Deriving Maxwell's Equations from Field Tensor (Griffith 4ed)

In summary, In summary, the main message of this post is that you can use SI units for relativistic electrodynamics as long as you leave at least one physical dimension unspecified.
  • #1
omega_minus
72
11
Hello,
I am reading Griffith's "Introduction to Electrodynamics" 4ed. I'm in the chapter on relativistic electrodynamics where he develops the electromagnetic field tensor (contravariant matrix form) and then shows how to extract Maxwell's equations by permuting the index μ. I am able to follow the argument for the μ=0 case as it appears straightforward: He takes the spatial derivative of the corresponding matrix element and out pops Gauss' Law, etc. But when the time derivative term is not zero (μ≠0) there always appears an additional factor of 1/c on that term. I can see in the end this term is required to get the corresponding equation, but I don't understand where it comes from. For example when he takes the time derivative of the F10 term (the -Ex/c term) he ends up with a factor of -1/c2.
I'm a grad student in electromagnetics engineering and we didn't learn this formalism but I enjoy studying special relativity so I'm making an effort to understand it. I'm not enrolled this summer (no professors to bother...) so I was hoping someone here might point me in the right direction. Google has not been much help and the book didn't seem to address it before. The answer is probably staring me in the face...
Thanks in advance for any help.
 
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  • #2
I do not have Griffith in front of me, but the most likely explanation is that ##x^0 = ct## so ##\partial_0 = (1/c) \partial_t##. Then one can always question the sanity in working in units where ##c \neq 1##, which just obstructs the physics in my opinion.
 
  • #3
@Oridruin, thanks for the reply. It makes sense when you put it like that. Of course having c=1 is likely convenient to physicists but we engineers aren’t used to it (speaking for myself here). It does the same thing to my brain that setting the permittivity and permeability to unity does; it makes unit analysis difficult if you don’t know there are constants with units “hiding“ in the equation. I’m guessing if you’re used to it, it is easier. Thanks again for your help!
 
  • #4
omega_minus said:
It does the same thing to my brain that setting the permittivity and permeability to unity does; it makes unit analysis difficult if you don’t know there are constants with units “hiding“ in the equation.
You can go back to SI units at any time by doing dimensional analysis. You can do dimensional analysis as long as you leave at least one physical dimension. It is just that time and length have the same physical dimension when you work in units with c=1. Note that this c is dimensionless!

Regardless of units, that was not the main message of my post.
 
  • #5
Ok, I see. I didn’t realize time and space had the same dimension in natural units, I just thought the units’ differences were somehow just being ignored. As to the content of your original post, it cleared the matter up in a single sentence. Thanks again!
 
  • #6
omega_minus said:
I didn’t realize time and space had the same dimension in natural units
That’s fine, it is a big hurdle for many to get over. Still, it is the most natural choice in relativity as spacetime is an affine space with a pseudo metric and it therefore makes at least some amount of sense to use the same dimension for all of its directions. To me, using different dimensions for time and space is like using different dimensions for height and horizontal distance. They may seem very different (at least when you are flying), but they really are the same. Then you might want to use different units for either (e.g., meters for height and km for horizontal distances) and the resulting quotient between the units is some arbitrary number that is just the conversion factor between the two (e.g., 0.001 km/m). Same thing in spacetime, you can still use seconds and meters, but if you have a length in meters and want to convert it to seconds you need to divide by the convertion factor c = 3e8 m/s.
 
  • #7
Orodruin said:
You can go back to SI units at any time by doing dimensional analysis. You can do dimensional analysis as long as you leave at least one physical dimension. It is just that time and length have the same physical dimension when you work in units with c=1. Note that this c is dimensionless!

Regardless of units, that was not the main message of my post.
I've still quibbles with the SI. I'm right now preparing my lectures for my teacher-students lectures. One is special relativity and quantum mechanics. For the special-relativity part, I want to end with a very brief explanation of the manifest covariant form, and that's why I just thought about, how to work this out in the SI (it's a night-mare sorting all the ##\epsilon_0## and ##\mu_0## factors).

In my opinion, for this audience indeed the use of the SI units is mandatory. There's not a single high-school textbook, where they use Gauß (or even my favorite rationalized Gauß, i.e., Heaviside-Lorentz) units. That's why in the last semester I taught electrodynamics in the SI. Of course, now I'd like to do also a very short glimpse on the covariant formulation of electrodynamics at the end of the section on special relativity, and indeed of course there the SI units hit me. The only feasible way I see is to finally define ##F_{0 \mu} = \pm E_{\mu}/c## (##\mu \in \{1,2,3\}##) and ##F_{\mu \nu} = \pm \epsilon_{\mu \nu a} B^a## (don't nail me for the signs, which are not so important here). Then the inhomogeneous Maxwell equations in covariant form read
$$\partial_{\mu} F^{\mu \nu} = \mu_0 j^{\nu},$$
with the ugly arbitrary conversion factor ##\mu_0## to convert the unnatural SI unit for electric charge, the Coulomb or rather Ampere seconds. That's the way Sommerfeld does it in his famous Lectures on Theoretical Physics (however in the ##\mathrm{i} c t## convenction, which I find totally inacceptable in the 21st century).

Of course you could as well use ##F_{0 \mu}=\pm E_{\mu}## and ##F_{\mu \nu} = \pm c \epsilon_{\mu \nu \rho} B^{\rho}##, then ending up with
$$\partial_{\mu} F^{\mu \nu} = \frac{1}{\epsilon_0} j^{\nu}.$$
This latter possibility, however, is not seen in the literature anywhere.

Only going to "natural units", i.e., setting ##\mu_0=\epsilon_0=1##, which implies also ##c=1##, you get formulae in natural rationalized Heaviside-Lorentz units.
 

1. What is the significance of Maxwell's equations in the field of physics?

Maxwell's equations are a set of four fundamental equations that describe the behavior of electric and magnetic fields. They are crucial in understanding the fundamental principles of electromagnetism and have led to many important technological advancements, such as the development of radio, television, and other forms of communication.

2. How are Maxwell's equations derived from the field tensor in Griffith's 4th edition?

In Griffith's 4th edition, Maxwell's equations are derived from the field tensor using the principles of vector calculus and the laws of electromagnetism, such as Gauss's law, Ampere's law, and Faraday's law. The field tensor is a mathematical construct that simplifies the equations and allows for a more elegant and concise representation of electromagnetism.

3. What is the relationship between the field tensor and the electric and magnetic fields?

The field tensor is a mathematical representation of the electric and magnetic fields. It combines the electric and magnetic fields into one tensor, making it easier to manipulate and solve equations involving these fields. The components of the field tensor correspond to the electric and magnetic fields in different directions.

4. Are Maxwell's equations consistent with other fundamental laws of physics?

Yes, Maxwell's equations are consistent with other fundamental laws of physics, including Newton's laws of motion and Einstein's theory of relativity. They have been extensively tested and validated through experiments, and they accurately describe the behavior of electric and magnetic fields in various situations.

5. What are some real-world applications of Maxwell's equations derived from the field tensor?

Maxwell's equations derived from the field tensor have many practical applications, such as in the design and optimization of electronic devices, the development of wireless communication technologies, and the study of electromagnetic waves and their interactions with matter. They are also used in fields such as optics, quantum mechanics, and particle physics.

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