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Solutions for A Short Introduction to Mathematical Concepts in Physics, 1st Edition Napolitano (All Chapters included) CA$42.83   Add to cart
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Solutions for A Short Introduction to Mathematical Concepts in Physics, 1st Edition Napolitano (All Chapters included)
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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...

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  • February 24, 2024
  • 77
  • 2023/2024
  • Exam (elaborations)
  • Questions & answers

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  • solutions
  • manual
  • answer guide
  • a short introduction
  • 1st edition
  • napolitano
  • 9781032404301
  • exercises
  • solutions manual
  • mathematical concepts in physics
  • 1e

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Solutions to Exercises in
A Short Introduction to
Mathematical Concepts in Physics


Jim Napolitano

November 1, 2023



Complete Chapter Solutions Manual
are included (Ch 1 to 9)




** Immediate Download
** Swift Response
** All Chapters included

,Contents

1 Solutions 5

2 Solutions 13

3 Solutions 19

4 Solutions 37

5 Solutions 47

6 Solutions 53

7 Solutions 67

8 Solutions 73

9 Solutions 79

, Chapter 1

Solutions

(1) The answers are all straightforward:
z12 = (1 + i)2 = 1 + 2i − 1 = 2i
z1∗ = 1 − i
z1 z2∗ = (1 + i)(3 + 4i) = 3 + 3i + 4i − 4 = −1 + 7i
√ √ √
|z1 | = 1 + 1 = 2 and |z2 | = 9 + 16 = 5
√ √
|z1 z2 | = |(1 + i)(3 − 4i)| = |3 + 3i − 4i + 4| = |7 − i| = 49 + 1 = 5 2
= |z1 ||z2 |

(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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