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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 ...
mma's user avatar
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3 votes

Question on special relativity

No, there is no way to get a satisfactory theory using $$\tag1(c\Delta t)-\sqrt{(\Delta x)^2+(\Delta y)^2+(\Delta z)^2}=(c\Delta t^\prime)-\sqrt{(\Delta x^\prime)^2+(\Delta y^\prime)^2+(\Delta z^\...
naturallyInconsistent's user avatar
1 vote

Question on special relativity

Using $\Delta$anything is going to increase confusion. Relativity is about measuring the same events in different coordinates. So for light propagating a distance $L$, you have 2 events: Emission (Tx):...
JEB's user avatar
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7 votes

Question on special relativity

With some algebra, we can see eq. 1 and 2 are equivalent, \begin{align} &c\Delta t - \sqrt{(\Delta x)^2+(\Delta y)^2+(\Delta z)^2} = 0 \;\;\;\; (1) \\ \Rightarrow\;\;\; &c\Delta t = \sqrt{(\...
Aiden's user avatar
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Interpretation of degenerate metrics

My favorite example of the occurrence of degenerate metrics in general relativity is on null surfaces. Take for example Minkowski spacetime with Cartesian coordinates $(t,x,y,z)$. Then the plane $t-x =...
Níckolas Alves's user avatar
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Interpretation of degenerate metrics

When the determinant of the components of the metric is zero, then one cannot write its inverse-metric. This makes the metric tensor "degenerate". With a non-degenerate metric tensor, one ...
robphy's user avatar
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Derive Minkowski metric from Lorentz transformation

The problem is over here in the RHS: $\mathbf{x'^T x} = \mathbf{(Lx')^T (Lx)}$ This is not what is meant when we say that dot product of a vector remains invariant between reference frames. You don't ...
morpheus's user avatar
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2 votes

What happens if we differentiate spacetime with respect to time?

I think what you're reaching for is the four vector velocity of an object. That is, differentiating it's position with respect to time. You started with a four dimensional vector (position) and ...
StephenG - Help Ukraine's user avatar
2 votes

Physical intuition for the Minkowski space?

I don’t think that people can give you intuition. Intuition is something that you develop through experience. Since we basically live non-relativistic lives, such experience is gained through working ...
Dale's user avatar
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1 vote

Is the FRW metric, based on spatial homogeneity and isotropy, rotationally and translationally invariant? If so, how?

If what you're confused about is why the metric seems to depend on the coordinates and thus might not be translationally or rotationally invariant, look at the Minkowski metric (flat spacetime) in ...
controlgroup's user avatar
2 votes
Accepted

Is the FRW metric, based on spatial homogeneity and isotropy, rotationally and translationally invariant? If so, how?

Roughly speaking, in addition to Einstein equations, the (spatial) FLRW metric is constructed by assuming that, at fixed time, in the Riemannian manifold defining space: (a) metric properties are ...
Valter Moretti's user avatar
1 vote
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Weyl transformation of induced metric

The definition of the induced metric is $h(X,Y) := g(X,Y)$ if $X,Y$ are in tangent to $\partial M$. Therefore $$\tilde{h} = \Omega^2 h\:.$$ Then we can see $h_{ab}$ and $\tilde{h}_{ab}= \Omega^2 h_{ab}...
Valter Moretti's user avatar
1 vote
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Checking inverse metric and Christoffel symbols for the Kerr metric against references

UnkemptPanda wrote: "both sources seem to disagree by a factor of 2." The factor of 2 is wrong in some sources where they forgot that the crossterms are added twice in the ds² of the line ...
Yukterez's user avatar
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0 votes

Saddle Shaped Universe

It is simply an analogy (and not a very precise one) to illustrate that even in our familiar 3D space there are surfaces with negative curvature at all points. What is drawn in the illustration may be ...
gandalf61's user avatar
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3 votes

Photonic black holes

For the sake of argument, let us simplistically take the photon energy $E=hf$, and see what kind of black hole we get. The Schwarzschild radius of a given parcel of energy $E$, which has an equivalent ...
RC_23's user avatar
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-1 votes

Photonic black holes

"According to electromagnetic theory, the rest mass of photon in free space is zero and also photon has non-zero rest mass, as well as wavelength-dependent. The very recent experiment revealed ...
Ritzthephysibeast's user avatar
-2 votes

Photonic black holes

A photon in a box has mass and pressure. Put it in an impossibly small box and it'll theoretically make a black hole.
John Doty's user avatar
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1 vote

Homogeneous and Isotropic But not Maximally Symmetric Space

OP's question is closely related to their preceding one. I have actually provided a near-complete answer to the present question there, although it appears that OP's question predates the clarifying ...
Bence Racskó's user avatar
2 votes

Homogeneous and Isotropic But not Maximally Symmetric Space

You are correct, a spacetime need not be maximally symmetric to be homogenous and isotropic. Isotropy and homogeneity are restrictions on the spatial structure of the universe, which lead to spatial ...
CompassBearer's user avatar
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Deriving the Minkowski Metric from homogeneity of space-time and the isotropy of space

The Wikipedia page (at least as it stands when I read it today) says that "it follows" from the assumptions stated in the question that the spacetime interval between two events 1 and 2 is $$...
Brick's user avatar
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4 votes

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 ...
Bence Racskó's user avatar
2 votes

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 ...
Kurt G.'s user avatar
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0 votes

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 ...
Nayeem1's user avatar
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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 ...
Aiden's user avatar
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2 votes
Accepted

What is the determinant of the Wheeler-DeWitt metric tensor constructed from spatial metrics in ADM formalism?

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(\...
Qmechanic's user avatar
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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 ...
William Elderfield's user avatar
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 ...
Dale's user avatar
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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{...
Aiden's user avatar
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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 ...
Chauhan's user avatar
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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 ...
Octavius's user avatar
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4 votes

Conformal equivalent to Schwarzschild metric

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 $...
magma's user avatar
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