The speed of light with an accelerating observer.

[postermmxvi]

You are one light year away from Earth, at rest in Earth's frame of reference. A flash of light is emitted from Earth. You begin to accelerate toward Earth. Does it still take exactly one year for the light to reach you from the time it was emitted?

 

[Dale]

Does it still take exactly one year for the light to reach you from the time it was emitted?

In what reference frame?

 

[postermmxvi]

Sorry, I left out an important detail :x Well, I want to know if it takes a year from the perspective of the accelerating observer. I know he isn't in an inertial reference frame anymore. I have a fairly decent understanding of special relativity, but very little understanding of general relativity. I guess a better way to phrase it would be, does the accelerating observer pass through a year of time between when the light was emitted and when it reached him? I really want to understand the answer, but I don't even know if it is yes or no. I want to say it is yes, because the speed of light is constant - but I know I have only learned that in the context of special relativity for *inertial* frames (and our observer doesn't stay in an inertial frame). If the answer is no, then I will probably have lots of confused questions that I will spare everyone from until I can form a coherent thought :P

 

[Mike_Fontenot]

You are one light year away from Earth, at rest in Earth's frame of reference. A flash of light is emitted from Earth. You begin to accelerate toward Earth. Does it still take exactly one year for the light to reach you from the time it was emitted?

No.

Here's a place to start, if you really want to pursue this issue.

https://www.physicsforums.com/showpost.php?p=3217917&postcount=305 .

Mike Fontenot

 

[ghwellsjr]

Sorry, I left out an important detail :x Well, I want to know if it takes a year from the perspective of the accelerating observer. I know he isn't in an inertial reference frame anymore. I have a fairly decent understanding of special relativity, but very little understanding of general relativity. I guess a better way to phrase it would be, does the accelerating observer pass through a year of time between when the light was emitted and when it reached him? I really want to understand the answer, but I don't even know if it is yes or no. I want to say it is yes, because the speed of light is constant - but I know I have only learned that in the context of special relativity for *inertial* frames (and our observer doesn't stay in an inertial frame). If the answer is no, then I will probably have lots of confused questions that I will spare everyone from until I can form a coherent thought :P

As Mike said, the answer is no.

Twice you have said that the accelerating observer is no longer in an inertial frame and implied that you need to have a better understanding of General Relativity in order to handle this type of scenario but that is not true. Einstein's Special Relativity handles accelerating objects and observers just as well as moving and stationary objects and observers. In fact, in his 1905 paper, at the end of section 4, he analyzes a constantly accelerating clock from the frame of a stationary clock. This, by the way, is the origin of the Twin Paradox.

Einstein's 1905 paper is limited to inertial reference frames, not inertial objects and observers.

 

[Dale]

Well, I want to know if it takes a year from the perspective of the accelerating observer.

No, it takes less. Let's use units of years for time and light-years for space. The worldline (in an inertial reference frame) of a uniformly accelerating observer is given by:
$$(sinh(g\tau)/g,cosh(g\tau)/g)$$

where g is the proper acceleration, and tau is the proper time. For t=tau=0, the accelerating observer in this parameterization starts at x=1/g and accelerates in the positive x direction. So the worldline of the light pulse in your scenario would be:
$$(t,1/g+1-t)$$

Setting these two worldlines equal and solving for t and tau gives:
$$t=\frac{2+g}{2+2g}$$
$$\tau=ln(1+g)/g$$

Both of which are always less than 1 year.

 

[94JZA80]

Dale,

i also concluded that it should take less than a year, but i did so visually and not mathematically. i was wondering if i was visualizing the situation correctly, or if i may have accidentally come to the right conclusion for all the wrong reasons, and i thought you might be able to confirm or deny this.

i first started thinking about the fact that simultaneity is relative, and that the only external observers who could tell if the observer does in fact start accelerating directly toward a light source at the very instant that it starts beaming are those who lie in the plane that makes up the set of all points equidistant from the observer and the light source at t = 0. only such an external observer could confirm that the observer started accelerating at the instant that the light source turned on.

regardless of that moment of clarity, i realized that it was not a necessary piece of the puzzle. and i went on to simply assume that the observer does in fact start accelerating directly toward the light source at the very instant it starts beaming toward him. now if the source and the observer are exactly 1 light year apart, then the only way for the observer to witness exactly 1 year pass between t = 0 and seeing the source is if he remains motionless with respect to the source for that entire year. that is, he cannot accelerate toward the source or he'll see it before an entire year passes by. likewise, he cannot accelerate away from the source or it'll take somewhat more than 1 year for the source's light beam to reach him. am i on the right track here?

TIA,
Eric

 

[Dale]

Yes, that sounds right to me also.