# Tag Info

1

All three are "correct", and all three refer to mass-energy equivalence discovered by Einstein. Equations (2) and (3) are algebraically identical, and are generalizations of (1). Equation (1) only takes into account an object's rest mass, whereas equation (2) also takes into account the momentum $p$ of the object, and (3) takes into account velocity $v$. ...

1

The photons released are individually massless, but all of them together have an effective mass equal to the original masses of the particle and antiparticle; see my answer here. This isn't some mathematical abstraction either -- you can put the photons in a reflective box and weigh it, and it'll have extra weight. It's safe to say that the phrase ...

0

2) Does it include energy of electron due to electric field too? We do not know. If the electron was composed of charged elementary parts, its apparent inertial mass would be greater than sum of inertial masses of the parts. This is because such parts will act on each other with electromagnetic forces and in accelerated motion, sum of these internal ...

1

There is the concept of the self energy of the electron that can be described by its electric field. The model, which goes back to classical physics, can be described as a spherically distributed charges (that equal the charge of the electron) at infinity and the work required to bring the charges to the radius of the electron. The calculation would show the ...

0

To begin with, I can't think of any reason why an increase in the mass of an electrically charged object should affect the magnitude of the electric field it produces. You can easily check that $E=kq/r^2$ doesn't depend on $m$. Moving charges do produce magnetic fields, but this is a different matter (and the magnitude of the magnetic field has nothing to do ...

3

Yes, both the internal potential energy and the internal kinetic energy of a bound system (in the rest frame of its center of mass) contribute to the bound system's inertial mass according to $E=mc^2$. For a paper discussing the evidence that this is true for internal kinetic energy in particular, see Kinetic Energy and the Equivalence Principle.

2

Theoretically, the answer is yes. However, looking at the practicality of the situation, the answer is no. The photons can not be contained inside the box unless they are either 1. Created inside the box itself, or 2. They are trapped beforehand, and then brought inside the box. In 1., they do not add to inertia because, they are created using energy ...

6

Yes, mass and energy are equivalent. A more competent relativist might be able to give you the complete description, but to first order you can say that the mass of an object is simply the total energy in its volume divided by c^2. That mass is equivalent to the inertial mass by the weak equivalence principle, which is a cornerstone of GR. That is to say, ...

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.

1

Einstein's equation does not indicate how mass can be converted into energy. It only says that mass and energy are equivalent. Nobody has as yet found a way to convert mass directly and completely into energy. The best that can be done at present is nuclear fission or fusion reactions. Fission is already well established in the nuclear energy industry. ...

3

You're mixing up rest mass and relativistic mass (which is a concept that should be taken out and shot, and mostly has). $E^2=p^2c^2+m^2c^4=\gamma^2m^2c^4$ where $m$ is the rest mass. None of your equations are using the rest mass, so you should go through and replace $m$ with $\gamma m$ in your equations, in which case you'll get the meaningless equation: ...

0

You are asking about micro black holes. Some hypotheses involving additional space dimensions predict that micro black holes could be formed at energies as low as the TeV range, which are available in particle accelerators such as the LHC (Large Hadron Collider). Popular concerns have then been raised over end-of-the-world scenarios (see Safety of ...

1

Physics does not answer "why" questions at the level of basic laws and postulates, i.e. "why" this postulate. Newtons gravitational law : Newton's law of universal gravitation states that a particle attracts every other particle in the universe using a force that is directly proportional to the product of their masses but also inversely proportional to ...

2

Given the context of the question, the fact that it seems to be about $E=mc^2$ specifically, and that the OP says he's having a hard time understanding it, i'm going to try and give a simple answer in plain english without a load more complicated formulae. I am no physicist, and although the concept may not be that easy, the formula is pretty simple, maybe ...

2

I will try to answer this question with my basic understanding of special relativity: Is matter condensed energy? It kind of is, but a better way to phrase it would be that everything that has energy, (behaves like it) has mass. Imagine you have a hollow box with the insides covered with perfect mirrors and you put it on a scale. If you shone a light ...

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 ...

13

does this equation mean masses are just condensed energy? No, it means that mass is just another form of energy, just like heat, motion, electric attraction, etc. For example, the energy of a charged sphere is $$E=\frac{3}{5}\frac{Q^2}{R}$$ This equation doesn't mean that charge is just condensed energy; it means that charged objects have energy. ...

6

The two are indeed related. The relativistic expression for the total energy is: $$E^2 = p^2c^2 + m^2c^4 \tag{1}$$ where $p$ is the relativistic momentum: $$p = \frac{mv}{\sqrt{1 - v^2/c^2}}$$ and $m$ is the rest mass. If the object isn't moving then $p=0$ and equation (1) becomes: $$E = mc^2$$ which is of course the well known expression for the ...

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