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Natural Science Forum / Physics / Particle Physics / August 2006THE COMPOSITION OF VELOCITIES
I speak of Einstein's composition of velocities theorem. It gives an erroneous result. It can easily be tested by using the conservation of momentum. To make it quick and easy, I shall use an example: the addition of .75c to .75c. The theorem yields .96c. To be consistent, we must be certain that both velocities are truly .75c -- and we must also be certain that the final momentum represents the correct amount and that it also represents a conserved amount. To accomplish this, we set up a two case scenario. We declare m = 1 and c= 1. This makes computation simpler with no jeopardy to veracity. CASE 1 ( )-------.75c--------------><-------.75c--------------( ) p=1.13 p= 1.13 This certifies that we have two velocities of .75c. Now the complication here is that according to custom, the final momentum is zero. So to acquire a meaningful momentum resultant, we simply change the perspective and declare the right hand system as inertial and all the velocity and momentum to reside in the left hand system. This also guarantees that the resultant momentum is a conserved quantity. According to Einstein the answer is Case 2. CASE 2 ----------------------.96c--------------------------->( )( ) p=3.43 Next, we note that we have changed nothing between Case 1 and Case 2 except the viewpoint. Therefore, in order maintain conservation, the final momentum must be the sum of the momenta in Case 1, i.e., 2.26. But it isn't. Therefore, the answer is wrong. According to my theory, written up in my book, the correct answer is .915c. ******************************************** Vertner H. Vergon : ( )-------.75c--------------><-------.75c--------------( ) : p=1.13 p= 1.13 [quoted text clipped - 12 lines] : : But it isn't. Therefore, the answer is wrong. Suppose that instead of Einstein's formulas, you use Newton's. Instead of a relativistic velocity, assume that in the first case your two masses are moving toward each other at a velocity of 1 m/s. What do you get for the momentum in each of the two cases? ----- Richard Schultz schultr@mail.biu.ac.il Department of Chemistry, Bar-Ilan University, Ramat-Gan, Israel Opinions expressed are mine alone, and not those of Bar-Ilan University ----- "You don't even have a clue about which clue you're missing." > : ( )-------.75c--------------><-------.75c--------------( ) > : p=1.13 p= 1.13 [quoted text clipped - 24 lines] > ----- > "You don't even have a clue about which clue you're missing." VERGON Do you? Case I: 0 Case II: 2 gr m/s In case you can't handle relativity, I have included here a Newtonian example. -------------------------- Shalom From: Randy Poe - view profile Date: Mon, Aug 7 2006 2:02 pm Email: "Randy Poe" <poespam-t...@yahoo.com> Groups: sci.physics.relativity vertver...@msn.com wrote: > Randy Poe wrote: > > vertver...@msn.com wrote: > > > Harry wrote: > > > VERGON > > > That was the complication I mentioned. The standard method of > > > calculation gives > > > 0 versus 1.92. Thus simply shifting the viewpoint destroys the > > > conservatrion. That may make sense to you, but it doesn't to me. > > It doesn't make sense to you that an object can be moving > > with respect to one observer but stationary with respect to > > another? > > - Randy > VERGON > Of course that makes sense but what has that to do with my proposition? > I'm saying that if there is a certain amount of momentum in a closed > system -- and if conservation exists -- then the momentum will be > conserved regardless of the observer's viewpoint. But why should the momentum be the SAME regardless of the observer's viewpoint? Consider a single object. Suppose it has velocity v and momentum p in one frame of reference. Let us suppose the value of p is 1.92 in that frame. Momentum conservation says that the momentum will remain 1.92 unless there is a force acting. >From another reference frame, the momentum of this object is 0. Momentum conservation says that the momentum of the object will remain 0 unless there is a force acting. Those are the numbers from above. You claim it makes no sense to you that the momentum can be 1.92 in one frame and 0 in another. You think somehow that "momentum conservation" means an object with p=1.92 can not have p=0 in another frame. "Momentum conservation" doesn't mean the momentum in one frame is the same as the momentum in another frame. - Randy Reply » From: vertver...@msn.com - view profile Date: Tues, Aug 8 2006 9:51 am Email: vertver...@msn.com Groups: sci.physics.relativity Randy, An excellent answer to my post. And I must admit it gave me a hard time for awhile but I think I have the answer. If I understand you correctly, when the observer changes frames he changes the velocity between frames and therefore the momentum. Let's apply that to the case at hand. In Case I the observer is at the point of impact, and the momentum is 0 because the momentum vectors are equal and opposite. In case II the observer is at point B (the right hand mass) and all the velocity is in mass A. The question is, what is that velocity? And, collaterally, what is the momentum of impact? If we determine the momentum of impact, we can calculate the velocity of A. I submit the following: Firstly, we have the physical situation of two masses impacting. Changing the perspective of the observer does not alter the physical happening. The two masses impact the same in either case. The problem is to determine the momentum of the impact in Case II. Now momenta are vectors. And the reason the momentum of Case I is 0 is because one vector is positive and the other negative. By changing the perspective to mass B, both vectors become positive. So we add them together. So now we have the momentum of impact from which we can calculate the velocity - all of it in A. Einstein's calculations (in this case) yields .96c, the momentum of which is too high. My calculations yield .915c, which creates the momentum of impact. Let me elucidate on this MO. Let's switch (temporarily) to Newtonian mechanics. We have two locomotives (same weight and speed) on the same track approaching each other. They collide. The momentum of collision is 0. If each locomotive weighed 75 tons and was going 100 miles per hour the momentum of each would be the same - only one would be positive and the other negative. Combined speed, 200 miles/hour. Now the collision is observed from one of the locomotives - and it is standing still while the other is approaching at 200 miles per hour. The momentum of impact is the same as in the prior case. The difference between the Newtonian case and the relativistic is the velocities in SR are added differently and the momentum is subject to the Lorentz transform --BUT THE IMPACT IS INVARIANT IN BOTH THE NEWTONIAN AND RELATIVISTIC CASES. Another way of looking at it is --- in your case, the change in frames means a change in the velocity of the observer. In my case there is a change in VIEWPOINT of the observer. Both viewpoints are at rest, no motion. :> Suppose that instead of Einstein's formulas, you use Newton's. Instead :> of a relativistic velocity, assume that in the first case your two masses :> are moving toward each other at a velocity of 1 m/s. What do you get for :> the momentum in each of the two cases? : If each locomotive weighed 75 tons and was going 100 miles per hour the : momentum of each would be the same - only one would be positive and : the other negative. Combined speed, 200 miles/hour. In this case, the total momentum is zero. (75)(100) + (75)(-100) : Now the collision is observed from one of the locomotives - and it is : standing still while the other is approaching at 200 miles per hour. : The momentum of impact is the same as in the prior case. In this case, the total momentum is not zero. (75)(200) + (75)(0) I suggest that you go back to a basic book about mechanics and learn something about frames of reference. ----- Richard Schultz schultr@mail.biu.ac.il Department of Chemistry, Bar-Ilan University, Ramat-Gan, Israel Opinions expressed are mine alone, and not those of Bar-Ilan University ----- "You don't even have a clue about which clue you're missing." > :> Suppose that instead of Einstein's formulas, you use Newton's. Instead > :> of a relativistic velocity, assume that in the first case your two masses [quoted text clipped - 22 lines] > ----- > "You don't even have a clue about which clue you're missing." Vergon I suggest you take your own advice PLUS take smart pills so you can understand what you read. Changing the POINT OF VIEW is different than changing frames. Changing frames means to change the velocity due to the change in velocity of the observer IN CHANGING VIEWPOINT THERE IS NO CHANGE IN VELOCTY only the change in DIRECTION. This very clever maneuver was necessary to produce a momentum that would test the correctness of a velocity ADDITION. You can't test the momentum of a resultant velocity if the momentum is ZERO. Oi Vey, some people just don't get it. You're from the department of chemistry --- do you teach it or are you the janitor? Shalom Vert |
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