Sunday, January 30, 2011

Mastering Physics: ± Understanding Lorentz Transformations

Learning Goal: To be able to perform Lorentz transformations between inertial reference frames.
Suppose that an inertial reference frame S' moves in the positive x direction at speed u with respect to another inertial reference frame S. In classical physics, the Galilean transformations relate the coordinates measured for an event in frame S to the coordinates measured for the same event in frame S'. Assuming that both frames have the same origin (i.e., at t=t'=0, x=x'=0), the Galilean transformations take the following simple form:
t'=t, \quad x'=x-ut,
y'=y, \quad z'=z.
The Galilean transformations are not valid at very large speeds. To transform between inertial frames when u is close to the speed of light c, we need to use the Lorentz transformations of special relativity. Again, assuming that both frames have the same origin, the Lorentz transformations take the following form:
t'=\frac{t-\frac{u}{c^2}x}{\sqrt{1-\frac{u^2}{c^2}}}, \quad x'=\frac{x-ut}{\sqrt{1-\frac{u^2}{c^2}}}
y'=y, \quad z'=z.
These equations become more manageable with the introduction of the quantity
\gamma = \frac{1}{\sqrt{1-\frac{u^2}{c^2}}}=(1-u^2/c^2)^{-1/2},
so that the Lorentz transformations become
t'=\gamma\left(t-\frac{u}{c^2}x\right), \quad x'=\gamma(x-ut),
y'=y, \quad z'=z.
Often, the space-time coordinates for an event will be given in the form (t,x,y,z), or just (t,x) when the y and z coordinates are not important.

Part A
Consider an event with space-time coordinates (t=2.00\;{\rm s},\; x=2.50\times 10^{8}\; {\rm m}) in an inertial frame of reference S. Let S' be a second inertial frame of reference moving, in the positive x direction, with speed 2.70\times10^{8}\;{\rm m/s} relative to frame S. Find the value of gamma that will be needed to transform coordinates between frames S and S'. Use c = 3 \times 10^8\; {\rm m/s} for the speed of light in vacuum.
Express your answer to three significant figures.

  gamma  = 2.294

Part B
Suppose that S and S' share the same origin; that is, at t=t'=0, x=x'=0. Using the gamma you calculated in Part A, find x', the x coordinate of the event in frame S'.
Express your answer in meters to three significant figures.

  x'  = −6.650×108
  \rm m
Part C
Now find t', the t coordinate of the event in frame S'.
Express your answer in seconds to three significant figures.

  t'  = 2.87
  \rm s
Suppose that you are stationary with respect to an inertial reference frame Z. A spaceship flies by you in the positive x direction with speed v. Let Z' be the frame of reference associated with the spaceship; that is, the ship is stationary with respect to Z'. The frames Z and Z' have the same origin at t=t'=0. The proper length of the ship (the length of the ship as measured in the ship's frame, Z') is l_0. In other words, a passenger on the ship measures the back of the ship to be at x'=0 and the front to be at x'=l_0.
Part D
Find the gamma factor that should be used to transform between frames Z and Z'.
Express your answer in terms of v and c.

  gamma  = \frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

Part E
At time t=0, in your frame of reference Z, you measure the back of the spaceship to be at x=0 and the front of the ship to be at x=l. Find an equation relating the length that you measure l to the ship's proper length l_0.

Express your answer in terms of l_0 and gamma.

  l  = \frac{l_{0}}{{\gamma}}

You should recognize this as the equation for length contraction. The time dilation equation can also be found from the Lorentz transformations.

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