Tag Info

New answers tagged

5

For doing matrix multiplication, yes, we usually put the dummy indicies together, but this is (assuming one is not working over $\mathbb{H}$) convention really. Don't forget that this is shorthand Einstein convention. When you write $$ \mathbf{A} \cdot \mathbf{B} = A_{ij} B_{jk} = \sum_{j=1}^n A_{ij} B_{jk} $$ you are writing a shorthand version in ...


1

"... my questions is now one of why and how (mathematically) do we obtain the Minkowski Metric Signature. More specifically the one element with a different sign." Well, If you check Einstein's "Relativity: The Special and General Theory", you will find in the Appendix I (just before equation (10)) that Einstein simply started off with the ...


0

The only real reason to introduce $ict$ coordinates is to stress the similarity (for didactic purposes I guess) between Lorentz transformation and orthogonal rotations in more used-to Euclidian space. Note that Minkowski pseudo-Euclidian space obtains exactly "normal" Euclidian form if complex time is introduced,namely: metric signature becomes $++++$: ...


1

In general, the expression for the metric and the expression for the coordinates need to work together to give you the correct line element. So the following combinations all have the same line element $ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$: $g_{\mu\nu}=diag(1,1,1,1)$ with $x^{\mu}=(ict,x,y,z)$ or $g_{\mu\nu}=diag(-1,1,1,1)$ with $x^{\mu}=(ct,x,y,z)$ or ...


7

as you wrote, the spacetime invariant can be expressed as: $$ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$$ and from that we normally get: $$ds^2=-c^2dt^2+dx^2+dy^2+dz^2$$ This is not because of some arbitrary imaginary time unit, this is because the metric ($g_{\mu\nu}$) is a diagonal matrix with the coefficients of each term of the $ds^2$ equation: ...


7

If you define $x^0=ict$, then I assume one takes $x_\mu=x^\mu$ so that the metric is actually $\eta_{\mu\nu}=\text{diag}(1,1,1,1)=\delta_{\mu\nu}$, i.e. you're dealing with a Euclidean metric. Then $$ds^2=\delta_{\mu\nu}dx^\mu dx^\nu$$ gives the usual outcome : $$ds^2=-c^2dt^2+d\vec{x}^2$$ The usual conventions are as follows: Option one: One defines ...


0

This is more a comment than an answer. The condition for the conjecture to hold also require that the connection is torsion free. If you have a metric connection, but is not torsion free, you can build in an holonomic frame the torsion tensor: \begin{equation} T^{k}_{ij}=\Gamma^{k}_{ij}-\Gamma^{k}_{ji} \end{equation}


1

The dynamics of a large class of mechanical systems can be described as a geodesic motion in some ambient space. This is the essence of the Kaluza-Klein theory. The basic and most elementary example is the case of a charged particle in $3D$ coupled to a magnetic field which can be described as a neutral particle geodesically moving in a background metric ...


1

I) OP talks about minimizing curves (rather than higher dimensional objects) so let us concentrate on point mechanics (as opposed to field theory) with Lagrangian $L$ (rather than Lagrangian density ${\cal L}$). We conventionally call the curve parameter time $t$, although it doesn't have to correspond to any physical time variable. Let us for simplicity ...


1

You don't need to assume that the path of least action is the path taken. You can show it from Newton's laws. See http://www.damtp.cam.ac.uk/user/tong/dynamics/two.pdf The path of least action is the path for which $F = ma$ holds at each point. This is the geodesic. This is the shortest path through space-time. You get this path from the Lagrangian. You ...


2

You can derive the equations of motion (equations of geodesics) for a particle in curved spacetime by using the Lagrangian $$L = \frac{1}{2} \sum_{\mu,\nu} g_{\mu\nu} \frac{dx^\mu}{dt} \frac{dx^\nu}{dt},$$ so the answer is yes. You could regard the configuration manifold as the manifold, it need not be the physical spacetime. I would like to clarify that ...


2

A Penrose diagram of a metric $g_{ab}$ is used to represent the conformal structure of $g_{ab}$. Generally light rays move at $\frac{\pi}{4}$ from the upward vertical and the spacetime considered is spherical symmetric. The metric, $\overline{g_{ab}}$, on the Penrose diagram satisfies: $\overline{g_{ab}}=\Omega^{2} g_{ab}$. This implies that timelike ...



Top 50 recent answers are included