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== 5.8#h(0.6em)Relativistic Momentum


  Momentum is a central concept in physics. The broadest form of Newton’s second law is stated in terms of momentum. Momentum is conserved whenever the net external force on a system is zero. This makes momentum conservation a fundamental tool for analyzing collisions . Much of what we know about subatomic structure comes from the analysis of collisions of accelerator-produced relativistic particles, and momentum conservation plays a crucial role in this analysis.


  #figure(figph[A photo of a football player tackling an opponent.], alt: "A photo of a football player tackling an opponent.", caption: [Momentum is an important concept for these football players from the University of California at Berkeley and the University of California at Davis. A player with the same velocity but greater mass collides with greater impact because his momentum is greater. For objects moving at relativistic speeds, the effect is even greater.])


  The first postulate of relativity states that the laws of physics are the same in all inertial frames. Does the law of conservation of momentum survive this requirement at high velocities? It can be shown that the momentum calculated as merely #math.equation(block: false, alt: "p → equals m the fraction d x → over d t ,")[$arrow(p) = m frac(d arrow(x), d t) ,$] even if it is conserved in one frame of reference, may not be conserved in another after applying the Lorentz transformation to the velocities. The correct equation for momentum can be shown, instead, to be the classical expression in terms of the increment #math.equation(block: false, alt: "d τ")[$d τ$] of proper time of the particle, observed in the particle’s rest frame:


  #math.equation(block: true, alt: "p →, equals m the fraction d x → over d τ equals m the fraction d x → over d t the fraction d t over d τ; equals m the fraction d x → over d t the fraction 1 over the square root of 1 minus u squared / c squared; equals the fraction m u → over the square root of 1 minus u squared / c squared equals γ m u → .")[$arrow(p) & = m frac(d arrow(x), d τ) = m frac(d arrow(x), d t) #h(0.1em) frac(d t, d τ) \  & = m frac(d arrow(x), d t) #h(0.2em) frac(1, sqrt(1 − u^(2) "/" c^(2))) \  & = frac(m arrow(u), sqrt(1 − u^(2) "/" c^(2))) = γ m arrow(u) .$]
  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Relativistic Momentum]


    #strong[Relativistic momentum] #math.equation(block: false, alt: "p →")[$arrow(p)$] is classical momentum multiplied by the relativistic factor #emph[γ]:


    #math.equation(block: true, alt: "p → equals γ m u →")[$arrow(p) = γ m arrow(u)$]
    where #emph[m] is the #strong[rest mass] of the object, #math.equation(block: false, alt: "u →")[$arrow(u)$] is its velocity relative to an observer, and γ is the relativistic factor:


    #math.equation(block: true, alt: "γ equals the fraction 1 over the square root of 1 minus the fraction u squared over c squared .")[$γ = frac(1, sqrt(1 − frac(u^(2), c^(2)))) .$]
]


  Note that we use #emph[u] for velocity here to distinguish it from relative velocity #emph[v] between observers. The factor #math.equation(block: false, alt: "γ")[$γ$] that occurs here has the same form as the previous relativistic factor #math.equation(block: false, alt: "γ")[$γ$] except that it is now in terms of the velocity of the particle #emph[u] instead of the relative velocity #emph[v] of two frames of reference.


  With #emph[p] expressed in this way, total momentum #math.equation(block: false, alt: "p sub tot")[$p_("tot")$] is conserved whenever the net external force is zero, just as in classical physics. Again we see that the relativistic quantity becomes virtually the same as the classical quantity at low velocities, where #emph[u]/#emph[c] is small and #math.equation(block: false, alt: "γ")[$γ$] is very nearly equal to 1. Relativistic momentum has the same intuitive role as classical momentum. It is greatest for large masses moving at high velocities, but because of the factor #math.equation(block: false, alt: "γ ,")[$γ ,$] relativistic momentum approaches infinity as #emph[u] approaches #emph[c] . This is another indication that an object with mass cannot reach the speed of light. If it did, its momentum would become infinite—an unreasonable value.


  #figure(figph[This is a graph of the relativistic momentum as a function of speed. The function and its slope are zero at u=0, both increase with u, and the function has a vertical asymptote at u=1.0 c], alt: "This is a graph of the relativistic momentum as a function of speed. The function and its slope are zero at u=0, both increase with u, and the function has a vertical asymptote at u=1.0 c", caption: [Relativistic momentum approaches infinity as the velocity of an object approaches the speed of light.])


  The relativistically correct definition of momentum as #math.equation(block: false, alt: "p equals γ m u")[$p = γ m u$] is sometimes taken to imply that mass varies with velocity: #math.equation(block: false, alt: "m sub var equals γ m ,")[$m_("var") = γ m ,$] particularly in older textbooks. However, note that #emph[m] is the mass of the object as measured by a person at rest relative to the object. Thus, #emph[m] is defined to be the rest mass, which could be measured at rest, perhaps using gravity. When a mass is moving relative to an observer, the only way that its mass can be determined is through collisions or other means involving momentum. Because the mass of a moving object cannot be determined independently of momentum, the only meaningful mass is rest mass. Therefore, when we use the term “mass,” assume it to be identical to “rest mass.”


  Relativistic momentum is defined in such a way that conservation of momentum holds in all inertial frames. Whenever the net external force on a system is zero, relativistic momentum is conserved, just as is the case for classical momentum. This has been verified in numerous experiments.


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
What is the momentum of an electron traveling at a speed 0.985#emph[c]? The rest mass of the electron is #math.equation(block: false, alt: "9.11 times 10 to the power −31 kg .")[$9.11 #h(0.2em) × #h(0.2em) 10^(−31) "kg" .$]


    
    #solutionbox[
Substitute the data into the given equation: #linebreak() 
#math.equation(block: false, alt: "p equals γ m u equals the fraction m u over the square root of 1 minus the fraction u squared over c squared equals the fraction open parenthesis 9.11 times 10 to the power −31 kg close parenthesis open parenthesis 0.985 close parenthesis open parenthesis 3.00 times 10 to the power 8 m/s close parenthesis over the square root of 1 minus the fraction open parenthesis 0.985 c close parenthesis squared over c squared equals 1.56 times 10 to the power −21 kg-m/s.")[$p = γ m u = frac(m u, sqrt(1 − frac(u^(2), c^(2)))) = frac(( 9.11 #h(0.2em) × #h(0.2em) 10^(−31) #h(0.2em) "kg" ) ( 0.985 ) ( 3.00 #h(0.2em) × #h(0.2em) 10^(8) #h(0.2em) "m/s" ), sqrt(1 − frac(attach(( 0.985 c ), t: 2), c^(2)))) = 1.56 #h(0.2em) × #h(0.2em) 10^(−21) #h(0.2em) "kg-m/s."$]
]
]


  
    
=== Summary


    - The law of conservation of momentum is valid for relativistic momentum whenever the net external force is zero. The relativistic momentum is #math.equation(block: false, alt: "p equals γ m u ,")[$p = γ m u ,$] where #emph[m] is the rest mass of the object, #emph[u] is its velocity relative to an observer, and the relativistic factor is #math.equation(block: false, alt: "γ equals the fraction 1 over the square root of 1 minus the fraction u squared over c squared .")[$γ = frac(1, sqrt(1 − frac(u^(2), c^(2)))) .$]
- At low velocities, relativistic momentum is equivalent to classical momentum.
- Relativistic momentum approaches infinity as #emph[u] approaches #emph[c]. This implies that an object with mass cannot reach the speed of light.


  
  
    
=== Conceptual Questions


    
      How does modern relativity modify the law of conservation of momentum?


    
    
      Is it possible for an external force to be acting on a system and relativistic momentum to be conserved? Explain.


    
    #solutionbox[
Yes. This can happen if the external force is balanced by other externally applied forces, so that the net external force is zero.
]


  
  
    
=== Problems


    
      Find the momentum of a helium nucleus having a mass of #math.equation(block: false, alt: "6.68 times 10 to the power −27 kg")[$6.68 #h(0.2em) × #h(0.2em) 10^(−27) #h(0.2em) "kg"$] that is moving at 0.200#emph[c].


    
    #solutionbox[
#math.equation(block: true, alt: "4.09 times 10 to the power −19 kg times m/s")[$4.09 #h(0.2em) × #h(0.2em) 10^(−19) #h(0.2em) "kg" · "m/s"$]
]


    
      What is the momentum of an electron traveling at 0.980#emph[c]?


    
    
      (a) Find the momentum of a #math.equation(block: false, alt: "1.00 times 10 to the power 9 -kg")[$1.00 #h(0.2em) × #h(0.2em) 10^(9) "-kg"$] asteroid heading towards Earth at 30.0 km/s. (b) Find the ratio of this momentum to the classical momentum. (Hint: Use the approximation that #math.equation(block: false, alt: "γ equals 1 plus open parenthesis 1 / 2 close parenthesis v squared / c squared")[$γ = 1 + ( 1 "/" 2 ) v^(2) "/" c^(2)$] at low velocities.)


    
    #solutionbox[
a. #math.equation(block: false, alt: "3.000000015 times 10 to the power 13 kg times m/s;")[$3.000000015 #h(0.2em) × #h(0.2em) 10^(13) #h(0.2em) "kg" · "m/s;"$] b. 1.000000005
]


    
      (a) What is the momentum of a 2000-kg satellite orbiting at 4.00 km/s? (b) Find the ratio of this momentum to the classical momentum. (Hint: Use the approximation that #math.equation(block: false, alt: "γ equals 1 plus open parenthesis 1 / 2 close parenthesis v squared / c squared")[$γ = 1 + ( 1 "/" 2 ) v^(2) "/" c^(2)$] at low velocities.)


    
    
      What is the velocity of an electron that has a momentum of #math.equation(block: false, alt: "3.04 times 10 to the power −21 kg times m/s")[$3.04 #h(0.2em) × #h(0.2em) 10^(−21) #h(0.2em) "kg" · "m/s"$]
? Note that you must calculate the velocity to at least four digits to see the difference from #emph[c].


    
    #solutionbox[
#math.equation(block: true, alt: "2.988 times 10 to the power 8 m/s")[$2.988 #h(0.2em) × #h(0.2em) 10^(8) #h(0.2em) "m/s"$]
]


    
      Find the velocity of a proton that has a momentum of #math.equation(block: false, alt: "4.48 times 10 to the power −19 kg times m/s .")[$4.48 #h(0.2em) × #h(0.2em) 10^(−19) #h(0.2em) "kg" · "m/s" "."$]