# Paragraph from a Stephen Hawking's Book (Uncertainty Principle)

This is from a Stephen Hawking's Book (Uncertainty Principle). I would like to know how does the Uncertainty Principle impose a limit on the size. What does he mean?

From Hawking's book:

It turns out that the uncertainty principle imposes a limit on the size of something. After a little bit of calculation, one finds that for a given mass of an object, there is a minimum size. This minimum size is small for heavy objects, but as one looks at lighter and lighter objects, the minimum size gets bigger and bigger. This minimum size can be thought of as a consequence of the fact that in quantum mechanics objects can be thought of either as a wave or a particle. The lighter an object is, the longer its wavelength is and so it is more spread out. The heavier an object is, the shorter its wavelength and so it will seem more compact. When these ideas are combined with those of general relativity, it means that only objects heavier than a particular weight can form black holes. That weight is about the same as that of a grain of salt. A further consequence of these ideas is that the number of configurations that could form a black hole of a given mass, angular momentum, and electric charge, although very large, may also be finite. Jacob Bekenstein suggested that from this finite number, one could interpret the entropy of a black hole. This would be a measure of the amount of information that seems irretrievably lost, during the collapse when a black hole is created.

• This is very confusing and should have been edited, as it mixes up classical (general relativity) and quantum( heisenberg uncertainty) in a mixed up way, giving the impression that small mass particles are spread all over the place in quantum mechanics. This is wrong, it is only the probability of detecting a quantum mechanical particle that is spread out, not the particle itself, and that probability is small,as the validated .standard model of particle physics works perfectly with point particles. We do see electrons in detectors. – anna v Nov 22 '18 at 9:39
• I found by searching this arxiv.org/abs/1808.05121 , which clarifies the difference of quantum with classical, but in deriving the Hawking temperature using the Heisenberg uncertainty goes to velocities larger than c. I hope somebody working on this subject replies to this. – anna v Nov 22 '18 at 9:41
• This v>c mathematically can happen in quantum field theory diagrams, on virtual particles, still one has to be talking of probabilities, not space energy or mass waves – anna v Nov 22 '18 at 9:59
• Whole paragraph is here. "If one measures speed, then its position is undetermined. In practice,this means it is impossible to localise anything. Suppose you want to measure the size of something,then you need to figure out where the ends of this moving object are. You can never do this accurately, because it will involve making a measurement of both the positions of something and its speed at the same time. In turn, it is then impossible to determine the size of an object. All u can do is to say that the Uncertainty Principle makes it impossible to say precisely what the size of something" – Harry Nov 22 '18 at 10:04
• Yes, but it does not say that particles are spread out, it just gives an envelope which is consistent with the solution of quantum mechanical equations for the specific problem, as the heisenberg uncertatinty is incorporated with the commutators algebra of quantum mechanic theory.. And the spread out is in probability of detection. Look at this bubble chamber picture of very small qm particles there is a whole list from electrons to ... – anna v Nov 22 '18 at 10:13

To understand this you need to appreciate that particles are not point masses, that is they are not simply little balls of matter. Instead the particle is more like a fuzzy cloud that is spread out over some region of space. This cloud doesn't have a fixed size because it depends on how the particle interacts with its surroundings. What the uncertainty principle tells us is that there is a link between the size of the cloud and the energy of the particle.

Note: the argument I'm going to use here is qualitative rather than quantitative so don't take it too literally. I'm trying to give you an understanding for why forming a black hole requires a certain mass, and this necessarily means simplifying the calculation, though hopefully not oversimplifying it. Also don't worry too much about the factors of $$2$$ etc appearing in the equations - my argument is so approximate that factors of $$2$$ can be safely ignored.

Suppose we start with a particle that has spread out over a large volume. This happens naturally to particles if we keep them isolated, so to make our super-fuzzy particle we just have to leave it alone for a bit. The uncertainty principle tells us:

$$\Delta x \Delta p \ge \frac{\hbar}{2} \tag{1}$$

or for our purposes I'm going to rearrange this to:

$$\Delta p \ge \frac{\hbar}{2\Delta x} \tag{2}$$

For our fuzzy particle the uncertainty in its position, $$\Delta x$$, will be about the size of the cloud that is, if the particle is spread out over some region of size $$d$$ then we can't pin down its position to an accuracy of greater than $$d$$. So for our particle the uncertainty in the moment will be inversely proportional to $$d$$:

$$\Delta p \ge \frac{\hbar}{2d} \tag{3}$$

The reason this matters is that when we confine a particle to some region of size $$d$$ we have to put in energy to do it. This is because any confined particle had a zero point energy and to confine the particle we have to supply that energy. We can estimate this energy because the energy and momentum of a particle are related by the equation:

$$E = \frac{p^2}{2m}$$

And for our confined particle the value of the momentum $$p$$ will be around the momentum uncertainty that we calculated in equation (3) so the energy we have to put in to confine our particle to a region of size $$d$$ is:

$$E = \frac{\Delta p^2}{2m} = \frac{\hbar^2}{8d^2m} \tag{4}$$

Now the problem is that energy and mass are related by Einstein's famous equation $$E=mc^2$$, and if we put in too much energy that energy will be used to create a new particle. So when we are trying to confine our particle if we put in an energy $$E \gt mc^2$$ what happens is we create a new particle. Instead of confining our particle we end up with two particles. Put in more energy and we get three particles, and so on. This means there is a limit to how much we can confine a particle because putting in more energy just creates new particles instead of confining our first particle.

We can get an idea of how much we can confine a particle by taking our equation (4) and setting the energy equal to $$mc^2$$ so we get:

$$mc^2 = \frac{\hbar^2}{8d^2m}$$

which rearranges to:

$$d = \frac{\hbar}{2\sqrt{2}cm} \tag{5}$$

If we take an electron with mass $$9.11 \times 10^{-31}$$ kg and plug this into our equation we get $$d = 1.4 \times 10^{-13}$$ m so the smallest size we can compress an electron to is about $$0.14$$ picometres. If we try to make it any smaller the energy we put in will just be used to create another electron.

Note that $$d$$ is inversely proportional to $$m$$, so heavier particles can be confined to a smaller space. For example if we take a proton with a mass of about 2000 times larger than the electron mass we could shrink it down to around $$0.0001$$ picometres.

The point of all this is that to make a black hole we have to shrink the particle down to its Schwarzschild radius, which is given by:

$$r_s = \frac{2Gm}{c^2}$$

For an electron this is size is absurdly small. It comes out as about $$10^{-57}$$ m and that is far, far smaller than the minimum size we worked out above. So the bottom line is that we simply cannot compress an electron small enough to turn it into a black hole. But we know that the more massive a particle is the smaller we can compress it, so what mass would a particle need to have to form a black hole. Well all we have to do is take our equation (5) for the size and replace $$d$$ by the expression for the Schwarzschild radius and this gives:

$$\frac{2Gm}{c^2} = \frac{\hbar}{2\sqrt{2}cm}$$

and if we rearrange this to get the mass it gives:

$$m = \sqrt{\frac{\hbar c}{4\sqrt{2}G}}$$

And if we plug in the values for $$\hbar$$, $$c$$ and $$G$$ we get:

$$m \approx 10^{-8} \textrm{kg}$$

And this is our key result because this is the smallest mass that can be compressed enough to form a black hole. In fact this is a very special mass and it is called the Planck mass. The Planck mass is actually:

$$m_P = \sqrt{\frac{\hbar c}{G}}$$

but as I said in the introduction, my argument is so approximate that the factor of $$4\sqrt{2}$$ in the denominator can be ignored.

This has been a long and complicated journey, so let me summarise the main points:

• the uncertainty principle means that as we compress a particle its energy increases

• there's a limit to how much we can compress a particle because if the energy increases too much we just create more particles instead of compressing the original one. The minimum size gets smaller as the particle gets heavier.

• to make a black hole we need to compress the particle to smaller than its Schwarzschild radius

• so to make a black hole we need a particle sufficiently massive that its minimum size is smaller than its Schwarzschild radius. That mass turns out to be the Planck mass.

Hawking is equating the different length scales associated.

In quantum mechanics, the characteristic length scale is decided by the mass of the particle. It is known as Compton wavelength ($$\lambda_c$$), and roughly speaking, it sets the length scale where relativistic quantum field theory becomes important for the description of physics; because if you try to localize a particle such that the uncertainty in position is smaller than $$\lambda_c$$, the the uncertainty in momentum becomes high enough to (from $$E^2 = m^2 c^4 + p^2 c^2$$) induce uncertainty in energy of the order of the mass of the particle. This means that at such small length scales, and equivalently, high energy/momentum scales, we must take special relativity and particle production/annihilation into account. The Compton wavelength is given by:

$$\lambda_c = \frac{\hbar}{mc}$$

In general relativity, the characteristic length scale is also decided by mass, but is directly proportional to it. It is known as Schwarzschild radius $$r_S$$. The event horizon of a static and spherically symmetric object lies at $$r=r_s$$ and is given by:

$$r_s = \frac{2 G m}{c^2}$$

Substituting $$r_s$$ for $$\lambda_c$$ and barring a factor of $$\sqrt{2}$$, we can write the characteristic mass scale for such a situation in terms of the three fundamental constants of the three fundamental theories: $$c$$ of special relativity, $$G$$ of general relativity and $$\hbar$$ of quantum mechanics, and it is known as Planck mass:

$$m_p = \sqrt{\frac{\hbar c}{G}}$$

So the Schwarzschild radius of a black hole of mass equal to the Planck mass is its Compton wavelength.

The number of microscopic states of a black hole with given macroscopic characteristics (like charge, mass and angular momentum) can be written in terms of the area $$A$$ of the event horizon. This is known as black hole entropy and is given by:

$$S = \frac{kAc^3}{4G\hbar}$$

where $$k$$ is the Boltzmann's constant.

The issue that black holes are not entirely 'black' and end up radiating because of particle production when quantum mechanics is involved has led to many debates and open questions that fall under the umbrella term of black hole information paradox.