Johnny5

You know what, this is extraordinarily complicated, becase I am not explaining what I do, but rather trying to answer your question. Real clocks can fall in and out of sync, and I am fundamentally aware of this. But dt in the formulas of physics, is a theoretical aid. You are complicating things, furthermore, you should not try to map elements from R to R. The universe changes state discretely. Let S1 denote the first moment in time. Let S2 denote the second moment in time. Let S3 denote the third moment in time. And let S4 denote the fourth moment in time. Now, consider changes of state of the universe. Define them as follows: DT1 = S2 - S1 DT2 = S3 - S2 DT3 = S4 - S3 Answer me this, "what is the amount of time which elapsed by the fourth moment in time?" You are NOT to assume that each state change represents an equivalent amount of time. I can explain. Suppose that: DT1 = DT2 = DT3 = DT Then you can say the answer is 3 DT. But that is a supposition, one which I do not know the truth value of. Actually, I think it's false, because of something to do with acceleration. But you tell me what you think.

I am not introducing anything to do with a pre-conceived notion of time. I was just relying upon that which follows from the operational definition of 'before'. An "amount of time" becomes a difference in clock readings. This is in fact what we do in practice, so that is why it is there. Here is the definition of dt which you asked for... Let S n denote an arbitrary moment in time. Let S denote a state of the universe. dS = S n+1 - S n Let there be a clock in some arbitrary frame, which can map natural numbers onto S, in a 1-1 fashion, regardless of whether or not the clock is in an inertial or non-inertial reference frame. dt = (n+1) - n = 1

Well first off, I said that you can operationally define before. When you do so, you will encounter something which occurred to umm, what the heck was his name... the river guy... you cannot step twice into the same river... oh yeah Heraclitus i think was his name. What Heraclitus realized, is that the universe will never be in the same state twice. Here... use this operational definition... then you will see... Go to the moon, and drop a feather from a height h above the surface of the moon. You are well aware of two distinct moments in time, the one where you release the feather, and the one at which the feather collides with the surface of the moon. So you define the binary relation BEFORE, using that experiment... the free-fall experiment, made famous by Galileo. So the domain of discourse, is the set of "moments in time" Your binary relation takes this form: X before Y The X,Y are elements of the domain of discourse. So that the meaning of "moment in time" comes from that experiment which operationally defines "before." That's what I mean. So there is no need to define "moment in time" mathematically, because you can know what X,Y denotes from the Galileo experiment, which defines the binary relation before, on the set of moments in time. As for where Heraclitus' comment fits into all this... Is the following statement true or false: $X: X before X Read, "There is at least one X, such that X before X" or can translate into English as follows: "There is at least one moment in time X, such that X before X." I will try. Let S denote the set of states of the universe. Let X, Y be elements of S. Let it be true that X before Y. Additionally, let it be the case that X,Y are adjacent moments in time, as previously described. Let there be a clock, which maps natural numbers onto each state, such that there is a 1-1 mapping. The following notation could be used to refer to elements of the domain of discourse: S = {S1, S2, S3, ...} Each Sn, refers to a different configuration of matter in this universe. You asked for a mathematical definition of dt. I am certaint that means that you want to incorporate numbers into the definition of dt. I have chosen the natural number system, because the clock is digital. The clock ticks out elements of the following set: N={1,2,3,4,5,...} But they are just readings on this clock. Let Sm, Sn, denote arbitrary elements of S. An "amount of time as measured by this clock " would be defined as follows: Dt = Sn - Sm Where necessarily, Sm before Sn. So to have a time differential, one merely needs m,n to be consecutive natural numbers. To finish the definition, you can just use first order logic which I will do. I have NOT claimed that this clock ticks at a constant rate, only that it ticks once and only once, for each moment in time.

Why can't v be greater than c?

What part didn't make sense?

This is totally incorrect. In the time dilation formula, the symbol v is the relative speed between the centers of inertia of two things. So... V could equal zero, in which case the two things aren't moving relative to each other, we would say they are at rest relative to one another. Or, v could be 10 meters per second, as in the case we have here, where two clocks (skinny and fat clock) are running the hundred meter dash. You can answer your own question using the time dilation formula. v is the relative speed. There is no mathematical reason you cannot put v=0 in that formula. One problem would be in the case of division by zero error. Mathematically, that would occur if the relative speed v, between the centers of inertia of the two clocks was equal to c. In the case where v=c you have division by zero error UNLESS... the numerator is simultaneously zero. According to the formula, if the relative speed v exceeds the speed of light, then "amount of time" becomes imaginary, which is nonsensical.

Yes the time dilation formula. Dt = Dt`(1-vv/cc)-1/2 Where v is the relative speed between fat clock and skinny clock, which is 10 meters per second if i do recall, and c = 299792458 meters per second. That's the formula right?

Unfortunately the image didn't come up, so now what?

No problem. Let A, B denote two moments in time. If there is a moment in time M, such that either A before M and M before B OR B before M and M before A Then it is NOT the case that A and B are adjacent moments in time. On the other hand, if there is no moment in time M such that A before M and M before B OR B before M and M before A And additionally, not (A simultaneous B) then A,B are adjacent moments in time. You can now use first order logic to symbolize this appropriately. Let me try it. Let the domain of discourse be the set of moments in time. Let A,B be elements of the domain of discourse. We can easily prove trichotomy now. Trichotomy: A simultaneous B or A before B or B before A. Definition: In the following definition, let it be the case that A before B. A,B are adjacent moments in time if and only if not [$ M (A before M and M before B) ].

Is that the one Chinese astronomers observed? I remember that from history class.

Prove this. I expect to see the time dilation formula utilized. PS: I also know that the third frame is unnecessary because all three frames are inertial, so there's a bit of redundancy (which ok but just not necessary), so just use the two frames, the rest frame of skinny clock, and the rest frame of fat clock, which were stipulated to both be inertial reference frames. Let v denote their relative speed. (which happens to be 10 meters per second, everyone is using the same rest units for distance and time measurments have to say that somewhere)

NIce and slow. You are in the spectator stand with a stopwatch, and there is a race between fat clock A, and skinny clock B. Skinny clock B is going to win the race, because he can run faster, and you who are in the stands can see that clearly. The race is the 100 meter dash, and there is a tape measure, with marks on it, and the tape measure is at rest relative to you, and you are just sitting in the stand, with your stopwatch. Now, skinny clock and fat clock start the race neck and neck, at the point zero on the tape measure. But thereafter, skinny clock is ahead. At the moment the race begins, skinny clock reads zero, and fat clock also reads zero. Then you fire the starting gun, and they both take off simultaneously in your frame of reference. OK SO... lol Let the tape measure be along the x axis of your frame. The rest frame of each clock, was stipulated by swanson to be inertial. That means that in either of their frames of reference, you are moving in a straight line at a constant speed, and vice versa. So each of these clocks is moving in a straight line at a constant speed in your frame, and you are moving in a straight line at a constant speed in either of their frames. Skinny clock can run down the track at a speed of 20 meters per second. Fat clock can only run down the racetrack at a speed of ten meters per second. Here are their velocities using vector notation, as defined in your frame: Vs = 20 i^ Vf = 10 i^ Now, their relative speed is 10 meters per second, so the clocks are moving apart in your frame, at a rate of 10 meters per second. Now, let skinny clocks rest frame be such that his x axis is parallel to your x axis, and let fat clocks rest frame be such that his x axis is also parallel to your x axis, and therefore parallel to skinny's x axis as well. Here is what fat clock sees: Fat clock sees skinny moving with velocity 10i^ in his frame. Here is what skinny clock sees in skinny's rest frame: Skinny sees fat clock moving away from him with velocity -10i^. So all three of you agree, that the relative speed between the two clocks is 10 meters per second. The way I've set things up, you can formulate true statements in any of the three frames, that you wish.

It is infinitessimal, as t2 and t1 are supposed to be adjacent moments in time, in some chosen frame of reference.

I think I get it all, but just to be sure. Let us define frame 1, to have its origin permanently located at the center of mass of Clock A. Let frame 1 currently be inertial, which means that Newton's laws are currently true statements in frame 1. Let us define frame 2, to have its origin permanently located at the center of inertia of Clock 2. Let frame 2 currently be inertial, which means that Newtons laws are currently true statements in frame 2. Now, let there be a third frame (which you are at rest in), which is stipulated to forever be an inertial frame. Now, let both clocks be moving through the coordinates of the third frame. If they have the same velocity vectors in the third frame, then they are at rest relative to each other. That's not what swansont wants. The two clocks need to be in relative motion, with the relative speed being denoted by v. Ok so... Let clocks A,B currently be free of external forces. Therefore, if they are at rest in the third frame, then they will remain at rest in the third frame. If they are moving in the third frame, then they will continue to move in a straight line at a constant speed in the third frame, until such moment in time as they are acted upon by an external force. Can you be more precise as to the direction the two objects are moving? Just so that I can have a clear picture of the trajectory of both clocks in the third frame please. Thank you

That looks like waves in the water. What is that?

That is the kind of person I need to speak to. I have some simple questions that someone with that kind of expertise could answer. And NASA should come out with some videos of experiments done in space, for those of us not fortunate enough to go into space.

I am willing to follow this out with you, so that I can see what it is you want to demonstrate. But explain things more. Apparently there are three frames?

Let there be two clocks in relative uniform motion, and let them be identical in every way. Let them each be located at the origin of an inertial reference frame. (This means that Newton's laws are true in either frame of reference. Additionally, let them be in deep space, free of all external forces. I claim that if the two clocks are synchronous at one moment in time, they must remain synchronous.

The logic is correct, regardless of whether or not I disagree with SR. Additionally, I disagree with SR. Regards

Explain gravitational acceleration to me, when you get a chance. Thank you

This is impossible.