By Waner S.

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**Example text**

2 for several paths. ) 1 2 3 4 1 Path C: x = ct; x = x = 0, x = t (a photon traveling along the x -axis in E4). 2 (a) applies. Let D be the change-of-basis i matrix to the (other) inertial frame x– ; Dki = ∂x–i , ∂xk so that T—i = DkiTk D121 D = 31 D 41 D1 D 12 D 22 D 32 D 42 D 13 D 23 D 33 D 43 D 14 D 24 D 34 D 44 c 0 . 2, 1 2 1 2 1 2 2 1 2 (T— ) + (T— ) + (T— ) - c (T— ) = 0, so that (cD11 + D14)2 + (cD12 + D24)2 + (cD13 + D34)2 - c2(cD14 + D44)2 = 0 … (*) If we reverse the direction of the photon, we similarly get (-cD11 + D14)2 + (-cD12 + D24)2 + (-cD13 + D34)2 - c2(-cD14 + D44)2 = 0 …(**) Noting that this only effects cross-terms, subtracting and dividing by 4c gives D11D14 + D12D24 + D13D34 - c2D14D44 = 0; that is, “column 1, column 4‘ = 0 = “e1, e4‘.

Thus, very few frames are going to be Lorentz. Physical Interpretation of a Lorentz Frame What the definition means physically is that an observer in the x–-frame who measures a particle traveling at light speed in the x-frame will also reach the conclusion that its speed is c, because he makes the decision based on (I), which is equivalent to (II). In other words: A Lorentz frame in locally Minkowskian space is any frame in which light appears to be traveling at light speed, and where we measure length in the usual way.

Show that Dij, is a type (0, 2) tensor. ] ∂2xj– 9. What is wrong with the following “proof” that ∂xh∂xk = 0 regardless of what smooth functions x–j(xh) we use: ∂2xj– ∂ ∂x–j = Definition of the second derivative ∂xh ∂xk ∂xh ∂xk ∂ ∂x–j ∂x–l = l k h Chain rule ∂x– ∂x ∂x ∂2xj– ∂x–l = l k h Definition of the second derivative ∂x– ∂x ∂x ∂2xj– ∂x–l Changing the order of differentiation = k l h ∂x ∂x– ∂x ∂ ∂x–j ∂x–l Definition of the second derivative = k l h ∂x ∂x– ∂x ∂ ∂x–l ∂x–j = k ©li h Since l = ©li ∂x ∂x ∂x– =0 Since ©li is constant!