New answers tagged differential-geometry
0
I'm not asking for a definition of a tangent vector. I'm asking what criterion you can use to decide whether a certain object can be described as a tangent vector. For example, how do we know in this coordinate-free context that the four-momentum can be described as a vector, but the magnetic field can't?
If I understand your clarification correctly, ...
0
Mathematicians have their axioms to define what a vector is, physicists start with a vector as a physical quantity that has a magnitude and a direction. Or at least, this is how Feynman defines it in volume 1, 11-4 of his lectures on physics. These two properties belong to the object and can't possibly depend upon the coordinates used to label them.
Edit:
...
3
There are 4 common definitions of tangent vectors, some of which make use of coordinates only casually or even not at all.
Definition via transformation laws
There's a somewhat technical one preferred by some physicists (those who value calculation rules over geometric insight - shut up and calculate, you probably know the type): A vector is just an ...
1
From MTW's "Gravitation" (via Google Books):
Updated answer to edited question:
For example, how do we know in this coordinate-free context that the
four-momentum can be described as a vector, but the magnetic field
can't?
I'm reminded of a relevant section from "A First Course in General Relativity" by Schutz. In section 4.4 on the ...
3
Honestly, this coordinate-free GR stuff (Winitzki's pdf in particular) looks like GR as would be taught by a mathematician--very similar to do Carmo's text on Riemannian geometry. In classic (pseudo-)Riemannian geometry, vectors are defined as derivatives of affine parameterized curves, covectors as either maps on vectors to scalars or as gradients of scalar ...
1
I suppose $f$ is just an arbitrary scalar function on the manifold. I'm not well-versed with the concept of Ricci flow, so I'll try to give a simple operational answer. I also don't understand what exactly you're looking for.
The Ricci scalar $R$ roughly represents the amount of energy stored in spacetime (as curvature). The dilaton is a scalar field which ...
0
Let us denote
\begin{align}
\xi_1 = (0,1), \qquad \xi_2 = (-e^x, e^x/t), \qquad \xi_3 = (e^{-x}, e^{-x}/t)
\end{align}
Each of these killing vectors leads to a conserved quantity
\begin{align}
c_1 &= \dot x_\mu\cdot (\xi_1)^\mu = \dot x t^2 \\
c_2 &= \dot x_\mu\cdot (\xi_2)^\mu = \dot t e^x +\dot x te^x \\
c_3 &= \dot x_\mu\cdot ...
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The main question has already been answered by joshphysics. For the remaining, rescale the variables as
$$ U~:=~cu, \qquad V~:=~cv. $$
The two equations becomes
$$ \dot{U}^2 ~=~U^4+U^2, \qquad \dot{V}~=~U^2 ,$$
with full solution
$$U(t)~=~\pm {\rm csch}(t-t_0), \qquad V(t)~=~\coth(t-t_0)+V_0. $$
OP's sought-for equation now follows from
$$(V-V_0)^2 ...
1
The FLRW metric can be static, this is the solution that Einstein concocted before Hubble observed the expansion of the universe. The only way that Einstein could make his equations static was by introducing the infamous cosmological constant $\Lambda$. The general FLRW metric has the form
$$
\text{d}s^2 = -c^2\text{d}t^2 + a(t)\left[\frac{\text{d}r^2}{1 - ...
4
I prefer to use Killing vectors and conservation laws to solve stuff like this, so let's analyze the problem using Killing vectors, and see if the results agree with your Euler-Lagrange equations.
Notice that the metric is invariant under translations of $v$. The associated killing vector is $\partial_v$ which in turn gives the following conserved ...
1
I'm not sure what OP exactly is requesting, but OP's equation follows e.g. from the general fact that for an arbitrary 2D surface, the Ricci tensor
$$ R_{\mu\nu} ~\propto~g_{\mu\nu} $$
is always proportional to the metric tensor $g_{\mu\nu}$. This is basically a consequence of that in 2D the Riemann curvature tensor is complete determined by the scalar ...
2
I) Pragmatically speaking, the most important property of $\sqrt{-g}$ for model building purposes, is not per se the fact that $\sqrt{-g}d^{4}x$ measures the volume element of a 4-dimensional Parallelepiped with infinitesimal edges $dx^0, \ldots, dx^3$.
II) A more important property is that $\sqrt{-g}d^{4}x$ transforms as a scalar (i.e. is invariant) under ...
0
Obviously, you are talking about the derivation in Landau-Lifshitz book. I admit that this is not the clearest explanation of what is going on. (On the other hand that is not the first and not the last place like that in Landau-Lifshitz...)
Anyway, I don't really see what is your problem -- just use Stokes' theorem from (6.19):
$$\Delta A_i = \frac12\int ...
0
I've made a small document featuring fluid dynamics equations in terms of vector-valued differential forms. The document with information on any further developments can be found on my page.
1
The metric is spherically symmetric. This means that angular momentum of the system is conserved (you can show this directly using the metric by computing the three killing vectors associated with spatial rotation and their corresponding conserved quantities) and therefore that the motion is contained to lie in a plane. If the motion is in a given plane, ...
0
I think there is a method that I believe is rather simple. Take a look:
There is a thing called 'normal Riemann coordinates'. In this coordinates the metric is expanded around the origin, and the coefficients of expansion are expressed in terms of the Riemann tensor. I suggest that you read about them and check whether the coordinates described below are ...
0
Given that your metric is diagonal, it simplifies a lot these calculations. However, the Riemman tensor is such an object...
First, start with the Christoffel Symbols
$$ \Gamma^i{}_{k\ell}= {1 \over 2} g^{im} (g_{mk,\ell} + g_{m\ell,k} - g_{k\ell,m})$$
Note that $g_{im}=0$ for $i \neq m$ so it simplifies to
$$ \Gamma^i{}_{k\ell}= {1 \over 2} g^{ii} ...
3
To add a bit of fluff to twistor59's answer, let's take a bird's eye view of Riemannian geometry.
The Riemannian metric gives us the notion of lengths and angles as well as the concept of straight lines (geodesics).
Any submanifold inherits these notions from the ambient space, made explicit via the first fundamental form, which makes the submanifold is a ...
7
For the first fundamental form - if you've got two vectors tangent to $\Sigma$, and $\Sigma$ is embedded in $M$, and $M$ has a metric, just use the embedding to consider the vectors as living tangent to $M$ and use $M$'s metric to compute their inner product.
For the second fundamental form, basically, if you imagine a two surface $\Sigma$ embedded in ...
1
As soon as you get something like $\delta_{bd}$, alarm bells should ring, as this is not a tensor.
The inverse metric $g^{ac}$ is defined by the identity
$$
g^{ac}g_{cb} = \delta^a_b
$$
If you plug this into your expression (and use the fact that $g$ is symmetric), you will obtain the correct equation.
1
Great question!
I would say that "force is a one-form" is a statement that has some truth to it, but it's somewhat context-dependent.
In any context where you have a metric, you can freely convert back and forth between vectors and one-forms, and the distinction between them becomes uninteresting. Examples of such contexts include relativistic spacetime ...
2
Perhaps some insight into this problem can be gleaned from relativity. In relativity, the EM field is represented by a two-form (i.e. with 6 components). When this two-form is fed a current to act upon, this yields a force--or rather, four-force, so you get power as well as force.
What does this mean? Well, you can feed the four-force any timelike unit ...
0
It can be show easily by the next reasoning.
$$
DA_{i} = g_{ik}DA^{k},
$$
because $DA_{i}$ is a vector (according to the definition of covariant derivative).
On the other hand,
$$
DA_{i} = D(g_{ik}A^{k}) = g_{ik}DA^{k} + A^{k}Dg_{ik}.
$$
So,
$$
g_{ik}DA^{k} + A^{k}Dg_{ik} = g_{ik}DA^{k} \Rightarrow Dg_{ik} = 0.
$$
So, it isn't a condition, it is a ...
1
Unlike classical electromagnetism, General Relativity is highly nonlinear--this means that the gravitational field can serve as its own source. A consequence of this fact is that fields decidedly do not superpose, and you can get all sorts of effects even from vacuum relativity. The most notable of these effects are things such as Brill waves and Geons, ...
1
One way of phrasing this is that the nonlinearity of the equations means that you can't say that the sum (or difference) of two solutions to the Einstein field equations is also a solution. So even taking the delta with respect to the Minkowski metric, as you would like to do, isn't allowed (unless you work perturbatively as John alluded to).
2
The metrics don't simply add together as you suggest. In fact there is no known solution for the metric when you have two point masses (thought there are approximate solutions). If there was it would make calculating the motion of binary black holes a great deal simpler than it currently is. The curvature has to be calculated numerically.
0
The trickiness is what you mean by a spherical metric. What you've written down is the metric of flat space in spherical coordinates, which can be thought of as a warped product of the flat minkowskian two space $(t,r)$ with the unit sphere. This space is equivalent to the normal $(t,x,y,z)$ coordinates of standard special relativity under a coordinate ...
2
A couple of preliminaries:
(1) The Schwarzschild metric is not just the metric for a black hole. It's the exterior metric for any spherically symmetric, nonrotating gravitating body. For example, it's a very good approximation to the earth's metric, since the earth is nearly spherical and is not rotating at relativistic speeds.
(2) Let's take units with ...
2
Metric signature is a coordinate-invariant notion. Given a metric, one computes the number of positive and negative eigenvalues that it has, and this gives its signature. For a diagonal metric, like the metric
$$
ds^2 = dr^2 + r^2 d\theta^2
$$
both diagonal components are positive, so the metric has precisely two positive eigenvalues, and its signature ...
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