# Tag Info

15

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 ...

13

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 ...

10

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 ...

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

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 ...

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: $$S = -\frac{1}{2} \int \mathrm{d}\xi\ \left( \sigma(\xi) \left(\frac{\mathrm{d}X}{\mathrm{d}\xi}\right)^2 + ... 6 Expanded on Michael Brown's comments, you seem to have confused dt with d\tau. They are not the same. It is true that ds^2 = dt^2, but also true that ds^2 = g_{tt} dt^2 + \ldots (other terms omitted). What this means physically is that, even for two events that take place at the same location but different coordinate times (according to a given ... 6 You're assuming that the Kruskal–Szekeres (U,V) coordinates have to be defined in terms of the Schwarzschild (r,t) coordinates, but there is nothing special or fundamental about the Schwarzschild coordinates. General covariance says that we can use any coordinates we like. If the K-S coordinates had been the ones originally chosen by Schwarzschild, then ... 6 For a<0, it's a possible metric for spacetime. For a>0, it's a possible metric for a 4-dimensional Euclidean space. For a=0, it's degenerate, and in many cases it's not possible to work with a degenerate metric, e.g., the machinery of general relativity requires that the metric be nondegenerate. It doesn't matter whether a has a particular ... 6 The answer to your question is affirmative in the following sense: In the Riemann normal coordinates at p the coefficients of the Taylor expansion of the metric g_{ij}(x) are polynomials in the Riemann tensor at p and its covariant derivatives at p. [Assuming the proof in this random thing I googled[a] is correct, starting at (5.1)]. I think this ... 6 You need to be very careful. The \mathbf{e}_i are vectors, so they have a Lorentz index: \mathbf{e}_i^\mu. When you write$$\mathbf{e}_i \cdot \mathbf{e}_j$$you actually mean$$g_{\mu \nu} \mathbf{e}_i^\mu \mathbf{e}_j^\nu$$where g_{\mu \nu} is the flat Minkowski metric (not the Euclidean metric). Once you know this, it's straightforward to check ... 6 If you compute |(dx^+)^2 - (dx^-)^2|, you will not find  |(dx^0)^2 - (dx^3)^2|. So, you cannot obtain x^+,x^- (even with a different normalization) from x^0,x^3 by a Lorentz transformation. None of the coordinates x^+,x^- is time-like, or space-like, they are both light-like, and the metrics is 2 dx^+dx^-= (dx^0)^2 - (dx^3)^2 5 I recently re-derived these equations with all the dimensionful constants in place. Your last statement in the "Edit" is correct: T_{00} = \rho_{E}\,c^{2} = \rho\,c^{4}. It's easy to lose track of factors of c in calculations like this; the usual culprit is mixing up t and x^{0} = c\,t, and \partial_t and \partial_0 = c^{-1}\,\partial_{t}. For ... 5 The expression A^{\mu}B_{\mu} simply means that$$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}$$The metric simply tells you have how the components of a vector and its dual vector ... 5 An easy way to see that they are distinct is to consider what happens upon raising (or lowering) all indices. For example, upon lowering,$$ T_{ab}{}^{cde} $$becomes T_{abcde}, whereas$$ T_{a}{}^{cd}{}_{b}{}^{e} $$becomes T_{acdbe}, and similarly$$ T_{a}{}^{cde}{}_{b} $$becomes$$ T_{acdeb}. $$You need to "slant" the indices so as to keep track ... 5 Vibert is, of course, completely correct. I'm gong to propose a slightly more geometric version of what he says. The minkowski metric tensor is given by:$$ds^{2} = g_{ab}dx^{a}dx^{b} = -dt^{2} + dx^{2} + dy^{2} + dz^{2}$$Now, knowing that v = \tanh\phi, it is easy enough to show that the Lorentz transformations are given by:$$\begin{align} t' ...

5

Although the question is about Minkowski space, I think it may be helpful to consider the more general case first. It's not possible to distinguish past and future simply based on the metric. In fact, there are spacetimes that are not even time-orientable. This is similar to the way in which a Mobius strip is not an orientable surface. So the same ...

4

1) OP is asking about the use of the word flat metric. It means a pseudo-Riemannian metric (of arbitrary signature) whose corresponding Levi-Civita Riemann curvature tensor vanishes. 2) However, the word Euclidean space may potentially cause confusion among mathematicians and physicists. For a mathematician an Euclidean space is always an affine space, ...

4

Actually, in the context of general relativity, $c$ has no (physical) unit. More precisely, $c$ is meter per second. Meter is a measure of length. Second is a measure of time. In GR we unified space and time, and hence a meter and a second are different units of measurement for the "same thing". The number $c$ is a pure scalar that is just a conversion ...

4

Edit edit: as has been pointed out, I was incorrect to say $\partial_t = \partial_{t'}$ and so on. Serves me right for trying to look at it by inspection instead of being rigorous. Nevertheless, I do think cylindrical coordinates simplifies the problem somewhat. Recall the cylindrical line element: $$ds^2 = -dt^2 + dr^2 + r^2 \, d\phi^2 + dz^2$$ Now, ...

4

The problem arose when you wrote $ds^2 = g_{00} d\tau^2$. Generally one of your coordinates $x^\mu$ will be timelike, and the others spacelike, but the timelike one is not in general the proper time of someone whose spatial coordinates are not changing. That is, $t \neq \tau$. Using your sign convention,1 $d\tau^2 = ds^2$, so the (arbitrarily large) lapse in ...

4

There are two manifolds that are involved in string propagation. The spacetime in which the string propagates. The worldsheet of the string itself. The fields $X^\mu$ are embedding coordinates of the worldsheet in the spacetime manifold. This means that for each point $(\sigma^1, \sigma^1)$ on the worldsheet, $X^\mu(\sigma^1, \sigma^2)$ gives the ...

4

You are incorrect to suppose that this spacetime is curved. In fact, up to some conditions on the coordinate ranges, this is simply a piece of Minkowski spacetime. Let me put it in this form: $$ds^2 = dt^2 - t^2(d\psi^2 + \sinh^2\psi\,d\Omega^2)\text{,}$$ where $d\Omega^2 = d\theta^2 + \sin^2\theta\,d\phi^2$ is the metric for a unit $2$-sphere, and we can go ...

4

Since the metric and inverse metric are related by $$g^{\mu\lambda}g_{\lambda\nu} = \delta^\mu_\nu$$ taking the variation of both sides gives $$\delta g^{\mu\lambda}g_{\lambda\nu} + g^{\mu\lambda}\delta g_{\lambda\nu} =0$$ or in other words $$\delta g_{\mu\nu} = -g_{\mu\rho}g_{\nu\sigma}\delta g^{\rho\sigma}$$ It follows that there is a ...

4

A variation of a tensor is always a tensor and the formula for the value above doesn't show otherwise. What you probably find surprising is that $\delta g_{\mu\nu}$ and $\delta g^{\rho\sigma}$ are not related to each other by simply raising the indices $\mu,\nu$ or lowering the indices $\rho,\sigma$. Indeed, they're not related in this way. In this case, ...

4

The basic premise behind general relativity is the equivalence principle, the idea that an object moving in an accelerating (non-inertial) reference frame is indistinguishable from one moving under the influence of a gravitational field. (As an aside, Einstein's original proofs of time dilation, length contraction, even $E=mc^2$ don't involve any calculus - ...

4

Is there somewhere in the paper that they say that the third derivative vanishes, or invoke its vanishing as an approximation? In general, you can't make tensorial objects by differentiating the metric. To get a tensor by differentiating a tensor, you have to take a covariant derivative. But the covariant derivative of the metric vanishes identically. ...

4

Writing : $x_1 = \sin \theta \cos \phi$, $x_2 = \sin \theta \sin \phi$, $x_3 = \cos \theta$ The unit radius $2$-sphere metrics is $ds^2=(d\theta^2 + \sin^2 \theta ~d\phi^2)$ We are going to use the stereographic projection : $\large z = \frac {x_1+ix_2}{1-x_3}$ This gives : $z = cotg(\theta/2) ~e^{i \phi}$ So,  dz = \frac{1}{2}(\frac{-1}{\sin^2 ...

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