Johnny5

I can't understand it, it's in spanish. Do you speak spanish? If not why in the world would you read that? As for the other things you are trying to say, inertial mass isn't supposed to be a function of time. Do you know the differential calculus? If so, I can show you in what sense mass isn't a function of time.

SR totally aside, because I don't want to get into that with you here, do you regard an 'accelerating frame' as 100% equivalent to one in which inertial forces are being felt? Forgive that, if it's not totally clear. As for the Foucalt pendulum, I have yet to understand the experiment. I need to understand gyroscopic motion, i know that much. I read that Foucalt is actually the one who first developed a gyroscope, and was using it, in conjunction with his giant pendulum, to draw inferences about the earth's rotation. Let me see if i can find the pictures I was looking at, of some very old gyroscopes: Here are a few pictures of some old gyroscopes, and a brief article on Foucault: Gyroscopes

What do you consider to be an inertial frame?

Unfortunately' date=' latex isn't working. Let's see... The simplest way to do this, is to treat the event set theoretically. i will have to explain, because I am one of few who do this. Here is your given information: Cliff height above ground=63 meters Initial speed= 8 meters per second. Initial direction of projectile motion=straight up Start the clock at the moment you release the projectile. There is a moment in time at which the object comes to rest in the frame. That is one subevent, and lasts for time T1. There is a second subevent, which begins when the first subevent ends, and ends when the object strikes the ground. Denote the amount of time of the second subevent by T2. The whole is the union of the parts. So the time of the whole event is T1+T2. First compute T1, then compute T2, then add them to obtain the answer. Here is the kinematical formula for constant acceleration: D= v[sub']0[/sub]t + 1/2 a t2 The acceleration due to gravity has magnitude 9.8 meters per second squared, and its direction is opposite to the direction of motion. So for subevent one we have: D= 8t - (9.8/2) t2 Now, as you can see this is one equation in one unknown. The height that the object rises wasn't given, nor was the time of flight t. So we need a second equation in the same two unknowns, and then we can solve for T1, which is the amount of time between when the stone is released, to the moment in time at which it comes to rest in the frame. Here is a link to the kinematical equations for constant acceleration: Look and you will see this one; D = (vi+vf)t/2 We know the initial speed is 8 m/s And we know the final speed is 0 m/s Keep in mind that speed is a strictly non-negative quantity (distance traveled divided by time of travel). Ok so... Using the formula above we have: D = 8t/2=4t And we also have: D= 8t - (9.8/2) t2 So we have two equations in two unknowns, so we can now solve for both unknowns. It is the time of the first subevent that we are concerned with, so we need to eliminate D. By direct substitution we have: 4t= 8t - (9.8/2) t2 9.8/2 = 4.9 therefore 4t= 8t - 4.9t2 Hence: 4.9t2+ 4t-8t=0 Hence: 4.9t2+ -4t=0 Factor out a t to obtain: t(4.9t -4) =0 One root is zero, which cannot be an amount of time, because an amount of time is a strictly positive quantity, hence we want the other root, namely the one for which: 4.9t -4 =0 So: 4.9t = 4 So: t = 4/4.9 = .81 seconds So the amount of time of the first subevent is .81 seconds. T1=.81 seconds Now, we have to compute the amount of time for the second subevent. During this part of the whole event, the object falls to the earth from higher than 63 meters. Let D be the height the object rose above the cliff during the first subevent. Then the distance which the object falls during the second part, is 63+D. We need to know that in order to compute the time of the second subevent T2. We already know this: D =4t So T1=.81 hence D = 4(.81) = 3.26 meters H = 63 + 3.26 = 66.26 The amount of time of the second subevent satisfies the following equation: H = v0t + (9.8/2) t2 Now during this part of the whole event, the initial speed is zero. The magnitude of the acceleration hasn't changed, but the direction of motion has. So during the first subevent, the center of mass of the object was decelerating in the frame, during the second subevent, the center of mass is accelerating in the frame. So the initial speed is zero, hence: H = (9.8/2) t2 Hence: 66.26 = (9.8/2) t2 Hence: 66.26 = 4.9 t2 Hence: 13.52 = t2 Hence: 3.67 = t So the time of the second subevent is 3.67 seconds: T2 = 3.67 seconds And T1 = .81 seconds So T=T1+T2 = 3.67+.81 = 4.48 seconds Unless I made an error. You can compare with other peoples calculations. Regards PS: The thing about breaking the whole event into mutually exclusive and collectively exhaustive parts, is because: 1. It leads to the right answer. 2. It will help you analyze problems in special relativity theory.

As you can see, Mach had a leaning to the natural number system {1,2,3,4,...}. Whether or not a given quantity of physics can be represented by natural numbers is a good question. The concept of pi seconds does seem meaningless. But then again, the concept of a quantum unit of time has difficulties which have yet to be overcome. But scientists have to explain reality using numbers. They also use their natural language. But numbers are the basis of the "hard sciences." Without numbers, scientists would never have been able to develop computers. So even if you cannot model reality using numbers... you sure as hell can change it by using numbers. Merely look at the existence of a computer, and you will see that's true. Regards

I wasn't planning to complexify things by disscussing how SR holds up to acceleration. Right now I just wanted to see whether or not Dr Swanson claimed that V is the relative speed between two frames. He did.

It is best to show what I mean symbolically, rather than debate set theory. I only ran through his argument once, for a week or so, found a weak point, and haven't looked at it again. I could just show you what I mean, and then you could critique it.

Let me make sure I copied everything down correctly, here is your exact question again: I will use a diagram this time, to represent the given information: M..........X....................m X represents my position, Mx will represent my mass. The distance from me to m is twice the distance from me to M. M=2m FL = (2m)(Mx)/R2 FR =(m)(Mx)/(2R)2 FL = 4(2m)(Mx)/4R2 FL = 4(2m)(Mx)/(2R)2 FL = 8(m)(Mx)/(2R)2 FL = 8FR The answer is above is correct. The gravitational force to the left is 8 times greater than the gravitational force to the right. I see your point. Center of mass is linear, and gravitation is inverse r^2. I could just have known that from the formulas for both. Let me comare things. Here is the formula for the center of mass of an N body system. MR = S miri The indice ranges from i=1 to i=N, in the summation. M is the total system mass, and mi is the mass of the ith body. R is the position vector of the center of mass in some arbitrary reference frame, and ri is the position vector of the ith body. So: M = S mi Let us choose our frame to be the CM frame, then in particular, R=0... the zero vector. So, in the center of mass frame, the following statement is true: M0 = S miri Hence: 0 = S miri Now, the system is a closed three body system hence the following statement is true in the CM frame: 0 = m1r1+m2r2+m3r3 Using the original symbols for the three masses we have: 0 = Mr1+Mxr2+mr3 Now, my position happens to be at the center of mass of the system, that is, r2=0, but this should be inferrable. Let me not a-priori assume that my position is the center of mass of the system, but let me formulate true statements from a frame in which the center of mass of Mx is the origin. Therefore, for sure r2=0, and it should be inferrable that this frame happens to be the CM frame of the system. Since r2=0 it follows that: 0 = Mr1+mr3 0 = 2mr1+mr3 0 = 2r1+r3 -2r1=r3 The previous statement was stipulated to be true, hence the center of mass of Mx is the CM of the frame.

I didn't know that the concept of hypersphere is used to explain redshift. I certaintly don't believe that is necessary. At this point, i would focus upon the definition of 'hypersphere' most importantly, does the idea have direction of motion constant, yet you return to where you were.

This came up in post 12 of this thread, and rather than having the topic there branch, I thought i just might move my question to another post. Here is what arose in the other thread: Here is what i would ask you about. You say that if special relativity holds then you are not in an accelerating frame. Here is my question. Could you run through an explanation of the time dilation formula of SR for me. Specifically, I want to see how you explain each of the letters in the formula, be they constants, or variables. I suppose I am most focused on v. Dt` = Dt (1-v^2/c^2)^(-1/2)

I didnt do any calculation i just sort of blurted out the answer. Let me see... You say i have a mass of 2M on my left and a mass of M on my right. but the mass of 2M is twice as far away from me as the M mass. Let me use the mathematics, its the only way for me to be certain of the answer. denote my mass by m The gravitational force on me to the left is given by: F = (2M)(m)/R^2 and the gravitational force on me to the right is given by: F = (M)(m)/r^2 and the distance from me to the mass on my right is 200 million miles and the distance from me to the mass on my left is 100 million miles. therefore R=2r

mgh is not an exact formula. This is why i want to use gravitational field theory for everything.

Flywheel

Can we discuss this further? Introduce the mathematical tools necessary to discuss it.

Ok let me think about what you are saying. By the way you should know me by now, I don't play games. I just don't like contradictions. So let me slowly sift through what you have said up there. Ok, firstly you say look up the theorem. Well i have done that. I will do it again, then quote it below. Fundamental theorem of algebra Gauss' proof of the FTA Here is another link: Fundamental theorem of algebra Here is how that site explains it: Notice how the above site says, "NOT NECESSARILY DISTINCT' Ahem...

Thank you, now we are getting to the bottom of things.

This is what I want to think about. Agreement on the value of pi for example. I want to have access to a thorough knowledge of rotation.

This is very funny. Why don't you prove the theorem Matt? You are the professional mathematician, not I.

But you say 'roots' in the plural Matthew. yet there can be only one. x^2-4x+4=0 I don't think it's childish by the way. I am serious, there is only one root. Draw an X axis, and a Y axis, plot the graph of x^2-4x+4 ok let that have been done. So here is the function which you have plotted on Maple, or Mathematica, or what have you: y(x) = x^2-4x+4 Now, look at the plot and tell me how many times the curve intersects the x axis? One or two?