Speed of Light: Does Relative Motion Affect It?

lederhosen
Messages
9
Reaction score
0
does the fact we are moving at 250km per second around our galaxy and 300km per second around other larger galaxys? have any effect on the speed of light? sure they arent that great compared to the speed of light but does it?
 
Physics news on Phys.org
No effect whatsoever. The speed of light is a constant in all reference frames. Even if you are moving at 99.9999...% the speed of light, the speed of light is still c.
 
The basic assumption of special relativity is that there is no such thing as absolute velocity, only one thing relative to another. So if you are moving 250 km/sec around our galaxy, the galaxy is moving 250 km/sec around us, and the speed of light (in vacuo) is c no matter who is measuring.
 
dandy.
 
OK, so this has bugged me for a while about the equivalence principle and the black hole information paradox. If black holes "evaporate" via Hawking radiation, then they cannot exist forever. So, from my external perspective, watching the person fall in, they slow down, freeze, and redshift to "nothing," but never cross the event horizon. Does the equivalence principle say my perspective is valid? If it does, is it possible that that person really never crossed the event horizon? The...
ASSUMPTIONS 1. Two identical clocks A and B in the same inertial frame are stationary relative to each other a fixed distance L apart. Time passes at the same rate for both. 2. Both clocks are able to send/receive light signals and to write/read the send/receive times into signals. 3. The speed of light is anisotropic. METHOD 1. At time t[A1] and time t[B1], clock A sends a light signal to clock B. The clock B time is unknown to A. 2. Clock B receives the signal from A at time t[B2] and...
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...

Similar threads

Replies
25
Views
2K
Replies
10
Views
2K
Replies
42
Views
743
Replies
5
Views
1K
Replies
6
Views
446
Replies
6
Views
2K
Back
Top