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44

This is inevitably going to be an unsatisfactory answer because your question is vastly more complicated than you (probably) realise. I'll attempt an answer in general terms, but you have to appreciate this is a pale shadow of the physics that describes this area. Anyhow, Einstein was the first to spot that energy and mass were equivalent, and you've no ...


34

There is no controversy or ambiguity. It is possible to define mass in two different ways, but: (1) the choice of definition doesn't change anything about predictions of the results of experiment, and (2) the definition has been standardized for about 50 years. All relativists today use invariant mass. If you encounter a treatment of relativity that ...


33

When lasers cut something, they're only cutting in the sense that they're making atoms be not as attracted as they once were to each other. When you get down to the nitty-gritty details, it is not really the same as mechanical cutting. Remember that lasers shoot photons, and when photons hit atoms, they excite electrons. If you excite these electrons ...


26

A particle's rest mass never changes. It's mass is a natural constant, and one of the numbers which uniquely identifies it (like its spin). On the other hand, the invariant mass of the atomic system does increase as the electron becomes excited, bringing the atom into a higher energy state. In that sense, the atom (not the electron) gets "heavier" because of ...


26

This is really an extended comment to Geoffrey's answer, so please upvote Geoffrey's answer rather than this. The mass of a hydrogen atom is $1.67353270 \times 10^{-27}$ kg. If you add the masses of a proton and electron together then they come to $1.67353272 \times 10^{-27}$ kg. The difference is about 13.6eV, which is the ionisation energy of hydrogen ...


22

There are two explanations possible stemming from the fact that the definition of $m$ in the formula is ambiguous. Well, perhaps Einstein had only one of the two meanings in mind in his original paper but I'm afraid I wouldn't know that as I haven't read it. By that as it may one should recall that special relativity mixes space and time and therefore also ...


20

Yes, the total mass of a battery increases when the battery is charged and decreases when it is discharged. The difference boils to Einstein's $E=mc^2$ that follows from his special theory of relativity. Energy is equivalent to mass and $c^2$, the squared speed of light, is the conversion factor. I would omit the scenario I. If the lithium is leaking from ...


20

Firstly, mass is energy. There is no distinction at all. Not all of the "mass" of an atom comes from protons/neutrons/electrons, for example, there is a chunk that comes from potential energy — this chunk actually dominates over the mass of the constituents. However, we can reword your question to be "What stops nucleons from turning into other forms ...


18

MSalters already said "yes". I would like to expand on that by computing the change. Let's take a 10 kg cannon ball, made of lead. Heat capacity of 0.16 J/g/K means that in dropping from 1000 K to 100 K it has lost $10000\cdot 900 \cdot 0.16 \approx 1.4 MJ$. This corresponds (by $E=mc^2$) to a mass of $1.6 \cdot 10^{-11} kg$ or one part in $6\cdot 10^{11}$. ...


18

You've probably heard of Einstein's famous equation: $$ e = mc^2 $$ This states that mass and energy are equivalent, and indeed the LHC turns energy into matter every day. So to find the mass equivalent to an electron volt just convert eV to Joules and divide by $c^2$. 1 electron volt = $1.60217646 \times 10^{-19}$ joules, so 125 GeV is: $$ 125 \times ...


17

I am assuming that by "energy" you mean photons. So you want to transform protons into photons. It is not possible. It would violate several conservation laws - mainly the charge conservation (protons are positively charged), but also baryon number conservation. The antiparticle is necessary to cancel these quantum charges to make the transition possible.


16

Of course, it does, since: $$\frac{\partial E}{\partial t} = \frac{\partial }{\partial t} \left(m \cdot c^2 \right) $$ Very little, though


16

Because time is accumulating, to calculate the time lapse, you integrate. The elementary time interval transforms like mass. The difference is that the total time lapse is done by "summing" over all elementary intervals. For the mass, you don't do this. For mass: $$m=\frac {m_0} {\sqrt{1-\frac{v^2}{c^2}}}$$ For time: $$dt=\frac {dt_0} ...


15

Both are correct, within the domains for which they are correct. More seriously, the general relation $$ E^2 = m^2c^4 + p^2c^2$$ holds for all objects, whether they have mass or not, whether they are moving or not. The special case $E = mc^2$ is for $p = 0$, i.e. objects which do not move, as you said. The special case $E = pc$ is for objects which have ...


15

The definition of an antiparticle is dependent on having the opposite quantum numbers of the particle so that they can annihilate, i.e. the sum of the conserved quantum numbers are zero. Thus the answer by @mpv is adequate. The implication of your question is then: is baryon number conservation a strict law or an emergent law that may be violated at some ...


14

First, the non-relativistic equation $$ E= mc^2 + \frac{mv^2}{2} $$ is equivalent to its second power, $$ E^2 = (mc^2)^2 + m^2 c^2 v^2+ \frac{m^2v^4}{4} $$ If $v/c\ll 1$, then the last term is much smaller than the previous two, and the first two terms on the right hand side are equivalent to the correct relativistic $$ E^2 = (mc^2)^2+ (pc)^2 $$ which ...


14

You can find the shortest and easiest derivation of this result in the paper where it was released by Einstein himself (what better reference can you find?) in 1905. It is not the main paper of Special Relativity, but a short document he added shortly afterwards. A. Einstein,Ist die Trägheit eines Körpers von seinem Energieinhalt Abhängig?, Annalen der ...


14

To answer the question simply, $E=mc^2$. Energy is a manifestation of mass, and mass is a manifestation of energy. In a fusion or fission process, the total "energy" of the system remains constant, it just changes shape. By "energy" I mean the totality of the already present energy, and the bound energy of the mass that takes part in the reaction.


13

Cutting is a process when you deliver energy to break chemical bonds in material that you cut. When you use a saw, you deliver mechanical (kinetic) energy that converts into kinetic energy of particles of the thing you cut, so they can get out of the thing. Laser is just another way to deliver such energy, since the a photon has enough energy to break some ...


12

The answer is the there is some reduction in mass whenever energy is released, whether in nuclear fission or burning of coal or whatever. However, the amount of mass lost is very small, even compared to the masses of the constituent particles. A good overview is given in the Wikipedia article on mass excess. Basically, the mass of a nucleus will in general ...


11

As you may know, photons do not have mass. Relating relativistic momentum and relativistic energy, we get: $E^2 = p^2c^2+(mc^2)^2$. where $E$ is energy, $p$ is momentum, $m$ is mass and $c$ is the speed of light. As mass is zero, $E=pc$. Now, we know that $E=hf$. Then we get the momentum for photon. Note that there is a term called effective inertial ...


11

The conversion between mass and energy isn't even really a conversion. It's more that mass (or "mass energy") is a name for some amount of an object's energy. But the same energy that you call the mass can actually be a different type of energy, if you look closer. For example, we say that a proton has a specific amount of mass, about $2\times 10^{-27}\text{ ...


11

The reason $c$ is important is not because it is the speed of light. It's important because it is a universal conversion factor between time and distance. If you have a certain amount of time $t$, you can calculate the corresponding amount of distance by multiplying it by $c$. Note: I'm not talking about the distance any particular object travels in the ...


11

It is annoying to talk about mass-energy conversion, because it is too often misinterpreted to mean that energy doesn't weigh on a scale before it "turns into matter". So I will preface the answer by saying that if you heat up a gas, the heat weighs on a scale, if you burn some paper and let the heat escape, the escaping heat makes the combustion products ...


11

Every relationship between mass and energy will contain two factors of velocity, for dimensional consistency. In special relativity we have the more exact relationship $$ E^2 - p^2c^2 = m^2c^4 $$ where the momentum $p$ is $$ p = \frac{mv}{\sqrt{1-v^2/c^2}}. $$ You can do a little algebra to show that the total energy is always $$ E = \frac{ ...


11

In short, no, it is not a coincidence, they are related. Namely, you may derive the kinetic energy as the first order approximation to the relativistic energy. We have, $$ E_0 = mc^2 $$ as you say correctly. Then $$ E = \gamma m c^2 = \left( 1 - \frac{v^2}{c^2}\right)^{-\frac{1}{2}} m c^2 $$ or using a binomial expansion $$ E \simeq \left( 1 + ...


10

Capital $\mathrm{C}$, in upright font, is the symbol for the coulomb. Lowercase $c$, italicized, is the speed of light in vacuum. Thanks to Einstein's equation, we can switch between mass and energy ($\mathrm{MeV}$ is a unit of energy) by using factors of $c^2$, and sometimes it's more convenient to know the energy equivalent of a particle's mass rather than ...


10

If I'm reading rightly, I think your main question is: Why does only a small percentage of rest mass turn into energy [even for fusion]? It's because the universe is very strict about a certain small set of conservation rules, and certain combinations of these rules make ordinary matter extremely stable. Exactly why these rules are so strictly observed ...


10

No, it wouldn't change due to the energy conservation law or mass conservation law – these are the same laws in relativity where $E=mc^2$. The nuclear energy extracted from the nuclei by fission would be converted to other forms of energy inside the box – ultimately the thermal energy. High temperature means that particles are moving at higher speeds and ...


10

$E_0 = m_0 c^2$ is only the equation for the "rest energy" of a particle/object. The full equation for the kinetic energy of a moving particle is actually: $$E-E_0 = \gamma m_0c^2 - m_0c^2,$$ where $\gamma$ is defined as $\gamma = \frac{1}{\sqrt{1 - (v/c)^2}}, $ where $v$ is the relative velocity of the particle. An "intuitive" answer to the question can ...



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