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Gravity and Quantum Mechanics — The Simple Version

In general relativity, gravity is no longer considered a real force like the Lorentz force, but a fictitious one like the Coreolis force or centrifugal force in a rotating reference frame. To get rid of the force of gravity locally, all one has to do is switch to a reference frame that is accelerating at the same rate that objects in free fall are accelerating. The force of gravity term will then disappear from Newton’s second law.

Some claim that the same trick doesn’t work in quantum mechanics. This is not true. They just weren’t trying hard enough. Start with the wave equation for a particle in a constant gravitational field (Planck’s constant has been set to 1 and the units of mass are measured in seconds per metre squared as a result. See here.):

$-\frac{1}{2m} \psi_{zz} + mgz\psi = i\psi_t$ (1)

z is height and $V(z) = mgz$ is the gravitational potential energy. Thus in this coordinate system, objects fall in the -z direction. Now we’ll define new coordinates, u, such that when an approximately classical particle is in free fall in the old coordinate system, it is stationary in the new one. This implies $z = u - \frac{g}{2}t^2$. We’ll call our new wavefunction $\phi$. Now, we need to establish a relationship between the old wavefunction and the new one. A good first guess would be

$\psi(z, t) = \phi \left( z + \frac{g}{2} t^2, t \right) .$

There is a problem with this, though. Recall that given a pre-existing wavefunction $\psi$, to change its momentum by p one multiplies $\psi$ by $e^{i p z}$. There is no term like this in our first guess, which would mean that the particle has the same momentum in both reference frames. This is definitely not true. Thus an extra as-yet-unknown factor of $e^{i \gamma(z, t)}$ is needed, where $\gamma(z, t)$ must be determined. Our new coordinate transformation for the wavefunction is

$\psi(z, t) = e^{i \gamma(z, t)} \phi \left( z + \frac{1}{2} g t^2, t \right) .$ (2)

We would like to find a $\gamma$ such that $\psi$ obeys the free fall wave equation (1) and $\phi$ obeys the free particle wave equation,

$-\frac{1}{2m} \phi_{uu} = i \phi_t .$ (3)

To do this, first we replace $\psi$ in (1) with the right side of (2), resulting in the following:

$-\frac{1}{2m} \phi_{uu} - \frac{i}{m} \gamma_z \phi_u + \left( \frac{1}{2m} \left( \gamma_z^2 - i \gamma_{zz} \right) + mgz \right) \phi = i \phi_t + igt\phi_u - \gamma_t \phi$

The left and right sides of (3) are present in the above expression, but there are a bunch of extra terms. These terms will all cancel if

$-\frac{i}{m} \gamma_z = igt$ (4) and $\frac{1}{2m} \left( \gamma_z^2 - i \gamma_{zz} \right) + mgz = -\gamma_t$ (5).

Thus (3) is true if $\gamma$ is subject to conditions (4) and (5). Condition (4) implies $\gamma = -mgtz + c(t)$. Replacing $\gamma$ with this in condition (5) results in

$c = -\frac{mg^2}{6} t^3 .$

This means

$\gamma = -mgtz - \frac{mg^2}{6} t^3 .$

Including the $\gamma$, final formula for converting from $\phi$ to $\psi$ is

$\psi(z, t) = e^{-imgtz - \frac{img^2}{6}t^3} \phi \left( z + \frac{g}{2} t^2, t \right)$ (6).

Thus we have found a transformation of the original free fall wave equation to the free particle wave equation in complete analogy to the classical case. Now that we have $\gamma$, what is its significance? As was mentioned before, a factor of $e^{ipz}$ out in front of a wavefunction shifts the particle’s momentum by p. Such a factor is indeed present in (6). In our case, the difference in momentum from $\phi$ to $\psi$ is -mgt. This makes perfect sense.

But what about the other factor, $e^{-\frac{img^2}{6} t^3}$? It was necessary for the math to work out, but does it mean anything? It is just a constant of unit modulus out in front of the wavefunction. Thus it has no meaning at all and the state mentioned in (6) is completely equivalent to

$\psi(z, t) = e^{-imgtz} \phi \left( z + \frac{g}{2} t^2, t \right)$.

I was first shown this neat trick by Andy Randono in his 4th year undergraduate general relativity at the University of Waterloo. If you’re reading, thanks for teaching it to us!

Categories: Physics

Fields of Physics, Constants, and Units

This subject has many fields. Each field is identified by the set of laws that are asserted to be true. I consider classical mechanics the most basic one. There are very many assumptions made in classical mechanics. Objects are not too dense. They are not too small. Things don’t move too fast, nor do they move too slowly. The more any one of these assumptions is broken, the more the effects of other fields of physics need to be taken into account. Thus new fields of physics are entered. For example, if the objects involved in a certain problem are the size of molecules, accurate predictions of their behaviour cannot be made with classical mechanics. The field of quantum mechanics is required.

In entering a new field of physics and declaring its laws relevant to the problem at hand, there are often one or more physical constants that are immediately and frequently required. For example, in entering into the quantum realm, immediately Planck’s constant, $\hbar$, is needed and it is used frequently. Upon entering the special relativistic realm, the speed of light, c, is needed to do just about anything and it is used incessantly, just like the quantum realm’s $\hbar$. In fact, c is used so commonly, many of us decided that while we are solving a problem in this field, we would do something that one would think we are not allowed to do: set c equal to 1. I don’t mean 1 metre per second, obviously! So, what do I mean, then? Well, let me elaborate:

$\begin{matrix} c = 1 \\ 300\times10^6 \frac{m}{s} = 1 \\ 300\times10^6 m = 1 s \end{matrix}$

Thus setting c to 1 implies that metres must be the same units as seconds! That is, seconds are also a measure of space and metres are also a measure of time. One metre of time is the amount of time it takes light to travel one metre. One second of space is the distance light travels in a second. The really interesting part of this is that this isn’t just a neat mathematical trick. It is a reflection of the unification of space and time into spacetime. It says that in a sense, space is time.

Under this convention, the constant of proportionality between our two units of measurement of spacetime, metres and seconds, $\left( 300\times10^6 \right)$ is quite artificial and arbitrary. It is as arbitrary as the constant of proportionality between metres and inches. The artificial constant of proportionality gives rise to an artificial unit for measuring time, seconds; a unit that can be replaced with the more natural metres. Alternatively, metres could be considered the artificial unit for measuring space, making seconds the natural replacement.

This happens a lot throughout physics. When a repeatedly-used constant is set to 1, an artificial unit of measurement may be expressed in terms of more fundamental units. This table demonstrates this process in four different fields of physics.

field artificial constant(s) artificial unit natural replacement unit
Classical Mechanics none* none* none*
Thermal Physics k K J
Quantum Mechanics $\hbar$ kg $\frac{s}{m^2}$
Special Relativity c s m
Electromagnetism $\varepsilon_0, \mu_0$ ** C $\sqrt{Jm}$

*There would be artificial constants, only this field was known while SI units were first being created, so all constants of proportionality were defined 1. It doesn’t have to be this way. For example: artificial units for force. Let’s call them something brutally imperial like “slugs”. Yes, that sounds good. Now, let’s give it a ridiculously variable imperial definition: let one slug (1 S) be the amount of force it takes to crush a watermelon. Newton’s second law then becomes F = W m a, where W is Cletus’s constant. $W = 0.00012 \frac{S s^2}{kg m}$

**$\varepsilon_0$ is set to 1, but $\mu_0$ must be set to $\frac{1}{c^2}$ for Maxwell’s equations to still work.

There are also many times where a problem requires treatment with theory from many fields. If the frequently-used constants of the subfields are set to 1, then several units of measurement get natural replacements. Below is a table of some common combinations. Replacing kelvin with joules has no effect on any other units, so thermal physics can be included in any of the combined fields below.

field artificial constants (artificial unit, natural replacement)
EM & SR $c, \mu_0, \varepsilon_0$ $\left( C, \sqrt{kgm} \right)$ (s, m)
QED (EM, QM, & SR) $\hbar, c, \mu_0, \varepsilon_0$ (C, 1) $\left( kg, m^{-1} \right)$ (s,m)
Grand Unified (EM, QM, SR, GR) $\hbar, c, G, \mu_0, \varepsilon_0$ no units!

Note that in QED, electric charge is unitless and the only units left are metres! In the extreme case, including gravity and normalising G, there are no more units!

Categories: Physics