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I am trying to derive relativistic momentum in a manner similar to that of Max Planck in "The Principle of Relativity and the Fundamental Equations of Mechanics". I begin with a frame of reference $S$ with coordinates $x$, $y$, $z$, and $t$. In that frame of reference a charged particle of charge $\epsilon$ travels along the $x$-axis with a velocity $v$. There is another frame $S'$ with coordinates $x'$, $y'$, $z'$ and $t'$, which is moving at a constant velocity $v$ along the axis, that is, it initially moves together with the charged particle. A purely electric field is present in S' ($\vec{E}=[E_x, E_y ,E_z]$), exerting a force on the charge. Since in the frame $S'$ the charge is not moving, the equations of motion reduce to Newton's classical equations of motion: $$m\dfrac{d^2x'}{dt'^2}=\epsilon E_x'$$ $$m\dfrac{d^2y'}{dt'^2}=\epsilon E_y'$$ $$m\dfrac{d^2z'}{dt'^2}=\epsilon E_z'$$ Transforming these equations of motion into $S$ coordinates, using the transformation of the electromagnetic field (in Gaussian Units) and the Lorentz transformations: $$m\gamma^3 \dfrac{d^2x}{dt^2}=\epsilon E_x$$ $$m\gamma^2 \dfrac{d^2y}{dt^2}=\epsilon \gamma\left(E_y-\frac{v}{c}B_z\right)$$ $$m\gamma^2 \dfrac{d^2z}{dt^2}=\epsilon \gamma\left(E_z+\frac{v}{c}B_y\right)$$ Where $$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ And a magnetic field is present in S such that $\vec{B}=[B_x, B_y, B_z]$. If we make it so that the force present of the $S$ frame will be the Lorentz force due to the Electromagnetic Field is equivalent to the electric force in the $S'$ frame, but not necessarily equal. The components of the Lorentz Force are: $$F_x=\epsilon E_x+\frac{\epsilon}{c}(v_yB_z-v_zB_y)$$ $$F_y=\epsilon E_y+\frac{\epsilon}{c}(v_zB_x-v_xB_z)$$ $$F_z=\epsilon E_z+\frac{\epsilon}{c}(v_xB_y-v_yB_x)$$ Since the field in $S'$ is purely electric, electromagnetic field in $S$ is given by $$E_x=E_x'$$ $$E_y=\gamma E_y'$$ $$E_z=\gamma E_z'$$ $$B_x=0$$ $$B_y=-\gamma \frac{v}{c}E_z'$$ $$B_z=\gamma \frac{v}{c}E_y'$$ With these and the fact that in the $S$ frame the charge has no $y$ or $z$ conponents of velocity, we get that the Lorentz Force equations in $S$ are: $$F_x=\epsilon E_x$$ $$F_y=\epsilon \left(E_y-\frac{v}{c}B_z\right)$$ $$F_z=\epsilon \left(E_z+\frac{v}{c}B_y\right)$$ With these, the original equations of motion can be transformerd into $$m\gamma^3 \dfrac{d^2x}{dt^2}=F_x$$ $$m\gamma \dfrac{d^2y}{dt^2}=F_y$$ $$m\gamma \dfrac{d^2z}{dt^2}=F_z$$ From the $y$ and $z$ components I get the usual 'Relativistic Mass' formula $$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$ But I get a different 'mass' in the $x$-direction, one of $$m=\frac{m_0}{\left(\sqrt{1-\frac{v^2}{c^2}}\right)^3}$$ similar to what Einstein got in his original 1905 paper on Special Relativity using a different definition of force. Where is it that I went wrong?

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  • $\begingroup$ You can't use the non-relativistic Newton's Second Law in relativistic situations: en.wikipedia.org/wiki/Relativistic_mechanics#Force $\endgroup$ – probably_someone Jul 14 '17 at 17:12
  • $\begingroup$ But in the $S'$ frame of reference the relativistic laws of motion reduce to Newton's laws since the particle is not moving relative to it. Einstein did it, Planck did it, and Planck arrived at the correct answer. $\endgroup$ – user140323 Jul 14 '17 at 17:20
  • $\begingroup$ You should get a different factor in the x-direction. One of the reasons calling $\gamma m$ the "relativistic mass" is unhelpful. If you must talk about changing mass (I'm one of the ones who recommend against it), then $\gamma m$ is the "transverse mass" and $\gamma^3 m$ is the "longitudinal mass". $\endgroup$ – dmckee Jul 14 '17 at 17:31
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As the link I provided in the comments explains, the force on an object has different proportionalities to mass and acceleration in the longitudinal and transverse directions. Specifically,

$$\mathbf{F} = \gamma^3 m_0\mathbf{a_{||}}+\gamma m_0\mathbf{a_{\perp}}$$

which is derived from $\mathbf{F} = \frac{d\mathbf{p}}{dt}$ by letting $\mathbf{p}=\gamma m_0\mathbf{v}$ and taking the product rule when differentiating $\gamma\mathbf{v}$.

As is plain to see, you obtained the correct relations, so there is no problem.

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  • $\begingroup$ Oh, dear, I did not take the time to actually read through it. I apologize, and thank you. $\endgroup$ – user140323 Jul 14 '17 at 17:55

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