I am making fundamental research and I need someone to verify the transformations of SI units system given below. It should be easy for everyone with physical background. I am not sure however that I do not have mistake. Under SI transformation I mean SI units and the universal constants will change when the 3 basic units in SI - m,sec and kg are converted acoording to following : m' = a * m - the new unit for distance sec' = b * sec - the new unit of time kg' = d * kg - the new unit for mass where a,b and d are constants I am interested first only from the first transformation m' = m*a, I had found the following units and fundamental constants convertions: <PRE> J - joul N - newton F - force A - amper C - coulumb I - current mu - permeability of vacuum (permeability) eps - permittivity of vacuum (permittivity) c - speed of light in vacuum h - Planck constant e - elementary charge k - Boltzmann constant 1. Change of unit for distance - m' = a * m -------------- units -------------------------- J' = kg m^2 / sec^2 = a^2 J N' = aN F' = F/a A' = aA C' = A'.s' = aC I' = I / a mu' = a*mu eps' = a*eps --------------- constants ------------------------ h' = h / a^2 c' = c / a e' = e / a k' = k / a ^2 </PRE> Are my calculations correct?
1. Change of unit for distance - m' = a * m -------------- units -------------------------- J' = kg m^2 / sec^2 = a^2 J Yes. N' = aN Yes. F' = F/a No. A' = aA I believe an ampere is a fundamental unit, so No. C' = A'.s' = aC No. I' = I / a The unit for a current is an ampere, this is just like your Newton/Force thing. mu' = a*mu has fundamental units (kg m) / s<sup>2</sup> A<sup>2</sup>) So Yes. eps' = a*eps has fundamental units (A<sup>2</sup> s<sup>4</sup>) / (kg m<sup>3</sup>) So No. --------------- constants ------------------------ h' = h / a^2 has units J*s so, it should be multiplied by a<sup>2</sup> not divided. c' = c / a again, backwards. e' = e / a you mean a coulumb? That has fundamental units A*s k' = k / a ^2 has the units J/K so No.
Thank you for reply. I know what is farad and that current is measured in amperes, you misunderstood me. If we change unit for distance meter, and make it for example 2 meters how will change all units, variables and constants in physics? c' = c / a - this would be the new constant for the speed of light in vacuum, because if for SI the light passes X meters per second, in SI' the light will passes twice as little from the new meters because the new meter is twice as big as the old meter, then c' = c/2 [new meters] / second think again?!!
It seems like you are confusing what "a" is. You defined "a" as m'/m in your OP but seem to define "a" as m/m' here.
No, a is 2 in the example above and in the first post too: [ new meter ] = a * [old meter] [ new velocity ] = [ old velocity ] / a [ new meters] / [second] c' = c / a physical quontities changes reverse from units.
[m'] = a * [m] a = [m'/m] c' [m'/s] = c [m/s] * a [m'/m] c' [m'/s] = c * a [m'/s] You have it backwards.
Aer, if I change unit for mass from 1 kg to 2 kg [kg' = 2kg] how much will weight a 5 kg tube with water in kg'?
1 [kg'] = 2 [kg] 1/2 [kg'/kg] = 1 5 [kg] = 5 [kg] * 1/2 [kg'/kg] 5 [kg] = 2.5 [kg'] What is your point? (I tried to show every step explicitly - do you disagree with any of the above)
Did you see that new mass = old mass / a, Please Register or Log in to view the hidden image! physical properties changes reversely from their units, if m' = am, new distance = old distance / a, if Amper' = a amper, I' = I/a and so on. Guys, the topic here is simple and ideal for exercising your ground physics. It will help further in more complex topics. I am still waiting someone to validate or prove wrong the posted above relations.
Nope, a = 1/2. Like I said - you are confusing the definition of a. If you define m' = a * m, then a = m' / m = 1 m' / 2 m = 1/2 [m'/m]. Think about it. Holy crap, you don't even know how to do a simple unit conversion even after I showed you how to do it properly. How do you ever expect to do go on to more complex topics? I already did it for you. As I suspected - you were just looking for someone to tell you "looks good to me" and move on. However, half the things you did are retarded and your continued insistence that you were right just goes to show you didn't want anyone to point out any possible errors. Why are you here?
Please Register or Log in to view the hidden image!Please Register or Log in to view the hidden image! lets go in this again. physical properties changes reversely from their units, You had found out it by yourself. In the above example a = 2, m' = ma, 5 kg water = 5 kg' water / a Unlike you I am thinking about what I am posting.
You are never clear on your defintion of "a", is "a" the unit conversion (i.e. m'/m) or is "a" the scale factor? From the first set of conversion you did, it was apparent that you meant "a" to be the former. But with the last set of unit conversion, it appears you are defining it as the latter. You need to be consistent, either pick one or the other. All of my answers assumed a has the units [m'/m]. The definition you are using above assumes "a" is a scaling factor, that is you are not really attributing a unit conversion value to it, because your unit conversion value would in that case be 1/a [m'/m]. See?
I don't know how to put it more clear - a = m'/m , suppose that the ISO change the unit for distance and replace all the meters in the world with a*meters = meters' units. Then distance between two santimeters would be a times bugger. What would be the value of c - the speed of light in vacuum in the new SI' system?
assuming m' and m are units, it doesn't make sense to say a = m'/m, unless you mean "a" has the units of m'/m. In which case, 2 "meters" = 1 "new meter" (i.e. 2m=m'), then "a" would be 1/2 m'/m. That is as clear as it gets.
you are confusing unnecesary my very very simple question, a is dimensionless because m and m' are measuring one and the same, a do not has units it is a float type constant. Just imagine how the unit for meter, which is a rod keeped in the agency for standartization is replaced with a longer one. What would be the new velocity of light if the new rod is longer a times the original rod?
Then in that case, let's go back to your original conversions. We'll only look at three of them: from the first set: [J'] = [kg m'^2 / s^2] = [kg m^2 / s^2] * (1/a [m' / m])^2 = [kg m^2 / s^2] / a^2 [m'/m]^2 = [J] / a^2 from the first set: [N'] = [kg m' / s^2] = [kg m / s^2] * 1/a [m' / m] = [kg m / s^2] / a [m'/m] = [N] / a from the second set: c' [m' / s] = c [m / s] * 1/a [m'/m] Thus, either your first set was wrong and your second set is right, or you first set is right and your second set is wrong.
you do not make difference between units and oservables, for unitts - J,N,A ..., the transfornation is : m' = ma [J'] = [kg m'^2 / s^2] = [kg (m'*a)^2 / s^2] = [J] * a^2 [N'] = [N] * a however for constants or any fixed quatity the transformation is the reverse because it do not change with changing SI! For example the water in the tube will not becomes smaller because we had change measuring unit. Then c' [quantity] [m'/sec] = c[quantity] [m/sec] ---> c' [value] = c[value] / a try to see the physics not blindly using unit conversions, if the unit for distance m is changed with m' which is 2 times bigger, more energy would be needed to displace 1 kg of mass at distance 1 meter' because m' is longer, then J' should be bigger then J. the unit is quantity which is dependent from measuring units system, a fixed quantity - distance between two vilages for example do not change, then its value in SI' will be reversed to the change of the unit
You are a pain in the ass. Look at your "a" now. It is back to having the dimensions of [m'/m] so now we conclude for velocity c: c [m/s] * a [m'/m] = c' [m'/s] thus: c' = c * a, not c / a
