Johnny5

Thank you Dr Swanson, I will re-perform the calculation once again. Probably monday Dr. Kind regards Set centripetal force equal to gravitational force, and period is one sidereal day. That'l keep till monday.

Actually no, I used beforeness.

I am just reading about the Sagnac effect now, it will be interesting to see how long it takes me to understand it. Apparently, in 1913 French scientist 1869-Georges Sagnac-1926 came up with a way to modify the Michelson Morely experiment. Here is one source which describes the "Sagnac effect", and the relevent quote: Article on Sagnac effect Here is an article from the encylopedia Brittanica: Article2 on Sagnac Effect Ok, so the "Sagnac Effect" is one method of measuring rotation. That I consider useful and important. Today, optical gyroscopes (based on the Sagnac Effect) are replacing mechanical gyroscopes, because the optical gyroscopes have no moving parts. So the "circle is turning." The device is spinning, with some angular velocity. "Photon's do not arrive at emission point simultaneously" Well of course not, because the circle is turning. Let me find a picture of it. Here is a picture: Article 3 on Sagnac effect This is interesting: Everyone's thoughts on this appear somewhat scattered. Somewhere in the universe is the center of mass of the universe. Set up a rectangular coordinate system, with that point permanently fixed at the origin. Now, there are stars everywhere, in that frame. If the frame is spinning wildly out of control, then to an observer whose location is fixed in the frame, it will appear as if the stars are spinning out of control. And this will show up in his calculation of the total kinetic energy of the universe. But in just the right frames, the centers of mass of various galaxies obey Galileo/Newtons Law of Inertia. All such frames will be inertial reference frames. Now, the quote above, says that Sagnac rotated his turntable at two revolutions per second. Then it says that he "found the expected effect." Then the caption makes reference to "absolute rotation." Here is something interesting about Physicist H.E. Ives: Article making reference to Ives Cross with this source: Right now, I am reading the source which you said was excellent: Article Ok I am focusing on the mathematics, which is accompanied by figure 2. The Circular Sagnac Effect They use the symbol t0 for the time taken for a photon to travel around the circular path, when the motor which causes the device to rotate is off. I have a good book (Numerical Methods For Scientists And Engineers-1962) on the finite discrete difference calculus, written by R.W. Hamming, and the difference calculus would use Delta t, for an amount of time. So I am just going to switch that to: Dt0 So this denotes an amount of time, which is measured by a clock at rest in a frame attached to the surface of the earth, and the measurement is made when the device isn't spinning in the frame. For whatever it is worth, the Hamming code is named after him. Richard Hamming His book is really good, and influenced me mathematically so to speak. So now, let there be a coordinate system at rest in this frame, with its origin at the center of the circle, and the circle lying in the XY plane of the frame. Now, the photon will have some speed in the frame. Denote that speed by v. I deliberately did not choose c to denote the speed. So v is the speed of a photon, as defined in this frame of reference and no other. Now, let us presume that the tangential speed of the photon is uniform in this frame. The tangential speed of the photon is constant, but the velocity of the photon in this frame is changing. The photon is travelling in a circle, with radius R. So the photon has a nonzero centripetal acceleration, related to v, and to R as follows: ac = v2/R Distance around the circle = 2 pR And speed = distance traveled/travel, hence: v = 2 pR/Dt0 Assuming that the speed of the photon isn't zero in this frame, we can divide both sides by v to obtain: Dt0 = 2 pR/v The article then makes reference to the time measured aboard the spinning disc t`. I need to think about this carefully. From the diagram, the photon which moves against the rotation, reaches the detector before the photon which moves with the rotation, assuming that the two photons began their circular journey simultaneously, and have the same speed in the earth frame. Let the constant angular speed of the device in the earth frame be represented by w. Let the tangential speed of the device in the earth frame be denoted by vt. The tangential speed of a point fixed on the rotating disc is related to the angular speed in the earth frame, and the radius of the disc as follows: vt = wR Let t1=0 denote the moment in time, at which the two photons begin their circular journey. Later, the device will have rotated through some angle =q, at which moment in time, the photon moving against the direction of rotation will be back at the starting point(the detector). Call this moment in time t2. So this is an event, marked by two moments in time t1,t2. I will represent this event by the notation [t1,t2]. Now, during the event [t1,t2], the total distance which the photon traveled in the earth frame, is less than 2 pi r. Focus on the circle in figure 2(labeled "Whole apparatus turning at w clockwise"). There are two sectors to focus on. One bounded by circular arc SS`, and a larger sector bounded by circular arc SS``. Call the photon moving counterclockwise photon A, and call the photon moving clockwise photon B. So the total distance traveled by photon A, during event [t1,t2] is given by: 2p R - Rq. Where theta is the angle subtended by arc SS`. Now, let the amount of time which this took, as measured by a clock at rest in the earth frame be denoted by: Dt1 And a small amount of time later, photon A will have traveled a full circle, in the earth frame. Let us denote this amount of time by: Dt2 Lets add the two times together to obtain: Dt1 + Dt2 And this is the amount of time, that we previously denoted by: Dt0 Therefore: Dt0 = Dt1 + Dt2 And we already defined the speed of photon A in this frame as: v = 2 pR/Dt0 Now, denote the moment in time at which Photon A has traveled a distance of 2 pi R in the earth frame, as t3. So t1 before t2 and t2 before t3. Now, if the two photons have the same speed in the earth frame, then at moment in time t3, photon B will also have traveled a distance of 2 pi R in the earth frame, and so the two photons will be right back next to each other simultaneously. But at moment in time t3, the disc advanced slightly, because it is spinning in the earth frame, and so there will be a still later moment in time t4, at which moment photon B will finally coincide with the detector (which is fixed to the spinning disc). So the whole phenomenon can be broken into parts. [t1,t2] union [t2,t3] union [t3,t4] t1 corresponds to the moment in time at which the photons are simultaneously emitted. t2 corresponds to the moment in time at which photon A passes the detector. t3 corresponds to the moment in time at which photon A and photon B pass each other. t4 corresponds to the moment in time at which photon B passes the detector.

Actually' date=' I didn't prove it here. Swansont would remember that. I wanted to see what he would say about me claiming to prove it here. I expected him to say that I didn't prove it, all I proved here was that: If the speed of light is the same in all inertial frames then (the length contraction formula is true and the time dilation formula is true). I actually haven't shown anyone here how to falsify the antecedent, because it is ridiculously easy, but for some reason PhD's don't know how to do it. Regards PS: The answer in how to falsify the antecedent, is obvious on logical grounds alone. Here is the link to the statement which I did prove here: See post 79: discussion of light clock In that post, you will see the following highly important statement proven: Theorem: If the speed of a photon is the same in any inertial reference frame then (Dt` = Dt/ [1 - (V/c)2 ]1/2 AND L = L0 [1 - (V/c)2 ]1/2) QED So, the question now, is can the antecedent be falsified deductively? The answer is yes, and I explained to them how, but it seems to have been over their heads. I will just re-explain, though you must decide for yourself. We have a statement of the following form, which we are convinced is true: If A then (B AND C). Therefore, the following statement must be true on logical grounds alone: If A then B. Also, the following statement must be true on logical grounds alone: If A then C. In our case here, B is the time dilation formula, and C is the length contraction formula, and A is the fundamental postulate of the theory of special relativity. Now, the statement about time dilation confuses everybody, so disregard it, and focus only upon the length contraction formula. If you can now construct an argument, whose first and only assumption is that the length contraction formula is true, and which reaches contradiction, you can then negate your only assumption. Thus, you will conclude, deductively, that the Lorentz-Fitzgerald length contraction formula is false. That having been accomplished, you already knew that the following statement was true: If the fundamental postulate of SR is true then the Lorentz-Fitzgerald length contraction statement is true. The previous statement is logically equivalent to the contrapositive which is: If the Lorentz-Fitzgerald length contraction formula is false, then the fundamental postulate of SR is false. So now, since you just concluded that the Lorentz formula is false, you can conclude that the fundamental postulate of SR is false. And then furthermore, you can now finally conclude that the time dilation formula is false as well. So all these conclusions rest upon your being able to falsify the Lorentz contraction formula, and nothing else. Regards Here is the secret to doing it... Forget about the Lorentz formula, and focus on all formulas which state that length of bodies in relative uniform motion contract. We have two metersticks which have identical lengths, when at rest relative to one another. We can make them coincide as follows: AXXXXXXXXXXXXXXXXXXXXB A`XXXXXXXXXXXXXXXXXXXB` The ends of the rulers coincide when they are at rest relative to one another. Now, suppose that the length of a ruler is dependent upon its speed v, in someone elses frame of reference, inertial or otherwise. Let ruler AB be at rest in an inertial reference frame. Let ruler A`B` be in uniform relative motion to ruler AB, from left to right, and make only one assumption, which is that it's length is a function of its speed v in ruler AB's frame. (since AB is in an inertial frame, and A`B` is moving at constant speed through an inertial frame, it follows that the rest frame of A`B` is also inertial) This can be proven, and in another post somewhere I proved it. (Also, ruler AB cannot be rotating, but I've stipulated that). So there is translational motion of the center of mass of ruler A`B` in the rest frame of AB only, there is no rotation of either ruler, in the rest frame of the other. Assume the formula, which is a function of v, says that ruler A`B` should shorten, in the rest frame of AB. Therefore, eventually there will come a moment in time, at which the state of the universe will be as follows: ______AXXXXXXXXXXXXXXXXXXXXB A`XXXB` Then later: ______AXXXXXXXXXXXXXXXXXXXXB ______A`XXXB` Then later: ______AXXXXXXXXXXXXXXXXXXXXB ________________________A`XXXB` And this analysis was carried out in the rest frame of ruler AB, which is inertial by stipulation. So there is a necessary order to moments in time. There is no way in hell that B` coincided with B before A` coincided with A. From the state diagrams, the order of moments in time can be marked as follows: There is a moment in time, at which B` coincides with A. That is followed by a later moment in time, at which A` coincides with A, and then that is followed by a moment in time at which B` coincides with B. [A` would have coincided with A simultaneously to B` coinciding with B, had it not been for our single assumption that length contracts as a function of relative speed v] yet we must reason under that assumption for awhile. So this is the order of things: A` coincides with A before B` coincides with B. Now, frame switch into the other inertial frame. You are now analyzing the relative motion of the two rulers, from the rest frame of ruler A`B`. Using whatever formula for length contraction you concocted, you must also use it here, hence the length of ruler AB is shorter. So in this frame ruler AB is moving from right to left, and we have this: _________________________AXXXXB A`XXXXXXXXXXXXXXXXXXXXB` Then later we have this: ___________________AXXXXB A`XXXXXXXXXXXXXXXXXXXXB` Then later we have this: AXXXXB A`XXXXXXXXXXXXXXXXXXXXB` Therefore, the order of moments in time is as follows: B` coincided with B before A` coincided with A. So the following two statements are true, in global reality: X before Y and Y before X. The state diagrams should be sufficient to convince you, that there is an error in any formula which has length dependent upon relative speed v, in the manner here, where the contraction isn't due to an applied force. Now, all you have to do, is construct a mathematical proof. Which can be done. Regards again. PPS: If either of the rulers is being subjected to a force, then it's rest frame would be non-inertial, and its length would change, but that was not the case here. Furthermore, length of a real ruler depends upon temperature, but again, we can stipulate that the rulers have the same temperature throughout their passing by each other, where that temperature is measured by a thermometer in their rest frame.

I have performed an extremely thorough analysis of the special theory of relativity, and the inescapable analytical conclusion is that both the Lorentz contraction formula, and the time dilation formula are false in all inertial reference frames. I even showed you how to prove it in some other thread. Don't be mad at the messenger. Kind regards Dr. Swanson

How so? I don't think so. Regards

Yes I just ordered it, will have it in two weeks. Thank you

The question is, "can we now find the specific numbers r1,r2,r3, from the three equations immediately above?" Suppose that we add two of the three equations together, like so: [math] 2r_2 + 2r_3 = 2 (-3-r_1) [/math] And now divide both sides by two, to obtain: [math] r_2 + r_3 = -3-r_1 [/math] It now follows that: [math] r_1+r_2 + r_3 = -3 [/math] Thus, the sum of the roots corresponds to the product of the leading coefficient and the constant term in the original formula [math] 3n^3-9n^2+15n-1 =0 [/math] I don't know if there is a theorem for this, but it seems familiar for some reason. Ideally, we want three equations in three unknowns, so that we can then solve the system using matrices. If instead, we subtract 2r3 from 2r2, we obtain: [math] 2r_2-2r_3 = 2\sqrt{(-3+r_1)^2- \frac{4}{3r_1}}} [/math] And now, division of both sides by 2 yields: [math] r_2-r_3 = \sqrt{(-3+r_1)^2- \frac{4}{3r_1}}} [/math] If we now square both sides we obtain and expression which is devoid of the radical sign: [math] (r_2-r_3)^2 = (-3+r_1)^2- \frac{4}{3r_1}} [/math] If we multiply both sides by r1 we obtain: [math] r_1(r_2-r_3)^2 = r_1(-3+r_1)^2- 4 [/math] Now, the RHS is a cubic in r1, whereas the LHS is going to be quadratic in r2,r3. I am going to try to make use of this fact. If it leads nowhwere, then I'm going to give up. [math] r_1(r_2-r_3)^2 = r_1[9+r_1^2-6r_1]-4 [/math] [math] r_1(r_2-r_3)^2 = 9r_1+r_1^3-6r_1^2-4 [/math] From an earlier line of work we know that: [math] r_1^3-3r_1^2+5r_1-1 = 0 [/math] Solving for r1 cubed gives: [math] r_1^3 = 3r_1^2-5r_1+1 [/math] Therefore it follows that: [math] r_1(r_2-r_3)^2 = 9r_1+(3r_1^2-5r_1+1 )-6r_1^2-4 [/math] Combining like terms on the RHS leads to: [math] r_1(r_2-r_3)^2 = -3r_1^2+4r_1-3 [/math] Which leads to: [math] -3r_1^2+4r_1-3 - r_1(r_2-r_3)^2 =0 [/math] Therefore: [math] -3r_1^2+(4 - (r_2-r_3)^2)r_1 -3 =0 [/math] The previous equation is quadratic in r1, so in principle, if we knew r2, and r3, then we could solve it for r1 using the quadratic formula. Divide both sides of the equation above by negative three to obtain: [math] r_1^2+ ( \frac{-4 + (r_2-r_3)^2}{3} ) r_1 +1 =0 [/math] Now, to lighten up on the reading, define xi such that: [math] \xi \equiv( \frac{-4 + (r_2-r_3)^2}{3} ) [/math] And rewrite the equation above as: [math] r_1^2+ \xi r_1 +1 =0 [/math] The roots of the equation above, by the quadratic formula are: [math] \text{root 1} = \frac{-\xi + \sqrt{\xi^2 - 4}}{2} [/math] And [math] \text{root 2} = \frac{-\xi - \sqrt{\xi^2 - 4}}{2} [/math] Now, r1 is a root of the principle equation which was: [math] 3n^3-9n^2+15n-1 =0 [/math] And we cannot have two solutions for r1, as the above roots, labeled 'root 1' and 'root 2' suggest. So one of them will lead to an explicit contradiction. Hence root 1=root 2 if and only if [math] \sqrt{\xi^2 - 4} = 0 [/math] Which is true if and only if [math] \xi^2 - 4 = 0 [/math] Which is true if and only if [math] \xi^2 = 4 [/math] Which is true if and only if [math] \xi = 2 [/math] Which is true if and only if [math] \frac{-4 + (r_2-r_3)^2}{3} = 2 [/math] Which is true if and only if [math] -4 + (r_2-r_3)^2 = 6 [/math] Which is true if and only if [math] (r_2-r_3)^2 = 10 [/math] Which is true if and only if [math] r_2-r_3 = \sqrt{10} [/math] And this result is new. We now have two equations in three unknowns. [math] \text{Equation 1}: r_1+r_2 + r_3 = -3 [/math] [math] \text{Equation 2}: 0r_1+ r_2-r_3 = \sqrt{10}[/math] Now, we just need one more equation in the same three unknowns, and we can use the techniques of linear algebra, to find the unique solution.

That's the famous formula for energy quantization. It is equivalent to what Bohr got, using a planetary model. Have you ever gone through the solution of the Schrodinger PDE before?

Found something on it. Hermitian operators And here is the part about simultaneous observables: Simultaneous observable quantities There you see the notation I was talking about. Anyways, I think the position operator r does commute with itself. But i have to run through that notation again, to prove so.

At first they thought it was bird crap messing it up. They never tell you these things in the history books.

I'm going to have to prove this now. Regards

Dr swanson, can you show me this mathematically? The momentum operator, from memory is: [math] -i \hbar \frac{\partial}{\partial x} [/math] Or do it in three dimensions. Show me mathematically, how to say that the position operator does not commute with itself. For the position of the electronal node in the hydrogenic rest frame, we have the variable r. There is some neat symbolism, though I've forgotten it, to show this. Something about Hermitian operators, and simultaneous observables. It's coming back but very slowly. It seems to me that you have something like this; [r,r*] for the position operator, and [math] [-i \hbar \frac{\partial}{\partial x},i \hbar \frac{\partial}{\partial x}] [/math] For the momentum operator, and you are checking to see if the operators are commutative. I think in the case of the position operator r, it does commute with itself. Yes or no?

Law of conservation of momentum The total momentum of a closed system is constant in time. Mathematically: [math] \vec P_i = \vec P_f [/math] Initial system momentum equals final system momentum. The first thing you have to do, is choose an inertial reference frame, in which to observe the collision. Ball A, and ball B have an inertial mass of 5 kilograms each: [math] M_A = 5 kg [/math] [math] M_B = 5 kg [/math] Now, you have ball A colliding with a stationary ball B. This means that in the particular inertial reference frame you choose, the initial speed of ball b is zero, so that ball B is at rest in the frame, at least initially. On the other hand, ball A has to initially have a speed of 4 meters per second. So the frame I am going to carry out the analysis in, is going to have ball B initially located at the origin of the frame, and ball A is going to be moving upwards, along the y axis, with speed 4 meters per second. So before collision, the total momentum of ball B in this frame is zero. This fact follows since the magnitude of its momentum is equal to its mass, times its speed, and initally its speed is 0, hence the magnitude of its momentum is initially zero, hence its vector momentum is initially zero, in this frame. On the other hand, ball A has a nonzero speed, and a nonzero inertial mass, at least initially, so its initial momentum is nonzero, and will be represented by: [math] \vec P_A = M_A \vec v_A = M_A v_A \hat j = 5*4 \hat j = 20 \hat j[/math] And likewise the initial momentum of ball B in this inertial frame is: [math] \vec P_B = M_B \vec v_B = 5*\vec 0 = \vec 0 [/math] The intitial total system momentum is equal to the sum of the momentum of its parts: [math] \vec P_i = \vec P_A + \vec P_B = 20 \hat j + \vec 0 = 20 \hat j [/math] So as ball A creeps closer to ball B, the total system momentum does not change. But eventually there is a collision. It cannot have been a head on collision, because after the collision ball A moves diagonally in the second quadrant of the XY plane, and ball B moves diagonally in the first quadrant of the XY plane, each ball at a 45 degree angle with the Y axis(which is the original line of motion of ball A). The remainder of the problem is simple. Simply express the momentum of the parts, (draw the vector diagram for this), and lastly set the sum of the momentum of the parts Pf.(after collision) equal to the original momentum Pi. As a vector the momentum of ball A after collision is given by: [math] \vec P^\prime_A = M_A \vec v^\prime_A [/math] Now, I am assuming that the mass of ball A did not change during the collision, and that the mass of ball B didn't change during the collision. As for the final velocity vector of ball A, you have it as 45 degrees from the Y axis. So break its final velocity vector into horizontal and vertical components. [math] \vec v^\prime_A = -v_x_A \hat i + v_y_A \hat j [/math] and do the same thing for the final velocity vector of ball B, you have it as 45 degrees from the Y axis, but in the first quadrant of the XY plane, so: [math] \vec v^\prime_B = v_x_B \hat i + v_y_B \hat j [/math] By the way, speed is a strictly positive quantity. So now I am going to draw a picture of this, although unfortunately I cannot put it here. The initial momentum in the i^ direction was zero, and this cannot have changed therefore: [math] \vec 0 = -M_Av_x_A \hat i + M_Av_x_B \hat i[/math] Dividing both sides by MAi^ we obtain: [math] 0 = -v_x_A + v_x_B [/math] From which it follows that: [math] v_x_A = v_x_B [/math] Now, the initial total system momentum (before collision) was 20 kilogram meters/second in the j^ direction, and this cannot have changed, therefore: [math] 20 \hat j = M_Av_y_A \hat j + M_Bv_y_B \hat j [/math] Dividing both sides by j^ we obtain: [math] 20 = M_Av_y_A + M_Av_y_B [/math] And the mass of each ball just happens to be the same... namely 5 kilograms, therefore: [math] 20 = 5v_y_A + 5v_y_B [/math] Dividing both sides by 5 kilograms, we obtain: [math] 4 = v_y_A + v_y_B [/math] Now, although you cannot see the vector triangles which I drew, I can explain them to you. There is one vector triangle in the first quadrant, and its a right triangle, one leg is VxB, and the other leg is VyB, and both interior angles are 45 degrees. Therefore we have: [math] tan 45 = \frac{v_y_B}{v_x_B}=1 [/math] And in the second quadrant there is a similiar right triangle, satisfying: [math] tan 45 = \frac{v_y_A}{v_x_A}=1 [/math] From which we may infer that: [math] v_y_A = v_x_A [/math] [math] v_y_B = v_x_B [/math] And we already knew that: [math] v_x_A = v_x_B [/math] [math] 4 = v_y_A + v_y_B [/math] So we now have enough information, to figure out the actual speeds of each ball. The speed of ball A is given by: [math] V_A =\sqrt{(v_x_A)^2+(v_y_A)^2} [/math] And the speed of ball B is given by: [math] V_B =\sqrt{(v_x_B)^2+(v_y_B)^2} [/math] Now, [math] 4 = v_y_A + v_y_B [/math], and [math] v_y_A = v_x_A [/math] and [math] v_x_A = v_x_B [/math], therefore Now, [math] 4 = v_y_A + v_y_B [/math], and [math] v_y_A = v_x_B [/math], therefore Now, [math] 4 = v_x_B + v_y_B [/math] Therefore: [math] 4 - v_y_B = v_x_B [/math] Therefore: [math] V_B =\sqrt{(4 - v_y_B)^2+(v_y_B)^2} [/math] And similarly we find that: Since [math] v_x_A = v_x_B [/math] and [math] v_x_B =v_y_B [/math] we have: [math] v_x_A = v_y_B [/math] And since: [math] 4 = v_y_A + v_y_B [/math] we have: [math] 4 = v_y_A + v_x_A [/math] From which it follows that: [math] 4 - v_y_A = v_x_A [/math] So that: [math] V_A =\sqrt{(4 - v_y_A)^2+(v_y_A)^2} [/math] Now, conservation of momentum isn't the only conservation law of physics used in analyzing collisions, there is also conservation of energy. And the basic idea is the same, that the total energy of the system before collision, is equal to the total energy of the system after collision. Using conservation of energy, we will generate some more equations, in the same unknowns, allowing us to solve for the speed. Conservation of system energy The total energy of an object is a frame dependent quantity. E = T + U E is the total energy of the object, T is the kinetic energy of the object, and U is the potentially kinetic energy of the object. Let us imagine that the collision is taking place in space, where there is no friction, and furthermore, that neither of the balls involved in the collision are spinning in the inertial frame before collision. Thus, initially, the potentially kinetic energy U of each ball is zero. Thus, any energy which they have is purely translational kinetic energy T. I am going to use the classical formula for kinetic energy, from Newtonian mechanics, if one wishes to use relativistic mass, the calculations become needlessly complex. Here is the formula from Newtonian mechanics, for kinetic energy T: [math] T = \frac{mv^2}{2} [/math] m is the inertial mass of the object, and v is its speed in the inertial frame. Now, ball b is initially at rest in the frame, so that means that its speed v, is equal to zero initially. After collision its speed is greater than zero in the frame. As for ball A, initially, its speed is 4 meters per second in the frame, and after collision, its speed is something greater than zero, but less than 4 m/s. So the initial total system energy is just the kinetic energy of ball A, there are no other contributions to the total system energy in the frame before collision. Thus, the initial total energy E of the system is: [math] E_i = \frac{16M_A}{2} [/math] (The number 16 is the initial speed squared 4*4=16) And by the law of conservation of energy, this must equal the final system energy after collision Ef. Now, the whole final system energy is equal to the sum of its parts. Let us presume that after collision neither ball is rotating (this is unrealistic, but it will simply further calculations). The only way it would be literally true that after collision neither ball is rotating, would be for the collision to be head on. But if that was the case, then the balls would not depart at 45 degree angles. At any rate, we already introduced symbols for the final speeds of both balls in the inertial frame, after collision. The final speed of ball A in the frame was denoted by: [math] V_A [/math] The final speed of ball B in the frame was denoted by: [math] V_B [/math] Hence the kinetic energy of ball A in the frame after collision is: [math] \frac{M_A(V_A)^2}{2} [/math] And the final kinetic energy of ball B in the frame after collision is: [math] \frac{M_B(V_B)^2}{2} [/math] Since we presume that none of the initial energy was converted into rotational energy, the final total system energy is simply: [math] E_f = \frac{M_A(V_A)^2}{2}+\frac{M_B(V_B)^2}{2} [/math] And by the law of conservation of energy, the final system energy must equal the initial system energy, hence: [math] E_i = \frac{16M_A}{2} = E_f = \frac{M_A(V_A)^2}{2}+\frac{M_B(V_B)^2}{2} [/math] From which it follows that: [math] \frac{16*5}{2} = \frac{5(V_A)^2}{2}+\frac{5(V_B)^2}{2} [/math] Multiplying both sides of the previous equation by 2/5, leads to: [math] 16 = (V_A)^2 +(V_B)^2 [/math] And the previous equations we had were: [math] V_B =\sqrt{(4 - v_y_B)^2+(v_y_B)^2} [/math] [math] V_A =\sqrt{(4 - v_y_A)^2+(v_y_A)^2} [/math] Squaring both sides of the previous two equations leads to: [math] (V_B)^2 = (4 - v_y_B)^2+(v_y_B)^2 [/math] [math] (V_A)^2 = (4 - v_y_A)^2+(v_y_A)^2 [/math] From which we can now infer that: [math] (V_A)^2 +(V_B)^2 = (4 - v_y_A)^2+(v_y_A)^2 + (4 - v_y_B)^2+(v_y_B)^2 = 16 [/math] Recall that: [math] 4 = v_y_A + v_y_B [/math] Hence: [math] v_y_A = 4 - v_y_B [/math] and [math] v_y_B = 4 - v_y_A [/math] Therefore: [math] (4 - v_y_A)^2+(v_y_A)^2 + (v_y_A)^2+(4 - v_y_A)^2 = 16 [/math] From which it follows that: [math] 2[(4 - v_y_A)^2+(v_y_A)^2] = 16 [/math] From which it follows that: [math] (4 - v_y_A)^2+(v_y_A)^2 = 8 [/math] From which it follows that: [math] 16+(v_y_A)^2-8v_y_A +(v_y_A)^2 = 8 [/math] From which it follows that: [math] 2(v_y_A)^2- 8v_y_A + 8 = 0 [/math] Let X denote the unknown quantity we are looking for, hence [math] X \equiv v_y_A [/math] So we can rewrite the equation a little nicer as: [math] 2X^2- 8X + 8 = 0 [/math] And we can divide both sides by 2 to obtain [math] X^2- 4X + 4 = 0 [/math] Which can be easily factored. Any rational roots of the previous equation will be of the form p/q where p is one of the factors of the constant term(which is 4), and q is one of the factors of the leading coefficient(which is now 1, because we divided by 2). The positive factors of 4 are found from: 1*4 2*2 and the negative factors are found from: -1*-4 -2*-2 We now have to find which pair sums to the coefficient of the linear term (-4x). The answer is (-2)+(-2) = -4 Thus, we have the following true statement: [math] X^2- 4X + 4 = (X-2)(X-2)=0 [/math] So we have one repeated root, which counts as two roots in a theorem supposedly proven by Carl Friedrich Gauss 4 different times, the first of which was his dissertation for his doctorate (1799). Replacing X by [math] v_y_A [/math] we have: [math] (v_y_A-2)^2=0 [/math] From which it follows that: [math] v_y_A-2 =0 [/math] therefore: [math] v_y_A=2 [/math] So since: The following code was used to generate the image you clicked on: [math] (V_A)^2 = (4 - v_y_A)^2+(v_y_A)^2 [/math] It follows that: [math] (V_A)^2 = (4 - 2)^2+2^2 = 4+4 = 8 [/math] From which it follows that: [math] V_A = \sqrt{8} = \sqrt{(2)^2 \cdot 2} =\sqrt{(2)^2} \cdot \sqrt{2}=2\sqrt{2} [/math] Radical two is approximately equal to 1.4, and two times 1.4 is equal to 2.8. And similarly, you will find that the speed of ball 2 in the frame, after collision is 2.8 meters per second. You should check that for yourself. Regards

In the case of a hydrogen atom, free of all external forces, you have an electron in a potential given by: [math] U® = \frac{1}{4\pi \epsilon_0} \frac{Q_1Q_1}{r} [/math] And Schrodinger's equation is: [math] \frac{- \hbar^2}{2 \mu} \nabla^2 \Psi +U\Psi = i \hbar \frac{\partial \Psi}{\partial t} [/math] Where mu is the reduced mass of the electron. So we can perform the analysis in the center of mass frame, of the proton-electron system. Now, suppose that an electron is in some energy state, above the ground state, and you seek to measure its position. Is position a well defined observable quantity? And if not, then you are going to use probability theory. So my question is, how do you go from the boundary conditions for a hydrogen atom in free space, to knowing the number of elements in the sample space, in order to justify the treatment of the electron's position as a random variable. If that didn't make sense, then I will solve the TISE for Hydrogen, to make the question clearer. You aren't going to exactly follow where I am going with this, so to be crystal clear: How many possible places are there for an electron in hydrogenic atom in free space to be at the next moment in time? One, or more than one?

So by taking the limit as k approaches infinity of f(k), I am in effect performing the ratio test.

Even the usage of discrete distribution functions, still leads to conceptual problems. Just a dumb question... how is one to know how many elements are in the sample space? Regards