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1

This is my first problem, as the modulus of a vector shouldn't be negative. First, while there are many useful properties of introductory linear algebra you should keep in mind with GR, thinking in Cartesian terms with positive definite matrices simply has to go. Vectors in relativity can very much have negative norm. Even though it's not often done in ...


5

Let's start at the beginning: The setting for relativity - be it special or general - is that spacetime is a manifold $\mathcal{M}$, i.e. something that is locally homeomorphic to Cartesian space $\mathbb{R}^n$ ($n = 4$ in the case of relativity), but not globally. Such manifolds possess a tangent space $T_p\mathcal{M}$ at every point, which is where the ...


0

You're so close to the answer I'm not sure how to nudge you along without basically just giving you the answer. I'll try anyways. There are some troubling conceptual mistakes you've made in an otherwise straightforward derivation. For starters : The reason I don't set $E_0 x_0$ to zero is because there is no $1/x$ term; otherwise I'd be able to make the ...


4

I think the confusion arises from indistinct usage of the term scalar fields when referring to a physical field, and when referring to scalar functions. I will try to answer your question clarifying separately these elements: Scalar field vs Scalar function Complex numbers' polar representation What this parametrization means... Scalar field vs Scalar ...


11

Why is there no curvature outside this spherically symmetric, non-rotating, uncharged body that still has mass? I suspect you're getting confused by the fact that the Ricci tensor $R_{\mu\nu} = 0$ and therefore the scalar curvature $g^{\mu\nu}R_{\mu\nu} = 0$. This is always the case in regions of space where the stress-energy tensor is zero. The ...


4

Use $$ \begin{align} \frac{a_{23}}{a_{33}} & = \frac{ -\cos b \sin a}{\cos a \cos b} = -\tan a \\ \frac{a_{13}}{\sqrt{ a_{23}^2 + a_{33}^2 }} & = \frac{\sin b}{\cos b \sqrt{\sin^2 a +\cos^2 a}} = \tan b \\ \frac{a_{12}}{a_{11}} & = \frac{-\cos b\sin c}{\cos b \cos c} = -\tan{c} \end{align} $$ a = atan2(-a23,a33) b = atan2(a13, ...


1

Now if I consider a particular set of rotation (say X first, then Y , then Z), with the corresponding Tait-Bryan angles --- a,b and c. My rotation matrix will be the following ... Look at that array. It's of the form $$\begin{array}{ccc} \cos(b) \cos(c) & -\cos(b) \sin(c) & \sin(b) \\ \cdots & \cdots & -\cos(b) \sin (a) \\ \cdots & ...


1

I don't think I fully understand how the Eulers angles were defined, my intuition to these derivation should be defined like this, 1. rotate about z axis 2. rotate about y axis, 3 rotate about x axis. But I can't convince myself why I'm wrong. I hope any experts can point me out. The sequence depicted in the original post involved A rotation by angle ...


1

The rigid body that's being rotated is arbitrary; it need not be rotationally symmetric. To use Euler angles, we assume that rotation is executed by the body being rotated (say, an airplane with flaps, or spacecraft with thrustors) instead of by some external manipulator. So for all intermediate rotations, we have to consider the perspective of the body. ...


2

This 3D problem (v2) is related to this Phys.SE post. A general strategy to deal with quantization in any curved coordinate system (in any dimension) is outline there. The answer is that the appropriate quantization of the classical Hamiltonian for a free particle in spherical coordinates is $-\frac{\hbar^2}{2m}$ times the Laplacian $\nabla^2\equiv \Delta$ ...


1

This is essentially the same effect that you get in special relativity as the velocity approaches the speed of light. If you take a clock and accelerate it towards the speed of light then it will run slowly. If you could get the clock to the speed of light (which you can't of course) then it would stop completely. To use your words for any infinitesimally ...


1

It seems to me that a good starting point is the geodesic equation: [...] This apparently refers to some particular (image of) curve $\gamma$; indeed to some particular time-like curve $\gamma$ for which $$\int_{\gamma} d \tau = \Delta \tau \mid_{\gamma} ~ \gt 0.$$ Given two (not necessarily distinct) (images of) time-like curves $\gamma$ and $\psi$ ...


1

I think it is better to argue from the curvature tensor $R_{ab}{}^\mu{}_\nu$. It is defined by $$R_{ab}{}^\mu{}_\nu x^\nu = (\nabla_a \nabla_b - \nabla_b \nabla_a)x^\mu$$ so it tells you the degree to which covariant derivatives along the $a$ and $b$ axes do not commute. You can see formally from this that curvature requires two dimensions, so it does not ...


1

In order for a manifold to be curved (intrinsic curvature), it must be of dimension $\geq 2$, which means that at least two principal curvatures must be non-zero, since Gauss's curvature is the product of them. This cannot be done with only one curved basis vector or dimension, as you put it; actually, there's no way to define intrinsic curvature in ...


2

Well, by the presentation you give, you're going to have $\frac{d^{2}x^{i}}{d\tau^{2}}\neq 0$, because you have those $\Gamma_{0i}{}^{j}$ terms. For instance ${\ddot y} + 2\frac{\dot a}{a}{\dot y}{\dot t} = 0$ (I abuse notation and mean the obvious things with dots, but obviously, $a = a(t(s))$ and $y=y(s)$) The condition you want is ...


0

In the drawing you give, it appears that the 'down' direction is the direction of positive displacement, i.e., the $x$ unit vector points towards the bottom of the page. Thus, a positive acceleration $a = \frac{d^2x}{dt^2}$ is towards the bottom of the page in the direction of increasing $x$. If you calculate a negative value for $a$, you know the ...


0

In general it doesn't make sense to talk of curvature being only in space or only in time. The geometry of a spacetime is described by the metric. Normally we start with some distribution of matter/energy and solve the Einstein equations to calculate the metric. Alternatively you can start with the desired metric and use the Einstein equations to work out ...


2

The acceleration always points in the same direction as the force. That's because Newton's second law tells us that: $$ \vec{F} = m\vec{a} $$ where the force $\vec{F}$ and the acceleration $\vec{a}$ are both vectors. So to work out which direction the acceleration is just ask which direction the force is. In the example you've given consider whether the ...


6

Your intuition is correct that you should be able to use this formula in any coordinate system. Indeed, you have presumably done problems with the scalar potential where you have a similar formula that you can evaluate in any coordinate system. However you have to be more careful when defining what you mean by this integral in the vector case in general ...


1

First off, I'm not entirely sure of what you are asking, or what you are thinking of as curvatur. There are certainly coordinate systems which are non-euclidean that are not considered to be "curved." For instance standard cylindrical coordinates have zero Riemann Curvature, but they are "curvy looking." My take though is that in GR you have a coordinate ...


3

Comments to the question (v5): In General Relativity (GR), the notation $x^{\mu}$, $\mu=0,1,2,3,$ usually denotes some (local) coordinates of a (spacetime) manifold $M$. Note that $x^{\mu}$ does in general not transform as a $(1,0)$ (contravariant) tensor in the sense that $$ x^{\prime \nu}~=~\frac{\partial x^{\prime \nu}}{\partial x^{\mu}} x^{\mu} ...


8

A physicist would write your first equation $x^a = x^\mu e_\mu^a$. The notation $x^a$ is invariant in your terminology. The $a$ is an abstract index. It is ostensibly not supposed to be thought of as ranging over a set of numerical values, but is just a marker that indicates that $x$ is a vector (i.e., rank 1,0 tensor.) Similarly for each $\mu$, $e^a_\mu$ is ...



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