Hot answers tagged

50

[5/3 - Extended the answer, made some corrections, and responded to John Duffield's comment] This is actually the paradox that led Einstein to General Relativity. Consider a special case: An electron and positron are at the Earth's surface. Bring them together and they annihilate, creating gamma rays (which is very energetic light). The gamma rays travel up ...


46

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 ...


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 ...


41

Let's take the carbon nucleus as a convenient example. Its mass is $1.99 \times 10^{-26}$ kg and its radius is about $2.7 \times 10^{-15}$ m, so the density is about $2.4 \times 10^{17}$ kg/m$^3$. Your density is ten orders of magnitude too high. The Schwarzschild radius of a black hole is given by: $$ r_s = \frac{2GM}{c^2} $$ and for a mass of $1.99 \...


36

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 enough,...


35

The answer is no. The simplest proof is just the principle of relativity: the laws of physics are the same in all reference frames. So you can look at that 1-kg mass in a reference frame that's moving along with it. In that frame, it's just the same 1-kg mass it always was; it's not a black hole.


31

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 ...


26

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 ...


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 (...


23

Yes! In fact, this kind of phenomenon is very common. For example, the mass of a proton is much greater than the sum of the masses of the constituent quarks. Much of the extra mass comes from the gluons that bind the quarks together; like photons, gluons are massless, but they contribute to the inertia.


22

If I ruled the world, I would ban the phrase "pure energy" in contexts like this. There's no such thing as pure energy! When particles and antiparticles annihilate, the resulting energy can take many different forms -- one of the basic principles of quantum physics is that any process that's not forbidden (say, because of violation of some sort of ...


22

As far as the theory goes, you are absolutely correct, the (negative) binding energy between atoms in a molecule contributes to the total mass of that molecule, so a stable molecule is less massive than the sum of the masses of its constituent atoms. However (as you yourself calculated), the mass difference is absolutely tiny, and as far as I know, it has ...


21

At tree level, a matter-antimatter annihilation reaction doesn't just produce gamma rays, nor can you exclude neutrinos in the final state. Even the simplest such reaction can—given enough energy—produce a variety of particle-pairs. However, those pairs are subject to two subsequent processes: If the particles are not stable, they will decay towards ...


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 ...


20

Your guess at the solution to this paradox is correct. "Pumping energy up" to the space station, regardless of the method you choose, would require an input of at least the amount of energy you would gain in kinetic energy on the way down. This is just a variation on the impossible perpetual motion machine concept. In practice, you would not only not gain ...


19

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 10^...


19

A lot of different forms, but mostly kinetic energy. A good table is given at Hyperphysics. The energy released from fission of uranium-235 is about 215 MeV. This is divided into: Kinetic energy of fragments (heat): ~168 MeV Assorted gamma rays: ~15-24 MeV Beta particles (electrons/positrons) and their kinetic energy: ~8 MeV Assorted neutrons and their ...


18

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 ...


18

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 ...


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 $6\cdot 10^{11}$. ...


17

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} {\sqrt{1-\frac{v^2}{...


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.


17

The famous equation $E = mc^2$ is actually just a special case of the relativistic equation for the total energy: $$ E^2 = p^2c^2 + m^2c^4 \tag{1} $$ where $p$ is the relativistic momentum and $m$ is the (constant) rest mass: $$ p = \frac{mv}{\sqrt{1 - v^2/c^2}} $$ For an object that isn't moving $p=0$ and equation (1) becomes: $$ E = mc^2 $$ which is ...


16

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 ...


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

The problem is that the two calculations have hardly anything to do with one another - so it's no wonder you don't get the same result. The electron volt, as you say, measures the work you need to move an electron across a potential difference of one volt. On the other hand, if you want to calculate the mass of an electron using $E=mc^2$, what you need is ...


15

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.



Only top voted, non community-wiki answers of a minimum length are eligible