# Tag Info

11

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: ...

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 ...

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 ...

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 ...

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 ...

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 $2 dx^+dx^-= (dx^0)^2 - (dx^3)^2$

6

Why don't we chose a random point in the void present between galaxy clusters and define it as the absolute origin? Why don't we choose a random bachelor and define him as married? The point is this: if you understand the concept bachelor, you know that there is no married bachelor. And, if you understand the term absolute origin, you know that ...

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 ...

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

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

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 ...

5

There isn't any difference here between using curved spacetime and curvilinear coordinates. While one could tell "true" curvature in spacetime by, say, calculating scalars like $R^{\alpha\beta\gamma\delta} R_{\alpha\beta\gamma\delta}$ or $R^\mu{}_\mu$ and seeing if they vanish, the fact is your differential operator doesn't care. Put another way, all you ...

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 ...

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

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. ...

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 ...

4

The comic is making a joke out of reference frames, it is making no such remark about the accuracy of the physical statement that "trains rotate earth." If trains did rotate earth, then two trains coming at each other at the same time would cause earth to stop rotating. However, in your own personal reference frame, the train appears to rotate earth as you ...

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