An Introduction to Thermal Physics
An Introduction to Thermal Physics
1st Edition
ISBN: 9780201380279
Author: Daniel V. Schroeder
Publisher: Addison Wesley
Question
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Chapter A.6, Problem 24P

(a)

To determine

To Write: The formula for the total zero-point energy of all the modes of the field inside the box in terms of a triple integral over the mode numbers in the x, y, and z direction.

(a)

Expert Solution
Check Mark

Answer to Problem 24P

  EV=2π2c20ω3dω

Explanation of Solution

Formula used:

Total energy within the box can be written as:

  E=2 n x n y n z ε( n ) =2ε( n )d n xd n yd n z

Where, n is positive integernx,ny,nz is a triplet of positive number.

Factor of 2 is for the two-spin orientation for each (n)

Calculation:

Wavelength can be given as:

  λn=2Ln

So, speed of light:

  c=ω2πλnc=ω2π2LnC=ωLnπn=ωLcπ...(1)

  An Introduction to Thermal Physics, Chapter A.6, Problem 24P

So, E=20n maxdn0πdθ0π/2dϕn2sinθε(n)

From equation (1)

  E=2 L cπdω0πdθ0 ω max ω 3dωE=2(L cπ)( L cπ)2(π2)0 ω 3dωE=L32c2π20 ω 3dωEL3=2c2π20 ω 3dωEV=2c2π20 ω 3dω

Conclusion:

The formula for the total zero-point energy of all the modes of the field inside the box is

   EV=2c2π20ω3dω .

(b)

To determine

To Show: The Planck length indeed has units of length.

To Calculate:Numerical value of Planck length.

(b)

Expert Solution
Check Mark

Answer to Problem 24P

  1.61×1035m

Explanation of Solution

Given:

Planck Length = Gc3

Formula used:

Dimension formula ofG=[M1][L3][T2]=[M][L2][T1]c=[M0][L1][T1]

Calculation:

Dimension formula for given expression can be calculated as:

   G c 3 = ( [ M 1 ][ L 3 ][ T 2 ] )( [M][ L 2 ][ T 1 ] ) [ M 0 ][ L 3 ][ T 3 ] G c 3 =( [ L 2 ] 1/2) G c 3 =[L]

Planck length can be calculated as:

   G c 3 = 6.674× 10 11 ×1.053× 10 34 27× 10 24 G c 3 =2.6028× 10 70 G c 3 =1.61×1035m

Conclusion:

It is proved that the Planck length indeed has units of length and its magnitude is 1.61×1035m .

(c)

To determine

To Estimate: The energy per unit volume in empty state and equivalent mass density.

(c)

Expert Solution
Check Mark

Answer to Problem 24P

  2.52×10123 J/m3

  2.8×10106 kg/m3

Explanation of Solution

Formula used:

  EV=2π2c20ω3dω

Where, is the reduced Planck’s constantc is the speed of lightω is the angular velocity

Calculation:

  EV=2π2c20 ω 3dωEV=2π2c2×ω max44EV=ω max48π2c2

But,

  c=ω2πλλ=Lp

Therefore,

  EV=8π2c2×( 2πc L p )4EV=2π2c2Lp4EV=2×1.051× 10 34× ( 3× 10 8 )2 ( 3.14 )2 ( 1.65× 10 35 )4EV=2.52×10123 J/m3

Now, mass density can be calculated asm=EVc2m=2.52× 10 123  ( 3× 10 8 )2m=2.8×10106 kg/m3

It is much greater than the average mass density of ordinary matter in the universe.

Conclusion:

Average mass density is 2.8×10106 kg/m3 .

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