# Help me understand special relativity...

This occurred to me the other day, and has been bugging me since then. Everyone knows that Einstein’s special relativity centres around mass-energy equivalency, expressed by the famous formula e=mc^2 (can’t work out how to do superscript!)

But is this formula complete in itself, or is there a hidden constant? Both options seem equally odd to me, a guy with only high school physics.

Let’s say I’ve got 1kg of matter and it’s converted entirely into energy. Would that mean that:
a) I end up with 9 x 10^16 Joules of energy. (The formula’s complete in itself.)

or

b) I end up with an amount of energy that is proportional to mc^2, but not exactly the same ie k x 9 x 10^16 Joules. (The formula has a hidden constant, and should be better expressed as e = kmc^2)

If a), why do the units work so perfectly? It couldn’t be a coincidence, so why do the values of energy, mass and speed work so well together?

If b), why don’t they show the hidden constant?

They don’t work perfectly. In SI units, c = 299,792,458m/s. There’s the constant.

I see I’ve phrased my question badly - I’m aware that c isn’t exactly 3 x 10^ m s^-1, but failed to put that in correctly. Given Gorillaman’s correction, is option a) the correct one? That is, 1kg of mass equates to about 8.99 X 10^16 Joules?

If so, why does it work without a constant? I presume it isn’t a coincidence, so what is it about the nature of the SI units for mass, speed and energy mean that this simple calculation works?

You’re thinking of it backward.

The amount of energy in an object is proportional to the mass of that object.
E ≈ M

To turn a proportion into an equality, one introduces a constant.

The constant that is introduced is C, which must be squared so that the units come out correctly.

C is not a hidden constant: it’s right there smack out in public, as it should be.

Superscripts are done by {sup}2{/sup}, except that square brackets are used instead of the curly ones.

Buy the book Relativity And Common Sense by Herman Bondi. It’s published by Dover and is probably available through Amazon or other on line book sources.

It should give you what a relativistic layman needs to get a good idea of what it’s all about.

I think he’s asking, “How did they figger out that C is a constant involved in calculating mass and energy?”

I’m obviously not expressing myself well here, sorry for the confusion.

The SI units with which the formula e=mc[sup]2[/sup] operates are J, kg and m s[sup]-1[/sup]. These units were developed before Einstein came up with special relativity. Correct so far?

So how does the formula e=mc[sup]2[/sup] come up with the exact answer in Joules? How come a concept that is beyond what these units were designed to deal with is expressed in such a way that it’s a perfect fit for these units? Why does the answer come out in that exact number of Joules?

I presume that if you substituted other units, eg calories, lb and yards per second, the formula wouldn’t work. Obviously, these units are old fashioned and aren’t used by most scientists.

If nothing else, here’s my calculator for E=mc²
www.1728.com/einstein.htm

The famous E = mc[sup]2[/sup] equation falls out naturally in the derivation of relativistic kinetic energy, which in turn comes from the definition of force as a change in relativistic momentum. I don’t know whether that’s how Einstein found the result originally, but it’s not a bad bet.

Specifically you end up with E[sub]k[/sub] = (γ–1)mc[sup]2[/sup] = γmc[sup]2[/sup] – mc[sup]2[/sup], which has a term dependent on velocity — the first one, with γ — and a term independent of velocity — a constant quantity that’s intrinsic to the mass. The first term is the total energy of the mass; the second is the rest energy, and essentially expresses the mass-energy equivalence. I can’t find a good Web link for this derivation just now, so you’ll have to take my word for it, or better yet consult a physics textbook. I’m refreshing my own memory here with Tipler’s Modern Physics, my old university text. Any lengthy treatment of SR should have it though.

As to how the c[sup]2[/sup] gets in there in the first place, it’s a consequence of the Lorentz transformation equations, which include c[sup]2[/sup] in their mapping of space-time coordinates from one reference frame into another. And the basis for that comes from the fact that objects travelling at velocity c (like photons) have that exact same velocity in all reference frames.

Crap, That calculator won’t let me use exponents.

How do I calculate for the Planck Mass? :dubious:

Correct. These SI units — except the second of course, which is ancient — were defined by the French in the post-Revolution era. All long before Einstein was born.

The Joule is defined to be one kg-m[sup]2[/sup]/s[sup]2[/sup], so it’s not too surprising that the equivalent energy of a given mass would turn out to be the mass’s magnitude times the square of some velocity. Not that you could possibly derive E = mc[sup]2[/sup] from the definition of a Joule of course. In principle someone could have suspected that was the case, more or less out of thin air, but I don’t think anyone did before Einstein published his seminal SR paper.

To actually derive E = mc[sup]2[/sup], you really need to start with the Lorentz transformations. Like all proper physics equations, they express relationships between quantities without regard to any particular choice of units.

True, but this is a general rule of discipline needed when plugging values into any physics equation. You must use compatible units, otherwise you get numerical nonsense. For example in the equation F = ma, you will get Newtons of force only if you use kilograms for your mass and meters-per-second[sup]2[/sup] for your acceleration. You’re not really free to use slugs and yards-per-week[sup]2[/sup] — or anyway if you do, you’re result will have a bizarre and nameless unit of force that no one ever uses.

The same discipline is needed when using the CGS variant of the SI system, as opposed to the more familiar MKS. In CGS, the fundamental units are centimeters, grams, and seconds, and the units of force and energy are dynes and ergs respectively. If you compute E = mc[sup]2[/sup] from a mass expressed in grams, and c[sup]2[/sup] in cm[sup]2[/sup]/s[sup]2[/sup], your result will automatically be in ergs (and not Joules).

Not with lbs because that’s not a mass unit, but if you had said, perhaps, slugs, then the formula would work just fine. For example, if we take

m=10 slugs
c=1.80x10[sup]12[/sup] furlongs/fortnight

we find

E=3.24x10[sup]25[/sup] slug-furlong[sup]2[/sup]/fortnight[sup]2[/sup]

The unit “slug-furlong[sup]2[/sup]/fortnight[sup]2[/sup]” is a perfectly valid, albeit obtuse, unit of energy. If it were used often enough, it might get a name – perhaps the “Finch”. 1 Finch is equal to 4.03650685x10[sup]-7[/sup] Joules.

A proper equation doesn’t care about the units you use – the left-hand side will still equal the right-hand side.
Regarding c: Relativity wants us to treat space and time on equal footing. Unfortunately, humans have had thousands of years of practice treating space and time as quite different things, and they’re not about to stop. If you do treat them on equal footing, then you can make statements like:

“A massless particle (like a photon) will travel one unit through space for each unit of travel through time.”

When they’re treated this way, there are no c’s in any equation anywhere. For example: E=m is the complete equation. But if we want to talk about distance and time in different units, then we are forced to make statements like:

“A massless particle (like a photon) will travel 299,792,458 meters through space for each second of travel through time.”

The ratio of these numbers (299,792,458 meters/1 second) must show up throughout the equations to “undo” mankind’s desire to treat space and time differently.

That’s exactly what I was looking for, thankyou. The very definition of the Joule makes it likely that an equation involving mass and velocity would have a result in Joules.

I will devote my life to popularising the use of the “Finch” as an alternate to the ridiculous Joule. I ask you all to join me in my quest.

Everyone is wrong! Einstein’s theory does not centre around the mass-enrgy equivalency, it centres around the two postulates of special relativty:

1. The laws of physics are the same in all inertial frames.

2. The speed of light is constant in all inertial frames.

(neither of these two postulates hold for non-inertial frames, though it doesn’t mean that special relativty cannot handle non-inertial frames).

The equivalency of mass and energy is an important derived result, though it should be said that mass and enrgy are not completely equiavlent by convention.

The formula is for the rest enrgy of the particle and does not take into acount the kinetic enrgy, so the formula for a particles total energy (in some inertial frame) is:

E[sup]2[/sup] = m[sup]2[/sup]c[sup]4[/sup] + p[sup]2[/sup]c[sup]2[/sup]

where p is momentum.

In special relativty the mass of a particle is defined as it’s mass in it’s momentarily co-moving reference frame, this means that it is a Lorentz scalar and invariant under a Lorentz transformation.

Option a) would be correct (but you have to ask what do you mean b convert to energy and will momentum be conserved as well?).

c is a propertyb of the geometry of spacetime and much more fundamental than simply the norm of a photon’s velocity vector in an inertial frame. The result is derived, so you have to understand how it is derived in order to understand why it works so well, but you can also see by dimensional analysis that if there were another constant ‘k’ it would have to be dimensionless and that a constant of proportionality between mass and enrgy musthave units of speed squared.

I realise this is your sixth post, and not your first, but in case no one has said it before: welcome aboard, Complex Conjugate.

Enola Straight
The calculator WILL let you use exponents. Ever see how scientfic notation is used in Excel™ ? Well, that is exactly the way it is done for that calculator.
Examples: 1.2E+32 ; and for negative exponents 1.2E-11

As far as the Planck Mass is concerned, the value at the NIST site is:
2.176 45 x 10[sup]-8[/sup] kg
http://physics.nist.gov/cgi-bin/cuu/Value?plkm

Sorry if this sounds nitpicking but if you needed to input that number, you could have input .0000000217645 and obtained an answer of 1.9561e+9 joules. Granted it’s a little awkward but not that difficult. Of course, if you are working in the range of a reciprocal googol, then exponents would be the better choice.

Well, I hope I have helped you out.

No, you’re still a bit off.

The very definition of a constant is one that is picked with the correct set of units that conforms to the units used to measure the other elements in the equation to form an equality.

The relationship here is not between mass and velocity. The relationship is between mass and energy. It was Einstein’s insight that a constant with the units of velocity would show the equivalence between the two.

Einstein’s equation has nothing to do with joules. They’re just the measurement that the international scientific community happened to have standardized on.

C can be any number whatsoever as long as the units it is expressed in form a velocity. It could be (approximately) 300,000,000 meters/second or 300,000 kilometers per second or 186,000 miles per second or whatever it is in furlongs per fortnight or in parsecs per sidereal day.

The constant in the equation is not even a true velocity since it is squared. Again, the constant is merely a way of getting the numerical relationship equal and the units all neat and even.

What’s interesting here is exactly what Complex Conjugate said. “The speed of light is constant in all inertial frames.” Nobody had realized this before, and this insight led to the rest of relativity theory.

By convention c is often set to a dimensionless 1, the reasons for this are twofold; firstly any choice of units are arbiatry so setting c as 1 is perfectly accpetable and the chioce simplifies equations, secondly it means that the components of Lorentz vectors have the same units (time is measured in units of length in this choice of natural units).