Transforming Maxwell's Equations in Special Relativity.

[ObsessiveMathsFreak]

What is the simplest derivation of the transformation rules for Maxwell's equations in special relativity?

I'm working through Einstein's original 1905 paper(available here), and I'm having trouble with the section on the transformation of Maxwell's equations from rest to moving frame.

The paper proceeds as follows:

Let the Maxwell-Hertz equations for empty space hold good for the stationary system K, so that we have

where (X, Y, Z) denotes the vector of the electric force, and (L, M, N) that of the magnetic force.

If we apply to these equations the transformation developed in § 3, by referring the electromagnetic processes to the system of co-ordinates there introduced, moving with the velocity v, we obtain the equations

I do not understand just where this second set of equations comes from.

The only derivations of this transformation that I have been able to find involve either potentials, four-vectors, or both. However since four-vectors had not been invented in 1905, and because the statement of the transformation is so blunt, it seems that Einstein is using or appealing to a simpler method for finding the transformation.

So my question is: What is the simplest derivation of the transformation rules for Maxwell's equations in special relativity? Is Einstein appealing to a pre-existing Lorentz method here, or is there a trick to accomplish all of this quickly? Or does the text here simply belie the true work involved in the derivation?

 

[Mentz114]

They come from "the transformation developed in § 3,", one supposes.

There are some recent threads in this forum discussing the derivation of that transformation from first principles. It turns out be non-trivial. If you are what your moniker suggests, you'll enjoy them, obsessively.

 

[ObsessiveMathsFreak]

Could you give links to the relevant threads, for future googlers if nothing else.

 

[vanhees71]

Einstein's derivation is much more complicated than necessary, because he had not the elegant mathematical formulation by Minkowski at hand, which makes everything way easier.

You can either go via the electromagnetic four-potential $(A^{\mu})=(\Phi,\vec{A})$ (in good old Gaussian or Heaviside-Lorentz units, which are way more natural in this context than the SI), which transform as a vector field under $\mathrm{O}(1,3)$:
$$A'^{\mu}(x')={\Lambda^{\mu}}_{\nu} A^{\nu}(x)={\Lambda^{\mu}}_{\nu} A^{\nu}(\Lambda^{-1} x'),$$
or via the gauge independent Faraday tensor $F_{\mu \nu}=\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}$ which transforms as a 2nd-rank tensor field:
$$F'^{\mu \nu}(x')={\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} F^{\rho \sigma}(\Lambda^{-1} x').$$
Here $\Lambda$ is a $\mathrm{O}(1,3)$ matrix, i.e., a matrix fulfilling
$$\eta_{\mu \nu} {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma}=\eta_{\rho \sigma}, \quad (\eta_{\mu \nu})=\text{diag}(1,-1,-1,-1).$$

 

[ObsessiveMathsFreak]

Einstein's derivation is much more complicated than necessary, because he had not the elegant mathematical formulation by Minkowski at hand, which makes everything way easier.

But he doesn't give a derivation, and seems to rely on existing principals. How exactly did he go about it without four-vectors, etc?

 

[Mentz114]

Could you give links to the relevant threads, for future googlers if nothing else.

https://www.physicsforums.com/showthread.php?t=651640

https://www.physicsforums.com/showthread.php?t=600613

This is a good post in the first linked thread

Didn't the paper that Ben mentioned in another thread, http://arxiv.org/abs/physics/0302045, go through all this?

The assumptions that that paper made were (skimming)

* replacing v with -v must invert the transform
* isotropy
* homogeneity of space and time

with a few tricks along the way:
* adding a third frame
* noting that x=vt implies x'=0

The result was pretty much that there must be some invariant velocity that was the same for all observers. (THere were some arguments about sign of a constant before this to establish that it was positive). The remaining step is to identify this with the speed of light.

 

[ObsessiveMathsFreak]

To clarify, I'm not talking about the derivation of Lorentz transforms. I'm asking about the transformation of Maxwell's equations in special relativity. How are these derived without using 4-vectors?

 

[jtbell]

I suspect the first step is to expand the derivatives in the un-transformed Maxwell equations using the chain rule for partial derivatives, e.g.

$$\frac{\partial X}{\partial t} =
\frac{\partial X}{\partial \tau} \frac{\partial \tau}{\partial t} +
\frac{\partial X}{\partial \xi} \frac{\partial \xi}{\partial t} +
\frac{\partial X}{\partial \eta} \frac{\partial \eta}{\partial t} +
\frac{\partial X}{\partial \zeta} \frac{\partial \zeta}{\partial t}$$

where derivatives like $\partial \tau / \partial t$ are calculated using the (Lorentz) transformation from section 3. Many of these derivatives are zero if we assume v is along the x-axis as Einstein appears to do. Then collect terms so as to produce the transformed equations in terms of derivatives like $\partial X / \partial \tau$. This would be the brute-force method. There may be a more elegant method.

 

[Fredrik]

What is the simplest derivation of the transformation rules for Maxwell's equations in special relativity?
...
The only derivations of this transformation that I have been able to find involve either potentials, four-vectors, or both. However since four-vectors had not been invented in 1905, and because the statement of the transformation is so blunt, it seems that Einstein is using or appealing to a simpler method for finding the transformation.

It seems unlikely that there's a simpler method than to rewrite them in the form listed to the right of "tensor calculus" here.

Einstein must have used a more primitive method, but it was hardly simpler.

 

[ObsessiveMathsFreak]

I suspect the first step is to expand the derivatives in the un-transformed Maxwell equations using the chain rule for partial derivatives, e.g.

That was the first thing I tried, but I couldn't see how it would work.

Perhaps it is based on the galilean/linear transformations of the equations. Does anyone have a reference for the raw Newtonian/Galilean transformation of Maxwell's equations?

 

[Bill_K]

Einstein must have used a more primitive method, but it was hardly simpler.

It has always baffled me why anyone would care how Einstein did it 100 years ago. We've come a long way since then, and any current book on GR will convey to the reader a a simpler explanation and a better understanding.

 

[Fredrik]

It has always baffled me why anyone would care how Einstein did it 100 years ago. We've come a long way since then, and any current book on GR will convey to the reader a a simpler explanation and a better understanding.

I have always found it odd too. I understand that many will be curious about how Einstein did it, but we seem to be getting more questions about that, than about the best ways to do these things with the tools that we have at our disposal now. That's what really surprises me.

 

[ObsessiveMathsFreak]

I suspect the first step is to expand the derivatives in the un-transformed Maxwell equations using the chain rule for partial derivatives, e.g.

That was the first thing I tried, but I couldn't see how it would work.

Eventually I managed to figure this out. It turn out that you do use the chain rule, but you also have to use the EM conditions to derive the result.

For the sake of making the thread a potentially useful resource, I'll give the result for the first of the equations, but I'll use primed coordinates instead of greek ones, and labeling the EM field using the usual notations. Hopefully the steps are clear for anyone reading.

In Gaussian units, the first component of $\nabla \times \mathbf{B} =\frac{1}{c}\frac{\partial \mathbf{E}}{\partial t}$ is explicitly
$$\begin{align} \frac{1}{c}\frac{\partial E_x}{\partial t}= & \frac{\partial B_z}{\partial y}-\frac{\partial B_y}{\partial z} \end{align}$$
To transform this, firstly we need the chain rule. Since the transformations in this case are
$$\begin{align} t' = & \gamma \left( t-\frac{v}{c^2}x \right) \\ x' = & \gamma \left( x-vt \right) \\ y' = & y \\ z' = & z \end{align}$$
where $\gamma = 1/\sqrt{1-v^2/c^2}$ is the $\beta$ in the original paper above. It follows using the chain rule that
$$\begin{align} \frac{\partial }{\partial t}\equiv & \gamma \left( \frac{\partial }{\partial t'} - v \frac{\partial }{\partial x'}\right) \\ \frac{\partial }{\partial x}\equiv & \gamma \left( \frac{\partial }{\partial x'} - \frac{v}{c^2} \frac{\partial }{\partial t'}\right) \\ \frac{\partial }{\partial y}\equiv & \frac{\partial }{\partial y'} \\ \frac{\partial }{\partial z}\equiv & \frac{\partial }{\partial z'} \end{align}$$
So we can just make these substitutions in the equations.

Now, before transforming the "curl" equation above, we first need to use and transform the the free space EM condition that $\nabla \cdot \mathbf{E} = 0$. A representation of $\frac{\partial E_x}{\partial x'}$ is given using this by
$$\begin{align} &\frac{\partial E_x}{\partial x}+\frac{\partial E_y}{\partial y}+\frac{\partial E_z}{\partial z}=0\\ &\gamma \left( \frac{\partial E_x}{\partial x'}-\frac{v}{c^2}\frac{\partial E_x}{\partial t'} \right)+\frac{\partial E_y}{\partial y'}+\frac{\partial E_z}{\partial z'} =0 \\ \Rightarrow &\frac{\partial E_x}{\partial x'}=\frac{-1}{\gamma} \left( \frac{\partial E_y}{\partial y'}+\frac{\partial E_z}{\partial z'} \right)+\frac{v}{c^2}\frac{\partial E_x}{\partial t'} \end{align}$$

This step gives us what we need. Now applying both the chain rules and this result to the curl equation gives
$$\begin{align} \frac{\gamma}{c} \left( \frac{\partial E_x}{\partial t'} - v \frac{\partial E_x}{\partial x'}\right)=& \frac{\partial B_z}{\partial y}-\frac{\partial B_y}{\partial z} \\ \frac{\gamma}{c} \left( \left( 1+\frac{v^2}{c^2} \right)\frac{\partial E_x}{\partial x'} + \frac{v}{\gamma} \left( \frac{\partial E_y}{\partial y'}+\frac{\partial E_z}{\partial z'} \right) \right)=& \frac{\partial B_z}{\partial y}-\frac{\partial B_y}{\partial z} \end{align}$$
And since $1- v^2/c^2=1/\gamma^2$, it follows that
$$\frac{1}{c}\frac{\partial E_x}{\partial t'}= \gamma \frac{\partial }{\partial y'} \left[ B_z - \frac{v}{c}E_y \right] - \gamma \frac{\partial }{\partial z'} \left[ B_y + \frac{v}{c}E_z \right]$$
which is what was required. For one equation at least; I imagine the other three are fall out the same.

I don't know if there's a concise way to do this for all three equations in the curl at once, but the same applies to the both applicable Maxwell's equations in free space. This is straightforward, but long and did need more than just the mathematics to accomplish. The text didn't belie subtlety, just slogwork. Anyway, hopefully this is useful to someone in future.

It has always baffled me why anyone would care how Einstein did it 100 years ago. We've come a long way since then, and any current book on GR will convey to the reader a a simpler explanation and a better understanding.

One way or another might be more straightforward, but I find it's always useful to see something done in multiple ways.

 

[jtbell]

That was the first thing I tried, but I couldn't see how it would work.

When transforming the first equation of the untransformed set, I think you also need to take into account Gauss's Law for the electric field (in empty space), which in Einstein's notation reads:

$$\frac {\partial X}{\partial x} + \frac {\partial Y}{\partial y} + \frac {\partial Z}{\partial z} = 0$$

Transform the derivatives similarly, and combine the result with what you get from transforming the first equation. I've done enough of the algebra to convince myself that you get the correct derivatives in the final equation, but I haven't nailed down all the constants.

For some of the equations, you probably need to use Gauss's Law for magnetism instead.

I spent half an hour composing a reply showing some of the equations explicitly. Then I lost it all when I hit a wrong key, and I'm in no mood to do it all again.

(Aha, you posted your reply while I was going through all that! It looks like we hit upon the same solution.)

 

[dan_b_]

I found a great article which shows the omitted steps in Einstein'
derivation of the E-M fields transformations. I am pretty sure this is
what you are/were looking for. It's what I was looking for too.

Check out the article at:
h--p://fds.oup.com/www.oup.com/pdf/13/9780199694037.pdf
p41-47 (if you go by the pdf page numbers)
or p 146 to 150 (if you go by the page numbers marked on each page)

Cheers

 

[Gaba gaba]

Transforming Maxwell´s Equations in Special Relativity

Hello, I´m a new member of the Forum - glad to be here . I´m a brazilian man, so, I would like to apologize for any errors concerning the English language (grammar, spelling etc). I´m also studying the Einstein´s Relativity paper (1905) in my Masters degree course. My doubt (or question) is directed mainly to ObssessiveMathsFreak or, for anyone who can help me! Also I ask your pardon due to any mistake regarding the use of symbols, quotes, etc. The Lorentz-Einstein first transformed equation was skillfully done by ObssessiveMaths. But, it has been a long time since I study calculus. My question is: HOW you applied the "chain rule" in your LAST TWO equations in order to get the final result? When I say LAST TWO equations, I´m referring the last two BEFORE the final expression.

P.S.: I know how to apply the "chain rule" for simple derivatives, but I have forgotten the most "advanced" methods; or, may be, I´m confused.

 

[Gaba gaba]

Transforming Maxwell´s Equations in Special Relativity

I found a great article which shows the omitted steps in Einstein'
derivation of the E-M fields transformations. I am pretty sure this is
what you are/were looking for. It's what I was looking for too.

Hi, dan_b_!

I have tried to access this article address, as you said. But it returned a 404 error.
Please, do you know the article name and the author? I have remote access, here in Brazil,
to some good journals when logged in (via my institution).
I understood the ObssessiveMaths deduction only partly, and have been confused
in the final steps - concernig the "chain rule" application -, at the final equations. You could read
my last message (above). Perhaps, you can help me. If you have references to the article mentioned,
I would be grateful.



Gabriel.

 

[Fredrik]

Gaba gaba, there appear to be some typos in the part you're asking about:

This step gives us what we need. Now applying both the chain rules and this result to the curl equation gives
\begin{align}
\frac{\gamma}{c} \left( \frac{\partial E_x}{\partial t'} - v \frac{\partial E_x}{\partial x'}\right)=& \frac{\partial B_z}{\partial y}-\frac{\partial B_y}{\partial z} \\
\frac{\gamma}{c} \left( \left( 1+\frac{v^2}{c^2} \right)\frac{\partial E_x}{\partial x'} + \frac{v}{\gamma} \left( \frac{\partial E_y}{\partial y'}+\frac{\partial E_z}{\partial z'} \right) \right)=& \frac{\partial B_z}{\partial y}-\frac{\partial B_y}{\partial z}
\end{align}
And since $1- v^2/c^2=1/\gamma^2$, it follows that
$$\frac{1}{c}\frac{\partial E_x}{\partial t'}= \gamma \frac{\partial }{\partial y'} \left[ B_z - \frac{v}{c}E_y \right] - \gamma \frac{\partial }{\partial z'} \left[ B_y + \frac{v}{c}E_z \right]$$

The first line is just the x component of the equality in this quote:

In Gaussian units, the first component of $\nabla \times \mathbf{B} =\frac{1}{c}\frac{\partial \mathbf{E}}{\partial t}$ is explicitly

To get from the first line to the second, he's just using the result in this quote to rewrite the second term in the first line:

\begin{align}
\Rightarrow &\frac{\partial E_x}{\partial x'}=\frac{-1}{\gamma} \left( \frac{\partial E_y}{\partial y'}+\frac{\partial E_z}{\partial z'} \right)+\frac{v}{c^2}\frac{\partial E_x}{\partial t'}
\end{align}

The second line appears to contain two typos. I think it should be
$$\frac{\gamma}{c} \left( \left( 1-\frac{v^2}{c^2} \right)\frac{\partial E_x}{\partial t'} + \frac{v}{\gamma} \left( \frac{\partial E_y}{\partial y'}+\frac{\partial E_z}{\partial z'} \right) \right)= \frac{\partial B_z}{\partial y}-\frac{\partial B_y}{\partial z}$$ Then he's using the equality in this quote to rewrite the left-hand side of the first line:

And since $1- v^2/c^2=1/\gamma^2$, it follows that

And he "applies the chain rule" by using these two results to rewrite the right-hand side of the second line:

$$\begin{align} \frac{\partial }{\partial y}\equiv & \frac{\partial }{\partial y'} \\ \frac{\partial }{\partial z}\equiv & \frac{\partial }{\partial z'} \end{align}$$

So he ends up with this:

$$\frac{1}{c}\frac{\partial E_x}{\partial t'}= \gamma \frac{\partial }{\partial y'} \left[ B_z - \frac{v}{c}E_y \right] - \gamma \frac{\partial }{\partial z'} \left[ B_y + \frac{v}{c}E_z \right]$$
which is what was required. For one equation at least; I imagine the other three are fall out the same.