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19

That's a very good question. While it may seem "natural" that the world is ordered like a vector space (it is the order that we are accustomed to!), it's indeed a completely unnatural requirement for physics that is supposed to be built on local laws only. Why should there be a perfect long range order of space, at all? Why would space extend from here to ...


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

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


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

Why do we need coordinate free in the first place? Let me tell you about a related experience I have had teaching students. When I ask them to define the scalar product then the vast majority will write down something along the lines of $$ \vec{v}\cdot \vec{w} = v_x w_x + v_y w_y + v_z w_z \qquad (1) $$ i.e. a coordinate-based description. However, ...


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

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 ...


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 ...


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 ...


6

Here's what's really going on. In classical field theory, a basic set of objects that we often consider are scalar fields $\phi:M\to \mathbb R$ where $M$ is a manifold. Now we can ask ourselves the following question: Is there some natural notion of how a scalar field defined on a given manifold "transforms" under a coordinate transformation? I claim ...


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

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

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

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

The Christoffel symbols do not transform under any representation. The reason for this is that they do not transform linearly, which puts them out of the game altogether. The transformation law is $$ \tilde \Gamma^{\mu}_{\nu\kappa} = {\partial \tilde x^\mu \over \partial x^\alpha} \left [ \Gamma^\alpha_{\beta \gamma}{\partial x^\beta \over \partial \tilde ...


5

This is a great question, with three great answers, and here I am, a bit late to the party. There are two crucial things which the above answers don't seem to address, so I am going to try to give a really simple explanation at those levels. Locality / Manifolds I'm going to come close to giving you one technical definition of a manifold, following ...


5

Let me first tell you that what you read is very vague. In GR the laws of physics are assumed to be independent of the observer. An observer is represented by the a reference frame the observer uses to measure physical phenomena. There are a set of coordinate transformations that relates observables for different observers. Say for example the speed of a ...


5

There are two mathematical concepts which are both called vector. The first one, the vector from linear vector space is the basic "multicomponent object" which you seem to mainly talk about. The second notion of a vector is of a member of the so-called "tangent bundle" of a manifold. The second notion is the one which is defined equivalently with the ...



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