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It is experimentally known that the equation of motion for a charge $e$ moving in a static electric field $\mathbf{E}$ is given by: $$\frac{\mathrm{d}}{\mathrm{d}t} (\gamma m\mathbf{v}) = e\mathbf{E}$$ Is it possible to show this using just Newton's laws of motion for the proper frame of $e$, symmetry arguments, the Lorentz transformations and other additional principles? |
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It pretty much is that simple. However, if for some reason you'd like an explicitly relativistic formulation, take a look at the Lorentz force law: $$\frac{{\mathrm d}p^\mu}{{\mathrm d}\tau} = ev_\nu F^{\mu\nu}$$ For the derivative with respect to coordinate time that you want, we need to multiply through by $dτ/dt = 1/\gamma$. But for a constant electric field in Cartesian coordinates, the only nonzero components of $F^{\mu\nu}$ are $F^{0a} = -F^{a0}$ for a = 1,2,3, which are the electric field components. Thus, only the $v_0 = \gamma$ term can contribute, canceling factor brought in by time dilation factor. QED. |
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I think this is a lot simpler than you suspect. It's really just Newton's 2nd law, and recognising the concept of momentum. $$\frac{d}{dt} (\gamma mv) = \frac{dp}{dt} = F = eE$$ (Since classically the electric field $E$ is defined as the force $F$ divided by the elementary charge $e$.) |
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The "firstest" principle I know of is the principle of least action. So let us start with it. The action for a charged particle in external electromagnetic field is: $$S=\int_a^b\left(-mc\,ds-\frac{e}{c}A_\mu dx^\mu\right)$$ The integration have to be along the worldline of a particle with endpoints $a$ and $b$. Now, substituting: $$ds=\sqrt{c^2-v^2}dt,\quad A_\mu=(\phi,0,0,0),\quad dx^0=c\,dt$$ $$S=\int_{t_a}^{t_b}\left(-mc\sqrt{c^2-v^2}-e\phi\right)dt$$ The expression in curly brackets is just a Lagrangian, so you just do ordinary Euler-Lagrange stuff. $$L = -mc\sqrt{c^2-v^2}-e\phi,\quad \frac{\partial L}{\partial \mathbf{v}}= \gamma m \mathbf{v}\,\quad \frac{\partial L}{\partial \mathbf{r}}=e\mathbf{E}$$ $$\frac{d}{d t}\frac{\partial L}{\partial \mathbf{v}} = \frac{\partial L}{\partial \mathbf{r}} \quad \Rightarrow \quad \boxed{\frac{d}{dt}(\gamma m \mathbf{v}) = e\mathbf{E}} $$ |
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This is obtained from the equations of motion in either Lagrangian or Hamiltonian form. For instance the Hamiltonian equation of motion are $$\frac{\mathrm{d}}{\mathrm{d}t} \mathbf{r} = \frac{\partial H}{\partial \mathbf{p}}$$ $$\frac{\mathrm{d}}{\mathrm{d}t} \mathbf{p} = - \frac{\partial H}{\partial \mathbf{r}}$$ The Hamiltonian for a massive relativistic particle with charge $e$ in an external scalar potential $\phi$ is $$H = \sqrt{m^2c^4 + p^2c^2} + e \phi$$ Substituting the Hamiltonian in the equations of motion we obtain $$ \frac{\mathrm{d}}{\mathrm{d}t} \mathbf{r} = \mathbf{v} = \frac{\mathbf{p}}{m \gamma}$$ $$\frac{\mathrm{d}}{\mathrm{d}t} \mathbf{p} = - e \left( \frac{\partial \phi}{\partial \mathbf{r}} \right) $$ Substituting the first into the second and denoting $\mathbf{E} \equiv -{\partial \phi}/{\partial \mathbf{r}}$ $$\frac{\mathrm{d}}{\mathrm{d}t} (m \gamma \mathbf{v}) = e \mathbf{E}$$ |
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No it is not possible to derive the full relativistic law, because in the newtonian limit electrostatics and gravitostatics are identical, but the electrostatic potential is completely different from the gravitational potential in relativity. The easiest generalization of Newton's nonrelativistic law: $${dp\over dt} = e \nabla(\phi)$$ is to have a four dimensional scalar Higgs-type field $\phi$, then the relativistic equation of motion for a particle with a coupling e to the Higgs is $$ {d\over d\tau}( (m+e\phi(x)) p_\mu) = e \partial_\mu \phi $$ In the limit of nearly constant $\phi$ (or small e) and small velocities, this reproduces the inverse square law, and the force is the gradient of the Higgs. To get EM, you have to assume that $\phi$ is a zero component of a four-vector. This is the standard case, and I won't talk about it. The other noteworthy case is when $\phi$ is the zero-zero component of a tensor. Then the full special relativistic equation of motion is the linearized geodesic equation $${d \over d\tau} v_{\mu} = (\partial_\mu h_{\nu\sigma} - {1\over 2}\partial_\mu h_{\mu\nu} ) v^{\mu}v^{\nu}$$ And this is yet another generalization, linearized General Relativity, that reproduces the inverse square behavior for slow velocities, from the gradient of $h_{00}$. You can rule out possibilities 1 and 3 by the observation that there is repulsion between like charges, but this requires Maxwell's equations, and you asked to do it from the force law only. |
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If I understand your question correctly, you can show it by the next way. First of all, use some expressions from Special relativity, which are Lorentz transformations for force $\mathbf F $, radius-vector $\mathbf r$ and speed $\mathbf v$ ($\mathbf u$ is the speed of inertial system): $$ \mathbf r' = \mathbf r + \Gamma \mathbf u \frac{(\mathbf u \cdot \mathbf r)}{c^{2}} - \gamma \mathbf u t = \mathbf r + \Gamma \mathbf u \frac{(\mathbf u \cdot \mathbf r)}{c^{2}} \quad(t = 0) \quad \Rightarrow r'^{2} = r^{2} + \frac{(\mathbf u \cdot \mathbf r)^{2}\gamma^{2}}{c^{2}}, $$ $$ (\mathbf u \cdot \mathbf r') = \gamma (\mathbf u \cdot \mathbf r), \quad (\mathbf v' \cdot \mathbf r') = \frac{(\mathbf r \cdot \mathbf v)}{\gamma (1 - \frac{(\mathbf v \cdot \mathbf u)}{c^{2}})} - \gamma (\mathbf r \cdot \mathbf u), $$ $$ \frac{\mathbf F}{\gamma(1 - \frac{(\mathbf v \cdot \mathbf u)}{c^{2}})} = \mathbf F' + \gamma \frac{\mathbf u (\mathbf F' \cdot \mathbf v')}{c^{2}} + \Gamma \mathbf u \frac{(\mathbf u \cdot \mathbf F')}{c^{2}} \qquad (.1). $$ Secondly, use (.1) for Coulomb's law. You can make it, because it doesn't have an information about speed of interaction, which can be proved by thought experiment with two charges, binded by the stiff spring at the rest state. So, $$ \frac{\mathbf F}{\gamma(1 - \frac{(\mathbf v \cdot \mathbf u)}{c^{2}})} = \frac{Qq}{r'^{3}}\!{\left[\mathbf r' + \gamma \mathbf u \frac{(\mathbf r' \cdot \mathbf v')}{c^{2}} + \Gamma \mathbf u \frac{(\mathbf u \cdot \mathbf r')}{c^{2}}\right]} = $$ $$ = \frac{Qq}{r'^{3}}\!{\left[\mathbf r + \Gamma \mathbf u \frac{(\mathbf u \cdot \mathbf r)}{c^{2}} + \gamma \frac{\mathbf u}{c^{2}} \left(\frac{(\mathbf u \cdot \mathbf r)}{\gamma(1 - \frac{(\mathbf u \cdot \mathbf v)}{c^{2}})} - \gamma (\mathbf u \cdot \mathbf r)\right) + \Gamma \gamma \mathbf u \frac{(\mathbf u \cdot \mathbf r)}{c^{2}}\right]} = $$ $$ = \frac{Qq}{r'^{3}}\!{\left[\mathbf r + \Gamma \mathbf u \frac{(\mathbf u \cdot \mathbf r)}{c^{2}}(1 + \gamma) - \frac{\mathbf u}{c^{2}}\gamma^{2}(\mathbf u \cdot \mathbf r) + \gamma \frac{\mathbf u}{c^{2}}\frac{(\mathbf v \cdot \mathbf r)}{\gamma(1 - \frac{(\mathbf u \cdot \mathbf v)}{c^{2}})}\right]} = | \Gamma (1 + \gamma) = \gamma^{2} | = $$ $$ = \frac{Qq}{r'^{3}}\!{\left[\mathbf r + \gamma^{2} \frac{\mathbf u}{c^{2}}(\mathbf u \cdot \mathbf r) - \frac{\mathbf u}{c^{2}}\gamma^{2}(\mathbf u \cdot \mathbf r)) + \gamma \frac{\mathbf u}{c^{2}}\frac{(\mathbf v \cdot \mathbf r)}{\gamma(1 - \frac{(\mathbf u \cdot \mathbf v)}{c^{2}})}\right]} = $$ $$ = \frac{Qq}{r'^{3}}\!{\left[\mathbf r + \frac{\mathbf u (\mathbf v \cdot \mathbf r)}{c^{2}(1 - \frac{(\mathbf u \cdot \mathbf v)}{c^{2}})}\right]} = |\mathbf u (\mathbf v \cdot \mathbf r) = \left[ \mathbf v [\mathbf u \times \mathbf r ] \right] + \mathbf r (\mathbf u \cdot \mathbf v)| = $$ $$ = \frac{Qq}{r'^{3}}\left[ \mathbf r + \frac{[\mathbf v \times \frac{[\mathbf u \times \mathbf r]}{c^{2}}]}{1 - \frac{(\mathbf v \cdot \mathbf u)}{c^{2}}} + \frac{\mathbf r \frac{(\mathbf u \cdot \mathbf v)}{c^{2}}}{1 - \frac{(\mathbf v \cdot \mathbf u)}{c^{2}}}\right] = $$ $$ = \frac{Qq}{r'^{3}}\left[ \frac{[\mathbf v \times \frac{[\mathbf u \times \mathbf r]}{c^{2}}]}{1 - \frac{(\mathbf v \cdot \mathbf u)}{c^{2}}} + \frac{\mathbf r \left(\frac{(\mathbf v \cdot \mathbf u)}{c^{2}} + 1 - \frac{(\mathbf v \cdot \mathbf u)}{c^{2}}\right)}{1 - \frac{(\mathbf v \cdot \mathbf u)}{c^{2}}}\right] = \frac{Qq}{r'^{3}}\left[ \frac{[\mathbf v \times [\mathbf u \times \mathbf r]]}{c^{2}\left(1 - \frac{(\mathbf v \cdot \mathbf u)}{c^{2}}\right)} + \frac{\mathbf r}{1 - \frac{(\mathbf v \cdot \mathbf u)}{c^{2}}}\right]. $$ After that, using $\mathbf r'^{3} = \left(r'^{2}\right)^{\frac{3}{2}} = (r^{2} + \gamma^{2}\frac{(\mathbf r \cdot \mathbf u)^{2}}{c^{2}})^{\frac{3}{2}}$, we can assume, that $$ \frac{\mathbf F}{\gamma(1 - \frac{(\mathbf v \cdot \mathbf u)}{c^{2}})} = \frac{qQ}{(r^{2} + \gamma^{2}\frac{(\mathbf r \cdot \mathbf u)^{2}}{c^{2}})^{\frac{3}{2}}}\!{\left[\frac{[\mathbf v \times \frac{[\mathbf u \times \mathbf r]}{c^{2}}]}{1 - \frac{(\mathbf v \cdot \mathbf u)}{c^{2}}} + \frac{\mathbf r}{1 - \frac{(\mathbf v \cdot \mathbf u)}{c^{2}}}\right]} \Rightarrow $$ $$ \Rightarrow \mathbf F = \frac{q Q \gamma}{(r^{2} + \gamma^{2}\frac{(\mathbf r \cdot \mathbf u)^{2}}{c^{2}})^{\frac{3}{2}}}\!{\left[\mathbf r + \left[\mathbf v \times \frac{[\mathbf u \times \mathbf r]}{c^{2}}\right]\right]} \qquad (.2). $$ Using designations $$ \mathbf E = \frac{Q\gamma \mathbf r}{\left(r^{2} + \frac{\gamma^{2}}{c^{2}}(\mathbf u \cdot \mathbf r)^{2} \right)^{\frac{3}{2}}}, \quad \mathbf B = \frac{1}{c}[\mathbf u \times \mathbf E], $$ (.2) can be rewrited as $$ \mathbf F = q\mathbf E + \frac{q}{c}[\mathbf v \times \mathbf B]. $$ Of course, you know, that $$ \mathbf F = \frac{d}{dt}(\frac{m \mathbf v }{\sqrt{1 - \frac{v^{2}}{c^{2}}}}) = \frac{m\mathbf v (\frac{\mathbf v \cdot \mathbf a}{c^{2}})}{\left(1 - \frac{v^{2}}{c^{2}}\right)^{\frac{3}{2}}} + \frac{m \mathbf a}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}. $$ |
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