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Stretched string and the Einstein cross

A geodesic is not the path followed by a stretched string. It is the path followed by a thrown rock. You can clearly see the difference on Earth. A stretched string is straight. The trajectory of a ...
mmesser314's user avatar
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R-W Metric and null geodesic path of photon

In any metrics in general relativity, the photon geodesic is light-type by definition. Consequently, the scalar product of an elementary displacement of the photon $ \overrightarrow{ds}.\...
Cornelius Fyla's user avatar
1 vote

Equation for the curve of a free falling particle in Kruskal diagram

This is not intended to be a formal answer. \i just wanted to post the diagrams I referred to in the comment as I feel they may be useful to the discussion. This first one is from Mathpages; This ...
KDP's user avatar
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5 votes

Is the definition of geodesics different if the electromagnetism is added to GR?

In presence of external forces the motion is non-geodesic, as explained in other answers. One way to see it is that the full action of a point particle in a curved background that is coupled to ...
Bronstein's user avatar
1 vote

Is the definition of geodesics different if the electromagnetism is added to GR?

No, the geodesic equation is the analog of $x''=0$ (N2L with no external force). With the presence of electromagnetism, the equations of motion become as you have stated and is the analog of $x''=F/m$ ...
Toyesh Jayaswal's user avatar
8 votes
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Is the definition of geodesics different if the electromagnetism is added to GR?

Geodesics are still the same: they are the paths that a particle in free fall follows. However, charged particles are no longer in free fall. They are subject to the Lorentz force. This is described ...
Níckolas Alves's user avatar
2 votes
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How does this canonical transformation on a Schwarzschild black hole work?

We restrict phase space to the equatorial plane $\theta=\frac{\pi}{2}$ and $p_{\theta}=0$. We consider the Schwarzschild solution $$\begin{align} ds^2~=~&-f(r)dt^2+\frac{dr^2}{f(r)}+r^2(d\theta^2+...
Qmechanic's user avatar
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Necessity of equivalence principle

It's possible to take different assumptions to start formulating general relativity, but any assumptions which reproduce the standard theory will imply the equivalence principle, because it's just ...
11zaq's user avatar
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Geodesic equations and Carter constant: how can I derive the right equations?

In my opinion there is a mistake related to the derivative of $ S $. For any type of particle, you have $ S = -\frac{1}{2}\kappa\lambda-Et+L\phi+S_r(r)+S_\theta(\theta) $ with $ \kappa=-1 $ for time-...
Cornelius Fyla's user avatar
4 votes

Necessity of equivalence principle

There are several equivalence principles, the weak, Einstein and strong EPs, and general relativity is based on the Einstein equivalence principle. This states: the weak EP is valid. The outcome of ...
John Rennie's user avatar
11 votes

What mistake did Einstein make in 1911 when he miscalculated the light deviation?

The 1911 paper can be found here in German and there is an English translation here. The GR prediction is roughly $\displaystyle \frac{2GM}{R}\left(\frac{1}{v^2}+\frac{1}{c^2}\right)$ radians, and he ...
benrg's user avatar
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