# Tag Info

2

If the curve is a geodesic then in the coordinate system of an observer moving along the geodesic coordinate time and proper time are the same. That's because in the freely falling observer's coordinates $dx = dy = dz = 0$ and therefore $ds^2 = -c^2dt^2 = -c^2d\tau^2$. This makes proper time a natural way of parameterising the curve because it's just the ...

6

A metric structure $g$ and a symplectic structure $\omega$ are two very different structures, although sometimes they can co-exist in a compatible way. Unlike a symplectic structure, there are no Jacobi-like identity and no Darboux-like theorem for a metric structure. There exists a unique torsionfree metric connection $\nabla$ on a pseudo-Riemannian ...

1

So, in classical mechanics, we know nothing of this strange "spacetime" and its metric. We know only that systems are described by $n$ continuous generalized coordinates $q^i$ with a certain range, and we take the manifold $\mathcal{M}$ consisting of all possible different $\vec q$ as our starting point. Note that there is no metric, no form, nothing on ...

6

To calculate the Hubble constant we need the a scale factor, $a(t)$. This is a measure of how much the universe has expanded. We take the scale factor to be unity at the current moment, so if $a = 2$ that means the universe has expanded twice as much as it has right now. Likewise $a = 0.5$ means the universe had expanded only half as much as it has right ...

3

I'm pretty sure I know the answer to this question, even though this question provides very little context for what the various tensors are, it gives a couple expressions without including the important surrounding equation for context, and it includes an equation with a typo. First of all, as it stands, the first equation has unbalanced indices. I assume ...

1

If I remember well, this section of Peskin & Schroeder uses the Wick rotation to solve integrals of type $$\int \frac{1}{k^n + ...} \cdot ...$$ Where $k^n = k \cdot k \cdot k ... (\rm n \, times)$. By performing the Wick rotation we suddenly get a spherically symmetric problem which enables us to use the well known tricks for such a case. But notice ...

3

The natural choice is actually $\mathcal{L}=\text{something}$, the reason being is that the $\sqrt{-g}$ term is naturally paired with the volume form $d^4x$. Even before considering curved spacetime, consider non Cartesian coordinates. For example spherical coordinates $$dt\,dx\,dy\,dz = r^2 \sin\theta\ dt\,dr\,d\theta\,d\phi$$ Where does that term $r^2 ... 2 Indeed, if no values of$a$and$b$work for across different sets of indices, then the forms are not equivalent. In fact, these two forms are not equivalent even under the restriction of the metric being diagonal (and thus are not equivalent under a general metric). The diagonal case is easy to analyze, and you gave a good set of indices to do it: ... 4 When we say that they are unit vectors, we mean that the proper length is equal to one. The proper lengths of the two vectors are $$\gamma_{ab} t^a t^b=1,\quad \gamma_{ab}n^a n^b=1$$ and should be equal to one, i.e.$1\to 1$, at all times. (In the Minkowski signature, one of these squared lengths is minus one, but that won't change anything about the text ... 1 You may think to a inner product$(T, T')$:$(T,T') = \int d^2\sigma \sqrt{g} \,T. T'$where$T$and$T'$are tensors of equal rank, and$T.T'$corresponds to a contraction on the tensor indices. Now, your expression is simply, that, for all the Diff-Weyl variations$\delta g_{DW}$, and for all the moduli variations$\delta g_M$, you have :$(\delta ...

Top 50 recent answers are included