Acceleration in Special Relativity

[Fascheue]

TL;DR

What (if any) examples are there of acceleration problems being solved with SR?

Can special relativity handle acceleration? I believe the answer is yes, but I don’t recall dealing with any acceleration problems when I took SR. I remember using the time dilation, length contraction and Lorentz transformation equations. These equations all assume constant motion iirc.

 

[anuttarasammyak]

As an example, as for space rockets accelerating in a same pace, i.e. space pilots keep feeling constant 'gravity' force, SR gives their time dilation and rocket length contraction observed by Earth people changing with Earth time.

More in general for any motion of a body in IFR, SR gives its time dilation and length contraction observed in the IFR. Because any motion is expressed as time integrated infinitesimal parts of translation motions that may have infinitesimal different velocities. Integration of individual SR effect on these parts give the answers.

 

[PeterDonis]

Can special relativity handle acceleration? I believe the answer is yes

Yes, as long as spacetime is flat (i.e., there are no gravitating masses present).

I don’t recall dealing with any acceleration problems when I took SR

Not all SR courses cover such problems, true. IMO they should, to make clear that it can be done. Simple relativistic rocket problems (e.g., the ones discussed here) would be enough.

 

[anuttarasammyak]

Another simple example is rotation in IFR. After one turn of Merry-go-round of time T on the Earth, parents on the Earth find children has grown up shorter time of
T\ \sqrt{1-v^2/c^2}=T\ \sqrt{1-r^2\omega^2/c^2}
as if v is of translation motion though it is tangential. Here r is radius and $\omega$ is angular velocity of merry-go-round.

 

[PeroK]

Summary:: What (if any) examples are there of acceleration problems being solved with SR?

Can special relativity handle acceleration? I believe the answer is yes, but I don’t recall dealing with any acceleration problems when I took SR. I remember using the time dilation, length contraction and Lorentz transformation equations. These equations all assume constant motion iirc.

Someone posted this homework problem yesterday:

https://www.physicsforums.com/threads/express-this-trajectory-in-terms-of-proper-time.994906/

 

[vanhees71]

Of course can SR handle acceleration. Why shouldn't it?

I think the most simple example is the motion of a particle in an electromagnetic field, neglecting the radiation reaction. One should always start writing the mechanics in manifestly covariant form, using four-vectors. With the proper time $\tau$ of the particle you get
$$m \ddot{x}^{\mu}=K^{\mu},$$
where dots are derivatives wrt. proper time and $K^{\mu}$ is the socalled Minkowski force, which obviously is a four-vector since $m$ is a scalar (invariant mass of the particle) and $x^{\mu}$ is a four-vector and $\tau$ a scalar.

Now obviously
$$\dot{x}_{\mu} \dot{x}^{\mu}=c^2=\text{const}. \qquad (*)$$
Taking the derivative of this equation wrt. to $\tau$ you get
$$\ddot{x}^{\mu} \dot{x}_{\mu}=0 \; \Rightarrow\; K^{\mu} \dot{x}_{\mu}=0.$$
The most simple equation of motion is thus given by using an antisymmetric 2nd-rank tensor field $F_{\mu \nu}(x)$ and writing
$$K^{\mu}=\frac{q}{c} F^{\mu \nu} \dot{x}_{\nu},$$
where $q$ is another scalar. This is indeed the Minkowski force for a particle in an electromagnetic field, which is given by the antisymmetric tensor field (which has 6 components, corresponding to the electric and magnetic field components of the electromagnetic field):
$$F^{j0}=E_j, \quad F^{jk}=-\epsilon^{jkl} B_l.$$
Here $j,k,l \in \{1,2,3\}$.

Due to the given constraints following (*), you only need to consider the spatial part of the equation of motion. The time component then is automatically fulfilled. So let's write the spatial part of the equations first:
$$m \ddot{x}^j=\frac{q}{c} F^{j\nu} \dot{x}_{\nu}=\frac{q}{c} (F^{j0} c \dot{t}-F^{jk} \dot{x}_k) = \frac{q}{c} (E^j \gamma + \epsilon^{jkl} B_l \dot{x}_k)$$
or in vector notation
$$m \ddot{\vec{x}}=q \gamma\left (\vec{E} +\frac{\vec{v}}{c} \times \vec{B} \right).$$
Here I used $\gamma=\mathrm{d} t/\mathrm{d} \tau=1/\sqrt{1-\vec{\beta}^2}$ with $\vec{\beta}=\vec{v}=\mathrm{d}_t \vec{x})$.

Writing $\vec{p}=m \gamma \vec{v}$ you get
$$m \dot{\vec{p}}=q \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right)$$
as expected.

You can check that the time component is automatically fulfilled and is just the work-energy theorem.

 

[George Jones]

Summary:: What (if any) examples are there of acceleration problems being solved with SR?

Can special relativity handle acceleration? I believe the answer is yes, but I don’t recall dealing with any acceleration problems when I took SR.

Not all SR courses cover such problems, true. IMO they should, to make clear that it can be done. Simple relativistic rocket problems (e.g., the ones discussed here) would be enough.

The special relativity section of the mechanics course that I took in second-year covered this in its special relativity section. As an example, we calculated how much time would on an astronaut's watch during a trip to the centre of our galaxy at constant acceleration $g$. Can't remember if the astronaut stopped at the galaxy's centre, or if it was a fly-by.

 

[etotheipi]

$$\dot{x}_{\mu} \dot{x}^{\mu}=c^2=\text{const}. \qquad (*)$$
Taking the derivative of this equation wrt. to $\tau$ you get
$$\ddot{x}^{\mu} \dot{x}_{\mu}=0 \; \Rightarrow\; K^{\mu} \dot{x}_{\mu}=0.$$

I have a maths question, how did you take the derivative there? I tried$$d_{\tau} (\dot{x}_{\mu} \dot{x}^{\mu}) = \ddot{x}_{\mu} \dot{x}^{\mu} + \dot{x}_{\mu} \ddot{x}^{\mu} = 0$$Then I try to use $g^{ab}g_{bc} = \delta^{a}_c$, but in doing so I repeat the summation index $\mu$ four times (which is a sign of a mistake )$$d_{\tau} (\dot{x}_{\mu} \dot{x}^{\mu}) = \dot{x}_{\mu} \ddot{x}^{\mu} + \ddot{x}_{\mu}g^{\mu \nu} g_{\mu \nu} \dot{x}^{\mu} = 2\dot{x}_{\mu} \ddot{x}^{\mu} = 0$$ $$\dot{x}_{\mu}\ddot{x}^{\mu} = 0$$

 

[PeroK]

I have a maths question, how did you take the derivative there? I tried$$d_{\tau} (\dot{x}_{\mu} \dot{x}^{\mu}) = \ddot{x}_{\mu} \dot{x}^{\mu} + \dot{x}_{\mu} \ddot{x}^{\mu} = 0$$

For any four-vectors $\mathbf a$ and $\mathbf b$:
$$\mathbf a \cdot \mathbf b = a_{\mu}b^{\mu} = a^{\mu}b_{\mu} = g^{\mu \nu}a_{\nu}b_{\mu} \dots$$

 

[Ibix]

I find it helpful to break these things down into single elements and then muck around with those single elements. Also, note that the $\mu$ are not free indices, so you can change labels to suit yourself. So write $\dot x^\mu=g^{\mu\nu}\dot x_\nu$ and $\ddot x_\mu=g_{\mu\rho}\ddot x^\rho$. That gives you $\dot x^\mu\ddot x_\mu=g^{\mu\nu}\dot x_\nu g_{\mu\rho}\ddot x^\rho=\delta^\nu_\rho\dot x_\nu\ddot x^\rho=\dot x_\nu\ddot x^\nu=\dot x_\mu\ddot x^\mu$.

I won't tell you how many times I had to edit that before I got all the indices to match...

 

[etotheipi]

For any four-vectors $\mathbf a$ and $\mathbf b$:
$$\mathbf a \cdot \mathbf b = a_{\mu}b^{\mu} = a^{\mu}b_{\mu} = g^{\mu \nu}a_{\nu}b_{\mu} \dots$$

Fair enough, thanks! Actually that does make sense, since the inner product should be symmetric $a \cdot b = b \cdot a$

I find it helpful to break these things down into single elements and then muck around with those single elements. Also, note that the $\mu$ are not free indices, so you can change labels to suit yourself. So write $\dot x^\mu=g^{\mu\nu}\dot x_\nu$ and $\ddot x_\mu=g_{\mu\rho}\ddot x^\rho$. That gives you $\dot x^\mu\ddot x_\mu=g^{\mu\nu}\dot x_\nu g_{\mu\rho}\ddot x^\rho=\delta^\nu_\rho\dot x_\nu\ddot x^\rho=\dot x_\nu\ddot x^\nu=\dot x_\mu\ddot x^\mu$.

Thanks, I see - I was being too frugal with my indices.

I won't tell you how many times I had to edit that before I got all the indices to match...

I mean, what's so tricky about...

\dot x^\mu\ddot x_\mu=g^{\mu\nu}\dot x_\nu g_{\mu\rho}\ddot x^\rho=\delta^\nu_\rho\dot x_\nu\ddot x^\rho=\dot x_\nu\ddot x^\nu=\dot x_\mu\ddot x^\mu

 

[Ibix]

I mean, what's so hard about...

\dot x^\mu\ddot x_\mu=g^{\mu\nu}\dot x_\nu g_{\mu\rho}\ddot x^\rho=\delta^\nu_\rho\dot x_\nu\ddot x^\rho=\dot x_\nu\ddot x^\nu=\dot x_\mu\ddot x^\mu

It's all Greek to me. Apart from the xs and the gs, of course.

 

[etotheipi]

It's all Greek to me. Apart from the xs and the gs, of course.

You know, sometimes I really wish there was a "groan" emoji

 

[Ibix]

Thanks, I see - I was being too frugal with my indices.

I have to admit I struggle with it sometimes. I find that sometimes thinking about what it looks like with explicit sums helps. Sometimes just putting in brackets and doing things in painful one-step-at-a-time detail (with liberal use of copy-paste) helps:$$\begin{eqnarray*}
a_\mu b^\mu&=&\left(a_\mu\right)\left(b^\mu\right)\\
&=&\left(g_{\mu\rho}a^\rho\right)\left(b^\mu\right)\\
&=&\left(g_{\mu\rho}a^\rho\right)\left(g^{\mu\sigma}b_\sigma\right)\\
&=&\left(g_{\mu\rho}g^{\mu\sigma}\right)a^\rho b_\sigma\\
&=&\left(\delta^\sigma_\rho a^\rho\right) b_\sigma\\
&=&a^\sigma b_\sigma\\
\end{eqnarray*}$$

 

[DrGreg]

I have to admit I struggle with it sometimes. I find that sometimes thinking about what it looks like with explicit sums helps. Sometimes just putting in brackets and doing things in painful one-step-at-a-time detail (with liberal use of copy-paste) helps:$$\begin{eqnarray*}
a_\mu b^\mu&=&\left(a_\mu\right)\left(b^\mu\right)\\
&=&\left(g_{\mu\rho}a^\rho\right)\left(b^\mu\right)\\
&=&\left(g_{\mu\rho}a^\rho\right)\left(g^{\mu\sigma}b_\sigma\right)\\
&=&g_{\mu\rho}a^\rho g^{\mu\sigma}b_\sigma\\
&=&\delta^\sigma_\rho a^\rho b_\sigma\\
&=&a^\rho b_\rho\
\end{eqnarray*}$$

For what it's worth, there's a slightly faster method:$$
a_\mu b^\mu
= \left( a^\rho g_{\rho \mu} \right) b^\mu
= a^\rho \left( g_{\rho \mu} b^\mu \right)
= a^\rho b_\rho\
$$

 

[Fascheue]

Of course can SR handle acceleration. Why shouldn't it?

I think the most simple example is the motion of a particle in an electromagnetic field, neglecting the radiation reaction. One should always start writing the mechanics in manifestly covariant form, using four-vectors. With the proper time $\tau$ of the particle you get
$$m \ddot{x}^{\mu}=K^{\mu},$$
where dots are derivatives wrt. proper time and $K^{\mu}$ is the socalled Minkowski force, which obviously is a four-vector since $m$ is a scalar (invariant mass of the particle) and $x^{\mu}$ is a four-vector and $\tau$ a scalar.

Now obviously
$$\dot{x}_{\mu} \dot{x}^{\mu}=c^2=\text{const}. \qquad (*)$$
Taking the derivative of this equation wrt. to $\tau$ you get
$$\ddot{x}^{\mu} \dot{x}_{\mu}=0 \; \Rightarrow\; K^{\mu} \dot{x}_{\mu}=0.$$
The most simple equation of motion is thus given by using an antisymmetric 2nd-rank tensor field $F_{\mu \nu}(x)$ and writing
$$K^{\mu}=\frac{q}{c} F^{\mu \nu} \dot{x}_{\nu},$$
where $q$ is another scalar. This is indeed the Minkowski force for a particle in an electromagnetic field, which is given by the antisymmetric tensor field (which has 6 components, corresponding to the electric and magnetic field components of the electromagnetic field):
$$F^{j0}=E_j, \quad F^{jk}=-\epsilon^{jkl} B_l.$$
Here $j,k,l \in \{1,2,3\}$.

Due to the given constraints following (*), you only need to consider the spatial part of the equation of motion. The time component then is automatically fulfilled. So let's write the spatial part of the equations first:
$$m \ddot{x}^j=\frac{q}{c} F^{j\nu} \dot{x}_{\nu}=\frac{q}{c} (F^{j0} c \dot{t}-F^{jk} \dot{x}_k) = \frac{q}{c} (E^j \gamma + \epsilon^{jkl} B_l \dot{x}_k)$$
or in vector notation
$$m \ddot{\vec{x}}=q \gamma\left (\vec{E} +\frac{\vec{v}}{c} \times \vec{B} \right).$$
Here I used $\gamma=\mathrm{d} t/\mathrm{d} \tau=1/\sqrt{1-\vec{\beta}^2}$ with $\vec{\beta}=\vec{v}=\mathrm{d}_t \vec{x})$.

Writing $\vec{p}=m \gamma \vec{v}$ you get
$$m \dot{\vec{p}}=q \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right)$$
as expected.

You can check that the time component is automatically fulfilled and is just the work-energy theorem.

Can you dumb down the four-vector stuff for me? Can you not solve these problems with F = dp/dt, where p is the standard 3-dimensional xyz momentum vector?

The only familiarity I have with with - what I assume are - four vectors, is solving problems using spacetime intervals. They had a fourth time dependent element iirc.

 

[etotheipi]

Can you not solve these problems with F = dp/dt, where p is the standard 3-dimensional xyz momentum vector?

Yeah! Well, actually I don't know, I read somewhere that the four-force only really works for electromagnetic forces [and that people tend to use Lagrangians anyway, because apparently forces in SR are disgusting ], but if you play around with the formulae (using @vanhees71's notation)$$K^{\mu} = \frac{dP^{\mu}}{d\tau}$$if you just take the spatial part, i.e. the spatial part of the four force is $\gamma \vec{f}$ (with $\vec{f}$ the 3-force) and the spatial part of the 4-mentum is $\vec{p} = m\vec{u} = \gamma m\vec{v}$ then on division by $\gamma$ you get$$\vec{f} = \frac{1}{\gamma} \frac{d\vec{p}}{d\tau} = \frac{d\vec{p}}{dt} = \frac{d(\gamma m \vec{v})}{dt}$$

 

[PeterDonis]

Can you dumb down the four-vector stuff for me? Can you not solve these problems with F = dp/dt, where p is the standard 3-dimensional xyz momentum vector?

If you pick a specific inertial frame and stick to it, then yes, you can solve these problems using $F = dp / dt$ in that frame. But your results will be specific to that frame. (For example, the equations in the relativistic rocket FAQ I linked to in post #3 are derived using $F = dp / dt$, under the assumption that the acceleration felt by the crew of the rocket is constant, using an inertial frame in which the Earth--the rocket's starting point--is at rest.)

The big advantage of learning the 4-vector method is that it gives you the equations in frame-independent form; you can then easily apply them to any frame.

The only familiarity I have with with - what I assume are - four vectors, is solving problems using spacetime intervals.

Spacetime intervals aren't quite the same as 4-vectors, but they are similar.

 

[vanhees71]

Can you dumb down the four-vector stuff for me? Can you not solve these problems with F = dp/dt, where p is the standard 3-dimensional xyz momentum vector?

The only familiarity I have with with - what I assume are - four vectors, is solving problems using spacetime intervals. They had a fourth time dependent element iirc.

That is the final equation in my previous posting and in @etotheipi 's posting #17.

Of course for other interactions than the em. field it's a bit more complicated, but in principle you can build interactions via any kind of fields (usually one considers scalar and vector fields only though).

You can always write the EoM. in terms of a Lagrangian that is equivalent to a Poincare invariant action. To make it manifestly covariant you also want to have it also invariant under changes of the world-line parameter, which leads to the idea to look for a Lagrangian that's 1st-order homogeneous in the $\dot{x}^{\mu}$ (where the dot now refers to the derivative wrt. the arbitrary world-line parameter). For the em. field this naturally leads to
$$L=-m c^2 \sqrt{\dot{x}^{\mu} \dot{x}_{\mu}}-\frac{q}{c} A_{\mu}(x) \dot{x}^{\mu}.$$
For a scalar field you can use
$$L=-mc^2 \sqrt{\dot{x}^{\mu} \dot{x}_{\mu}}-g \sqrt{\dot{x}^{\mu} \dot{x}_{\mu}} \Phi(x).$$
A special choice then always is to use some inertial reference frame and use the corresponding (coordinate) time as the world-line parameter. Then you get a Lagrangian and EoMs in the usual form as 2nd-order coupled ODE's of the spatial components $\vec{x}$ wrt. this inertial frame.

 

[etotheipi]

Figured since there was some discussion of index notation above I may as well tag another similar question onto this thread... when I try to find the momentum conjugate to $\dot{x}^{\mu}$ (in cartesian coordinates) at one point you need to differentiate the sum $-\eta_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}$ with respect to $\dot{x}^{\mu}$. Taking the advice above about renaming dummy indices as you like, I tried$$\frac{d}{d\dot{x}^{\mu}} (-\eta_{\rho \sigma} \dot{x}^{\rho} \dot{x}^{\sigma}) = - \eta_{\rho \sigma} \left(\frac{d\dot{x}^{\rho}}{d\dot{x}^{\mu}}\dot{x}^{\sigma} + \frac{d\dot{x}^{\sigma}}{d\dot{x}^{\mu}} \dot{x}^{\rho}\right)$$I'm not sure whether Einstein notation dictates that I need to show the $\eta_{\rho \sigma}$ distributed with each term? Next, evidently those two terms in the brackets will be exactly the same, since $\sigma$ and $\rho$ run over the same indices so$$\frac{d}{d\dot{x}^{\mu}} (-\eta_{\rho \sigma} \dot{x}^{\rho} \dot{x}^{\sigma}) = -2 \eta_{\rho \sigma} \left(\frac{d\dot{x}^{\rho}}{d\dot{x}^{\mu}}\dot{x}^{\sigma} \right)$$and in the sum inside the brackets, $\frac{d\dot{x}^{\rho}}{d\dot{x}^{\mu}} = \delta^{\rho}_{\mu}$ (in this case) which implies$$\frac{d}{d\dot{x}^{\mu}} (-\eta_{\rho \sigma} \dot{x}^{\rho} \dot{x}^{\sigma}) = -2\eta_{\mu \sigma} \dot{x}^{\sigma}$$That seems like quite a long process to work out a simple derivative, so I wondered if there is a faster way (or some tricks of the trade that people know to make this less cumbersome)? The notes I was using didn't show any steps for the differentiation, so I guess they managed to do it in their head somehow?

 

[PeroK]

Figured since there was some discussion of index notation above I may as well tag another similar question onto this thread... when I try to find the momentum conjugate to $\dot{x}^{\mu}$ (in cartesian coordinates) at one point you need to compute $\frac{d}{d\dot{x}^{\epsilon}} (-\eta_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu})$.

I think you need a partal derivative there. And you can't have three $\mu$ indices.

 

[etotheipi]

I think you need a partal derivative there.

Oh, right yes, because $\mathcal{L} = \mathcal{L}(x^{\mu}, \dot{x}^{\mu})$.

 

[PeroK]

so I wondered if there is a faster way (or some tricks of the trade that people know to make this less cumbersome)? The notes I was using didn't show any steps for the differentiation, so I guess they managed to do it in their head somehow?

Using the metric tensor is the trick. Take, for example, the Proca Lagrangian:
$$\mathcal L = -\frac 1 4 F^{\mu \nu}F_{\mu \nu} + \frac {m^2}{2}A^{\nu}A_{\nu}$$ Where $$F^{\mu \nu} = \partial^{\mu}A^{\nu} - \partial^{\nu}A^{\mu}$$ We need to calculate: $$\frac{\partial \mathcal L}{\partial A_{\nu}} \ \ \text{and} \ \ \frac{\partial \mathcal L}{\partial (\partial_{\mu}A_{\nu})}$$ You can battle through that, noting that $\mathcal L$ is a function of sixteen partial derivatives and four vector components. Or, the quicker way is to use the metric tensor to rewrite the terms as: $$A^{\nu}A_{\nu} = \eta^{\nu \rho}A_{\rho}A_{\nu}$$ And also use a different dummy index, replacing $\nu$ with $\alpha$: $$\frac{\partial \mathcal L}{\partial A_{\nu}} = \frac{m^2}{2}\eta^{\alpha \rho}\frac{\partial}{\partial A_{\nu}} (A_{\rho}A_{\alpha}) = m^2A^{\nu}$$ Using the same approach, you can show that: $$\frac{\partial \mathcal L}{\partial (\partial_{\mu}A_{\nu})} = -F^{\mu \nu}$$
And, yes, a number of texts just quote the answer but don't indicate how the differentiation is best attempted.

 

[etotheipi]

Okay I tried the next one, not sure if I've mucked up but I define $C_{\mu \nu}:= \partial_{\mu} A_{\nu}$ and then $\partial^{\mu} A^{\nu} = \eta^{\mu \sigma} \eta^{\nu \rho} C_{\sigma \rho}$, and by extension$$F^{\mu \nu} = \eta^{\mu \sigma} \eta^{\nu \rho}(C_{\sigma \rho} - C_{\rho \sigma})$$ and also$$F_{\alpha \beta} = \delta^{\sigma}_{\alpha} \delta^{\rho}_{\beta} (C_{\sigma \rho} - C_{\rho \sigma}) = C_{\alpha \beta} - C_{\beta \alpha}$$then $$\frac{\partial}{\partial C_{\alpha \beta}} \left( -\frac{1}{4} F^{\mu \nu} F_{\mu \nu}\right) = -\frac{1}{4} \eta^{\mu \sigma} \eta^{\nu \rho} \frac{\partial}{\partial C_{\alpha \beta}} \left( C_{\sigma \rho} C_{\mu \nu} - C_{\rho \sigma} C_{\mu \nu} - C_{\sigma \rho} C_{\nu \mu} + C_{\rho \sigma}C_{\nu \mu} \right)$$Does that look okay so far? I'll try and finish it when I get back

 

[PeroK]

Okay I tried the next one, not sure if I've mucked up but I define $C_{\mu \nu}:= \partial_{\mu} A_{\nu}$ and then $\partial^{\mu} A^{\nu} = \eta^{\mu \sigma} \eta^{\nu \rho} C_{\sigma \rho}$, and by extension$$F^{\mu \nu} = \eta^{\mu \sigma} \eta^{\nu \rho}(C_{\sigma \rho} - C_{\rho \sigma})$$ and also$$F_{\alpha \beta} = \delta^{\sigma}_{\alpha} \delta^{\rho}_{\beta} (C_{\sigma \rho} - C_{\rho \sigma}) = C_{\alpha \beta} - C_{\beta \alpha}$$then $$\frac{\partial}{\partial C_{\alpha \beta}} \left( -\frac{1}{4} F^{\mu \nu} F_{\mu \nu}\right) = -\frac{1}{4} \eta^{\mu \sigma} \eta^{\nu \rho} \frac{\partial}{\partial C_{\alpha \beta}} \left( C_{\sigma \rho} C_{\mu \nu} - C_{\rho \sigma} C_{\mu \nu} - C_{\sigma \rho} C_{\nu \mu} + C_{\rho \sigma}C_{\nu \mu} \right)$$Does that look okay so far? I'll try and finish it when I get back

Yes, although I don't know that it will help to have multiplied out the two terms.

 

[etotheipi]

Ohh, alright, I'm being slightly thick again. In that case,$$\begin{align*}\frac{\partial}{\partial C_{\alpha \beta}} \left( -\frac{1}{4} F^{\mu \nu} F_{\mu \nu}\right) &= -\frac{1}{4} \eta^{\mu \sigma} \eta^{\nu \rho} \frac{\partial}{\partial C_{\alpha \beta}} \left((C_{\sigma \rho} - C_{\rho \sigma})(C_{\mu \nu} - C_{\nu \mu})\right) \\

&= -\frac{1}{4} \eta^{\mu \sigma} \eta^{\nu \rho} \cdot 2(C_{\sigma \rho} - C_{\rho \sigma})(\delta^{\alpha}_{\mu} \delta^{\beta}_{\nu} - \delta^{\alpha}_{\nu} \delta^{\beta}_{\mu})\\

&= - \eta^{\mu \sigma} \eta^{\nu \rho} (C_{\sigma \rho} - C_{\rho \sigma}) = -F^{\mu \nu}\end{align*}$$Yeah... I think I need a bit more practice . Thanks for the example!

 

[vanhees71]

I think, it's easier to work directly with the variations in such cases of multi-indexed quantities:
$$\delta (F^{\mu \nu} F_{\mu \nu})=2 F^{\mu \nu} \delta F_{\mu \nu}=4 F^{\mu \nu} \delta (\partial_{\mu} A_{\nu})$$
Now doing an integration by parts for the action,
$$S[A]=\int \mathrm{d}^4 x \left (-\frac{1}{4} F^{\mu \nu} F_{\mu \nu}-j^{\nu} A_{\nu} \right),$$
you get
$$\delta S[A]=\int \mathrm{d}^4 x \delta A_{\nu} \left (+\partial_{\mu} F^{\mu \nu} - j^{\nu} \right )\stackrel{!}{=}0 \; \Rightarrow \; \partial_{\mu} F^{\mu \nu} = j^{\nu}.$$
These are the inhomogeneous Maxwell equations as desired.