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Chapter 26 Learning Objectives

*Note: Type-setting limitations do not allow for arrows on the vector labels, so we have used boldface only.

Concepts and Skills to Review

  • Inertia (Section 2.3)
  • Relative velocity (Section 4.7)
  • Kinetic energy (Section 6.3)
  • General law of energy conservation (Section 6.7)
  • Conservation of momentum; collisions in one dimension (Sections 7.4 and 7.7)

Summary

  • The two postulates of relativity are:
         (I) The laws of physics are the same in all inertial frames.
         (II) The speed of light in vacuum is the same in all inertial frames.
  • The speed of light in vacuum in any inertial reference frame is
     c = 3.00 × 108 m/s 
  • Observers in different reference frames disagree about the time order of two events (including whether the events are simultaneous) if there is not enough time for a signal at light speed to travel from one event to the other.
  • The Lorentz factor occurs in many relativity equations.
     <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif:: ::/sites/dl/free/0070524076/58009/image26_2.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif"> (1.0K)</a> (26-2)
  • In time dilation problems, identify the two events that mark the beginning and end of the time interval in question. The "clock" is whatever measures this time interval. Identify the reference frame in which the clock is at rest. In that frame, the clock measures the proper time interval Δt0. In any other reference frame, the time interval is longer:
     Δt = γΔt0(26-3)
  • In length contraction problems, identify the object whose length is to be measured in two different frames. The length is contracted only in the direction of the object's motion. If the length in question is a distance rather than the length of an actual object, it often helps to imagine the presence of a long measuring stick. Identify the reference frame in which the object is at rest. The length in that frame is the proper length L0. In any other reference frame, the length L is contracted:
     <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif:: ::/sites/dl/free/0070524076/58009/image26_4.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif"> (0.0K)</a> (26-4)
  • Velocities in different reference frames are related by:
     <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif:: ::/sites/dl/free/0070524076/58009/image26_5.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif"> (1.0K)</a> (26-5)
    The subscripts in vBA mean the velocity of B as measured in A's reference frame. Eq. 26-5 is written in terms of the components of the three velocities along a straight line. The components are positive for one direction (your choice) and negative for the other. If A moves to the right in B's frame, then B moves to the left in A's frame:
    vBA = -vAB.
  • The relativistic expression for momentum is
     p = γmv(26-6)
    With relativistic momentum, it is still true that the impulse delivered equals the change in momentum (FnetΔt = Δp), but Fnet = ma is not true: the acceleration due to a constant net force gets smaller and smaller as the particle's speed approaches c. Thus it is impossible to accelerate something to the speed of light.
  • The rest energy E0 of a particle is its energy as measured in its rest frame. The relationship between rest energy and mass is:
     E0 = mc2(26-7)
    Kinetic energy is
     K = (γ - 1)mc2(26-8)
    Total energy is rest energy plus kinetic energy:
     E = γmc2= K + E0(26-9)
  • Useful relations between momentum and energy:
     <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif:: ::/sites/dl/free/0070524076/58009/image26_10.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif"> (0.0K)</a> (26-10)
     (pc)2 = K2 + 2KE0(26-11)
     <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif:: ::/sites/dl/free/0070524076/58009/image26_12.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif"> (0.0K)</a> (26-12)







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