Johnny5

Oh you were talking to the thread starter. Geeze... Hold on Kingjewel: i got i - 5 j- 7k + t(9i-15j-12k) Kingjewel: <1+9t,-15t-5,-12t-7> Now I got... R = <1+9t,15t-5,12t-7> So these answers are not equivalent. Therefore one of us has made an error. I don't remember making one. Nonetheless, i will go back and double check. Given by thread starter: A (i-5j-7k) and B (10i+10j+5k). P is a point on AB A,B are position vectors. They go from the origin of the frame (0,0,0) to a field point (x,y,z). P1=(1,-5,-7) P2=(10,10,5) vector from P1, to P2 = <10-1,10--5,5--7>= <9,15,12> R = <1,-5,-7> + t<9,15,12> R = <9t+1,15t-5,12t-7>

No its not. Your answer and the one I gave should be equivalent. Let me check. Here is what you wrote' date=' which is probably equivalent:

You certainly do know the position vector of a point which lies on the line. In fact you know two: A=<1,-5,-7> B=<10,10,5> The other thing you need to know is a vector which gives the direction of the line. So now, i have to think. I need to take the given information and come up with the direction. Done. You have two points, and you can construct a vector from one point to the other by definition. The resulting vector will lie on the line. Here are your two points: Point A = (1,-5,-7) Point B = (10,10,5) The vector from point A to point B is given by: AB= <10-1,10-(-5),5-(-7)> = <9,15,12> = 9i + 15j + 12 k So this may actually be all you were asked to find in part a. Nonetheless, we can continue on, and find the formula for the whole line. The position vector of any point on the line, in this frame, is given by: R = a + tb (notation at the link i posted) Look carefully at their diagram. For their a, you are going to use: A= <1,-5,-7> And you already have a vector which lies on the line namely: AB = 9i + 15j + 12 k So you can use this for their b. You could first turn the vector above into a unit vector, but that's not necessary because t is just a scaling factor. So you are now ready to write the answer. There is a line, in some reference frame, its not moving in the frame. The position vector of any point on this straight line is given by: R = <1,-5,-7> + t <9,15,12> R = <1,-5,-7> + <9t,15t,12t> R = <1+9t,15t-5,12t-7>

Ok... I'll keep that in mind. I gathered that much from the link you posted. It is highly abstract information though. Without an experiment firmly in mind, it's difficult to understand the point of them.

I read the article posted by Dr. Mattson, and not suprisingly understood extradinarily little. I have a book on Quantum mechanics written by Hans C. Ohanian, which is where I first encountered creation/annihilation operators. I think the impediment to me learning how to use them, is that I don't see what their utility is.

Regarding irrational numbers, or regarding complex numbers?

You both seem to think the parallel axis theorem can be used here. Do you have an explanation of how that gets used? I understand the focus of the post is using three conservation laws. Conservation linear momentum Conservation of angular momentum Conservation of energy The conservation laws give you the equations, you need as many equations as you have unknowns, and then you can solve for whatever it is you are looking for. But I don't see how the parallel axis theorem fits in.

Ok thanks. It sure is a cool looking picture. Regards

I would like to say something. As algebra developed, so too did the notation. We have now arrived at simple expressions such as: 3x^3-2x^2+8 = 0 9x-5=0 x^2+2x-1=0 And so on. Such polynomials have a degree n; (the highest power of x in the expression). The complex numbers emerge right when we consider second degree polynomials. It doesn't matter whether or not the coefficients are rational irrational, or complex, my point is this. From the desire to have n solutions to an nth degree polynomial equation (a few examples listed above) came the set of complex numbers. So, whatever the answer is to my original question, it lies in my previous sentence.

What particles are they counting?

Do you have a diagram of this somewhere? Maybe provide a link to a similiar problem? I think it's a good problem.

If we can find a counterexample to the statement, then the answer is false. From calculus, you know that the derivative of any constant is zero. So suppose that the electric potential at the field point is 4. The partial derivative of 4 with respect to x is zero, the partial derivative of 4 with respect to y is zero, and the partial derivative of 4 with respect to z is zero. Thus, if v=4 then we have the following true statement: -Ñ 4 = ¶4 /¶x i + ¶4 /¶y j + ¶4 /¶z k = 0 So, the voltage at the field point can be a nonzero constant, and the electric field there will be zero. So it isn't necessary that the potential at the field point be zero, in order for the electric field there to be zero. What is necessary for the E field to be zero there, is for the potential to be a spatio-temporal constant. I have just read Dr. Swanson's post, and that is what allows you to intuitively know that the answer to question one is false, without doing all the mathematics. Also, you can use that fact to reduce the number of mesh equations you have, when you are working with Kirchoff's laws, in order to generate equations in v,i. You can set the ground to be zero, wherever you want in a circuit, and things will still work out. I doubt you want to get that involved here. Here, you simply want to know a quick way to get at the right answer. But if you already know electrodynamics, then you will have run through the theory at least once, and should already know that you can add an arbitrary constant to the voltage, and your electric field at a point doesn't change. The reason is quite mathematical. Suppose that the following scalar function is the potential at a point: V(x,y,z) = e/4pe0r Where r = (x^2+y^+z^2)1/2 The electric potential above, is the potential due to a single electron, with electric charge e, located at the origin of reference frame S. You can compute the electric field at an arbitrary field point, by taking the gradient. The answer you get will be the same as if the electric potential was: V(x,y,z) = e/4pe0r + C Where C is any constant. Which is what Dr. Swanson alluded to. So after all that, you can be certain that the answer to question a... is false. I see that Dr. Mattson has answered the others, so if I have nothing further to add, I will leave things here. Without reading his answer to b, you should be able to figure out the answer. Suppose that v=0. Take your partials, and you will get E=0. The answer to c is not so easy to understand as the answer to b. Right now I am trying to think of a realistic potential function v(x,y,z), which has a value zero at some point, yet the electric field at the point is nonzero. I have already read Dr. Mattsons answer, and he got me thinking of a paraboloid. Here is the formula for it: z = x2/a2 + y2/b2 The above formula, leads to the shape of a satellite dish, a circular paraboloid. As you can see from the scalar function, when x=y=0, z is also equal to zero. So suppose you have this, due to some charge distribution somewhere: v = x2/a2 + y2/b2 You can see that the potential function is zero, when x=y=0. Let's look at the gradient of v. Ñv = 2x/a2 + 2y/b2 As it turns out, this implies that when x=y=0, that the electric field is zero too. But you should now be able to see the following: Had the exponents of x,y been 1, instead of zero, then after the partial derivatives were taken, we would have had a constant for an answer. So, suppose that you have the following potential function: v(x,y,z) = 3x+2y+5z Clearly, you have v(0,0,0) = 3*0+2*0+5*0=0 So at the point (0,0,0) in a frame, the voltage is zero, but now take the partials to obtain: gradient v = 3i^+2j^+5k^ Which is clearly nonzero everywhere. I'm not saying that the above function for v corresponds to any realistic configuration of charge distribution throughout the universe, but it does make Dr. Mattson's point, which is that there are mathematical functions which constitute counteraxamples to part © here. ¶ alt-0182 Ñ alt-0209

Isn't the total gravitational field at the center of the universe necessarily zero? Maybe he could distinguish it that way.

I just read about this in another thread, and I have a question. Look at the following picture. What are the units on the Y axis. They say DN/s Thanks

You stay technical Severian, I didn't fully understand your answer, but I'm sure others did. I don't see how the commutator fits in really, but Tom Mattson will get around to explaining the commutator, and then I will look at your answer again. Personally, my take on particles is that we are far from understanding them. But i think half of the fun is in trying to, and of course the other half is understanding them. Perhaps I can be slightly more specific. Some particles are very short-lived. Say a lifetime of a nanosecond, others a femtosecond. Now, in the case of a free neutron, i think the average lifetime is around 11 minutes, or maybe 14, i forget. But, and this is my take on it, to say we understand particle physics, would require us to understand these particle lifetimes (transition times). As I recall, there is a long formula, but it doesn't work well. Right now I am still focused on this:

Dr. Swanson, How does special relativity deal with what you call syntonization? In my paper on relativity, I analysed what you call 'syntonization' actually quite a bit. Maybe I will go into some of that analysis. Gamma appeared in my formulas, as a difference in clock rates, regardless of whether or not SR was correct.

If you take away their imaginary number i, they will not know what to do.

Well then, if you think you can do a better job, then by all means have at it. You have SR formulas, and the concept of a Hypershperical universe. Mix them together and what do you have? PS: As an aside, I take it you have a PhD, but just to be sure, do you? If you do, then I can refer to you as Dr. Mattson; which seems appropriate. Don't think I've forgotton about the topic of vector division, and the elegant mathematical analysis which you gave. And we still have your question in quantum mechanics, about observables. I remember you showing me the basics of bra/ket notation, and analysis. I don't forget anything which makes sense... Dr. Mattson. Kind regards, as always.

So, i believe the answer is 151 joules. I'm going to go think about it for a few. ___________________________________________________________ Well if i use that formula, together with Dr. Swanson's, the answer is 151 Joules. They have: E = T times theta(in radians) So that torque would have units of energy per radian. I still have to think about this a bit though. Something about spinning... I need to develop a new kind of algebra to handle this. It has to do with the truth value of statements about spinning frames of reference. Well that's what I am thinking about anyways. This goes back to Geistkiesel's post on the Sagnac effect, and a problem I was working on there. It has to do with rotation being absolute, not relative. Something about the coriolis force is coming to mind. Think about a spinning top, or a spinning wheel axis, or what have you. There are a couple of things you must do, in order to mathematically analyze that which spins. First you have to identify a spin axis. Then you must attach that axis to the thing which spins. Now, in the rest frame of the object, the change in angle is zero. So change in angle, necessarily involves what I am now thinking about. The non-relativistic character of that which spins. Think about that merry go round ride, where your back is against a circular wall, then the whole thing starts spinning, then the floor drops out, and you don't fall. I've been on one before, long ago, and I still remember something quite vividly... I moved my head towards the center of the spinning ride, just to see what would happen. Instantly, I got nauseous. I moved my head back to where it was, and it went away. The fact that I was spinning, is absolute. You cannot say that I was at rest, and the rest of the universe was spinning. That's incorrect. From a purely kinematical point of view, you can, but not a dynamical one. And so in that rest frame, (the spinning one), my change in angle was zero. Of course that frame was a non-inertial one. What I am concerned about, is the mathematical analysis only. In Dr. Swanson's answer, the angle theta is introduced, but little is said about what frame to define it in. And then of course we use units of radians, when in fact 2 times pi is a pure number. Rotation