# Tag Info

14

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: ...

12

In hindsight, here is a short proof. The metric $g_{\mu\nu}$ is the flat constant metric $\eta_{\mu\nu}$ in both coordinate systems. Therefore, the corresponding (uniquely defined) Levi-Civita Christoffel symbols $$\Gamma^{\lambda}_{\mu\nu}~=~0$$ are zero in both coordinate systems. It is well-known that the Christoffel symbol does not transform as a ...

12

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 ...

11

What you're asking about is the existence of surfaces of simultaneity. In SR, surfaces of simultaneity can be defined by measurement procedures such as Einstein synchronization, and they turn out to depend on one's frame of reference. In GR it gets a lot tougher to do this. We don't even have global frames of reference. It turns out that what you need in ...

11

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 ...

10

It sounds like you're interested in when a spacetime admits a Cauchy surface. The answer is that every spacetime that is globally hyperbolic has this property. This was proved by Geroch in 1970 (article here, see Section 5). This includes most of the textbook relativistic spacetimes --- Schwarzschild, Kerr, FLRW, and many others. But there are some ...

9

There is an effortless way, if you accept geometrical reasoning. You know, that $T = \frac 1 2 m \vec v^2 = \frac 1 2 m \lvert \vec v \rvert^2$. Furthermore, spherical coordinates are orthogonal, therefore you can just write: $$\lvert \vec v \rvert = \sqrt{v_\phi^2 + v_\theta^2 + v_r^2}$$ Geometrically, one easily finds: $v_r = \dot r$, $v_\theta = r \dot ... 9 If you define$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 ... 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 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 ... 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 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 ... 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 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 ... 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 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 ... 7 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 ... 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 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 ... 6 Here I just want to mention that there exists a direct proof in$1+1$dimensions using elementary arguments. Let the two coordinate patches$U_x$and$U_y$(which are, say, both convex sets in$\mathbb{R}^2$, containing the origin) have light-cone coordinates$x^{\pm}$and$y^{\pm}$, respectively. The metric reads $$dy^{+}dy^{-} ~=~ dx^{+}dx^{-}.$$ ... 6 This axiom is a "smoothness" property of manifolds. You could define a manifold without this property simply as a space which is locally Euclidean. However, then something which looks smooth in one chart may be highly non-smooth in another chart. For example a smooth curve in one chart may be discontinuous in another chart (since the representation in 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 ... 6 The two solutions are different because they have different boundary conditions. In the first case, the equation is indeed $$\frac{\partial^2u}{\partial t^2} = c^2 \nabla^2 u = c^2 \left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}\right) u.$$ Here though we specify$u(t,x=x_0)$to be some value ... 5 Let's first assume that the scalar product that is preserved has positive signature to show the main idea. Also, you say you don't want to assume homogeneity but this is already implicit in your equation since to form intervals differences of space-time points are used so we might as well take one of those points to be zero of a vector space (equivalently, ... 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$\vec r= (x,y,z)in cartesian coordinates, then \begin{align} \int d^3r\, \vec r\,\rho(r) &= \left(\int d^3 r\, x\,\rho(r),\int d^3 r\, y\,\rho(r),\int d^3 r\, z\,\rho(r) \right) \end{align} Now, we simply note the ... 5 There isn't any difference here between using curved spacetime and curvilinear coordinates. While one could tell "true" curvature in spacetime by, say, calculating scalars likeR^{\alpha\beta\gamma\delta} R_{\alpha\beta\gamma\delta}$or$R^\mu{}_\mu$and seeing if they vanish, the fact is your differential operator doesn't care. Put another way, all you ... 4 In general, the expression for the metric and the expression for the coordinates need to work together to give you the correct line element. So the following combinations all have the same line element$ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$:$g_{\mu\nu}=diag(1,1,1,1)$with$x^{\mu}=(ict,x,y,z)$or$g_{\mu\nu}=diag(-1,1,1,1)$with$x^{\mu}=(ct,x,y,z)\$ or ...

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 ...

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