B Is E=mc^2 a Bidirectional Equation in Physics?

[saddlestone-man]

TL;DR

We know that matter can be converted into energy, but can energy be converted into matter?

Hello All

Does the most famous and elegant equation in physics imply bidirectionality in the way it describes the relationship between energy and mass?

There are many examples of matter being converted into energy (stars, bombs, reactors, etc), but can energy be converted into matter? At the point of the Big Bang, I presume this is what happened, but has it happened since, and has it been reproduced in the laboratory?

best regards ... Stef

 

[Orodruin]

Summary:: We know that matter can be converted into energy, but can energy be converted into matter?

Hello All

Does the most famous and elegant equation in physics imply bidirectionality in the way it describes the relationship between energy and mass?

There are many examples of matter being converted into energy (stars, bombs, reactors, etc), but can energy be converted into matter? At the point of the Big Bang, I presume this is what happened, but has it happened since, and has it been reproduced in the laboratory?

best regards ... Stef

There is no such thing as converting matter into energy. Energy is a property of matter, not a thing in and of itself. Energy may be converted between different forms of energy, including the energy related to the mass of an object. In the case of a nuclear reactor, some of the energy that was in the form of mass of the fuel is converted into kinetic energy of the fission products, which is then converted into thermal energy of the surrounding water, heating it up and using it to drive a generator that transforms it to electrical energy. But nowhere in the process was there a conversion of something to energy, the energy was the same all the time, just converted from one form to the other.

 

[PeroK]

Technically, energy is a property of matter. E.g. an object may have kinetic or potential energy. The equation $$E = mc^2$$ implies that an object at rest has a certain amount of energy depending on its mass, with $c^2$ as the conversion factor.

The total energy of a closed process is conserved, but not the total rest mass of the particles - which can increase or decrease. For example, a particle can decay into two particles with a total rest mass of less than the original. (Heuristically, rest mass has been converted to kinetic energy.) Alternatively, two particles in collision can produce a single particle of greater rest mass than the combined rest masses of the original particles. (Heuristically, kinetic energy has been converted to rest mass.)

So, yes, these processes work both ways.

PS I see I was beaten to it.

 

[Delta2]

implies that an object at rest has a certain amount of energy depending on its mass, with c2 as the conversion factor.

What form of energy is that, kinetic, potential, gravitational potential, electric potential, thermal, chemical or what exactly energy is it that the object at rest has.

 

[PeroK]

What form of energy is that, kinetic, potential, gravitational potential, electric potential, thermal, chemical or what exactly energy is it that the object at rest has.

Rest energy!

 

[Delta2]

Rest energy!

Seriously is this term scientifically established?

 

[DrStupid]

Seriously is this term scientifically established?

It doesn't matter as long as everybody knows what it means. If you don't like 'rest energy', how about internal energy?

 

[Delta2]

Ok but there is something I don't understand, where does the rest mass goes in the case of a nuclear reaction if we cannot say that it is converted to energy. @PeroK what do you mean when you say "heuristically" rest mass is converted to energy

 

[saddlestone-man]

Many thanks for the replies ... which don't quite address my question.

If the energy is released from an amount of matter by a nuclear explosion (or maybe more simply by nuclear decay), can the energy somehow be re-packed to form the original matter. I realize than tracking down all the released energy may be difficult, so how about any old energy you can collect being re-formed into matter?

best regards ... Stef

 

[Ibix]

Seriously is this term scientifically established?

It can be lots of different things - we don't have to care about the details. For example, a mirrored box full of photon gas has electromagnetic energy, elastic stresses, the binding energy of the atoms of the box, their intrinsic masses, and probably other things. And as photons are absorbed and reflected you may see shifts from one form to another. But for $E=mc^2$ we don't care about the details.

 

[Ibix]

can the energy somehow be re-packed to form the original matter.

Not exactly. Photon-photon interactions are incredibly hard to do. But you can certainly do stuff like take a couple of low mass particles and slam them together at high speed, and you may get high mass low speed particles out - energy to mass (at least in some senses). CERN does that all the time.

 

[Delta2]

It can be lots of different things - we don't have to care about the details. For example, a mirrored box full of photon gas has electromagnetic energy, elastic stresses, the binding energy of the atoms of the box, their intrinsic masses, and probably other things. And as photons are absorbed and reflected you may see shifts from one form to another. But for $E=mc^2$ we don't care about the details.

No Ibix, its another thing all these different forms of energy that are in the box, and another thing the "rest" energy that the box has BECAUSE it has rest mass.
Seriously this might prove to be one of the biggest misconceptions i have from the few i know about relativity. I thought E=mc^2 meant that if something of rest mass m is converted to some of the known forms of energy, then the energy is equal to mc^2.

 

[Ibix]

I thought E=mc^2 meant that if something of rest mass m is converted to some of the known forms of energy, then the energy is equal to mc^2.

It does. But the mass of the object before it is converted can be traces back to all the different forms of energy I mentioned (and probably some I forgot).

 

[Delta2]

But the mass of the object before it is converted can be traces back to all the different forms of energy I mentioned (and probably some I forgot).

The mass of the object before it is converted is rest mass and remains rest mass what it has to do with the other forms of energy?? It has to be converted first...

 

[saddlestone-man]

Another question about this equation:

Why did Einstein believe that its form was E=mc^2 and not say E=0.9mc^2, or any other constant in front of the righthand side? Any constant would maintain the dimensional accuracy, so why is it 1?

How long after the equation was published was there experimental evidence of sufficient accuracy to confirm that the constant was 1, and would such an experiment be a good way to determine c?

best regards ... Stef

 

[Ibix]

The mass of the object before it is converted is rest mass and remains rest mass what it has to do with the other forms of energy?? It has to be converted first...

I'm not sure I understand your problem.

A lump of lead has a certain rest mass, $m$, which can also be stated as having rest energy $E$, equal to $mc^2$. If I react it with an equal mass of anti-lead, it will be converted into photons. Photons have zero rest mass, but their four momenta add like any other and add to a four momentum which has zero momentum and energy $2mc^2$ - so the individual photons' energies add to $2mc^2$.

The rest mass/rest energy of the system is unchanged, but the masses of the components have gone from $2m$ to zero. None of this changes the fact that the mass of lead is largely traceable to the energy of the strong force binding the nuclei together.

 

[Ibix]

Why did Einstein believe that its form was E=mc^2 and not say E=0.9mc^2, or any other constant in front of the righthand side?

He proved it in a paper in 1905. Essentially he worked out the implications of his new physics on a moving sphere emitting radiation uniformly in its rest frame. Its momentum changes as it emits, which means either its mass changes or its velocity changes. Its velocity can't change because it can't be accelerating in its initial rest frame, so its mass must be changing. Actually doing the calculation comes out to the energy emitted being $c^2$ times the mass loss.

 

[SiennaTheGr8]

Mass is nothing but the sum of all the energy inside an object as measured in the object's rest frame. That's what Einstein showed. Weighing an object is measuring its rest-energy.

 

[Delta2]

My problem is that I don't find it right when we say a rest mass m has rest energy E=mc^2, rather i find it correct to say that rest mass m is equivalent to rest energy mc^2.

 

[DrStupid]

My problem is that I don't find it right when we say a rest mass m has rest energy E=mc^2, rather i find it correct to say that rest mass m is equivalent to rest energy mc^2.

I do agree. However, isn't that just semantics? I don't see a risk for misunderstandings.

 

[Delta2]

No it isn't just semantics. When we say for example a charge q has electric potential energy or a mass m has gravitational potential energy we don't mean that if the charge q is reduced and converted to energy it will gives us that much electric potential energy. But when we say a rest mass m has rest energy mc^2 we mean this thing that if the mass m is reduced it will give us that much energy (of one of the known forms).

 

[Dale]

If the energy is released from an amount of matter by a nuclear explosion (or maybe more simply by nuclear decay), can the energy somehow be re-packed to form the original matter.

In principle, yes. All of the reactions in quantum mechanics can go either direction.

 

[etotheipi]

People throw around the term 'energy' and forget that it is just a number associated with a system. If a Lagrangian possesses a translational symmetry $L(x^{\mu} + \epsilon G^{\mu}, \dot{x}^{\mu} + \epsilon \dot{G}^{\mu}) = L(x^{\mu}, \dot{x}^{\mu})$ then $\dfrac{d}{d\lambda} \left( \dfrac{\partial L}{\partial \dot{x}^{\mu}} G^{\mu} \right) = 0$ along the worldline of the particle. For translational symmetry $G^{\mu} = \delta^{\mu}_{{\mu}_0}$ and the conserved quantity is $\mathcal{Q}_{{\mu}_0} = mu_{{\mu}_0}$ where ${\mu}_0 \in \{ 0,1,2,3 \}$. Then in inertial co-ordinates you have $-mu_0 = mu^0 = E = \text{constant}$. [At rest $u^0 = 1$ and $E=m$.] Meanwhile letting $\mu_0 = 1,2,3$ results in momentum conservation.

Another way to see it: to find the Lagrangian of a free particle, the heuristic argument is that since timelike straight lines of $M^{1,3}$ achieve maximum proper time between two events then $S \propto (\tau_2 - \tau_1)$. The proportionality constant must be negative so that maximum proper time $\iff$ minimum of action, and factor of $m$ gives the proportionality constant dimensions of energy so $L = -m\sqrt{-g_{{\mu} {\nu}} \dot{x}^{\mu} \dot{x}^{\nu}}$. Of course in inertial co-ordinates $\dot{x}^0 = 1$ and $\dot{x}^i = v^i$ thus $L = -m\sqrt{1-\delta_{ij} v^i v^j}$. Write the energy function $h = \dfrac{\partial L}{\partial \mathbf{v}} \cdot \mathbf{v} - L = \dfrac{m}{\sqrt{1-\delta_{ij} v^i v^j}}$ which has at $h(v=0) = m$; it's just to say that the mass $m$ is the energy of a particle at rest.

 

[saddlestone-man]

My problem is that I don't find it right when we say a rest mass m has rest energy E=mc^2, rather i find it correct to say that rest mass m is equivalent to rest energy mc^2.

When a mass at rest is turned into its equivalent energy (say massless photons), where does the gravity go to?

 

[etotheipi]

Firstly, you must never say something like 'mass is turned into equivalent energy'. Maybe you have a particle initially at rest with energy $m$, which disintegrates into two photons of total energy $m$. The point is that the energy is preserved. Secondly, it's not clear to me what it means to say where does the gravity go?

 

[Delta2]

Firstly, you must never say something like 'mass is turned into equivalent energy'. Maybe you have a particle initially at rest with energy $m$, which disintegrates into two photons of total energy $m$. The point is that the energy is preserved. Secondly, it's not clear to me what it means to say where does the gravity go?

First, you use the word disintegrate, he uses the word turned into equivalent energy here it is just semantics e hehe.

Second I think he means the energy of the gravitational field that surrounds mass m (having mass m as the source). Is that included in the mc^2 or it is totally different thing? I don't know I am not expert in relativity.

 

[stevendaryl]

No it isn't just semantics. When we say for example a charge q has electric potential energy or a mass m has gravitational potential energy we don't mean that if the charge q is reduced and converted to energy it will gives us that much electric potential energy. But when we say a rest mass m has rest energy mc^2 we mean this thing that if the mass m is reduced it will give us that much energy (of one of the known forms).

I think that it’s better to think of mass of an object as the same thing as the energy of the object as measured in its rest frame, rather than thinking in terms of mass being converted into energy. It already is energy.

A proton, for example, is believed to be made up of three quarks. The mass of the proton is just the sum of all forms of energy of those quarks and their interactions, including kinetic energy. There is no “conversion” necessary (other than the historical fact that mass and energy were measured in different units).

 

[etotheipi]

Second I think he means the energy of the gravitational field that surrounds mass m (having mass m as the source). Is that included in the mc^2 or it is totally different thing? I don't know I am not expert in relativity.

It's definitely not included in the $mc^2$

You need to consider the four momentum of the matter + gravitational field. If the gravitational field is constant then the $P^0$ component is easy to write down,$$P^0 = - \frac{1}{c} \int \sqrt{-g}(T^1_1 + T^2_2 + T^3_3 -T^0_0) dV$$If the gravitational field is not uniform, the expression is a bit more complicated. You can take a look at Landau & Lifshitz volume 2 for details, specifically chapter §101.

 

[stevendaryl]

Here’s an illustration of “converting mass into energy” that maybe explains why I consider that phrase misleading:

Suppose you have an object that has a certain mass, $M$, and it is at rest. What you don’t know is that it is actually two objects, each of mass $m$ connected by a rope, and they are orbiting around each other. The mass of the composite object counts both the masses of the parts, plus the kinetic energy of the parts (plus whatever energy is associated with the rope).

If somebody cuts the rope, the two smaller objects will fly apart. You will find that the two masses don’t add up to the mass of the original object. Ignoring the energy of the rope, for simplicity, you would find that $M c^2 \approx 2 mc^2 + KE$, where $KE$ is the sums of the kinetic energies of the two smaller objects.

So you might say: “Some of the mass of the original object was converted into kinetic energy.” That’s sort of right, and sort of wrong. The original mass already counted the kinetic energy of the parts. No conversion happened.

 

[hutchphd]

One should also mention that the supposed "conversion" of mass to energy is not a mysterious feature limited to quarks or nuclear reactions. It is true (or not) for all processes: the nuclear ones are where the energies get large enough to beg the question "where the heck did all that energy come from?"