Hot answers tagged mass-energy
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 ...
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.
11
$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 ...
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
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 ...
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$$
Now, factoring out a common $m^2 c^4$ from both terms on the RHS in anticipation of the ...
7
As you may know, photons do not have mass.
Relating relativistic momentum and relativistic energy, we get:
$E^2 = p^2c^2+(mc^2)^2$.
where $E$ is energy, $p$ is momentum, $m$ is mass and $c$ is the speed of light.
As mass is zero, $E=pc$.
Now, we know that $E=hf$. Then we get the momentum for photon.
Note that there is a term called effective inertial ...
7
This equation is incredibly generic and describes many phenomena outside of nuclear phenomena.
For example: place the the following setup in a box (a spring and some bars) and weigh them:
.
Now, loosen the spring and repeat. You should measure a smaller mass because you've removed some of the energy.
In reality you couldn't possibly measure the ...
6
If I'm reading rightly, I think your main question is:
Why does only a small percentage of rest mass turn into energy [even for fusion]?
It's because the universe is very strict about a certain small set of conservation rules, and certain combinations of these rules make ordinary matter extremely stable. Exactly why these rules are so strictly observed ...
6
Yes, everything generates a gravitational field, whether it is massive or massless like a photon.
The source of the gravitational field is an object called the stress-energy tensor. This is normally written as a 4 x 4 symmetric matrix, and the top left entry is the energy density. Note that mass does not appear at all. We convert mass to energy by ...
5
It is the convention of setting the velocity of light $c=1$ that allows for this, the natural units, otherwise it is $\mathrm{GeV}/c^2$
The rest mass energy connection $$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$ ...
4
This is a common question that applies to chemical reactions, batteries, and other common forms of energy conversion. The short answer is yes, but of course the change is negligible.
In a chemical reaction you have a set of reactants and a set of products. If you were to take the mass of the reactants and sum them up, you would find them to be more than ...
4
Yes.
I feel like there should be more of an explanation, but it's pretty straightforward. A blackbody absorbing energy will increase in mass. The absolute amount of increase is pretty miniscule, but it is not zero.
Since you ask about an object that does not also radiate energy, a blackhole might be a decent analogy. So, does a blackhole increase in ...
4
Einstein's equation doesn't have a "proof" because it's not a mathematical theorem. It's a physical theory that is overwhelmingly supported by experimental data. So you could say that the "proof" is in the mountains of experimental results that agree with the theory.
To understand Einstein's motivation for developing the theory of relativity, as well as ...
4
The more general equation is: $$E=\sqrt {(mc^2)^2+(pc)^2}$$
where p is momentum
$E=mc^2$ is a special case when momentum of system is zero. This popular equation says following things:
Energy has all properties of mass. It means, energy also posses momentum. And, it can be influenced by gravity. It can create gravity, too.
Mass has all properties of ...
4
The relation $E=mc^2$ only works for particles at rest, which is evidently not the case for photons. In the general case, the relation is
$$E^2=m^2c^4+p^2c^2$$
for a particle with momentum $p$. (Note, though that the momentum is not necessarily $p=mv$ as in the newtonian case! See for instance If photons have no mass, how can they have momentum?)
For a ...
4
Yes. Matter (assuming it has mass) can be created from energy using the mass-energy equivalence. And, hence this principle got its name. For instance, take pair production where an energetic photon can create an electron-positron pair.
4
Your symbolic manipulations are correct, but the relations you write down do not properly describe Newton's second law in the context of special relativity.
In the context of special relativity, the relativistic momentum of a particle is defined as
$$
\mathbf p = \gamma m \mathbf v, \qquad \gamma = (1-\mathbf v^2/c^2)^{-1/2}
$$
Using this definition, ...
3
No, the room will stay the same weight.
I'm guessing that your question is really about the process of burning fat to create energy. You mention that the man loses water, but of course this just condenses on the walls of the room so no water is lost.
Strictly speaking it isn't fat that the man is burning to create the energy. Fat is used for energy storage ...
3
According to Special Relativity the relativistic energy for a particle is: $E^2= m^2c^4+p^2c^2$
The invariant quantity under relativistic transformations is the rest mass $m$ of the particle.
For a photon $m=0$
Using some simple algebra it is found $E=pc$ for a photon.
You will see this preserves the frequency and energy relationship.
The error in the ...
3
While it is true that the mass will increase your requirement of perfect absorption without radiation cannot be satisfied so easily. Total absorption would require a black surface at 0 K. Every surface would, according to the Planck's law and the Stefan Boltzmann law radiate thermal energy with $\propto T^4$. This limits the temperature rise to the surface ...
3
From your question and your comments on Google+, it appears you think there is a problem with having the increased energy stored in the gravitational field since the gravitational field is really just curved space-time. That is not a problem, curved space time does have an energy density and in fact can cause additional curvature of space time. That is why ...
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