EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Chapter 41, Problem 40P

(a)

To determine

The time difference between two neutrinos to cover the distance.

(a)

Expert Solution
Check Mark

Explanation of Solution

Introduction:

Neutrinos travel at the speed nearly equal to the speed of light. Time required is the distance covered by an object travelling with some speed.

Write the expression for time taken by first neutrino.

  t1=xu1

Here, x is the distance covered, u1 is the speed of first neutrino and t1 is the required time.

Write the expression for time taken by second neutrino.

  t2=xu2

Here, u2 is the speed of second neutrino and t2 is the required time.

Write the expression for time difference.

  Δt=t2t1

Substitute xu1 for t1 and xu2 for t2 in above expression.

  Δt=xu2xu1Δt=x(u1u2)u1u2

Since, the speed of neutrino is nearly equal to the speed of light; therefore, u1u2c2 .

Substitute c2 for u1u2 and Δu for u1u2 in above expression.

  Δtx(Δu)c2

Here, Δu is the difference in speed, c is the speed of light and Δt is the time difference.

Conclusion:

Thus, the time difference between two neutrinos to cover the distance is Δtx(Δu)c2 .

(b)

To determine

The ratio of speed of neutrino to the speed of light.

(b)

Expert Solution
Check Mark

Explanation of Solution

Introduction:

Neutrinos travel at the speed nearly equal to the speed of light. Time required is the distance covered by an object travelling with some speed.

Write the expression for theory of relativity.

  E=mc21(u/c)2

Here, m is the rest mass, u is the speed of mass, c is the speed of light and E is the energy of rest mass.

Rearrange the above expression in terms of u/c .

  uc=(1(mc2E)2)1/2

Expand the above expression through binomial expansion.

  uc=112(mc2E)2

Conclusion:

Thus, the ratio of speed of neutrino to the speed of light is uc=112(mc2E)2 .

(c)

To determine

The speed difference and time difference between two neutrinos.

(c)

Expert Solution
Check Mark

Explanation of Solution

Given:

The distance of supernova is 170,000cy .

The energy of first neutrino is 20MeV .

The energy of first neutrino is 5MeV .

The mass of neutrino is 2.0eV/c2

Formula used:

Write the expression for theory of relativity.

  E=mc21(u/c)2

Here, m is the rest mass, u is the speed of mass, c is the speed of light and E is the energy of rest mass.

Rearrange the above expression in terms of u .

  u=c(1(mc2E)2)1/2

Expand the above expression through binomial expansion.

  u=c(112(mc2E)2)

Write the expression for speed difference.

  Δu=u1u2Δu=c(112(mc2E1)2)c(112(mc2E2)2)

Here, u1 is the speed of first neutrino, u2 is the speed of second neutrino, E1 is the energy of first neutrino, E2 is the energy of second neutrino and Δu is the difference in speed.

Simplify the above expression.

  Δu=c(mc2)2(E12E22)2E12E22   ....... (1)

Write the expression for time taken by first neutrino.

  t1=xu1

Here, x is the distance covered and t1 is the required time.

Write the expression for time taken by second neutrino.

  t2=xu2

Here, t2 is the required time.

Write the expression for time difference.

  Δt=t2t1

Substitute xu1 for t1 and xu2 for t2 in above expression.

  Δt=xu2xu1Δt=x(u1u2)u1u2

Since, the speed of neutrino is nearly equal to the speed of light; therefore, u1u2c2 .

Substitute c2 for u1u2 and Δu for u1u2 in above expression.

  Δtx(Δu)c2   ....... (2)

Here, Δt is the time difference.

Calculation:

Substitute 2.0eV/c2 for m , 20MeV for E1 and 5MeV for E2 in equation (1).

  Δu=c((2.0eV/c2)c2)2((20MeV)2(5MeV)2)2(20MeV)2(5MeV)2=7.5×1014c

Substitute 170,000cy for x and 7.5×1014c for Δu in equation (2).

  Δt(170,000cy)(7.5×1014c)c2=(1.275×108y)(31.56Ms1y)=0.40s

Conclusion:

The speed difference between two neutrinos is 7.5×1014c ; the time difference between two neutrinos is 0.40s .

(c)

To determine

The speed difference and time difference between two neutrinos.

(c)

Expert Solution
Check Mark

Explanation of Solution

Given:

The distance of supernova is 170,000cy .

The energy of first neutrino is 20MeV .

The energy of first neutrino is 5MeV .

The mass of neutrino is 20eV/c2

Formula used:

Write the expression for theory of relativity.

  E=mc21(u/c)2

Here, m is the rest mass, u is the speed of mass, c is the speed of light and E is the energy of rest mass.

Rearrange the above expression in terms of u .

  u=c(1(mc2E)2)1/2

Expand the above expression through binomial expansion.

  u=c(112(mc2E)2)

Write the expression for speed difference.

  Δu=u1u2Δu=c(112(mc2E1)2)c(112(mc2E2)2)

Here, u1 is the speed of first neutrino, u2 is the speed of second neutrino, E1 is the energy of first neutrino, E2 is the energy of second neutrino and Δu is the difference in speed.

Simplify the above expression.

  Δu=c(mc2)2(E12E22)2E12E22   ....... (1)

Write the expression for time taken by first neutrino.

  t1=xu1

Here, x is the distance covered and t1 is the required time.

Write the expression for time taken by second neutrino.

  t2=xu2

Here, t2 is the required time.

Write the expression for time difference.

  Δt=t2t1

Substitute xu1 for t1 and xu2 for t2 in above expression.

  Δt=xu2xu1Δt=x(u1u2)u1u2

Since, the speed of neutrino is nearly equal to the speed of light; therefore, u1u2c2 .

Substitute c2 for u1u2 and Δu for u1u2 in above expression.

  Δtx(Δu)c2   ....... (2)

Here, Δt is the time difference.

Calculation:

Substitute 20eV/c2 for m , 20MeV for E1 and 5MeV for E2 in equation (1).

  Δu=c((20eV/c2)c2)2((20MeV)2(5MeV)2)2(20MeV)2(5MeV)2=7.5×1012c

Substitute 170,000cy for x and 7.5×1012c for Δu in equation (2).

  Δt(170,000cy)(7.5×1012c)c2=(1.275×106y)(31.56Ms1y)=40s

Conclusion:

The speed difference between two neutrinos is 7.5×1012c ; the time difference between two neutrinos is 40s .

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