# Tag Info

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I think you're misinterpreting what the no-hair theorems say. There's a no-hair theorem for stationary electrovac solutions. It applies to electrovac solutions, not to solutions containing any matter field you like; in fact, there are known counterexamples if you include certain types of matter fields. Also, it's a theorem about stationary solutions. A ...

0

I've since found out that where I'm going wrong is that I don't need to find the absolute derivative. I've been told on another physics forum that this problem is framed in terms of Riemann normal coordinates that makes it OK to assume $\frac{D^{2}\xi^{\mu}}{dt{}^{2}}=\frac{d^{2}\xi^{\mu}}{dt{}^{2}}$. Apparently, this is because the distance the cars travel ...

2

For the sake of the explanation I will assume you mean a gas bubble in a liquid*. David Hammen names a few conditions for a bubble to be spherical, in fact you could summarize these all as: for a bubble to be spherical the surface tension has to dominate over other forces (per unit length). If surface tension is indeed dominant than the pressure in the ...

4

Rising bubbles of air in a liquid oftentimes are anything but spherical. These bubbles have haphazard shapes because they are rising and because they are interacting with other nearby bubbles. The combination of drag, turbulence, and mutual interactions prevents those bubbles from taking on a nice, simple spherical shape. Here's a rather non-spherical ...

1

The thing you'll notice about a sphere is that it's symmetrical. very symmetrical. No matter how you rotate it, it looks the same. the surface tension pulls the surface of the bubble into a shape that has even surface tension over the entire bubble. The shape with even surface tension is a sphere. a sphere has the smallest possible surface area for an ...

2

In the context of general relativity it is often stated that one of the main purposes of tensors is that of making equations frame-independent. Question: why is this true? Actually this isn't quite true. General relativity doesn't have frames of reference (except locally, which is trivially true because GR is the same as SR locally). A better way of ...

-1

Nature exists withouth mathematics. Mathematics is just a language tool (a way of visualizing things about how nature behaves). Observations of nature show us that things try to flow to especific regions of space. By definitions, this regions are described as lower potentials (the more potential, the more capacity of doing things). Gradient of the potential ...

2

Incompleteness of a coordinate system is not a canonical definition as that, for instance, of geodesical (in)completeness. It simply means that the domain of the coordinate system does not cover the whole manifold (and perhaps there are several inequivalent extensions of the initial manifold represented by the given domain of the coordinate system). If a ...

2

As Wikipedia says, a great circle is a circle formed by the intersection of a sphere and a plane that passes through the center of the sphere. The great circles parametrized by $\theta =\tau$ and $\phi =\text{const}$ are not all the great circles. They are only the `vertical' great circles, that is, the great circles formed by intersection with a plane ...

1

A full 3D model might be too complex for this case, because then you need to know everything about the wall, and the tree, and the interaction between the two in great detail. I think a simplified approach might be more suitable. I'm not sure if this is oversimplified, but let me take a swing at it: If we assume the tree is supported by both the ground and ...

3

Ghostly Lie algebra cohomology Let $\mathfrak{g}$ be our Lie algebra and $V_\rho$ a representation space with representation map $\rho : \mathfrak{g} \to \mathrm{End}(V_\rho)$. $V_\rho$ is, by the action through the representation, naturally a $\mathfrak{g}$-module (people missing the ring structure in $\mathfrak{g}$ - just embed it into the universal ...

6

A metric structure $g$ and a symplectic structure $\omega$ are two very different structures, although sometimes they can co-exist in a compatible way. Unlike a symplectic structure, there are no Jacobi-like identity and no Darboux-like theorem for a metric structure. There exists a unique torsionfree metric connection $\nabla$ on a pseudo-Riemannian ...

1

So, in classical mechanics, we know nothing of this strange "spacetime" and its metric. We know only that systems are described by $n$ continuous generalized coordinates $q^i$ with a certain range, and we take the manifold $\mathcal{M}$ consisting of all possible different $\vec q$ as our starting point. Note that there is no metric, no form, nothing on ...

3

I'm pretty sure I know the answer to this question, even though this question provides very little context for what the various tensors are, it gives a couple expressions without including the important surrounding equation for context, and it includes an equation with a typo. First of all, as it stands, the first equation has unbalanced indices. I assume ...

1

To calculate explicitly the curvature and geodesic equations for the conical spacetime you need an explicit metric. The metric $ds^{2}=dr^{2}+r^{2}d\phi^{2}$ describe a conical spacetime in the range of definition of the coordinates $(r,\phi)\in (0,\infty)\times [0,2\pi-\alpha)$. You can notice that this metric describe a flat spacetime in the domain of ...

6

I think that it is only necessary to use the cyclic identity. Contracting both sides with the Levi-Civita, we should have $$0 = (R_{abcd} + R_{adbc} + R_{acdb}) \varepsilon^{abcd} \tag{1}.$$ Let $S = R_{abcd}\varepsilon^{abcd}$. Then $R_{adbc}\varepsilon^{abcd} = -R_{adbc}\varepsilon^{acbd} = R_{adbc}\varepsilon^{adbc} = S$ where the last step is renaming ...

9

I was always told that to find whether or not a field is conservative, see if the curl is zero. This is almost always true, but not always true. I have now been told that just because the curl is zero does not necessarily mean it is conservative. Correct! To illustrate what's going on, let's do an example. Conside the following vector field: ...

2

I) The (Lagrangian) canonical conjugate momentum $$\tag{1} p_i ~:=~\frac{\partial L(q,\dot{q},t)}{\partial \dot{q}^i}$$ transforms as a one-form/co-vector $$\tag{2} p_i~=~ p^{\prime}_j \frac{\partial f^j(q,t)}{\partial q^i}$$ under (possibly time-dependent) position coordinate transformations $$\tag{3} q^i~\longrightarrow~ q^{\prime j}~=~f^j(q,t)$$ ...

2

The objects that we call vectors in Euclidean space are, in more general situations, actually two possible different types of objects. If you change coordinates ($q_i\to q_i'$), the components change with different rules, according to how the Jacobian $\Lambda_{ij} = \frac{\partial q'_i}{\partial q_j}$ (using your notation) appears. 1) Vectors (or tangent ...

1

It may be a mistake in Eq.4. Your function $T(r,z)$ does not depend on $\xi$ and has nothing in principle to do with triangles. Since you are only interested in values of $T$ for $(r,z)$ pairs that satisfy Eq.1, you could write $T$ as a function of $r$ and $\xi$, OR as function of $z$ and $\xi$. However, until you do, the partial derivative of $T$ with ...

1

The stationary action principle and the Euler-Lagrange (EL) equations are very broad and general constructions. The field variables in the variational principle could in principle map into some generic manifold $M$. On the other hand, Euler-Poincare (EP) equations appear in the special situation where the manifold is a Lie group $M=G$, and the action is ...

3

The main difference between Hamilton and Lagrange/Newton mechanics is, that it happens directly on the phase space, i.e. any point on your manifold already fully determines the state of your system. Intuitively, you realize this by specifying position and momentum coordinates. On a mathematical level, the world we see is some smooth manifold (a priori not ...

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