# Extending the Lorentz force equation to accommodate SR

1. Apr 21, 2013

### MarkovMarakov

I would be very grateful if someone would kindly explain this generalization of the Lorentz force law to the special relativity domain. I am not entirely sure if what I have jotted down is exactly as the speaker intended to convey. But here is what I have got. Please bear with me.

>Classically, the Lorentz force law is $$m\frac{d^2x}{dt^2}=q(E+v\times B)$$. We want to reformulate this relativistically. Noticing that $\frac{dt}{d\tau}\approx 1$, we could postulate that the relativistic version of the equation has a factor of $\frac{dt}{d\tau}$ multiplying E. But the force now depends on velocity linearly classically, so the only possibility hat is consistent with both the classical limit and SR is $$m\frac{d^2x^\mu}{d\tau^2}=qF^\mu{}_\nu \frac{dx^\nu}{d\tau}$$

Here are some of the things I don't understand:

1. What does it mean to reformulate it relativistically? As far as I understand, it means we want an equation that holds at high velocities but reduces to the Lorentz equation in the Newtonian limit. OK. But what does that mean mathematically? How does one go about generalizing equations to fit SR?
2. Why does $\frac{dt}{d\tau}\approx 1$ in the classical case suggest multiplying $\frac{dt}{d\tau}$ in front of $E$?
3. How do they come about with $$m\frac{d^2x^\mu}{d\tau^2}=qF^\mu{}_\nu \frac{dx^\nu}{d\tau}$$? I suppose the linear-dependence of the force on velocity in the classical limit suggests the RHS has to be linear in $dx\over d\tau$. But what is so SR about it? How is it consistent with/accommodate SR?

Thank you very much!

2. Apr 21, 2013

### WannabeNewton

Hi Markov. We don't need to make any such assumptions. We can derive the equations of motion for the charged particle interacting with an electromagnetic field straight from a variational principle. See here: https://www.physicsforums.com/showpost.php?p=4336855&postcount=5

To reformulate things relativistically can mean, in the context of SR, to make things lorentz covariant (or at least make it more apparent). If you can take an equation and write it down in terms of lorentz covariant quantities, then if it holds in one inertial frame it will hold in all inertial frames which is what we desire in SR. For example, the lorentz force law written as $a^{b} = u^{a}\partial_{a}u^{b} = \frac{q}{m}F^{b}{}{}_{c}u^{c}$ is clearly lorentz covariant because it is written down in terms of lorentz co-variant quantities i.e. the 4-velocity $u^{a}$, the flat space-time derivative operator $\partial_{a}$ associated with the minkowski metric $\eta_{ab}$, and the electromangetic field strength tensor $F_{ab}$.

3. Apr 21, 2013

### MarkovMarakov

Thanks, WannabeNewton :) This is very helpful!

4. Apr 21, 2013

### WannabeNewton

Perhaps a related example would be helpful as well? The classical form of Maxwell's equations read $\nabla \times B - \partial _{t}E = j, \nabla\cdot E = \rho, \nabla\times E + \partial _{t}B = 0, \nabla\cdot B = 0$. If we can rewrite these equations in terms of lorentz covariant quantities then we can immediately conclude that they hold in all inertial frames if they hold in any given inertial frame (so in particular the equations will transform covariantly under lorentz transformations).

We can start by writing the four equations in terms of components as $\epsilon^{ijk}\partial_{j}B_{k} - \partial_{0}E^{i} = j^{i}, \partial_{i}E^{i} = j^{0}, \epsilon^{ijk}\partial_{j}E_{k} + \partial_{0}B^{i} = 0, \partial_{i}B^{i} = 0$ where the 4-current density $j^{\mu} = (\rho,j^{i})$ has been used. I'll leave the details to you but it is straightforward to then take the electromagnetic field strength tensor $F_{\mu\nu}$, which is defined in terms of the electric and magnetic field, and rewrite the above equations in terms of the 4-current density as $\partial^{\mu}F_{\mu\nu} = j_{\nu}, \partial_{[\alpha}F_{\mu\nu]} = 0$. We have taken maxwell's equations and written them in an extremely elegant form that also immediately conveys the fact that maxwell's equations are lorentz covariant because the equations have been written down entirely in terms of lorentz covariant quantities.

5. Apr 21, 2013

### HomogenousCow

Ultimately aren't we still just guessing what the equations of motion might be, I mean how did we know what the interaction term should be like?

Last edited: Apr 21, 2013
6. Apr 21, 2013

### WannabeNewton

Well yes there is a bit of guesswork, but we just take the simplest possible lorentz scalar that involves the particle's 4-velocity as well as the electromagnetic field and see if it works. I find it much more satisfying than other approaches, such as the one alluded to by the OP.

7. Apr 21, 2013

### HomogenousCow

Just a side question, in the generally covariant formulation of maxwells equations, how does one handle back reaction from the materials? i.e. the permativity and permability tensors (linear and non linear)

8. Apr 21, 2013

### WannabeNewton

9. Apr 21, 2013

### HomogenousCow

No as in, in "non-relavistic" electromagnetism we have the electric prrmativity and magnetic permabilities, which describe the macroscopic response of the material, how does one describe this in the generally covariant form of maxwells equations (those that are in your signature)? I mean in general, whether we have linear reaction or nonlinear.
On another note regarding the back reaction articles you mentioned, why can't we just simply abandon the notion of a point particle, it is obviously an artifical concept, should we not simply stick to realistic matter distributions? (i.e. couple the maxwells equations to the continuum mechanics ones)

Last edited: Apr 21, 2013
10. Apr 21, 2013

### WannabeNewton

It seems like you are asking for a covariant formulation of Maxwell's equations in matter. I'm not familiar with the covariant formulation of the equations in such media but take a look here: http://www.sciencedirect.com/science/article/pii/0016003278901205 and here is a general overview http://en.wikipedia.org/wiki/Covari..._electromagnetism#Covariant_objects_in_matter Sorry I couldn't be of much further use regarding this. Thankfully, the usual Maxwell's equations in matter still work fine in the down to earth cases :D

11. Apr 24, 2013

### Agerhell

Now formally, even classically:
$$\frac{d}{dt}(m\bar{v})=q(\bar{E}+\bar{v} \times \bar{B})$$.

Assuming $dm/dt=0$ you get the expression you wrote above. However, in special relativity, the mass can be percieved as implicitly time-dependent, (it varies with velocity that varies with time) so that you can replace $m$ in the expression I wrote above with $m\gamma$ where:
$$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$
and m is still considered invariant (the time-dependence lies in $\gamma$). Then you get, after solving a differential equation:
$$\frac{d\bar{v}}{dt}=\frac{q}{\gamma^3}(\bar{E} \cdot \hat{v})\hat{v}-\frac{q}{\gamma}((\bar{E}\times\hat{v})\times\hat{v} + ((\bar{v}\times\hat{B})\times\hat{v})\times\hat{v})$$

(Perhaps this can be written somewhat simpler, I split the acceleration up in components along and orthogonal to the velocity vector...)

I would assume that this is the most obvious extension of the Lorentz law into the special relativistic domain that you can use for calculating the acceleration of charged particles.

12. Apr 24, 2013

### jfy4

I would think that the lattice shape of the material already breaks lorentz symmetry, and so there won't be a formulation of E&M that is lorentz invariant in matter, since if you were sitting inside the matter, it isn't lorentz invariant.