There are no space integrals - it is a (gauge fixed) action for point particles, proportional to the length of the worldline in D dimensional target space plus some interaction terms. The domain is one dimensional, as for the ordinary relativistic point particle.
Sorry for the confusion
I
Hi all,
If I take an action involving two point particles coupled together by a delta function contact interaction is it possible to carry out the variation with respect to the fields? For e.g.
S = \int dt \frac{1}{2} \dot{x}^{2} + \int \int dt \dot{x}(t) \delta^{D}\left(x(t) - y(t')\right)...
Ok great, thank you very much :-)
Two minor questions...
Presumably finding geodesics is non trivial if you use the plane plus identifications...or do you extend the fundamental domain and place image points for your end point and find the geodesics for those, then wrapping them back to the...
Hi - thank you so much for the replies :-D
So in the two cases above (assuming that the plane was initially a piece of ℝ2) are you saying that the two methods lead to different induced metrics? In the case of the pullback onto a surface we end up with
ds2 = (1 + a.cos(v))2du2 + a2dv2
in...
Hi all,
Perhaps I'm asking the wrong question but I am wondering about the relationship between different definitions of, for the sake of argument, the torus.
We can define it parametrically (or as a single constraint) and from there work out the induced metric as with any surface.
But...
Ok, that relation looks like it might help, and I'll give the half angle formulae a go...
What is the correct way to handle to transform of the x dependent part of the wavefunction? Do I keep -imt and substitute the new t, or do I go for -ip.x with the new x and p?
Thanks
Hi guys,
I'm currently struggling to show something my lecturer told us in class. We have that
\Psi\left(x\right) \rightarrow S\left(L\right)\Psi\left(L^{-1}x\right)
under a Lorentz transform defined
L = exp\left(\frac{1}{2}\Omega_{ij}M^{ij}\right)
with
S\left(L\right) =...