New answers tagged

7

Actually the result is more strong: Given a timelike geodesic $\gamma$ and a point $p \in \gamma$, there is a neighborhood $U \ni p$ equipped with coordinates, $x^0,x^1,x^2,x^3$ such that in the portion of $\gamma$ included in $U$, exactly along $\gamma$, the derivatives of the metric vanish in the said coordinates. Equivalently the Christoffel symbols $\...


0

Here is what I understand: if you have a particle at state $|x \rangle$, active translating it by $a$ means moving the particle to state $ | x + a \rangle$. Passive transformation means you keep the particle in the same place, and change the coordinate by new variable $x = x' + a$ (note that the coordinate system is translated backwards $-a$). I am not very ...


0

Yes, this is usually what is meant by a negative longitude, it's just a convenient way to express a range that happens to straddle the zero-point. And in the context of the HiGAL survey you were looking at, this is precisely what is meant.


1

What your book meant was, "When applying the equations of motion, it is important that the acceleration of a particle be measured by an observer traveling in a reference frame that is either fixed (in space) or translates with a constant velocity'. The n-t coordinate system is fixed in space, and is not traveling along with the particle. The reference ...


2

The geodesic equation is given by, \begin{equation} \frac{\mathrm d^2}{\mathrm d\tau^2}x^{\mu}+\Gamma^{\mu}_{\lambda\sigma}\frac{\mathrm dx^{\lambda}}{\mathrm d\tau}\frac{\mathrm dx^{\sigma}}{\mathrm d\tau}=0 \end{equation} which is a set of 4 equations for $x^{i}$. $\Gamma^{i}_{jk}$ tells us about the curvature of the space time which can be written in ...


2

A metric defines a flat space within an open neighborhood $U$ if and only if the Riemann tensor $R$ vanishes over that neighborhood. So you simply have to calculate $R$ and check that it vanishes in the neighborhood. The only if part of the assertion is clear. The if part is probably more interesting to you and indeed a simple proof of the if shows you (in ...


6

I'm not sure if it helps you with your students, but maybe gives you some background: I guess the underlying reason for orthogonal basis vectors is that you are implicitly using a euclidean metric that will just have diagonal values. These would e.g. be $$g_\mathrm{\mu\nu, ~euclidean}=I=\pmatrix{1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1} \...


3

We don't always use orthogonal coordinate frames. For example working with three phase motors it's sometimes convenient to work with a three axis coordinate system in a plane. Convenience, simplicity set aside, the main reason we most often work with orthogonal reference frames is the concept of dimension. We can express an n-dimensional linear system as a ...


0

Here are a few examples of simple notation you could adopt. Rotations in $d$ dimensions can be represented using $d\times d$ matrices (call them $R$) such that $R^TR$ is the identity matrix. If your initial vector that describes a 'positive' direction is $v$, then you can express directions relative to $v$ as $R v$, where $R$ is one of these rotation ...


0

I suppose you are asking about locally inertial frames? Chapter 2.4: Postulate (2) of general relativity implies that at each point of spacetime it is possible to choose locally inertial coordinates: $\xi^m$ Say you have coordinates $x^\mu$ and you want to transform to inertial coordinates $\xi^m$ which are in locally inertial frame $ds^2=\eta_{m ...


1

In the context of GR and the equivalence principle, given a Lorentzian manifold $(M,g)$, the following comments seem relevant: If the (Levi-Civita) Riemann curvature tensor does not vanish in a point $p\in M$, then there does not exist a neighborhood $U \subseteq M$ of $p$ (and a coordinate system defined on $U$) such that the metric $g_{\mu\nu}$ becomes ...


0

The hypersurface at t = 0 as shown in the diagram is not the observers plane of instantaneous, contemporaneous events constituting the present. The speed of light is as instantaneous as anything gets because at the speed of light no time passes. (Therefore the present is relative.)


0

There's a plain distinction between a reference system (to use a more contemporary designation) and a coordinate system: a coordinate system is a reference system together with an (one-to-one) assignment of a coordinate value ($n$-tuple) to each element. Expressed more formally, a reference system to begin with is constituted by a set of distinguishable, ...


1

As the saying goes, a picture is worth a thousand words. Below is a time-spatial axis diagram of the time and x-axis in two frames. One is at rest relative to the blue, and the other is Lorentz boosted to some velocity. What is the plane of simultaneity is dependent upon the frame that you are on


4

Quite clearly the answer to this is that no, it does not. In particular, consider two inertial observers moving (in flat spacetime) relative to one another. We know that neither of these two observers is more privileged than the other: the laws of physics are the same for each of them and so on. Yet they will draw different hypersurfaces of simultenaity ...


4

The hypersurface of simultaneity does not represent the present. It is just a plane cut through spacetime. If you changed your own state of motion this would tilt the plane by some angle. So the notion of "now" as the hypersurface of simultaneity would depend on your state of motion, which is of course not meaningful. In fact, the notion of "now" itself is ...


0

I would work out your barycentric co-ord first, then compare them with a matrix subtraction and normalisation (Most efficient in code). This would give you a vector multiplier to from 0-1 (percentage) per component. Next work out the maximum velocity in each dimension given the maximum offset. The simply times your normalised components by this velocity.



Top 50 recent answers are included