# Mass in terms of energy

Given the relationships between mass and energy in relativity, and given that particles with mass can be created given energy over the threshold energy, and vice-versa, can we say that mass is simply an extremely dense form of energy? Or is there a deceptive parallel between the two?

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Mass, like energy, is just a physical quantity. A number with units. Special relativity says they are related but that's all there is to it. So no, it doesn't make any sense to say what you are saying. –  Marek Jun 8 '11 at 8:03

I'll offer my take on this, which is very specific to me. As an alternative to the vague notion that matter and energy are interchangeable I claim the following

• All energy has mass
• All mass has energy

To specify further, I will describe a process commonly thought as changing matter into energy, then a process commonly thought of as changing energy into mass, then show that neither of them "loose" either matter of energy in favor of creating the other.

Start off with a nuclear reactor. During operation the nuclear energy is changed into heat and electrical energy. Say that this is done in a system that is thermally isolated. It doesn't matter specifically what form that energy takes, because should it go into batteries, the weight of the batteries would increase, should be be stored as heat, the weight of the medium holding the heat increases. This energy exists either in chemical bonds and kinetic motion respectively. Both increase the mass of the system.

For changing energy into mass, we can look at particle creation as you mention. In order to create the particle, that energy had to exist previously, and sure enough, whatever reservoir held the energy before it was used experienced a decrease in mass corresponding with the movement of "energy".

A recent question I wrested with was Explain how (or if) a box full of photons would weigh more due to massless photons. Even though photons are massless, if they are somehow confined they will increase the mass of the system they are a part of. This is because no matter what transitions occur, the measured mass within a boundary that doesn't exchange mass or energy will remain constant. Likewise, the energy of that system will remain constant. This is in spite of the fact that matter-energy transitions are apparently occurring.

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sorry, but not all energy has mass: take a single photon. Its mass is zero but it has energy. Mass is the "length" of the four vector, and in a system of particles, one sums the energies, one sums the momenta, and takes the dot product of the four vector to find the mass. The relationship is not linear. en.wikipedia.org/wiki/Four-vector .-1 –  anna v Jun 8 '11 at 6:15
@anna Although semantics are semantics you misunderstand what I wrote. I never said "rest mass", not once. Every reference to mass I made was relativistic mass, sorry if this wasn't clear. I think we've agreed other places the photons exhibit relativistic mass. The four vector is peripheral to this topic, my post is about systems thinking. You also don't help to explain the fact that a majority of "rest mass" is just relativistic mass of something with a smaller rest mass on the quantum level. –  Alan Rominger Jun 8 '11 at 13:00
@zassounotsukushi A single photon has both rest mass and relativistic mass 0, and it has energy. If you have more than one photon, their four vectors can add up and create a relativistic non zero mass. The four vector is at the centre of the topic. –  anna v Jun 8 '11 at 16:28
@anna Ok, requiring a 4 vector sum is perfectly fine. But a photon, or a group of photons, could be balanced by a recoil momentum from ordinary matter too, like the relativistic rocket problem. The only point I want agreement on is that an objectively defined "system" maintains both its total value of mass and energy at all times, and that these two values are, in fact, the same value. –  Alan Rominger Jun 8 '11 at 16:39
I know this is kind of late, but the common convention here (and among most physicists, as far as I know) is that the word "mass" refers to rest mass, not relativistic mass, unless otherwise specified. –  David Z Jun 22 '11 at 20:39

The concept of a four vector is an extrapolation to an extra dimension of the three vector.

Say that you have two three vectors, the three momenta of two particles. Each vector has a length: p1, p2. When you add them vectorially, the length of the vectors do not add linearly. It will depend at what angle they are added. For example, if p1=p2 and the addition is head on, the length of the resultant vector will be 0.

It is the same with four vectors where the mass is the measure, what you get from the dot product, of the four vector. If you read the link above you will see that a system of four vectors will have masses which will depend on the values of the summed energies and the summed three vector momenta, in no way linear. When $E^2=P^2$ (in a system where $c=1$) the mass is zero, as happens with the photons.

When you have a system of massive particles the lowest mass you can have is the sum of the rest masses, if all particles are at rest. These are the masses that enter the $E=mc^2$ (in a system where $c=1$) which says that masses have a minimum energy content. It comes naturally in the four vector representation, since at rest, the three momentum vectors are zero, and only energy exists, thus the "length" of this four vector, which is its mass, is also its energy.

We can say that mass is a form of energy, depending on the coordinate system where the particles are studied. There is no density concept, just "length".

Energy is a four vector component. Mass is a scalar identifying the length of the four vector.

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As it pertains to energy, it seems the crux of your argument would be the trigonometric interpretation of $E=(pc)^2+(m_0 c^2)^2$. –  Alan Rominger Jun 8 '11 at 13:06
Sorry, I'm a little confused. Mass is the length of which four vector? In an object's frame of reference p is 0 all right, and if we take the length of four-momentum, we have $\gamma m c^2$, where $\gamma$ is 1, and since $c=1$ for simplification, you can say $E = m$. But neither hold if we're in a reference frame where the object's 3-momentum is nonzero, or if it is a massless particle like the photon, in which case E != m (rest mass). I don't think I fully understand you, can you elaborate further? Thank you! (Also, I've learned about four vectors already, but thanks anyways for the link.) –  wrongusername Jun 9 '11 at 16:38
@Zassounotsukushi should that be a $E^2$? –  wrongusername Jun 9 '11 at 16:48
@wrongusername yes it should be. good eye. –  Alan Rominger Jun 9 '11 at 16:56
Each vector has a characteristic scalar "length" given by the square root of the dot product of the squares of the vector components. In this case there is an imaginary i in front of the momentum components in the definition of the dot product and that is how one gets E2-p2=m2 in units c=1. Vector addition ends up in a single vector and its "length" is the effective mass of the system whose vectors were added. This length is invariant in linear transformations. The single photon's rest mass is E2-p**2 =0 because both E=hnu and p=hnu (see en.wikipedia.org/wiki/Photon ) –  anna v Jun 9 '11 at 17:50

I don't really know what your question actually addresses how ever my guess is that you have a problem relating the energy to mass but you can say that energy IS mass there is no deference between the two except how you perceive them and in fact most subatomic particles have their mass in energy units and the way you would be able to understand it is by studying quarks now quarks get their mass from the Higgs field and that is a very very very small number and so it isnt the main reason we have mass so what's the main reason you may ask?

Well the main reason is that quarks hate to be alone they just cant stay alone at all at they follow the white color rule which is a rule that helps us study quarks and it simply says that quarks like the color white how ever they are dont have that color instead they are blue ,red,and so on... and because of that they come in pairs and threes and there are even fours but how ever for simplistic explanation we'll talk about two quark particles and to follow the whit rule it should have the green quark and with it the magenta (these two together=white) so these two bind together using force (energy) and because force doesn't just go to no where so where does it go? well you guessed it, it is translated into what we call mass and therefore mass is energy and energy is mass.

I hope that this simplified model actually helps you understand the way it is.

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