# Tag Info

1

This is my first problem, as the modulus of a vector shouldn't be negative. First, while there are many useful properties of introductory linear algebra you should keep in mind with GR, thinking in Cartesian terms with positive definite matrices simply has to go. Vectors in relativity can very much have negative norm. Even though it's not often done in ...

5

There is theory that light cone shape does not depend on the reference frame in which it is viewed. So why we draw light cones near black hole differently? In general relativity, frames of reference are local, not global. Each of the light cones in your diagram corresponds to a certain local frame of reference. An observer using that frame of reference ...

1

Light travels along paths with a metric interval of zero. In flat spacetime this would be drawn as a light cone with a 45 degree opening angle in a standard Minkowski space time diagram.Things get a bit weirder in GR when spacetime is curved by mass/energy. In GR, the concept of an invariant speed of light only applies locally in non-accelerating frames of ...

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

Congratulations, you made me look into this for the last hour! And, unfortunately, I believe the answer is: Nope We are looking for a Ricci-flat Riemannian symmetric space, since your isometry group is a Lie group. I spent some time trying to construct the Ricci-flat manifold from the irreducible symmetric spaces given there, but couldn't figure out a good ...

6

Your second method is correct. To compare, say, the magnetic field with what you find in Jackson, you really need to realize that there's an assumption that you have unit basis vectors there, and that the cross product is actually a hodge dual (which will invoke factors of the square root of the determinant of the metric). These will make direct ...

0

The second form is the way in which the metric was written in the age of Kaluza and Klein. Why? out of embarrassment. If you keep the scalar field when considering the action you get an scalar. That was an undesired feature those times and that's why they hid it making it constant (actually they made it equal to 1). Now, is there a reason why the first is ...

1

Notation: I will use overdot for differentiation with respect to $\tau$, overtilde for partial differentiation with respect to $x^0 = t$, and prime for partial differentiation with respect to $x^1 = r$. (Edit: removed overloading of $\lambda$, sorry.) I assumed a general $\nu = \nu(t,r)$; reading the question more carefully, they're functions of $r$ only, ...

2

I don't think performing a Lorentz rotation/translation will get you anywhere. $T_{\mu\nu}$ and $F_{\mu\nu}$ being tensors automatically makes them frame independent so performing a Lorentz transformation gives you back the same tensor-equation by defininion (though with different components). I will assume that you mean having a large scale EM field ...

3

The formulation you seek is gauge theory. It is not completely analogous to changing the metric of spacetime, but many similarities can be seen. In this, we take as our starting point a certain gauge group $G$ (In the case of EM, $\mathrm{U}(1)$), which will induce symmetries of our theory, just as the Lorentz group of special relativity is the symmetry of ...

3

I guess you are asking about the difference between distances in Euclidean and Minkowskii space. In a "Euclidean spacetime diagram" the distances $ds^2_E=c^2dt^2+dx^2$ would correspond to the lines you draw on the diagram. In Minkowskii space, the lines you draw on the diagram might correspond to particle paths, but they do not correspond to the interval ...

2

Tensors in abstract mathematics are just functions with linear arguments. In abstract index notation, the placement of indices--up vs. down--tells you whether that particular argument should be a vector or covector. For example: $$T^{\mu \nu} \equiv T(\text{covector}, \text{covector})$$ Since both the indices are up, it means both the first and second ...

1

It seems to me that a good starting point is the geodesic equation: [...] This apparently refers to some particular (image of) curve $\gamma$; indeed to some particular time-like curve $\gamma$ for which $$\int_{\gamma} d \tau = \Delta \tau \mid_{\gamma} ~ \gt 0.$$ Given two (not necessarily distinct) (images of) time-like curves $\gamma$ and $\psi$ ...

1

I think it is better to argue from the curvature tensor $R_{ab}{}^\mu{}_\nu$. It is defined by $$R_{ab}{}^\mu{}_\nu x^\nu = (\nabla_a \nabla_b - \nabla_b \nabla_a)x^\mu$$ so it tells you the degree to which covariant derivatives along the $a$ and $b$ axes do not commute. You can see formally from this that curvature requires two dimensions, so it does not ...

1

In order for a manifold to be curved (intrinsic curvature), it must be of dimension $\geq 2$, which means that at least two principal curvatures must be non-zero, since Gauss's curvature is the product of them. This cannot be done with only one curved basis vector or dimension, as you put it; actually, there's no way to define intrinsic curvature in ...

2

Well, by the presentation you give, you're going to have $\frac{d^{2}x^{i}}{d\tau^{2}}\neq 0$, because you have those $\Gamma_{0i}{}^{j}$ terms. For instance ${\ddot y} + 2\frac{\dot a}{a}{\dot y}{\dot t} = 0$ (I abuse notation and mean the obvious things with dots, but obviously, $a = a(t(s))$ and $y=y(s)$) The condition you want is ...

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