Forgive me if this has been asked before, but…I really don’t get the “C” part of that famous equation. Yes, it’s the speed of light, but the speed of light is totally relative. The result of the equation depends on what you consider the speed of light to be
(Huh?) Let me put it this way…
Let’s say the speed of light is 186,000 miles per second, which it is (or maybe it’s kilometers, whatever.) This makes C² equal 34,596,000,000 miles per second. That sounds reasonable.
But what if you define the speed of light as one light year per year? That’s also a perfectly valid value for the speed of light. But by that definition, C² is still one light year per year.
And, if we go even further, what if we define the speed of light as 1/2 of a light year every 6 months. If you square that, it becomes 1/4 of a light year every 6 months. That’s smaller than the original value!
So how are we supposed to compute C squared? Am I missing something here? Or is this a flaw in the famous equation?
The problem is that you really can’t put stuff like this down in word problem form.
In this case C is an absolute, for that was the theory that Einstein believed… and most science still uses.
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I’ve got a good friend, who’s biggest nightmare is that he will die, get to the gates of heaven, and upon being told that he led a mostly good life, will be told that there’s one last question… if train A leaves the station at 11:30, and…
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You can measure c in furlongs per century if you like, and the equation is still valid. But if you measure it something odd, such as miles per second, you’ll get 34,596,000,000 miles squared per second squared. Multiply this by the mass of an object that weighs 10 slugs and you’ll get 345,960,000,000 slug-miles squared per second squared, which are valid units of energy. Trouble is, no one’s ever heard of these units.
To make it meaningful to most people, keep a consistent measurement scheme, such as the meters-kilogram-seconds one. This way, mass in kilograms times the speed of light squared (meters squared per second squared) gives units of kg*m^2/s^2, otherwise known as Joules.
Anything that measures energy can be represented as mass*distance^2/time^2. For anther example, potential energy is mass times height times the acceleration of gravity (meters per second squared), so which is
Your average speed on a trip can be calculated as v=d/t, where v=your average speed, d=the distance you traveled, and t=the time it took you to make the trip.
If I went 100 miles in 2 hours, my average speed was 100/2, or 50 miles per hour.
If I went 100 feet in 2 seconds, my average speed was 100/2, or 50 feet per second.
Miles per hour and feet per second are both valid, acceptable units for speed. If you took the 100-mile trip in the first example above, and expressed it in feet (it’s 528000 feet), you could use the same equation (v=d/t) to get 264000 feet per hour, which is also a valid unit for speed. When you put numbers into an equation, you have to put the units (feet, miles, seconds, or whatever) in there as well. They multiply and divide (and, if appropriate, cancel one another out) just like numbers do.
In fact, if you have a complicated equation, checking the units of your result is a good way of catching mistakes. If the units of the answer don’t come out correctly, it means you have made a mistake somewhere. Some equations require certain units in order to work properly, but E=mc2 is not one of them. (This is common among equations that use temperature-- they usually require absolute temperature, and the Fahrenheit and Centigrade scales most people use in daily life are not absolute.)
For the E=mc2 equation, any mass and velocity units you care to use will produce a value for E in units of energy, but some units of energy are more commonly used than others. (Just like you wouldn’t say your average speed from Washington to Richmond was 650 inches per second. You’d use miles per hour, or km per hour.)
Basically, it’s no different than any real-world calculation - if you use different units, the result is a different number with different units attached.
The units of measurement are as important as the actual numerical value. 186,000 miles per second, squared, is 34,596,000,000 miles-squared per second-squared. The units are squared right along with the number. As the article scr4 linked to touches upon, particle physicists often work with units such that the speed of light is 1, for convenience.
Come on, someone quit fooling around with the guy, and give him the straight dope. Even the mailbag article ignores the answer, and takes a condescending tone of “you couldn’t possibly understand, you mere human nonphysicist.”
E is for Erg, a unit of measure of energy. I forgot what M is, I seem to recall it stands for amu (atomic mass units). Einstein didn’t pick these letters out of a hat, he had specific physical units in mind.
No he did not, the equations work out in any units you choose. M and C can be in any unit you want, and the result would be the enrgy in whichever units you chose for M and C. If you use meters per second for C and kilograms for M, you end up with energy in terms of kg m[sup]2[/sup] s[sup]-2[/sup]. This unit is also called Joules because it’s a common unit for energy. If you use cm/s for C and grams for M, you end up with g cm[sup]2[/sup] s[sup]-2[/sup], which is ergs. If you use slugs for M and miles per hour for C, you end up with energy in terms of slug mile[sup]2[/sup] hour[sup]-2[/sup], which doesn’t have another name because people usually don’t use that as a unit of energy. Still, it’s a perfectly valid unit of energy.
E = energy
m = mass
c = speed o’ light in a vacuum (supposedly…it’s a constant)
They can be in any units (i.e. E could be Joules, ergs, etc and c could be meters/second, miles/hour, or any other distance/time). The units are part of the numbers, and the math works on them too.
c is a very large number, whatever the units. Therefore, c squared is insanely huge. Which means that you can get a whole lot of energy out of very little mass.
Good work, scr4. Another way to think about this is that the equation is unitless, because the units cancel out on each side. Let’s start with something simple and work up to it:
1 mile = 1.61 km
This equation has units, because the units are not the same on each side. Now let’s multiply the right side of the equation by 1, which we all agree is legal. Let’s specify 1 as ([sup]1 mile[/sup]/[sub]1.61 km[/sub]) The ‘1.61 km’ part cancels out, and we get
1 mile = 1 mile.
Now the units are the same on each side, so the units of miles cancel out and we get 1 = 1, a unitless equation.
The same thing happens with any other equation that is unitless, for example F=ma. What this equation really describes is a fundamental relationship between Force, Mass, and Acceleration. The result of this is you can specify Force in terms of units of mass * distance * time[sup]-2[/sup]. Which means no matter what units of mass, distance, and time you choose, you get a meaningful result. You can say something like “I exerted 15 femto solar mass AU years[sup]-2[/sup] on the shopping cart”. This means that you exerted 15 * 10[sup]-15[/sup] of the force required to accelerate the sun at a rate of 1 AU/year per year.
Let’s look at two seemingly unrelated equations that are also unitless and see what other underlying relationships they reveal:
Kinetic Energy (T) = [sup]1[/sup]/[sub]2[/sub]mv[sup]2[/sup]
Potential Energy § = mgh
Both of these equations describe energy, does it make sense that one is mv[sup]2[/sup] while the other is mgh? The units seem really screwy!
Let’s look at the first. Velocity is measured in terms of distance/time, so the first equation is telling us that
T = [sup]1[/sup]/[sub]2[/sub]massdistance[sup]2[/sup]*time[sup]-2[/sup]
Now the second. ‘g’ is the acceleration due to gravity, so it is measured in distance/(time[sup]2[/sup]). ‘h’ is the height of the object, so it is also a distance:
P = mass * distance * time[sup]-2[/sup] * distance = mass * distance[sup]2[/sup] * time[sup]-2[/sup]
So, even though these equations look very different, they tell us the same thing! Energy is in units of a mass times the square of a velocity. So, E=mc[sup]2[/sup] tells us exactly the same thing in terms of units. Where things get interesting is the values of the constants - the [sup]1[/sup]/[sub]2[/sub] in the kinetic energy equation, the ‘c’ in the mc[sup]2[/sup], etc. The units don’t matter at all because you can always convert from one unit to another.
Gunslinger writes:
This is only true if you are willing to convert units. If you choose c in units of miles/hour, and m in slugs, then E is in terms of slug mile[sup]2[/sup] hour[sup]-2[/sup], and to get Ergs or Joules you need to convert.
The definition of a Joule is the amount of work needed to exert a force of one Newton over a distance of 1 meter. The definition of a Newton is the force required to accelerate a mass of 1 kg by 1 ms[sup]-2[/sup]. So, a Joule is in terms of kg m[sup]2[/sup]s[sup]-2[/sup]. TO convert from slug mile[sup]2[/sup] hour[sup]-2[/sup] to kg m[sup]2[/sup]s[sup]-2[/sup] you need to know the relationships beween hours and seconds, between miles and meters, and between slugs and kilograms. THEN you can get the results in Joules.
So, while the equation itself is unitless, as soon as you plug in numbers with units, the other side of the equation is tied to those units.
since the translation that I’ve found (the libraries closed early today, so I can’t look up the original paper at the moment, but I doubt it was changed) has “L” used for energy.
Besides, the erg is the cgs unit of energy, equal to one g*cm[sup]2[/sup]/s[sup]2[/sup], with no amu involved. So clearly the letters themselves don’t directly stand for units.
Now, we look at the paper. After going through the equations, right before the end of the paper, he states (translation and formatting as found here):
Looks like a simple example more than anything else. If he’d had units in mind they would’ve been expressed earlier in the paper (where none at all are mentioned when he first begins discussing energy) and, for that matter, in the first relativity paper.
As Karen somewhat indicated in the Mailbag article, physicists tend to leave out units (other than checking for consistency, e.g. not having masslength[sup]2[/sup]/time[sup]2[/sup] on one side of an equation and masslength/time[sup]2[/sup] on the other) until they start applying it to something “real”. Einstein would have been no exception; I highly doubt he began any Gedankenexperimente with “now what would I see if I traveled at three times ten to the tenth centimeters per second…”
Thanks for the link and the explanation, Philbuck. I agree he must have used the cgs for real-world calculations, but no physicist would have claimed that a certain law would be valid only in one set of units. That’s like saying “speed is distance divided by time, provided you measure distance in miles and time in hours.”
By the way, the quoted article is not the original relativity article but a shorter follow-up that focused on the e=mc[sup]2[/sup] and its consequences. The original article is Annalen Physik 17:891. Our library doesn’t have Ann. Physik that old (or maybe it’s locked up somewhere) - anybody know if it’s online somewhere? It can’t still be copyrighted.
I was only drinking last night and I’m not hung over this morning so I think my math skills may be somewhat useful.
Your basic calculations here are flawed. You can’t square the distance without squaring the time - or rather, you need to divide the distance by the time and then square the result. You can’t square one factor and leave the other one untouched.
The calculation should look like this:
E = m * (d/t)²
What you’re doing is this:
E = m * (d²/t)
which is incorrect.
Hope that clears things up as regards the approach in the OP.
Or, if we want to use the units that God intended, mass is measured in Planck masses, length in Planck lengths, and time in Planck times. Then we have that c=L[sub]P[/sub]/t[sub]P[/sub] , and E[sub]P[/sub] = M[sub]P[/sub] * (L[sub]P[/sub]/t[sub]P[/sub])[sup]2[/sup]. Easy!
