Question 26.4: Urgent Course Correction Required! GOAL Apply the concept of...
Urgent Course Correction Required!
GOAL Apply the concept of relative velocity in relativity.
PROBLEM Suppose that Alice’s spacecraft is traveling at 0.600c in the positive x-direction, as measured by a nearby Earth-based observer at rest, while Bob is traveling in his own vehicle directly toward Alice in the negative x-direction at velocity -0.800c relative the same Earth observer. What’s the velocity of Alice according to Bob?
STRATEGY Alice’s spacecraft is the object of interest that both the Earth observer and Bob are tracking. We’re given the Earth observer’s velocity measurements and wish to find Bob’s measurement. Use the relative velocity equation for relativity, Equation 26.7,
v_{\mathrm{AB}}={\frac{v_{\mathrm{AE}}~-~v_{\mathrm{BE}}}{1\,~-~\,{\frac{v_{\mathrm{AE}}v_{\mathrm{BE}}}{c^{2}}}}} [26.7]
with v_{\mathrm{AB}} corresponding to the measurement of Alice’s velocity made by Bob and v_{\mathrm{AE}} is the measurement of Alice’s velocity according to the Earth observer. Notice that the velocity of Bob’s frame is in the negative x-direction, so v_{\mathrm{BE}} < 0.
Write Equation 26.7:
v_{\mathrm{AB}}={\frac{v_{\mathrm{AE}}~-~v_{\mathrm{BE}}}{1\,~-~\,{\frac{v_{\mathrm{AE}}v_{\mathrm{BE}}}{c^{2}}}}}
Substitute values:
v_{\mathrm{AB}}={\frac{0.600c~-~(-0.800c)}{1~-~{\frac{(0.600c)(-0.800c)}{c^{2}}}}}={\frac{1.400c}{1~-~(-0.480)}}= 0.946c
REMARKS Notice that care was taken to use the correct signs. Common sense might lead us to believe that Bob would measure Alice’s velocity as 1.40c, but as the calculation shows, Bob measures Alice’s velocity as less than that of light.