I Can Crafts Reach 2c Relative to Me?

  • I
  • Thread starter Thread starter Abbott287287
  • Start date Start date
  • Tags Tags
    Relative
Abbott287287
Messages
6
Reaction score
0
TL;DR Summary
Separation speed
Simple question for you guys. Thanks in advance for the help.

If you are standing on earth with a watch and hypothetical crafts to your left and right instantly take off 180 degrees apart at the speed of light, at the 1 second mark on YOUR watch they will be 372,000 miles apart. From YOUR perspective, it seems they should be moving apart at 2c. You measure the distance they traveled at its exactly as above. You were stationary and your watch clicked one second. It seems they had to be moving apart at 2c, but that would be the upper limit as to how fast they can move away from each other from your perspective. No laws are being broken as each craft is moving at c, not above it. From each crafts perspective I know it would be totally different. Is this correct or not? If its wrong, please explain how.
VERY much appreciated!
 
Physics news on Phys.org
That's correct, the upper limit for separation speed is ##2c##. Note that a massive spacecraft can't actually reach ##c## in your reference frame, but can get arbitrarily close to ##c##.
 
  • Like
  • Love
Likes FactChecker, Dale and Abbott287287
Much appreciated. Its being argued on a different forum. :smile:
 
Abbott287287 said:
and right instantly take off 180 degrees apart at the speed of light,
That’s not possible, but we can fix this without changing the basic sense of your question by specifying that their speed relative to you is not ##c## but something like ##.99c##.

If one is moving to left at speed ##v\lt c## and the other is moving to the right at the same speed, you will observe them separating at speed ##2v \lt 2c##.

However, their speed relative to one another will be ##2v/(1+v^2)## which is less than ##c##. Google for “relativistic velocity addition” to see how I get that result.

(And note that I am measuring time in seconds and distances in light-seconds so that ##c=1##. This is a handy and very common, almost universal, trick for not cluttering the equations up with a whole bunch of ##c## and ##c^2## factors)
 
I asked a question here, probably over 15 years ago on entanglement and I appreciated the thoughtful answers I received back then. The intervening years haven't made me any more knowledgeable in physics, so forgive my naïveté ! If a have a piece of paper in an area of high gravity, lets say near a black hole, and I draw a triangle on this paper and 'measure' the angles of the triangle, will they add to 180 degrees? How about if I'm looking at this paper outside of the (reasonable)...
From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...
Back
Top