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1 vote

How do I know if a motion is 1 dimensional or 2 dimensional?

the main thing take away: physics does not care a about your coordinates. So if the thing is moving on: $$ \vec r(t) = t\cos{\theta}\hat e_1 +t\sin{\theta}\hat e_2 $$ thats linear motion in a plane. ...
JEB's user avatar
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6 votes

How do I know if a motion is 1 dimensional or 2 dimensional?

Both you and your teacher can be correct depending on what you mean by "dimension." In everyday experience, we generally consider dimensionality to be the number of independent parameters ...
Roger Yang's user avatar
7 votes
Accepted

How do I know if a motion is 1 dimensional or 2 dimensional?

Let's say you have the following motion (the red arrow) It looks 2D on this plot, since its motion changes both the $x$- and $y$-coordinates. However, if you redefine the axes like so: Then the ...
Allure's user avatar
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4 votes

How do I know if a motion is 1 dimensional or 2 dimensional?

One dimensional motion is any kind of motion that happens on a line. There are many ways to define this. For example, you could say that the position vector of the particle is always $\vec{r}(t)=r(t)\...
agaminon's user avatar
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1 vote

Boundary conditions on transition maps on general relativity

There's nothing fancy going on with transition maps. What a mathematician means by a "transition map", in the setting of general relativity, is nothing more or less than a coordinate change ...
Lee Mosher's user avatar
1 vote

Trying to understand a visualization of contravariant and covariant bases

Given the metric tensor for the basis vectors $$A = \begin{bmatrix}\mathbf e_1 \cdot \mathbf e_1 & \mathbf e_1 \cdot \mathbf e_2 \\ \mathbf e_2 \cdot \mathbf e_1 & \mathbf e_2 \cdot \mathbf ...
Antoni Parellada's user avatar
1 vote
Accepted

What kind of object is a function in the context of gauge theory?

The terminology used in physics texts is often a bit imprecise, and thus the word "scalar" is somewhat ambiguous. Often if a function $\phi(x)$ of the coordinates is a "scalar", it ...
Bence Racskó's user avatar
2 votes

Cone vs. small circle parallel transport

This is in a way a Riemann-geometric analogue of the Aharonov--Bohm effect in quantum mechanics. Consider a manifold $M$ equipped with a linear connection $\nabla$. It is well-known that the parallel ...
Bence Racskó's user avatar
3 votes

GR and Riemann Surfaces -- does the complex plane have anything to do with it?

This question seems mainly spurred by a conflation of a Riemann surface (which is a 1-dimensional complex manifold), and a (pseudo)Riemannian manifold (which is the underlying geometry of GR).
Qmechanic's user avatar
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0 votes

What actually is Boyer-Lindquist coordinates?

For me, the main difference between spherical and BL coordinates is related to the Cartesian coordinates which are as follows: spherical coordinates $$ x = r\sin\theta\cos\varphi $$ $$ y = r\sin\...
Cornelius Fyla's user avatar
0 votes

Cone vs. small circle parallel transport

The path the vector you are describing takes on the cone is a closed circle while the cone is rolled, but it is no longer a closed circle once the cone is sliced and flattened!
mike1994's user avatar
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6 votes
Accepted

Is it possible to understand in simple terms what a Symplectic Structure is?

At the most rough level possible, a symplectic structure (geometrically) is an even-dimensional manifold together with a preferred choice of two-dimensional planes which, taken together, span the ...
11zaq's user avatar
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0 votes

General Relativity in a Differential Geometry concept

Definition: We say that a subset $\mathcal{X}$ of $\mathcal{A}$ is an elementary surface if there exists a parametrization $\gamma$: $\mathcal{U}$ $\longrightarrow$ $\mathcal{A}$ (*) (called ...
The Tiler's user avatar
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0 votes

Geometrical interpretation of gauge fields of spin other than 2

``Geometry'' is a very vague term. The usual gauge theories (Yang-Mills theory) is about connections on vector bundles, see https://en.wikipedia.org/wiki/Gauge_theory_(mathematics). If you include ...
John's user avatar
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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

Confusion over what constitutes a uniform gravitational field in relativity

a. The Rindler coordinates are sometimes referred to as a "uniform gravitational field" because they have the property of not exhibiting any tidal-forces, like a classical uniform ...
NaiDoeShacks's user avatar
1 vote

What's the difference? $\nabla_\mu e_\nu=\Gamma_{\mu \nu}^\rho e_\rho~\text{ and }~\partial_\mu e_\nu=\Gamma_{\mu \nu}^\rho e_\rho~?$

The expression $\partial_\mu \mathbf{e}_\nu$, at face value, does not really make sense in curved space due to the vectors being in different tangent spaces. In order to make it work, vectors need to ...
Vincent Thacker's user avatar
0 votes

Does (covariant) divergence-freeness of the stress-energy tensor ${T^{\mu\nu}}_{;\nu}=0$ follow from the Bianchi identity?

${G^{\alpha\beta}}_{;\beta} = 0$ is an identity, but ${T^{\alpha\beta}}_{;\beta} = 0$ is valid only `on-shell'.
Sean's user avatar
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1 vote

Does (covariant) divergence-freeness of the stress-energy tensor ${T^{\mu\nu}}_{;\nu}=0$ follow from the Bianchi identity?

The law of local conservation of $T$ is consequence of the law of motion of the matter part of the total action matter + curvature $$I[g]+ J[\phi, g]$$ These equations of motion for the matter arise ...
Valter Moretti's user avatar
0 votes

Does (covariant) divergence-freeness of the stress-energy tensor ${T^{\mu\nu}}_{;\nu}=0$ follow from the Bianchi identity?

I haven't read Dirac's book specifically, but usually authors prove the Einstein equations are consistent in part by showing that the divergence on both sides is 0. The construction of stress-energy ...
Brick'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
0 votes
Accepted

Question about Young-Laplace equation proof

The observed physical property of surface tension, as tension acting on bodies in the interface, is that this is due to a set of forces that act on all elements of the body that are on the boundary ...
Ján Lalinský's user avatar
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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1 vote

Maximizing proper time with parabolic trajectory in uniform gravitational field

The uniform gravitational field is oriented in the $-z$-direction, with acceleration $g$. Clock A is at rest at $z=0$ and reads time $t_A$. Clock B is allowed to move in three-dimensional space $(x, y,...
Aiden's user avatar
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