Speed of Propagation of Forces in Newtonian Mechanics Abstract: We apply established knowledge of Newtonian mechanics regarding propagation of forces in bodies. We construct special situation to test whether same force can act twice at same point during specific time interval due to delay in propagation of decompression waves carrying information regarding release of force. If "extra" force of same magnitude is generated due to propagation delay, then it violates law of conservation of energy. Force is separated from system, which is in state of static equilibrium, and speed of compression waves due to mechanical deformation of system is measured from point of separation of force to supporting point of system. We find that although decompression waves propagate through system, information regarding separation of force from system can not propagate through decompression waves with speed of sound in system. Introduction: In Newtonian mechanics, any body, which has mass, can not be perfectly rigid. Every body has elastic properties to some extent. (1) Hence in Newtonian mechanics, when we apply force, at one point of body, it causes deformation of body and this mechanical deformation and information regarding applied force is communicated to other points of body through compression waves which propagate with speed equal to speed of sound in that body. (2) In Newtonian mechanics, it is also documented that when we release applied force, again it causes mechanical deformation of body and this information regarding mechanical deformation i.e. release of force is communicated to other points of body through decompression waves which propagate with speed of sound in that body. We construct special situation to test the validity of second case. We set out to prove that information regarding separation or release of applied force to body can not propagate through decompression waves. We do not attempt to prove existence or non-existence of decompression or compression waves or it's speed or it's way of propagation. Description of Special case. C------------------B-----------------------E| |S----------------- |------------------------S | |S------------------| -----------------------S | |S------------------|------------------------S| |S------------------|control pointA---------S| |S-------------Hangig Body ---------------S| D-----------------G------------------------F --plate GM------>||<---plate GM----------- ====Table====M================ ============================= As shown in above figure, square shaped system of lines DCBEFG is placed on table in such a way that it is supported by small plate GM at mid-point M of table. Vertical line AB is attached to point B of system. At point A, a body is attached by very small string and control mechanism at point A can lift or release the body. The body attached to point A is just hanging right over point G at negligible height from point G. No part of system is touching the table except at point G, which is supported by plate GM. Lines AB, BC, CD, DG, GF, FE, BE, plate GM, body, table etc can be made of any material, any shape which is convenient for understanding. The whole system is in state of static equilibrium. Let mass of system, excluding body, be m and mass of body be m'. Let length of lines ABCDG and ABEFG be L each. Let speed of compression wave in material of system be c. As mass of system and body is m+m', point M on table is supporting weight (m+m')g. At t = 0, we begin to release the body very slowly using control mechanism at point A and at t = t' the body is completely supported by plate GM on table at point M and is in state of static equilibrium. That means there is no contact between body and control mechanism at point A whatsoever. We call this time interval t' as release time. As the body is still under gravitational field, gravity is acting on this body and hence weight m'g is acting on point G and through plate GM, at point M of table at t = t'. As the body touches point G and hence will be supported by table through plate GM. This will cause release in strain at point A and this information will begin to be transmitted through decompression waves starting from point A, to B, C, D and to point G on left side of system. On right side of system, decompression will propagate from point A to B, E, F and to point G. As length of segment ABCDG is L and length of segment ABEFG is also L, decompression waves will reach point G after time interval t where t = L/c. Length L of each segment is such that total transmission time of decompression waves t is far greater than total release time of body t'. Hence t >> t'. Let difference between total transmission time and release time be T = t-t'. We call this difference T as time delay. During this time delay T, the point G and M has absolutely no information about events happening at point A of system. Even though the weight m'g of body is acting on point M during delay time T, point M has absolutely no way to "know" that the body was earlier hanging at point A of system. This point M will only come to know about it after decompression waves reaches point M after time interval t. Until that moment this body will be treated as second body which is brought from outside the system. So we find that during this time delay T, weight = (m+m')g + m'g = (m+2m')g is acting on point M of table even though mass of system and body is just (m+m'). Suppose the weight (m+m')g is not enough to break down table at point M. But weight (m+2m')g is enough to break the table. Hence the table will break down during time delay T. This is in sharp contrast to law of conservation of energy because we find that extra force m'g is generated due to propagation delay of compression waves. In this way, as we use this extra force to break the table, we can use it to do any "extra" work we want. Example: Suppose, instead of body, it is you hanging at point A of system. Let your mass be 100 Kg and that of system be 100 Kg. You are hanging right over point G in such a way that distance between your feet and point G is just fraction of millimeter. So weight acting on point M of table is (100 +100)g N. (1) Now at t = 0, you leave point A and jump outside table on ground. As your mass is 100 Kg, force equal to 100g N is separated from system. This will cause release of strain and mechanical deformation at point A. of system. In Newtonian mechanics, this information will propagate across transmission lines of system. Let it take 10 seconds to reach these waves to point M of table. So during these 10 seconds, gravitational force (100+100)g N is still acting and point M of table is still exerting upward force of same magnitude because point M has no information regarding separation of your mass and gravitational force. (2) But you change your intentions and at t = 0, instead of jumping outside table, you jump on point G of system, which is just fraction of millimeter from your feet. Even if you release point A of system and standing on point G, gravity is still acting on you. As your mass is 100 Kg, gravitational force equal to 100g N is acting on point G and hence point M. But the moment you leave point A of system, information regarding separation of your force begin to propagate on its long journey of 10 seconds. Central idea is, how the point M of table will know during these 10 seconds that your body was previously hanging at point A and it should not experience your weight twice because it is already experiencing it? The only way to know it in Newtonian mechanics is decompression waves. These waves are taking long path to propagate and reach to point M. Before these waves reach point M, information regarding weight of your body is taking very short path through your feet directly to point M of table. So during these 10 seconds, weight 200g N is acting on point M of table just like it was acting at t = 0. But during these 10 seconds, your weight 100g N is also acting on point M of table. So during these 10 seconds, total weight acting on point M of table is 200g + 100g = 300g N. Suppose table can break down only after applying weight of 300g N at point M. Then we see no reason why the table will not break down during these 10 seconds. In this way we can break any body which is supposed break at 300g N force by using just 200 Kg mass. This is in sharp contrast of law of conservation of energy. Analysis: First Case: Suppose initially, the point M is supporting only weight of system mg. When we first attach the body, the compression waves beginning from point A will take time to reach to point M of table. So during this propagation delay, the point M is balancing only weight mg and not (m+m')g N. But in such situation, even if we use that weight m'g which is propagating across system and weight mg at point m of table to do work, it does not violate law of conservation in any way. There is no extra weight or force generated in this first case. Second Case: In our special case, we get extra force m'g generated due to propagation delay. And we get enough time to use it. Point M has no way to know that extra force is generated and it should not break. So first case does not violate law of conservation of energy, but second case does violate law of conservation of energy. I agree that when we apply force or pressure to surface of elastic body, be it gas, liquid or solid, the force or pressure is transmitted from one molecule to another and not instantaneously. But opposite is NOT true. When the force or pressure is applied and both bodies are in state of static equilibrium, then if one of the source of force moves in opposite direction of application of force, then this information is communicated to other force "instantaneously" and NOT from one molecules to another. This is specifically what I am saying. If I transmitt force F = ma from point A to point B and even if takes time = t to reach to point B, there is no harm. But suppose the force is reached to point B and exerting on point B. In this case point B is also exerting equal and opposite force F= -ma to balance it. A/\/\/\/\/\/\/\/\/\/\/\/\->B Now If source of force at point A moves away and if it takes same time = t to reach this information to point B, then during this time interval t, source of force at A can take short route to reach point B which takes time t' where t' << t. A------------------------->B So during this propagation delay of force, this source of force at A can exert twice force on point B even if it do not have capacity to do that. Using this extra generated force, it can do "free" work and this free work can lead to another free work.... How can this extra generated force be used to create extra energy? Do you think that the table will break down? If yes then this means that table experienced 300g N force in downward direction and it could not exert upward force of same magnitude to balance it. So it just gave in. We change the situation. Now consider that instead of table, the point M of system is supported by a small rope which is passed on a pulley. The point M of system and the whole system is at height h from ground. Other end of rope is tied to stones a mass of 299 Kg. Initially the mass of system and body is 200 kg. So gravity is exerting force 200g N on system+body and force 299g N on stones. So 200g N force will never be able to lift 299g N weight. But using the idea I have given in that file, you can generate extra force of 100g N at point M of which is tied to rope. So the point M of rope is experiencing 300g N force in downward direction. This extra force will be communicated through short rope over pulley to stones of weight 299g N. And this 300g N force can lift 299g N weight in upward direction during propagation delay. Suppose during this propagation delay, system+body has lifted the stones to height h and system+ body has come down to ground and resting on it. Now we can remove stones of mass 98 Kg at height h and we can use just 201g N force to lift the system+body of weight 200g N at it's previous height h. Again we follow same technic, we generate extar force and again we lift stones weighing 299g N to height h. We remove stones of mass 98 Kg and.......we jus go on doing that. What extra energy I am spending to lift these stones of mass 98+98+98... Kg to lift to height h? Perhaps you will say that every time I will have to lift body to attach it to control point A of system. But I can esily afford to lift body of mass 100 Kg to lift to negligible height d << h from point G. I will have to spend energy 100*g*d but I am getting energy 98*g*h. As h >> d, hence 98*g*h >> 100*g*d In any case I get extra energy. I will chose propagation delay in such a way that the force acting on point m of system+body remains 300g N until the stones reach to desired height h. Conclusion: We do not conclude that propagation wave does not propagate across system.We merely conclude that information regarding separation or release of applied force, be it gravitational force or mechanical force, does not propagate through decompression waves with speed of sound in system. Hence we prove that second case of Newtonian mechanics, as stated out in introduction part above, does not hold true. -Abhi. Abhijit B Patil, India. Email: sciphysics@yahoo.co.in Homepage: http://www.geocities.com/sciphysics
Not feeling well to reply myself. But anyway, here we go... In Newtonian mechanics, any body, which has mass, can not be perfectly rigid. Every body has elastic properties to some extent. (1) Hence in Newtonian mechanics, when we apply force, at one point of body, it causes deformation of body and this mechanical deformation and information regarding applied force is communicated to other points of body through compression waves which propagate with speed equal to speed of sound in that body. (2) In Newtonian mechanics, it is also documented that when we release applied force, again it causes mechanical deformation of body and this information regarding mechanical deformation i.e. release of force is communicated to other points of body through decompression waves which propagate with speed of sound in that body. This experiment is proposed to test the validity of second case. This experiment is initially intended to prove that information regarding separation or release of applied force to body can not propagate through decompression waves with speed of sound. We use very long transmission line made of any material in which information regarding release of applied force is supposed to travel in accordance with established knowledge of physics. The force is applied to starting point A of transmission line and "end point B of transmission line is exactly at same level of starting point A" (where force is applied) of transmission line. We apply established knowledge of physics to calculate propagation delay time taken for information regarding release of applied force to travel from starting point A to end point B of transmission line. If during this propagation delay time, end point B of transmission line does not move in space, then established knowledge of physics is correct. But if end point B of transmission line moves in space during this propagation delay time, then we show that information regarding separation or release of applied force to body does not propagate in accordance with established knowledge of physics. We use high-speed video camera focused on starting point A and end point B of transmission line and electronic clock in the background. We check every frame of video film in slow motion to see whether end point B of transmission line is displaced in space during propagation delay time or not. If end point B of transmission line moves in space during this propagation delay time, then we go on proving validity of following principle. "When the system is in state of static equilibrium and if we release applied force to a body, this information regarding release of applied force is transmitted to every point of body instantaneously". -Abhi.
It is well established that all bodies fall with same gravitational acceleration g regardless of mass of body. But when we remove applied upward force to body which is in state of static equilibrium, so that it falls with free fall acceleration, it is thought that this information regarding removal of applied upward force is communicated to every point of that body through compression or decompression waves with speed of sound in that body. We set out to prove that information regarding separation or release of applied force to body has no relation to propagation of decompression or compression waves in body falling with free fall acceleration. Deriving equation of speed of sound in falling body: We take a rod of length L, area A and mass m. We place it on ground in vertical position. G force F = mg is acting on area A of rod. Due to this force, the rod undergoes deformation and its length is decreased by L' which is given as L' = (F*L / A*E) Where F = gravitational force applied, L = length of rod, A = area of rod on which gravitational force is applied, E = Young's modulus. Now when we leave support of this rod so that it falls freely, this information regarding release of support i.e. release of upward force propagate from lower end to upper end molecule by molecule with speed of sound c. Speed of sound(c) in bulk media is given as c = sqrt[bulk modulus / density] = sqrt( B/d) So it takes time t = L/c to reach to upper end. But during this time interval t, lower end is falling with free fall acceleration g. So during time interval t, lower end covers distance h in space which is given as h = 0.5gt^2 As lower end has fallen by height h until information reaches to upper end, that means the rod has regained its original length L. So the distance h through which lower end has fallen must be equal to deformation L'. Hence L' = h (FL / AE) = 0.5gt^2 substituting t = L/c, we get (FL / AE) = 0.5g(L/c)^2 = 0.5gL^2 / c^2 Hence c^2 = (0.5gL^2*AE) / FL As F = mg c^2 = 0.5gL^2AE / mgL c^2 = 0.5(LA)E / m Here L*A = length od rod * area of rod = volume of rod = V Hence c^2 = 0.5V*E /m c^2 = 0.5 E / (m/V) As m/V = density of medium =d, we get c^2 = 0.5 E /d Hence speed of sound in rod is, c = sqrt(0.5)*sqrt (E/d) c = 0.707* sqrt(E/d) = 0.707*sqrt[bulk modulus / density] We must get standard equation of speed of sound in such situation if deformation L' is equal to height h through which rod falls in time interval t = L/c. But we do not get standard equation of sound. Compressibility or tension in body has no relation to propagation of information regarding separation of applied force: Now consider this rod is placed vertically on ground. Let its original length in absence of force be L. As we place this rod vertically, it undergoes deformation and its length is decreased by L'. Its lower end is compressed maximun and upper end minimum. Now we take two magnets, M1 and M2. These two magnets are fixed to upper and lower end of this vertical rod. South pole of magnet M1 is at lower end and south pole of other magnet M2 is at upper end. So south-south poles are repelling each other. Due to this upper magnet M2 fixed to upper end of rod will be pushed upward exerting upward force on upper end of rod. Due to this, rod will undergo tension. Let this upward force acting on upper end of rod be exactly equal to downward G force mg. This will cancel the deformation(compressibility)caused by G force and rod regains its original length L. Now molecules at lower end are NOT compressed. If we remove support from lower end of rod, how this information will propagate to upper end of rod? -Abhi.
We do not conclude that propagation wave does not propagate across system. An object with a newly applied/changing force will experience compression/tension waves. The object under a newly applied force does not need to instantly apply an equal and opposite force. Looking at the object as a whole, the center of mass will start to move. Then it will be compensated for. There is no need for the information to instantly reach the other end of the object so it can counter-act. The table is not supplying some mystical force but IS bending in. Then it will unbend slightly from a 'resoration force'. We merely conclude that information regarding separation or release of applied force, be it gravitational force or mechanical force, does not propagate through decompression waves with speed of sound in system. Well, in almost any case we deal with on earth, gravitational force doesn't require any compression waves because it is acting on every atom. It is the impact that causes waves to be created. As for mechanical force (such as the above impact), waves are generated. Multiple wave velocities can exist after an impact. You can have longitudinal, transverse, surface waves (Rayleigh), and shock waves. I'm pretty sure that the first 3 travel at the speed of sound, although it may be measured as lower because of the path they take. Shock waves however, travel faster then sound. I'll try and dig up some deformation wave photos and speed measurements if you want. Hence we prove that second case of Newtonian mechanics, as stated out in introduction part above, does not hold true. This has been heavily tested, and is an integral part of how they figure ut a material's properties. Regardless, I fail to see the logic of where you proved that it is wrong.
Originally posted by Abhi ...when we remove applied upward force... it is thought that this information regarding removal of applied upward force is communicated to every point of that body through compression or decompression waves with speed of sound in that body. ok We set out to prove that information regarding separation or release of applied force to body has no relation to propagation of decompression or compression waves in body falling with free fall acceleration. good luck We take a rod of length L, area A and mass m. We place it on ground in vertical position. G force F = mg is acting on area A of rod. Due to this force, the rod undergoes deformation and its length is decreased by L' which is given as L' = (F*L / A*E) This is an estimate, however, usually it is close enough. The real problem however is that the force in this case is not applied at the end of the rod. It is applied over the entire length. Now when we leave support of this rod so that it falls freely this information... propagates from lower end to upper end molecule by molecule with speed of sound c. c = sqrt[bulk modulus / density] = sqrt( B/d) So it takes time t = L/c to reach to upper end. The speed of sound in a stressed material need not be constant. But during this time interval t, lower end is falling with free fall acceleration g. So during time interval t, lower end covers distance... h = 0.5gt^2 As lower end has fallen by height h until information reaches to upper end, that means the rod has regained its original length L. So the distance h through which lower end has fallen must be equal to deformation L'. The relief of the compression is not always done in a single wave. The top will start moving as soon as the wave front hits it. There is no need for the length to go back to L before movement begins. Compressibility or tension in body has no relation to propagation of information regarding separation of applied force How do you suggest it is propagated?
One Question Persol, I don't see it changing the overall picture you painted but I do question your following reply: As I understood him he is talking about the compression load. While gravity is applied over the entire length, the compression is compounded toward the lower end of the rod. He is saying the release of the compression must propagate up to the top.
That equation is derived to be used with a force sitting on top of the rod, and assumes the same dL/dX over the entire length. This is not the case, as the bottom does compress more. With gravity, you don't have the same propogation waves as with impact/pressure... because the force is applied on every single atom at the exact same time (or at the very least at c).
