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Relativistic Equation Questions

  1. Jun 17, 2012 #1

    CAF123

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    I understand that the relativistic eqn that applies to all particles in all frames of reference and that works for both massless and massive particles is E^2 = (p^2)(c^2) + (m^2)(c^4).
    I then attempted a small question:
    Deduce the de Broglie wavelength of thermal neutrons from a nuclear reactor which have k.E ~(3/2)kT.
    I have the correct answer and know how to solve the problem (which is why I did not post in homework forum), however, I found the answer by finding k.E by the above formula and then equating this to (1/2)mV^2 and solving for v. After finding v, I found p = mv and therefore the de Broglie wavelength.

    My question is: when I instead use the relativistic eqn at the top (which according to my notes works in all frames of references) I yield a negative under the square root. Why is this so? Why aren't things consistent?

    For reference, I used E = 6.1 x10^-21 J in both attempted methods.

    Many thanks
     
  2. jcsd
  3. Jun 17, 2012 #2
    It might be easier to find the non-relativistic p=mv this way
    [tex] pc=m_ov \cdot c=m_o \beta c^2 = 2\beta(\frac{1}{2}m_o c^2) = 2\beta \cdot KE [/tex]
    [added] The fully relativistic pc is
    [tex] pc=mv \cdot c=m\beta c^2 = \beta \gamma \cdot m_o c^2 [/tex]
     
    Last edited: Jun 17, 2012
  4. Jun 17, 2012 #3

    CAF123

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    What does beta stand for?
    I am still not sure, however, why the results would be inconsistent.
     
  5. Jun 17, 2012 #4
    beta is the ratio [itex] \beta = \frac{v}{c} [/itex] where v is the velocity. γ is the ratio of the total energy (kinetic plus rest mass) divided by the rest mass.
     
  6. Jun 17, 2012 #5

    CAF123

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    Thanks. Have you any ideas why both my methods are giving inconsistent results?
     
  7. Jun 17, 2012 #6
    [tex] (pc)^2=E^2-(m_oc^2)^2=(KE+m_oc^2)^2-(m_oc^2)^2 = KE^2+2KE \cdot m_oc^2 + (m_oc^2)^2 -(m_oc^2)^2[/tex]
    [tex] (pc)^2= KE^2+2KE \cdot m_oc^2 [/tex]
    At very low energies the first term can be dropped, yielding
    [tex] (pc)^2= 2KE \cdot m_oc^2 [/tex]
    so
    [tex] pc=\sqrt{2 \cdot KE \cdot m_oc^2} [/tex]
    All energies including pc have to be in same units.
    [tex] m_ov \cdot c=\sqrt{2 \cdot KE \cdot m_oc^2} [/tex]
    [tex] v=\frac {\sqrt{2 \cdot KE \cdot m_oc^2}}{m_oc} = \frac {c \cdot\sqrt{2 \cdot KE \cdot m_oc^2}}{m_oc^2} = \frac {c \cdot\sqrt{3 kT }}{\sqrt{m_oc^2}} [/tex]
    where [itex] KE=\frac{3}{2}kT [/itex]
    So v is c times a dimensionless ratio.

    [added] For neutrons, moc2 = 939 MeV = 1.50 x 10-10 Joules.
     
    Last edited: Jun 17, 2012
  8. Jun 18, 2012 #7

    CAF123

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    Ok, thanks a lot. I now have the correct answer using both methods.
     
  9. Jun 18, 2012 #8
    There is another way:
    [tex] (pc)^2=E^2-(m_oc^2)^2=\left(E+m_oc^2\right)\left(E- m_oc^2\right) [/tex]
    At non-relativistic energies E - moc2 = KE and E + moc2 = 2moc2 so this becomes
    [tex] (pc)^2=2\cdot m_oc^2\cdot KE [/tex] etc.
     
  10. Jun 19, 2012 #9

    CAF123

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    I've thought of one question: if we are dealing with non-relativistic scales, why does E = K.E + mc^2 and not simply E= K.E?
    Is this a necessary condition that makes the relativistic eqn applicable?
     
  11. Jun 19, 2012 #10
    In relativistic equations, E is always the sum of the kinetic plus rest mass energies, so the non-relativistic approximations (when KE<<moc2) are E - moc2 = KE, and E+moc2 = 2moc2.
     
  12. Jul 11, 2012 #11
    Calculate the relalitivistic mass-rest mass ratio, for an electron whose velocity is 20% of the velocity of light??? can anyone answer this for me? thanks
     
  13. Jul 11, 2012 #12

    CAF123

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    Just use the relativistic eqn which relates relativistic mass to rest mass.
    (the one with rest mass/sqrt(1-(v^2/c^2)) - sorry not proficient with latex.

    V = 0.2c.
    Substitute with rest mass = 9.11 10 ^-31 kg.
    Then calculate ratio: relativistic mass/rest mass.
     
  14. Jul 11, 2012 #13
    can you show me how to solve it? and what is the answer?
     
  15. Jul 12, 2012 #14

    CAF123

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    Are you familiar with the eqn I transcribed in the text?
    If so, just use that formula to find the relativistic mass of the electron.
    We know rest mass : ~ 9.11 x 10^-31kg.
    We know v: 0.2c , c is speed of light in vacuum (~ 3x 10^8 m/s)
    Substitute this into the formula, perhaps calculating the parts under the square root independently so as to not confuse your calculator.
    You should get a relativistic mass of ~ 9.30 x 10^-31 kg. This makes sense: it is travelling faster and so by relativity, there should be an increase in its mass and this is what is found.
    To find ratio, simply divide relativistic mass by rest mass. Should get about 1.02. Yes?
     
  16. Jul 12, 2012 #15
    thank you for helping me but... i really dont get it... hehehe..

    can you show me step by step solution of the problem? how did you get 1.02? is that the relalitivistic mass-rest mass ratio?
     
  17. Jul 12, 2012 #16

    jtbell

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    We don't do people's homework for them. Please re-post your question in one of the "Homework & Coursework Questions" forums ("Introductory Physics" is probably the most appropriate one), and show your work so far, even if you're pretty sure it's wrong. People there will try to identify specific errors and give you hints to proceed further.
     
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