# I Lorentz invariant integration - where is the error?

1. Mar 2, 2017

### Demystifier

Let $j^{\mu}(x)$ be a Lorentz 4-vector field in Minkowski spacetime and let $\Sigma$ be a 3-dimensional spacelike hypersurface with constant time of some Lorentz frame. From those I can construct the quantity
$$Q=\int_{\Sigma} dS_{\mu}j^{\mu}$$
where
$$dS_{\mu}=d^3x n_{\mu}$$
and $n_{\mu}$ is the unit timelike vector normal to $\Sigma$. The quantity $Q$ is a Lorentz scalar. Since $j^{\mu}$ is a Lorentz vector, it follows that $dS_{\mu}$ must also be a Lorentz vector. But $n_{\mu}$ is also a Lorentz vector, so $dS_{\mu}$ can be a Lorentz vector only if $d^3x$ is a Lorentz scalar. Yet, $d^3x$ is not a Lorentz scalar, leading to a contradiction.

Where is the error?

There is a similar "paradox" with 4-momentum defined as
$$P^{\nu}=\int_{\Sigma} dS_{\mu}T^{\mu\nu}$$

Last edited: Mar 2, 2017
2. Mar 2, 2017

### Orodruin

Staff Emeritus
$dS_\mu$ is equal to $n_\mu d^3x$ only if $n_\mu = (1,0,0,0)$, i.e., in the frame where your chosen spatial coordinates are within the surface.

3. Mar 2, 2017

### Orodruin

Staff Emeritus
To elaborate on that. If you pick the spatial coordinates of the surface to be coordinates in a frame where $S$ is not a surface of simultaneity, you will get an additional factor that takes care of your invariance. In general, using parameters $\xi^1$, $\xi^2$ and $\xi^3$ for the surface, we have
$$dS_\mu = \varepsilon_{\mu\nu\sigma\rho} \frac{\partial x^\nu}{\partial \xi^1}\frac{\partial x^\sigma}{\partial \xi^2}\frac{\partial x^\rho}{\partial \xi^3} d^3\xi.$$

4. Mar 2, 2017

### Orodruin

Staff Emeritus
As an example, take the surface $t = vx$ parametrised with the coordinates $x$, $y$ and $z$. This would lead to $\partial t/\partial x = v$ and therefore
$$dS_0 = \varepsilon_{0123} d^3x = d^3x, \quad dS_1 = \varepsilon_{1023} \frac{\partial t}{\partial x} d^3x = -v\, d^3x$$
with the other elements being equal to zero. It should be clear that this is not a unit vector multiplied by $d^3x$.

5. Mar 2, 2017

### Demystifier

Exactly! For that reason, if $x^{\mu}$ is the spacetime coordinate, it is very misleading to write $d^3x$ for integration over $\Sigma$. Instead, one should write $d^3q$, where $q^i$ are coordinates on the hypersurface with induced metric
$$\gamma_{ij}=\frac{\partial x^{\mu}}{\partial q^i} \frac{\partial x^{\nu}}{\partial q^j} g_ {\mu\nu}$$
With such notation, it is more clear that one should not attempt to make a Lorentz transformation of $q^i$.

6. Mar 2, 2017

### Orodruin

Staff Emeritus
Yes, I would agree with this. But many physicists are notoriously sloppy with notation. I try to write $dS_\mu$ in most cases.

You actually never need to refer to the induced metric. What you need is the volume form $\eta$ on Minkowski space (which does depend on the metric) and the 3-form to integrate over the surface is $i_j\eta$. Your defined quantity becomes
$$Q = \int_\Sigma i_j \eta$$

7. Mar 2, 2017

### Demystifier

I am starting to appreciate the coordinate free notation that mathematicians prefer. With coordinates and indices one can easily be mislead in an attempt to use in 3-space the quantities originally defined in 4-space.

8. Mar 4, 2017

### davidge

Why $n_\mu$ has components $(1,0,0,0)$ when the coordinates are within the surface? Would this mean that at the surface both $x,y$ and $z$ are equal to zero? Why?

EDIT: I just realized that that is because $n_\mu$ is a unit vector, orthogonal to the hyper-surface of constant time.