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Q & A: Mixing classical and relativistic equations

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Most recent answer: 07/08/2016
Hello, I have recently found a strange deduction, it's just come so sudden that I don't even know where it comes from,I understand these following three equations well as I did high school physics,e=w=f*s e=m*c^2 f=m*a=m*s/t^2 because they are equal, can I therefore say m*c^2=f*s=m*a*s=m*s^2/t^2 hence c^2=(s/t)^2 therefore, s=c*t, since c is a constant s≣t When I come up with this conclusion, I simply cannot tell why and how as it appears in the end to me that space is equivalent to time, and therefore c is just like an exchange rate of scale of measurement like that from kilometre to mileI understand that in some physics, time and space are together considered as one, (x0, x1, x2, x3) to replace (t, x, y, z) but this proportion I derived above just seems ridiculous, because should it be correct, speed is therefore unit-less, like a direction as up and down cancels out and acceleration would just be -1 power of time or space aloneIf later make it into the definition of gravitation,it would�ve become, F1=F2=G*(m1*m2)/s^2m1*a=G*(m1*m2)/s^2 1/(s*c^2)=G*m2/s^2c^-2=G*m2/s and after e=m*c^2 e=s/G, since G is a constant e≣s I'm so confused, because after I have double checked it with GR, as I understand GR formulae are very complicated, I don't see GR formulae have ever exceed classical formulae in terms of quality, all those factors & constants and tensor are simply corrections quantitatively�I hope you could answer the questions for me!Thanks!
- Jack Samuels (age 19)
New York City, NY, USA

One of your conclusions is very much an accepted fact: c is the unit conversion factor between standard distance and time units. In more fundamental units we use c=1 and you can treat all velocities as dimensionless quantities. The velocities are often called β in this notation. 

A lot of your reasoning is all scrambled up, however. The equation F=ma, for example, is only an approximation that works for things moving slowly in the chosen reference system. Combining that approximate formula, incorrect in the general case, with ones from special relativity can lead to nonsense.

I'm puzzled by your remarks comparing the "quality" of GR with classical formulae. Classical Newtonian gravity is simply incorrect, although it's a useful approximation. GR is accurately confirmed by a large array of precise measurements. One of the most dramatic recent ones was the observation of gravity waves from merging black holes.

Mike W.


(published on 07/08/2016)

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