First of all it seems best to explain why the term [math]\frac{\partial \mathcal{L}}{\partial \dot{q}}[/math] is important when describing world lines.

Consisder a simple spacetime interval as:

[math]d\tau^2 = dt^2 - d\vec{x}^2[/math]

Where we have set [math]c=1[/math] in this case. You actually calculate the length of a worldline by taking into consideration the integral

[math]L(W) = \int_W d\tau[/math]

You can, it was shown to me a while ago now, that worldines can be written in terms of time by the chain rule. Doing so, you can rewrite the time derivatives as dots on your variables and can end up with

[math]L(W) = \int_{t_0}^{t_1} \sqrt{1 - ||\dot{x}||^2}\ dt[/math]

From here, you would calculate the Langrangian by simply multiplying mass into the equation, so we would have

[math]\mathcal{L} = -M\sqrt{1 - ||\dot{x}||^2}[/math]

Now in my equation, we have been using the generalized velocity, and can be freely exchanged now to make the above equation into

[math]\mathcal{L}(\dot{q}\dot{q}) = -M\sqrt{1 - \dot{q}\dot{q}}[/math]

Now, the canonical momentum part in my equation can be written as

[math]\frac{\partial \mathcal{L}}{\partial \dot{q}} = \frac{M\dot{q}}{\sqrt{1 - \dot{q}\dot{q}}}[/math]

This is relativistic and is incomporated as one can see, into the idea of worldlines. Now, in my equation, I decided to multiply the momentum with distance. Of course, this was just the quantum action [math]\hbar[/math], but ignoring that fact for now, we wish to calculate the distance really as a displacement of all the particles in the universe [math]d_i[/math] using Barbour's approach. Doing so, will require an integral.

Taking the integral of the equation, which ''cuts up'' or ''slices'' a worldline for a particle, then the distance will be small [math]\delta d[/math] for a particle which is the way elluded to in the OP for how to calculate displacement of particles instead of distance exactly.

Remembering that

[math]\frac{\partial \mathcal{L}}{\partial \dot{q}} = \frac{M\dot{q}}{\sqrt{1 - \dot{q}\dot{q}}}[/math]

then my equation

[math]\frac{\partial \mathcal{L}}{\partial \dot{q}_i} d_i \nabla^2\psi = \dot{m}\psi[/math]

Can actually be rewritten (including the integral this time) as

[math]\int \dot{m}\psi\ dt = \int (\frac{M\dot{q}}{\sqrt{1 - \dot{q}\dot{q}}}) \delta d_i \nabla^2\psi\ dt[/math]

So as you can see, the integral not only ''cuts'' up the worldline of a particle appropriately, but making the interval short enough will ensure that your distance is really just a very small displacement on a system of particles.

**Edited by Aethelwulf, 13 July 2012 - 02:50 AM.**