- Requirements for invariant force laws
- The relativistic law of motion of a charge particle in an electromagnetic field
- Putting the speed of light back in
- Transformation properties of the Faraday tensor
Requirements for invariant force laws
To correct the Newtonian equations of motion - or force laws - so that they are the same in all reference frames, we need to write them in terms of four-vectors - meaning we must write them in the form
All invariant force laws must satisfy certain conditions:
The proper force must be normal to the proper velocity.
That is,
The law must tend to the Newtonian limit in non-relativistic situations.
Given a particular force, when velocities are small compared to the speed of light, the proper force must approximate the Newtonian version of the law,
The law must be the same in all reference frames.
Finally, the law must be invariant under Lorentz transformations. Given any other (primed) reference frame, we must have
This is the basic premise of the exercise - to convert non-relativistic equations of motion into relativistic ones.
The relativistic law of motion of a charge particle in an electromagnetic field
We shall show that the invariant version of Lorentz's law of motion of a charge particle in an electromagnetic field is
Thus, in a coordinate frame,
Nb. There is (at least, I think) one other tensor that satisfies the conditions, and we shall need to define it when reformulating Maxwell's equations in the next lecture.
Force normal to velocity
The first condition is easily satisfied, because being anti-symmetric means
Newtonian limit
Now, the left-hand side of the force equation - in the coordinate frame - is
Thus,
The other spatial components follow the same pattern, so we can write
Dividing everything by , we seem to recover the Newtonian Lorentz force. However, there is a hidden factor of in the momentum (using for the index to save confusion with the mass ),
Now, if is small compared to the speed of light, then , and
Since
Invariance
The last, and most important requirement for this formulation is that is the same in all reference frames. To show that this is indeed the case, we need a result from the next section, which tells us how covariant four-vectors transform.
In short, if are the components of a given Lorentz transformation matrix and the covariant components of some four-vector, then
We want to show that the equation,
Thus, we have shown that the four-vector force law for charged particles in electromagnetic fiields satisfies the three requirements of an invariant force law.
Putting the speed of light back in
We can abuse matrix notation somewhat and write the equations more succinctly in terms of the three-velocity , the electric field and the magnetic tensor ,
Nb. Covariant four-vectors form column-vectors (as do contravariant four-vectors).
Setting means we multiply the time components of the two four-vectors by ,
But what about the electromagnetic tensor (Faraday tensor)? We need to ensure that the equations still tend to the Newtonian limit as gets small compared to . We can, naively, just use the current version and then see how we need to correct it,
In order to get the correct answer for the Newtonian momentum, we need to divide the electric field vector by the speed of light in the Faraday tensor. That is,
The spatial components haven't changed (except for of course), but the time component on the right-hand side has been divided by . We can show that this works out correctly by multiplying both sides of the first equation by to get
Thus, the invariant version of Lorentz's law of motion of a charge particle in an electromagnetic field, even when , is
Transformation properties of the Faraday tensor
Given we know the components of the Faraday tensor in one frame, we would like to know its components in another (primed) frame, moving at speed relative to the first, along their common -axes - the usual setup. Also, let's go back to .
Nb. This is not the speed, of the particle in the Lorentz force - we are not considering the equation now, just the tensor.
Thus, since it's a rank-two tensor, the Faraday tensor in the same way as a product of two four-vectors,
Now, explicitly, a single four-vector transforms as
Suppose we wanted to know how the -component of the electric field transforms. Well, which transforms in the same as the product of two vectors,
So, using the same pattern,
This means that two observers moving at any relative speed along their common -axes will measure exactly the same value for the -component of the electric field.
How about the -component of the electric field? transforms in the same way as
This means that, for the same two observers, the relative speed is now important, because the -component of the electric field in one frame gets mixed up with the -component of the magnetic field in the other frame.
The full set of transformations are