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Down quarks force vs ten tonnes truck force

  1. Aug 13, 2012 #1
    I have calculated that the strong repulsive force between two down quarks is 25.3N.
    A book that I am reading said that the force of attraction between two quarks is 'equivalent to the weight of a ten tonne truck’, whose force is equals to 98000N. How can I compare the values of these two forces to explain the significance of the relative magnitude of the strong force. Many thanks.
     
  2. jcsd
  3. Aug 14, 2012 #2

    CAF123

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    I believe what you have calculated is the electrostatic repulsion, not the strong force attraction.
    Did you use [tex] F = k \frac {q_1 q_2}{r^2}, [/tex] where [itex] k = \frac {1}{4\piε_o} [/itex]?
     
  4. Aug 14, 2012 #3
    yes, I used that equation to calculate the force of repulsion but with –ke = 9.0 x 10^9 N m^2 C^-2. How can I compare the values of this repulsion force between the two down quarks with the gravitational force of a 10 tonnes tanck in order to explain the significance of the relative magnitude of the strong force?
     
  5. Aug 14, 2012 #4

    CAF123

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    That's right, [itex] \frac{1}{4πε_o} [/itex] has magnitude [itex] 9 [/itex] x [itex] 10^9 [/itex]. [itex] 25.3 N [/itex] is not the strong force attraction between two down quarks, it is the magnitude of electrostatic repulsion. This is why this number and 98000N are not very comparable.

    Direct calculation of the strong force is not too easy:
    See http://home.fnal.gov/~cheung/rtes/RTESWeb/LQCD_site/pages/calculatingtheforces.htm

    I think perhaps what the question wants you to recognise that at such small distances, of the order of [itex] fm [/itex], the nuclear scale, this force is incredibly huge. (in fact, if one tries to pull quarks apart, in order to give isolated quarks, you will find you can't. Instead when the potential energy is enough will result in a quark-antiquark pair)
     
  6. Aug 14, 2012 #5
    Thanks caf123, that's was very helpful!
     
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