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1

Because if your manifold is not differentiable (and even then, at least $C^3$), you end up doing non-linear distribution theory and having to use Colombeau algebras, and trust me, you do not want that.


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 ...


0

As Danu said, indices are not "natural" part of tensor fields, they are just a pretty outdated formalism of dealing with them. Unfortunately or fortunately, it is also a pretty well-working and efficient formalism, at least in general relativity. Electrodynamics and classical mechanics would be better off using differential form notation imo... Anyways, the ...


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 ...


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 ...


1

Since $\rho$ is already a summed-over dummy index in the first equation, we can't introduce it again. Instead, multiply both sides of the first equation by $g_{\sigma\mu}$: \begin{align} g_{\sigma\mu} \Gamma^\mu_{\nu\lambda} & = -\frac{1}{2} g_{\sigma\mu} g^{\mu\rho} (\partial_\nu g_{\lambda\rho} + \partial_\lambda g_{\rho\nu} - \partial_\rho ...


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 ...


0

The shortest world line (geodesic path) is given by the GR coordinate system. GR has $x_0$ (ct), $x_1$ (x), $x_2$ (y), $x_3$ (z). Often in GR, c is set to 1 (c = 1). Distance squared ($ds^2$) is given by $$ds^2 = -dt^2 + dx^2 + dy^2 + dz^2$$. If one second passes, the shortest distance is to stay in the same spatial location, that is $$ds^2 = -dt^2$$ ...


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

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

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.


1

Mathematicians call lots of things maps, but this concept is really no different from a scalar field. Given a point (which is an element of $M$), the scalar function returns a number (an element of $\mathbb R$). A chart maps an open set $U \to \mathbb R^n$--for each point on $U$, there is a corresponding point in $\mathbb R^n$. That's different from $U ...


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 ...


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 ...


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

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 ...


0

The way the equations are presented seems unnecessarily obscure, as there are only two equations that matter: $$ \frac{d^2r}{dt^2} = \frac{-4M^2+2Mr+(r-5M)r^3}{r^3}\,\dot{\phi}^2 $$ $$ \frac{d^2\phi}{dt^2} = \frac{2(-3M+r)}{(2M-r)r} \, \dot{r}\dot{\phi} $$ These come from the geodesic equation expressed using coordinate time. So you start at some ...


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 ...


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 ...


0

Equation 42.3 states that if you have a sphere with a mass $M$ inside it then the radius excess is given by: $$ \Delta r = \frac{GM}{3c^2} $$ This equation does not follow from preceeding equations. It is introduced as a new fact. To derive equation 42.3 you need to compute the proper radial distance using the metric. This is easier to do than you might ...


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 ...


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 ...


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 ...


2

Basically think of it this way. Take the original equation $\tau$ = $\int$ $f(x)$ $d\lambda$--------(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 $f(x)$ = ...


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 ...


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 $\Gamma^r_{\phi\phi}$.


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 ...


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 ...


0

In modern mathematical terminology, a functor is called covariant when it preserves the direction of the morphisms, contravariant if it reverses it. For a given differentiable map between manifolds (of which a special case would be open sets within the same manifold), the derivative is a map between the associated tangent bundles. This defines a covariant ...


0

Actually, you don't have to use proper time for a parametrisation of the Euler-Lagrange-Equations / geodesic equations in GR. Just take any parametrisation you want. However, if you solve the equations and use initial conditions for a time/light/space-like-path, that geodesic will stay time/light/space-like with $$ g_{ij}V^iV^j = \text{const.} $$ over the ...


0

Imagine you are on the northern hemisphere on Earth (assuming it's a perfect sphere). Now go north with constant speed: you can just go straight north you don't need to steer. Now go east with constant speed: this is something different now, in order to remain on the same latitude circle you must steer northwards constantly. If you don't see why, try to ...


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 ...


0

The momentum is a covector because it is a gradient, and gradients are always covariant. It does what it says on the tin. However, you are right that this is a subtle point and it's not particularly clear at first sight. For a lagrangian of the form $L=T-V$ with $V$ independent of $\dot q$, the canonical momentum is given by $$ p=\frac{\partial L}{\partial ...


0

I think that calculating the Riemann tensor manually is not particularly illuminating, but if you really want to do it, then why ask for help from us and not from a book? It's quite probable that any advice we may give you comes from a book anyway. Having said that, the most powerful tensor manipulation package for mathematica is xAct. It requires some solid ...


0

As derivatives, the Lie and covariant derivatives involve comparing tensors at different points on the manifold. They differ in the prescription given for comparing the tensors at two different points. The key concept with a covariant derivative $\nabla_\xi = \xi^a\nabla_a$ is parallel transport. It is defined so that as you move along a geodesic in the ...


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 ...


0

Maybe it is useful to list a few packages which help you evaluating the Riemann tensor: RGTC Easy (Mathematica) GRTensorII Easy (Maple and limited version for Mathematica) xAct Hard (Mathamatica) If you want a fast calculation I would recommend using RGTC for Mathamtica and GRTensorII for Maple. If you want some special features (manipulation of large ...


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

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 ...


-1

I think the normal is always time-like because when you slice your space-time you do it in such a way such that the normal vector to this hyper-surface is time-like. Thus, time components of the original metric are absent in the induced metric. Which reference are you reading from?


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 ...


0

Smack me down if I am off base, however because Landau-Lifshitz pseudotensor behaves as a tensor only with respect to restricted coordinate transformations, It would be considered as a part of the jet bundle within the manifold when used to this end only? Again, let me know if I am off base, and I will delete my answer.


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

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

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 ...



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