The matrix F ′ , and to verify the transformation equations (15.146) for the electromagnetic fields.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 15, Problem 15.105P

To determine

The matrix F′, and to verify the transformation equations (15.146) for the electromagnetic fields.

Expert Solution & Answer

Answer to Problem 15.105P

The matrix F′=[0γ(B3−βE2/c)−γ(B2+βE3/c)−E1/c−γ(B3−βE2/c)0B1−γ(E2−βcB3)cγ(B2+βE3/c)−B10−γ(E3+βcB2)cE1/cγ(E2−βcB3)cγ(E3+βcB3)c0], and the transformation equations are verified.

Explanation of Solution

Write the matrix form of F.

F=[0B3−B2−E1/c−B30B1−E2/cB2−B10−E3/cE1/cE2/cE3/c0] (I)

Here, F is the four tensor.

Write the matrix form of standard Lorentz boost Λ.

Λ=[γ00−γβ01000010−γβ00γ] (II)

Here, Λ is the matrix form of standard Lorentz boost.

From the above equation Λ˜

Λ˜=[γ00−γβ01000010−γβ00γ] (III)

Since F′=ΛFΛ˜, from equation (I), (II), and (III) the matrix representation of F′ is given as

F′=[γ00−γβ01000010−γβ00γ][0B3−B2−E1/c−B30B1−E2/cB2−B10−E3/cE1/cE2/cE3/c0][γ00−γβ01000010−γβ00γ]=[0+0+0−γβE1/cγB3+0+0−γβE2/c−γB2+0+0−γβE3/c−γE1/c+0+0+00−B3+0+00+0+0+00+B1+0+00−E2/c+0+00+0+B2+00+0−B1+00+0+0+00+0−E3/c+00+0+0+γE1/c−γβB3+0+0+γE2/cγβB2+0+0+γE3/cγβE1/c+0+0+0][γ00−γβ01000010−γβ00γ]=[−γβE1/cγB3−γβE2/c−γB2−γβE3/c−γE1/c−B30B1−E2/cB2−B10−E3/cγE1/c−γβB3+γE2/cγβB2+γE3/cγβE1/c][γ00−γβ01000010−γβ00γ]

The above equation can be written as

F′=[−γ2βE1/c+0+0+γ2βE1/c0+γB3−γβE2/c+0+00+0−γB2−γβE3/c+0γ2β2E1/c+0+0−γ2E1/c−γB3+0+0+γβE2/c0+0+0+00+0+B1+0γβB3+0+0−γE2/cγB1+0+0+γβE3/c0−B1+0+00+0+0+0−γβB2+0+0−γE3/cγ2E1/c+0+0−γ2β2E1/c0−γβB3+γE2/c+0+00+0+γβ2+γE3/c+0−γ2βE1/c+0+0γ2βE1/c]=[0γ(B3−βE2/c)γ(−B2−βE3/c)γ2(β2−1)E1/cγ(−B3+βE2/c)0B1γ(βB3−E2/c)γ(B2+βE3/c)−B10γ(−βB2−E3/c)γ2(1−β2)E1/cγ(−βB3+E2/c)γ(βB2+E3/c)0] (IV)

The above can be also written as

F′=[0B′3−B′2−E′1/c−B′30B′1−E′2/cB′2−B′10−E′3/cE′1/cE′2/cE′3/c0] (V)

By comparing (IV) and (V), write the expression for B′3.

B′3=γ(B3−βE2/c) (VI)

By comparing (IV) and (V), write the expression for B′2.

−B′2=−γ(B2+βE3/c)B′2=γ(B2+βE3/c) (VII)

By comparing (IV) and (V), write the expression for B′1.

B′1=B1 (VIII)

By comparing (IV) and (V), write the expression for E′1.

−E′1/c=E′/cE′1=E1 (IX)

By comparing (IV) and (V), write the expression for E′2.

−E′2/c=−γ(E2−βcB3)E′2=γ(E2−βcB3) (X)

By comparing (IV) and (V), write the expression for E′3.

−E′3/c=−γ(E3+βcB2)E′3=γ(E3+βcB2) (XI)

The results obtained in equation (VI), (VII), (VIII), (IX), and (X) is same as the transformation equation in Equation (15-146).

Conclusion:

Therefore, the matrix F′ is derived as

F′=[0γ(B3−βE2/c)−γ(B2+βE3/c)−E1/c−γ(B3−βE2/c)0B1−γ(E2−βcB3)cγ(B2+βE3/c)−B10−γ(E3+βcB2)cE1/cγ(E2−βcB3)cγ(E3+βcB3)c0], and the transformation equations are verified.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

A ball is thrown with an initial speed v, at an angle 6, with the horizontal. The horizontal range of the ball is R, and the ball reaches a maximum height R/4. In terms of R and g, find the following. (a) the time interval during which the ball is in motion 2R (b) the ball's speed at the peak of its path v= Rg 2 √ sin 26, V 3 (c) the initial vertical component of its velocity Rg sin ei sin 20 (d) its initial speed Rg √ sin 20 × (e) the angle 6, expressed in terms of arctan of a fraction. 1 (f) Suppose the ball is thrown at the same initial speed found in (d) but at the angle appropriate for reaching the greatest height that it can. Find this height. hmax R2 (g) Suppose the ball is thrown at the same initial speed but at the angle for greatest possible range. Find this maximum horizontal range. Xmax R√3 2

An outfielder throws a baseball to his catcher in an attempt to throw out a runner at home plate. The ball bounces once before reaching the catcher. Assume the angle at which the bounced ball leaves the ground is the same as the angle at which the outfielder threw it as shown in the figure, but that the ball's speed after the bounce is one-half of what it was before the bounce. 8 (a) Assuming the ball is always thrown with the same initial speed, at what angle & should the fielder throw the ball to make it go the same distance D with one bounce (blue path) as a ball thrown upward at 35.0° with no bounce (green path)? 24 (b) Determine the ratio of the time interval for the one-bounce throw to the flight time for the no-bounce throw. Cone-bounce no-bounce 0.940

A rocket is launched at an angle of 60.0° above the horizontal with an initial speed of 97 m/s. The rocket moves for 3.00 s along its initial line of motion with an acceleration of 28.0 m/s². At this time, its engines fail and the rocket proceeds to move as a projectile. (a) Find the maximum altitude reached by the rocket. 1445.46 Your response differs from the correct answer by more than 10%. Double check your calculations. m (b) Find its total time of flight. 36.16 x Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. s (c) Find its horizontal range. 1753.12 × Your response differs from the correct answer by more than 10%. Double check your calculations. m

Answer 2

Question

Chapter 15, Problem 15.105P

To determine

The matrix F′, and to verify the transformation equations (15.146) for the electromagnetic fields.

Expert Solution & Answer

Answer to Problem 15.105P

The matrix F′=[0γ(B3−βE2/c)−γ(B2+βE3/c)−E1/c−γ(B3−βE2/c)0B1−γ(E2−βcB3)cγ(B2+βE3/c)−B10−γ(E3+βcB2)cE1/cγ(E2−βcB3)cγ(E3+βcB3)c0], and the transformation equations are verified.

Explanation of Solution

Write the matrix form of F.

F=[0B3−B2−E1/c−B30B1−E2/cB2−B10−E3/cE1/cE2/cE3/c0] (I)

Here, F is the four tensor.

Write the matrix form of standard Lorentz boost Λ.

Λ=[γ00−γβ01000010−γβ00γ] (II)

Here, Λ is the matrix form of standard Lorentz boost.

From the above equation Λ˜

Λ˜=[γ00−γβ01000010−γβ00γ] (III)

Since F′=ΛFΛ˜, from equation (I), (II), and (III) the matrix representation of F′ is given as

F′=[γ00−γβ01000010−γβ00γ][0B3−B2−E1/c−B30B1−E2/cB2−B10−E3/cE1/cE2/cE3/c0][γ00−γβ01000010−γβ00γ]=[0+0+0−γβE1/cγB3+0+0−γβE2/c−γB2+0+0−γβE3/c−γE1/c+0+0+00−B3+0+00+0+0+00+B1+0+00−E2/c+0+00+0+B2+00+0−B1+00+0+0+00+0−E3/c+00+0+0+γE1/c−γβB3+0+0+γE2/cγβB2+0+0+γE3/cγβE1/c+0+0+0][γ00−γβ01000010−γβ00γ]=[−γβE1/cγB3−γβE2/c−γB2−γβE3/c−γE1/c−B30B1−E2/cB2−B10−E3/cγE1/c−γβB3+γE2/cγβB2+γE3/cγβE1/c][γ00−γβ01000010−γβ00γ]

The above equation can be written as

F′=[−γ2βE1/c+0+0+γ2βE1/c0+γB3−γβE2/c+0+00+0−γB2−γβE3/c+0γ2β2E1/c+0+0−γ2E1/c−γB3+0+0+γβE2/c0+0+0+00+0+B1+0γβB3+0+0−γE2/cγB1+0+0+γβE3/c0−B1+0+00+0+0+0−γβB2+0+0−γE3/cγ2E1/c+0+0−γ2β2E1/c0−γβB3+γE2/c+0+00+0+γβ2+γE3/c+0−γ2βE1/c+0+0γ2βE1/c]=[0γ(B3−βE2/c)γ(−B2−βE3/c)γ2(β2−1)E1/cγ(−B3+βE2/c)0B1γ(βB3−E2/c)γ(B2+βE3/c)−B10γ(−βB2−E3/c)γ2(1−β2)E1/cγ(−βB3+E2/c)γ(βB2+E3/c)0] (IV)

The above can be also written as

F′=[0B′3−B′2−E′1/c−B′30B′1−E′2/cB′2−B′10−E′3/cE′1/cE′2/cE′3/c0] (V)

By comparing (IV) and (V), write the expression for B′3.

B′3=γ(B3−βE2/c) (VI)

By comparing (IV) and (V), write the expression for B′2.

−B′2=−γ(B2+βE3/c)B′2=γ(B2+βE3/c) (VII)

By comparing (IV) and (V), write the expression for B′1.

B′1=B1 (VIII)

By comparing (IV) and (V), write the expression for E′1.

−E′1/c=E′/cE′1=E1 (IX)

By comparing (IV) and (V), write the expression for E′2.

−E′2/c=−γ(E2−βcB3)E′2=γ(E2−βcB3) (X)

By comparing (IV) and (V), write the expression for E′3.

−E′3/c=−γ(E3+βcB2)E′3=γ(E3+βcB2) (XI)

The results obtained in equation (VI), (VII), (VIII), (IX), and (X) is same as the transformation equation in Equation (15-146).

Conclusion:

Therefore, the matrix F′ is derived as

F′=[0γ(B3−βE2/c)−γ(B2+βE3/c)−E1/c−γ(B3−βE2/c)0B1−γ(E2−βcB3)cγ(B2+βE3/c)−B10−γ(E3+βcB2)cE1/cγ(E2−βcB3)cγ(E3+βcB3)c0], and the transformation equations are verified.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 15, Problem 15.105P

To determine

The matrix F′, and to verify the transformation equations (15.146) for the electromagnetic fields.

Expert Solution & Answer

Answer to Problem 15.105P

The matrix F′=[0γ(B3−βE2/c)−γ(B2+βE3/c)−E1/c−γ(B3−βE2/c)0B1−γ(E2−βcB3)cγ(B2+βE3/c)−B10−γ(E3+βcB2)cE1/cγ(E2−βcB3)cγ(E3+βcB3)c0], and the transformation equations are verified.

Explanation of Solution

Write the matrix form of F.

F=[0B3−B2−E1/c−B30B1−E2/cB2−B10−E3/cE1/cE2/cE3/c0] (I)

Here, F is the four tensor.

Write the matrix form of standard Lorentz boost Λ.

Λ=[γ00−γβ01000010−γβ00γ] (II)

Here, Λ is the matrix form of standard Lorentz boost.

From the above equation Λ˜

Λ˜=[γ00−γβ01000010−γβ00γ] (III)

Since F′=ΛFΛ˜, from equation (I), (II), and (III) the matrix representation of F′ is given as

F′=[γ00−γβ01000010−γβ00γ][0B3−B2−E1/c−B30B1−E2/cB2−B10−E3/cE1/cE2/cE3/c0][γ00−γβ01000010−γβ00γ]=[0+0+0−γβE1/cγB3+0+0−γβE2/c−γB2+0+0−γβE3/c−γE1/c+0+0+00−B3+0+00+0+0+00+B1+0+00−E2/c+0+00+0+B2+00+0−B1+00+0+0+00+0−E3/c+00+0+0+γE1/c−γβB3+0+0+γE2/cγβB2+0+0+γE3/cγβE1/c+0+0+0][γ00−γβ01000010−γβ00γ]=[−γβE1/cγB3−γβE2/c−γB2−γβE3/c−γE1/c−B30B1−E2/cB2−B10−E3/cγE1/c−γβB3+γE2/cγβB2+γE3/cγβE1/c][γ00−γβ01000010−γβ00γ]

The above equation can be written as

F′=[−γ2βE1/c+0+0+γ2βE1/c0+γB3−γβE2/c+0+00+0−γB2−γβE3/c+0γ2β2E1/c+0+0−γ2E1/c−γB3+0+0+γβE2/c0+0+0+00+0+B1+0γβB3+0+0−γE2/cγB1+0+0+γβE3/c0−B1+0+00+0+0+0−γβB2+0+0−γE3/cγ2E1/c+0+0−γ2β2E1/c0−γβB3+γE2/c+0+00+0+γβ2+γE3/c+0−γ2βE1/c+0+0γ2βE1/c]=[0γ(B3−βE2/c)γ(−B2−βE3/c)γ2(β2−1)E1/cγ(−B3+βE2/c)0B1γ(βB3−E2/c)γ(B2+βE3/c)−B10γ(−βB2−E3/c)γ2(1−β2)E1/cγ(−βB3+E2/c)γ(βB2+E3/c)0] (IV)

The above can be also written as

F′=[0B′3−B′2−E′1/c−B′30B′1−E′2/cB′2−B′10−E′3/cE′1/cE′2/cE′3/c0] (V)

By comparing (IV) and (V), write the expression for B′3.

B′3=γ(B3−βE2/c) (VI)

By comparing (IV) and (V), write the expression for B′2.

−B′2=−γ(B2+βE3/c)B′2=γ(B2+βE3/c) (VII)

By comparing (IV) and (V), write the expression for B′1.

B′1=B1 (VIII)

By comparing (IV) and (V), write the expression for E′1.

−E′1/c=E′/cE′1=E1 (IX)

By comparing (IV) and (V), write the expression for E′2.

−E′2/c=−γ(E2−βcB3)E′2=γ(E2−βcB3) (X)

By comparing (IV) and (V), write the expression for E′3.

−E′3/c=−γ(E3+βcB2)E′3=γ(E3+βcB2) (XI)

The results obtained in equation (VI), (VII), (VIII), (IX), and (X) is same as the transformation equation in Equation (15-146).

Conclusion:

Therefore, the matrix F′ is derived as

F′=[0γ(B3−βE2/c)−γ(B2+βE3/c)−E1/c−γ(B3−βE2/c)0B1−γ(E2−βcB3)cγ(B2+βE3/c)−B10−γ(E3+βcB2)cE1/cγ(E2−βcB3)cγ(E3+βcB3)c0], and the transformation equations are verified.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 15, Problem 15.105P

To determine

The matrix F′, and to verify the transformation equations (15.146) for the electromagnetic fields.

Expert Solution & Answer

Answer to Problem 15.105P

The matrix F′=[0γ(B3−βE2/c)−γ(B2+βE3/c)−E1/c−γ(B3−βE2/c)0B1−γ(E2−βcB3)cγ(B2+βE3/c)−B10−γ(E3+βcB2)cE1/cγ(E2−βcB3)cγ(E3+βcB3)c0], and the transformation equations are verified.

Explanation of Solution

Write the matrix form of F.

F=[0B3−B2−E1/c−B30B1−E2/cB2−B10−E3/cE1/cE2/cE3/c0] (I)

Here, F is the four tensor.

Write the matrix form of standard Lorentz boost Λ.

Λ=[γ00−γβ01000010−γβ00γ] (II)

Here, Λ is the matrix form of standard Lorentz boost.

From the above equation Λ˜

Λ˜=[γ00−γβ01000010−γβ00γ] (III)

Since F′=ΛFΛ˜, from equation (I), (II), and (III) the matrix representation of F′ is given as

F′=[γ00−γβ01000010−γβ00γ][0B3−B2−E1/c−B30B1−E2/cB2−B10−E3/cE1/cE2/cE3/c0][γ00−γβ01000010−γβ00γ]=[0+0+0−γβE1/cγB3+0+0−γβE2/c−γB2+0+0−γβE3/c−γE1/c+0+0+00−B3+0+00+0+0+00+B1+0+00−E2/c+0+00+0+B2+00+0−B1+00+0+0+00+0−E3/c+00+0+0+γE1/c−γβB3+0+0+γE2/cγβB2+0+0+γE3/cγβE1/c+0+0+0][γ00−γβ01000010−γβ00γ]=[−γβE1/cγB3−γβE2/c−γB2−γβE3/c−γE1/c−B30B1−E2/cB2−B10−E3/cγE1/c−γβB3+γE2/cγβB2+γE3/cγβE1/c][γ00−γβ01000010−γβ00γ]

The above equation can be written as

F′=[−γ2βE1/c+0+0+γ2βE1/c0+γB3−γβE2/c+0+00+0−γB2−γβE3/c+0γ2β2E1/c+0+0−γ2E1/c−γB3+0+0+γβE2/c0+0+0+00+0+B1+0γβB3+0+0−γE2/cγB1+0+0+γβE3/c0−B1+0+00+0+0+0−γβB2+0+0−γE3/cγ2E1/c+0+0−γ2β2E1/c0−γβB3+γE2/c+0+00+0+γβ2+γE3/c+0−γ2βE1/c+0+0γ2βE1/c]=[0γ(B3−βE2/c)γ(−B2−βE3/c)γ2(β2−1)E1/cγ(−B3+βE2/c)0B1γ(βB3−E2/c)γ(B2+βE3/c)−B10γ(−βB2−E3/c)γ2(1−β2)E1/cγ(−βB3+E2/c)γ(βB2+E3/c)0] (IV)

The above can be also written as

F′=[0B′3−B′2−E′1/c−B′30B′1−E′2/cB′2−B′10−E′3/cE′1/cE′2/cE′3/c0] (V)

By comparing (IV) and (V), write the expression for B′3.

B′3=γ(B3−βE2/c) (VI)

By comparing (IV) and (V), write the expression for B′2.

−B′2=−γ(B2+βE3/c)B′2=γ(B2+βE3/c) (VII)

By comparing (IV) and (V), write the expression for B′1.

B′1=B1 (VIII)

By comparing (IV) and (V), write the expression for E′1.

−E′1/c=E′/cE′1=E1 (IX)

By comparing (IV) and (V), write the expression for E′2.

−E′2/c=−γ(E2−βcB3)E′2=γ(E2−βcB3) (X)

By comparing (IV) and (V), write the expression for E′3.

−E′3/c=−γ(E3+βcB2)E′3=γ(E3+βcB2) (XI)

The results obtained in equation (VI), (VII), (VIII), (IX), and (X) is same as the transformation equation in Equation (15-146).

Conclusion:

Therefore, the matrix F′ is derived as

F′=[0γ(B3−βE2/c)−γ(B2+βE3/c)−E1/c−γ(B3−βE2/c)0B1−γ(E2−βcB3)cγ(B2+βE3/c)−B10−γ(E3+βcB2)cE1/cγ(E2−βcB3)cγ(E3+βcB3)c0], and the transformation equations are verified.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 15, Problem 15.105P

To determine

The matrix F′, and to verify the transformation equations (15.146) for the electromagnetic fields.

Expert Solution & Answer

Answer to Problem 15.105P

The matrix F′=[0γ(B3−βE2/c)−γ(B2+βE3/c)−E1/c−γ(B3−βE2/c)0B1−γ(E2−βcB3)cγ(B2+βE3/c)−B10−γ(E3+βcB2)cE1/cγ(E2−βcB3)cγ(E3+βcB3)c0], and the transformation equations are verified.

Explanation of Solution

Write the matrix form of F.

F=[0B3−B2−E1/c−B30B1−E2/cB2−B10−E3/cE1/cE2/cE3/c0] (I)

Here, F is the four tensor.

Write the matrix form of standard Lorentz boost Λ.

Λ=[γ00−γβ01000010−γβ00γ] (II)

Here, Λ is the matrix form of standard Lorentz boost.

From the above equation Λ˜

Λ˜=[γ00−γβ01000010−γβ00γ] (III)

Since F′=ΛFΛ˜, from equation (I), (II), and (III) the matrix representation of F′ is given as

F′=[γ00−γβ01000010−γβ00γ][0B3−B2−E1/c−B30B1−E2/cB2−B10−E3/cE1/cE2/cE3/c0][γ00−γβ01000010−γβ00γ]=[0+0+0−γβE1/cγB3+0+0−γβE2/c−γB2+0+0−γβE3/c−γE1/c+0+0+00−B3+0+00+0+0+00+B1+0+00−E2/c+0+00+0+B2+00+0−B1+00+0+0+00+0−E3/c+00+0+0+γE1/c−γβB3+0+0+γE2/cγβB2+0+0+γE3/cγβE1/c+0+0+0][γ00−γβ01000010−γβ00γ]=[−γβE1/cγB3−γβE2/c−γB2−γβE3/c−γE1/c−B30B1−E2/cB2−B10−E3/cγE1/c−γβB3+γE2/cγβB2+γE3/cγβE1/c][γ00−γβ01000010−γβ00γ]

The above equation can be written as

F′=[−γ2βE1/c+0+0+γ2βE1/c0+γB3−γβE2/c+0+00+0−γB2−γβE3/c+0γ2β2E1/c+0+0−γ2E1/c−γB3+0+0+γβE2/c0+0+0+00+0+B1+0γβB3+0+0−γE2/cγB1+0+0+γβE3/c0−B1+0+00+0+0+0−γβB2+0+0−γE3/cγ2E1/c+0+0−γ2β2E1/c0−γβB3+γE2/c+0+00+0+γβ2+γE3/c+0−γ2βE1/c+0+0γ2βE1/c]=[0γ(B3−βE2/c)γ(−B2−βE3/c)γ2(β2−1)E1/cγ(−B3+βE2/c)0B1γ(βB3−E2/c)γ(B2+βE3/c)−B10γ(−βB2−E3/c)γ2(1−β2)E1/cγ(−βB3+E2/c)γ(βB2+E3/c)0] (IV)

The above can be also written as

F′=[0B′3−B′2−E′1/c−B′30B′1−E′2/cB′2−B′10−E′3/cE′1/cE′2/cE′3/c0] (V)

By comparing (IV) and (V), write the expression for B′3.

B′3=γ(B3−βE2/c) (VI)

By comparing (IV) and (V), write the expression for B′2.

−B′2=−γ(B2+βE3/c)B′2=γ(B2+βE3/c) (VII)

By comparing (IV) and (V), write the expression for B′1.

B′1=B1 (VIII)

By comparing (IV) and (V), write the expression for E′1.

−E′1/c=E′/cE′1=E1 (IX)

By comparing (IV) and (V), write the expression for E′2.

−E′2/c=−γ(E2−βcB3)E′2=γ(E2−βcB3) (X)

By comparing (IV) and (V), write the expression for E′3.

−E′3/c=−γ(E3+βcB2)E′3=γ(E3+βcB2) (XI)

The results obtained in equation (VI), (VII), (VIII), (IX), and (X) is same as the transformation equation in Equation (15-146).

Conclusion:

Therefore, the matrix F′ is derived as

F′=[0γ(B3−βE2/c)−γ(B2+βE3/c)−E1/c−γ(B3−βE2/c)0B1−γ(E2−βcB3)cγ(B2+βE3/c)−B10−γ(E3+βcB2)cE1/cγ(E2−βcB3)cγ(E3+βcB3)c0], and the transformation equations are verified.

Answer 6

Chapter 15, Problem 15.105P

To determine

The matrix F′, and to verify the transformation equations (15.146) for the electromagnetic fields.

Expert Solution & Answer

Answer to Problem 15.105P

The matrix F′=[0γ(B3−βE2/c)−γ(B2+βE3/c)−E1/c−γ(B3−βE2/c)0B1−γ(E2−βcB3)cγ(B2+βE3/c)−B10−γ(E3+βcB2)cE1/cγ(E2−βcB3)cγ(E3+βcB3)c0], and the transformation equations are verified.

Explanation of Solution

Write the matrix form of F.

F=[0B3−B2−E1/c−B30B1−E2/cB2−B10−E3/cE1/cE2/cE3/c0] (I)

Here, F is the four tensor.

Write the matrix form of standard Lorentz boost Λ.

Λ=[γ00−γβ01000010−γβ00γ] (II)

Here, Λ is the matrix form of standard Lorentz boost.

From the above equation Λ˜

Λ˜=[γ00−γβ01000010−γβ00γ] (III)

Since F′=ΛFΛ˜, from equation (I), (II), and (III) the matrix representation of F′ is given as

F′=[γ00−γβ01000010−γβ00γ][0B3−B2−E1/c−B30B1−E2/cB2−B10−E3/cE1/cE2/cE3/c0][γ00−γβ01000010−γβ00γ]=[0+0+0−γβE1/cγB3+0+0−γβE2/c−γB2+0+0−γβE3/c−γE1/c+0+0+00−B3+0+00+0+0+00+B1+0+00−E2/c+0+00+0+B2+00+0−B1+00+0+0+00+0−E3/c+00+0+0+γE1/c−γβB3+0+0+γE2/cγβB2+0+0+γE3/cγβE1/c+0+0+0][γ00−γβ01000010−γβ00γ]=[−γβE1/cγB3−γβE2/c−γB2−γβE3/c−γE1/c−B30B1−E2/cB2−B10−E3/cγE1/c−γβB3+γE2/cγβB2+γE3/cγβE1/c][γ00−γβ01000010−γβ00γ]

The above equation can be written as

F′=[−γ2βE1/c+0+0+γ2βE1/c0+γB3−γβE2/c+0+00+0−γB2−γβE3/c+0γ2β2E1/c+0+0−γ2E1/c−γB3+0+0+γβE2/c0+0+0+00+0+B1+0γβB3+0+0−γE2/cγB1+0+0+γβE3/c0−B1+0+00+0+0+0−γβB2+0+0−γE3/cγ2E1/c+0+0−γ2β2E1/c0−γβB3+γE2/c+0+00+0+γβ2+γE3/c+0−γ2βE1/c+0+0γ2βE1/c]=[0γ(B3−βE2/c)γ(−B2−βE3/c)γ2(β2−1)E1/cγ(−B3+βE2/c)0B1γ(βB3−E2/c)γ(B2+βE3/c)−B10γ(−βB2−E3/c)γ2(1−β2)E1/cγ(−βB3+E2/c)γ(βB2+E3/c)0] (IV)

The above can be also written as

F′=[0B′3−B′2−E′1/c−B′30B′1−E′2/cB′2−B′10−E′3/cE′1/cE′2/cE′3/c0] (V)

By comparing (IV) and (V), write the expression for B′3.

B′3=γ(B3−βE2/c) (VI)

By comparing (IV) and (V), write the expression for B′2.

−B′2=−γ(B2+βE3/c)B′2=γ(B2+βE3/c) (VII)

By comparing (IV) and (V), write the expression for B′1.

B′1=B1 (VIII)

By comparing (IV) and (V), write the expression for E′1.

−E′1/c=E′/cE′1=E1 (IX)

By comparing (IV) and (V), write the expression for E′2.

−E′2/c=−γ(E2−βcB3)E′2=γ(E2−βcB3) (X)

By comparing (IV) and (V), write the expression for E′3.

−E′3/c=−γ(E3+βcB2)E′3=γ(E3+βcB2) (XI)

The results obtained in equation (VI), (VII), (VIII), (IX), and (X) is same as the transformation equation in Equation (15-146).

Conclusion:

Therefore, the matrix F′ is derived as

F′=[0γ(B3−βE2/c)−γ(B2+βE3/c)−E1/c−γ(B3−βE2/c)0B1−γ(E2−βcB3)cγ(B2+βE3/c)−B10−γ(E3+βcB2)cE1/cγ(E2−βcB3)cγ(E3+βcB3)c0], and the transformation equations are verified.

Answer 7

Chapter 15, Problem 15.105P

To determine

The matrix F′, and to verify the transformation equations (15.146) for the electromagnetic fields.

Answer 8

Expert Solution & Answer

Answer to Problem 15.105P

The matrix F′=[0γ(B3−βE2/c)−γ(B2+βE3/c)−E1/c−γ(B3−βE2/c)0B1−γ(E2−βcB3)cγ(B2+βE3/c)−B10−γ(E3+βcB2)cE1/cγ(E2−βcB3)cγ(E3+βcB3)c0], and the transformation equations are verified.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Write the matrix form of F.

F=[0B3−B2−E1/c−B30B1−E2/cB2−B10−E3/cE1/cE2/cE3/c0] (I)

Here, F is the four tensor.

Write the matrix form of standard Lorentz boost Λ.

Λ=[γ00−γβ01000010−γβ00γ] (II)

Here, Λ is the matrix form of standard Lorentz boost.

From the above equation Λ˜

Λ˜=[γ00−γβ01000010−γβ00γ] (III)

Since F′=ΛFΛ˜, from equation (I), (II), and (III) the matrix representation of F′ is given as

F′=[γ00−γβ01000010−γβ00γ][0B3−B2−E1/c−B30B1−E2/cB2−B10−E3/cE1/cE2/cE3/c0][γ00−γβ01000010−γβ00γ]=[0+0+0−γβE1/cγB3+0+0−γβE2/c−γB2+0+0−γβE3/c−γE1/c+0+0+00−B3+0+00+0+0+00+B1+0+00−E2/c+0+00+0+B2+00+0−B1+00+0+0+00+0−E3/c+00+0+0+γE1/c−γβB3+0+0+γE2/cγβB2+0+0+γE3/cγβE1/c+0+0+0][γ00−γβ01000010−γβ00γ]=[−γβE1/cγB3−γβE2/c−γB2−γβE3/c−γE1/c−B30B1−E2/cB2−B10−E3/cγE1/c−γβB3+γE2/cγβB2+γE3/cγβE1/c][γ00−γβ01000010−γβ00γ]

The above equation can be written as

F′=[−γ2βE1/c+0+0+γ2βE1/c0+γB3−γβE2/c+0+00+0−γB2−γβE3/c+0γ2β2E1/c+0+0−γ2E1/c−γB3+0+0+γβE2/c0+0+0+00+0+B1+0γβB3+0+0−γE2/cγB1+0+0+γβE3/c0−B1+0+00+0+0+0−γβB2+0+0−γE3/cγ2E1/c+0+0−γ2β2E1/c0−γβB3+γE2/c+0+00+0+γβ2+γE3/c+0−γ2βE1/c+0+0γ2βE1/c]=[0γ(B3−βE2/c)γ(−B2−βE3/c)γ2(β2−1)E1/cγ(−B3+βE2/c)0B1γ(βB3−E2/c)γ(B2+βE3/c)−B10γ(−βB2−E3/c)γ2(1−β2)E1/cγ(−βB3+E2/c)γ(βB2+E3/c)0] (IV)

The above can be also written as

F′=[0B′3−B′2−E′1/c−B′30B′1−E′2/cB′2−B′10−E′3/cE′1/cE′2/cE′3/c0] (V)

By comparing (IV) and (V), write the expression for B′3.

B′3=γ(B3−βE2/c) (VI)

By comparing (IV) and (V), write the expression for B′2.

−B′2=−γ(B2+βE3/c)B′2=γ(B2+βE3/c) (VII)

By comparing (IV) and (V), write the expression for B′1.

B′1=B1 (VIII)

By comparing (IV) and (V), write the expression for E′1.

−E′1/c=E′/cE′1=E1 (IX)

By comparing (IV) and (V), write the expression for E′2.

−E′2/c=−γ(E2−βcB3)E′2=γ(E2−βcB3) (X)

By comparing (IV) and (V), write the expression for E′3.

−E′3/c=−γ(E3+βcB2)E′3=γ(E3+βcB2) (XI)

The results obtained in equation (VI), (VII), (VIII), (IX), and (X) is same as the transformation equation in Equation (15-146).

Conclusion:

Therefore, the matrix F′ is derived as

F′=[0γ(B3−βE2/c)−γ(B2+βE3/c)−E1/c−γ(B3−βE2/c)0B1−γ(E2−βcB3)cγ(B2+βE3/c)−B10−γ(E3+βcB2)cE1/cγ(E2−βcB3)cγ(E3+βcB3)c0], and the transformation equations are verified.