This question is not about a moving clock, but about the frame of reference.

@nonymous
Messages
4
Reaction score
0
I found out that Proxima Centauri is 4.2 light years away from earth. If someone was on a voyage to this star via space ship, would this person measure the time to be shorter than 4.2 years if their spaceship was traveling 99.99999 percent the speed of light, assuming that the 4.2 years is measured on earth? Wouldn't this make the voyage a lot less daunting for the traveler?
Doesn't a moving clock move slower?
 
Last edited:
Physics news on Phys.org
Yes, this is one of the prospects for possible interstellar travel.
 
What if he shines a laser pointer out of the front windshield? Does he see it escaping him at the speed of light? I think yes. However the observer sees it just overtaking the ship barely. This seems too bizarre...
 
Everyone always agrees, regardless of reference frame, that light is always going the same speed, the speed of light. It is from this single postulate (and one other smaller one) that all of Special Relativity is derived.
I would suggest this as a place to read up on special relativity and its implications: http://en.wikipedia.org/wiki/Introduction_to_special_relativity
 
@nonymous said:
What if he shines a laser pointer out of the front windshield? Does he see it escaping him at the speed of light? I think yes. However the observer sees it just overtaking the ship barely. This seems too bizarre...
Actually, you cannot see light escaping from you. Once it has left the laser pointer, it's gone and you'll never see it again or have any awareness of its progress, unless it hits something and illuminates it and reflects back to you. Then what you see is a result of the light making a round trip from your laser pointer, to an object, and back to you. So the only way you can measure the speed of light is to have it reflect off of something and measure how far away that something is (with a ruler) and measure how long the round trip took (with a clock or other timing device) and then you can calculate the "average" speed of light during the round trip. But you have no idea whether the light took the same time to get to the object as it did to get back to you.

The second postulate of Special Relativity defines those two times to be equal for any inertial measurement and this is the basis for a Frame of Reference. So when we say that the light propagates away from a high speed spacecraft at c, we mean that it is defined to be traveling at that speed according to the rest frame of the spacecraft and according to the same definition of a different rest frame for the earth, it is also traveling at c. It's not the least bit bizarre once you grasp what a Frame of Reference is.

Einstein's 1905 paper introducing relativity is a good place to learn about this.
 
...would this person measure the time to be shorter than 4.2 years...

You mean "measure the [local] time aboard their own spaceship shorter than 4.2 years..." right? That's because they travel a shorter distance for the trip than was initially measured from earth...not because the person sees her own clock ticking differently.

Time on Earth passes as normal...but is not observed (measured) as such from the
fast moving spaceship. Only when Earth clocks are compared with spaceship clocks upon return would observers recognize different elapsed times.
 
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...
Back
Top