Recent content by Pentaquark5

Thanks everybody for your help! I got thrown off by a Hint that said "use coordinates associated with right-going and left-going null geodesics." which made it sound much more complicated than it was.

Hi, how can I prove that any 2-dim Lorentzian metric can locally be brought to the form $$g=2 g_{uv}(u,v) \mathrm{d}u \mathrm{d}v=2 g_{uv}(-\mathrm{d}t^2+dr^2)$$ in which the light-cones have slopes one? Thanks!

Let us assume a "toy-metric" of the form $$ g=-c^2 \mathrm{d}t^2+\mathrm{d}x^2+\mathrm{d}y^2+\mathrm{d}z^2-\frac{4GJ}{c^3 r^3} (c \mathrm{d}t) \left( \frac{x\mathrm{d}y-y\mathrm{d}x}{r} \right)$$ where ##J## is the angular-momentum vector of the source. Consider the curve $$ \gamma(\tau)=(x^\mu...

Ah, thank you I don't know what went wrong in my head there. And you are of course correct in noting that ##\mathcal{M}## is pseudo Riemannian. Thanks!

On Riemannian manifolds ##\mathcal{M}## the covariant derivative can be used for parallel transport by using the Levi-Civita connection. That is Let ##\gamma(s)## be a smooth curve, and ##l_0 \in T_p\mathcal{M}## the tangent vector at ##\gamma(s_0)=p##. Then we can parallel transport ##l_0##...

Right. My bad. But do you see any argument as to why the identity above should be generally zero, then?

The Christoffel symbols vanish in Minkowski space, so this would hold for flat spacetime. Unfortunately, I need the more general form where the Christoffel symbols are non-zero. Thus, I do not believe the covariant derivative of the four velocity is generally zero, no?

Homework Statement My textbook states: Since the number of particles of dust is conserved we also have the conservation equation $$\nabla_\mu (\rho u^\mu)=0$$ Where ##\rho=nm=N/(\mathrm{d}x \cdot \mathrm{d}y \cdot \mathrm{d}z) m## is the mass per infinitesimal volume and ## (u^\mu) ## is...

Ok, sorry, but I still don't get it. When you refer to missing terms, I assume you mean the product rule in the following equation?$$\frac{\mathrm{d}}{\mathrm{d}s} \left( \frac{\partial \mathcal{L}}{\partial \dot{x}^\mu} \right)=\frac{\mathrm{d}}{\mathrm{d}s} \left(\dot{x}^\beta g_{\mu...

Thanks! I think I get it now!

Ok, I just realized my indecess are off. $$\ddot{x}^\beta \underbrace{g^{kl}g_{\mu\beta}}_{\delta^k_{\;\mu} \delta^l_{\; \beta}}=\frac{1}{2} g^{kl} \partial_\mu g_{\alpha \beta}(x^\mu)\dot{x}^\alpha \dot{x}^\beta$$ $$ \ddot{x}^l = \frac{1}{2} g^{kl} \partial_\mu g_{\alpha...

Homework Statement The Lagrange Function corresponding to a geodesic is $$\mathcal{L}(x^\mu,\dot{x}^\nu)=\frac{1}{2}g_{\alpha \beta}(x^\mu)\dot{x}^\alpha \dot{x}^\beta$$ Calculate the Euler-Lagrange equations Homework Equations The Euler Lagrange equations are $$\frac{\mathrm{d}}{\mathrm{d}s}...

Thank you all for your help, I believe I understand now!

I guess the issue is that I don't know how to calculate ##g(\dot{\gamma},\dot{\gamma})(s)##. From what I know ##g(X,Y)=g_{ij}\, dx^i(X) \otimes dx^j(Y)## for ##X,Y## arbitrary vector fields. How does this definition lead to the Lagrangian above?

I've recently read in a textbook that a geodesic can be defined as the stationary point of the action \begin{align} I(\gamma)=\frac{1}{2}\int_a^b \underbrace{g(\dot{\gamma},\dot{\gamma})(s)}_{=:\mathcal{L}(\gamma,\dot{\gamma})} \mathrm{d}s \text{,} \end{align} where ##\gamma:[a,b]\rightarrow...