Johnny5

Ok lets see if i remember the axioms of a group. Someone correct me if i am wrong, or incomplete in the answer. Let X be a binary operation on a group. Let A,B,C denote arbitrary elements of a group. Axiom 1: Closure under the operation X. Axiom 2: Operation X is associative. That is: AX(BXC)=(AXB)XC Hence usage of parenthesis are not necessary, since AXBXC is unambiguous. Axiom 3: There is group element 1, such that: 1XA=A Axiom 4: For any A, which is an element of a group, there is an element A* which is also an element of the group, such that: AXA*=1 And if the operation X is commutative the group is Abelian. --------------------------------------------------------------- Any errors in the above?

How do figure that?

subgroup of O(3). Group in the mathematical sense? O(3) is the set of all 3x3 'orthogonal matrices' (reason for the letter O) under maxtrix mult. Ok, so we are talking about a set of matrices. O(3) is a set of matrices, a set of 3X3 matrices. Additionally they have to be orthogonal. What's that mean again? What makes a 3X3 matrix orthogonal? S stands for special, ok... So SO(3) is a subset of O(3). Precisely, set SO(3) contains all the 3x3 orthogonal matrices of O(3) which have a determinant of +1, and only such matrices. Followed you, and thank you so much Tom... again. Kind regards

What is SO(3)?

I would be willing to, but it's been a long time since i've solved these kinds of problems. You will have to tell me what the definitions are. What is a ring? What is a simple ring? What is an ideal? I don't remember what a module is, or a submodule. I wouldn't mind at least trying to solve the problem, or watching others help you solve it.

Consider the bicycle wheel experiment described at the following site. Suppose that you get the wheel spinning at a rate of 10 revolutions per second. I think thats reasonable. Then you let go of the handle. The handle is going to trace out a circle, in the appropriate frame. denote the distance from the center of mass of the wheel assembly, to the point where it is attached to the chain by R. Suppose that you measure the time it takes the wheel axis to complete one rotation as 6 seconds, and to a reasonable approximation w is constant during the first 'orbit' Using your measurement of w, and of R, what is the mass of the bicycle wheel? Introduce whatever you need to solve the problem. Regards

Yes, I meant the earth station frame, which was built.

Well it is an alternative approach to finding the answer. I didn't intend to hijack the thread, just offer an alternative, since you can find the expansion of ln(1-x) identically. Anyways, I just happened to know the answer to this off the top of my head, so i thought i would share. Regards

In my response, i was thinking of the center of mass frame, when they come back to meet. But each is moving at a constant speed in that frame, and hence both of those are inertial frames as well. But I am assuming that as they approach each other, there is a point midway between them, which represents where they will eventually both be simultaneously, and that this is the CM of the system. So, we could say that there are three inertial frames one could refer to, in this problem. I meant the CM frame, I should have said that. Regards

That one's easy... Start off with this... Integral 1/U du = ln U + C Now, do what you have to to make the C=0 So you are going to write an indefinite integral instead... ln 1 = 0 IntegralU=1U=U(x) 1/U du = ln U(x) now make the following substitution: U(x) = 1+x dU=dx Thus you have: Integralx=0U=1+x 1/(1+x) dx = ln (1+x) So the RHS is what you want explicitely, and you can use the LHS to figure out the series for the expression for the RHS. Write the integrand so that you can use Newton's binomial series expansion formula: Integralx=0U=1+x (1+x)-1 dx = ln (1+x) Now, you can rewrite the integrand as a series switch the summation sign and the integral sign, and then integrate term by term.

No, I don't think you understand. In one frame a specific photon has one speed, in another frame the same photon has a different speed. Speed is relative.

Ok then you're fine. Maybe I should say more than just that... Suppose your speculation is correct. Then i can switch reference frames, and the same photon is moving faster than .99999c. Speed is relative.

It's a shame that they were banned. I liked discussing the Sagnac effect with them. Until they came along, id never even heard of it. Their posts yesterday left much to be desired, but the individual obviously spent a lot of time learning physics... to be ignored so easily. Anyways, i think its a loss rather than a gain to the forum. It's not like that many people in the world are interested in physics to the point where they are familiarized with the Stern-Gerlach experiment. Well anyways thats my two cents. I don't think permanent bans should be come to so hastily. People should get warnings, and then temporary bans, and then if they continue, make them permanent. but to be so quick about it, is a bit well... Regards

I understand your question. You are trying to understand things from an SR perspective. Here is the best way to answer you. If SR is correct then it is necessary for the mass of a photon to be zero, hence the photon could travel at c, it wouldnt have to be slightly less. But this answer is predicated on the "if SR is correct". But also note that many conceptual problems are created by the basic assumption of SR, which is that the speed of light is c in all inertial frames. Go back to the John and Jim example, and explain to me (within the context of SR) the resolution.

That is correct. Now does that make sense to you?

Thank you Dr. Swanson Kind regards

A couple of things. 1. There is no such thing as a universal speed limit. You can go as fast as you want, relative to some starting point, within propulsion system limitations. 2. c is by definition 299792458 meters per second, i think this was done in the 80's youd have to check. There is no uncertainty whatsoever in the number. Instead, the value of the meter is what is uncertain, as well as the second. You can call the above number "the speed of light" but that is irrelevant. Suppose, for the sake of argument, that a photon moves by you at speed c. And that John is sitting right next to you. How fast will the photon pass by John? Now, Jim is moving away from both you and john, at 10,000 miles per hour, in a spaceship. And that photon is headed Jim's way. How fast will the photon pass Jim? Regards

Severian, and Swansont: You are both encountering difficulties which arise from the use of probability theory in the uncertainty relations. Not everyone who reads this thread will understand how to go from the uncertainty principle to probability, and if they cannot do that, they cannot even have an opinion on whether or not the uncertainty relations are true, much less meaningful. Lest anyone challenge this, what is it about the mathematics of probablity, that led some physicists to postulate the many worlds interpretation of QM?

Severian can you go into this slightly? I've never studied QFT but i have studied QM analysis of hydrogen. Thank you

Tom what exactly is the Hamiltonian. I've tried several times to learn Lagrangian Dynamics' date=' and I know that the Hamiltonian can be written in terms of the Lagrangian, but I cannot seem to remember either, well we'll see in a moment I will check. I think the main reason I haven't learned is because I can't see the utility. One of the authors started out by discussing holonomic and non-holonomic constraints, and sceleromic(sp?) and non sceleromic constraints. And lost me, and then this same author also discussed degrees of freedom, and I didn't quite follow that either. Another thing that bugged me, is that this same author said that it is believed that both Lagrangian, and Hamiltonian approach to dynamics are equivalent to Newtonian Dynamics, but the author wasn't certain. That was a bit well didn't make me want to keep reading. Let me see if I even remember the Lagrangian. I don't think I will get this right, and I'm not gonna google for the answer. Let E denote total energy of a system. Let U denote potential energy, let T denote kinetic energy. E = T + U The Lagrangian I think is umm L = T - U Maybe? I'm gonna leave that as my first guess, and now I'll google. Ok, here is Wolfram on the Lagrangian. Yeah ok, so I remembered it, but I cannot think of any instance where I would use it. Thanks

I'm reading through this right now. You might want to know that you only have six hours to edit your post. Right now I have read up to here: I think you mean: If t' = 0 there is no motion, If t' > 0 motion is assured. You might want to check for any other tiny errors, since you only have six hours to edit your post. I'm still reading. Regards PS: Ok I've read the whole thing through once. Now, I am going back carefully. Consider where you write the following. To make sure I follow, here you are talking about there being a point, a place in space, which is "spatially invariant" for some "change in the universe." In other words a place which doesn't move, is that what you mean by "spatially invariant" or do you mean something else? I presume you are saying this is connected to the idea of "absolute rest," which of course is the subject of this thread in the first place. You are saying, i think, that you need to show that there is a place in space that doesn't move, yet other things did, because dt>0. Correct me if I'm wrong, or just add comments. Just trying to follow you here. Next you say this: You are now describing the linear Sagnac experiment, and the paragraph above is related to the attached diagram. To paraphrase you: There is an emitter at rest in space, in an inertial frame of reference, and it is about to emit two photons simultaneously in this inertial frame, one to the left, the other to the right (see diagram). Actually, I'm not sure if your emitter is at rest or not. It now seems like you intend the point P of space to be at rest, quite independently of the emitter, and that what happens is that there comes a point in time where two photons are emitted simultaneously, and that happened at location P of space, and you have location P at absolute rest relative to the center of the universe. Am i right here? From the looks of your diagram, the experimental setup is moving from left to right, relative to point P, which is at absolute rest. I understand now. Next you say: Ok so... there is a photon moving to the left wrt P, and after time t has passed in the appropriate frame, it strikes a mirror and is reflected 180 degrees, so that it is now moving back towards P. Now, the frame in which t is measured is stipulated to be an inertial reference frame. Therefore, by one of the postulates of SRT, it must be the case that the speed of this photon will be measured to be c, in the IRF. Using the postulate of SR that the formulas of physics are the same in all inertial frames, it must be the case that the definition of speed formula is true, hence: c = D/t Thus, the distance traveled by the photon in time t is ct, as you say. Now, during the time it took the leftgoing photon to go from point P to the mirror, the rightmoving photon traveled identical distance, in identical time, since by postulate speed of all photons in any IRF is c. There is only one frame here, so that amount of time is t. So let me carefully track the events here, in order. Let t0 denote the moment in time at which the photons were emitted, and let t1 denote the moment in time at which the leftgoing photon strikes the mirror. So the following event took time t, in the appropriate inertial frame: [t0,t1] During that event, the leftgoing photon traveled a distance ct relative to P, and also, the rightgoing photon traveleed a distance ct relative to P, as in your diagram 1. The initial state was labeled 0. Now, it appears to me that the experimental device is traveling to the right, relative to point P, with uniform constant speed v, throughout the experiment. So at the moment in time at which the leftgoing photon strikes mirror L, namely moment in time t1,the rightgoing photon does not strike mirror R simultaneously, because both mirrors have advanced some nonzero distance vt, to the right relative to P. You say: So there is a third moment in time at which the photon reflected from mirror L returns to the point P in space. Denote this instant by t2. So, during the event [t1,t2], the photon which struck mirror L must travel an identical return distance, in an identical time. This is important to see. Moment in time t2 corresponds to your diagram labeled 2. That is the instant at which the photon which struck mirror L coincides with point P. Certainly, the speed of the photon must be c, since the rest frame of P is inertial, and we have assumed einstein postulate 1. Thus, the time of event [t1,t2] is also t, like during the first half of its trip away from P, and its speed on the return journey is again c, therefore the distance traveled by that photon on the return trip satisfies: ct=D All as you say. The very next thing you say is this: Correct... AND IF the initially rightmoving photon hit mirror R simultaneously to the moment in time the other photon hit mirror L then it would simultaneously arrive at P with the other photon, yet the photons did not hit the mirrors simultaneously in the P frame, because the experimental device was moving to the right with nonzero speed v in that frame. The photon that initially was moving right strikes mirror R after the other photon struck mirror L, in the inertial rest frame P. Next, you say this: Right. At the moment in time that photon L has returned to P, photon L is now a distance 2vt from the physical midpoint of the Sagnac device. That's simple to see from your diagrams. Its also conceptually easy to see. The time it took that photon to hit mirror L was amount of time t, t measured in rest frame P. The time it took that photon to travel back to P was amount of time t, as measured in rest frame P. Total time of travel of Sagnac device t+t=2t. The speed of the Sagnac device was constantly v, during event [t0,t2]. Hence, during this event, the total distance traveled by the Sagnac device in rest frame P is given by: v = D/2t Hence D=2vt. Right. The next thing you say is this: I think you made an error here. I believe you meant to say "located small distance vt from the R clock." Correct me if I'm wrong. After photon R has traveled a distance ct away from P, it is at that moment in time, a distance vt away from mirror/clock R. You say 2vt, but i think you meant vt. I could be wrong, I'm not sure what you said here just yet. Ok, its been about ten minutes and I still don't get the previous paragraph. So right here is where I got lost. I can see from your diagrams, exactly what's going on though. They were drawn perfectly. Let me just try to keep going. At the moment in time photon L strikes mirror L, photon R is a distance ct away from point P, a distance vt away from mirror R, and a distance ct-vt away from the physical midpoint of the Sagnac device.