Strauss PDEs 2e: Section Exercise 4 Page 1 of 5

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1 Strauss PDEs 2e: Section Exercise 4 Page 1 of 5 Exercise 4 Lorentz invariance of the wave equation) Thinking of the coordinates of space-time as 4-vectors x, y, z, t), let Γ be the diagonal matrix with the diagonal entries 1, 1, 1, 1. Another matrix L is called a Lorentz transformation if L has an inverse and L 1 Γ t LΓ, where t L is the transpose. a) If L and M are Lorentz, show that LM and L 1 also are. b) Show that L is Lorentz if and only if mlv) mv) for all 4-vectors v x, y, z, t), where mv) x 2 + y 2 + z 2 t 2 is called the Lorentz metric. c) If ux, y, z, t) is any function and L is Lorentz, let Ux, y, z, t) ulx, y, z, t)). Show that U xx + U yy + U zz U tt u xx + u yy + u zz u tt. d) Explain the meaning of a Lorentz transformation in more geometrical terms. Hint: Consider the level sets of mv).) Solution Start by writing the matrix of Γ Γ Notice that Γ is its own inverse and its own transpose. Part a) Γ Γ 1 t Γ Suppose L and M are Lorentz. Then, by the given definition, L 1 Γ t LΓ and M 1 Γ t M Γ In order to show that LM is Lorentz, multiply M 1 and L 1 together. M 1 L 1 Γ t M Γ)Γ t LΓ) Use one of the basic properties of the inverse on the left side: LM) 1 M 1 L 1. Use the fact that matrix multiplication is associative on the right side. ΓΓ is the identity matrix. LM) 1 Γ t M ΓΓ) t LΓ LM) 1 Γ t M t LΓ Use one of the basic properties of the transpose: t LM) t M t L. Therefore, LM is a Lorentz transformation. LM) 1 Γ t LM)Γ

2 Strauss PDEs 2e: Section Exercise 4 Page 2 of 5 Now it will be shown that L 1 is Lorentz. L 1 Γ t LΓ Γ t L)Γ Take the transpose of both sides and apply its basic property twice. Note that t Γ Γ and t t L) L. Premultiply both sides by Γ. Postmultiply both sides by Γ. Since L L 1 ) 1, we have Therefore, L 1 is a Lorentz transformation. Part b) t L 1 ) t Γ t L)Γ t Γ t Γ t L) t Γ t t L) t Γ t L 1 ) ΓLΓ Γ t L 1 ) ΓΓLΓ) ΓΓ)LΓ LΓ Γ t L 1 )Γ LΓ)Γ LΓΓ) L L 1 ) 1 Γ t L 1 )Γ. Suppose that L is Lorentz and let x 1 x, x 2 y, x 3 z, and x 4 t so that series can be used. Then v x 1, x 2, x 3, x 4 ). The aim is to show that mv) mlv). mv) x 2 + y 2 + z 2 t 2 i1 j1 x i x j Because L is Lorentz, it s true that L 1 Γ t LΓ. Premultiply both sides by Γ and postmultiply both sides by L to obtain ΓL 1 L ΓΓ t LΓL, or Γ t L Γ L. t L Γ L) ij x i x j i1 j1

3 Strauss PDEs 2e: Section Exercise 4 Page 3 of 5 In parentheses is the product of three 4 4 matrices. To get the ij-component, i and j will be the leftmost and rightmost indices, and the indices of summation will be connected in between them. mv) t L) ik L lj x i x j i1 j1 k1 l1 The transpose has the effect of switching the indices. ) L ki L lj x i x j i1 j1 k1 l1 The limits of summation are constant, so the sums can be arranged however we like. ) L ki x i L lj x j k1 l1 i1 As L is a 4 4 matrix and v is a 4 1 matrix, the terms in parentheses are the k- and l-components of Lv. k1 l1 mlv) Lv) k Lv) l Now that it has been shown that mv) mlv), the first part of the proof is complete. For the second part, suppose that mv) mlv). The aim is to show that L is Lorentz. We have and working backwards) mlv) mv) x 2 + y 2 + z 2 t 2 k1 l1 k1 i1 j1 x i x j Lv) k Lv) l L ki x i ) j1 L lj x j l1 i1 j1 ) L ki L lj x i x j i1 j1 i1 j1 i1 j1 k1 l1 k1 l1 t L Γ L) ij x i x j. t L) ik L lj x i x j

4 Strauss PDEs 2e: Section Exercise 4 Page 4 of 5 Because mv) mlv), This implies that i1 j1 x i x j i1 j1 Γ t L Γ L. t L Γ L) ij x i x j. Premultiply both sides by Γ and postmultiply both sides by L 1. ΓΓL 1 Γ t L Γ LL 1 LL 1 and ΓΓ are equal to the identity matrix, so L 1 Γ t L Γ, which means L is Lorentz. Therefore, L is Lorentz if and only if mlv) mv) for all 4-vectors v x, y, z, t). Part c) Let x 1 x, x 2 y, x 3 z, and x 4 t so that series can be used. Then v x 1, x 2, x 3, x 4 ). u xx + u yy + u zz u tt i1 j1 x i x j ulv) Apply the chain rule to differentiate ulv). i1 j1 k1 l1 2 ulv) Lv) k Lv) l x i x j Replace ulv) with Uv) U and write the expression for the multiplication of 4 4 matrix L with 4 1 matrix v. 2 ) ) U L km x m L lm x m x i1 j1 k x l x i x k1 m1 j m1 l1 The derivative of a sum is the sum of the derivatives. i1 j1 k1 l1 2 U m1 L km x m x i ) m1 L lm x m x j Only if m is equal to i in the first sum will x m /x i be equal to 1. Similarly, only if m is equal to j in the second sum will x m /x j be equal to 1. The Kronecker delta symbol represents this. i1 j1 k1 l1 2 U m1 L km δ mi) m1 L lm δ mj ) )

5 Strauss PDEs 2e: Section Exercise 4 Page 5 of 5 δ mi sifts m i from the first summand in parentheses, and δ mj sifts m j from the second one. u xx + u yy + u zz u tt i1 j1 k1 l1 2 U L ki ) L lj ) The limits of summation are constant, so the sums can be arranged however we like. L ki L lj 2 U k1 l1 i1 j1 In order to connect the inner indices, use the transpose of L. L ki t L) jl 2 Uv) k1 l1 i1 j1 Because L is Lorentz, it s true that L 1 Γ t LΓ. Premultiply both sides by L and postmultiply both sides by Γ to obtain LL 1 Γ L Γ t LΓΓ, or Γ L Γ t L Consequently, the term in square brackets is the kl-component of the matrix product L Γ t L, which is equal to Γ. k1 l1 x k x l Uv) U xx + U yy + U zz U tt Therefore, the wave equation is invariant under Lorentz transformations. Part d) u xx + u yy + u zz u tt U xx + U yy + U zz U tt The Lorentz metric of a 4-vector mv) can be written as follows. mv) x 2 + y 2 + z 2 t 2 x, y, z, it) x, y, z, it) x, y, z, it) 2 We see from this form that mv) represents the magnitude of v x, y, z, t) in space-time. Unlike the magnitude of a regular 3-vector x x, y, z), which is x 2 x 2 + y 2 + z 2, it can be negative. The magnitude of x only depends on its distance from the origin. That is, it is invariant under rotations in space: Rx x for any rotation matrix R. By comparison, since mlv) mv), L can be thought of as an imaginary rotation in space-time. In the theory of special relativity, for example, it is a Lorentz transformation that relates the space-time coordinates of two observers moving relative to one another at constant velocity. The extent of rotation that occurs in space-time is related to the observers relative velocity.