# Tag Info

## Hot answers tagged coordinate-systems

9

No, it really is arbitrary. The reason we use the right hand rule today (although it may have been chosen for different reasons of convenience in the past) is simply that our coordinate system of choice is right-handed. Mathematically, this means that we define the directions of the axes so that you have to use the right-hand rule to evaluate this cross ...

8

The question seems to conflate many different things: the invariance of a mathematical quantity (usually a scalar such as $ds^2$ for the separation of two events in special relativity) covariance of tensors (the values of components of tensors may be calculated from those in another frame but they're not the same thing) universality of equations in ...

7

The tensor equations you mention are not invariant, they are covariant. Big difference. Both are differential equations, which transform linearly under nonlinear transformations from one manifold to another because they are differential equations at a point. The nonlinear transformation from one manifold to another induces a linear transformation of the ...

6

You should think of this by timestepping Newton's laws--- if you know the positions and velocity and one instant, you know the force, and the force determines the acceleration. This allows you to determine the velocity and an infinitesimal time in the future by $$v(t+dt) = v(t) + dt F/m$$ $$x(t+dt) = x(t) + dt v$$ You then find the position and ...

6

If you're sitting outside the event horizon watching a clock fall in, you will never see the clock reach the event horizon. You will see the clock slow as it approaches the horizon and you'll see it running slower and slower. However there is no sense in which time stops at the event horizon. You can wait as long as you want, and you'll see the clock creep ...

5

This depends on what you mean by "pass from one manifold to another". In General Relativity one generally considers a single manifold $\mathcal{M}$ and diffeomorphisms $\phi: \mathcal{M} \rightarrow \mathcal{M}$. I think the idea you are trying to get at is that if you consider a geometry on $\mathcal{M}$, that is a pair $(\mathcal{M} , g)$ where $g$ is a ...

5

There are two equivalent descriptions$^1$ of the reduced two-body problem with a central potential $V(r)$: In an inertial frame with no fictitious forces: Here $\frac{1}{2}\mu r^{2}\dot{\theta}^{2}$ is the angular part of the kinetic energy. In a rotating frame following the reduced particle with fictitious forces and only 1D radial kinematics: Here ...

5

You're assuming that the Kruskal–Szekeres (U,V) coordinates have to be defined in terms of the Schwarzschild (r,t) coordinates, but there is nothing special or fundamental about the Schwarzschild coordinates. General covariance says that we can use any coordinates we like. If the K-S coordinates had been the ones originally chosen by Schwarzschild, then ...

4

This is a common confusion students have with coordinate systems when first learning SR. They hear us say things like "the time according to observer Bob" or "in the reference frame of Alice" or "boost to the rest frame of Charlie". To be clear, these are short hand. Something isn't literally/physically "in" one frame and not "in" another frame. When we ...

4

Even in curved spacetime, you can perform a coordinate transformation at any location ("move to a freely falling frame") such that your metric is locally flat , and takes the form $$ds^2 = -c^2 dt^2 + dx^2 + dy^2 + dz^2$$ If you consider a null trajectory where $ds^2$ is set to 0, then the above equation is the statement that "the ...

4

This is just the standard measure for integration given in spherical coordinates. If you write the momentum vector $\mathbf{p}$ in terms of the magnitude $p$ and polar angles you can write $$\begin{array}{lcl} \int\mathrm{d}^{3}\mathbf{p} &\equiv& ... 4 Just use the Jacobian of the coordinate system transformation. If your Cartesian coordinates are \mu and \nu and your cylindrical coordinates are \mu', \nu', then there is a Jacobian {f_\mu}^{\mu'} that allows you to write$$F^{\mu' \nu'} = F^{\mu \nu} {f_\mu}^{\mu'} {f_\nu}^{\nu'}$$where the Jacobian is given by$${f_\mu}^{\mu'} = ...

4

The fundamental difficulty here is that If two elements are orthogonal, it means that measuring one component does not give any information about the other. is incorrect. Orthogonality mean exactly that the inner product between the two things is zero. If $a \cdot b = 0$ then $a$ and $b$ are said to be orthogonal. In a Cartesian space this has the ...

3

All these (infinitesimal) transformations act locally on the world sheet; the strings are extended but physics is (and symmetry transformations and compensations needed to restore a gauge-fixing condition are) still local on the world sheet when interpreted properly. The transformation of individual fields may be computed as the commutators (or ...

3

It depends on whether the coordinates are given globally or locally. In Classical Mechanics, we usually work with a system of coordinates which are global, i.e., they work everywhere. (Usually, not always. You have to look at the context to see which is intended.) Even if they are generalised coordinates. Now in fact, even if they weren't global, it is ...

3

Choosing an appropriate coordinate system often vastly simplifies a problem. Anyone who wants to solve a problem expediently will try to find a coordinate system that simplifies the problem. If your professors told you that physicists do not do this, then your professors told you a falsehood.

3

As dmckee wrote, the term "symmetry" has a fully uniform meaning. It is not used ambiguously in any way and for the same reason, it is not overused. Symmetries are really important in physics and that's why they're used so often. (We also use "symmetries" with various well-defined adjectives such as "global", "local/gauge", "approximate", "broken", ...

3

It is special relativity which only works in inertial frames. General relativity is more general exactly because it can handle accelerated frames. You can really work in any reference frame you like, but transforming into a frame where the earth is stationary and the sun orbits it would result in all sorts of weird virtual forces. You could do it, but it ...

3

$$r^2 = x^2 + y^2 + z^2$$ What does that make $r^{-n}$? As for converting $\hat{r}$, the position vector can be written $$\vec{r} = r \hat{r}$$ in spherical coordinates, but it can also be written $$\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$$ in rectangular coordinates. Therefore, the two are equal. $$r \hat{r} = x\hat{i} + y\hat{j} + z\hat{k}$$ ...

3

You use a computer--- you pick the initial position and velocity, then you find the force (in x,y coordinates) and therefore the acceleration a, then you pick a small timestep $\epsilon$, and you add $\epsilon a$ to $v$, and $\epsilon v$ to x, and repeat. Once you find the solution, you make $\epsilon$ smaller until it stops changing, and this is the answer. ...

3

The well known property of the harmonic coordinates is that the covariant divergence of a vector field and the d'Alambertian of a scalar field take a particularly simple form: $$D_{\mu}A^{\mu} \rightarrow g^{\mu\nu}\partial_{\mu}A_{\nu},\\ g^{\mu\nu}D_{\nu}D_{\mu}\phi \rightarrow g^{\mu\nu}\partial_{\mu} \partial_{\nu}\phi.$$ The harmonic condition $$... 3 Does this simply mean that any sound theory expressed in K, should be able to withstand a transfer to another system Z and still hold "true"? Or is there more to "uniform translation relatively to K"? The 'uniform translation ...' part is crucial. "K' moving in uniform translation relatively to K" means that the relative velocity between them is ... 3 skymandr has covered a useful application of cylindrical coordinates. There are many problems with this kind of symmetry. However, I must differ on one thing: the \hat \phi unit vector doesn't enter into a position vector at all. Any general position vector can be written as \vec r = \rho \hat \rho + z \hat z. This is because the direction from the ... 3 According to the blog Jacobs Physics: Resources for teachers and students of introductory physics: It's perfectly acceptable, and sometimes desirable, not to begin an axis at zero. And: Students will attempt to demand a hard-and-fast rule about scaling graphs from the origin, but such a rule does not exist. The scaling of a graph depends on the ... 3 Consider an equation that you have written in Cartesian coordinates x,y,z which gives you these coordinates as a function of time or something like that. Then you want to write it in, say, cylidrical coordinates. Then you write something like:$$ x=r\cos\phi\\ y=r\sin\phi\\ z=z $$and plug it into your equation. Now just forget the word coordinates. It is ... 2 http://arxiv.org/abs/0803.4441 The Kustaanheimo-Stiefel transform turns a gravitational two-body problem into a harmonic oscillator, by going to four dimensions. In addition to the mathematical-physics interest, the KS transform has proved very useful in N-body simulations, where it helps handle close encounters. Yet the formalism remains somewhat ... 2 The Universe doesn't come equipped with any "clocks at each moment" that would immediately recognize different "moments". Indeed, the laws of physics - except for cosmology - are totally invariant with respect to translations in time, which is just a different way of saying that there is no way, even in principle, to find out whether an event occurred at ... 2 Radius of curvature is governed by a = v^2/r. The radius of curvature thus calculated is good at that instant only, since 'v' will continue to increase; and, if 'a' remains constant, change 'r'. The 'a' in the equation is the component of total acceleration which is normal to the velocity vector, or sin(20^o)(8m/s^2) 2 1) The first derivation is correct. When you make the second one for \dot x written as \dot x = \dfrac{d x}{d t} you make an error when using full derivatives instead of partial ones. In the latter case you would have obtained:$$ \dot x=\dfrac{d x}{d t}=\dfrac{\partial x}{\partial \theta}\dfrac{d \theta}{d t} +\dfrac{\partial x}{\partial \phi}\dfrac{d ...

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