# Invariant Momentum

1. Jul 14, 2006

### MetaJoe

Hi, All,

First time post, and this is quite possibly a very basic question: Is there a way to describe a particle's momentum such that the momentum itself is Lorentz invariant? The reason I am asking is this: As I understand it, if for example an electron and a positron were to collide and thus annihilate, such annihilation must (among other things) conserve momentum. What I'm looking for is a way to describe this momentum as it "carries though" the annihilation in such away that it is Lorentz invariant. Thank you so much!

MetaJoe

2. Jul 14, 2006

### pmb_phy

Hi Al and welcome to the forum.

Since the total 3-momentum of any closed system is conserved it follows that the total 4-momentum is also conserved. The magnitude of this 4-vector (the "invariant mass") is Lorentz invariant. For details please see

http://www.geocities.com/physics_world/sr/invariant_mass.htm

Pete

3. Jul 14, 2006

### pervect

Staff Emeritus
If an electron is at rest, it will have no momentum. If an electron is moving, it will have a non-zero momentum.

The difference between a moving electron and a stationary electron is just a Lorentz boost.

Therfore we do not expect the momentum of an electron to be Lorentz invariant - we expect it to change from zero when it is at rest to a non-zero value when we "boost" it.

The length of the enregy momentum 4-vector is an invariant as Pete says, however - it is the electron's rest mass.

In addition to the invariant rest mass, one can also write for a system a set of equations that represents the conservation of momentum, expressed in terms of the stress-energy tensor. These are known as the continuity equations.

The stress-energy tensor treats matter as a fluid, not as a collection of point particles. Therfore one sees laws that are similar to the laws of hydrodynamics, rather than laws written for a set of discrete particles.

The total momentum of a continuous system can be represented by

$$P^i = \int_V T^{i0} dV$$

where dV is an infinitesimal volume element expressed as a vector, and $T^{ij}$ is the stress-energy tensor.

A vector-valued volume element is just a 4-vector that is perpendicular to all spatial vectors in the volume element, and has a magnitude that's proportional to the volume.

The continuity equations

$$\frac{\partial T^{i0}}{d x^0} + \frac{\partial T^{i1}}{d x^1} + \frac{\partial T^{i2}}{d x^2} + \frac{\partial T^{i3}}{dx^3} = 0$$

can be regarded as a set of 4 equations (i=0,1,2,3) which represent the local conservation of energy and momentum.

The above equations are written for an orthonormal cartesian coordinate system. (Note that in such a cartesian coordinate system, the vector-valued volume element dV is just the time vector multipled by the volume element).

For an arbitrary coordinate system, one must replace the partial derivative $\frac{\partial}{\partial x^j}$ with the covariant derivative $\nabla_j$

This gives the continuity equation in general coordinates as

$$\nabla_a T^{ab} = 0$$

Last edited: Jul 14, 2006
4. Jul 16, 2006

### MetaJoe

Thank you very much. These were quite helpful responses.