(

Application of these ideas to the black body radiator• we assume that a cavity with a small hole acts as a perfect black body, since any radiation (light) that strikes the hole is ‘absorbed’… so whatever light comes out of the hole is characteristic of black body radiation• if one ‘counts standing wave modes’ in a cavity of volume V, it may be shown that the number of modes in frequency range df is expressed as a probability-like distribution

• When this is normalized, it is easy to show A = 1/kT

• for a system to possess an energy E when in an environment at temperature T, the probability that its energy is within dE is expressed by the Boltzmann factor: dEAedEE kTE)(

• average energy is therefore kTdEEekT

Eb

a

kTE /1

dffc

VdffdNdf 2

3

8)(: is intervalfrequency in tiny modes ofnumber

The wrong black body theory and its repair

• this ‘energy-per-frequency’ function (f), unfortunately, diverges to infinity as f ∞, and so does its integral over all frequencies, which would be the total energy in the cavity!• both problems are collectively termed the ultraviolet catastrophe!• to fix the problem, Planck proposed that the mode energy E is restricted to only be an integer times a constant times the frequency: E = nhf, and the value of h will be determined by the data• thus, the mode energy is quantized so now are statistics are those of a discrete distribution: we say that each mode contains an integer # of photons appropriate to its frequency

• Now we make an incorrect but reasonable assumption: assume that each mode can have any energy E at all, as long as the average in the mode is that average energy kT (thus, consistent with the Boltzmann probability), so that the amount of energy in frequency range df is (f) kT df:

dfkTfc

Vdff 2

3

8)(

Implications of quantizing a mode’s energy• again we invoke Boltzmann, and say that the probability for a mode to have energy E is nhfEAep kTnhf

E where/

• now we can find the average energy of a mode as usual:

• of course, we must normalize this discrete probability distribution:

1 1 10

/

0

/

n

nkThf

n

kTnhfE eAeAp

kThfkThf

n

nkThf

n

n

kThf

eAe

exx

x

fe

//

0

/

0

/

11

11 ||for

1

1

:sum behaved- wellandcommon a is this,any for 10 since

kTnhfkThf

E eep //1 • so the probability to have energy E is

0

//

0

//

0

1

1

n

kTnhfkThf

n

kTnhfkThf

nE

neehf

eenhfEpE

The UV catastrophe has vanished!

• drops to zero for small f like f3

kThfkThfkThf

kThf

n

nkThf

n

n

eee

een

xx

xnx

//2/

/

0

/

20

11

1

1

1 ||for 1

:sum behaved- wellandcommon a is too this

• assembling all the pieces we arrive at the Planck distribution

df

e

f

c

Vh

dfee

ehffc

Vdff

kThf

kThfkThfkThf

1

8

11

11

8)(

/

3

3

///2

3

kThfe /

• drops to zero for large f like