Solutions for A Short Introduction to Mathematical Concepts in Physics, 1st Edition Napolitano (All Chapters included)
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Course
Math
Institution
Math
Complete Solutions Manual for A Short Introduction to Mathematical Concepts in Physics, 1st Edition by Jim Napolitano ; ISBN13: 9781032404301. (Full Chapters included Chapter 1 to 9)....
1. Basic Concepts
2. Infinite Series
3. Ordinary Differential Equations
4. Vector Calculus and Partial Diffe...
(2) [P V ] = [Force]L−2 L3 = M LT −2 ]L−2 L3 = M L2 T −2 = [Energy]. From the ideal gas law,
since N is dimensionless, kT must have the dimensions of energy. The SI unit of energy is
Joule, so the units of k are Joule/K.
(3) First find [k] = [Force]/[Distance] = M LT −2 L−1 = M T −2 . (a) Write ϵ = mx k y Az ,
so
M L2 T −2 = M x M y T −2y Lz = M x+y T −2y Lz
and y = 1, z = 2, and x = 0. That is, the energy scale is ϵ = kA2 . You’ll recall that
the potential energy of a harmonic oscillator with amplitude A is kA2 /2 so the scale is the
correct energy to within a factor of two. (b) Write ϵ = mx k y ℏz , so
M L2 T −2 = M x M y T −2y L2z M z T −z = M x+y+z T −2y−z L2z
and z = 1, 2y + z = 2y + 1 = 2 so y = 1/2, and x + y + z = x + 3/2 = 1 so x = −1/2.
Therefore ϵ = ℏ(k/m)1/2 . In quantum mechanics you will learn that the energy spacing in
the harmonic oscillator is indeed ℏω where ω = (k/m)1/2 .
(4) The calculation is straightforward:
τ = mx ℓy g z
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