Conservation of Kinetic Energy
Date: 6/8/96 at 3:56:22 From: PB Subject: The conservation of kinetic energy under a translational transformation Dear Sir, I am a mechanical engineer from Sweden, and I got stuck with a theoretical questions in physics. The formula mv^2/2 + mgh = constant is the energy conservation formula in physics, and is said to be invariant for all observers who move with constant velocity relative to one another. Suppose that I am an observer who moves with the body: then v = 0. So how is this formula conserved? John Mendelsson
Date: 6/8/96 at 7:5:23 From: Doctor Anthony Subject: Re: The conservation of kinetic energy under a translational transformation The energy conservation formula refers to a frame of reference. If the frame of reference is the body itself, then v=0, h=0 and this simply makes the constant equal to 0. Total energy is still conserved. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: 6/8/96 at 9:55:26 From: PB Subject: Re: The conservation of kinetic energy under a translational transformation I think the formulation of my question does not reflect my thoughts. I give another example: the work energy theorem: I Fds = mv^2/2 where "I" is the integral sign (I cannot produce it with my software). Suppose that another observer, B, who is moving 5 m/s faster than I am, watches the work done and measures m(v+5)^2/2-mv^2/2, which is not equal to mv^2/2, whereas I Fds is still the same.... or isn't it ?
Date: 6/13/96 at 21:36:48 From: Doctor Tom Subject: Re: The conservation of kinetic energy under a translational transformation Energy is NOT conserved under a change of reference frame; your example is a perfect demonstration of that. When you change reference frames, some things are conserved, and some things are not. One of the most interesting things in physics studies is which things are conserved, and which are not. For example, in Newtonian physics, at a fixed time, two observers of two particles will observe the same distance between them, but in Einstein's relativistic physics, they will not. However, in Einstein's physics, they will observe the same "interval" between events, where the "interval" includes a term involving the observed time of the events. Two Newtonian observers would notice that light travels at different speeds in different reference frames. Einsteinian observers will see no difference. And the real world appears as the Einsteinian observers would predict. -Doctor Tom, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: 6/22/96 at 10:2:9 From: peter Subject: Re: The Conservation of Kinetic Energy under a Translational Transformation In a previous response, you explained that the ds (velocity) used in determining work done depends on the reference frame; my mistake was that I thought that ds is constant, which it is not in this case, since here the particle accelerates due to a force F, i.e. has not const velocity. But since you are a mathematician, you ought to be able to prove it purely mathematically, not just make it plausible. Actually I feel that all Newtonian mechanics could be made pure math, i.e. formulated as axioms, although I do not know how this is done, nor have I seen any book introducing these ideas...
Date: 6/24/96 at 14:13:34 From: Doctor Tom Subject: Re: The Conservation of Kinetic Energy under a Translational Transformation My example can be made mathematically rigorous - I just find it much easier to understand using physical intuition. You can certainly prove that the laws of physics will appear the same in different Galilean frames from first principles. I've never found it useful to work physics problems from "axioms", although I suppose it might be possible to write down those axioms. I have always had a wonderful "mathematical intuition", but my "physical intuition" has never been as good, and I've had to struggle to get what little I have. I know lots of physicists who are exactly the opposite. I suspect the world's best physicists (and mathematicians) are those with strong intuitions in both areas. -Doctor Tom, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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