Consider a meterstick at rest in a given inertial frame ( make this the Other Fr

Question

Consider a meterstick at rest in a given inertial frame ( make this the Other Frame) oriented in such a way that it makes an angle ?1 with respect to the x1 direction in that frame. In the Home Frame, the Other Frame is observed to move with a velocity of ? in the +x direction.

(a)  Keeping in mind that the distances measured parallel to the line of relative motion are observed to be Lorentz contracted in the Home Frame while distances measured perpendicular to the line of motion are not, show that the angle ? that this meter stick will be observed to make with the x direction in the Home Frame is given by:

arctan((tan?1)/sqrt(1-?^2))

(b) What would the length of the meter stick be as measured in the Home Frame?

(c) Assume that the meterstick makes an angle of 30 degrees with the x1 direction in the Other Frame. How fast would that frame have to be moving with respect to the Home Frame for the meterstick to be observed in the Home Frame to make an angle of 45 degrees with the x direction?

Explanation / Answer

a)

x2 = x1/gamma = sqrt(1 - (B)^2) x1

y2 = y1

theta2 = arctan(y2/x2)

theta2 = arctan(y2/(sqrt(1 - (B)^2) x1))

theta2 = arctan((y2/x1)/sqrt(1 - (B)^2)))

theta2 = arctan((tan(theta1))/sqrt(1 - (B)^2))

b)

L = sqrt(x2^2 + y2^2)

L = sqrt((sqrt(1-(B)^2) x1)^2 + y1^2)

L = sqrt((1-(B)^2) x1^2 + y1^2)

L = sqrt(-B^2 x1^2 + x1^2 + y1^2)

x1^2 + y1^2 = 1

==> L = sqrt(-B^2 x1^2 + 1)

x1 = 1 * cos(theta1) = cos(theta1)

==> L = sqrt(1 - B^2 (cos(theta1))^2)

c)

theta2 = arctan((tan(theta1))/sqrt(1 - (B)^2))

45 = arctan((tan(30))/sqrt(1 - (B)^2))

tan(45) = (tan(30))/sqrt(1 - (B)^2)

sqrt(1 - (B)^2) = 0.5774

==> B = 0.816 c = 2.45 x 10^8 m/s

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