I'm having trouble figuring out the relationship between time dilation in special relativity and time dilation in general relativity.

I think it might have something to do with accelerations in SR. I know that two inertial observers in SR will view each others clocks as time dilated by the exact same factor.

What happens when one observer is accelerating? At an instant do they view each others clocks as dilated by the same amount, or does an asymmetry come into play, where the accelerating observers clock is viewed as slow by the inertial observer and the intertial observers clock is viewed as fast by the accelerating observer.

If I can get an answer to the above question I think I'd be able to understand the assymetry in time dilations in GR.

Thanks to all who reply.

I know that two inertial observers in SR will view each others clocks as time dilated by the exact same factor.

What happens when one observer is accelerating? At an instant do they view each others clocks as dilated by the same amount, or does an asymmetry come into play, where the accelerating observers clock is viewed as slow by the inertial observer and the intertial observers clock is viewed as fast by the accelerating observer.

When you accelerate you change your standard for simultaneity. As a result according to your coordinates strange standard of time you will reckon that clocks far in the direction of your acceleration run fast. So lets say you accelerate for a short time away from earth then drift at constant velocity away for a long time. Then accelerate for a short time toward the earth. Then drift at the same speed as befor back toward the earth, then stop relative to the earth back home. And lets say your twin remained home for the journey. You will observe your twin to be time dilated for both the outward and inward constant speed portions of the trip, but you will reckon that he undergoes rapid advancement in age when you are far away accelerating back toward the earth. You can describe your accelerated perspective a couple of ways. First knowing special relativity alone you can acknoledge that you changed standards of simultaneity durring acceleration to accound for your time coordinate describing him as suddenly advancing in age. Second you can look at your time standard under acceleration as being equivalent to undergoing gravitational time dilation in the presence of gravitation. When accelerating toward the earth you feel the floor push you toward the earth and so you perceive the earth to be *above you*. When something is low near a gravity source its time dilates and relative to it the time of something higher runs fast. So you perceive the earths time to run fast durring acceleration and explain it as gravitational time dilation through equivalence.

To calculate something's gravitational time dilation with respect to your coordinate time in general you write out the equation for ds^2 according to your coordinates and then integrate ds with respect to your coordinate time applying it to the path that it took. s will then be the total proper time elapsed. For this scenario the equation for ds for the accelerated observer is fairly well known. You could also just use the coordinate transformation equations for accelerated Vs inertial frame observers which is somewhat less well known for the case on nonconstant proper acceleration.

First knowing special relativity alone you can acknoledge that you changed standards of simultaneity durring acceleration to accound for your time coordinate describing him as suddenly advancing in age.

I think this is the answer I'm looking for. I'm trying to understand the assymetrical time dilation in GR in terms of SR. I know that the force you feel while undergoing constant acceleration in a spaceship is exactly the same as the force you are feeling on earth, since in both cases you are simply accelerating with regard to your light cone.

If there is a clock on the earth and a clock at an altitude 2 miles up, the clock on the earth is accelerating more with regard to its light cone than the clock 2 miles up, since they are stationary with regard to each other but spacetime is more curved closer to the the mass of the earth.

For an analogous situation in SR, I can imagine a clock accelerating towards another clock, at rest, at a constant acceleration. The accelerating clock would be analogous to the one closer to the mass of the earth. The accelerating clock would view the rest one as running fast.

If what I wrote was correct, I have another question. When you say it seems to be running fast because the observers standard of simultaneity is changing, is this a different effect than the Doppler shift? So would it seem to be running fast due to Doppler shift and because the observers standard of simultaneity is changing?

Thanks for the reply.

When you say it seems to be running fast because the observers standard of simultaneity is changing, is this a different effect than the Doppler shift? So would it seem to be running fast due to Doppler shift and because the observers standard of simultaneity is changing?

Thanks for the reply.
The effect can be encorporated into the Doppler shift equation, yes. A subtlety few realise is that relativistic Doppler shift actually is ordinary Doppler shift. The reason the equations look different is that you replace the actual emission frequency according to your frame in the equation for your percieved emmission frequency with the actual emmission frequency according to the proper frame. You would do the same thing in writting the equation for Doppler shift according to the accelerated observer using the accelerated time dilation formula except that you'll also have to be careful about relative simultaneity. An accelerated observer will observe an inertial frame clock to be time dilated by
<FONT FACE=SYMBOL>g</FONT>dt = (1 + <FONT FACE=SYMBOL>a</FONT>x<B>'</B>/c<SUP>2</SUP>)dt<B>'</B>
where <FONT FACE=SYMBOL>a</FONT> is the accelerated observer's proper acceleration. The primed frame is the accelerated frame.