Unleash Your Curiosity: An average is an average unless Einstein calls it something else

In past series, I’ve directly challenged Einstein’s theory of relativity, which many have found insightful. Unsurprisingly, others reject a direct challenge to one of their most deeply–held beliefs. This leads to the question: Is it possible for someone with such a deeply–held belief to be receptive to ideas that directly challenges their belief. Fortunately, the answer is yes! I believe that scientists have an innate curiously; one where they will explore and examine things that pique their interest. This series is intended to do exactly this: The goal is to present you with several interesting “insights” which you can independently explore and confirm. If at any time you say to yourself “Hmmm?”, then I will ask you to unleash your curiosity!

In Part 1 we are simply going to look at how to calculate an average (arithmetic mean), ξ, of two numbers, s and t, where their average is found using the addition mean equation:

ξ = (t + s) / 2.

Consider a specific example where a laser of wavelength x’ is emitted from a stationary location and a moving vehicle is approaching or receding from this location at velocity v. The corresponding Doppler equations are:

c * x’ / (c – v) and c * x’ / (c + v).

Using the addition mean equation, their average is:

ξ = c^2 * x’ / (c^2 – v^2).

Does this average define the position or a location coordinate of the moving vehicle?  You should agree that the answer is no.

Alternatively, the average Doppler equation can be found using the mathematically equivalent, but not widely–used, subtraction mean equation:

ξ = t – ½ * (t – s).

It is important to understand that the absolute value of the expression ½ * (t – s) is called the half–difference.

Now I will ask you to use the subtraction mean equation to find the average using the following steps:

  1. Factor out the “c” variable from the numerator in the original Doppler equation so that you begin with x’ / (c – v), which we’ll use at “t”, and x’ / (c + v), which we’ll use as “s”. You will multiply “c” into the final equation at the end.
  2. Find the actual half–difference and replace the half–difference in the subtraction mean equation, resulting in: t – vx’ / (c^2 – v^2).
  3. Replace the “t” in the subtraction mean equation with x’ / (c – v) and simplify.
  4. Multiply the result by “c” and confirm that you have arrived at the same average equation: ξ = c^2 * x’ / (c^2 – v^2).

While finding the average using the subtraction mean equation involves more steps, you will arrive at the same answer that is found using the addition mean equation. Now for a critical question: If you use the subtraction mean equation to find an average, but fail to recognize that you’ve done so, do you get to redefine the equations and expressions any way that you choose? You should agree that the answer is no.

Please keep the subtraction mean equation’s final equation and intermediate expressions in mind as you (re)read section 3 of Einstein’s 1905 paper: On the Electrodynamics of Moving Bodies. Hint: Einstein mathematically performs every step I asked you to perform (above) to find the average, but fails to recognize his use of the subtraction mean equation.

If during your review of Einstein’s paper you said “Hmmm?”, then you will find the remaining insights in the series interesting.


Steve_aes-114Steven B. Bryant is a researcher and author who investigates the innovative application and strategic implications of science and technology on society and business. He holds a Master of Science in Computer Science from the Georgia Institute of Technology where he specialized in machine learning and interactive intelligence. He also holds an MBA from the University of San Diego. He is the author of DISRUPTIVE: Rewriting the rules of physics, which is a thought–provoking book that shows where relativity fails and introduces Modern Mechanics, a unified model of motion that fundamentally changes how we view modern physics. DISRUPTIVE is available at Amazon.com, BarnesAndNoble.com, and other booksellers!

Images courtesy of Pixabay.