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

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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.

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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.

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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.

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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.

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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.