Johnny5

You know what, no two real solid objects touch, because of the repulsion of electrons at the surface of each object, yet there is still friction.

Ok, you are saying that the gravitational force is not causing a mechanical effect that slows down the clock. I disagree. Now what? There has got to be some kind of gravitational effect on the innards of any clock. They (the interior parts of the clock) are being pulled down, this increases friction.

Are you certain that is true?

Yes, this is the classical definition of the derivative.

Actually, i don't want to declare that: [math] \Delta f(x) = f(x+h) - f(x) [/math] That's already been defined by others, I didn't say I want to do it that way. I really just want to see what you have to say on the matter. As a matter of fact, i define the infinitessimal difference operator using first order logic, but thats my own business. I didn't say that I wanted to set dt=1, you did. Plus you are ignoring whether you take the limit from the left, or the right. All I wanted to do, was to answer your question, as to how to prove the following fact: [math] \frac{d(A+B)}{dx} = \frac{dA}{dx}+ \frac{dB}{dx} [/math] Now I'm just trying to understand what your obection is to the method of proof I used. I just want to prove the fact rapidly, using only algebra. I know that I can do it quickly and elegantly using the infinitessimal difference. And you saw it, I removed dx from the denominator first.

Let [math] dt = 1 [/math] There you go, now it's a number. For the purposes of doing physics, with t representing time, I would think that you need units. One state change. One time unit. Something like that. hence [math] dt = 1 TU[/math] TU = time unit, one state change. At some moment in time, the universe is in state S1. At the very next moment in time, the universe is in state S2. Things moved, the state changed, there was a change of state. From a purely abstract point of view... [math] \Delta S = S2-S1 [/math] If S2=S1 then nothing moved relative to anything else, and time hasn't passed so dt>0 if and only if DS > 0

Time can only flow in one direction. In other words, dt is strictly positive. So specifically... [math] dt = t_2 - t_1 [/math] Where t1 is the value of the time coordinate of some three dimensional rectangular coordinate system, at some moment in time, and t2 is the value of the time coordinate of that three dimensional rectangular coordinate system, at the very next moment in time. By saying that its at the very next moment in time, we ensure that: 1. not (t1 = t2) So we can divide by dt, and we ensure that: 2. t2 > t1 So that dt is nonnegative. I have analyzed dt, as much as is possible. It used the concept of subtraction.

I know it's an operator, and I know it's not a fraction. I don't see why I cannot divide whatever I wish by dt, given that its 1. nonzero But at any rate, why doesn't it make sense from a strictly mathematical point of view?

In theory, they would be touching yes, but lets make this problem slightly more realistic, let there be electrons located at the surface of both spheres. Now what? wouldn't there be an electrostatic force of repulsion? So that the spheres wouldn't really be touching?

So how am I to interpret h? From finite discrete difference calculus, we have: Definition: [math] \Delta f(x) = f(x+h) - f(x) [/math] where Delta is the difference operator, and h is the step size.

"differential of a sum equals the sum of the differentials" d(A+B+C+D+E) = dA+dB+dC+dD+dE That kind of thing. [math] d [\Sigma_{n=1}^{n=N} Q_n ] [/math] The differential of the sum' date=' from n equals one to n equals N, of Q sub N. [math'] \Sigma_{n=1}^{n=N} d[Q_n] [/math] The sum from n equals one to n equals N of the differential of Q sub N. when you see the capital Greek letter sigma written that way, it just means repeated addition, and is called summation notation. But you knew that, but for anyone reading this who didn't...

Well dave, you are missing the limit concept, you have to take the limit as h goes to zero don't you? And h is called the "step size" in the finite discrete difference calculus right?

It goes until friction stops it. But would there be friction if the spheres were perfectly rigid, and truly spherical? Or wouldn't the surfaces be so smooth, that there would be no friction?

Right. Now, suppose that I give one of them a nudge. What's gonna happen?

Theorem: [math] \frac{d(A+B)}{dt}} = \frac{dA}{dt} + \frac{dB}{dt} [/math] To prove the theorem, it suffices to prove the following statement is true: [math] d(A+B) = dA + dB [/math] It suffices, because if the previous statement is true, and dt is nonzero, then we can divide both sides by dt, to obtain the theorem we are attempting to prove. Let [math] Q = A+B [/math] Thus, we are attempting to prove that: [math] dQ = dA + dB [/math] Definition: [math] dQ \equiv Q2-Q1 [/math] Where we have used subtraction to define the differential of quantity Q. Q1 is the value of Q at one moment in time, and Q2 is the value of Q at the very next moment in time. The amount of time which has passed is called a time differential, and is denoted as dt. By definition: [math] d(A+B) \equiv (A_2 + B_2) - (A_1 + B_1) [/math] Where A1 is the value of quantity A at some moment in time, and B1 is the simultaneous value of quantity of B. A2 is the value of A at the very next moment in time, and B2 is the simultaneous value of B. Now, we can just use the field axioms of algebra at will, to prove what we want to. In particular, using just those axioms, we can prove conclusively that: [math] (A_2 + B_2) - (A_1 + B_1) = (A_2 - A_1) + (B_2 - B_1) [/math] Now, by definition A2-A1 is dA, and B2 - B1 is dB, therefore: [math] (A_2 + B_2) - (A_1 + B_1) = dA + dB [/math] Thus, we have shown this: [math] d(A+B) = dA + dB [/math] Now, divide both sides of the statement above, by dt, to obtain: [math] \frac{d(A+B)}{dt}} = \frac{dA}{dt} + \frac{dB}{dt} [/math] which is the theorem. QED The result is generalizable too. Suppose that you have: d(A+B+C) Let D=B+C So you have d(A+D) By the theorem just proven you know this: d(A+D) = dA+dD Therefore: d(A+D) = dA+dD = dA + d(B+C) By the theorem just proven you know that: d(B+C) = dB + dC Hence, if you know the theorem just proven, then you can figure out that: [math] d(A+B+C) = dA + dB + dC [/math] In which case you can learn this: [math] d [\Sigma_{n=1}^{n=N} Q_n ]= \Sigma_{n=1}^{n=N} d[Q_n] [/math] Dave, if you object to this method of proof, please let me know why. Thank you

In what frame? You can't just say something is accelerating, without explaining what frame it is accelerating in.

The two are in vaccum, and in an inertial reference frame, and nothing else in the universe is exerting a force on either object. The only force either object experiences is the gravitational attraction of the other. So I don't see why it would depend on where the second is located. The second is at rest, on top of the first. Does it start rolling, without a nudge? I am assuming both things are symmetrically shaped. They are rigid spheres.

Are you implying that gravity isn't a force?