My Last Love

I should be over her by now.  I'm not.

Have you heard the expression . . .

"I don't know what to do with myself"?

Well, that describes me to a T.

Playing with i pi

We learn in math that e to the i pi equals -1.  [e in the base of the natural logarithim, i is the square root of minus 1, and pi is the ratio of the area of a circle to the radius squared.]  Squaring both sides gives e to the i pi squared equals 1.  Now e to the i pi squared equals e to the 2 i pi and 1 equals e to the 0.  Have I proven, falsely, that 2 i pi equals 0 so that i equals 0?  Actually not.  What I have proven (again falsely) is that 2 pi equals 0.  It's like saying that since sine of 2 pi equals sine of 0, then 2 pi equals 0.  And if you think in terms of angles, the angle 2 pi is the angle 0.  This is because sine of x is a cyclic function of x. Well, it turns out that e to the i x is also a cyclic function of x for e to the i x equals cos x plus i sin x.  In other words, it is a trigonomic function. 

I Was There

I bought a book today by David Weinberger titled "Too Big to Know" which seems to be of the "Oh, Wow!" genre.  He spent almost a page expounding on how big a number sextillion is.  What really caught my attention was "the most popular corporate computers of the 1950s the IBM 650.  With only only seventy-five of these machines installed anywhere . . . ."   "The 650 cost your company $500,000, equivalent to $4 million in 2011 dollars.  It's got its own room, its own fleet of maintenance folks, its own dress code:  white lab coats, please."  Well, I happened to be working with computers at an insurance company when we got our IBM 650.  "Only seventy-five" seems inconsistent with "most popular."  It didn't cost the company $500,000 because IBM didn't sell them, they leased them.  Ours didn't have its own room and the maintenance was by one IBM employee who wasn't there full time and as for white lab coats, nary a one.

There was a footnote:
5. Leonard Dudley, Information Revolution in the History of the West (Edward Elgar Publishing, 2008), p. 266.

Useless Information

Recent posts discussed the amount of matter equivalent to the energy need to accelerate a Toyota Corrola.  Enen if my calculations are correct, which is doubtful, it is useless information. Obviously there is nothing in an automobile which can turn so small amount of matter into equivalent energy.  This is done elsewhere in a nuclear reactor by the fission of radioactive isotopes.  Even there, only a very small fraction of the mass becomes energy. The number of protons/neutrons remains the same. (This was true even in the atomic bomb.)  To find out how much, you need to compare the atomic weight of the isotope to the total atomic weight of the decay products.  [Way beyond me.]

Then there is a series of processes to convert heat to electricity and transmit to where it needs to be to charge the battery. There are energy losses along the way.

Oops!  The Corolla is not an electric car.

Silly Example

What is the equivalent energy in grams required to accelerate a car to sixty miles per hour in 15 seconds.

Data:
The car weighs 2800 pounds.
One pound equals 453.59 grams.
One mile equals 1.6 kilometers.
The speed of light is 300,000 kilometers per second.

Calculation:
(2800 x 453.59 x 1.6 / (60 x 9)) x 10 ^ -10

= 3.768 x 10 ^ -7

= 0.0000002768 grams

Next blog will cover why this is useless information.

Dimensions

I was taught in university science classes that the dimensions on the left side of an equation should match the dimensions on the right.  I had trouble applying this to Einsteins equation because energy is expressed in so many ways (joules, foot-pounds, kilowatt hours).  I finally figured out that the right way to look at it is energy = force times distance and force = mass times acceleration and acceleration = distance divided by time squared.  Combining these gives energy = mass times distance squared divided by time squared, which, of course, matches the dimensions on the right side of the equation.

My next blog will give an example.