The Curvature of Space and Acceleration Confusion

[drewb]

Like many of these forum dwellers, I've been reading the Elegant Universe and I've hit a fit of confusion. So I've got a couple of questions.

In the book, it is explained that accelerated motion results in the warping of space and time (I'm thinking specifically of his example of the rigidly rotating disc). It is also later explained that gravity --while causing accelerated motion-- IS the warping of space and time. So it is this warping of space that causes the body of matter to accelerate toward its counterpart. That's where I get lost. If acceleration from gravity is caused by the warping of space itself, and acceleration itself causes space to warp, then why isn't it self-sustaining? Why doesn't acceleration cause its own warping of space, and this warping of space cause more acceleration, etc? I'm sure you can see the loop that I'm in ;)


Another question:

As explained to me, if we were in our own little compartment, oblivious to all outside happenings, we would not in any way be able to tell if our compartment was moving or static, OK, good. Similarly, if were in our compartment, we would not be able differentiate acceleration from gravity (eg. the compartment simply sitting on the ground) from a 9.8 m/s^2 rocket pack acceleration in space. So, my question is this: Why does a gravitational field (say, Earth's) only cause a minuscule time dilation? If we were on a rocket for 50 years accelerating at 9.8 m/s^2, we'd (ignoring the speed limit) be going 15 386 000 000 m/s. That kind of acceleration would make anything outside of a significant gravitational field infinitely (or just vastly) "older" than that within.

As my logic here is fairly simple, I'd imagine you guys have a quick and easy answer to give. ^^

 

[PAllen]

Your first question is too much about 'physics poetry' rather than physics. Physics poetry is trying to make physical theories that are precise mathematically 'comprehensible' non mathematically using vague, imprecise analogies. Greene does a lot of this, not always successfully. So I have no comment on this part.

However, your second question is really good one that causes plenty of confusion even among physics students.

The proper analogy for the difference in clock rate on the ground versus a tall tower is not an accelerating clock versus an inertial clock. Instead it is a clock on the back end of a long, accelerating rocket versus the front end. The front end is higher in the pseudo-gravity well, and will have time difference equal to the gravitational situation.

Note that in flat space (no significant masses nearby), all time dilation effects are really due to speed. The rocket acceleration case can be derived by noting that for the rocket to maintain constant length from the point of view of the back of the rocket (equivalent to tower on Earth maintaining constant height), the front must always be moving slightly slower than the back, and this accounts for the back having time dilation relative to the front. Don't worry if you don't understand this part. If you want to read more about it, google 'Bell spaceship paradox'.

 

[bcrowell]

In the book, it is explained that accelerated motion results in the warping of space and time (I'm thinking specifically of his example of the rigidly rotating disc). It is also later explained that gravity --while causing accelerated motion-- IS the warping of space and time.

If "warping of space and time" means spacetime curvature, as measured by the Riemann tensor, then it's not true that "accelerated motion results in the warping of space and time." What causes curvature is matter. Flat spacetime as described by an accelerated observer is still flat. E.g.: http://en.wikipedia.org/wiki/Rindler_coordinates

 

[drewb]

The proper analogy for the difference in clock rate on the ground versus a tall tower is not an accelerating clock versus an inertial clock. Instead it is a clock on the back end of a long, accelerating rocket versus the front end. The front in is higher in the pseudo-gravity well, and will have time difference equal to the gravitational situation.

I'm not sure that is relevant to the question (correct me if I'm wrong). I'll rephrase my question.

We know that Δt' = Δt / (root[1 - (v/c)^2]) ...

so, if you were in a rocket accelerating at 9.8 m/s^2 for 50 years... you'd be going extremely fast. 99.999% speed of light (I'm just throwing out a number; you'd be approaching the speed of light anyway). As v approaches c, the denominator gets smaller, and, thus Δt' gets huge. Δt in this scenario would be a stationary observer, say, floating around when mr. Δt' flies by on his rocket. Huge time dilation.

ok, so now, a separate situation: On Earth, where we've been accelerating at 9.8 m/s/s for a very, very long time, there is very little time dilation when comparing an observer on the Earth with someone floating in outer space, say, 30 million miles away.

My question is, since we consider both acceleration scenarios equivalent (there's no way to tell the difference between the rocket and gravity acceleration, if I'm understanding this right), why would there be only a small time dilation in the second scenario and a massive one in the first? Both environments (in their separate scenarios, the rocket and the ground on the Earth) have an equal acceleration, but a different time dilation.

 

[drewb]

If "warping of space and time" means spacetime curvature, as measured by the Riemann tensor, then it's not true that "accelerated motion results in the warping of space and time." What causes curvature is matter. Flat spacetime as described by an accelerated observer is still flat. E.g.: http://en.wikipedia.org/wiki/Rindler_coordinates

So the curvature of spacetime from gravity is different than the "curving of space and time"? I'll give you the scenario the book gives -- and some quotes (I know, it's not a textbook, and I'm no physicist).

"And so Einstein realized that the familiar geometrical spatial relationships codified by the greeks, relationships that pertain to "flat" space figures like a circle on a flat table, do not hold from the perspective of an accelerated observer. Of course, we have discussed only one particular kind of accelerated motion, but Einstein showed that a similar result--the warping of space--holds in all instances of accelerated motion. In fact, accelerated motion not only results in a warping of space, it also results in an analogous warping of time." - Brian Green, Elegant Universe

The scenario the book gives is with a rigidly rotating disc and two observers. One observer, Jim, is sitting on a radial strut of the disc (so on a radial beam through the middle of the circle), and the other observer, Slim, is sitting on the edge of the disc. Slim's ruler (he's measuring the circumference) undergoes Lorentz contraction, as it is pointing in the direction of the motion, while the length of Jim's ruler (he's measuring radius) is not contracted since it's perpendicular to the direction of motion. When they compare their measurements, they find that 2(pi)r =/= circumference, since the circumference is longer and the radius is the same as it originally was. The solution Einstein (or, rather, Greene crediting Einstein of) gives is a curvature of space. He gives pictures of a circle with a constant radius sitting on top of a sphere or a saddle. The circumference, in the sphere case the circum. of the circle is smaller than when sitting on a flat surface, and the circumference in the case of the saddle is longer than the flat circle.

So that is the type of curvature I'm referring to. I expected that it was the same as the curvature from gravity.

 

[PAllen]

I'm not sure that is relevant to the question (correct me if I'm wrong). I'll rephrase my question.

We know that Δt' = Δt / (root[1 - (v/c)^2]) ...

so, if you were in a rocket accelerating at 9.8 m/s^2 for 50 years... you'd be going extremely fast. 99.999% speed of light (I'm just throwing out a number; you'd be approaching the speed of light anyway). As v approaches c, the denominator gets smaller, and, thus Δt' gets huge. Δt in this scenario would be a stationary observer, say, floating around when mr. Δt' flies by on his rocket. Huge time dilation.

ok, so now, a separate situation: On Earth, where we've been accelerating at 9.8 m/s/s for a very, very long time, there is very little time dilation when comparing an observer on the Earth with someone floating in outer space, say, 30 million miles away.

My question is, since we consider both acceleration scenarios equivalent (there's no way to tell the difference between the rocket and gravity acceleration, if I'm understanding this right), why would there be only a small time dilation in the second scenario and a massive one in the first? Both environments (in their separate scenarios, the rocket and the ground on the Earth) have an equal acceleration, but a different time dilation.

That is exactly what I answered, but you don't seem to understand my answer. The long accelerating clock relative to an inerttial clock is not analogous to gravitational time dilation on Earth's surface relative to higher altitude. The rocket would, indeed, develop arbitrarily large time dilation, over time, compared to the inertial clock. The correct equivalence principle analog of the Earth surface situation is the one I gave - different heights along a very long accelerationg rocket.

 

[drewb]

Oh, ok then, I didn't understand. ^^

Thank you