Personally I've gotten sick of Tach's blatant denial, deliberate obfuscation, and "demands" others do all his work for him. Since the weekend has started for me and I
haven't gone skiing, here's how one can calculate the energies in the top and bottom halves of the wheel, using the approach I suggested in my [POST=2848590]very first post in this thread[/POST]: transforming the stress-energy tensor (cue charges that Tach had
days to do this, and failed). Note that by "top" and "bottom" halves I mean what everyone with the possible exception of Tach (who seems to like keeping this ambiguous) means: everything above the axle, or above the strip of paper mentioned in the OP. So specifically, to calculate the energy of the top half, I'll just be integrating the energy density everywhere above the axle, which in the coordinates I'll be using, means integrating everywhere over
z > 0.
Before tansforming the stress-energy tensor, we actually need something to transform, so...
Stress-energy tensor in the axle frame
I'll map this coordinate system with the "primed" coordinates
x'[sup]0[/sup] =
t',
x'[sup]1[/sup] =
x', and
x'[sup]2[/sup] =
z'. The idea is that the wheel will be moving in the
x direction in the ground frame, and
z (=
z') is the vertical coordinate, with the axle at the origin, the ground at
z = -
R, and the top of the wheel at
z = +
R.
Let the wheel have a linear total (kinetic + rest) energy, or "relativistic mass" (since I'll be setting
c = 1) density of $$\rho$$ and be rotating in the axle frame with angular frequency $$\omega$$, so the velocity of a point on edge of the wheel is given by $$v_{x} = \omega z$$, $$v_{z} = -\omega x$$.
The
Stress-energy tensor is a unified way of describing energy, momentum, and momentum flux well suited to relativity (because the components of this tensor transform in a simple way under general Lorentz transformations). The $$T^{00}$$ component is the energy density in the frame under consideration, $$T^{0i} = T^{i0}$$ is linear momentum, and the components $$T^{ij} = T^{ji}$$ represent momentum
flux. In our case, these components are given by
$$
\begin{align}
T'^{00} \,&=\, \rho \delta(r - R) \,, \\
T'^{10} \,&=\, \rho \omega z \delta(r - R) \,, \\
T'^{20} \,&=\, - \rho \omega x \delta(r - R) \,, \\
T'^{11} \,&=\, \rho \omega^{2} z^{2} \delta(r - R) \,, \\
T'^{21} \,&=\, - \rho \omega^{2} x z \delta(r - R) \,, \\
T'^{22} \,&=\, \rho \omega^{2} x^{2} \delta(r - R) \,,
\end{align}
$$
where $$r^{2} = x'^{2} + z'^{2}$$, and the Dirac delta is included because I'm considering an infinitely thin wheel rim as an idealisation. Integrating $$T'^{00}$$, one can check that the total energy is $$2 \pi R \rho$$ as one would expect.
Transformation to ground frame
The energy density is the new component $$T^{00}$$ in the ground frame, related to the axle frame by the (inverse) Lorentz boost
$$
\begin{align}
t \,&=\, \gamma (t' + vx') \,, \\
x \,&=\, \gamma (x' + vt') \,, \\
z \,&=\, z' \,.
\end{align}
$$
In general we could consider any boost of any velocity, though for the ground frame, $$v = \omega R$$.
In order to transform the stress-energy tensor, we have to transform the components, which transform according to $$T^{\mu\nu} = \Lambda^{\mu}_{\;\rho} \Lambda^{\nu}_{\;\lambda} T'^{\rho\lambda}$$ where ($$\Lambda^{\mu}_{\;\rho}$$ are the matrix elements of the Lorentz boost above), and re-express the resulting quantity in terms of the ground frame coordinates. In our case, re-expressing things in terms of the ground frame coordinates just means re-expressing $$r$$ in the Dirac delta as $$r^{2} = \gamma^{2} (x - vt)^{2} + z^{2}$$, And we're only interested in calculating the transformed energy density, given in terms of non-zero contributions by
$$
T^{00} \,=\, \bigl(\Lambda^{0}_{\;0}\bigr)^{2} T'^{00} \,+\, 2\Lambda^{0}_{\;1} \Lambda^{0}_{\;0} T'^{10} \,+\, \bigl(\Lambda^{0}_{\;1}\bigr)^{2} T'^{11} \,.
$$
Substituting and simplifying, one finds
$$
T^{00} \,=\, \gamma^{2} \rho (1 + v \omega z)^{2} \delta(r - R)
$$
with $$r^{2} = \gamma^{2} (x - vt)^{2} + z^{2}$$. Notice that we already see the asymmetry appear in the energy distribution, since $$(1 + v \omega z)^{2}$$ is higher in the top half of the wheel (where
z is positive) than in the bottom half (
z negative). But of course that won't be enough for Tach, so...
Energy of the upper and lower halves
This means calculating the inegral of $$T^{00}$$ seperately over
z >0 (for the upper half) and
z < 0 (for the lower half):
$$
\begin{align}
E_{\mathrm{upper}} \,&=\, \int_{-\infty}^{\infty} \mathrm{d}x \int_{\;0}^{\infty} \mathrm{d}z \, T^{00} \,, \\
E_{\mathrm{lower}} \,&=\, \int_{-\infty}^{\infty} \mathrm{d}x \int_{-\infty}^{0} \mathrm{d}z \, T^{00} \,.
\end{align}
$$
The easiest way to calculate these is to express things in terms of $$r$$ and an angular variable $$\theta$$ defined such that
$$
\begin{align}
\gamma(x - vt) \,&=\, r \cos(\theta) \,, \\
z \,&=\, r \sin(\theta) \,.
\end{align}
$$
Then the measure changes to $$\mathrm{d}x \mathrm{d}z = \frac{r}{\gamma} \mathrm{d}r \mathrm{d} \theta$$. For $$E_{\mathrm{upper}}$$, one finds
$$
\begin{align}
E_{\mathrm{upper}} \,&=\, \int_{0}^{\infty} \frac{r}{\gamma} \mathrm{d}r \int_{0}^{\pi} \mathrm{d}\theta \, T^{00} \\
\,&=\, \int_{0}^{\infty} \frac{r}{\gamma} \mathrm{d}r \int_{0}^{\pi} \mathrm{d}\theta \, \gamma^{2} \rho \, \bigl(1 \,+\, v \omega r \sin(\theta)\bigr)^{2} \, \delta(r - R) \\
\,&=\, \gamma R \rho \int_{0}^{\pi} \mathrm{d}\theta \, \bigl(1 \,+\, v \omega R \sin(\theta)\bigr)^{2} \,.
\end{align}
$$
Getting a result is just a matter of evaluating the integral, which isn't difficult. In general,
$$
\int \mathrm{d}\theta \, \bigl( 1 \,+\, \alpha \sin(\theta) \bigr)^{2} \,=\, \Bigl( 1 \,+\, \frac{\alpha^{2}}{2} \Bigr) \theta \,-\, 2 \alpha \cos(\theta) \,-\, \frac{\alpha^{2}}{4} \sin(2\theta) \,+\, C \,.
$$
Using this, the final result is
$$
E_{\mathrm{upper}} \,=\, \gamma \pi R \rho \Bigl( 1 \,+\, \frac{\omega^{4} R^{4}}{2} \Bigr) \,+\, 4 \gamma \omega^{2} R^{3} \rho \,,
$$
where I've finally substituted in $$v = \omega R$$. Similarly, for the lower half, the result is
$$
E_{\mathrm{lower}} \,=\, \gamma \pi R \rho \Bigl( 1 \,+\, \frac{\omega^{4} R^{4}}{2} \Bigr) \,-\, 4 \gamma \omega^{2} R^{3} \rho \,.
$$
Clearly, $$E_{\mathrm{upper}} > E_{\mathrm{lower}}$$ except in the case where $$\omega = 0$$, with a difference of $$E_{\mathrm{upper}} - E_{\mathrm{lower} = 8 \gamma \omega^{2} R^{3}$$.
All this just to show something that was pretty much self evident to everyone except Tach right from the beginning: in the ground frame, more of the energy associated with the rolling wheel is concentrated above the axle (and strip of paper mentioned in the OP) than below.
Cue Tach's last ditch attempt to redefine the problem...
