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In a traingle the proper length of each side equals a. Find the perimeter of this triangle in the reference frame moving relative to it with a constant velocity V along one of its (a) bisectors, (b) sides. Investigate the results obtained at V lt lt c and Vrarr c, where c is the velocity of light. |
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Answer» SOLUTION :(a) In the FRAME in which the traingle is at rest the space COORDINATES of the vertices are `(000)`, `(asqrt3/2, +a/2, 0)(asqrt3/2, -a/2, 0)`, all MEASURED at the same time t. In the moving frame the CORRESPONDING coordinates at time `t^'` are `A: (vt^', 0, 0), B: (a/2sqrt3sqrt(1-beta^2)+vt^', a/2, 0)` and `C: (a/2sqrt3sqrt(1-beta^2)+vt^', -a/2,0)` The perimeter P is then `P=a+2a(3/4(1-beta^2)+1/4)^(1//2)=a(1+sqrt(4-3beta^2))` (b) The coordinates in the first frame are shown at time t. The coordinates in the moving frame are,  `A: (vt^', 0, 0), B: (a/2sqrt(1-beta^2)+vt^', asqrt3/2, 0), C: (asqrt(1-beta^2)+vt^', 0, 0)` The perimeter P is then `P=asqrt(1-beta^2)+a/2[1-beta^2+3]^(1//2)xx2=a(sqrt(1-beta^2)+sqrt(4-beta^2))` here `beta=V/c` |
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