B The relationship between mass, light and energy

[alan123hk]

I found that I had an inaccurate understanding of their relationship earlier. After rethinking, I try to express their relationship in a simple and understandable way

Nuclear reactions convert energy from form of mass to form of light, and both energy and momentum are conserved during the transition.
Although the mass is reduced in this process, it does not convert the mass into energy. This is because when energy exists in the form of mass, mass and energy are equivalent, and energy is already associated with mass. They exist at the same time, and mass can be understood as a measure of the energy associated with it.

Please correct me if the above description is wrong.

 

[Delta2]

My opinion is that mass is a form of energy (we seem to agree on this, don't know if we should call it massial or massitic energy) and it is just that our human senses perceive it different from other forms of energy, e.g infrared photons energy which we perceive as heat.

 

[Drakkith]

Although the mass is reduced in this process, it does not convert the mass into energy. This is because when energy exists in the form of mass, mass and energy are equivalent, and energy is already associated with mass.

The issue is that the mass of the system isn't reduced. Put a uranium atom in a box and measure the mass of the box before and after the atom has decayed. You'll find that as long as nothing has left the box, including any photons, the mass is the same. Energy isn't a form of mass and the two are not equivalent in the sense of their definitions and uses. They are different but related things. The connection is that if you move X amount of mass into or out of a system, you also move Y amount of energy, and vice versa. The 'equivalence' talked about is a mathematical relationship given by the equation $e=mc^2$.

 

[Delta2]

Energy isn't a form of mass and the two are not equivalent in the sense of their definitions and uses

In the realm of mainstream physics yes but in my opinion, mass, electric charge, the color in quantum chromodynamics are all forms of energy. It is not a coincidence that we talk about conservation of mass (in classical physics), conservation of electric charge and conservation of energy.

 

[alan123hk]

don't know if we should call it massial or massitic energy) and it is just that our human senses perceive it different from other forms of energy, e.g infrared photons energy which we perceive as heat.

Aside from how people feel about them, there seem to be some differences.
Infrared photons are massless, and heat refers to the motion of particles with mass.

 

[alan123hk]

Energy isn't a form of mass

If I say that light (photons) is a form of energy, mass is also a form of energy, am I right?

The issue is that the mass of the system isn't reduced. Put a uranium atom in a box and measure the mass of the box before and after the atom has decayed. You'll find that as long as nothing has left the box, including any photons, the mass is the same

So can I say that when the uranium atom decays, some massless particles are released, and the total mass of the system is reduced when these particles have not yet turned back into matter?

 

[Ibix]

So can I say that when the uranium atom decays, some massless particles are released, and the total mass of the system is reduced when these particles have not yet turned back into matter?

Fundamentally, the problem is that "mass" is slightly ambiguous here. In relativity mass is not additive, so the mass of A and B is not necessarily the mass of A plus the mass of B. So when you talk of the "mass of the system", do you mean the sum of the masses of the components or the mass you'd measure if you put the thing on a scale? Typically you'd mean the latter (as @Drakkith does in the post to which you were responding), but you seem to be thinking of the former.

Mathematically, what happens is that you can add the four momenta of the components and that total is conserved. So if Uranium alpha-decays to Thorium emitting a photon then considering their four momenta, $p$, you find $$\begin{eqnarray*}
p_U&=&p_{Th}+p_\alpha+p_\gamma\\
|p_U|&=&|p_{Th}+p_\alpha+p_\gamma|\\
&\neq&|p_{Th}|+|p_\alpha|+|p_\gamma|
\end{eqnarray*}$$because the modulus of a sum of vectors is not in general equal to the sum of the moduli. Remember that $|p|=m$ (give or take a factor of $c$) and you can see that the total mass is conserved but the component masses are not. (Edit: it's been a while since I studied nuclear physics and the decay $\mathrm{U}\rightarrow \mathrm{Th}+ \alpha+\gamma$ may not be right - the relativistic point stands.)

So when people say something like "mass has been converted to energy" they mean something like "the sum of the rest masses of the particles has decreased and, in the rest frame of the original nuclide, some of the decay products have kinetic energy". Less reliable sources will tell you that "photons are pure energy", but that's Star Trek science and should be ignored. However, @Drakkith is correct to say that if you have uranium atoms in an insulated box on a weighing scale you will not be able to detect their decay from any change in the scale reading because the system mass does not change (at least, not until heat starts to leak through the insulation).

 

[alan123hk]

@Ibix Thanks for your reply, it was informative and gave me multiple directions for further thinking.

 

[Drakkith]

In the realm of mainstream physics yes but in my opinion, mass, electric charge, the color in quantum chromodynamics are all forms of energy. It is not a coincidence that we talk about conservation of mass (in classical physics), conservation of electric charge and conservation of energy.

Why energy? Why wouldn't you say all of these forms of mass?

If I say that light (photons) is a form of energy, mass is also a form of energy, am I right?

I don't like the phrase 'form of energy'. Energy is a quantity that we can measure and calculate, but beyond that it is ephemeral, formless, and something that's only found in the math. Should a single mathematical relationship linking two things somehow make one a form of the other? Should we say momentum is a form of velocity, since moving objects always have both a velocity and momentum? Or that momentum and velocity are forms of mass? Or forms of energy? There are mathematical equations linking all of these things to each other, yet we don't usually call one of them a form of another except when it comes to mass and energy.

Perhaps I am wrong and the relationship between mass and energy is 'special' in some way that only a physicist could explain. But, for now, I'd simply say that light has energy, not that it is a form of energy.

Honestly it's not really about being right or wrong, but being accurate. It's 'right' to say that light is a form of energy in the sense that it's a simplification of a more complicated and nuanced topic. So if you take anything away from this thread, let it be the knowledge that there isn't a single, unambiguous answer and almost no one is going to try to call you out for saying that light or mass is a form of energy.

So can I say that when the uranium atom decays, some massless particles are released, and the total mass of the system is reduced when these particles have not yet turned back into matter?

You need to define what your system is before and after the decay event. If we define our system to be 'everything released from the decay, including massless particles", then no, the mass is not reduced. It is the same. If we say our system after the decay is only particles with mass, then yes, the mass has been reduced.

If this is confusing then remember that a photon can carry energy (and thus mass) between one system and another. A dust particle in space is more massive (has more mass) after absorbing a photon than before, despite the photon having no mass.

 

[Delta2]

Why energy? Why wouldn't you say all of these forms of mass?

Well tbh I don't know, Energy is a more general concept in physics (potential energy, kinetic energy, heat energy, chemical energy, EM field energy e.t.c.) than mass.

 

[vanhees71]

Aside from how people feel about them, there seem to be some differences.
Infrared photons are massless, and heat refers to the motion of particles with mass.

Rather than heat, I'd say internal energy (in the thermodynamic sense). Thermal radiation ("Planck radiation") also has internal energy although it's described by a massless (quantum) field, the electromagnetic field.

As has been stressed several times in this and other recent threads in this forum, from a relativistic point of view, the mass term in the Hamilton density of the fields is just one part of the total field energy. Both, inertia and the sources of the gravitational field, are not due to the mass but due to the energy-momentum-stress distribution of the fields describing matter and radiation.

 

[Dale]

In the realm of mainstream physics yes but in my opinion, mass, electric charge, the color in quantum chromodynamics are all forms of energy.

This forum is about “the realm of mainstream physics”, not your personal opinion.

 

[alan123hk]

If we define our system to be 'everything released from the decay, including massless particles", then no, the mass is not reduced.

a photon can carry energy (and thus mass) between one system and another. A dust particle in space is more massive (has more mass) after absorbing a photon than before, despite the photon having no mass.

Thanks for explaining the point, I seem to have started to understand something.

I now think the first case is preferable, that is "Define our system to be 'everything released from the decay, including massless particles", then no, the mass is not reduced"

This is consistent with the special theory of relativity that the rest mass of the system is invariant and conserved in the four vectors of energy and momentum.

 

[Dale]

Nuclear reactions convert energy from form of mass to form of light, and both energy and momentum are conserved during the transition.
Although the mass is reduced in this process, it does not convert the mass into energy.

Nuclear reactions can be complicated, so let me walk you through a simpler reaction: the annihilation of a positron and an electron. I will use units where $c=1$ and where mass, energy, and momentum are all measured in units of $\mathrm{eV}$.

Energy and momentum form a four-vector called the four-momentum: $\mathbf{P}=(E,\vec p)$. Then the mass of an isolated system is $m^2 = E^2-p^2$.

If the velocities of the electron and positron are negligible then before the reaction we have the four-momentum of the electron is $\mathbf{P_-}=(511,0)\mathrm{\ keV}$ and the four-momentum of the positron is $\mathbf{P_+}=(511,0)\mathrm{\ keV}$. Before the reaction the mass of the system is $1022 \mathrm{\ keV}$ and the mass of each lepton is $511 \mathrm{\ keV}$.

Now, after the reaction energy and momentum is conserved, so the momenta of the resulting photons must be equal and opposite, and since each photon is individually massless each has $E=|\vec p|$. So the four momenta of the two photons are $\mathbf{P_A}=(511,511) \mathrm{\ keV}$ and $\mathbf{P_B}=(511,-511) \mathrm{\ keV}$. So the system mass remains $1022 \mathrm{\ keV}$, but the sum of the masses of the parts of the system is $0 \mathrm{\ keV}$

So it is not entirely accurate to say that the mass of the system is reduced. The sum of the masses of the parts of the system is reduced. The mass of a system is always greater than or equal to the sum of the masses of the parts of the system. This is basically a version of the triangle inequality.

 

[hutchphd]

Nuclear reactions convert energy from form of mass to form of light

Why is it the fashion to invoke "nuclear energy" to discuss this topic?? It is true for ordinary chemical reactions and every other manipulation of energy that the books must balance.

 

[alan123hk]

let me walk you through a simpler reaction

Thank you very much for your detailed explanation. What you describe is the core problem that I have been confused about. I think you have solved my doubts with irrefutable logical relationships and clear mathematical equations.

 

[alan123hk]

The most amazing and wonderful things are as follows.

Although each photon is individually massless , but the total energy and momentum of the system after the reaction is $\mathbf{P_A}+\mathbf{P_B}=(1022,0)\mathbf{ keV}$. Now the energy of the system is $1022~$and the momentum of the system is $0$, then the system mass has no change as follows..
$$m^2 = E^2-p^2~~\Rightarrow~~m=E~(c=1)~~\Rightarrow~~m=1022 \mathrm{\ keV}$$This is equal to the mass of one electron and one positron added together. Also note that this is calculated from the definition of the system and equations, although the sum of the masses of the parts of the system does decrease.

 

[alan123hk]

A very interesting and informative video, although I don't fully understand it as it still seems a bit contradictory to me.

The Real Meaning of E=mc²​

 

[jbriggs444]

A very interesting and informative video

I was favorably impressed by the video. He had appropriate weasel words in most the places where he was oversimplifying. He used the modern notion of invariant mass and nimbly refrained from calling it that. He got into the notion that mass is not additive and went in depth into examples of both mass surplus and mass deficit. And he pointed out that mass is a property.

I was not fond of the final question since the idea of gravitational potential energy in the context of Special Relativity is something that does not really fly.

Listening for errors, the only one I spotted was the implicit use of the Bohr model of the atom. But it wouldn't be fair to call him out for that.

 

[vanhees71]

Well, afaik there is no clear notion of "gravitational potential energy" within General Relativity either. This is a very tough problem, already known since the very beginning of GR. It lead to Noether's famous work about symmetries and conservation laws of 1918.

 

[alan123hk]

I'm a bit lost, there seems to be some contradiction here, can we perform additions of different masses in a system? Does this mean we can perform additions of different masses if they represent different energies? Or something else?

 

[jbriggs444]

I'm a bit lost, there seems to be some contradiction here, can we perform additions of different masses in a system? Does this mean we can perform additions of different masses if they represent different energies?

You can't add masses. But as long as you stick to a single reference frame, you can add energies.

You can (almost always) find a reference frame in which the system of interest has zero momentum. You can call it the system's rest frame, but "center of momentum" frame is more accurate.

An object's mass is equal to its total energy (divided by c squared) in its center of momentum frame.

Energy is not invariant. It changes depending on the reference frame that you choose. A baseball considered in its rest frame has less total energy than a baseball considered in a different frame of reference. In any other frame, it has non-zero kinetic energy.

If you divide a system into pieces, the mass of each piece will be equal to the energy of that piece in the center of momentum frame for that piece. If you add the energies... you've just tried to add non-invariant quantities drawn from a bunch of different reference frames. That's like adding apples, bananas and pears. The total is not meaningful.

If you are going to add non-invariant quantities, you have to pick a reference frame first.

Edit:

There is a corner case where the definition of "mass" as a system's total energy in its rest frame fails: a massless particle moving at the speed of light (for instance, a pulse in a well collimated light beam). Such an entity has no center of momentum frame, so the definition fails.

There is an alternate definition of mass that succeeds: $$m^2 = \frac{E^2}{c^4} - \frac{p^2}{c^2}$$or$$m=\sqrt{\frac{E^2}{c^4}-\frac{p^2}{c^2}}$$If $m=0$ then $E=pc$. And vice versa.

If you chase this down, you can motivate it with four-vectors. Take a system and pick a reference frame. Any reference frame. Write down the three components of the system momentum and add a fourth component for its energy. This is the system's energy-momentum four-vector. The components of this vector will not be invariant. They will obviously depend on the choice of reference frame.

Use units where c=1 and take the magnitude of the energy-momentum four-vector using a "magnitude" rule that treats the contribution of the momentum as negative and energy as positive.$$|\vec{V}| = \sqrt{{V_e}^2 - {V_x}^2 - {V_y}^2 - {V_z}^2}$$By no coincidence, this matches the formula for mass above.

This number will turn out to be independent of reference frame. No matter what frame you choose, it will be the same number.

Mass is the invariant magnitude of an object's energy-momentum four-vector.

 

[alan123hk]

You can't add masses. But as long as you stick to a single reference frame, you can add energies.

Thank you for your reply pointing out basic facts and how to avoid misunderstanding.

I am trying to understand it in the simplest possible way.

In a frame of reference apparently only the masses of stationary objects can be added together.

$E_{totat}= E_1+E_2=\sqrt{ {P_1}^2c^2+{M_{01}}^2c^4} +\sqrt{ {P_2}^2c^2+{M_{02}}^2c^4} = M_{01} c^2+M_{02}c^2~~$, when the momentums of $M_{01}$ and $M_{02}~$ are both equal zeros, namely $P_1=0,P_2=0$

If the objects are moving,
$E_{totat}=E_1+E_2=\sqrt{ {P_1}^2c^2+{M_{01}}^2c^4} +\sqrt{ {P_2}^2c^2+{M_{02}}^2c^4} \neq \sqrt{ {(P_1+P_2)}^2c^2+{(M_{01}+M_{02)}}^2c^4}$
So obviously we can't add their masses together because the result is meaningless.

I hope the above explanation is correct.

 

[davenn]

In the realm of mainstream physics yes but in my opinion, ...

You keep saying " my opinion "
Sorry to be straight forward, but, your opinion is not relevant and should truly be left at the door

Physics and the understood law "laws" are what are relevant and that's what the Physics Forums deals with