# Tag Info

1

The proton is not fundamental. It is made up of quarks and gluons. It is these constituents that are colliding in the LHC to produce, in your example, a Higgs boson. The quarks and gluons only carry a fraction of the energy of the proton. In addition, the colliding gluons or quarks in general do not have the same momentum. Therefore some of the energy will ...

0

I lost one term, which is the one containing $\nabla^t(\partial_t)^r=g^{tt}\Gamma^r_{tt}=-M/r^2$, so this term is $$-\frac{1}{8\pi}\int r^2\sin\theta \nabla^t(\partial_t)^r d\theta d\phi=\frac{M}{8\pi}\int \sin\theta d\theta d\phi=M/2$$ Add this term to previous one and correct!

5

If you consider your photon as a point object, it cannot bend its own path. It will always travel on the ridge it creates, speaking in terms of curvature of space. The other idea is possible. Two photons having a momentum, attract each other, trapping each other, like a positronium (typical example for this behavior). In the model of relativity this is ...

1

Even burning in a closed container would result in a loss of mass via the electromagnetic radiation emitted (i.e. visible light and infrared), and as stated, it would be immeasurably small.

2

To understand the answer you should have understood well a point. Energy is conserved, mass is not. Moreover, masses possess an energy content pictured by the equation you wrote and thus this part of energy must be taken into account when writing the energy conservation law. For instance, suppose you have a particle at rest in your (inertial) reference ...

1

It is what it seems to be at first glance: matter intrinsically contains energy even if it is stationary and thus has no kinetic energy. This energy defined as $E=mc^2$ is called rest energy. All matter carry rest energy. Now, due to the large numerical value of c, even the lightest objects contain such immensely huge energy. But this is not energy that is ...

0

It means that this mass is available for transformation in some other kind of energy. It means that there is a type of energy that is only a function of a mass of a particle. There are processes in nature in which entire mass of a particle transforms in some other type of energy...like annihilation of particle and antiparticle.How to calculate this energy? ...

2

If a photon [is] massless then it must have no energy This is not the case. One way to think of mass is as nothing more than a convenient name for rest energy. Photons are indeed massless and thus have zero rest energy. This is not an issue because according to special relativity, they do not come with a rest frame. Please note that assuming we denote ...

0

No, a photon does not need to have mass to be able to interact with matter. In fact it is its energy which is important in interactions. For the photoelectric effect the incoming photon must contain enough energy to displace the electrons on the metal of the photodiode. The question of what are photons made of is a pretty deep and difficult question, one ...

1

When people claim that a photon is massless, they mean that a photon has zero rest mass. In special relativity, the formula for the energy of a particle with mass $m$ possessing a momentum $p$ is $$E = \sqrt{p^2c^2 + m^2c^4}$$ If we set $m = 0$ for a photon, we'll end up with $$E = pc$$ Here the momentum of a photon is described by quantum mechanics ...

1

Nuclear plants (and theoretically fusion plants) work with $e=mc^2$. For example: The mass of 2 Protons and 2 Neutrons is bigger than the mass of Helium, which consists of 2 Protons and 2 Neutrons. The difference is emitted as energy when Helium is made by the fusion of 2 Protons and 2 Neutrons (the actual reaction in the sun are a bit more complicated, ...

3

Particle--antiparticle annihilation events are direct evidence of the mass-energy correspondence. Michelson and Morley interferometric results support the absoluteness of the speed of light (or some more esoteric possible results), and building from that and the relativity principle you can arrive at the mass-energy correspondence rather indirectly.

2

According to Maxwell's theory of electromagnetism, a light pulse (or generic electromagnetic wave) carries momentum, which can be transferred to an absorbing surface hit by the pulse. This momentum transfer is known under the name 'radiation pressure'. Despite carrying momentum, light carries no mass. Yet a light pulse does carry energy. For a light pulse ...

-3

The total energy of a photon, the carrier of the electromagnetic force, is given by $E=hf$ where $h$ is Planck's constant and $f$ is the frequency of the light. So yes, if two EM waves have the same energy, they will have the same frequency and wavelength, meaning they have the same colour. Photons have no rest mass, but they do have a relativistic ...

0

this intrigues me since Einstein was wrong in his theory of relativity , first its a theory which by definition means "unproven hypothesis" and believe me its wrong. the math doesn't work. lets look at it I'm not going off half baked here iv'e really considered this and Einstein's theory is no longer a theory because of scientific rules to a theory demand ...

1

E=m*c**2 is not the defining equation of relativity. The theory is called special relativity and the equation is a derived part of the results of the theory. It is the result of Lorenz transformations on moving systems, which do take care of space and time in addition to energy.

1

In special relativity, mass / energy has no influence on spacetime. However, in general relativity, the curvature of spacetime is directly related to energy, or equivalently, mass. The Einstein field equation $$R_{\mu \nu} - {1 \over 2}g_{\mu \nu}\,R + g_{\mu \nu} \Lambda = {8 \pi G \over c^4} T_{\mu \nu}$$ includes the stress-energy tensor, which ...

1

I think this solves everything, it is slightly adapted from the wikipedia. "The closely related concept of matter conservation was found to hold good in chemistry to such high approximation that it failed only for the high energies treated by the later refinements of relativity theory, but otherwise remains useful and sufficiently accurate for most chemical ...

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

16

Of course, it does, since: $$\frac{\partial E}{\partial t} = \frac{\partial }{\partial t} \left(m \cdot c^2 \right)$$ Very little, though

2

Would the mass of burnt firewood be equal to the mass of firewood before burning? You won't get a good answer by simply looking at the "burnt firewood". The combustion is using oxygen from the air, and it is creating carbon dioxide and many volatilized materials that will disperse in the air. But we can imagine combustion happening in a box that is ...

0

If we add the energy and mass we will find that it adds to the total mass of the original wood. The total energy of the system is conserved, this will always be true. If you only look at the weird and burnt wood you will find a discrepancy. This is because some of the mass was lost in smoke. If you measured the mass of the smoke and the wood you will be in ...

-1

The mass of the original material equals the mass of the combustion product. You have to take the carbondioxide and the oxygen into account, not just wood and ash. The energy is stored chemicaly before released. Not every time energy is stored it is in form of additional mass. For the subject of mass loss have a look at weak interaction, Einstein is off ...

1

The law of the conservation of mass was superceded by the more general law of conservation of energy when it was realized that mass and energy were equivalent. Anyway, you are correct. The mass of the combustion products will always be less than the mass of the original materials. The difference being equivalent to the energy produced.

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