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i have a very basic question from school days. what does it mean to say an object is moving with uniform speed? it seems to me now that it should be an unit dependent concept. for example if speed is the derivative of distance traveled, i.e. $X'(t)$ , and I decide to measure distance on a new scale $F(X)$, a monotonic function of $X$, but not a linear multiple. Then, speed in that scale at a point t would be $F'(X(t))$. $X'(t)$ , which will not be constant as $F$ can be an arbitrary increasing function.

and if indeed "uniformness of speed" is a unit dependent concept, what does it mean to say, light travels at a constant speed ? also this would mean "uniform acceleration" is also a unit dependent concept. how is then, "acceleration due to gravity" a universal constant ? would it cease to be a constant if I measured acceleration in $\log(m/s^2)$ ?

Dan
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S B
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1 Answers1

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Neat trick.

Fortunately, you're not actually changing anything physical. The graph drawn in those units would not be a straight line but the particle would still move the same distance in any fixed amount of time, i.e. it would still be uniform velocity. So it might move 8 units of distance in a 1 second interval and 16 units in the next 1 second interval but the actual distance moved would be the same in both intervals of time.

What you're doing is more like a change of coordinates than a change of units.

knucklebumpler
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  • can you clarify with an example ? suppose at time points 1, 2 and 3, an object is at 5, 10 and 15 metres from the origin. measured in log this would be log(5), log(10) and log(15). do you agree that distance travelled is log(10)-log(5)=0.69 and log(15)-log(10)=0.40 which are not equal. Supposed "metres" did not exist and you only knew about log-meters. In what sense would you say "it moves equal distances in equal time" ? – S B Jul 01 '11 at 02:18
  • @S B, I would say that if I took a stick and laid it off between log(10) and log(5) and log(15) and log(10), I would be able to lay it down the same amount of times between each of them. This is the basic operative definition of distance. – knucklebumpler Jul 01 '11 at 02:26
  • @knucklebumper: sorry still confused. if the length of the stick is L. for what you say to be true, both these lengths have to be "r * L" for some r (r is the possibly fractional number of times you repeat the stick). But that means, both lengths have to be equal, but they are unequal (one 0.69 the other 0.40). – S B Jul 01 '11 at 02:33
  • @knucklebumper i think i understand a little what you mean by "not changing anything physical". so if I measure the distance that it travelled in the first time interval using a stick. then the same stick can be laid in the path traversed in the next time interval. so "assuming" that it had a uniform speed during the first interval, the second interval has the same speed. of course one can not use "distance traveled" by that same object in 1 second (the stick) as the unit for distance, so we need a common reference...continued. – S B Jul 01 '11 at 03:37
  • @knucklebumber continued...so we can define the distance traveled by light in one second as the unit of distance. then basically the object has a constant speed in light-year per second. change of unit like log is not meaningful. the length of the stick (used as standard of measure) in light-seconds would remain the same. does this make sense ? – S B Jul 01 '11 at 03:38
  • ok...i think i understand better now. indeed in the log-scale the object is moving at non uniform speed. all units of distance we have meter, feet etc are multiples of each other. if i started measuring in log, those would be the same distances as measured by a person located in a transformed co-ordinate system, which is a non-inertial reference frame with respect to me. so the speed is not constant in that frame. – S B Jul 01 '11 at 04:51
  • "do you agree that distance travelled is log(10)-log(5)=0.69 and log(15)-log(10)=0.40" No. Or rather, I can't speak for knucklebumper, but I strongly suspect that what you are measuring there is not actually distance; however, to tell you why it is not distance would take some thought. If you're curious you can bring this up in the chat room, since it is an interesting thing to think about. – David Z Jul 01 '11 at 05:02
  • @S B, by "you don't change anything physical" I mean that if you forget all about units and lay the stick end to end between log(10) and log(5) and log(15) and log(10) then you will find that the stick goes the same amount each time. In saying that both lengths have to be rL in your log units, you are treating the log units as regular units. The formula rL works because regular units are translation invariant. Your log units are not. If I was to measure the length of the stick at different points, I would get a different answers. – knucklebumpler Jul 01 '11 at 05:06
  • @David Z, Does my last comment make it clear why what's being measured is not distance? – knucklebumpler Jul 01 '11 at 05:10
  • @knucklebumpler: yeah, pretty much. I was thinking of translation invariance but couldn't put my finger on the specific concept. – David Z Jul 01 '11 at 05:13
  • @knucklebumper yes i think i understand...basically distance is always defined as fraction (number of unit lengths), which is being violated. – S B Jul 01 '11 at 05:32
  • @S B, pretty much. The important thing is that one unit of length here is the same thing as one unit of length there, i.e. it's translation invariant. If I tell you that you need to walk .01 log meters, that tells you nothing unless you also know where to start. .01 log meters starting from the coordinate .01 log meters is no big deal. .01 log-meters starting from the coordinate 10000000 log meters is a very big deal indeed! – knucklebumpler Jul 01 '11 at 20:21
  • @knuclebumpler i think now that in some sense its more a case of the co-ordinate transform (from X to log-X) not being allowed. the distance metric i am using is still (|X-Y|) the difference, which is translation invariant. so its a valid measure of speed in that other reference frame (co-ordinate system). – S B Jul 01 '11 at 20:32