In order to work with the relativity equation, surely we must know what system of units are being used. Are they in Metric, Imperial, SI units ISO or some other; or am I being realy dumb.
Can anyone go through a working example for me please?
2006-07-12 06:38:34 · 9 answers · asked by ? 2 in Science & Mathematics ➔ Physics
as long as you maintain a consistent approach to your units, it will work. stick with SI units or imperial measures.
one point though, c does not refer to "the speed of light". it refers to "the speed of light travelling through the medium that makes up the system you are calculating the value of E for".
2006-07-12 09:58:39 · answer #1 · answered by top_cat_1972 2 · 0⤊ 2⤋
They can be any units as long as they are of the proper dimensions. Energy can be in joules, ergs, BTUs, kilowatt-hours, etc. Mass can be in kilograms, grams, stones (not pounds as that is a unit of force). The speed of light can be km/s, m/s mph, knots, etc. However, c will need to be expressed based on the units chosen.
For example, c is defined as being 299792458 meters per second. Using this with a mass expressed in kilograms, you result in the units kg m^2/s^2, which is a joule. However, if one converts c to 29979245800 cm/s and usind mass in grams, the energy turns into g cm^2/s^2, which is an erg.
For other systems, it does get convoluted. However, energy is always a mass times the square of velocity. (This is derivative of defining work as force time distance.) One can wind up with a very odd energy unit, such as kg (mph)^2. Or similarly, the speed of light could be written in terms of the square root of energy divided by mass. If one has a conversion table, then one could has units of c written is some very odd units, such as BTU^(1/2) stone^(-1/2).
It's true that the mathmatics is easier using metric-type units, which is why scientists tend to use them. (Most these days use SI's kg-m-s system to produce joules, but some older scientists still use the g-cm-s system to produce ergs.) The point that I wish to convey is that the E=mc^2 equation doesn't care which units one uses, But the selection of a unit system can make life either easier or more complex.
2006-07-12 13:43:48 · answer #2 · answered by Ѕємι~Мαđ ŠçїєŋŧιѕТ 6 · 0⤊ 0⤋
We will do a dimensional analysis.
E=energy=units of work=force*distance=MLT^-2*L=ML^2T^-2
MC^2=MASS*VELOCITY^2=M*(LT^-1)^2=ML^2T^-2
It can be seen from dimensional analysis that units on either side match. What ever units units are used results will turn out to be the same as long as we use compatible units of one system.
2006-07-12 13:52:07 · answer #3 · answered by openpsychy 6 · 0⤊ 0⤋
equations know nothing of units. So long as the units are consistent (proper within any system) the answer is correct for that units system.
2006-07-12 13:40:53 · answer #4 · answered by Anonymous · 0⤊ 0⤋
All systems of units are equally valid, as long as you stay within the system. Try the chapter 'dimensions, dimensional analysis and similarity, section 1.1.4 of Barenblatt's 'Scaling, self-similarity and intermediate asymptotics' CUP 1996, ISBN 0521435226
2006-07-13 07:29:11 · answer #5 · answered by Sonia M 2 · 0⤊ 0⤋
For SI units:
E = energy in Joules
m = mass in kg
c = speed of light in metres per second
You should be able to do it from there.
2006-07-12 15:22:09 · answer #6 · answered by mesun1408 6 · 0⤊ 0⤋
SI units
c=meters/second
m=kilograms
e= joules
I am pretty sure but know they have to be in SI units.
2006-07-12 13:42:21 · answer #7 · answered by DoctaB01 2 · 0⤊ 0⤋
It doesn't matter which units you use as long as you are consistent and you make sure to convert. For instance,
joules=kilograms * (meters per second)^2 is okay
but joules = pounds * (miles per hour)^2 is not okay
the fix joules =pounds (kilograms per pound) * [(miles per hour)*{(meters per second) per (miles per hours)}^2
ya follow?
Typically we use SI though
2006-07-12 13:46:46 · answer #8 · answered by Nick N 3 · 0⤊ 0⤋
E = Joules, M = kilograms and C is (m/s)2
2006-07-12 13:42:58 · answer #9 · answered by Anonymous · 0⤊ 0⤋