| From: Yoda Oz | 11/02/99
16:21:33 |
| Subject: E=MC^2 | post id:
954 |
| What is the real meaning of this
equation? I know energy equals mass times the square of the speed of light constant, but what does it have to do with anything. My high school physics teacher told me that it had something to do with a nuclear reaction and the energy released when the reaction occurs. Is he right? | |
| From: Jeremy | 11/02/99
16:34:09 |
| Subject: re: E=MC^2 | post id:
957 |
| Yoda asks: "What is the real meaning of this equation? I know energy equals mass times the square of the speed of light constant, but what does it have to do with anything. My high school physics teacher told me that it had something to do with a nuclear reaction and the energy released when the reaction occurs. Is he right?" It means that mass and energy are really the same thing. This is an important statement. It tells us a lot about the relationship between that which we perceive as solid hard objects today on a macro scale as compared to radiation observed in the early universe. You see, if you look back way back into time by peering deep into space, all you see is a cosmic background radiation with tiny tiny irregularities. Somehow, then this radiation (energy) must have coalesced into stars, planets, and ultimately you and me. E=mc^2 has more than a tad to do with the explanation(s?) It also means that a stupendous amount of energy is bound up in things that zip around at (or near) the speed of light. This equation was the direct cause of nuclear bomb development - a realisation that caused Einstein (as a passivist) great anguish. It is also the basis of the understanding of nuclear reactions, electron ballistics calculations in CRT technology, Quantum theory, and heaps of other stuff. | |
| From: Maril | 11/02/99
16:36:03 |
| Subject: re: E=MC^2 | post id:
958 |
| yes the energy from a nuclear explosin is large (please do not say nu-killer ) therefore because the (C^2) speed of light is consant and a multiplier of the mass the mass must b large look at what they use in nuclear power and check out where those elements are on the periodic table :) | |
| From: Yoda Oz | 11/02/99
16:36:56 |
| Subject: re: E=MC^2 | post id:
960 |
| Could i have a better explanation
please, in simple English. | |
| From: Maril | 11/02/99
16:39:50 |
| Subject: re: E=MC^2 | post id:
961 |
| oki energy is changing mass is changing big elements make more energy when they are converted plutonium uranium are all at the bottom of the table and are the largest :) | |
| From: Halogen Fisk | 11/02/99
16:48:03 |
| Subject: re: E=MC^2 | post id:
964 |
| Hi
Yoda, Simply... The values in the formula are not as important as the fact the formula can apply at all. The important thing is that mass & energy are manifestations of the same thing, and can exist as one or the other. The formula does apply in nuclear fission reactions, but it's not the magic key to nuclear energy. Einstien's research was not directed specifically at subatomic physics, & much more work needed to be done after his discoveries before nuclear power (or bombs) could be produced. | |
| From: Nicole | 11/02/99
16:48:17 |
| Subject: re: E=MC^2 | post id:
965 |
| I could be trudling off down the
wrong lane but, I have no idea about E=MC^2. From what I can gather, it
means that energy is the amount of mass times the speed of light times
two. But you say that the heavier elements create more energy. So at what
point does, say, a pertoleum explosion equal that of a nuclear explosion?
It takes an awful lot of petrol to make an explosion quite the same size.
And it's mass would be heavier. I think. Nicole Nicole | |
| From: Maril | 11/02/99
16:57:15 |
| Subject: re: E=MC^2 | post id:
967 |
| petrol is a compound so its
bonding will affect the change of state (explosion) energy = mass * speed of light (squared) as halogen fisk said its not about the nuclear bombs but the whole concept that the two, energy and mass are equivicable in the formula. a lot of petrol would b needed but the explosions would be of a different nature nuclear is the very parts of the atoms that make up everything being torn apart by each other petrol just burns (changes into carbondioxide and water when its oxidized) very quickly | |
| From: Steve | 11/02/99
16:58:06 |
| Subject: re: E=MC^2 | post id:
968 |
| When an atom splits (say, in a
nuclear reactor) it breaks down to form two new atoms and releases a large
amount of energy. The total mass of the two new atoms is slightly less
than the mass of the original atom. The difference between the two amounts is given by the equation you've quoted. So, the energy released when the atom splits is equal to the difference between the masses times the speed of light squared. Does that help? | |
| From: Steve | 11/02/99
17:04:12 |
| Subject: re: E=MC^2 | post id:
971 |
| Nicole, As somebody else stated, the equation really shows that energy and mass are related. However, with reference to your question regarding petroleum explosions, the equation shows you the amount of energy that is released when an atom splits into two. Perhaps somebody with access to the right information could post the formula for the splitting of uranium (ie the mass of the uranium and the mass of the two resulting atoms - post split). When you add it up you will find that the sum of the two masses from the resulting masses is slightly smaller than the original mass of the uranium atom. The amount of energy produced by the resulting nuclear explosion is given by this equation. The explosion of petroleum produces a small amount of energy by comparison because it is a chemical reaction, rather the conversion of nuclear forces. Cheers Steve | |
| From: Chris (Avatar) | 11/02/99
17:56:08 |
| Subject: re: E=MC^2 | post id:
989 |
| E = m
c^2 The technical answer is that this equation does not strictly relate our everyday understanding of mass and energy, and actually causes a lot of confusion for people trying to use it in the context of relativity. The mass referred to above by the m symbol is relativistic mass, which is not really mass at all. This is the "mass" people are talking about when they proclaim that an object gets heavier as it approaches light speed - which, for want of a better word, is bollocks. To include real mass in the relationship we need to do a transform to the mass to convert it from relativistic mass energy to normal or "rest" mass. The transform is a math operation called a Lorentz transform. The resulting complete relation is E^2 = m^2 c^4 + p^2 c^2 where E is energy, m is rest mass, p is momentum and c is a constant equal to the speed of light in a vacuum. Now the use becomes more apparent. If I consider a proton in its rest frame (ie where its local velocity is zero) then p = 0 and the relation collapses back to E = mc^2. However this time we have defined the conditions and the mass unambiguously. Now I can use the relation as a mass - energy conversion for something like a fusion reaction, but not for imagining that a photon (for example) has mass. In the heart of young suns power is derived from nuclear fusion reactions. There are several chains for this type of Hydrogen fusion, but the net result is: 4 x H -> He + 2v + 2e+ where H is a proton (H nucleus), He is a helium nucleus, v is a neutrino and e+ is a positron (anti-electron). Now the mass of a helium nucleus is about 3.97 times the mass of a proton, so 0.03 times the proton mass is lost across the equation (we can disregard the positron and neutrino masses in this estimate). The proton rest mass is 1.67 x 10^-27kg. By Einstein's relationship the lost mass is liberated as photons. The energy E = mc^2 = 0.03 x (1.67 x 10^-27) x (2.99 x 10^8)^2 = 4.5 x 10^-12J. If you reacted 1 gram of protons in this way you would liberate 2.7 billion kilojoules of energy, and there are no real dangerous byproducts. Unfortunately we can't reproduce nuclear fusion on earth as yet, which is why we muck about with the less energy efficient fission (atom splitting) instead. Hope this helps! Chris | |
| From: James Richmond (Avatar) | 11/02/99
21:00:11 |
| Subject: re: E=MC^2 | post id:
1001 |
| A typical nuclear reaction
involving Uranium in a reactor or bomb is: 235U + n -> 140Xe + 94Sr + 2n A Uranium nucleus absorbs a neutron. This makes it so unstable that it splits into two pieces, a Xenon nucleus and a Strontium nucleus. Two neutrons are also released in the process. I'll come back to this later. The Xenon and Strontium nuclei are themselves very unstable, and so decay by a series of reactions to 140Ce (Cerium) and sup>94Zr (Zirconium). In atomic mass units the relevant masses are U 235.0439 Ce 139.9054 Zr 93.9064 n 1.00867 If we add up the masses of what we started with and subtract the total mass of what we end up with, we find a difference of 0.224 atomic mass units. This "missing mass" is converted to energy in the reaction, and is the number we plug into E=mc2 (in appropriate units) to calculate the energy released. This is a large amount of energy considering we are dealing with a single atom. In 1 gram of Uranium, there are about 1021 atoms (1 followed by 21 zeros!). Notice that at the start of the reaction there was one neutron, but at the end there were two. These two neutrons could cause nearby Uranium nuclei to split (or fission). We start with 1 atom splitting, which leads to 2 more splitting, then 4, then 8 and so on. This is known as a chain reaction and can be very nasty if uncontrolled. In a nuclear reactor, control is achieved by putting in rods which absorb neutrons without undergoing fission. The energy released in a nuclear reaction is huge. Witness the devastation caused by the single nuclear bomb dropped on Hiroshima in 1945. This bomb weighed about 1 tonne, but was nuclear powered. Its destructive force is often compared to how much TNT would be needed to produce the same effect - about 15,000 tonnes. The explosion of TNT is a chemical, as opposed to nuclear, reaction. And what have we learned from Hiroshima and Nagasaki? Today, there are enough nuclear weapons on Earth to destroy every city on the planet several times over. | |
| From: Martin Smith | 12/02/99
14:03:50 |
| Subject: re: E=MC^2 | post id:
1056 |
| Chris and James have as usual
giving you the good stuff. But Maril - I think you might be misunderstanding the fact that Uranium is heavy so gives off a lot of energy. The change in mass even in a Nuclear reaction is tiny (do the math). The Uranium doesn't dissapear - the vast majority of the mass remains (now as different elements). Uranium etc is used because its nucleus is fairly unstable. Look at hygrogen - the lightest element. When it fuses (as happens in the Sun) to make helium the release of energy is huge - the amount of mass gone is small (C squared is a big number). The resulting helium (plus anything else produced) is slightly lighter than two hydrogen atoms. When talking about petrol burning - you are right in saying it is a chemical reaction but that still has nothing to do with its relative atomic mass. Uranium can be involved in chemical reactions and the elements of the petrol molecule can be involved with nulclear reactions (that is how those elements came to be), and E=MC^2 holds for all. | |
| From: Dr. Ed G (Avatar) | 12/02/99
19:52:43 |
| Subject: re: E=MC^2 | post id:
1093 |
| Unfortunately we
can't reproduce nuclear fusion on earth as yet, which is why we muck about
with the less energy efficient fission (atom splitting)
instead. Minor correction... unfortunately we can. In fact it's relatively easy to reproduce fusion on Earth, all you need is a fission bomb (an A-bomb with either plutonium of uranium) to start the ball rolling, and a solid source of deuterium and tritium (lithium deuteride will do nicely). Okay, you need some pretty fancy ignition circuitry and materials engineering as well, but if you are a world superpower that's not a problem. The result is what is known as a thermonuclear device (a neutron or H-bomb). The vast majority of strategic nuclear warheads (there the one's that 'strategically' wipe a major city completely off the map) are fusion weapons because its much easier to concentrate a large amount of deuterium and tritium into a device (for that extra 'umpf' you need at the start of a global nuclear catastrophy), without the danger of them detonated before they leave the factory, than it is to pack a similar explosive power of uranium or plutonium. Such devices have been detonated HEAPS of times in the many tests conducted by France, England, China, the USA, and the USSR (Pakistan, India, and Israel are still mucking about with fairly rudimentary fission bombs as I understand although I could be mistaken). The problem, of course, is controlling the fusion reaction in a way that allows you to extract useful energy without vapourising your power station and staff. At a recent AINSE (Australian Institute of Nuclear Science and Engineering) conference in Canberra it was kinda depressing to learn that we are still some 30-50 years away from a commercial fusion plant, if it ever happens at all. However, fusion has been achieved in many research reactor so far, and one or two have even managed to pass the "break even" point (that's where the energy coming out is greater than the energy going in). Though this has not been achieved for more than a tiny fraction of a second - the plasma is simply too difficult to contain and as soon as it touches the walls it snuffs itself out almost immediately. It was interesting to attend a conference where many of the researchers don't expect to be alive when their efforts come to full fruition... perhaps our politicians can take learn a lesson in looking further than 1-3 years into the future from these people. Soupie twist, Ed G. | |
| From: Cass | 13/02/99
14:49:20 |
| Subject: re: E=MC^2 | post id:
1123 |
| James, You referred to rods that they stick in nuclear reactors to absorb neutrons. Do these rods have a finite life span? What are they made of? Do they constitude radioactive waste? Cass | |
| From: James Richmond (Avatar) | 15/02/99
12:20:56 |
| Subject: re: E=MC^2 | post id:
1214 |
| Control rods often contain
cadmium, which is a good neutron absorber. I'm not sure about the average
lifespan of the control rods, but I think that they would make a very
minor contribution to the waste material from the reactor. Again, I'm not
sure, but I think the control rods would become significantly radioactive
after a while. | |