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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 ...


8

It's sloppy language that is confusing you here. A Jacobian is not a transformation. The Jacobian of a transformation measures by how much the transformation expands or shrinks volume(/area/length/hypervolume/whatever) elements. Example: let $x' = 2x$. Then $dx = dx'/2$. The Jacobian is the $1/2$, meaning nothing more than "a unit of the $x'$ scale has a ...


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 ...


7

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 ...


7

If you define $x^0=ict$, then I assume one takes $x_\mu=x^\mu$ so that the metric is actually $\eta_{\mu\nu}=\text{diag}(1,1,1,1)=\delta_{\mu\nu}$, i.e. you're dealing with a Euclidean metric. Then $$ds^2=\delta_{\mu\nu}dx^\mu dx^\nu$$ gives the usual outcome : $$ds^2=-c^2dt^2+d\vec{x}^2$$ The usual conventions are as follows: Option one: One defines ...


7

as you wrote, the spacetime invariant can be expressed as: $$ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$$ and from that we normally get: $$ds^2=-c^2dt^2+dx^2+dy^2+dz^2$$ This is not because of some arbitrary imaginary time unit, this is because the metric ($g_{\mu\nu}$) is a diagonal matrix with the coefficients of each term of the $ds^2$ equation: ...


6

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 ...


6

If you compute $|(dx^+)^2 - (dx^-)^2|$, you will not find $ |(dx^0)^2 - (dx^3)^2|$. So, you cannot obtain $x^+,x^-$ (even with a different normalization) from $x^0,x^3$ by a Lorentz transformation. None of the coordinates $x^+,x^-$ is time-like, or space-like, they are both light-like, and the metrics is $2 dx^+dx^-= (dx^0)^2 - (dx^3)^2$


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

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 \begin{equation} ds^2 = -c^2 dt^2 + dx^2 + dy^2 + dz^2\end{equation} If you consider a null trajectory where $ds^2$ is set to 0, then the above equation is the statement that "the ...


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

The integral of a vector, as you have written it, is just shorthand notation for a vector of integrals. Concretely, if we write $\vec r= (x,y,z)$ in cartesian coordinates, then \begin{align} \int d^3r\, \vec r\,\rho(r) &= \left(\int d^3 r\, x\,\rho(r),\int d^3 r\, y\,\rho(r),\int d^3 r\, z\,\rho(r) \right) \end{align} Now, we simply note the ...


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

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 ...


4

A Fourier transform is a linear transformation between two particular bases, the point functions and the periodic functions. The vector space we are talking about here is the space of functions. A Jacobian matrix is a linear approximation for a general transformation. For example, you mention transforming from a Cartesian basis to a spherical basis. This ...


4

We take: $$x=r\cos\theta\cos\phi$$ $$y=r\cos\theta\sin\phi$$ $$z=r\cos\theta$$ Now, you know the definition of the gradient in spherical coordinates: $\vec{\nabla}=\frac{\partial}{\partial x}\hat{x}+\frac{\partial}{\partial y}\hat{y}+\frac{\partial}{\partial z}\hat{z}$ Now, we use the chain rule or each component. For instance, $$\frac{\partial}{\partial ...


4

If you express velocity in polar (planar) coordinates you get:$$\mathbf v = \dot r \hat r + +r\dot \theta \hat \theta,$$ so a correct expression for the kinetic energy would be:$$T=\frac{m}{2}(\dot r ^2 + r^2\dot \theta ^2).$$ To find the expression of the velocity in polar coordinates you can work in different ways. I'll suggest you one, very ...


4

GR doesn't have global frames of reference. No global or local frame of reference is needed in order to define coordinates in GR. Picking a local inertial frame is one possible way of defining coordinates locally, but it isn't the only way, and it doesn't typically suffice as a way of defining coordinates globally. In GR, coordinates are just labels for ...


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

My understanding was that relativistic physics can be expressed in any inertial coordinate system, but not arbitrary systems. No, both SR and GR are capable of dealing with accelerated frames of reference. GR doesn't even have global frames of reference, and it's not valid to think of a coordinate system as being the same thing as a frame of reference. ...


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

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 $$ ...



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