# Principle of relativity and point particle in electromagnetic field

Equaling relativistic version of 2nd Newton law to Lorentz force $$\frac{d}{dt} (\gamma_u m \mathbf{u}) = q(\mathbf{E} + \mathbf{u} \times \mathbf{B})$$ we can see that a point particle (of mass $m$ and charge $q$) in a electromagnetic field ($\mathbf{E},\mathbf{B}$) has acceleration $$\mathbf{a} = \frac{q}{\gamma_u m} \left[ \mathbf{E} + \mathbf{u}\times\mathbf{B} - (\mathbf{E} \cdot \mathbf{u}) \frac{\mathbf{u}}{c^2} \right]$$ where I don't use nor four vector formalism, wich I don't know (I speak about ordinary acceleration), neither Gauss units. Let's consider the usual $S'$ system (in motion at speed $v$ toward positive $x$). Lorentz transformations give $$\mathbf{u'} = \left( \frac{u_x - v}{1-\frac{u_x v}{c^2}} , \frac{u_y}{\gamma \left( 1-\frac{u_x v}{c^2} \right)} , \frac{u_z}{\gamma \left( 1-\frac{u_x v}{c^2} \right)} \right)$$ ($\gamma_u$ relative to speed of particle in $S$ should not be confused with $\gamma$ relative to the speed of $S'$) from wich we get the gamma factor of the particle by S' measurements. After many simplifications we get $$\gamma_{u'} = \frac{1}{\sqrt{1-\frac{u'^2}{c^2}}} = \gamma \gamma_u (1- \beta_{ux} \beta)$$ where $\gamma=\frac{1}{\sqrt{1-\beta^2}}$, $\gamma_u=\frac{1}{\sqrt{1-\beta_u^2}}$, $\beta=\frac{v}{c}$, $\beta_{u}=\frac{u}{c}$ and $\beta_{ux}=\frac{u_x}{c}$. But $$\mathbf{E}' = (E_x , \gamma (E_y - v B_z) , \gamma (E_z + v B_y))$$ $$\mathbf{B}' = \left(B_x , \gamma \left(B_y + \frac{v}{c^2} E_z \right) , \gamma \left(B_z - \frac{v}{c^2} E_y \right) \right) \label{trab}$$ Now, I expect that if $S'$ calculus acceleration using Lorentz force with the speed and fields measured by him, he should find the acceleration of $S$ transformed using Lorentz transformations, wich for example for $x$ component gives $a_x' = \frac{a_x}{\gamma^3 \left( 1 - \beta_{ux} \beta \right)^3 }$. But $\frac{q}{m}=\frac{q'}{m'}$, so $x$ components of $S$ and $S'$ should be related by (I use subscript $x$ to denote $x$ componentof vector in square brackets) $$\frac{1}{\gamma_{u'}} \left[ \mathbf{E'} + \mathbf{u'}\times\mathbf{B'} - (\mathbf{E'} \cdot \mathbf{u'}) \frac{\mathbf{u'}}{c^2} \right]_x = \frac{1}{\gamma^3 \left( 1 - \beta_{ux} \beta \right)^3 } \frac{1}{\gamma_u} \left[ \mathbf{E} + \mathbf{u}\times\mathbf{B} - (\mathbf{E} \cdot \mathbf{u}) \frac{\mathbf{u}}{c^2} \right] _x$$ wich can be arranged in this better way $$\gamma^2 { (1 - \beta_{ux} \beta)^2 } \ \left[ \mathbf{E'} + \mathbf{u'}\times\mathbf{B'} - (\mathbf{E'} \cdot \mathbf{u'}) \frac{\mathbf{u'}}{c^2} \right]_x = \ \left[ \mathbf{E} + \mathbf{u}\times\mathbf{B} - (\mathbf{E} \cdot \mathbf{u}) \frac{\mathbf{u}}{c^2} \right] _x$$ Substituting $\gamma$, $\beta$, $\beta_{ux}$ and primed quantities written above, I expect to find an identity but this doesn't work: what went wrong here?

Lorentz transformations are designed to relate the particle's properties in two different inertial frames. When you try to make a transformation into a non-intertial frame, e.g. when the Lorentz-factor $\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$ becomes time dependant with $v = v(t)$ then

$$\Lambda^\mu_{~~~\nu} \frac{d^2}{d\tau^2} x^\nu \ne \frac{d^2}{d\tau^2} \left(\Lambda^\mu_{~~~\nu}~ x^\nu \right)$$

If you transform your fields E and B with $\gamma = \gamma(t)$, you are calculating the acceleration in S and then transform it to S' (left hand side). This means you calculate the instantaneous acceleration measured by an observer in an inertial frame at that single instance of time in which the particle and the observer in S' travel at the same speed. So strictly speaking, that acceleration is not measured relative to the particle but relative to a different observer which is in an inertial frame.

If you use the right hand side, and equate it to the transformed Lorentz-force, you introduce so called fictitious forces, which you should know form newtonian-mechanics, e.g centrifugal forces. This is the closest you can get with special relativity in describing a frame which is moving with the accelerated particle.

• The motion of the particle is generic, but S and S' are both inertial frames (perhaps I did not make myself clear, but there aren't observer in motion with the particle). In don't understand your formula: I don't know tensor calculus (sooner or later I'll study it but anyway I strongly dislike 4-vector formalism, they doesn't match my liking, and I'd like to solve the problem without using them). I'm simply looking for a "proof" (so to speak) of relativistic version of 2nd Newton law (as far as I know first written by Planck). A way to say "that's why insert gamma factor here is a good idea!". – Fausto Vezzaro Oct 3 '16 at 19:55
• @FaustoVezzaro: The classical analogue would be a time dependant rotation matrix $R(t)$ then $R(t)\frac{d^2}{dt^2}\vec{x} = R(t)\vec{F}$ is different from $\frac{d^2}{dt^2}(R(t)\vec{x}) = R(t)\vec{F}$. The 2nd equation is what an "accelerated" frame would look like. The 1st is for many inertial frames (one for each t). If you transform (E,B) and equate it to $\vec{a}'$ you are doing sth. similar as the first equation. If you transform (E,B) and use the 2nd equation you get a different result for $\vec{a}'=R(t)\frac{d^2}{dt^2}\vec{x}$ – image Oct 3 '16 at 20:12
• @FaustoVezzaro: In simple words: For non-constant transformations, differentiation and transformation do not commute. Thus, the usuall formuals for acceleration transformation are the ones for the 1st equation of my above comment. I expect that when you write "Now, I expect that if S′ calculus acceleration", you are in fact using the 2nd equation and set $\vec{a}' = R(t) \vec{a}$ (in analogy to my comment above). – image Oct 3 '16 at 20:18
• This week-end I found the answer: simply that isn't true that the final equation doesn't work: it works. I shame (in my defense I say that the wrong calculus is due to a bug in wxMaxima software, I had to solve this very long calculations putting pen to paper...!) but at the same I'm happy of having show an original way to relativistic 2nd Newton law (a way very long with calculus but simple in concepts: Planck used lagrangians and other tools I can't understand in this contest). Now I ask to moderator if this answer has to be left or removed: I solved my problem and for me is the same. – Fausto Vezzaro Oct 9 '16 at 17:53
• You were the only one that answer and I really thank you for your interest, but I fear we aren't in the same wavelength, we think in a different way, I didn't understand your answer and probably you didn't understand my question, in which I calculus $a'_x$ both transforming $a_x$ and using $S'$ measurements and the first equation again (maybe my ugly english doesn't help in this sense). This two ways give the same results so first equation is consistent with Lorentz transformation. Anyway I write an answer. – Fausto Vezzaro Oct 9 '16 at 22:40

It isn't true that the final equation doesn't work: it work. The right side is $$E_x + u_y B_z - u_z B_y - \frac{E_x u_x + E_y u_y + E_z u_z}{c^2} u_x$$ While the left side is $$\gamma^2 { (1 - \beta_{ux} \beta)^2 } \left[ E'_x + u'_y B'_z - u'_z B'_y - \frac{E'_x u'_x + E'_y u'_y + E'_z u'_z}{c^2} u'_x \right]$$ where (I don't write $B'_x=B_x$ because we don't need it now)

• $E'_x = E_x$
• $E'_y = \gamma (E_y - v B_z)$
• $E'_z = \gamma (E_z + v B_y)$
• $B'_y = \gamma \left(B_y + \frac{v}{c^2} E_z \right)$
• $B'_z = \gamma \left(B_z - \frac{v}{c^2} E_y \right)$
• $u'_x = \frac{u_x - v}{1-\frac{u_x v}{c^2}}$
• $u'_y = \frac{u_y}{\gamma \left( 1-\frac{u_x v}{c^2} \right)}$
• $u'_z = \frac{u_z}{\gamma \left( 1-\frac{u_x v}{c^2} \right)}$
• $\gamma = \frac{1}{\sqrt{1-\beta^2}}$
• $\beta = \frac{v}{c}$
• $\beta_{ux} = \frac{u_x}{c}$

Doing substitution we see that both sides are the same.

In the calculus, they can be useful these intermediate steps: $$E'_x u'_x + E'_y u'_y + E'_z u'_z = \frac{E_x u_x + E_y u_y + E_z u_z + v (B_y u_z - B_z u_y - E_x) }{1-\frac{u_x v}{c^2}}$$ $$u'_y B'_z - u'_z B'_y = \frac{u_y \left( B_z - \frac{v E_y}{c^2} \right) - u_z \left( B_y + \frac{v E_z}{c^2} \right)}{1-\frac{u_x v}{c^2}}$$ I did by hand this check two times, I'm sure the last equation of my answer works. I checked $y$ component too (and surely $z$ is ok by symmetry).