# Tag Info

## New answers tagged mass-energy

1

What would occur if an electron at rest was accelerated to the speed of light? Any charged particle can be accelerated its speed increasing as more energy is supplied, but the limit of the speed is the speed of light. At a Lorentz factor ( = particle energy/rest mass = [104.5 GeV/0.511 MeV]) of over 200,000, LEP still holds the particle accelerator ...

1

Electrons can be accelerated to the speed of light (or practically to the speed of light). If you accelerate electrons to merely 5 MeV the velocity of 0.996 c where c is velocity of light, and yes if they are accelerated to that velocity they will emit gamma like radiation. Here I would like to clarify that the term gamma radiation is mostly used for the ...

4

No, in our own frame (comoving frame) we are not moving at all. Even though the CMB radiation is defining a preferred frame, to us it looks like the CMB is moving with respect to us - that's why we see a kinematic dipole in the CMB maps: http://scienceblogs.com/startswithabang/2013/06/28/our-great-cosmic-motion/

0

You are over complicating what appears to be your interest. You are asking in essence if there is an energy change in a particle as it moves in a gravitational field, i.e., a given spacetime metric. The answer is simple: except for singularities (i.e., what happens there, which is unknown, that's where any particle path ends in general relativity), a ...

0

The main reason is that lightest common particle/anti-particle pair is the electron/positron. When they collide, the overwhelmingly most likely result is emitting photons. The electron and positron weigh 511 kev, so when they annihilate, there is 1.02 MeV of energy. So each photon has 511 kev. Photons with this energy are gamma rays. Other photons like ...

1

It is possible. The rule is simple: the total energy (which is at least mc^2, m the sum of both masses) will be converted into other particles (photons or particle-anti-particle pair). So it is possible for a photon to pick up only a very few energy (for example by having a lot of photons, which is possible, even dough very unlikely). I think, it is ...

4

The amount of energy to emit is enormous ($2mc^2$ plus kinetic energy), so the emitted energies per photon will be at least the rest mass of the electron, which is half GeV (not counting some extreme and unlikely redshifts due to the collision point moving with respect to the observer). And that's well in the gamma spectrum.

4

To expand on a @ACuriousMind's comment, consider an electron-positron pair such that their center of mass is at rest. Since energy and momentum are conserved in the annihilation process, the resulting photons will be emitted in opposite directions with an energy $E = m_e = 511\ \text{keV}$ each (units such that $c = 1$). If the pair is instead in motion, one ...

11

In the field of PET (positron emission tomography) people tend NOT to call the product of the annihilation "gamma rays", but rather "annihiliation photons". While that may be a subtle distinction, the view is that a "gamma ray" is emitted by a nucleus, while a "photon" is a more general term used for a quantum of electromagnetic energy. But according to ...

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 ...

0

You have luck, the conversion is very simple. Do you remember $E=mc^2$? $c$ is the speed of the light, it is around 300 000 $\frac{km}{s}$. It is big. Its square is yet bigger. In SI, $\mathrm{1kg}$ of mass is $\mathrm{1kg}*(3*10^8\frac{m}{s})^2$ energy. It is $9*10^{16}J$. It is enough for a small city for a century. Unfortunately there is no way to ...

1

A fast moving mass will not become a black hole. See the existing question If a 1kg mass was accelerated close to the speed of light would it turn into a black hole? for a discussion of this. However suppose you take two fast moving 1kg masses moving in opposite directions and collide them. In the centre of mass frame you have the original rest masses, i.e. ...

1

I think both of the pre-existing answer violate (or at least play with the idea of violating) general relativity. General relativity is in a nuthell that gravity and accerlation cannot be distiguished. Just from that principle, we can solve this entire problem! I think the situation is quite simple afterall. The photon will bounce up and down, but ...

1

The reason that energy is usually conserved in most contexts is that Noether's theorem guarantees that energy is conserved in systems with time translational invariance. But the metric of the universe as a whole is (approximately) the Friedmann–Lemaître–Robertson–Walker metric, which does not have time translational invariance (more precisely, there does ...

1

This is one of those cases where you need to distinguish mass and weight. Mass (equivalently energy, by the famous $E=mc^2$) is defined in two ways. There's inertial mass, which relates the acceleration of a body to the force applied ($F=ma$), and there's the gravitational mass, which scales the gravitational force exerted by a body ($F=GMm/r^2$). The ...

1

The problem here is that the intuition about fields is different than for particles. And light and particles are pretty similar in that they behave like both or either one on occasion. Regarding the light as a stream of pretty localized photons, I would say the following: The light gives the flashlight a bit of recoil when it leaves. Say the flashlight lies ...

2

The minimum radius of a spherical object to be a black hole is given by : r = 2Gm/ (c^2) From this, I think we may be able to calculate the minimum density for the object to be a black hole, which is: d = (21/704)((c^6)/((G^3)(m^2)) (Assuming pi = 22/7) it is, d = (7.37 x 10^79 )/(m^...

2

Because glueballs have energy, and $E = m c^2$ says that energy is equivalent to mass. (Or another way to say it is that if you "zoom out" far enough that you can't see the constituent gluons that form the glueball, than you just lump all their energy into an effective glueball mass.) The energy can be thought of as just being the kinetic energy of the ...

5

If you had a gas of photons in a perfect cavity and these photons had energy $E~=~h\nu$, then for $N$ photons the cavity would have a mass $m~=~Nh\nu/c^2$ of photons. Glueballs as similar. The gluon carries two color charges (really color plus anti-color) and they can interact with each other. This forms a self-bound system that confines the massless gauge ...

4

Because in relativity the mass of a collection of particles is not necessarily the sum of the masses. Even two photons (treated as a unit) can have mass. Consider the total four-vector of a system with component four-vectors $(E,\hat{z}E/c)$ and $(E,-\hat{z}E/c)$. It has mass $(mc^2)^2 = (2E)^2$.

0

I haven't run the numbers, I assume JR is correct. I do know that the 'density' of a proton is much less than that required (isn't this obvious?) for gravitational collapse to occur (otherwise, it would). The only thing I want to add here is a cautionary note about our lack of understanding about quantum gravity. That is, as soon as you want to discuss ...

3

If you mean the mass of the hadrons, it is because of the energy/mass contributed by the gluons carrying the SNF and virtual quarks and antiquarks, which is high compared the three valance quarks. The valance quarks are the three "permanent" quarks that we think make up the proton and neutron. The Higgs field has a much lower effect on the mass of these ...

0

If $W$ is the weight of the block resting on a slant of angle $\theta$ w.r.t. horizontal then component of its own weight along the slant is $W\sin\theta$. If friction force acting on the block if $F=\mu W\cos\theta$ ($\mu$ is coefficient of friction) then the wind must exert a force greater than $F-W\sin \theta$ to make it move. So you must use some ...

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 $1.99 \... 0 As the question already beyond exact science, I will take a stab at it. There are three components here - Kinetic Energy (KE), Energy of gravitational waves (GW), and lost mass in merger. KE and GW - As the GW are ripples in space (time), they have to be generated by motion and/or its disruption (i.e stoppage at moment of merger). Therefore, the energy ... 2 Let me try to explain this by making an analogy with a simpler system i.e. a hydrogen atom. If you measure the mass of a hydrogen atom you find it is less than the mass of an electron plus the mass of a proton. In fact it is 13.6eV less. This happens because if you let a separated electron and proton fall together under their mutual electrostatic attraction ... 3 You've forgotten an important player in the system: the gravitational field. Here's a pretty argument that gravitational fields are physically meaningful objects that carry energy: imagine two masses accelerating towards each other from rest, from a great distance away. The rest energy of the system is$E_\text{rest} = (m_1+m_2)c^2$; the kinetic energy is ... 0 If we measure weight in units of$c^{2}$, e.g. instead of 10 kg we were to write$90 \cdot 10^{16}$kg m²/s², then the formula would read$E=m$. Proportionality constants are always a characteristic of the system of units used, not of the actual physical phenomenom. Indeed, in theoretical particle physics it is quite common to work in systems with$c=\hbar=1$... 2 In the context of$E=mc^2$, it's pretty irrelevant that$c$is the speed of light. It just so happens that photons, being massless, travel at the maximum speed$c$allowed by Einstein's theory of relativity. But$c\$ is a far more fundamental constant. It's the constant that links the 3 spatial dimensions to the dimension of time. Literally, in the case of ...

1

Actually, the critical mass is not affected by external gravitational fields. Gravity does not effect nuclear reactions. And, for the most part, gravity does not effect even chemical reactions either. Intermolecular forces (electromagnetic in nature) are vastly stronger than gravitational forces. Rather, the critical mass has to do with the effective ...

0

The mass is the same to a very good approximation: gravity is absurdly weak compared to the factors that influence this. One way that it might influence things was if the shape of the mass was significantly macroscopically distorted by gravity (ie a sphere might become flattened and you might therefore need slightly more mass). However metals are quite ...

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