Does E=mc2 Work with Different Metric Systems?

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Discussion Overview

The discussion revolves around the implications of using different metric systems when applying the equation E=mc². Participants explore how variations in units of measurement for mass, distance, and energy can affect calculations and the consistency of results across different systems, including SI and imperial units.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about how different metric systems yield different numerical outcomes when calculating energy using E=mc², questioning the validity of using "man-made numbers."
  • Another participant suggests performing calculations in both English and SI units to illustrate the differences in outcomes.
  • Some participants emphasize the necessity of using consistent units within any physical formula, noting that mixing units leads to nonsensical results.
  • A participant provides an example calculation using both SI and imperial units, showing that while the numerical values differ, the energy results remain consistent when properly converted.
  • There is a discussion about why certain units, like kilometers per hour, fit with kilograms but not with pounds or ounces, raising questions about the compatibility of different measurement systems.
  • Another participant explains that units in a system work due to well-defined conventions, and that creating a new system would require defining derived units accordingly.
  • Some participants argue about the consistency of units, with one asserting that a consistent set of units is essential for valid calculations, while another challenges this by suggesting that different units can still yield valid results if properly managed.
  • A participant reiterates that all components of the equation must match the units for energy being used, providing a breakdown of how units relate within the equation.

Areas of Agreement / Disagreement

Participants generally agree on the importance of using consistent units within a given system, but there is disagreement on the implications of using different systems and whether valid results can be achieved through careful conversions.

Contextual Notes

Some participants note the limitations of using archaic units and the potential confusion that arises from mixing different measurement systems, highlighting the need for clarity in unit definitions and conversions.

Jarfi
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Ok so I am confused with the face the different metric systems and sizes use different numbers for the same size f.ex. 1000 meters can be 1 km or 100 mph can be 135 km, And the mass can be told in pounds, kg, grams, and more. And energy in calories. So it seems to me that you can get different outcomes when calculating with e=mc2 if you use different systems, its like it doesn't seem right, they use the international metric system, so e is than actually just a number that is (300000)2 times the mass but what if you use a system from usa with (180000)2 times the mass, than you would get a totally different outcome, I don't know if you understand me, it's just that I don't understand how it can work out with man made numbers.
 
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Pick some quantities and calculate E in English units. Convert your quantities to SI units. Calculate E again. Post your results.
 
In any physical formula, you must use consistent units. If you have mass in kilograms and velocity in miles per hour, the calculated E in E = mc^2 will be nonsense.

In SI, to calculate E measured in joules, m must be measured in kilograms and c must be converted to meters per second.

In imperial units, to calculate E measured in foot-pounds, m must be in slugs and c must be converted to feet per second.
 
Jarfi said:
Ok so I am confused with the face the different metric systems and sizes use different numbers for the same size f.ex. 1000 meters can be 1 km or 100 mph can be 135 km, And the mass can be told in pounds, kg, grams, and more. And energy in calories. So it seems to me that you can get different outcomes when calculating with e=mc2 if you use different systems, its like it doesn't seem right, they use the international metric system, so e is than actually just a number that is (300000)2 times the mass but what if you use a system from usa with (180000)2 times the mass, than you would get a totally different outcome, I don't know if you understand me, it's just that I don't understand how it can work out with man made numbers.

As long as you stay within the same system, it doesn't matter, your answer will come out the same.

Example:

1kg mass
c=300,000,000 m/sec

gives you 9 x 10^16 joules.

converting to fps

1 kg = ~2.2 lbs
c= 990,000,000 ft/sec

gives you ~2.16 x 10^18 ft-poundals

If you convert joules to ft-poundals you will find there are 23.7 ft-poundals to a joule and 9 x10^16 joules = 2.16 x 10^18 ft-poundals. (More or less, I rounded out a bit which means you won't get an exact answer.

The trick is that when you do the calculation, all the units have to be in the same system

You use grams with centimeters, Kg with meters, etc. You can't use kg and ft together, for example.
 
SteamKing said:
In any physical formula, you must use consistent units. If you have mass in kilograms and velocity in miles per hour, the calculated E in E = mc^2 will be nonsense.

In SI, to calculate E measured in joules, m must be measured in kilograms and c must be converted to meters per second.

In imperial units, to calculate E measured in foot-pounds, m must be in slugs and c must be converted to feet per second.




This is what has been confusing me! Why does kmph
fit with kg and not pounds or ounces, if i where to create my own measuring system, how would i coordinate speed and energy to make it work out. Why do some measuring units work together but others dont? Why is it better to use foot with pounds than with kiloes?
 
Jarfi said:
This is what has been confusing me! Why does kmph
fit with kg and not pounds or ounces, if i where to create my own measuring system, how would i coordinate speed and energy to make it work out. Why do some measuring units work together but others dont? Why is it better to use foot with pounds than with kiloes?

Units in a system work because there exist a well defined conventions of what the derived units are. You could of course make a system that uses basic units form other systems but then you have to define other derived units. An example J=Kg*m^2/s^2 but Kg*ft^2/s^2 =X which is still a valid unit of energy, X=0.093J. Not much of a point of doing this though.
 
Jarfi said:
This is what has been confusing me! Why does kmph
fit with kg and not pounds or ounces, if i where to create my own measuring system, how would i coordinate speed and energy to make it work out. Why do some measuring units work together but others dont? Why is it better to use foot with pounds than with kiloes?

All the parts of your equation have to equal the unit for energy your using, for example Joules (J):
J= kg(m/s)2
In order to get the energy in Joules you have to have all of those components in your mathematical equation.
So if:
E= J and J= kg(m/s)2

Then:
E= kg(m/s)2
-------------------
If:
E= kg(m/s)2 and E= Mc2

Then:
kg(m/s)2= Mc2
--------------------
If:
kg(m/s)2= Mc2 and M=kg

Then:
kg(m/s)2= (kg)c2
--------------------
If:
kg(m/s)2= (kg)c2 and c= (m/s)

Then:
kg(m/s)2= kg(m/s)2

That's why you have to use kg for m and such. You still can solve an equation using other units but we don't have simple words for those units like Joules you'd have to write out the unit.
 
Janus said:
As long as you stay within the same system, it doesn't matter, your answer will come out the same.
I beg to differ.

As long as you stay within a consistent set of units, it doesn't matter.

SI comprises a consistent set of units (so long as you steer clear of archaic units such as calories). The standard English system, with mass expressed in pounds mass, force in pounds force, energy in BTUs is anything but consistent.

A consistent set of units has
  • 1 unit of energy = 1 unit of length * 1 unit of force
  • 1 unit of force = 1 unit of mass * 1 unit of acceleration
  • 1 unit of acceleration = 1 unit of velocity / 1 unit of time
  • 1 unit of velocity = 1 unit of length / 1 unit of time

An inconsistent set of units requires the use of values other than one in the above. For example, when force is expressed in pounds force, mass is expressed in pounds mass, and acceleration is expressed in ft/sec2 you get 1 unit of force = 1 unit of mass * 32.17404855643 units of acceleration.
There's a very good reason scientists prefer to work with consistent units. To steal a phrase from Steve Jobs, "It just works".
 
Energy(J)=Force(N)*Distance(m)=
distance(m)*(Δmomentum(kgms-1)/time(s))=Energy(kgm2s-2)Energy=Mass(kg)*(speedlight(ms-1))2=Energy(kgm2s-2)


See those units at the end of both? They are the same units. Therefore... the equation works.
 

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