Johnny5

This thread is for any experts in the general theory of relativity. Suppose that it is possible to accelerate like the starship enterprise does, where its going at warp 1, and then jumps to warp 7, but the inhabitants aren't tossed about, because they cannot feel any forces due to changes in their speed. Suppose that is possible, and has been done by mankind in the future. Would that existence of that kind of ship be experimental proof of the falsity of the principle of equivalence? A simple yes or no will do. Thank you in advance

My first line stopped you??? Umm, the first thing I did was explain the simplest experiment conceivable, that can measure the speed of light. So which specific line are you referring to?

I agree completely. If you assume your conclusion, then you have made a reasoning error, but I didn't assume that simultaneity was absolute. What I assumed was this... I assumed the Lorentz contraction formula is true. <---- that's what I assumed.

Well I have a reason for each thought. I hesitate to say belief, because... well I have a reason for what I think.

I think GR is wrong though. I could have told you that Newton's gravitational law cannot be correct. It's impossible that it is correct, becase as formulated, it's an action at a distance formula, and the speed of gravity isn't infinite, it is finite, and nonzero. But then that is why we have field theory now isn't it? It is there that we encounter the concept of "retarded potentials." A poor choice of words but whatever.

Martin, swansont's quote came from the thread just below this one, entitled, "Decisive answer please" The question was about the speed of gravitational force.

What is population biology? It sounds like what I think it is. Math + evolution is a good thing Evolution is a mathematical challenge, more than a biological one really.

Hmm, so they would start spinning like umm gears ? Let's say that there is friction between them. I am having some trouble picturing what happens. I'm not using math by the way. Let the two spheres have identical masses and radii. They are gravitationally attracting each other, and electrically repelling each other. They are in deep space away from all other external fields, other than their own. You are viewing the motion in the center of mass frame. Since the spheres are equally massive, and have identical radii, the center of mass of this two body system is located at the point of contact. Let the spheres be rigid. Right now, they are at rest in this frame. Someone then gives the one on top a gentle tap. Now what happens? I think conservation of energy comes into play here. Anyone? It will help to talk about what happens to rigid vectors drawn from the center of each sphere to a fixed point on the surface, because as Bob says, each sphere is going to start spinning in the center of mass frame. oh oh oh, and one more question before I forget, is the center of mass frame an inertial reference frame? Regards

That's the way it is. You can see it clearly from the argument which is being developed here.

This isn't right. The variable which supposedly causes time dilation is speed, not acceleration. Acceleration is a vector quantity, speed is a scalar quantity. Regards

How so? I haven't finished the argument yet.

With this problem, for some reason the confusion is never over... Kind regards PS: Someplace I called it the most complex problem in the history of theoretical physics... I wasn't kidding.

In another thread, which is now closed, swansont said this:

I didn't assume that simultaneity is absolute, actually I know what I did, and I know I didn't circularly reason. It will be advantageous to just say it... Suppose that you are an observer permanently stationed in inertial reference frame F. Now, there is a meterstick which you have which is made out of wood. It is your favorite meterstick, and you had it engraved with your name and put a nice coat of varnish on it and what not. So you know how long it is when it isn't moving relative to you. And your friend has an almost identical meterstick, varnished and all, but he has his name engraved on his. When these two rulers are placed side by side they have the exact same length, as far as sensory perception can tell. Now, you are stuck in this inertial reference frame F forever. Here comes the reasoning to be analyzed: Assumption 1: The Lorentz contraction formula is true, in any inertial reference frame. Since you are in an inertial reference frame, the Lorentz-contraction formula applies to any and all wooden rulers in your frame. It even applies to steel rulers, and titanium rulers as well, but you don't own one of them. So since the LCF is true in this frame which you are in, any and all rulers which are at rest in this frame, will have their maximum possible length, which is called their proper length, or rest length. Now, if for any reason whatsoever, one of these rulers is moving through your inertial frame F, instead of being at rest in F, its length will be less than it's proper length, since we have assumed that the LCF is true. Let us presume that your friends wooden meterstick with his name on it, is moving from left to right, along the x axis of reference frame F, with speed V as defined using the coordinates of frame F, and clocks at rest in frame F. If V is sufficiently large, your friends meter stick is now .25 meters, instead of one meter. Your meterstick with your name on it is at rest in your hand, and you are using it to measure the length of his meterstick when it passes by you. So here is where simultaneity problems arise, I tried to show it somewhere else, but its the same argument over and over again, so here it comes: Ruler 1: A__B Ruler 2:......A`____________B` Then later Ruler 1:......A__B Ruler 2:......A`____________B` Then later Ruler 1:........................A__B Ruler 2:......A`____________B` So because your friends ruler was length contracted, A` coincided with A before B` coincided with B. If instead we use the Galilean transformations, there would have been a moment in time, in your reference frame F where things would have been this way: Ruler 1:......A_____________B Ruler 2:......A`____________B` In other words the ends would have coincided simultaneously. But because we assumed that your friends ruler contracts if it has a nonzero speed V in frame F, then as your friends ruler passes by yours, it is impossible for the ends to coincide simultaneously. So there is a reductio ad absurdum proof that can be used (not circular reasoning), which can allow you to conclude that the LCF is false. I haven't shown it all yet. But I've shown enough where someone else can figure out the problems caused by assuming LFC is true.

Before going any further, did you follow my derivation of the time dilation formula? If you had an objection to any step, I want to hear it.

No it is still interesting. When they get the Latex fixed, I'll pick it up. If you aren't interested in it anymore fine... but I really still am, so I will finish it up. I need to work a problem using separation of variables again anyway, its been awhile since I've done one. So I am gonna finish this problem. Regards

Sarah, there is a diagram which helps in understanding the epsilon/delta definition of 'limit' in any calculus text. Have you seen it?

I concluded that the length contraction formula is false. Let two rulers have identical lengths when at rest with respect to one another. Let them be situated in a common inertial reference frame, and let the ends currently coincide with one another. Now, symmetrically push the rulers apart, then bring them to rest, so that they are still parallel, but separated in space by a trillion miles, and each had the same force exterted on it and so on. They are now both at rest in the original frame again, but are separated by a trillion miles. Now, make them head for each other by applying identical forces. Thus, they are now in relative motion, with relative speed V, in the original frame. Each is moving in a direction parallel to the length of the other, and they are eventually going to pass by one another. Think about things in the center of mass frame of the two rulers if you want. The speed of ruler 1 is V/2 towards the center of mass and its moving from left to right. The speed of ruler 2 is V/2 towards the center of mass and its moving from right to left. Ruler 1:A-----------B Ruler 2:___________A`----------B` There will come a moment in time, at which B will coincide with A`. That moment in time marks the beginning of an event, which ends when A coincides with A`. My question is, when A` coincides with A, does B` simultaneously coincide with B??? I say yes. Check the truth value of your statements in multiple frames. If simultaneity is absolute, then A` coincides with A simultaneously to B` coinciding with B in all frames of reference. Here swansont... Consider things from the rest frame of ruler two. Suppose the length contraction formula is true. Therefore, in this frame, Ruler one is length contracted, so that we can conclude that it's true that A coincides with A` before B coincides with B`. Now, consider things from the rest frame of ruler one. In this frame, Ruler two is length contracted, so that we can conclude that it's true that B` coincides with B before A` coincides with A. So we have concluded this: A` coincides with A before B` coincides with B, AND B` coincides with B before A` coincides with A. This conclusion came from assuming the length contraction formula is true. There's my problem. Regards