Capiert Posted June 2, 2016 Posted June 2, 2016 I don't know much about vectors. (Maybe you can help?) Perhaps it's what I don't know that counts? That little I know (& whether it's correct?) is: vectors have magnitude (amount) & direction. Negative (direction) is rotated 180 degrees (angle) in the opposite direction. Vectors are subdivided into 3 major axii (xyz), which are 90 degrees to each other (so Pythagorus's 90 degree triangle rule can be used, easily). Vectors can be added, graphically, maintaining their directions, not necessarily their positions. e.g. chained together, arrow tip point on end of the next vector, important is the angles (orientation) stay the same. Vectors can be multiplied, either: dot product, "simple multiplication" with a factor, called scalar, because it can globally change all (or any, single) vectors' amount, but not the direction, so. Scalars only change the magnitude (amount), not the angle (orientation). Scalar multiples change e.g. the size of a (vector triangle) shape, but not its shape or angles. Or Vector cross products use 2 coordinates input, to produce a 3rd coordinate output. At least that's what rumor(s I've heard) & intuition tells me. Vectors can be subtracted. That means the negative vector is rotated (angle) 180 degrees, & then stringed (=chained) together in the usual method (added or) (arrow's point tip onto the tail end of the previous vector (or arrow)). So knowing that. Is mass m a vector? No it's a scalar (having amount, no direction (yet)). I haven't seen negative mass (not to mention mass as only a number, but that's beside the point. We deal only with positive mass, mostly. So I'll assume it's a scalar. We can multiply mass with its speed v=v1-v0 (difference). (v1=final_speed. We can assume the initial_speed v0=0 is zero here for simplicity. But it could be any value we choose, if we leave the formula so.) Is the (mass's) speed v a vector? Well yes, wrt zero initial_speed, it has magnitude (amount) & direction forth (positive) or backwards (negative, wrt positive direction). Is their (m & v) product (momentum) a vector? Yes because the scalar mass (m) is multiplied by the vector speed (difference) v to give a different vector (momentum). The speed is scaled (up or down) by the mass. Let's look at the average speed va=(v0+v1)/2=v2/2. That's 2 vectors (v0 & v1) added together, result is a vector v2=v0+v1. That's simple vector addition. Ok divide (vector v2) by 2. Ops! We didn't mention vector division, please excuse me. Then let's multiply by the scalar 0.5, instead, that's a simple number, enough. Scalar (0.5) multiplied by a vector (v2) gives another vector, we call va, average_speed. Even if average_speed was not a vector, we could call it a scalar then, but it's not. va is really a vector. Ok. Scalar multiplied by a vector is a vector. What about a vector multiplied by a vector. Well, if they are both in the same direction, no problem, because we don't need vectors' direction for that. Are v & va in the same direction? Yes if initial_speed v0=0 is zero. What did we assume at first for simplicity? Right, let v0=0. Does a vector multiplied by a vector (in the same direction) remain a vector in the same direction (e.g. the magnitude of 1 of the vectors (e.g. va) is used as a scalar quantity) for the other vector e.g. the speed v. The problem would have been a lot easier, if va were (simply) only a scalar.) Anyway, I will assume, a vector multiplied by a (parallel*) vector remains a vector. 2 (parallel) vectors & a scalar, all 3 multiplied together gives a(other) vector. KE=m*v*va. KE is a vector. Sorry! I can only conclude, KE is a vector. * (If non parallel, it (the vectors' product, as new vector) may not remain pointed in the same direction, but it still remains "a vector", perhaps with a different: (orientation) direction & magnitude (amount).)
imatfaal Posted June 2, 2016 Posted June 2, 2016 So if a car going down the motorway from North to South has 700kJ of kinetic energy does a car on the opposite carriage way have -700kJ of kinetic energy? As both the cars started from being parked - with no kinetic energy the car travelling south must have somehow used its engine and petrol in its fuel tank to gain 700kJ of KE; but hang on, the car travelling North has 700kJ less energy than it started with - has the fuel tank refilled itself?
ajb Posted June 2, 2016 Posted June 2, 2016 You need to think what multiplication of vector means... you need to brush up on the basic notions in linear algebra -- I have written a book on this! (almost I have some lecture notes that are not polished)
Capiert Posted June 2, 2016 Author Posted June 2, 2016 (edited) Let us first get the orientation correct (standardized) & say northward is positive, & southward is negative.I will quote your text with that correction made. So if a car going up the motorway from South to North has 700kJ of kinetic energy does a car on the opposite carriage way (going down, North to South) have -700kJ of kinetic energy?Yes, (it is the same way as) how would you describe the momentum (while ignoring KE for the answer).The momentum would have similar polarities. We determine that (direction) based on our earth reference seen to us as stationary. For all practical purposes. We have created the math construct, not nature. It's our job to make some sense out of what we do, not natures. As both the cars started from being parked - with no kinetic energy the car travelling north must have somehow used its engine and petrol in its fuel tank to gain 700kJ of KE; but hang on,(to what? pun intended)the car travelling South has -700kJ (that's) less energy than it started with - has the fuel tank refilled itself?No, obviously not. You don't seem to be adhering to the rules of vectors. If KE is a vector, then you must also use those vector laws to explain its behaviour. KE is bidirectional. Dealing with it as though it is (only) a scalar is not acceptable. That won't work. (Chemical energy, is expressed as Gibb's Free energy, in "both polarities", for determining if a chemical reaction will happen spontaineously. Negative difference is spontaineous. E.g. burning carbon with oxygen is only spontaineous beginning at a specific hot temperature. The chemists seemed to have taken this (bipolar energy) concept further than you Physicists. It's not new, just for you maybe?) Look at it so, (I know it's not a good example because a lot more happens than I will simply say, but let me simply say it). Please change the energy of both cars so they are going very fast (MJ?, GJ?) each ~100km/h & opposite directions. If each car started near the (geo) pole(s), & a (thick) wall was at the equater, then each car would receive a dent when colliding against the wall. If there was no wall, & both cars collided together, then each car's dent would be, let's say double so deep. Their energies, would have added, similar to the way it would have added with momentum. There is really not much of a difference between the 2 concepts energy & momentum, because they both use the same 2 units, (m & v) but with different amounts, thus they are scaled differently, that's all. KE is simply scaled momentum. The scaling "factor" is va. KE=mom*va Mom is the vector. KE must obviously be a vector also. But va is also a vector, instead of scalar. I can't say it very well, but maybe you can get what I mean? Please restate it in your own words, I'm sure you'll do a better job at it, than me. Is that ok? You need to think what multiplication of vector means... you need to brush up on the basic notions in linear algebra -- I have written a book on this! (almost I have some lecture notes that are not polished)Yes. I said I do not know much about vectors. I need a few pointers. You may correct my assumptions. (There is not really much there.) I have only tried to make use of that little I know. For I know it is better to do much with less, than little with much. Edited June 2, 2016 by Capiert
swansont Posted June 2, 2016 Posted June 2, 2016 Vectors can be multiplied, either: dot product, "simple multiplication" with a factor, called scalar, because it can globally change all (or any, single) vectors' amount, but not the direction, so. Scalars only change the magnitude (amount), not the angle (orientation). Scalar multiples change e.g. the size of a (vector triangle) shape, but not its shape or angles. Or Vector cross products use 2 coordinates input, to produce a 3rd coordinate output. At least that's what rumor(s I've heard) & intuition tells me. ! Moderator Note There's not much point to this if you are going to make up the math as you go, and then make conclusions about the resulting physics. You have two options here. Either you can study up on the basics and ask questions as needed in the mainstream sections, or you can present something here in speculations which has the minimum rigor that we ask for, as described in the guidelines. What you can't do is continue to start threads that have no basis in any established field of study, and no rigor.
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