# Field Tensor and classical limits

I would be very grateful if someone would kindly explain this generalization of the Lorentz force law to the special relativity domain. 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?
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let me give some really really heuristic arguments as to why we end up with the last equation. Classically, we place time and space on different footings. similarly, the E and B fields are quite different. But under a Lorentz transformation, space is rotated into time and vice versa, and E into B and vice versa. So the equation we have should be place space & time, and E & B, on an equal footing. Hence, there should be some 'spacetime-derivative' multiplying E in the equation, which we postulate to be $dt/d\tau$... –  nervxxx Apr 21 '13 at 22:50
@nervxxx: I like your argument better as a plausibility argument for combining E and B into a single tensor, rather than as an a plausibility argument for the WP article's silly $dt/d\tau$ fudge factor. –  Ben Crowell Apr 21 '13 at 22:56
Thank you, @nervxxx ! :) –  Bess Apr 21 '13 at 22:59

Welcome to physics.SE!

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?

In SR, we have scalars, which don't change under a Lorentz transformation, vectors, which transform under the Lorentz transformation the same way as a displacement vector, and tensors, which have their own transformation law that involves applying the Lorentz transformation twice. Once you have built up some collection of scalars, vectors, and tensors, you can form new ones from them. For example, if you multiply the scalar mass $m$ ("rest mass") by the velocity four-vector $v$, you get another thing that transforms like a four-vector; it's the energy-momentum four-vector. It's not always obvious whether or not a particular quantity built like this is of interest. For example, Einstein didn't just assume that the energy-momentum four-vector was conserved; he made a separate argument, in his 1905 paper "Does the inertia of a body depend upon its energy-content?" After that, someone (maybe him, I don't know) reformulated the theory in terms of four-vectors.

1. Why does $\frac{dt}{d\tau}\approx 1$ in the classical case suggest multiplying $\frac{dt}{d\tau}$ in front of $E$?

It doesn't. That's just bad writing in the WP article -- and someone appears to have removed that text now.

1. 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?

What makes it compatible with SR is that it's built only out of Lorentz scalars ($m$, $\tau$, and $q$), Lorentz vectors ($x$) and Lorentz tensors ($F$). This doesn't guarantee that it's right, but it does guarantee that if one observer says that the equation holds, so will other observers.

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this answer is not wrong but I fear it is not very illuminating to the reader quite unfamiliar with SR... –  nervxxx Apr 21 '13 at 22:41
Thank you, Ben! :) –  Bess Apr 21 '13 at 22:46