# Tag Info

7

Density is a 3-form, since you would write it as $$\omega:=\rho\text dx\wedge\text dy\wedge\text dz.$$ In special relativity it remains (the time component of) a 3-form. More specifically you have a current density $J$ of the form $$J = \rho\text dx\wedge\text dy\wedge\text dz + J_x \text dt\wedge\text dy\wedge\text dz+ J_y \text dx\wedge\text dt\wedge\text ... 3 The theorem does not apply, as we do not have spherical symmetry. All we have is rotational symmetry about a preferred axis. In fact, the gravitational field outside a rotating object will be Kerr, which only reduces to Schwarzschild in the case of no rotation. Otherwise, there will be time-space terms in the metric, making it not static. Still, Kerr is ... 3 I'm not altogether sure what you are asking, but I suspect the following may help. To represent rotations, spins and vectors in SU(2) we work as follows. Rotations live in SU(2). Vectors (in the physicist's sense) live in the algebra \mathfrak{su}(2). The position vector (x,\,y,z) is:$$X =x\,\hat{s}_x+y\,\hat{s}_y+z\,\hat{s}_z = ...

3

There is a relatively fast approach to computing the Riemann tensor, Ricci tensor and Ricci scalar given a metric tensor known as the Cartan method or method of moving frames. Given a line element, $$ds^2 = g_{\mu\nu}dx^\mu dx^\nu$$ you pick an orthonormal basis $e^a = e^a_\mu dx^\mu$ such that $ds^2 = \eta_{ab}e^a e^b$. The first Cartan structure ...

3

The metric times the Kronecker delta gives $$g_{ab} \delta^a_c = g_{cb}$$ Since the Kronecker delta tells us to replace the $a$ indice with $c$. We do this for both terms in your equation, $$\frac{\partial L}{\partial \dot{x}^c} = \frac{1}{2} g_{cb} \dot{x}^b + \frac{1}{2} g_{ac} \dot{x}^a$$ and then rename the dummy indices (the indices that are summed ...

3

You may know about it already, but you can find an excellent account of Lagrangian Mechanics on manifolds in the book Mathematical Methods of Classical Mechanics by V. Arnold. Also to specifically address your question: $L:TM \rightarrow \mathbb{R}$ so that $L$ is a 1-form; ie $L \in \Omega^1 M$ which is neccessary to integrate over a ...

2

In the context of the mathematics of differential geometry, the concept of differential forms is made more precise. It is an invariant object, and does not transform under coordinate changes. In particular, they are not objects that 'inherently' come with indices, so a 2-form like the electromagnetic field tensor would be expressed by mathematicians as ...

2

We do not "swallow" the index. You must distignuish between the geometrical object and its components, and that is not unique to forms, but occurs for all vectors: If you have a vector $v\in V$, where $V$ is some vector space, it has no indices. It's just an element. Now, if you choose a basis $e_1,\dots,e_n$ of the space, you write $$v = v^\mu e_\mu$$ and ...

2

A very short answer might be that in general relativity, spacetime can be curved. To estimate how much it's curved, you need to be able to calculate the rate of change, that is done by differentiating the co-ordinate system you are using to map each region of spacetime you are dealing with.

2

In order to be able to even define a metric, you need tangent vectors, since these are the arguments to the metric, and to have tangent vectors you need differentiability.

2

(1) Of course bosons can be coupled to gravity. The graviton is a spin 2 field. (2) We want to use vierbien to couple fermions to gravity because the Dirac equation is formulated in Minkowski space and we want the behavior of fermions in a general spacetime to be locally like their behavior in Minkowski space. Therefore we use the local frame ...

2

Your professor is telling you something that is absolutely fundamental to a proper understanding of relativity. Suppose we draw out the trajectory of some object on a space time graph, we may get something like this: The path traced out by the object(the blue curve) is called the world line. The length of the world line, $s$, is equal to $c\tau$, where ...

2

The Schwarzschild metric describes the geometry of the spacetime containing a time independant spherically symmetric mass and nothing else. In other words the spacetime has to have existed unchanged for an infinite time and continue to exist unchanged for an infinite time, and there must be nothing else in the universe. Obviously there is no object in the ...

2

A lot of your question really boils down to how to switch coordinates, which already happens in $\mathbb{R}^n$, but usually purely for convenience. In General Relativity, you might be forced to changed coordinates. And there is also the physical implications of general coordinate transformations. This answer will try to address all these issues. Firstly, ...

2

First we sketch a proof that a timelike geodesic is a maximum of proper time. (We exclude saddle points for now.) Let $\gamma$ be a curve satisfying the geodesic equation, i.e. it is an extremum of proper time defined by $\tau[\gamma]:=\int\sqrt{-\langle\dot\gamma,\dot\gamma\rangle}\,\mathrm{d}t$. It is fairly simple to show that there always exists a curve ...

2

Basically, vectors are called contravariant because their components transform oppositely to the basis vectors: if our change of coordinates is such that $$\frac{\partial}{\partial x^i} = \frac{\partial y^j}{\partial x^i} \frac{\partial}{\partial y^j}$$ then if we have a vector $\mathbf{V}$, its components $V^i_x$ in the $x$ coordinates are related to its ...

2

$g^{\mu\nu}$ is the inverse of the metric $g_{\mu\nu}$; and the expression is for a general metric, not for the Minkowski metric. In 4 dimensions, if $g = \det(g_{\mu\nu})$ $$\det(\sqrt{-g} g^{\mu\nu}) = (\sqrt{-g})^4\det(g^{\mu\nu}) = \frac{g^2}{\det(g_{\mu\nu})} = g$$

2

1) The spacelike hypersurface has three spacelike directions tangent to it. Any vector that is normal to all three spacelike directions in the eneveloping space is necessarily timelike. Equivalently, the spacelike surfaces can be thought to be labeled by a function $\tau$ which gives the "time coordinate"'s value on those surfaces. the normal to the ...

2

The notion of derivative requires a notion of comparison. In a general manifold, tangent vectors at different points belong to totally different vector spaces (see footnote 1), so we must define a way of mapping one tangent vector to another tangent space that we shall take, by definition to be the the "invariant image" of the vector in the new tangent space ...

2

The short answer is that calculating the Riemann Tensor is a grind. It will take a while, no matter what way you do it. Presumably you're doing the Schwarzschild metric in the standard (Schwarzschild) coordinates, so you're aided by the fact that the metric tensor is diagonal. This means that $R^\alpha_{\beta \gamma \delta} = g^{\alpha \alpha}R_{\alpha ... 2 So the function defined by $$f(\theta,\phi)=(\sin\theta\cos\phi,\sin\theta\sin\phi, \cos\theta)$$ is not a coordinate chart but rather an embedding$\mathbb{S}^2\hookrightarrow \mathbb{R}^3$. The reason it cannot be a coordinate chart is as follows: A coordinate chart must be a homeomorphism from the open sets$\mathcal{U}_i$of the manifold$M$to ... 2 Actually the equations of motion one ends up with are not manifestly the same: If I let $$L_1 = \sqrt{g_{\mu \nu} \frac{dx^{\mu}}{dt} \frac{dx^{\nu}}{dt}}$$ one finds that the Euler-Lagrange equation is $$\frac{d^2 x^{\mu}}{dt^2} + \Gamma^{\mu}_{\nu\sigma} \frac{dx^{\nu}}{dt} \frac{dx^{\sigma}}{dt} = f(t) \frac{dx^{\mu}}{dt},$$ for a suitable function ... 2 Basically think of it this way. Take the original equation$\tau$=$\intf(x)d\lambda$--------(1) which in differential form becomes$d\tau$=$f(x)d\lambda$--------(2) after a little rearranging gives$\frac{d\lambda}{d\tau}$=$(f(x))^{-1}$--------(3) with the function$f(x)$in this case being equal to$f(x)$= ... 1 I think you have misunderstood the text slightly. In figure 17.7, figure (a) shows a general Newton-Cartan spacetime with random gravitational fields. The trajectories of the freely moving particle worldlines are curves, and there is no global transformation that can simultaneously make them all straight. Figure (b) shows the special case where the ... 1 Mathematically speaking they are the same operator. Usually we reserve the d'Alembertian for 3+1 dimensional spacetime (so in absence of curvature it takes the form$\partial_0^2 - \nabla^2$), while the Laplace-Beltrami operator is defined for an aribtrary dimensional manifold with arbitrary signature. The only possible difference is that sometimes (not ... 1 By "derived the Schwarzschild metric" I assume you mean calculating the exterior solution of the form $$ds^2 = -\left(1-\frac{2M}{r}\right)dt^2 + \left(1-\frac{2M}{r}\right)^{-1}dr^2 +r^2\Omega^2$$ Applications: Describing deflection of light by the sun Precession of the perihelia of the orbits of the inner planets Schwarzschild singularity and ... 1 You set$\rho$equal to one for no reason. In detail the expression for$\Gamma$is: $$\Gamma^1_{01}=\frac{1}{2}\sum_\rho g^{1\rho}\left[\frac{\partial g_{1\rho}}{\partial x^0} + \frac{\partial g_{\rho 0}}{\partial x^1} - \frac{\partial g_{01}}{\partial x^\rho}\right].$$ And since$g_{01}$equals zero (since your metric is diagonal), all four partial ... 1 The geodesic equation is $$\frac{d^2x^\mu}{dt^2} + \Gamma^\mu_{\nu \rho} \frac{dx^\nu}{dt}\frac{dx^\rho}{dt} = 0$$ The coefficient of$\dot{\phi}^2$you're seeing corresponds to$\Gamma^r_{\phi\phi}\$.

1

I think Wald gets it right and Gaul and Rovelli get it wrong. Active and passive have very different feels to them, since active is moving points and passive is just re-choosing coordinates. But the whole point of coordinates is that they describe the points very well, and so mapping the points is completely equivalent to unmapping the coordinates. There ...

1

Can these two pictures be connected in some way? Yes, that's why the Wikipedia spinor article features a picture of a Möbius strip: GNUFDL image by Slawekb, see Wikipedia The Mobius strip also features in the Mathspages Dirac's belt article where you can read that it's "reminiscent of spin-1/2 particles in quantum mechanics, since such particles must be ...

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