Einstein's equation and units

Factual Questions
1

E = mcc I know that’s supposed to be E = m c squared, but I don’t know how to do a superscript in UBB code. So E = mcc will have to do for now.

What are the units associated with Einstein’s equation? Joules, kilograms, meters, and seconds? British thermal units, long tons, rods, and fortnights? The last time I asked this question, I think I was in junior high school, and I was assured (by other junior high students) that it didn’t matter. I couldn’t figure that one out.

So I decided to ask again, hoping that it does matter, because if it doesn’t, my world view will need serious revision, something I can’t contemplate so hard on the heels of tax season.

Second, I have noticed the conspicuous lack of coefficients in Einstein’s equation. Heck, even kinetic energy (E = 0.5mvv) has a coefficent. Is this a major cosmic coincidence (assuming that units do matter, contrary to the opinions of a couple of 8th-graders)? I mean, they original came up with the definition of the energy, time, mass, and distance units without regard to one another, right? So, unless Einstein created a new unit to get rid of pesky coefficents, something really intense must have happened.

I hope I’m making sense here. It’s just weird that, of all the weird unit conversions I had to memorize, there wasn’t one for Einstein’s central equation. And I know mass-energy conversion is an intense concept in and of itself; it’s just double weird if it somehow knits together independently-derived units.

2

I don’t know if this is what you’re looking for, but if you just plug in your standard scientific units…

m — meter
s — seconds
kg — kilogram

Now you just have to get an equation to simplify the energy unit, Joules. Since energy is often used to do work, we can use the definition of work here:

W = F * d
(Work = Force * Distance)

Now plug that in, with M being mass and a being acceleration, all units are in parenthesis to avoid confusion (as opposed to variables)…

E = M * c * c
W = (kg) * (m/s) * (m/s)
F * d = (kg) * (m/s) * (m/s)
M * a * d = (kg) * (m/s) * (m/s)
(kg) * (m/s^2) * (m) = (kg) * (m/s) * (m/s)
(kg) * (m/s) * (m/s) = (kg) * (m/s) * (m/s)

So, there. Does that help? Basically the units for energy is Joules, which is just (kg*m^2/s^2)…

3

The units do not really matter, as long as the units are of the same system.

So lets just say we are using the old SI system. And lets further stick to the MKS (meter/kilogram/second) units, instead of the CGS (centimeter/gram/second) units.

E = m * c * c, where the mass is in kilograms, and the speed of light is in meters per second, or meters/second.

Therefore using dimensional analysis shows that E = kilogram * meter * meter / second / second, or kgmm/s/s.

Energy is a force occuring over a distance, which can also be defined as work. It just so happens that the unit of force in the SI-MKS system is the Newton. Remember that F=ma, or Force = mass * acceleration = kilogram * meter / second / second, which reduces to kg*m/s/s (since we can’t use superscripts).

This force acting over a distance gives energy or work. The SI-MKS unit of work is the Joule, which is a Newton * meter.

So a Joule looks like kgmm/s/s, which is exactly the units that we needed to match for E=mcc.

Therefore, using the SI-MKS system, the energy that corresponds to the famous equation is the Joule.

Using other systems, you can get this equation to work out for BTUs, calories, ergs or horsepower. It just matters that you stay consistent withtin the system for everything to work out.

But then you might have to figure the speed of light in furlongs per fortnight.

4

Yep. And the E of your kinetic energy has the same units as the E of Einstein’s formula (c, the speed of light, has the same units as v). Of course, you can convert those metric units to other systems, if you like.

rocks

5

Can you figure it in rods, fortnights, and long tons? Of course, but you have to be very, very careful about the dimensions (pounds, seconds, rods, etc.). It pays to be careful with the dimensions even when you are using only the SI units. Even professional physicists make errors when they neglect to pay attention to them.

What is the energy content of one long ton at rest with respect to the observer (on the surface of the earth)? A long ton (2240 lb) is not a unit of mass, but of weight. Since F=ma, m=F/a. The gravitational field strength on the earth (“acceleration due to gravity,” so called) is about 32 ft/s^2. So the long ton therefore has a rest mass of (2240/32) lb-s^2 / ft=70 lbs^2/ft. This figure is not in our system of units (rod-ton-fortnight), so we have to multiply it by a series of factors, each of which is equal to one:

(70 lb-s^2/ft)(16.5 ft / rod)[(1h/3600s)(1 day/24h)(1 fortnight/14 days)]^2= 7.910^-6 long tons(fortnights-squared)/rods. (The feet, seconds, hours, pounds, and days all cancel out.) This is a unit of mass.

Now what is the speed of light in rods per fortnight? The speed of light (186000mi/s) can be multiplied by any number of fractions each of which equals one, and the product is still the speed of light.

(186,000 mi/s )* (320 rods/mile)(3600s/hr)(24hr/day)(14 days/fortnight)=7.210^13 rods/fortnight. This is a unit of speed.

We are now ready to calculate E=mc^2. E=7.910-6(7.210^13)^2=4.110^22 (times whatever units we’re using). But what are the units we’re using? It is in units of (long ton-rods) since the fortnights cancel. This is a unit of energy. But a long ton-rod is not the same as a Btu. This is the reason that the SI has won out over the customary system in science. In the SI, a Newton-meter is exactly the same thing as a joule. So whatever units you are using for the mass and the speed, Einstein’s formula does give you the answer in units of energy. They just may not be the units of energy you want to use.

You could then multiply the answer you get by another factor. By now you should be able to guess that the factor will be equal to one. It turns out that one Btu divided by 4.910^8 long ton-rods is equal to one. So a long ton contains (4.110^22 / 4.9 ^10^8 ) Btu of energy = 8.4*10^13 Btu.

Work is the curse of the drinking classes. (Oscar Wilde)

6

7.2*10^13 rods/fortnight. It’s not just a good idea, it’s the law. :slight_smile:

It is too clear, and so it is hard to see.

7

Boris, this question has actually been posed to the Mailbag, and SDStaff Karen’s response will be posted in the next week or so.

8

I disagree with the idea that you have to stay within one system. It is entirely possible to cross from english to metric. In college physics, we answered a question about torque in newton*feet. What you must do is keep track of your unit types.

If I have an area of land two kilometers long and 3 miles wide, my area is 6 mile*kilometers.

9

Yeah, but “208 light fortnights” to the second nearest strar just sounds funny.

10

To get E=mc² type E=mc²

“It is not from the benevolence of the butcher, the brewer, or the baker that we expect our dinner, but from their regard to their own interest.” - Adam Smith

11

A formula doesn’t really care what units you are using…as long as you are consistent! In the case of e=mc^2, if the speed of light is in SI (metric) units, then simply be consistent and stick with SI units! It’s when you start mixing units that trouble starts (i.e.: NASA’s units conversion blunder).

“They’re coming to take me away ha-ha, ho-ho, hee-hee, to the funny farm where life is beautiful all the time… :)” - Napoleon IV

12

Actually, there is a conversion factor in Einstein’s equation: c² is, itself, the conversion factor for converting mass units into energy units. And yes, it does say a lot about the Universe that this conversion factor is also important for other reasons, i.e., being the speed of light.

“There are only two things that are infinite: The Universe, and human stupidity-- and I’m not sure about the Universe”
–A. Einstein

13

A lot of you have done more calculating than I really intended. The stuff about rods and fortnights was a joke, really.

I am pleased that this question is going to the Mailbag.

I’m still not sure I understand. Units don’t matter, as long as you’re consistent? That sounds fine, but I’m not sure about the overall consequences. Let’s say you measure energy in joules, mass in kilograms, and the speed of light in light-years per year. Please tell me there would need to be a coefficient there. Either that, or joules = kilograms, and it’s still to close to tax season for me to contemplate.

14

Boris, if you wanted to calculate your energy in Joules, which is in units of kg-m^2/s^2, you would have to convert your velocity from light-years/year to m/s. (Obviously, your kg unit for mass is just fine) I’ll use kinetic energy (K=0.5mv^2) as an example to better illustrate my point. Say you have a mass of 2 kg moving at a velocity of .3 light-year/year (stay out of the way!). Calculate the kinetic energy in Joules. You perform the conversion by multiplying (.3 light-year/year) X (9.46E15 m/light-year) X (1 year/3.15E7 sec) = 9E7 m/s*. The energy is then 0.52(9E7)^2 = 8.12E15 Joules.

Or, you could always say the the energy is 0.52 kg(0.3 light-year/year)^2 = 0.09 kg-light-year^2/year^2, but there is no energy unit designation for that.

*I realize that I over-complicated this a little bit in this example, but I wanted to give a general overview of unit conversion.

15

How would you apply that same logic to your previous posting of the formula for kinetic energy (E = 0.5mvv)? What if you measured v in light years per year? Would you then say that joules = 0.5 kilograms?

When they say to use consistent units, they mean if you’re using joules and kilograms, you have to use meters per second to measure velocity.

rocks

16

Strainger:
Not to be a nit-picker (okay, it is to be a nit-picker, I admit), the formula you give for the kinetic energy is only an appoximation at such high speeds. I suspect the answer you give is off by about 5% because of relativistic considerations.

17

Okay, I understand the “consistent units” bit now. I’m always wary of tacit units, since there are multiple units, sometimes even more than one metric unit (calorie and joule, for example) for the same thing. Now I know that Einstein had assumed use of SI units, which are apparently a fair bit more standard in the scientific world than in my rustic little corner.

What really bothered me was the supposed treatment of the speed of light as a “unitless” ratio, as if meters and seconds mutually cancelled. I don’t think anybody was treating it that way, but for a while I thought that people were.

18

Yeah, yeah, yeah, bibliophage, but look at the velocity units that Boris stuck me with.

Boris, uh, I gotta correct you on a couple of things.

No, it could be any standard of units. Let’s revisit E=mc^2.

SI units (kg-m-s):
c=3.0E8 m/s. Given a 1 kg mass results in E = 1 kg * (3.0E8m/s)^2 = 9E16 kg-m^2/s^2 = 9E16 Joules

CGS units (cm-gram-s):
c=3.0E10 cm/s. Given a 1 gram mass results in E = 1 gram * (3E10 cm/s)^2 = 9Ee20 gram-cm^2/s^2 = 9E20 ergs

BFT units (ft-slug-s):
c=9.8E8 ft/s. Given a 1 slug mass results in E = 1 slug * (9.8E8 ft/s)^2 = 9.6E17 slug-ft^2/s^2 = 9.6E17 ft-lb

The speed of light in this equation isn’t treated as a unitless ratio. As indicated above, it has units of length/time. It is a constant, but it still has units. Expressing the whole equation in units gives

E (mass*length^2/time^2) = m (mass) c^2 (length^2/time^2).

If you check the units, it’s all consistent.

19

http://boards.straightdope.com/sdmb/showthread.php?threadid=24116

What I’m trying to say, is that not everyone knows that many energy units are derived from kinetic energy (0.5mvv) equations to begin with. So just stick to the energy unit derived from the mass, distance, and time units that you’re using, and you’ll be fine.

20

No, he didn’t. You are confusing physical equations with engineering formulas.

Engineers commonly use formulas in which the variable quantities in the formula have specific units. There are therefore constants in the equation that depend on what units are used for everything else (the constants themselves have specific units associated with them). The constants are just precomputed conversion factors between the units of the input quantities and the units of the output quantity.

The constants in physical equations are generally either pure numbers that are part of the nature of the universe or mathematics, like 1/2, pi or e, or they represent some fundamental physical quantity like the speed of light, the gravitational constant, or Planck’s constant. These latter are represented symbolically to prevent their contribution to the relationship between variables from being lost. To do any computations with physical formulas, you have to reconcile units, including conversion factors, but to understand relationships, units are, in some sense, superfluous.

In the case of E = mc^2, you should think of the speed of light squared as the constant that gives the proportionality between energy and mass. The units of that constant will depend on the way we define mass and energy, but they will always fundamentally represent speed squared. If aliens in the Andromeda galaxy defined the concepts of mass, energy, and velocity the same way we do, they’d have the exact same formula we do (independent of their commonly used units for expressing those quantities, or their particular mathematical notation). Their numerical value for the speed of light would be different, but we’d still agree with them on the ratio between the energy content of a photon produced by a particular electronic transition (the sodium D line for instance), and the rest mass energy of an electron computed with Einstein’s formula, since that ratio is dimensionless.

Rick