10
$\begingroup$

The mass of a hydrogen is less than its constituent parts(proton/electron). The explanation given for this is the following: Youtube

For hydrogen, $m = m_{components} + m_{extra}$ where we can write $m_{extra} = E/c^2$ where $E$ is the PE and KE of the proton/electron interaction. It turns out while KE > 0, PE is much more < 0, so their sum becomes negative, so $m_{extra} < 0$.

What I have a difficulty to see is how PE can be negative. I know, if you treat that when proton and electron are infinity far away and we treat PE=0 for that case, then the more they get closer, PE decreases and becomes negative, but for our case, this seems to me no proof at all. I could just treat $PE=infinity$ when proton/electron are far away from each other and that way, in hydrogen, PE wouldn't be negative.

I just can't get how this proof is solid and how negative energy can exist in a way that it can decrease the mass of the hydrogen atom.

$\endgroup$
2
  • 5
    $\begingroup$ "I could just treat PE=infinity when proton/electron are far away..." No, you can't, since infinity is not a number. However, you certainly are welcome to set PE equal to any fixed number you would like when the proton/electron "are far away." (But you have to stick with the same convention going forward.) For example you could treat PE=42 Joules when the proton and electron are far away. The potential energy of the bound state would then be some number less than 42. (Just like it is some number less than zero in the usual treatment.) $\endgroup$
    – hft
    Jun 28, 2023 at 1:10
  • 2
    $\begingroup$ I don't think this video explains it well. It's not "because PE can be negative". What's negative is the change in energy ("negative" change just means that it was released when the atom formed - it's about process direction). It's just that you can set the zero level so that you don't have to deal with differences. $\endgroup$ Jun 28, 2023 at 21:28

5 Answers 5

24
$\begingroup$

In fact, there's a sense in which pretty much all Potential Energy is negative (for an attractive force).

But it's better to avoid thinking of positive and negative energy, and focus on thinking of higher and lower energy states, regardless of sign.

If you have a ball at the lip of a bowl-like valley and push it in, it will roll to the bottom, trading its potential energy for kinetic energy. If there are no losses due to friction or collisions, it will simply roll up the opposite side of the bowl, come all the way back, and roll away from the valley back onto flat ground. If, however, while inside the bowl the ball knocks into another ball and starts it moving, it has lost some of its energy by transferring it to the other ball, and the original ball will no longer have enough energy to escape. It will continue oscillating back and forth in the bowl forever.

This is exactly what happens to the electron and proton in your example. If a free electron and free proton meet without shedding any of their kinetic energy, they will simply bounce apart again and continue on their way. But sometimes when they meet, they release a photon that carries away some energy. Now the electron is trapped in the bowl so to speak of the proton's potential energy well, and the pair becomes an atom. This process is known as Photorecombination. The energy the electron had to lose via the photon to become trapped (and the same energy it would have to gain to break free again) is somewhat unhelpfully known as the binding energy of the hydrogen atom.

Now, because $E=mc^2$, the free electron and free proton have a combined energy (mass), which is higher than the total energy (mass) of the hydrogen atom when they are bound, because some energy was carried away by the photon. This is why 1 atom of hydrogen has lower mass than the free electron plus the free proton. There is no "negative energy" or "binding energy" at play. It is simply a lack of energy in the system in that state, compared to what it would have in a different state.

Note: there is a very subtle but important point here that I've glossed over – how is the rest energy of the free electron or proton defined in the first place? This is discussed very well here.

$\endgroup$
9
$\begingroup$

You can't just set PE = infinity when the electron and proton are far away because infinity is not a number. You need to set some finite number instead.

Let's say you take PE = 1 million when the electron and proton are far away. This works, but then you need to recalibrate everything, because it implies that an electron with 1 J of kinetic energy will not escape the proton (it will, since the ionization energy of hydrogen is only 13.6 eV). What you'd need to do is compare your 1 J of kinetic energy against the potential energy recalibrated so that the energy at infinity is the energy that's left when the electron escapes. That would make your hydrogen ground state have an energy of (1 million J - 13.6eV) joules. Which would still indicate you lose potential energy when the electron travels towards the proton from infinitely far away. Which would mean that the mass will decrease.

Setting PE = 0 at infinitely far away is a convention. You could set it to some other number, but it won't change the physics.

$\endgroup$
2
  • $\begingroup$ This comes down to the idea that if electron escapes nucleus in hydrogen, its PE is 0(by convention), but even if it escaped, that doesn't mean they're far away from each other, so potential energy of nucleus still must be attracting electron. but you say that we have PE = 0 for electron when it's infinitely far away. How can infinitely far away and escaping mean the same thing ? escaping doesn't mean they are separated now by infinitey. The distance is pretty much not big. $\endgroup$
    – Zaza Orji
    Jun 28, 2023 at 20:07
  • $\begingroup$ @Zaza the PE asymptotically approaches zero. Increasing the distance to an infinite value will only increase the PE to zero (from a negative value). That is not the same as the PE approaching infinity (or negative infinity). No matter what the distance is, the PE remains finite. As far as "escaping," you can set an arbitrary distance when the electron PE is at $-0.1$ eV, $-0.001$ eV, $-0.00000001$ eV (as you like), and say at that distance the electron has "basically escaped" $\endgroup$
    – RC_23
    Jun 28, 2023 at 21:14
7
$\begingroup$

What I have a difficulty to see is how PE can be negative.

Recall how Potential Energy (PE, here symbolized by $U$) is defined.

A potential energy $U$ is defined implicitly in terms of a force $\vec F$ via the equation: $$ \vec F(\vec x) = -\vec\nabla U(\vec x)\;. \tag{1} $$

Clearly, the potential energy is not defined uniquely via Eq. (1). Any other potential energy function that only differs from the first by a constant will result in the same physics.

For example, let: $$ V(\vec x) = U(\vec x) + C\;, $$ where $C$ is a constant (independent of spatial position).

Then we also have: $$ \vec F = -\vec \nabla V\;. $$

Therefore, you are free to add an arbitrary constant to your definition of potential energy in order to make the math easier.

Clearly, you can choose to make the potential energy at any given point positive, or negative, or zero, by definition. Thus, the sign of the potential energy is not necessarily of any significance.


Recall that the electrostatic force, due to a fixed particle of charge $Q$ at the origin, on a test particle of charge $q$ at position $\vec x$ is given by: $$ \vec F = \frac{Qq\vec x}{4\pi\epsilon_0|\vec x|^3}\;. $$

You are free to rewrite this force field $\vec F$ in terms of a potential energy field $U$, where $$ U(\vec x) = \frac{Qq}{4\pi\epsilon_0|\vec x|}\;, $$ which has the convenient property of approaching zero as $|\vec x|$ approaches infinity.

However, if you so please, you can also write the potential energy as: $$ V = \frac{Qq}{4\pi\epsilon_0|\vec x|} + C\;, $$ where $C$ is any constant value you would like it to be.


OK, so let's say we hate negative numbers for some reason and we want to try and make sure we never have to see any negative numbers by imposing a large (and physically irrelevant) C value.

Let's take $C=1000000$ so we can write: $$ V = \frac{Qq}{4\pi\epsilon_0|\vec x|} + 1000000 $$

Classically, the total energy of a proton/electron system when the particles are very far apart and at rest is: $$ E_1 = m_p c^2 + m_e c^2 + 1000000\;. $$

Classically, the total of a proton/electron system when the particles are a distance $d$ apart and at rest is: $$ E_2 = m_p c^2 + m_e c^2 + 1000000 - \frac{|e|^2}{4\pi\epsilon_0 d}\;. $$

Even though the sign of each individual energy ($E_1$ and $E_2$) might be positive, the energy difference is still negative, regardless of the value of C: $$ E_2 - E_1 = - \frac{|e|^2}{4\pi\epsilon_0 d} $$

$\endgroup$
1
$\begingroup$

Electromagnetic potential energy is the energy of the electromagnetic field, that is, $\int \frac12 (\mathbf E^2 + \mathbf B^2)$. Just like any other form of energy, it has (or is) mass. When you weigh a system, you are weighing its electromagnetic field along with everything else. When you pull oppositely-charged particles apart, the total energy in the electromagnetic field increases, so the system weighs more.

That's the principle, anyway. Making it work consistently is an open problem. For example, it's difficult to understand the tiny mass of the electron. The electric field of an electron integrated from a macroscopic distance to about 3 fm (called the classical electron radius) is already enough to account for all of the electron's mass. But 3 fm is a huge distance by electron standards; if the field really cut off at that distance, it would be noticeable in scattering experiments. Despite those problems, the electromagnetic field has to have mass according to everything we think we know about physics, so this basic picture of weighing the field must be to some extent correct.

$\endgroup$
-1
$\begingroup$

Think about what happens when a proton and electron combine in a hydrogen plasma. What you see is optical emission.

Now, use the unfashionable definition of mass: relativistic mass, as defined by the famous $E=mc^2$. Relativistic mass is a conserved quantity. The optical emission has energy, and thus carries mass away. Therefore, the bound atom has less mass than the sum of the masses of its constituents. The photons have carried away some of the mass in the electromagnetic field of the proton and electron.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.