- #1
nonequilibrium
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Griffiths says "unlike the first two, Newton's third law does not, in general, extend to the relativistic domain."
I don't understand, what is wrong with the following derivation?
So we take it as an experimental fact that momentum is conserved in isolated systems (for clarity, the momentum we're working with is the spatial part of the 4-vector momentum or momenergy).
Now imagine an isolated system consisting of two particles. Call [tex]\vec P[/tex] the momentum of the whole system, and [tex]\vec p_1[/tex] and [tex]\vec p_2[/tex] those of the respective particles in the system. By definition, [tex]\vec P = \vec p_1 + \vec p_2[/tex].
Differentiate both sides with respect to coordinate time, then by definition of ordinary force:
[tex]\frac{\mathrm d \vec P}{\mathrm d t} = \boxed{ \vec F_{\textrm{ext}}= \vec F_{12} + \vec F_{21}} = \frac{\mathrm d \vec p_1}{\mathrm d t} + \frac{\mathrm d \vec p_2}{\mathrm d t}[/tex].
Due to the experimental fact that [tex]\vec P = \textrm{constant} \Rightarrow \vec F_{\textrm{ext}} = 0[/tex] and thus as a result of the boxed equation: [tex]\vec F_{12} = - \vec F_{21} [/tex].
Did I make an error in the above derivation, or is it simply that Newton's 3rd law only fails for systems consisting of 3+ particles?
I don't understand, what is wrong with the following derivation?
So we take it as an experimental fact that momentum is conserved in isolated systems (for clarity, the momentum we're working with is the spatial part of the 4-vector momentum or momenergy).
Now imagine an isolated system consisting of two particles. Call [tex]\vec P[/tex] the momentum of the whole system, and [tex]\vec p_1[/tex] and [tex]\vec p_2[/tex] those of the respective particles in the system. By definition, [tex]\vec P = \vec p_1 + \vec p_2[/tex].
Differentiate both sides with respect to coordinate time, then by definition of ordinary force:
[tex]\frac{\mathrm d \vec P}{\mathrm d t} = \boxed{ \vec F_{\textrm{ext}}= \vec F_{12} + \vec F_{21}} = \frac{\mathrm d \vec p_1}{\mathrm d t} + \frac{\mathrm d \vec p_2}{\mathrm d t}[/tex].
Due to the experimental fact that [tex]\vec P = \textrm{constant} \Rightarrow \vec F_{\textrm{ext}} = 0[/tex] and thus as a result of the boxed equation: [tex]\vec F_{12} = - \vec F_{21} [/tex].
Did I make an error in the above derivation, or is it simply that Newton's 3rd law only fails for systems consisting of 3+ particles?