Coming down to Earth

Problem: an object initially at rest some distance above the Earth begins to fall toward the Earth. Show that the time it takes for the object to fall halfway to the Earth is about 9/11 of the time it takes to fall all the way to the Earth.

Solution: use conservation of energy. Let \(R\) be the initial distance of the object above the Earth. Then the total energy of the object in its initial state is its gravitational potential energy:

\(– \frac{G M_e m}{R}\).

Note that the factors in the numerator correspond to the Universal Gravitation constant, the mass of the Earth, and the mass of the object, respectively.

At some point \(r\) between the initial distance and the Earth, the object will have a different value of the potential energy, as well as a nonzero value of kinetic energy. The statement of conservation of energy becomes

\(\frac{1}{2} m \dot{r}^2 – \frac{G M_e m}{r} = – \frac{G M_e m}{R}\).

The time \(t_0\) the object takes in falling to some position \(r_0\) is then found by solving this differential equation:

\(t_0 = \frac{1}{\sqrt{2 G M_e}} \displaystyle \int_{r_0}^{R} dr \left ( \frac{1}{r} – \frac{1}{R} \right )^{- \frac{1}{2} } = \sqrt{\frac{R^3}{2 G M_e}} \displaystyle \int_{x}^{1} du \left ( \frac{1}{u} – 1 \right )^{- \frac{1}{2} }\),

where \(x = r_0/R\). The integral on the right hand side may be evaluated through an appropriate substitution (details left to the reader):

\(\displaystyle \int_{x}^{1} du \left ( \frac{1}{u} – 1 \right )^{- \frac{1}{2} } = \arccos{ \sqrt{x} } + \sqrt{x(1-x)}\),

where the branch of the inverse cosine is the principal branch.

The fraction of time it takes for the object to fall halfway to Earth is the ratio \(p\) of the above expression at \(x = 1/2\) to the same at \(x = 0\):

\(p = \frac{1}{2} + \frac{1}{\pi} \approx 0.818\),

which is about 9/11.


The expected value of a Mega Millions ticket

As I type this post, the Mega Millions jackpot projects to be $500,000,000 (annuity value – after-tax cash value is more like $220,000,000). A single ticket costs $1. Is it worth it?

By “worth it,” I mean that the expected value of a ticket is greater than zero. The expected value of a random variable \(W\) which represents the event of winning the Mega Millions lottery with the ticket is equal to, roughly, (Probability of winning)*(Value of winning) + (Probability of losing)*(value of losing). In shorter math notation, this looks like:


Here, \(E[W]\) is the expected value of \(W\), \(P[W]\) is the probability of \(W\), and therefore \(1-P[W]\) is the probability of NOT \(W\). Also, \(V\) is the value of winning – in this case, $500,000,000 or $220,000,000, and \(C\) is the value of losing: -$1.

What is this probability? The players are asked to pick 5 balls from a first set of 56, then 1 ball from a second set of 46, the first set and second sets being independent. The number of possibilities is \(46 \binom {56}{5} = 175,711,536\). The probability \(P[W]\) is then the reciprocal of this number.

Although this probability is indeed small, it would seem that a sufficiently large jackpot would make buying a ticket worth it. And, indeed, even with the after-tax cash value, the expected value is greater than zero:

\(E[W]=\displaystyle \frac{220,000,000}{175,711,536} – \displaystyle \frac{175,711,535}{175,711,536} \approx 0.25\).

This could be taken as an encouragement to purchase Mega Millions tickets; after all, you can expect about $0.25 of value for each ticket you buy. What a bargain! With a return on investment like that, why don’t investment houses just put their client’s money into Mega Millions tickets?

The reason is that return is wrong and based on a false assumption: no matter how many tickets got the correct number, the payout is the same. In fact, the payout is split evenly among the winners. This will of course lower the expected value of a ticket. But by how much?

Consider a population of \(N\) tickets, where yours is among the \(k\) winners. (Mazeltov.) That means there are \(k-1\) winners besides yourself among the rest of the remaining \(N-1\) tickets. What is the probability that there are \(k-1\) winners among \(N-1\) tickets?

We consider each event of a ticket being a winning ticket being independent from any other ticket being a winning or losing ticket. (This is clearly the case.) The probability of a particular set of \(k-1\) winners among the \(N-1\) tickets is then \(p^{k-1} (1-p)^{(N-1)-(k-1)}\). That is, when there are \(k-1\) winners, there are also \((N-1) – (k-1) = N – k\) losers.

That said, there are also many different ways to arrange the other \(k-1\) winners among the remaining \(N-1\) tickets. To be precise, there are \(\binom {N-1}{k-1}\) different arrangements of winners and losers, on top of you as a winner. (Not literally!) I imagine that my friends literate in statistics recognize the random variable \(W\) as having a binomial distribution.

Note that, when there are \(k\) winners, the jackpot is split evenly among the winners; that is, each winner gets \(V/k\), and not \(V\). The expected value of a ticket is then a sum over all possible values of \(k\):

\(E[W]= \displaystyle V \sum_{k=1}^{N} \frac{1}{k} \binom{N-1}{k-1} p^{k} (1-p)^{N-k} – C (1-p)\),

where \(p = P[W]\) as defined above. Note that there is an extra factor of \(p\) coming from your own ticket.

The above sum is very difficult to evaluate numerically for \(N \approx 100,000,000\), and approximations to normal or Poisson distributions do not apply. However, we can observe that we are comparing the value of the above sum to \(p\). We can see that the value of the sum is less than \(p\) because of the fact, well-known from binomial distributions, that

\(\displaystyle \sum_{k=1}^{N} \binom{N-1}{k-1} p^{k} (1-p)^{N-k} = p\).


\(\displaystyle \sum_{k=1}^{N} \frac{1}{k} \binom{N-1}{k-1} p^{k} (1-p)^{N-k} < p[/latex], because the factor [latex,size="-2"]\frac{1}{k} < 1[/latex]. In fact, a rough calculation (it turns out you can neglect anything beyond the 6th term) gives a value of about 0.374 for the above sum. Again, compare this to 1 using the crude (and incorrect) estimate. So, in multiplying the first term in the first expected value equation by about 0.374, we get an expected value of about -$0.53. That is, rather than gaining 25 cents, you should expect to lose about 47 cents for every dollar you spend on a Mega Millions ticket. Then again, we're all kind of suckers for that galactically small chance we could win, and I don't blame any of you for throwing away a few pennies chasing the dream. Update (3/31/12): I need to correct an assertion I made, and some numbers. The conclusion stands - in fact, in light of what has happened over the past 24 hours, the conclusion is even more stark. First of all, I posted some incorrect numbers that I have since corrected. For the above values of the number of tickets in circulation and probability of winning: [latex]\displaystyle \sum_{k=1}^{N} \frac{1}{k} \binom{N-1}{k-1} p^{k-1} (1-p)^{N-k} \approx 0.763[/latex]. This is the reduction factor on the jackpot, not 0.374 as I published before. The expected value of a ticket is then about -$0.05. Again, still a loser. That said, it turns out that 100,000,000 tickets was a gross underestimate; rather, 1,500,000,000 tickets were sold! For this number, the reduction factor is about 0.117, which gives an expected value of a ticket as -$0.85, an even worse value than my incorrect previous numbers show. Second, although I was right in asserting that the multiple winners are not governed by a Poisson distribution (a great explanation is here), I was incorrect in ignoring the Law of Rare Events, which states that, under a certain limiting behavior, the binomial sum approaches the Poisson sum that results from assuming a Poisson distribution. Further, the limiting behavior need not be rigorously enforced: a large enough sample and a small enough probability does the trick.

The mathematical statement of the Law of Rare Events in this context is

[latex]\displaystyle \sum_{k=1}^{N} \frac{1}{k} \binom{N-1}{k-1} p^{k-1} (1-p)^{N-k} \approx \sum_{k=1}^{\infty} \frac{(N p)^{k-1}}{k!} e^{-N p} = \frac{1 – e^{-N p}}{N p}\).

It turns out that this is a very good approximation out to many decimal places. So, a very simple formula for the expected value of a Mega Millions ticket is

\(\displaystyle E[W] = V \frac{1 – e^{-N p}}{N} – C (1-p)\).

Top Ten Twilight Zone Episodes

As I watch the annual marathon on Syfy, I figure it would be worthwhile to compile a Top Ten Twilight Zone episodes. Hardly original, I know, but having watched every episode multiple times, I feel I have some standing here. Anyway…

  1. Eye of the Beholder
  2. Number Twelve Looks Just Like You
  3. And When the Sky Was Opened
  4. On Thursday We Leave For Home
  5. The Obsolete Man
  6. Death Ship
  7. A Game of Pool
  8. The Howling Man
  9. Judgment Night
  10. The Masks

Computing square roots

Let’s say you want to take the square root of a real number \(a\) without a computer. How would you do it? How do you think a computer does it?

The only way I know any computer performs square roots practically is via the following recurrence:

\(s_1=1; s_n=\displaystyle \frac{1}{2} \left ( s_{n-1} + \frac{a}{s_{n-1}} \right )\),


\(\sqrt{a}=\displaystyle \lim_{n\to\infty}s_n\).

The recurrence derives from Newton’s Method of finding roots, as applied to the function \(f(x)=x^2-a\). But that is not the point; the point is the recurrence and how fast it converges to its goal. Typically, roots found via Newton’s Method exhibit quadratic convergence; that is, the error in an iteration is the square of the error of the previous iteration. It turns out that the above recurrence has an exact solution, and from this solution we can closely examine the convergence toward the above limit.

The way to see this solution is to set \(s_n=\sqrt{a} \coth{\theta_n}\), where the hyperbolic cotangent is

\(\coth{x}=\displaystyle \frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}\).

The hyperbolic cotangent satisfies a doubling formula:

\(\coth{2 x}=\displaystyle \frac{1}{2} \left ( \coth{x} + \frac{1}{\coth{x}} \right )\).

The above recurrence then takes the simplified form

\(\theta_{1}=\tanh^{-1}{\sqrt{a}}; \theta_{n}=2 \theta_{n-1}\).

The solution to the original recurrence then easily follows:

\(s_n=\sqrt{a} \coth{ \left ( 2^{n-1} \tanh^{-1}{\sqrt{a}} \right ) }\).

One slight complication: for \(a>1\), \(\tanh^{-1}{\sqrt{a}}\) is a complex number with imaginary part = \(\frac{\pi}{2}\). Because the recurrence involves a doubling of the argument, the imaginary part has no effect on the result. That said, it is more direct to write the result as

\(s_n=\sqrt{a} \coth{ \left ( 2^{n-1} \Re \left [ \tanh^{-1}{\sqrt{a}} \right ] \right ) }\),

where \(\Re \left [ z \right ]\) denotes the real part of \(z\).

On the surface, it seems silly to express this solution to the recurrence in terms of the limit that it approximates. That said, the goal in deriving this solution was to examine how it converges to the limit. Along these lines, consider the following approximation, valid for large arguments:

\(\coth{x} \approx 1+2 e^{-2 x}\).

The error at later stages of the recurrence is then about

\(\displaystyle \left | \frac{s_n}{\sqrt{a}} – 1 \right | \approx 2 \times 10^{- \left (\log_{10}{e} \right ) \left ( \Re \left [ \tanh^{-1}{\sqrt{a}} \right ] \right ) 2^{n}} \).

For each increment in \(n\), the error is the square of the previous error, as I mentioned above being a characteristic of root finding via Newton’s Method. The solution allows us to be even more specific. Because \(\log_{10}{e} \approx 0.4343\) and \(\Re \left [ \tanh^{-1}{\sqrt{a}} \right ] \approx 1\) for most values of \(a>1\), each iteration supplies slightly less than \(2^n\) decimal places of accuracy.

Note that this analysis only applies to square roots of real numbers. For complex square roots, the initial guess in Newton’s Method must be complex, and the solution of the recurrence is more complicated.

Hungry Goats

Problem: A goat is tied to the edge of a circular plot of grass by a length of rope. How long should the rope be so that the goat eats exactly half of the grass?

Solution: Let the radius of the plot of grass be \(R\) and the length of the rope be \(L\). The center of the circular plot is \(O\) and the goat is tied to the edge of the plot boundary at \(C\).

Note that the area representing the grass that the goat eats is the intersection of two offset circles: one being the plot of grass (green), the other defined by the area the goat can move given that it’s tied where it is (red). Let the points of intersection of the circles be \(A\) and \(B\).

This area of intersection looks difficult at first, but it is really two circular segments: one for the green circle, and one for the red. A circular segment is the area between a chord of a circle and the arc it bounds. The green segment lies between arc \(\widehat{ACB}\) and line \(\overline{AB}\). The area of this segment is the difference between the area of the sector bounded by the arc \(\widehat{ACB}\) and the lines \(\overline{OA}\) and \(\overline{OB}\), and the triangle \(\bigtriangleup{AOB}\).

Let the angle subtended by the arc \(\widehat{ACB}\) be \(2 \phi\). The area of the sector \(A_{GS}\) is given by \(A_{GS}=\frac{1}{2} R^2 (2 \phi)\), and the area \(A_{GT}\) of triangle \(\bigtriangleup{AOB}\) is given by \(A_{GT}=\frac{1}{2} R^2 \sin{2 \phi}\). The area of the green segment \(A_G\) is then \(A_G=\frac{1}{2} R^2 \left (2 \phi – \sin{2 \phi} \right )\).

Results are similar for the red segment. If the angle subtended by the arc \(\widehat{AOB}\) is \(2 \theta\), then the area of the red segment \(A_R\) is \(A_R=\frac{1}{2} L^2 \left (2 \theta- \sin{2 \theta} \right )\). The area of the grass eaten by the goat is the sum of the areas of the red segment and the green segment, \(A_R + A_G\).

This area depends on four parameters, \(R\), \(L\), \(\phi\), and \(\theta\). Of these, \(R\) is given, and \(L\) is what we are tasked to find in terms of \(R\). This leaves us to find the other two parameters in terms of \(R\) and \(L\).

We do this by noting that triangle \(\bigtriangleup{AOC}\) is isosceles. From this triangle, we see that \(L=2 R \sin{\frac{\phi}{2}}\) and \(2 \theta + \phi = \pi\). These two relations will allow us to get a single equation relating \(L\) to \(R\).

We begin by expressing the condition that the area the goat eats, \(A_R + A_G\), is one half of the area of the circular plot, \(\pi R^2\):

\(A_R + A_G = \frac{1}{2} R^2 \left (2 \phi – \sin{2 \phi} \right ) + \frac{1}{2} L^2 \left (2 \theta- \sin{2 \theta} \right ) = \frac{\pi}{2} R^2\).

Let \(\beta = \frac{L}{R}\). Then \(\phi = 2 \arcsin{\frac{\beta}{2}}\) and \(\theta = \frac{\pi}{2} – \arcsin{\frac{\beta}{2}}\). Diving both sides of the area equation above by \(R^2\), and plugging in the above relations, we get the following equation for \(\beta\):

\(4 \left ( 1 – \frac{\beta^2}{2} \right ) \arcsin{\frac{\beta}{2}} – 2 \beta \sqrt{1 – \frac{\beta^2}{4}} + \pi \left ( \beta^2 – 1 \right ) = 0\).

(Yes, I combined a lot of steps here, including some non-trivial trig simplifications. This stuff is really better off left to the reader.)

This equation cannot be solved in analytical closed form. I think this is what is unexpected from a cursory inspection of the problem. So, you need some tool to solve it. Fortunately, Wolfram Alpha will solve it just nicely for free. Go to the site and type in the following string into the box: “findroot[ 4 (1 – b^2/2) ArcSin[b/2] – 2 b Sqrt[1 – b^2/4] + Pi (b^2 – 1) , {b,1}]”. The result is that, to six significant figures, \(\beta \approx 1.15873\). That is, the length of the rope is about 15.9% larger than the radius of the circular plot.

NB Some of you may scoff at my use of a web tool to solve my equation, and wonder what happened to rigorous analysis. The truth is that the solution of such equations has become so routine and cheap that, unless you are demonstrating something special about the solution process, there is little value in going through all of the low-level details about solutions. I know this contradicts something I posted earlier today, but the important detail was the derivation of the above equation, from which we obtained the solution.

IYI (If you’re interested): This problem is structurally very similar to those found in certain optics applications. In particular, folks who model image formation in microscopes and similar systems deal with geometry just as in this problem.  Actually, a little more general.  Imagine the following problem: there is a circular plot of grass, and a goat is tied up somewhere with a rope of a given length.  How much of the grass can the goat eat?  Now imagine two goats, each tied with rope of the same length, but in different places.  Again, how much of the grass can the goats eat?

I published a complete solution to this problem, but in the context of imaging a transilluminated object with a microscope using an extended source.  The problem, of course, can get far more complicated in the optics context, and all analogies with goats and grass gets lost when we ask about aberrations and defocus.  If you have the stomach for this, I published another paper which dealt with this.


An interesting sum

Problem: compute, in closed form, the following sum:
\(S = \displaystyle\sum_{n=1}^{\infty} b_n 2^{n}\),
\(b_n = 2 + \sqrt{b_{n-1}} – 2 \sqrt{1 + \sqrt{b_{n-1}}}\),
and \(b_0 = 1\).

Solution: Make the following substitution:

\(b_n = (p_n-1)^2\).

Then we find that \(p_n^2 = p_{n-1}\) and \(p_0 = 2\). Therefore:
\(p_n = 2^{2^{-n}}\).

The sum desired then takes the form

\(S = \displaystyle\sum_{n=1}^{\infty} \left (2^{2^{-n}} – 1 \right )^2 2^n\).

It is not obvious how to go from here. It’s not even obvious from a cursory inspection that the sum converges. We can verify convergence by observing that

\(\displaystyle\left (2^{2^{-n}} – 1 \right )^2 2^n \sim (\log 2)^2 2^{-n} \; (n \to \infty )\).

Now that we know that the sum converges, we can continue. We could expand the sum to evaluate, but there is a diverging piece which would give us fits (the \(2^n\) piece). Knowing that the sum converges, and therefore the divergences cancel, we write

\(S = \displaystyle\lim_{m\to\infty} S_m\)
\(S_m = \displaystyle\sum_{n=1}^{m} \left (2^{2^{-n}} – 1 \right )^2 2^n\).

The trick to see here is that
\(\displaystyle\left (2^{2^{-n}} \right)^2 = 2^{2^{-(n-1)}}\).

\(S_m = \displaystyle\sum_{n=1}^{m}2^{2^{-(n-1)}} 2^{n} – 2 \displaystyle\sum_{n=1}^{m}2^{2^{-n}} 2^n + \displaystyle\sum_{n=1}^{m} 2^n\), or

\(S_m = 2 \displaystyle\sum_{n=0}^{m-1}2^{2^{-n}} 2^{n} – 2 \displaystyle\sum_{n=1}^{m}2^{2^{-n}} 2^n + 2^{m+1}-2\).

The terms in the first two sums all cancel except for the first term in the first sum and the last term in the last sum. We can then write \(S_m\) in closed form:

\(S_m = 2 – 2 \displaystyle\left (2^{2^{-m}} – 1 \right ) 2^m\).

Using the fact, recited above, that

\(\displaystyle\lim_{m\to\infty}\left (2^{2^{-m}} – 1 \right ) 2^m = \log 2\),

we can then write the solution as

\(S = 2 (1 – \log 2)\).

Of Burdens and Trophies; or, How I Got Here

I’m sure that everyone has someone in their family who was, at one time, at least a little famous. Maybe not a parent, child or sibling, but perhaps a cousin, or a uncle, or a great-aunt. What value does such proximity to fame hold? In most cases, about zero. So what if your second cousin, thrice removed, was in that TV commercial for Preparation-H; not only did that get you no free hemorrhoid relief, but it didn’t even garner you the additional attention you desired.

On the other hand, I’ve lucked out. No, my relative is not some actor. She’s not even very famous for that matter. But she was well-known in her day…at least in certain circles. Dairy circles, that is. But that’s not what concerns me. What is even better is that she authored an autobiography, called The Burden and the Trophy. For someone like me who is always trying to get more detail about the ancestors who came to these shores from the Old Country, this is a treasure trove.

Don’t get me wrong: the book isn’t exactly a bestseller.  That’s putting it mildly.  My Great-Aunt Maete was known for the Watertown Dairy farm and the big house in Watertown on Grove Street around which the farm was based.  Aunt Maete’s goals are pretty much summed up in the Foreword:

I trust when you turn the pages to read my story you will derive as much pleasure from it as I did writing it, and that you will be proud of your Grandma Shick. (Emphasis added.)

That is, I’m not sure she envisioned anyone beside her grandchildren reading the book.  Maybe she would be delighted that her great-nephew read the book 40 years after her death.  No matter: this book is really of interest to descendants of the characters in the book.  Which means, for me, my grandfather Leo (Lajzar in the Old Country), her younger brother by 18 years. (!)

(Sidebar: Maete was born in 1885, I was born in 1970, just after she died.  Her being my great-aunt may have you scratching your head; stay with me.  Maete was 18 years the senior of Grandpa Leo, born in 1903 [and 108 next week].  Leo was pretty old when my Dad was born in 1945, my Dad being the youngest of four by a country mile.  Maete’s grandkids were older than my Dad, her nephew.)

The book itself is really two books: growing up in the Old Country, and making a life for herself here in the Boston area.  As far as giving you an objective impression of the book as an autobiography, well…OK, I’ll try.  I think the strength of the book lies in its structure; the two separate parts represent the clean break Maete had when she left her family in Europe.  Maete manages to answer the basic questions a descendant of hers has: why did she leave Europe in 1908?  Why Boston?  I wish she said more about my grandfather – he is only mentioned three times – but he gets more of a billing than other siblings (9 kids in the Gordon family) and I am able to understand a bit of his life in conjunction with what my Dad has told me.

That said, despite what the publisher, Pageant Press, says all over the jacket, this book is really only of interest to her descendants and maybe to Watertown historians.  The book itself is very hard to plow through at first.  The writing seems subpar; I identified grammatical and usage mistakes throughout.  This is likely not Aunt Maete’s fault: she wrote the book in her mother tongue, Yiddish, and Pageant had the work translated by one Mary J. Reuben.  Ms. Reuben could have used a better editor for sure.  Other things, however, lie with the original writing.  The story is difficult to follow: Maete’s mother is pregnant a lot, for sure, but one can never be sure when she is or isn’t.  Siblings are introduced and are rarely heard from again unless something happens (one dies).  Times are very difficult to track: the chapters seem to spill beyond each other.  Only the fact that this story was about me in a way made me scour the text for logical connections; without such a fundamental motivation, I wouldn’t bother.

That all said, the book is worth its weight in gold – way more than that – to me for what it is: a document of my ancestral heritage.  OK, 1/4 of my ancestral heritage, but the one with my namesake.  This is a big deal.  You may have stopped reading long ago, I don’t know; nevertheless, I am going to put down what I have learned so far about my heritage.

First of all, some perspective on a topic that has always been confusing to me.  My grandmother Lillian (Liza in the Old Country), Leo’s wife, grew up in Vilna, which is now known as Vilnius, the capital of Lithuanian.  Yet Lillian spoke, with my grandfather, Polish. (Yiddish also, of course.)  This always confused the shit out of me and I never got a decent answer to my questions until recently, when the book forced me to think hard about Eastern Europe at the turn of the 20th Century.

So, Maete was born in a town called Lubyotka and her family settled in another, larger town called Vassilishok.  Now, a naive soul googles “vassilishok” and gets babkes.  Did it disappear off the map?  No, no, although, as you can imagine and I will mention below, it may as well have.  What I found via the site JewishGen is that Vassilishok was the Yiddish name for the current town Vasilishki, Belarus.  Belarus!  Leo was born here in Vasilishki, and he, as far as anyone knows, never spoke a word of Belorussian.  Not because he only spoke Yiddish; he spoke Polish, as did most Jews who, as opposed to stereotypes, did interact with the greater non-Jewish population.

The reason for the weird language thing is that, in the 19th and early 20th century, these areas (Vilnius and Vasilishki, as well as Belorussian cities like Grodno and Lida) were a part of Poland.  Poland, as some of you may know, was taken over by Germany and Russia between 1795 and 1945 more times than one cares to count.  The region we understand to make up Poland, Belarus, Lithuania and Ukraine today was the land under the Polish-Lithuanian Commonwealth.  It turns out that this Commonwealth was very progressive and allowed different peoples to flourish, Jews included.  Much of this area became heavily Jewish and Jewish life and culture thrived here.

The Commonwealth ceased to be in 1795.  The modern borders of Poland, Belarus, Lithuania and Ukraine were established after WWII.  In between, up until WWII, are details I will leave to my reading of Timothy Snyder’s The Reconstruction of Nations: Poland, Ukraine, Lithuania, Belarus, 1569-1999; let’s just say that national myths were created, and peoples and their respective languages dominated the land inside the borders.  None of this, of course, bade well for Jews living in this region; Snyder called this region of the Commonwealth the Bloodlands.

What I did learn about Vasilishki during WWII was this:

The ghetto was surrounded on May 9, 1942, and no one escaped.  On May 10 prisoners started to be taken out to the Jewish cemetery on the outskirts of town in groups of 60 people.  The graves were already prepared.  They ordered people to undress and stack their clothes in piles.  They were then pushed into ditches and shot.  This continued until May 11.  The total number of those shot over those two days was 2,159.  A total of 2,865 of all nationalities died in Vasilishki and surrounding region during the years of the occupation, and 598 were sent off to Germany.  A special group of the SS from Lida took an active role in the pogrom under the command of the headquarters officer Windish and his assistant Vasyukevich, together with the gendarmie and Gestapo of neighboring regions.

The Belarussian police, headed by commander Yezhevsky, and aided by Tubilevich, Vitold Schmigger and Nikolai Zhurun — who distinguished themselves with particular cruelty – provided significant aid to the Nazis.  The Aid Commission of the [acronym for a State Committee] of the USSR of the Vasilishki Region put together a list of 616 families from the peaceful population who were victims.  The majority were Jews.

Now, I’ve read loads of Holocaust horror stories, but when you know that your extended family was subjected to this inhuman Hell, it hits you like a gut punch.  When I found the above, I just stared at the text as a whole block, afraid to read the words for a while.  I know I lost distant family in The War, but placing a name on the town and seeing numbers like that makes my flesh crawl.  My task now is to find names; some of those doubtless will be Maete’s siblings and their kids – my cousins once removed.

Then I found this little gem:

In 1967 an obelisk to “Soviet citizens” was erected to those killed by the Fascists.

And people want to know why Jews wanted to get the flock out of Soviet Russia from the 1960’s on.

OK, back to Maete.  Her father (my great-grandfather) was a cattle trader named Mordechai Isaac Gordon; her mother was Chaye Reva der Brashkier Gordon.  Maete, as I alluded to above, was the oldest of 9 children.  (My grandfather Lajzar was the 8th.)  Anyway, the first part of the book details living conditions for Maete growing up.  They were the equivalent of middle class, which in absolute terms in America means dirt poor, but they had a house and could support an army of kids, so, as the Yiddish parable says, it could be worse.

Maete is “modern” in the sense that she knows she is a looker (although, from the blurry photo on the jacket, you’d be forgiven for wondering what happened in her old age; of course, I inherited these looks, so perhaps I should just shut up now), and wants to follow her heart.  This is, of course, the Old Country, so…never happen.  Maete is fixed up with Isaiah Shick, a native of Vassilishok just back from Buenos Aires, having worked there for a brother in his hat factory.  Maete is none too pleased at the arrangement when she learns that Isaiah is illiterate, but relents to preserve her family’s reputation.  (Actually, Maete breaks off the engagement, but her scheming parents make Maete and Isaiah godparents to my just-born grandfather; the godparent-to-my-future-grandpa-thing works its charm and the wedding is set for, like, 2 weeks after my grandpa’s bris.  Good G-d!)

Now, this is late 1903.  Maete and Isaiah are trying to figure out what to do; my great-grandfather Mordechai forbids them to go back to Argentina.  (A big “whew!” on my part.)  Maete and Isaiah start to plan to settle in Vilna.  But, history intervenes in the form of the Russo-Japanese War.  (Vassilishok, and Poland generally, is under the rule of the Tsar.)  Isaiah is at extreme risk of being drafted into the military to fight, so he quickly makes a plan to haul ass somewhere where he and Maete could plausibly start a life.  It turns out that Isaiah has yet another brother, Feive, who tells Isaiah that coppersmiths (which is Isaiah’s trade, I guess) are in high demand in Boston in the USA.  (A big shout out to the Russian and Japanese empires, as well as brother Feive.  Thank you all; without you, I would not be here.)  So, after an interlude in Paris, Isaiah travels to Boston, leaving his pregnant wife behind.

Maete and her new daughter stay back in Vassilishok for four years before they make the move to Boston.  (Yikes.)  Nonetheless, she comes; the reunited family settle in a shoddy apartment in the West End.  Maete is humbled by how much they will struggle even in modest circumstances on Isaiah’s pay, so she decides to go into business.  Based on her father’s trade, she decides to try dairy farming.  (!!)  After a humble beginning, Maete eventually moves the works out to Watertown, a ten-cent cab fare from the West End; the growing family moves to Watertown permanently as the business and family grow.  The book chronicles the struggles of running the farm in the face of vicious competition from the likes of Hood (who messed with her bottles) and myriad state regulations which made dairy farming an expensive business indeed.

What’s interesting to me, ultimately, is any information I can glean about my grandfather Leo/Lajzar.  (BTW my Hebrew middle name is Lazar; my grandfather was very much alive when I was born.  So what gives?)   I know that he and my grandmother essentially eloped in 1930 (Lillian’s mother had a stroke soon after, so I am told) and came to Boston, ostensibly because Maete had work for Lajzar.  I believe they arrived in 1931, but the text seems to imply that Lajzar was helping Maete’s son Hyman with milk routes in 1929 or so.  I don’t know, but this is why I found the book a little frustrating; some immigration records would help sort this out.

Anyway, Lajzar gets a third mention toward the end of the book as Maete talks about his taking on a major route in Roxbury and Dorchester because all of her customers from the West End have moved to those places.  This makes perfect sense, as my grandparents settled in Dorchester (well, “settled” is not a great word; my grandfather was somewhat ornery and moved the family a lot due to battles with landlords).

At long last, this takes me through the family history I have gleaned from reading this valuable book.  Thank you Auntie Maete for having the patience and guts to write a book that, while not exactly something to be paraded in the NYRB, is something which maintains a link between your generation and ours.  The book is a lifeline, and is your permanent imprint.


What, pray tell, will make Obama’s critics happy about his position vis-a-vis Israel?

Reading reactions from right-wing politicians and commentators to Obama’s Mideast speech yesterday, one wonders what it would take to make these guys happy.

Several of my friends yesterday immediately went ballistic. Over what? Mention of “1967 lines”, which of course became “pre-1967 borders”. Jeffrey Goldberg, as usual, clears up the confusion:

President Obama didn’t “insist” that Israel return to its 1967 borders. He said the 1967 borders should form the basis of negotiations, and that Israel and Palestine should swap land, land swaps that would bring settlement blocs and East Jerusalem Jewish neighborhoods into Israel proper.

BTW, Goldberg’s reaction to Romney is pitch perfect:

Obama has thrown Israel under a bus? Top officials of the Israeli defense ministry have been telling me, and other reporters, for a couple of years now that military cooperation between their country and America has never been better. Some bus. There are a lot of countries out there that would like to be thrown under simliar buses.

Perhaps this is the sort of reception Republicans expect Obama to have for Bibi:

White House Reception

[BTW I did not mean to imply that Bibi would destroy the White House on his visit.]

But seriously, what are we, as Americans, supposed to do? I love Israel and I worry for her long-term future. But it makes me sick to stand here and watch her leaders make poor decisions, time after time. I do know that the Palestinian leadership is no gift, especially as it has teamed up with the apocalyptic Jew-haters of Hamas. That said, Israel needs to draw borders and soon if it cares to be a Jewish democracy.

The Truth about Gandhi?

A new book about Gandhi has made some waves this week, not least because it exposes Gandhi as a bisexual, even as, as reactionary historian Andrew Roberts put it, a “sexual wierdo”. [For a classic exposé on Roberts, see this piece by Johann Hari.] Apparently, this book is already banned in Gujarat [Gandhi’s home state] and may be banned in the whole of India.

Personally, I don’t see what the big deal is. Yes, I know I have very liberal beliefs regarding this subject. But still, what exactly does it change? To me, it seems like Gandhi and his wife Kasturba become like FDR and Eleanor: a powerful couple but for all intents, not physically bonded.

Then again, my views on Gandhi are very complicated and still in flux. I loved the 1982 movie and who could not see it and want to learn more about this Great Soul? He absolutely was a leader of men and a great man of his time, and deserves to be hallowed by the people of India.

But, in my view, his achievements are way exaggerated. I do not want to minimize the achievement of being able to get the message out to all of the Indian people about the wrongs of the British. But the principle of ahimsa, or nonviolence, worked only in his special case, where there were 3000 Indians for every British in India during Gandhi’s time. The British further had no interest in exterminating the natives, just controlling them enough to get at the jewels of the land. Gandhi knew that if enough people were willing to take the blows and the bullets, then the British were in a no-win situation.

But even with those odds, Gandhi did not succeed in his mission. He was only able to seem like he drove the British out where, in fact, the British treasury and strength were completely sapped due to WWII. WWII permanently changed the political face of the earth and, although the British were on the winning side, they ultimately became amongst the long-term losers. Colonies in the Levant, Singapore and South Africa were breaking away. India was going as well, Gandhi or no Gandhi.

It’s not to say that Gandhi played no role in this. Gandhi helped define India as a secular, socialist state for the first 45 or so years of its existence. [Nehru would have much more of a role in hardening the clay formed by Gandhi.] That is, Gandhi helped India down the path of a terribly poor, third-world nation, unable to keep its most talented citizens from leaving to enrich nations like the US and the UK. Only recently has India begun to make use of its considerable manpower, although it still houses 1/3 of the world’s poor.

Worse, Gandhi was so blinded by ahimsa principle that he thought it could solve problems everywhere. Gandhi’s answer to the tragedy of the slaughter of Europe’s Jews was, of course, nonviolence. But did he not understand that the Nazis were bent on annihilation, not subjugation? They did care a bit for world opinion, but only in peacetime; the real killings took place during the war, and they were convinced that they were in the right. And, further, Jews made up, at most, 10% of any country – and even that is an anomoly, a typical figure was 2%. Greatly outnumbered, only wanted dead by a ruthless state, how was ahmisa to work in the face of incredible himsa? At some point, the whole thing is just silly.

Ahimsa further had no role to play in the controversy in Kashmir, a conflict that is active to this very day and could have greater consequences to the world than anything in the Middle East. I know little about the role Gandhi played in Kashmir, although it could not have been very great given its history. The only thing I could imagine Gandhi doing at this point is fasting, only nobody would care now.

That said, Gandhi remains a fascinating figure of 19th and 20th century history and I look forward to reading this new book.

Thomas Pakenham’s The Boer War

I have this odd fascination with England and her Empire. I also have this odd fascination with Germanic peoples and languages) Of course South Africa and her Afrikaners then are a natural subject toward which I tend to gravitate. Plus, I am a sucker for war and various other misery, so hell, why not learn about the Tweede Vryheidsoorlog, or the Boer War [actually, the Second Boer War].

I first learned about the Boer War from watching Breaker Morant. [NB One is not allowed to live freely in Australia without watching Breaker Morant, period. I think it is a condition for obtaining a visa.] Breaker Morant, of course, takes place during the Boer War in 1902, rather late in the game. The Boers have gone guerrilla on the British, who in desperation get their colonial subjects to pick up the slack. Breaker Morant his his chum Handcock [am I the only one to notice that this is a rather unfortunate name?] are part of an Australian regiment, the Bushveldt Carbineers. They get into a battle with some Boers who kill their leader, Captain Hunt. Morant and Handcock, as vengeance, kill a bunch of Boer prisoners and a German missionary. Morant and Handcock are arrested, tried and found guilty of murder, for which they are put to death by firing squad. Bloody evil Poms!

For all the rage this incident caused in Australia, it is but a single paragraph of the 700-page history weaved together by Thomas Pakenham, one of the finest popular works of history I have ever read. There are 3 reasons why Pakenham’s account stands out from so much of the other fine works I have read. First, his timing is impeccable. This book was written over 10 years in the 1970s, a period in which Pakenham was able to track down over 50 surviving veterans of the war. [The youngest veteran was 86 years old!] By getting access to these men and so much first-hand material, Pakenham is able to provide incredible detail in every facet of the agonizing story, especially in describing many of the battles.

Second, Pakenham does more than simply recite the facts and weave them into a narrative. Pakenham is able to make a case that the previous histories were biased and faulty. For example, Pakenham works hard throughout the work to resurrect the reputation of Sir Redvers Buller, who was blamed for the sorry state of the early part of the war and left South Africa with his reputation in tatters, despite a long string of victories. Pakenham details the infighting in the War Office and the Army and concludes that, in many cases, the British were their own worst enemy. Pakenham is also able to present the Boer perspective based on his primary sources.

Finally, Pakenham can bloody well write. This work of history reads like a novel. Because Pakenham takes a stand on certain topics, there are clear heroes and villains and this makes for a terrific popular history. Further, Pakenham provides just the right level of detail in his maps: every feature marked in the maps represent a significant point of discussion. Despite the length, this is a lean work presented with efficiency.

I cannot recommend this book more highly.  That is, unless you are Australian and believe Breaker Morant was sacrificed by Lord Kitchener on the altar of international intrigue.