When measurement standards for E, M, and C are all arbitrary human inventions, like “joules” and “meters per second” and “kilograms”
As far as I can tell, all of these measurement standards existed long before Einstein made his famous discovery, and were developed using completely unrelated methodology, so how did it happen that these standards fit perfectly into that equation?
I could understand it much more easily if a new standard measurement of E, M, or C was invented and used to apply to this equation, but how previous standards of measurement fit in it seems like magic to me.
The units used in the equation do not have to be joules, or kilograms, they just have to be internally consistent. A joule has merely been defined to be a kilogram - meter squared per second squared.
Right, but the units that it is in is arbitrary. We can say that it’s about 2.998e8 m/s, but we could just as easily say that it’s 1 light-second per second. We just wouldn’t be using joules to measure the energy of the object.
You’re saying (I think) “It could just as well be something like E = 5mc[sup]2[/sup], if we just specify a unit of energy that’s five times smaller.” For instance, one joule is 5 fifths of a joule, and who is to say we couldn’t have chosen “fifths of a joule” as our unit of energy.
The reason this isn’t the case is that the units of energy, mass and velocity aren’t all defined independently. Rather, the appropriate unit of energy depends on our choice of units for mass and velocity. For instance, the kinetic energy of an object is 1/2 times its mass times its velocity squared. For any choice of units for mass and velocity, there is an appropriate choice of unit for energy that makes this equation valid. And for that same choice of units, E = mc[sup]2[/sup] will be valid.
So sure, you could “break” one equation by choosing the wrong units (thus requiring a conversion factor like the number 5 I added above.) But in order to do that, you have to break all the other equations involving that quantity (by adding the exact same conversion factor). So in that case all you’ve really done is changed the meaning of the word “energy”, so that it instead means what we previously called “energy divided by five”.
The mathematical relationship between those things has absolutely nothing to do with what you measure them in.
I mean, 5 is half of 10 whether you count in base 10, base 2, Roman numerals, or pennies, right?
Nature is full of physical laws that happen to be relatively simple mathematically. Things fall in a way that be expressed as v=t^2. Force is equal to f=ma. Doesn’t matter how you count them.
C is a constant. So it can be expressed in whatever units are necessary to make the rest of the equation work.
Mass is kilograms.
Energy is joules.
1 joule is 1 kilogram * meter[sup]2[/sup] / second[sup]2[/sup].
So c[sup]2[/sup] = E / m or kilogram * meter[sup]2[/sup] / second[sup]2[/sup] / kilogram. The kilograms cancel out and means it equals meter[sup]2[/sup] / second[sup]2[/sup] .
But if c[sup]2[/sup] = meter[sup]2[/sup] / second[sup]2[/sup], then c = meter / second. And that’s obviously a speed.
If you measure mass in carats and the speed of light in furlongs per fortnight, than your energy unit had better be furlong-carats squared per fortnight per fortnight.
Think of a right triangle A[sup]2[/sup] + B[sup]2[/sup] = C[sup]2[/sup].
It does not matter if you are measuring in inches, mm, feet, meters, yards, or whatever as long as you measure all three side with the same units, you are golden.
Almost exactly what I was going to post (except you meant to say carat-furlong squared per fortnight squared). The carat furlong squared per fortnight squared is a perfectly valid unit of energy. It just happens to be one which isn’t used very often.
By the same token, if you measure your mass in grams, and your speed in centimeters per second, you’ll get an energy in grams centimeters squared per second squared. That’s also a perfectly valid unit of energy, just one that’s encountered a lot more often than the other one. It’s encountered so often, in fact, that we give it a name: One gram centimeter squared per second squared is called an erg.
To the OP - on your physics test, you will be perfectly justified in writing that
E = 0.3456 mc[sup]2[/sup]
If your professor marks it wrong, ask him who he is to impose his energy units on you.