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From what I'm understanding, the Speed of Light is roughly 186k MPS NO MATTER WHAT!

Then say if a bean of light shoot pass near a Black Hole (Outside the Event Horizon of course) 1k lightyear away to us, it will take 1k year for us to reach us.

But because of the Gravitational Time Dilation, a year near a Black Hole will be a thousand year for us, so for some observer near the Black Hole, the same light reached us in only a year! Thus it had traveled at the speed of 10c!

Then if the Speed of Light is STILL 1c for the observer near the Black Hole, then it would be seem to us only travel at .1c!

Theoretically the light will take a thousand years to reach the destination for both observers, but a thousand Black Hole years will be a million years for us!

Unless for the Black Hole observer, we are only "a light year" away!

I can't wrap my head around this! Somebody PLEASE be so kind and explain it for me!

Much much appreciated!!!

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marked as duplicate by John Rennie gravity Sep 5 at 17:57

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  • $\begingroup$ Please see en.wikipedia.org/wiki/Shapiro_time_delay $\endgroup$ – PM 2Ring Sep 5 at 7:58
  • $\begingroup$ So... to put it simply, the speed of light slowed down near a massive body!? $\endgroup$ – PiggyChu001 Sep 5 at 8:05
  • $\begingroup$ @PiggyChu001 No. The time taken for the light to travel from A to B is longer (from the point of view of a distant observer) if it passes close to a black hole, but the distance from A to B also increases, so the speed of light stay the same. $\endgroup$ – gandalf61 Sep 5 at 16:22
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Light always makes progress through any given region of space at the rate of one local light-second (approx 300 million metres) per local second. Owing to the warping of spacetime, when you integrate this over the path of the light beam, you can get many answers for the average speed along the path, depending on how you choose to define the total time elapsed between start event and end event.

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    $\begingroup$ Do you mean "light second"? $\endgroup$ – badjohn Sep 5 at 8:24
  • $\begingroup$ @badjohn Yes: thanks (edited accordingly). $\endgroup$ – Andrew Steane Sep 5 at 9:31

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