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I'm trying to wrap my head around how geodesics describe trajectories at the moment.

I get that for events to be causally connected, they must be connected by a timelike curve, so free objects must move along a timelike geodesic. And a timelike geodesic can be defined as a geodesic that lies within the light cone.

I want to know why exactly null geodesics define the light cone. Or, why null geodesics define the path of light.

Also, if there's a better explanation why matter follows timelike geodesics, that would also be welcome.

David Z
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P O'Conbhui
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2 Answers2

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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 $ds^2 = 0$, then the above equation takes the form

\begin{equation} cdt = \sqrt{dx^2 + dy^2 + dz^2}. \end{equation}

This is the statement that "the speed of light times the differential time interval, as measured by an observer in a freely falling frame at the location in consideration, is equal to the differential physical distance traveled along the trajectory, measured by that same observer." From Einstein's equivalence principle, this is precisely the way that light must behave.

kleingordon
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    If one is still unsatisfied by this answer, you will want to look into more details about the "derivation" of the Minkowski metric. –  May 12 '17 at 19:20
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    What is the proof that "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 " ?? I've been curious about WHY this is true, and have taken it for granted that it is. – Daddy Kropotkin Nov 04 '18 at 19:59
  • We can make a coordinate transformation locally such that my metric is flat which is fairly easy to obtain by taylor expansion around any event in spacetime. But I am still not satisfied why light should move on null geodesic because by the above argument we can conclude that only in a locally inertial frame light moves on null geodesic and hence speed of light is c. – Ashley Chraya Jun 11 '20 at 03:56
  • @RandomXYZ ds^2 is invariant, so once you determine that ds^2 for your trajectory is 0 in a local inertial frame, you know it's a null trajectory in any other local coordinate system that applies to your problem – kleingordon Jun 11 '20 at 20:19
  • @kleingordon yes , i got it. But the speed of light would not be c in other frame right? – Ashley Chraya Jun 11 '20 at 20:38
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    @RandomXYZ No, the speed of light is c in all frames. You just have to make sure to compute distance in curved space time using the metric, which will not be the same as simply summing the squares of some of the coordinate intervals – kleingordon Jun 11 '20 at 20:49
  • While the equation $c^2dt^2=dx^2+dy^2+dz^2$ is true along the null cone, there is no certainty that the left and right terms of the equation are constant with respect to different observers. In otherwords, the forms $c^2 dt^2$ and $dx^2+dy^2+dz^2$ are not Lorentz invariant (although their difference is invariant). Thus the form $c^2dt^2$ is not invariant, and neither is the "coefficient" $c^2$. So in what sense do the physicists consider this consistent with the supposed absolute velocity of light? Curves on the null cone do not have Lorentz invariant velocities except the zero velocity – JHM Jan 08 '22 at 21:23
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Suppose there is a flash event that we can represent as a light cone as the flash expands.

There are three shutters with detectors around this flash event. The shutters open and close only once and this is almost instantaneous. One shutter opens in spacetime outside the light cone. One Shutter opens in spacetime inside the lightcone and the third opens exactly on the edge of the light cone.

The first shutter misses the flash because it opens before the flash reaches the shutter. According to the spacetime diagram the separation between the flash and the shutter is spacelike because there is a distance of space between the shutter opening and the flash reaching the shutter. The second shutter misses the flash because the flash has already passed over it in the past. Thus the separation between the shutter and the flash is timelike. The shutter that catches the flash has no separation. There is no difference in space or time between the shutter opening and the flash entering the shutter. You could say the difference is Null. Thus you can think of it as a null geodesic.

Chen
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