# Tag Info

20

The connection is chosen so that the covariant derivative of the metric is zero. The vanishing covariant metric derivative is not a consequence of using "any" connection, it's a condition that allows us to choose a specific connection $\Gamma^{\sigma}_{\mu \beta}$. You could in principle have connections for which $\nabla_{\mu}g_{\alpha \beta}$ did not ...

15

A homogeneous cosmology is one in which there are no "special" places in the universe: at a given instant in time, the universe appears the same at every location (on large enough spatial scales). An isotropic cosmology is one in which there are no "special" directions: at a given instant in time, the universe appears the same in every direction (again, on ...

14

It means that the laws of physics are the same everywhere and the same in every direction. It is of fundamental importance as these symmetries give rise to conservation laws. The isotropy of the universe means that angular momentum is conserved; its homogeneity means that momentum is conserved. A similar symmetry, that the laws of physics are the same for ...

13

The significance of the metric: $$d\tau^2 = dt^2 - dx^2$$ is that $d\tau^2$ is an invarient i.e. every observer in every frame, even accelerated frames, will agree on the value of $d\tau^2$. In contrast $dt$ and $dx$ are coordinate dependant and different observers will disagree about the relative values of $dt$ and $dx$. So while it is certainly true ...

12

In hindsight, here is a short proof. The metric $g_{\mu\nu}$ is the flat constant metric $\eta_{\mu\nu}$ in both coordinate systems. Therefore, the corresponding (uniquely defined) Levi-Civita Christoffel symbols $$\Gamma^{\lambda}_{\mu\nu}~=~0$$ are zero in both coordinate systems. It is well-known that the Christoffel symbol does not transform as a ...

11

You tell if a space (or spacetime) is curved or not by calculating its curvature tensor. Or more unambiguously one of the curvature scalars (e.g. Ricci, or Kretschmann) since these don't depend on the coordinate system, but all of the information in the scalars is also contained in the Riemann tensor. It is not necessarily obvious whether a given metric is ...

11

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: ...

10

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 ...

9

By defintion, Minkowski space $\mathbb{R}^{p,q}$ must have signature $(p,q)=(1,d-1)$, with metric, $$ds^2 = -dt^2 +dx_1^2 + dx^2_2 + \dots$$ The signature $(+,+,\dots)$ corresponds to Euclidean space, which is obtained by a Wick rotation, $$t\to -i\tau$$ to imaginary time $\tau$, and the metric is modified, in the case of Wick rotated Minkowski space, to ...

9

I) Let us for simplicity discuss tensors in the context of (finite-dimensional) vector spaces and multilinear algebra. [There is a straightforward generalization to manifolds and differential geometry.] II) Abstractly in coordinate-free notation, the Kronecker delta tensor, or tensor contraction, is the natural pairing $$\tag{1} V \otimes ... 8 Each of the indices in a tensor have a particular left-right ordering. This ordering cannot be changed unless the tensor has some particular symmetry that permits it (or rather, that equates different components on interchange). The up-down positions of indices tells us about whether the index is associated with using a basis vector (up) or a basis ... 8 The covariant derivative is metric compatible, so \nabla_{\alpha} g_{\beta \gamma} = 0. This is the condition that the inner product is preserved under parallel transport. 8 An interesting question indeed :-) Yes, you can flip the overall sign of the Minkowski metric, and in fact a lot of physicists do this! The sign choice \operatorname{diag}(-1, 1, 1, 1) is conventional in fundamental quantum field theory and in quantum gravity, if I remember correctly, whereas \operatorname{diag}(1, -1, -1, -1) is conventional in particle ... 8 and Δs is the space-time interval. Actually, many (most?) will say that the spacetime interval is \Delta s^2. In other words, \Delta s^2 is not the squared interval; it is the symbol for the interval. Since this has been questioned in a comment, I provide some references below: Bernard Schutz writes in Gravity from the Ground up: An Introductory ... 8 On any manifold we can define the differential df of a scalar f. The differential is a 1-form: something that eats vectors and spits out scalars, or even less formally, something with one down index. We have the following formula for the differential,$$df = \frac{\partial f}{\partial x^i} dx^i$$(sum over i implied). You can write it in index notation ... 7 If I take the (corrected) formula :$$R= 2M + \sqrt{r(r-2M)}+2M\ln \left[ \sqrt{\frac{r}{2M} -1} + \sqrt{\frac{r}{2M}} \right] \tag{1}$$You have (if no error) : dR = \dfrac{dr}{\sqrt{ 1 - \dfrac{2M}{r}}}, so dR^2 = \dfrac{dr^2}{f(r)}, and this simplifies your metrics. However, you cannot invert the formula (1) to get r as an explicit function of ... 7 You are correct when you point out that any function of \Delta x^2 + \Delta y^2 + \Delta z^2 - \Delta t^2 will be constant and agreed on by all observers. So we could define \Delta s to be its cosine...if all we were interested in was getting an invariant. You are also right when you point out the dimensional issue. Measure time in ... 7 If you define x^0=ict, then I assume one takes x_\mu=x^\mu so that the metric is actually \eta_{\mu\nu}=\text{diag}(1,1,1,1)=\delta_{\mu\nu}, i.e. you're dealing with a Euclidean metric. Then$$ds^2=\delta_{\mu\nu}dx^\mu dx^\nu$$gives the usual outcome :$$ds^2=-c^2dt^2+d\vec{x}^2$$The usual conventions are as follows: Option one: One defines ... 7 If your solution is not a null geodesic then it is wrong for a massless particle. The reason you go astray is that Lagrangian you give in (1) is incorrect for massless particles. The general action for a particle (massive or massless) is:$$ S = -\frac{1}{2} \int \mathrm{d}\xi\ \left( \sigma(\xi) \left(\frac{\mathrm{d}X}{\mathrm{d}\xi}\right)^2 + ...

6

You're assuming that the Kruskal–Szekeres (U,V) coordinates have to be defined in terms of the Schwarzschild (r,t) coordinates, but there is nothing special or fundamental about the Schwarzschild coordinates. General covariance says that we can use any coordinates we like. If the K-S coordinates had been the ones originally chosen by Schwarzschild, then ...

6

The answer to your question is affirmative in the following sense: In the Riemann normal coordinates at $p$ the coefficients of the Taylor expansion of the metric $g_{ij}(x)$ are polynomials in the Riemann tensor at $p$ and its covariant derivatives at $p$. [Assuming the proof in this random thing I googled[a] is correct, starting at (5.1)]. I think this ...

6

If you compute $|(dx^+)^2 - (dx^-)^2|$, you will not find $|(dx^0)^2 - (dx^3)^2|$. So, you cannot obtain $x^+,x^-$ (even with a different normalization) from $x^0,x^3$ by a Lorentz transformation. None of the coordinates $x^+,x^-$ is time-like, or space-like, they are both light-like, and the metrics is $2 dx^+dx^-= (dx^0)^2 - (dx^3)^2$

6

Note that if you lower an index of the Kronecker delta, it becomes the metric: $\eta_{\mu\nu}\delta^{\mu}_{\rho}=\delta_{\nu\rho}=\eta_{\nu\rho}$ And in your last step you got a wrong index. It should be $\omega_{\rho\sigma}$, not $\omega^{\rho}_{\sigma}$. Then, the metric terms cancel and you neglect cuadratic terms. That should be enough to solve it.

6

You need to be very careful. The $\mathbf{e}_i$ are vectors, so they have a Lorentz index: $\mathbf{e}_i^\mu.$ When you write $$\mathbf{e}_i \cdot \mathbf{e}_j$$ you actually mean $$g_{\mu \nu} \mathbf{e}_i^\mu \mathbf{e}_j^\nu$$ where $g_{\mu \nu}$ is the flat Minkowski metric (not the Euclidean metric). Once you know this, it's straightforward to check ...

6

The expression $A^{\mu}B_{\mu}$ simply means that $$A^{\mu}B_{\mu}=A^{0}B_{0}+A^{1}B_{1}+A^{2}B_{2}+A^{3}B_{3}$$ Using the Minkowski metric with signature $(+---)$ you write this as $$A^{\mu}B_{\mu}=A^{\mu}\eta_{\mu\nu}B^{\nu}=A^{0}B^{0}-A^{1}B^{1}-A^{2}B^{2}-A^{3}B^{3}$$ The metric simply tells you have how the components of a vector and its dual vector ...

6

Here I just want to mention that there exists a direct proof in $1+1$ dimensions using elementary arguments. Let the two coordinate patches $U_x$ and $U_y$ (which are, say, both convex sets in $\mathbb{R}^2$, containing the origin) have light-cone coordinates $x^{\pm}$ and $y^{\pm}$, respectively. The metric reads $$dy^{+}dy^{-} ~=~ dx^{+}dx^{-}.$$ ...

6

Expanded on Michael Brown's comments, you seem to have confused $dt$ with $d\tau$. They are not the same. It is true that $ds^2 = dt^2$, but also true that $ds^2 = g_{tt} dt^2 + \ldots$ (other terms omitted). What this means physically is that, even for two events that take place at the same location but different coordinate times (according to a given ...

6

For $a<0$, it's a possible metric for spacetime. For $a>0$, it's a possible metric for a 4-dimensional Euclidean space. For $a=0$, it's degenerate, and in many cases it's not possible to work with a degenerate metric, e.g., the machinery of general relativity requires that the metric be nondegenerate. It doesn't matter whether $a$ has a particular ...

6

Your second method is correct. To compare, say, the magnetic field with what you find in Jackson, you really need to realize that there's an assumption that you have unit basis vectors there, and that the cross product is actually a hodge dual (which will invoke factors of the square root of the determinant of the metric). These will make direct ...

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