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18

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 ...


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 ...


8

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

When you write the five dimensional Kaluza-Klein metric tensor as $$ g_{mn} = \left( \begin{array}{cc} g_{\mu\nu} & g_{\mu 5} \\ g_{5\nu} & g_{55}\\ \end{array} \right) $$ where $g_{\mu\nu}$ corresponds to the ordinary four dimensional metric and $ g_{\mu 5}$ is the ordinary four dimensional vector potetial, $g_{55}$ appears as an ...


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 + ...


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 ...


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

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

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 $2 dx^+dx^-= (dx^0)^2 - (dx^3)^2$


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.


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

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 ...


5

What you said is only true if the hypersurface is space-like or time-like. If a non-null hypersurface is defined by $f(x) = $ constant, then the normal to the hypersurface is given by $$ n_\alpha \propto \partial_\alpha f $$ The fact that the hypersurface is non-null implies $$ g^{\alpha\beta} \partial_\alpha f \partial_\beta f = \varepsilon\neq 0 $$ ...


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 ...


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

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

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

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 ...



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