# Tag Info

16

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: $$125 \times ... 16 Yes, the total mass of a battery increases when the battery is charged and decreases when it is discharged. The difference boils to Einstein's E=mc^2 that follows from his special theory of relativity. Energy is equivalent to mass and c^2, the squared speed of light, is the conversion factor. I would omit the scenario I. If the lithium is leaking from ... 16 Because time is accumulating, to calculate the time lapse, you integrate. The elementary time interval transforms like mass. The difference is that the total time lapse is done by "summing" over all elementary intervals. For the mass, you don't do this. For mass:$$m=\frac {m_0} {\sqrt{1-\frac{v^2}{c^2}}}$$For time:$$dt=\frac {dt_0} ...

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.

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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

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 ...

10

$E = mc^2$ 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: $E = \gamma mc^2 - mc^2$ 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 be seen by ...

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 $2\times 10^{-27}\text{ ... 9 Take a nucleus of U-235 and determine its mass. Induce it to fission by firing a neutron at it. When it does so, collect all the pieces (except the extra neutron) and determine their total mass. You will find that all the pieces weigh just a hair less than original nucleus. The difference is the "binding energy", also previously known as the "packing ... 8 This is actually a more complex question than you might think, because the distinction between mass and energy kind of disappears once you start talking about small particles. So what is mass exactly? There are two common definitions: The quantity that determines an object's resistance to a change in motion, the$m$in$\sum F = ma$The quantity that ... 8 As noted by someone else, energy can be "converted" into mass e.g. via pair production. However, there is another example of this that you may be interested in: The mass of the matter you come into contact with on an everyday basis is almost entirely from protons and neutrons, which are roughly 2000x more massive than electrons. The proton, for example, is ... 7 Starting with your given equation, we add$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 ... 7 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 ... 7 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 ... 7 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 ... 7 It's certainly possible for a particle's mass to come partially from kinetic energy of massless particles; for example, about half of a proton's mass is the kinetic energy of its gluons. But the kind of mass that fundamental particles have, the kind that comes from the Higgs mechanism, doesn't appear to be of that kind. Maybe someday we will discover that it ... 6 This equation is incredibly generic and describes many phenomena outside of nuclear phenomena. For example: place the the following setup in a box (a spring and some bars) and weigh them: . Now, loosen the spring and repeat. You should measure a smaller mass because you've removed some of the energy. In reality you couldn't possibly measure the ... 6 Assume that we started with a dry cloth, use a dry iron (no steam) and don't initiate any chemistry (you don't want to burn the collar, after all). There are two way to look at this. On a macroscopic scale the cloth (taken as a whole) has gained internal energy, so it simply is more massive. No "transformation" is required. The energy exists in the more ... 6 Your kinetic energy formula is incorrect. The correct one for a massive particle is $$KE = m c^2\left[ \frac{1}{\sqrt{ 1 - v^2/c^2}} - 1 \right]$$ with $$p = \frac{ m v}{\sqrt{1 - v^2/c^2}}$$ Now, to describe a photon, we wish to take two limits$m \to 0$and$v \to c$. These limits must be taken properly. To be precise, we wish to take this limit in ... 6 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 ... 6 The equation, properly understood is: $$E = \gamma m c^2$$ where$m$is the invariant mass (or the deprecated "rest mass"). Now, for a photon, the invariant mass is zero. But this does not imply that$E$is zero since the Lorentz factor$\gamma \rightarrow \infty$as the speed goes to c. Thus, this equation has an indeterminate form for a massless ... 6 It is incorrect to say that the energy of a string directly gives us the mass of the particle. While it is true that more the oscillations on the string, higher the mass, the relation between the oscillations and the mass it not that of a simple proportionality. What's really happening is that the string has some energy$E$(due to oscillations on it) and a ... 5 It is the convention of setting the velocity of light$c=1$that allows for this, the natural units, otherwise it is$\mathrm{GeV}/c^2$The rest mass energy connection $$E^2=p^2+m^2$$ at rest then the mass is identified with energy in natural units. 5 To understand binding energy and mass defects in nuclei, it helps to understand where the mass of the proton comes from. The news about the recent Higgs discovery emphasizes that the Higgs mechanism gives mass to elementary particles. This is true for electrons and for quarks which are elementary particles (as far as we now know), but it is not true for ... 5 The Einstein's mass-energy relation,$E = mc^2$, gives the total energy content of the system. But this is not the energy we get from the object. When you annihilate an electron with a positron, both particles vanish so that the released energy is equal to the energy of the two particles according to Einstein's formula. But when you burn 1 kg of wood you ... 5 In order to answer this question, you should first ask yourself what you mean by "object". From an elementary particle perspective, every particle has a characteristic constant rest mass. These masses aren't thought to change, just like the charge of an electron doesn't ever change. So in this sense, the answer to your question is "no, you cannot accelerate ... 5 This is cool because$E=mc^2$can act as some sort of uncertainty relation; if you have a population of photons with energy$E$, they are engendered with a mass$\frac{E}{c^2}$, no matter what my intuition says. (Is that right?) No, it's not quite right. In relativity, it turns out that energy and momentum are parts of a single four-dimensional vector, ... 5 The equation$E=mc^2$equates rest energy to mass. There is a third symbol in this equation that represents the speed of light, but this is a universal constant. One can always select physical units such that this constant attains value unity. Regardless the system of units selected, up to a numerical proportionality constant, the equation$E=mc^2$... 5 The relation$E=mc^2$only works for particles at rest, which is evidently not the case for photons. In the general case, the relation is $$E^2=m^2c^4+p^2c^2$$ for a particle with momentum$p$. (Note, though that the momentum is not necessarily$p=mv\$ as in the newtonian case! See for instance If photons have no mass, how can they have momentum?) For a ...

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