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33

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 ...


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 ...


18

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.


17

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

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 ...


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} ...


16

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 ...


16

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

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

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 ...


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 ...


10

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{ ...


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 ...


10

Take a nucleus of U-235 and determine its mass. Induce it to fission by firing a neutron at it. When it does so, collect all the pieces (except the extra neutron) and determine their total mass. You will find that all the pieces weigh just a hair less than original nucleus. The difference is the "binding energy", also previously known as the "packing ...


9

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 ...


9

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 ...


9

As noted by someone else, energy can be "converted" into mass e.g. via pair production. However, there is another example of this that you may be interested in: The mass of the matter you come into contact with on an everyday basis is almost entirely from protons and neutrons, which are roughly 2000x more massive than electrons. The proton, for example, is ...


9

Yes, everything generates a gravitational field, whether it is massive or massless like a photon. The source of the gravitational field is an object called the stress-energy tensor. This is normally written as a 4 x 4 symmetric matrix, and the top left entry is the energy density. Note that mass does not appear at all. We convert mass to energy by ...


9

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{ ...


9

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 + ...


8

This is actually a more complex question than you might think, because the distinction between mass and energy kind of disappears once you start talking about small particles. So what is mass exactly? There are two common definitions: The quantity that determines an object's resistance to a change in motion, the $m$ in $\sum F = ma$ The quantity that ...


8

In the early days of special relativity it was noted that the mass of an object appeared to increase as the speed of the object approached the speed of light. It was common to see the notation $m_0$ used for the rest mass and $m$ for the relativistic mass. In this sense the equation $E = mc^2$ is always true. However the concept of relativistic mass is ...


8

Energy and matter are not the same. Matter is a type of thing, whereas energy is a property of a thing, like velocity or volume. So your premise is flawed. In particular: there's no such thing as "a solid state of energy" - hopefully it makes sense that a property of something does not have states energy is not represented by waves, though it is a property ...


8

The annihilation produces gamma photons, whose total energy sums up to the total energy $E_0=\sqrt{p^2 \,c^2 + m^2\,c^4}$ formerly contained in the matter / antimatter kinetic energy (the $p\,c$ term) and that "frozen" in rest mass (the $m\,c^2$ term). So energy is conserved. As for gravity, the Einstein field equations "can't tell the difference" between ...


8

Think of the laser process as being similar to melting a substance through a change in state. An analogy might be putting a hot wire on top of an ice cube. It makes a 'slice' by heat, not by cutting. It turns the solid state ice into water and gas which doesn't hold together the same anymore.


7

Strictly speaking it's a unit of energy. But using $m=\frac{E}{c^2}$, you can convert energy into mass. Operating, we get $1{\rm\,eV}/c^2 =1.78\cdot 10^{-36}\rm{\,kg}$. (The $c^2$ is usually ommited.)


7

Starting with your given equation, we add $p^2 c^2$ to both sides to get $$ E^2=m^2 c^4 + p^2 c^2$$ now using the definition of relativistic momentum $p=\gamma m v$ we substitute that in above to get $$E^2 = m^2 c^4 +(\gamma m v)^2 c^2=m^2 c^4 +\gamma^2 m^2 v^2 c^2$$ Now, factoring out a common $m^2 c^4$ from both terms on the RHS in anticipation of the ...


7

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 ...



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