Johnny5

No, I don't have to assume that's true, since I have an argument which leads to that conclusion. Kind regards

Thank you. "The trace of a matrix is the sum of its diagonal components." I wish to understand this question here: Now, a positive matrix is one which has a positive trace. In other words, let A denote a matrix, and let Tr(A) denote the sum of the elements along the main diagonal. That sum is a number. So the question is, "if that sum is greater than zero, does that imply that a geodesic "through it" is always longer than a geodesic through empty space-time" Well, somewhere Rev Prez you need to talk about two points in a frame. Can you please phrase your question using two points? (x1,y1,z1) (x2,y2,z2)

I guess so...

I'd rather talk to rev.

Indeed.

What is the definition of the trace of a matrix?

I guess I could do that, but where would I post it?

I have a few questions about this... The definition of "absolute frame of reference" is still unclear to me. Any frame of reference has it's stationariness, or non-stationariness only in relation to another frame. I would suggest using first order logic to define what you mean. Let the domain of discourse be frames. In fact, you can use multiple domains of discourse, and I could still follow. Let A denote the origin of one frame. Let B denote the origin of another frame. Let v denote the relative speed of A to B. All points in frame A are rest relative to the point A. All points in frame B, are at rest relative to the point B. But the point B can be moving through coordinates of frame A, and vice versa, with some relative speed v. You cannot say frame B is absolutely moving. You cannot say frame A is absolutely moving. You cannot say frame B is absolutely at rest. You cannot say frame A is absolutely at rest. Suppose that the relative speed v is equal to zero. There is still the possibility that frame B is spinning, in frame A. To take that possibility away, there need to be three non-collinear points in B which at rest in frame A. Then points in B are neither rotating or translating in frame A. Let that be the case. Then we can say frame B is at rest in frame A. We can also say that frame A is at rest in frame B. Now, suppose that v is nonzero. v is speed which cannot be negative, therefore in the case where v is nonzero, v is positive. So therefore, the point B is in relative motion to the point A. So that A is moving relative to B, and B is moving relative to A. In other words relative speed is a relation. Now, you say, "An absolute frame of reference is one which is stationary with respect to all movement." In order for me to understand you, you need to formulate the definition using first order logic, as I have already said. Then it will make sense to me. Kind regards

This question almost makes sense to me, and I know nothing of GR. Why is that? Revprez, if you don't mind my asking, how many dimensions is space? Also, what is a positive definite matrix? Kind regards

Yes, I considered that. But you know, questions lead to answers.

First swansont, and now you. What is an absolute frame?

It isn't long at all. I wrote two papers actually, the second a revised edition of the first. The second one included some material that wasn't in the first one. The proof was 2 pages, and the steps were algebraic.

Actually, I just followed your proof, and it convinced me that the limit is 2. A good proof must be: 1. Utterly convincing 2. Short I still have to look at Dapthar's. But I mean I remember that technique, multiplying by the conjugate, because the interior term vanishes. Ok, I just followed Dapthar's as well, and both were convincing. I would say this... some folks will follow one better than the other, so I would do both.

That's not an error. Here i will show you what I'm referring to... You can see this is false rather rapidly.

There are still a few errors Dapthar. Look at the sign of your 4x term. In one of the steps you switch from negative to positive by accident.

No problem Dapthar, but... Isn't the limit zero?

Dapthar this utterly confused me. Firstly the limit is towards negative infinity, not positive infinity. Second of all, in the denominator you have 1-1=0, which is division by zero error. Not two in the denominator. You seem to have written her problem down incorrectly. Let me give it a try: [math] \lim_{x\to - \infty} [x + \sqrt{x^2-4x+1}] [/math] So you want to know what happens to the value of the function you are taking the limit of, as x becomes increasingly negative. The first thing I would do, is replace x by -x, and take the limit as x goes to positive infinity. So you have this instead... [math] \lim_{x\to \infty} [-x + \sqrt{x^2+4x+1}] [/math] Intuitively, x^2 dominates over 4x. In other words, as x becomes increasingly huge, x^2 becomes enormous in comparison to 4x, so that we can neglect the addition of 4x to x^2, and certainly adding 1 doesn't alter the answer much either. Thus, we should expect the limit above to be equal to the following limit: [math] \lim_{x\to \infty} [-x + \sqrt{x^2}] = \lim_{x\to \infty} [-x + x] = \lim_{x\to \infty} [0] = 0 [/math] Now, of course this is not rigorous, but it's intuitive. And one other thing, the square root of x^2 is the absolute value of x.

That's correct Dave. You cannot prove a definition, definitions are stipulated to be true, so there would be no point to that. Dave said it correctly. I want to derive the formula for the inverse Fourier transform, not derive the definition of the Fourier transform.

Do you have the fortitude to follow my argument?

Do you realize how that looks to someone who doesn't know GR? If i had to sift through that to find an error, it could take me years. Let me ask you this. You say "we are solving for a metric." Explain that to me if you can. Maybe I'm not so blind after all.

But this example problem isn't how I arrived at the contradiction. I used barn/ladder. What I did is not in this thread, nor any other. But it essentially is barn/ladder. I just happen to have good logical form. Actually, there was some new physics in what I did, but I saw it as intuitive. I really didn't have to introduce it, but it simplified the explanation. I broke the whole event analysed, into two sub events, whose union was the whole.

Well I was discussing the "twin paradox," as that was the original poster's question. You are saying things wrong. You say, and I quote, "Once you accelerate you lose the symmetry." What you should say is this, "Once one of the ships is sujected to an external force, symmetry is lost". Force and acceleration are not the same thing. This is why I think superacceleration is possible, rather than impossible. They can both say they are stationary, and the other is moving. I didn't say they could not do that. But they cannot each say that the other person's clock is slowing down. For suppose they can... John,Jim Then in the one frame, the statement that Jim turned 40 when John turned 31 will be true, and in the other frame, the statement that, "John turned 40 when Jim turned 31" will be true. This is a temporal paradox. And it is caused by assuming that the time dilation formula is true in both frames.