vanhees71 said:
I'm not so sure about what you wrote about the transformation properties of the em. current.
Why is that? I started with the transformation law of a covariant vector field with respect to the
group of general coordinate transformations. That is
<br />
\bar{J}^{\mu}( \bar{x}) = \frac{\partial \bar{x}^{\mu}}{\partial x^{\nu}} \ J^{\nu}(x). \ \ (1)<br />
For general
infinitesimal coordinates transformation,
\bar{x}^{\mu} = x^{\mu} + \delta x^{\mu},
I wrote eq(1) as
<br />
\bar{J}^{\mu}( x + \delta x ) = J^{\nu}(x) \frac{\partial}{\partial x^{\nu}}\left( x^{\mu} + \delta x^{\mu}\right). \ \ (2)<br />
Since,
<br />
\frac{\partial \bar{x}^{\mu}}{\partial x^{\nu}} = \frac{\partial}{\partial x^{\nu}}\left( x^{\mu} + \delta x^{\mu}\right) = \delta^{\mu}_{\nu} + \partial_{\nu}\left( \delta x^{\mu}\right),<br />
I can rewrite eq(2) as
<br />
\bar{J}^{\mu}(x + \delta x ) = J^{\mu}(x) + J^{\nu}(x) \partial_{\nu}\left( \delta x^{\mu}\right). <br />
Since, \partial_{\nu}J^{\nu} = 0, I can write the above equation in the form,
<br />
\bar{J}^{\mu}(x + \delta x ) = J^{\mu}(x) + \partial_{\nu}\left( \delta x^{\mu}J^{\nu}(x) \right). \ \ (3)<br />
Expanding the LEFT HAND SIDE of eq(3), to first order in \delta x, leads to
<br />
\bar{J}^{\mu}( x + \delta x ) \approx \bar{J}^{\mu}(x) + \delta x^{\rho}\partial_{\rho}\bar{J}^{\mu}(x). \ \ (4)<br />
Now, I claim that,
\delta x^{\rho}\partial_{\rho}\bar{J}^{\mu}(x) \approx \delta x^{\rho} \partial_{\rho}J^{\mu}(x). \ \ (5)
Why? Well, the calculus of infinitesimals tells you so. If the two functions, \bar{J}(x) and J(x), differ by an infinitesimal \omega, i.e., if
\bar{J}(x) \approx J(x) + \mathcal{O}(\omega ),
then,
\omega \ \bar{J}(x) \approx \omega \ J(x) + \mathcal{O}( \omega^{2}).
Thus, to first order in \omega, we write
\omega \ \bar{J}(x) \approx \omega \ J(x),
which proves the claim in eq(5). The same is true for J(\bar{x}) and J(x), i.e. we can, to the first order, write
\omega \ J(\bar{x}) \approx \omega \ J(x). \ \ \ (R)
Ok, now putting eq(4) and eq(5) into eq(3) leads to
<br />
\bar{J}^{\mu}(x) - J^{\mu}(x) = \partial_{\nu}\left( \delta x^{\mu}J^{\nu}(x) \right) - \delta x^{\rho}\partial_{\rho}J^{\mu}(x). \ \ (6)<br />
To specialized eq(6) to infinitesimal Lorentz transformation,
\delta x^{\mu} = \omega^{\mu}{}_{\nu}x^{\nu},
I use the fact (which follows from \omega_{\mu\nu}= -\omega_{\nu\mu}) that
\partial_{\nu}\left( \delta x^{\nu}\right) = 0. \ \ (7)
Using eq(7) in eq(6), we arrive at the infinitesimal LORENTZ transformation of the CONSERVED vector current
<br />
\delta J^{\mu}(x) = \partial_{\nu}\left( \delta x^{\mu}J^{\nu}(x) - \delta x^{\nu}J^{\mu}(x) \right). \ \ (8)<br />
It's a four-vector field. So its transformation property is
$$\bar{J}^{\mu}(\bar{x})={\Lambda^{\mu}}_{\nu} J^{\nu}(x),$$
where
This Lorentz transformation follows from the general case of eq(1), when you specify
\frac{\partial \bar{x}^{\mu}}{\partial x^{\nu}} = \Lambda^{\mu}{}_{\nu}.
$$\bar{x}=\Lambda x$$
and \Lambda is the Lorentz-transformation matrix.
which means that YOU can write
\Lambda = \frac{\partial \bar{x}}{\partial x}.
For an infinitesimal Lorentz transformation you have
{\Lambda^{\mu}}_{\nu} =\delta_{\nu}^{\mu}+{\delta \omega^{\mu}}_{\nu} with \delta \omega_{\mu \nu}=g_{\mu \rho} {\delta \omega^{\rho}}_{\nu} antisymmetric.
YOU can write this as,
\frac{\partial \bar{x}^{\mu}}{\partial x^{\nu}} = \frac{\partial x^{\mu}}{\partial x^{\nu}} + \omega^{\mu}{}_{\nu},
or,
<br />
\omega^{\mu}{}_{\nu} = \frac{\partial}{\partial x^{\nu}}\left( \bar{x}^{\mu} - x^{\mu}\right) = \frac{\partial}{\partial x^{\nu}}\left( \omega^{\mu}{}_{\rho}x^{\rho}\right).<br />
So, YOU can write YOUR infinitesimal Lorentz transformation in the form
\bar{x}^{\mu} = x^{\mu} + \delta x^{\mu},
with
\delta x^{\mu} = \omega^{\mu}{}_{\nu}x^{\nu},
so that YOUR Lorentz condition, \omega_{\mu\nu}= - \omega_{\nu\mu}, become equivalent to
\partial_{\mu}\left( \delta x^{\mu}\right) = 0.
This gives
$$\bar{J}^{\mu}(\bar{x})=J^{\mu}(x)+{\delta \omega^{\mu}}_{\nu} J^{\nu}(x) = J^{\mu}(x')-{\delta \omega^{\rho}}_{\sigma} \bar{x}^{\sigma} \bar{\partial}_{\rho} J^{\mu}(\bar{x})+{\delta \omega^{\mu}}_{\nu} J^{\nu}(x).$$
This equation is correct but UGLY. This is why it doesn’t look like eq(8). OK, let us turn it into something BEAUTIFUL. First instead of YOUR \delta \omega, I will use \omega; it is the infinitesimal parameter, so there is no need to stick \delta in front of it.
As explained above, we write
<br />
\omega^{\mu}{}_{\nu} = \bar{\partial}_{\nu}\left( \omega^{\mu}{}_{\rho}\bar{x}^{\rho}\right) = \bar{\partial}_{\nu}\left( \delta \bar{x}^{\mu}\right),<br />
and eq(R) (remember the infinitesimal stuff),
\omega^{\mu}{}_{\nu} \ J^{\nu}(x) \approx \omega^{\mu}{}_{\nu} \ J^{\nu}(\bar{x}).
This transforms YOUR equation into
<br />
\bar{J}^{\mu}(\bar{x}) – J^{\mu}(\bar{x}) = - \delta \bar{x}^{\nu}\bar{\partial}_{\nu}J^{\mu}(\bar{x}) + J^{\nu}(\bar{x}) \bar{\partial}_{\nu}\left( \delta \bar{x}^{\mu}\right). \ \ \ (Y)<br />
Next, we use current conservation,
\bar{\partial}_{\nu}J^{\nu}(\bar{x}) = 0,
and the Lorentz condition
\bar{\partial}_{\nu} \left( \delta \bar{x}^{\nu}\right) = 0.
With these, eq(Y) becomes
<br />
\delta J^{\mu}(\bar{x}) = \bar{\partial}_{\nu}\left( \delta \bar{x}^{\mu}J^{\nu}(\bar{x}) - \delta \bar{x}^{\nu}J^{\mu}(\bar{x}) \right).<br />
Isn’t this eq(8)?
The current-conservation law (continuity equation), \partial_{\mu} J^{\mu}=0, ensures that the total charge
$$Q=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} J^0(x)$$
is a scalar (i.e., Lorentz invariant) quantity.
Well Sir, the whole thread was about proving that very statement.
Regards
Sam
