Relativistic Equation Questions

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  • #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
 

Answers and Replies

  • #2
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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]
 
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  • #3
CAF123
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What does beta stand for?
I am still not sure, however, why the results would be inconsistent.
 
  • #4
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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.
 
  • #5
CAF123
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Thanks. Have you any ideas why both my methods are giving inconsistent results?
 
  • #6
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[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.
 
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  • #7
CAF123
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Ok, thanks a lot. I now have the correct answer using both methods.
 
  • #8
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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.
 
  • #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?
 
  • #10
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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?
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.
 
  • #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
 
  • #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.
 
  • #13
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.

can you show me how to solve it? and what is the answer?
 
  • #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?
 
  • #15
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?

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?
 
  • #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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