Ok, I recently stumbled onto this page and must admit that I have been dissatisfied with the answers already given. They seem to be dodging the original question or saying "You only ask that question because you don't understand, if you understood you wouldn't have to ask the question!" Well gee, that doesn't really do a lot of good. So I will now attempt to answer the ORIGINAL question by using some elementary mathematics from relativity. My answer comes from A Modern Approach to Introductory Physics - Volume 1
by David J. Raymond at the New Mexico Institute of Mining and Technology. The text is freely available at http://physics.nmt.edu/~raymond/classes/ph13xbook/index.html
I will refer to Chapter 6, section 5 (Accelerated Reference Frames). Please note that I do not intend to derive all of Relativity, but rather just answer the question at hand, and assume that Nick is already somewhat familiar with the mathematics of relativity, or is willing enough to read the text which I have linked to learn. So let's look at the original question again:
How can something that is not moving be equated to something that is?
By the way Nick - VERY good question. It really gets at the heart of general relativity and the equivalence principle, and if you don't understand this, then you won't fully understand things you learn later on.
Anyway, let's consider you are in a car. The car starts accelerating forward with respect to the road. As a consequence, experience tells you that you will be pushed back into the seat as by a force. But what is pushing you back into the seat? There is nothing really forcing you. Then you reach a constant speed, and the force goes away. But then you reach a left turn. As the car turns left, you will feel "forced" towards the right. But again, nothing is really forcing you. Instead of "real" forces, these are referred to as "inertial" or "fictitious" forces. Common examples are the Coriolis effect and centripetal force. Now for some math:
Let's start with Newton's famous relation:
(1) F = ma
If x is the your position with respect to the street's origin, X is the position of the origin of the car's reference frame, and x' is YOUR position in the car (I know, it's kinda confusing) then we can say that
(2) x = x' + X
(adding a vector pointing from the street origin to the car origin to a vector pointing from the car origin to you will result in a vector pointing from the street origin to you)
When we take the second derivative we find of course that
(3) a = a' + A
We can substitute this into equation 1 and find thatF = ma' + mA
(4) F - mA = ma'
Remember that a' is the acceleration you experience in the car, and A is the acceleration OF the car with respect to the street.
So it seems that you cannot just apply F = ma
in an acclerated reference frame without taking into account an additional "force" equal to mA. This "force" is the inertial force mentioned above.
You might now say, sure, but the thing accelerating was still moving, so aren't you just dodging the question still. Well, I was just trying to set up SOME useful information in understanding general relativity. Because when you break it down, general relativity basically says that gravity is nothing more than an inertial force - a downward force that we feel because we are accelerating "upward". But CLEARLY we are not, right? I mean, as you said, we are not moving
upward, and since acceleration is the rate of change of the rate of change of motion, without motion there is no acceleration. BUT, I intend to show you, using simple mathematics from special relativity, that something CAN be accelerating away from something even though the distance between them is not changing. I will only show this for the simple 1 dimensional case, but it should illustrate the point.
To understand my explanation, you will need some rudamentary understanding of a spacetime diagram. I recommend reading Chapter 4 (Kinematics of Special Relativity) if you are unfamiliar with them (but I will let you do this on your own as I move on).
I will now be referring to the text (Chapter 6 Section 5) extensively.
The velocity of an object undergoing constant intrinsic acceleration in relativity is
(5) v = dx/dt = [at] / [1 + (at/c)^2]^0.5
an x(t) that satisfies this condition is
(6) x(t) = (c^2/a)[1 + (at/c)^2]^0.5
(you can check by finding dx/dt)
The interval OB (figure 6.5 in the text linked above) has length
(7) x(0) = c^2/a
(just plug t=0 into (6))
The line passing through OA (again fig 6.5) represents the line of simultaneity at point A. We can verify this by solving for [1 + (at/c)^2]^0.5 = x / (c^2/a)
from (6) and plugging into (5) to find
(8) v/c = ct/x
Let us now consider the invariant interval I (again, check ch 4 from the text)
(9) (I^2) = (x^2) - (c^2)(t^2)
(also known as the spacetime Pythagorean theorem due to it's resemblance to the famous (a^2) + (b^2) = (c^2)
, but note the minus instead of plus).
When we plug (6) into (9) we find
(10) I = [(x^2) - (c^2)(t^2)]^0.5 = [(c^4)/(a^2)]^0.5 = c^2/a
Here we reach an increadible result! Even though the object has undergone acceleration away from the origin, and time has transpired, the distance between the object and the origin is still EXACTLY the same! (Compare (10) with (7))
Combining everything we have said here:
-The object is undergoing constant acceleration away from the orign, but is NOT actually moving
-This acceleration will show up as an inertial force pulling the object towards the origin.
So here we have (hopefully rigorously and completely, but most importanly clearly) demonstrated that you CAN in fact have acceleration without actual movement, and it is easy to understand exactly what the equivalence principle is talking about. If anything was unclear, I encourage you to read through the entire text (it's very terse, but an excellent resource) or ask me, and I'll do my best to clarify things (my email is firstname.lastname@example.org). I understand that it is easy to get lost in equations and forget exactly what they mean, so it might take a few times reading through and will certainly require some critical thinking as to why certain steps are being made and what it all means. And if I were you, I'd get out a pencil and paper and transcribe the equations yourself and work through it all. Not only will you understand it better, but it should look more natural than the text on the computer. And if anybody is considering an education in physics, I can highly recomend the New Mexico Institute of Mining and Technology where Professor Raymond teaches the class that goes with this book (that he wrote). This material comes about halfway through your first semester there, so you really get hit with some interesting stuff early on, but Dr. Raymond is a great teacher who really strives to make his students understand. Anyway, I'm done here - just gonna go back and edit this, make sure it looks pretty and submit it! Enjoy!