# Tag Info

0

After reviewing the problem further, it looks like there is an error in the text. In fact, transforming to polar coordinates should be $P_\rho(x,y,P_x,P_y)=\dfrac{x*P_x+y*P_x}{(x^2+y^2)^{1/2}}$ and NOT $P_\rho(x,y,P_x,P_y)=\dfrac{x*P_x-y*P_x}{(x^2+y^2)^{1/2}}$ which was given.

0

Manifolds are defined such that locally they look like Euclidean space; this is why we call them smooth manifolds. A riemannian manifold is a manifold that locally has some inner product structure, ie a way of measuring length and angles. Lengths and angles are invariants, hence will have an invariant expression in terms of a local coordinate basis; and ...

0

In Riemannian geometry there is a beautiful theorem which states that a manifold with a symmetric connection is locally flat everywhere if and only if the curvature tensor vanishes. Therefore, in a locally flat coordinates such that $\Gamma_{jk}^i=0$, $g_{ij}$ is constant throughout the chart and a linear transformation can be used to diagonalize the metric ...

0

If $ds^2=\eta_{\alpha \beta}d \xi^{\alpha} d \xi^{\beta}$ were true for all points of space, we would have no curvature, hence no gravity! Take for example a sphere (the Earth), locally we can measure distances by $ds^2=dx^2+dy^2$, but this can't hold for two arbitrary points on the sphere. In fact, this coordinate system changes from point to point ...

0

Let $\mathcal{M}$ be the space time manifold, whose local charts (open sets) are described by $U_i$. A local coordinate frame $S_i$ is a map $\xi\colon U_i\mapsto \mathbb{R}^N$ such that $\xi(m) = (x_1,\ldots,x_N) \in \mathbb{R}^N, m\,\in U_i$. Let, moreover, $g$ be a $(0,2)$ rank tensor (the metric). A change of coordinates is any smooth invertible map ...

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$g_{\mu \nu}(x)$ means that $g$ is a function of location ($x$) --- so it varies across the manifold, which is the problem. I think that if $g \ne g(x)$, then necessarily $g = \eta$ ... Hopefully someone else can chime in on that.

-1

Actually, the assumption of a psuedo riemannian manifold doesn't require many tacit assumptions. Can you measure time and distances? Can you define a right angle? Ok, you now have a manifold equipped with a metric. Want to include time as a dimension? Now you have four dimensions. You can't turn around in time like you can in space, so you need the time ...

1

The velocity is downward, and the acceleration is downward. Whatever direction you choose, if you start with a velocity of zero the sign of both will be the same (if you throw the feather down, it will decelerate - so the acceleration will the "up". I don't think that is intended here). Whether the floor or the hand is zero in the coordinate system doesn't ...

2

The values of individual entries in the metric tensor depend on the coordinate system you choose. In the case of the FLRW metric there is a natural choice of coordinates called the comoving coordinates. In particular the comoving time has a very simple interpretation because it is equal to the proper time of a stationary observer, which obviously means ...

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I think we should start with the local form of Gauss's law $\nabla.\vec{E}=\rho$ Now $$\int \nabla.\vec{E}\,dv=\int \rho\,dv$$ Using Gauss's divergence theorem we have $$\int\vec{E}.\vec{ds}=q$$ I assume $\epsilon_{0}$ to be 1 but you can always put that back into this. I think this way of looking at it does not assume any coordinate dependence.. Ofcourse ...

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While I can't speak to the specific example listed, or the particular meaning of the $S$ term, there are examples of spacetimes that agree up until the singularity. First consider a spacetime that is topologically $\mathbb R^4$ with time being the radial coordinate and then for each time you get a three sphere where you then adjust the scale factor in the ...

2

A space train leaves Mars at 14:00pm and arrives on Earth at 19:45. The train moves at 0.001C and has 40Km of length. How long will it take for the whole train to arrive on Earth? - Disregard re-entry and friction. Nobody on Earth will say the train is leaving mars now. Same thing with the light, just it moves faster and is smaller than the train above. ...

18

A physicist, me for example, identifies events by choosing a set of coordinates. For example I have a clock that I use to record time and a ruler that I can choose to measure distance. This allows me to set up some coordinates $(t, x, y, z)$ so I can assign every event to some point in my coordinate system. If I received a laser pulse from Mars at 16:05 ...

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The continuity equation is $\frac{\partial \rho}{\partial t} + \nabla \cdot \mathbf{\rho \mathbf{v}} = 0$, Now you can substitute directly for $\nabla \cdot \mathbf{\rho \mathbf{v}}$ with the expression for divergence in spherical co-ordinates \$ \nabla \cdot \mathbf{A} = {1 \over r^2}{\partial \left( r^2 A_r \right) \over \partial r} + {1 \over ...

-3

"They call μrϕ˙2 the fictitious force or the centrifugal force. I'm quite hazy on my memory of non-inertial frames, but I was under the assumption that fictitious forces only appear in non-inertial frames." all "fictitious" forces still "transfer energy." Also anything that is an acceleration in math (distance/time^2) is indicative of energy transfer. ...

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