# Tag Info

### Null vector space in Minkowski space

The question is: What would the linearly independent null vectors of this space be? Since the null vectors do not form a vector space, the question only makes sense if "this space" is ...
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### Constant curvature on a sphere?

In Riemannian geometry, a constant curvature space means that the curvature tensor is of the form $$R_{ijkl}=k(g_{ik}g_{jl}-g_{il}g_{jk}),$$ where $k$ is a constant. This equivalently means that the ...
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### Constant curvature on a sphere?

It is true that scalar curvature is constant when all covariant derivatives of the Riemann tensor are zero: That assumption $$\tag1 \nabla_\mu R^\nu{}_{\eta\,\rho\,\sigma}=0$$ implies for the Ricci ...
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### Constant curvature on a sphere?

I got the answer. For curvature to remain constant the Riemann curvature tensor should be constant, meaning its covariant derivative along any direction must be zero. I ran EinsteinPy code and ...
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1 vote
Accepted

### A few doubts regarding the geometry and representations of spacetime diagrams

The dotted lines are null trajectories. You are correct there is no $T$-dependence in the slope of the light cones. Therefore, the slope of all dotted lines along a vertical (of constant $X$) is the ...
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Accepted

The Wheeler-DeWitt metric \begin{align} G~=~&G_{IJ}(\mathrm{d}y\odot\mathrm{d}y)^I\odot(\mathrm{d}y\odot\mathrm{d}y)^J\cr ~=~&G_{i_1i_2,j_1j_2}(\mathrm{d}y^{i_1}\odot\mathrm{d}y^{i_2})\odot(\... • 206k 2 votes ### How do you differentiate F^{αβ} with respect to g_{μν}? If you are going to differentiate L with the respect to the metric, L needs to be rewritten without the metric being implicitly used anywhere. Otherwise, you will not vary the entire dependence on ... 2 votes Accepted ### Confusion about local Minkowski frames x and t are just symbols. We like to use the symbol x to refer to the spatial coordinate and the symbol t to refer to the time coordinate. Doing so helps communication with other people as you ... • 103k 2 votes Accepted ### Confusion about timelike spatial coordinates For a vector \bf{V}, timelike, null and spacelike are defined \begin{align} \text{timelike:}&\;\;\;\;\mathbf{V}\cdot\mathbf{V} = g_{\mu\nu}V^\mu V^\nu < 0 \\ \text{null:}&\;\;\;\;\mathbf{... • 1,898 0 votes ### How to prove  g^{\mu\nu}\Lambda^{\rho}{}_{\mu}\Lambda^{\sigma}{}_{\nu}=g^{\rho\sigma}  for the inverse metric? Let me use condition{\Lambda^{-1}}^{\rho}_{}{\nu}= \Lambda_{\nu}{}^{\rho}$$We can raise \nu index left side as well as right side,$${\Lambda^{-1}}^{\rho\nu}= \Lambda^{\nu\rho}$$Now contract ... • 31 1 vote ### How to prove  g^{\mu\nu}\Lambda^{\rho}{}_{\mu}\Lambda^{\sigma}{}_{\nu}=g^{\rho\sigma}  for the inverse metric? The g_{ij} are the components of the metric tensor with respect to some local coordinate basis, which means they are scalars and can be imterchanged, i.e.$$g_{ij} g_{kl} = g_{kl} g_{ij} This is ...
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You have a misconception. Neither metric is degenerate. $det g$ should be $0$ somewhere in order for a metric to be degenerate. See here With your definition even any innocent looking 2D polar metric \$...