Gravity

Mr. Einstein and my GPS

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I promised to continue one of my previous posts and explain how Einstein’s theories of 1905 and 1915 together affect our GPS systems. If we hadn’t discovered relativity (special and general) by now, we’d have certainly discovered it by the odd behaviour of our clocks on the surface of the earth and on an orbiting satellite.

The previous post ended by demonstrating that the time interval between successive ticks of a clock at the earth’s surface $\Delta t_R$ and a clock ticking infinitely far away from all masses $\Delta t_{\infty}$ are related by the formula

$\Delta t_{R} = \Delta t_{\infty} (1 + \frac{ \Phi(R)}{c^2})$

The gravitational potential $\Phi(R)=-\frac{G M_E}{R}$ is a ${\bf {negative}}$ number for all $R$. This means that the time intervals measured by the clock at the earth’s surface is ${\bf {shorter}}$ than the time interval measured far away from the earth. If you saw the movie “Interstellar“, you will hopefully remember that a year passed on Miller’s planet (the one with the huge tidal waves) while 23 years passed on the Earth, since Miller’s planet was close to the giant Black Hole Gargantua. So time appears to slow down on the surface of the Earth compared to a clock placed far away.

Time for some computations. The mass of the earth  is $5.97 \times 10^{24} \: kg$, Earth’s radius is $R = 6370 \: km \: = 6.37\times 10^6 \: meters$ and the MKS units for $G = 6.67 \times 10^{-11} \: MKS \: units$. In addition, the speed of light $c = 3 \times 10^8 \frac {m}{s}$. If $\Delta t_{\infty} = 1 \: sec$, the clock on an orbiting satellite (assumed to be really far away from the earth) measures one second, the clock at the surface measures

$\Delta t_R = (1 \: sec) \times (1 - \frac {(6.67 \times 10^{-11}) \: \times \: (5.97 \times 10^{-24} )}{(6.37 \times 10^6 )\: (3 \times 10^8 )^2})$

this can be simplified to $0.69 \: nanoseconds$ less than $1 \: sec$. In a day, which is  $(24 \times 3600) \: secs$, this is $70 \times 10^-6 = 60 \: \mu \: seconds$ (microseconds are a millionth of a second).

In reality, as will be explained below, the GPS satellites are operating at roughly $22,000 \: km$ above the earth’s surface, so what’s relevant is the ${\bf {difference}}$ in the gravitational potential at $28,370 \: km$ and $6,370 \: km$ from the earth’s center. That modifies the difference in clock rates to $53 \: nanoseconds$ per second, or $46 \: microseconds$ in a day.

How does GPS work? The US (other countries too – Russia, the EU, China, India) launched several satellites into a distant orbit $20,000 \: - 25,000 \: km$ above the earth’s surface. Most of the orbits are designed to allow different satellites to cover the earth’s surface at various points of time. A few of the systems (in particular, India’s) have satellites placed in a Geo-Stationary orbit, so they rotate around the earth with the earth – they are always above a certain point on the earth’s surface. The key is that they possess rather accurate and synchronized atomic clocks and send the time signals, along with the satellite position and ID to GPS receivers.

If you think about how to locate someone on the earth, if I told you I was 10 miles from the Empire State Building in Manhattan, you wouldn’t know where I was. Then, if I told you that I was 5 miles from the Chrysler building (also in Manhattan), you would be better off, but you still wouldn’t know how high I was. If I receive a third coordinate (distance from yet another landmark), I’d be set.  So we need distances from three well-known locations in order to locate ourselves on the Earth’s surface.

The GPS receiver on your dashboard receives signals from three GPS satellites. It knows how far they are, because it knows when the signals were emitted, as well as what the time at your location is.  Since these signals travel at the speed of light (and this is sometimes a problem if you have atmospheric interference), the receiver can compute how far away the satellites are. Since it has distances to three “landmarks”, it can be programmed to compute its own location.

Of course, if its clock was constantly running slower than the satellite clocks, it would constantly overestimate the distance to these satellites, for it would think the signals were $emitted \: earlier$ than they actually were. This would screw up the location calculation, to the distance travelled by light in $0.53 \: nanoseconds$, which is $0.16$ meters. Over a day, this would become $14$ kilometers. You could well be in a different city!

There’s another effect – that of time dilation. To explain this, there is no better than the below thought experiment, that I think I first heard of from George Gamow’s book. As with ${\bf {ALL}}$ arguments in special and general relativity, the only things observers can agree on is the speed of light and (hence) the order of causally related events. That’s what we use in the below.

There’s an observer standing in a much–abused rail carriage. The rail carriage is travelling to the right, at a high speed $V$. The observer has a rather cool contraption / clock. It is made with a laser, that emits photons and a mirror, that reflects them. The laser emits photons from the bottom of the carriage towards the ceiling, where the mirror is mounted. The mirror reflects the photon back to the floor of the car, where it is received by a photo-detector (yet another thing that Einstein first explained!).

The time taken for this up- and down- journey (the emitter and mirror are separated by a length $L$) is

$\Delta t' = \frac{2 L}{c}$

That’s what the observer on the train measures the time interval to be. What does an observer on the track, outside the train, see?

She sees the light traverse the path down in blue above. However, she also sees the light traveling at the same (numerical) speed, so she decides that the time between emission and reception of the photon is found using Pythagoras’ theorem

$L^2 = (c \frac{\Delta t}{2})^2 - (V \frac {\Delta t}{2})^2$

$\rightarrow \Delta t = \frac {2 L}{c} \frac{1}{\sqrt{1 - \frac{V^2}{c^2}}}$

So, the time interval between the same two events is computed to be larger on the stationary observer’s clock, than on the moving observer’s clock. The relationship is

$\Delta t = \frac {\Delta t'}{ \sqrt{1 - \frac{V^2}{c^2}} }$

How about that old chestnut – well, isn’t the observer on the track moving relative to the observer on the train? How come you can’t reverse this argument?

The answer is – who’s going to have to turn the train around and sheepishly come back after this silly experiment runs its course? Well! The point is that one of these observers has to actively come back in order to compare clocks. Relativity just observes that you cannot make statements about ${\bf {absolute}}$ motion. You certainly have to accept relative motion and in particular, how observers have to compare clocks at the same point in space.

From the above, $1$ second on the moving clock would correspond to $\frac {1}{ \sqrt{1 - \frac{V^2}{c^2}} }$ seconds on the clock by the tracks. A satellite at a distance $D$ from the center of the earth has an orbital speed of $\sqrt {\frac {G M_E}{D} }$, which for an orbit $22,000$ km above the earth’s surface, which is $28,370 \: km$ from the earth’s center, would be roughly

$\sqrt { \frac {(6.67 \times 10^{-11} (5.97 \times 10^{-24})}{28370 \times 10^3} }\equiv 3700 \: \frac{meters}{sec}$

which means that $1$ second on the moving clock would correspond to $1 \: sec + 0.078 \: nanoseconds$ on the clock by the tracks. Over a day, this would correspond to a drift of $6 \: microseconds$, in the ${\bf {opposite}}$ direction to the above calculation for gravitational slowing.

Net result – the satellite clocks run faster by $40$ microseconds in a day. They need to be continually adjusted to bring them in sync with earth-based clocks.

So, that’s three ways in which Mr. Einstein matters to you EVERY day!

Arbitrage arguments in Finance and Physics

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Arbitrage refers to a somewhat peculiar and rare situation in the financial world. It is succinctly described as follows. Suppose you start with an initial situation – let’s say you have some money in an ultra-safe bank that earns interest at a certain basic rate $r$. Assume, also, that there is a infinitely liquid market in the world, where you can choose to invest the money in any way you choose. If you can end up with ${\bf {definite}}$ financial outcomes that are quite different, then you have an arbitrage between the two strategies. If so, the way to profit from the situation is to “short” one strategy (the one that makes less) and go “long” the other strategy (the one that makes more). An example of such a method would be to buy a cheaper class of shares and sell “short” an equivalent amount of an expensive class of shares for the same Company that has definitely committed to merge the two classes in a year.

An argument using arbitrage is hard to challenge except when basic assumptions about the market or initial conditions are violated. Hence, in the above example, suppose there was uncertainty about whether the merger of the two classes of shares in a year, the “arbitrage” wouldn’t really be one.

One of the best known arbitrage arguments was invented by Fischer Black, Myron Scholes and Robert Merton to deduce a price for Call and Put Options. Their argument is explained as follows. Suppose you have one interest rate for risk-free investments (the rate $r$ two paragraphs above). Additionally, consider if you, Dear Reader, own a Call Option, with strike price $\X$, on a stock price.  This is an instrument where at the end of (say) one year, you look at the market price of the stock and compute $\S - \X.$ Let’s say $X = \100$, while the stock price was initially $\76$. At the end of the year, suppose the stock price became $\110$, then the difference $\110 - \100 = \10$, so you, Dear Reader and Fortunate-Call-Option-Owner, would make $\10$. On the other hand, if the stock price unfortunately sank to $\55$, then the difference $\55 - \ 100 = - \45$ is negative. In this case, you, unfortunate Reader, would make nothing. A Call Option, therefore, is a way to speculate on the ascent of a stock price above the strike price.

Black-Scholes-Merton wanted to find a formula for the price that you should logically expect to pay for the option. The simplest assumption for the uncertainty in the stock price is to state that $\log S$ follows a random walk. A random walk is the walk of a drunkard that walks on a one-dimensional street and can take each successive step to the front or the back with equal probability. Why $\log S$ and not $S$? That’s because a random walker could end up walking backwards for a long time. If her walk was akin to a stock price, clearly the stock price couldn’t go below 0 – a more natural choice is $\log S$ which goes to $- \infty$ as $S \rightarrow 0$. A random walker is characterized by her step size. The larger the step size, the further she would be expected to be found relative to her starting point after $N$ steps. The step size is called the “volatility” of the stock price.

In addition to an assumption about volatility, B-S-M needed to figure out the “drift” of the stock price. The “drift”, in our example, is akin to a drunkard starting on a slope. In that case, there is an unconscious tendency to drift down-slope. One can model drift by assuming that there isn’t the same probability to move to the right, as to the left.

The problem is, while it is possible to deduce, from uncertainty measures in the market, the “volatility” of the stock, there is no natural reason to prefer one “drift” over the other. Roughly speaking, if you ask people in the market whether IBM will achieve a higher stock price after one year, half will say “Yes”, the other half will say “No”. In addition, the ones that say “Yes” will not agree on exactly by how much it will be up. The same for the “No”-sayers! What to do?

B-S-M came up with a phenomenal argument. It goes as follows. We know, intuitively, that a Call Option (for a stock in one year) should be worth more today if the stock price were higher today (for the same Strike Price) by, say $\1$. Can we find a portfolio that would decline by exactly the same amount if the stock price was up by $\1$. Yes, we can. We could simply “short” that amount of shares in the market. A “short” position is like a position in a negative number of shares. Such a position loses money if the market were to go up. And I could do the same thing every day till the Option expires. I will need to know, every day, from the Option Formula that I have yet to find, a “first-derivative” – how much the Option Value would change for a $\1$ increase in the stock price. But once I do this, I have a portfolio (Option plus this “short” position) that is ${\bf {insensitive}}$ to stock price changes (for small changes).

Now, B-S-M had the ingredients for an arbitrage argument. They said, if such a portfolio definitely could make more than the rate offered by a risk-less bank account, there would be an arbitrage. If the portfolio definitely made more, borrow (from this risk-free bank) the money to buy the option, run the strategy, wait to maturity, return the loan and clear a risk-free profit. If it definitely made less, sell this option, invest the money received in the bank, run the hedging strategy with the opposite sign, wait to maturity, pay off  the Option by withdrawing your bank funds, then pocket your risk-free difference.

This meant that they could assume that the portfolio described by the Option and the Hedge, run in that way, were forced to appreciate at the “risk-free” rate. This was hence a natural choice of the “drift” parameter to use. The price of the Option would actually not depend on it.

If you are a hard-headed options trader, though, the arguments just start here. After all, the running of the above strategy needs markets that are infinitely liquid with infinitesimal “friction” – ability to sell infinite amounts of stock at the same price as at which to buy them. All of these are violated to varying degrees in the real stock market, which is what makes the B-S-M formula of doubtful accuracy. In addition, there are other possible processes (not a simple random-walk) that the quantity $\log S$ might follow. All this contributes to a robust Options market.

An arbitrage argument is akin to an argument by contradiction.

Arguments of the above sort, abound in Physics. Here’s a cute one, due to Hermann Bondi. He was able to use it to deduce that clocks should run slower in a gravitational field. Here goes (this paraphrases a description by the incomparable T. Padmanabhan from his book on General Relativity).

Bondi considered the following sort of apparatus (I have really constructed my own example, but the concept is his).

One photon rushes from the bottom of the apparatus to the top. Let’s assume it has a frequency $\nu_{bottom}$ at the bottom of the apparatus and a frequency $\nu_{top}$ at the top. In our current unenlightened state of mind, we think these will be the same frequency. Once the photon reaches the top, it strikes a target and undergoes pair production (photon swerves close to a nucleus and spontaneously produces an electron-positron pair – the nucleus recoils, not in horror, but in order to conserve energy and momentum). Let’s assume the photon is rather close to the mass of the electron-positron pair, so the pair are rather slow moving afterwards.

Once the electron and positron are produced (each with momentum of magnitude $p_{top}$), they experience a strong magnetic field (in the picture, it points out of the paper). The law that describes the interaction between a charge and a magnetic field is called the Lorentz Force Law. It causes the (positively charged) positron to curve to the right, the (negatively charged) electron to curve to the left. The two then separately propagate down the apparatus (acquiring a momentum $p_{bottom}$) where they are forced to recombine, into a photon, of exactly the right frequency, which continues the cycle. In particular, writing the energy of the photons in each case.

$h \nu_{top} = 2 \sqrt{(m_e c^2)^2+p_{top}^2 c^2} \approx 2 m_e c^2$

$h \nu_{bottom} = 2 \sqrt{(m_e c^2)^2+p_{bottom}^2 c^2} \approx 2 m_e c^2 + 2 m_e g L$

In the above, $p_{bottom} > p_{top}$, the electrons have slightly higher speed at the bottom than at the top.

We know from the usual descriptions of potential energy and kinetic energy (from high school, hopefully), that the electron and positron pick up energy $m_e g L$ (each) on their path down to the bottom of the apparatus. Now, if the photon doesn’t experience a corresponding loss of energy as it travels from the bottom to the top of the apparatus, we have an arbitrage. We could use this apparatus to generate free energy (read “risk-less profit”) forever. This can’t be – this is nature, not a man-made market! So the change of energy of the photon will be

$h \nu_{bottom} - h \nu_{top} =2 m_e g L \approx h \nu_{top} \frac{g L}{c^2}$

indeed, the frequency of the photon is higher at the bottom of the apparatus than at the top. As photons “climb” out of the depths of the gravitational field, they get red-shifted – their wavelength lengthens/frequency reduces. This formula implies

$\nu_{bottom} \approx \nu_{top} (1 + \frac{g L}{c^2})$

writing this in terms of the gravitational potential due to the earth (mass $M$) at a distance $R$ from its center

$\Phi(R) = - \frac {G M}{R}$

$\nu_{bottom} \approx \nu_{top} (1 + \frac{\Phi(top) - \Phi(bottom)}{c^2})$

so , for a weak gravitational field,

$\nu_{bottom} (1 + \frac{ \Phi(bottom)}{c^2}) \approx \nu_{top} (1 + \frac{\Phi(top)}{c^2})$

On the other time intervals are related to inverse frequencies (we consider the time between successive wave fronts)

$\frac {1}{\Delta t_{bottom} } (1 + \frac{ \Phi(bottom)}{c^2}) \approx \frac {1}{\Delta t_{top}} (1 + \frac{\Phi(top)}{c^2})$

so comparing the time intervals between successive ticks of a clock at the surface of the earth, versus at a point infinitely far away, where the gravitational potential is zero,

$\frac {1}{\Delta t_{R} } (1 + \frac{ \Phi(R)}{c^2}) \approx \frac {1}{\Delta t_{\infty}}$

which means

$\Delta t_{R} = \Delta t_{\infty} (1 + \frac{ \Phi(R)}{c^2})$

The conclusion is that the time between successive ticks of the clock is measured to be much smaller on the surface of the earth vs. far away. Note that $\Phi(R)$ is negative, and the gravitational potential is usually assumed to be zero at infinity. This is the phenomenon of time dilation due to gravity. As an example, the GPS systems are run off clocks on satellites orbiting the earth at a distance of $20,700$ km. The clocks on the earth run slower than clocks on the satellites. In addition, as a smaller effect, the satellites are travelling at a high speed, so special relativity causes their clocks to run a little slower compared to those on the earth. The two effects act in opposite directions. This is the subject of a future post, but the effect, which has been precisely checked, is about 38 $\mu$seconds per day. If we didn’t correct for relativity, our planes would land at incorrect airports etc and we would experience total chaos in transportation.