I read somewhere that if you were traveling at say 90% of the speed of light, that it would take quite a bit less than 4 years to get to Proxima Centauri because of time dilation form your point of view, even though it’s 4 light-years away. But wouldn’t that mean that you would see light form earth take longer to from the Sun to Proxima Centauri than it would take you to get there? Now the answer is probably no, but why? Is it that distance dilation is having an effect? or that the whole size of a light year has changed? or am I just hopelessly lost here?
And on a side note, since light always appears to travel at the same speed, how could you possibly know that you are traveling at any percentage of that speed?
From your reference frame in the rocket, the distance is shortened for both you and for the light. Furthermore, from your point of view you’re not moving at all (rather, alpha Centauri would be moving towards you at .9 c), but the light is still zipping past you at the Speed of Light. No matter what reference frame you consider, the light will always beat you there.
The fundamental postulate of relativity is that the speed of light is a constant to all observers.
This means that if you in your spaceship are traveling at .9C away from our sun, you will see light from our sun arriving at your spaceship…at the speed of light. Similarly, you are traveling at .9C toward Proxima Centauri. So you see light from proxima centauri arriving at your spaceship…at the speed of light.
There is no such thing as “distance dilation”. In point of fact, the equations of general relativity can be solve two different ways. In one case, you consider distance to be a variable (in other words, the distance from A to B varies according to the path you take between A and B - which is to say: there is a distance dilation effect). Solving the equations this way leads to non-analytic solutions (look up the term if you don’t understand it) because as we all know the distance from A to B does NOT change depending on the path you take; the distance from A to B is the same regardless of the path you take, though the distance you travel will vary with your path.
Thus, this set of solutions is discarded because it is non-physical.
The other way you solve the equations is to assume that distance is constant but time is variable. Solving the equations this way leads to results that, while counter-intuitive, are sensible and analytic. Thus, these results are considered physically valid - the more so because there is experimental support for this.
Given that time is a variable and the speed of light is a constant to all observers, then what immediately drops out is that the wavelength of incident light varies according to the velocity difference between the source and the destination.
Thus, your spaceship at .9C sees light from the sun arriving at the speed of light, but it’s wavelength is greatly lengthened - in other words, it is shifted toward the red end of the spectrum (redshifted). Similarly, the light from proxima centauri arrives with a greatly reduced wavelength (blueshifted).
Another thing that drops out of this is that there is no universal time. It is literally meaningless to compare times between two different points in the universe because time flows at different rates in different parts of the universe due to relative velocities as well as relative gravitational potentials.
Because there is no universal time, and because the speed of light is a constant to all observers, people on the Earth and people on the traveling spaceship will not agree upon the distance between objects. People on the earth will measure the distance from the spaceship to proxima centauri as being a longer distance than is measured by people in the spaceship. However, because they also cannot agree on the wavelength of the light reaching them, all the math works out properly.
This is not a subject that can be easily discussed without mathematics. I suggest you look up redshifts, special relativity, and the twin paradox to learn more.
There’s no such thing as distance dilation simply because “dilation” refers to things getting longer. What happens with distances is contraction: They get shorter.
And again, that measurement is reference frame specific.
An observer on Earth will measure that spaceship as being foreshortened along its length because of its velocity relative to earth. An observer on the spaceship will observe no foreshortening at all.
I don’t mean to hijack this thread but I have a related quesiton which, once answered, would help the OP.
Question: what fraction of c do I need to travel at (with respect to some object, heading directly towards that object) for it to seem as though I am travelling 1 light year distance in exactly a years time from my point of view (ignoring any gravitational effects).
I just realised the question was not well formed, let me rather ask:
what fraction of c do I need to travel at (with respect to some object, heading directly towards that object, from an observers perspective when the observer is stationary with respect to the object) for it to seem as though I am travelling 1 light year distance in exactly a years time from my point of view (ignoring any gravitational effects).
Imagine two spaceships. One the normal sized you’re travelling from Earth to Proxima Centauri in, one a 4 ly monstrosity parked alongside your path. In the travelling spaceship’s reference frame, the monstrosity is the one moving at .9 c and the one being foreshortened. Just as the travelling path from Earth to Proxima Centauri is foreshortened.
In Earth, PC and the monstrosity’s reference frame, the travelling spaceship is the one getting shorter.
Assuming that that last term was supposed to be c^4 instead of c^2, to make the units work out right, the solution to that equation is v = ±c. Which is obviously incorrect.
Remember that the constant c^2 term isn’t velocity^2/time^2; it is actually velocity^4/time^4 because the “1” term it was multiplied onto already had dimensions of velocity^2/time^2 (it was one rest-frame light-year per one accelerated year, then got squared). So, as written, it is dimensionally correct.
And, since this is a 4th order equation, there are 4 roots. Only one of them will turn out to be the right one, and it should be obvious. This equation might have a doubled root. Not sure.
It’s just a synthetic division problem with coefficients 1, 0, -2c^2, 0, c^2. I just didn’t feel like doing it.
The world line is unchanged. The actual spacetime distance between the two points does not change - except, of course, due to their motion relative to each other. In the accelerated frame, time is changed and therefore space is changed to match.
I guess the basic point here is that it is not appropriate to talk strictly about spatial distances because, with no universal time, no two observers will agree on the spatial distance between two points. This does not mean the distance changes, it merely means that the observers have different perspectives.
Try this. You’re interested in travelling 1 ly from Earth, as measured in Earth’s reference frame, but timed in your travelling reference frame. Your time t at 0.707 c will be 0.707 t[sub]0[/sub] (Earth time).
Your mixed reference frame speed will then be distance in Earth’s frame / time in yours: (0.707 c * t[sub]0[/sub])/(0.707 t[sub]0[/sub]) = c
