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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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6 Answers 6

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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But now you need to derive the Hamiltonian from elementary principles to answer my question. Still +1 since your answer is very close to my conclusion given the Hamiltonian is the total energy: The Lorentz force law for a static electric field expresses the conservation of energy, with the change in field energy depending only on the change in postion of the charge. –  John McVirgo Nov 16 '12 at 22:10
Thank you! I would postulate the Hamiltonian --in reality I would postulate one more general and set $\mathbf{A}=0$ for your specific case--, because I do not know any "elementary principle" from which $H$ can be derived. Notice that Lagrangians $L$, actions $S$, Lorentz transformations, symmetries, conservation laws... all that can be derived from the Hamiltonian formalism. –  juanrga Nov 17 '12 at 11:33

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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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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This answer is awful nonsense, even though it might or might not be right (I didn't bother checking). You must not do this type of writing, it is obfuscatory and doesn't convey what you are actually doing. You are not using dot and cross products in private calculations, you are using a coordinate system where v is one of the axes, x is another and the component of a perpendicular to x is the third. Please use this explicit coordinate system when communicating, nobody has half an hour to translate your expressions. –  Ron Maimon Jul 24 '12 at 21:34
"...you are using a coordinate system where v is one of the axes, x is another and the component of a perpendicular to x is the third..." You're wrong. The coordinate system may be the way you want. –  PhysiXxx Jul 25 '12 at 21:11
If I understood your comment correctly. –  PhysiXxx Jul 25 '12 at 22:26
I think you understand my comment--- I am saying your way is obfuscated, impossible to follow, and is just generally incompetent. I read your calculation through quickly, you are just doing Lorentz transformations on E and B, and showing that the rest-frame Coulomb force, assuming a tensor transformation for E and B, leads to the Lorentz force law. This is a one-liner. I am sorry, but this is the type of intimidation that makes it impossible to explain physics, -1, even though your calculation is probably correct, for doing it in the worst possible way. –  Ron Maimon Jul 26 '12 at 1:13
I didn't use a special transformations for $\mathbf E, \mathbf B$. But these transformations can be derived analogically, of cource. I used that derivation, because it used minimum of postulates and tensor "things". I answered to the question correctly. No wonder that the author has chosen not the answer of the question even more then half year ago, hasn't he? –  PhysiXxx Jul 26 '12 at 8:35

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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By Newton's law, I meant his second law of motion, which will mean some editing in your answer ;) –  John McVirgo Sep 17 '11 at 20:41
I understood what you meant. That's what I wrote above. –  Ron Maimon Sep 17 '11 at 21:36
ok, you were comparing the gravitational and electric potential which is why I jumped to that incorrect conclusion. Your answer is too sophisticated for me, and I think a more reasonable answer is that the expression in the question comes from the conservation of relativistic energy: For the simple case of a charge moving along $x$, $d(\gamma mc^2) = eEdx\Rightarrow 1/v~d/dt(\gamma mc^2) = m\gamma^3 dv/dt= d/dt(m\gamma v) = eE$. A more complete answer would need more work, though. –  John McVirgo Sep 17 '11 at 23:20
Why is it "Higgs-type". $\phi$ is just a scalar-field. –  Killercam Jul 25 '12 at 9:13
@Killercam: This coupling between a scalar and a particle, the one that adjusts the mass in proportion to the field, is the same as a Higgs-like coupling that generates lepton masses in the particle formulation of field theory. This was just a motivating analogy--- I agree that $\phi$ is just a generic scalar without gauge charges. –  Ron Maimon Jul 25 '12 at 16:50

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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Too sophisticated? –  John McVirgo Jan 17 '11 at 1:12
@John: there is no such thing as being too sophisticated. It's okay to have answers of completely different level of sophistication so that everyone can pick one that suits them. –  Marek Jan 18 '11 at 10:15
lol, I love Stan's answers. They're so mathy. –  AlanSE Aug 11 '11 at 17:00
The problem with this answer is that it is assuming that which is to be derived. Once you know E is part of a tensor and p is part of a four vector, and e is a scalar, then the answer is obvious. The question wanted to derive these properties from symmetry arguments. –  Ron Maimon Sep 17 '11 at 3:14

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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@John: The end of your comment got cut off perhaps? –  Noldorin Jan 17 '11 at 1:27
@John: So... you're not gonna finish your point? –  Malabarba Jan 20 '11 at 19:55
Indeed, I'm waiting! –  Noldorin Jan 21 '11 at 1:56
I'm still thinking about this, on and off, when I have the time. The problem is trying to visualise the interaction of the charge with the EM field which looks hopeless without bringing in QED which I have no knowledge of. So for now, I think the expression is simply a statement of the conservation of either energy or momentum in the interaction. Eg. for the simple case along x, d(energy of moving charge) = eEdx, or d(momentum of moving charge) = eEdt. –  John McVirgo Jan 23 '11 at 23:46

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