Exactly What was On the Blackboard in the
1951 Film, The Day the Earth Stood Still ?
And What Do the Equations Mean?

(Content not attributed to anyone else, nor found on the
board in the film, is Copyright©2025 by Daniel B. Sedory.)

Note: This is an ongoing and currently incomplete project;
many comments about the equations must still be added!

In the 1951 film, The Day the Earth Stood Still, the main character "Klaatu" observes the blackboard of professor Barnhardt in his study containing various equations and a number of warnings to not erase or even touch it. Before moving on to the meaning of the equations, it must be noted that most of what you see on the blackboard had been developed by Samuel Herrick, an astrophysicist at UCLA at that time, who was hired as a science consultant for the film by producer Julian Blaustein.^[1]
"Herrick reasoned that an equation related to celestial mechanics would be most appropriate, specifically an equation related to his own work on the “three-body problem” in astronavigation."^[2]
Three quarters of a century later, I became interested in trying to understand the contents of that blackboard and to confirm whether all of the equations actually made sense; the following are my conclusions.

I have also transcribed all the equations from a single sheet included in the Samuel Herrick Papers on a separate page: Samuel Herrick's Mathematical Notes concerning the film, since they could be quite helpful in understanding what appears on the Blackboard.

Lastly, but of great importance: The equations on the board are given without any indication of what the variables actually represent! However, we already learned above, that the professor in the film would be someone with a knowledge of celestial mechanics, and from his conversation with the "Klaatu" character, they would be related to getting from planet to planet (or at least into a stable orbit about the earth). This would involve some now rather well-known equations from which we can assume the variables on the blackboard would agree with. 20 years later, Herrick wrote a textbook, titled, Astrodynamics; which might be helpful in this regard.^[3] But I plan on adding descriptions of them wherever they are first introduced.

The Blackboard

In the upper-left corner of the blackboard you can see^[4] what I've copied into the LaTex markup language here:

Figure 1

$\frac{d^{2}x}{d\tau^{2}}=\ddot{x}$ $=-\frac{x}{r_3}+m_3\left(\frac{x_{23}}{r_{23}^3}-\frac{x_{13}}{r_{13}^3}\right)$
$\ddot{y} =-\frac{y}{r_3}+m_3\left(\frac{y_{23}}{r_{23}^3}-\frac{y_{13}}{r_{13}^3}\right)$
$\ddot{z} =-\frac{z}{r_3}+m_3\left(\frac{z_{23}}{r_{23}^3}-\frac{z_{13}}{r_{13}^3}\right)$

The Greek letter tau “$\tau$” refers to time in these equations. The double dots above the x, y and z mean the same thing as taking the second order differential of x, y or z with respect to $\tau$ (time); as defined in the first line.

Commentary: What we have here is indeed related to the equations for the “Three-Body Problem” in which these three describe the accelerations in three dimensions of the mass of only the third body $m_3$ where $r_{12}$, $r_{23}$ and $r_{13}$ are the distances between each of two of the three bodies. So, the whole 3-body problem would need to include two other sets of equations similar to these for $m_1$ and $m_2$. But the form here is unusal compared to how the problem is normally written.

Below that, we have:

Figure 2

$Restrict$ $by:$

$r_{23} = r_{13} = const.$
$x = 0$ $y = r$ $z=z_{13}=z_{23}=0$

$Adopt:$ $m \equiv m_{3} / r_{13}^3 $

$Then:$ $\ddot{x} = 0$ $\ddot{y} = \ddot{r}$

$\ddot{r} = \displaystyle -\frac{1}{r^2} - mr$

By using those restricitions, much of what is written above (that is, the $\ddot{x}=$, $\ddot{y}=$ and $\ddot{z}=$) is reduced to only that which follows the "Then:" on the blackboard; including $\ddot{x} = 0$. Apparently, the fact that $\ddot{z}$ also becomes zero was considered so trivial that it doesn't even appear on the board. So the only equation of the first set of three to remain is:

$\ddot{y} = \ddot{r} = \displaystyle -\frac{r}{r_3}+m_3\left(\frac{r_{23}}{r_{23}^3}-\frac{r_{13}}{r_{13}^3}\right) $

But, even after employing $m \equiv m_{3} / r_{13}^3$, how does that result in what follows? The next line on the blackboard being:

$\ddot{r} = \displaystyle -\frac{1}{r^2} - mr$ (since it appears the whole $m_3$ term would become zero)!

This is only a 'placeholder' where at some point in the future, I will discuss how the equation above might have been obtained.

The equations continue below that with:

Figure 3

$2\dot{r}\ddot{r} = \displaystyle -\frac{2\dot{r}}{r^2} - 2mr\dot{r}$

But it's difficult to be certain of the left most side of the board for:

$r^2 = \displaystyle \left(\frac{dr}{d\tau}\right)^2 = \frac{2}{r} - \frac{1}{a} - mr^2 $

Where did the "$a$" in “$-(1/a)$” come from in that last equation?

Before continuing with the equations on the board, it must be noted that one of the fractions near the top middle of the board is missing its numerator:

Figure 4

After looking closely at the background erasures on the board and where something should have appeared (as pointed out in the image at the left), it's possible that the numerator had never been written on the board. However, we cannot completely rule out that someone did intentionally erase it.

In either case, after looking at the line which follows it, one can be quite certain that the missing part is: $\sqrt{r}\ dr$.

$\therefore M = \displaystyle\frac{k\sqrt{m_1 + m_2}}{(-a)^{3/2}}\ \left(t-\tau \right)$^[5] $= \displaystyle\int \frac{d\tau}{(-a)^{3/2}} = \int\frac{[\ \sqrt{r}\ dr\ ]}{(-a)^{3/2}\sqrt{\frac{2}{r}-\frac{1}{a}-mr^2}} $

We assume that “$k\ $” is some constant related to the Three-body problem, but it's not defined on the board.

Continuing from above we have:

(Note the yellow square where part of the numerator is missing)

Figure 5

$= \displaystyle\int \frac{\sqrt{r}\ dr}{(-a)\sqrt{r-2a+amr^2}} = \int\frac{\sqrt{r}}{(-a)}\frac{d}{\sqrt{r-2a}} + \frac{m}{2} \int\frac{r^{7/2}\ dr}{(r-2a)^{3/2}} $

$ \displaystyle\frac{1}{2}\ \int \frac{r^{7/2}\ [dr]}{(r-2a)^{3/2}} = -\ \frac{r^{7/2}}{\sqrt{r-2a}} + \frac{7}{2}\ \int\frac{r^{5/2}\ dr}{\sqrt{r-2a}} $ (Piece # 121)

$ \displaystyle\frac{7}{2}\ \int \frac{r^{5/2}\ dr}{\sqrt{r-2a}} = \frac{7}{6}\ r^{5/2}\ \sqrt{r-2a} + \frac{35a}{6}\ \int\frac{r^{5/2}\ dr}{\sqrt{r-2a}} $ ( " # 118)

$ \displaystyle\frac{35a}{6}\ \int\frac{r^{5/2}\ dr}{\sqrt{r-2a}} = \frac{35a}{12}\ r^{5/2}\ \sqrt{r-2a} + \frac{35a^2}{4}\ \int\frac{\sqrt{r}\ dr}{\sqrt{r-2a}} $ ( " # 118)

The following identities (written just below what you saw above) are used in the equations on the right side of the board:

Figure 6

$r = (-a)(\cosh F-1) $

$r - 2a = (-a)(\cosh F+1) $

$dr = (-a) \sinh F \ dF $
$= (-a)\sqrt{(\cosh F-1)(cosh F+1)} \ dF $

Followed by:

Figure 7

I have already confirmed that using the identities above, this is the correct conclusion:

$\therefore \displaystyle\int \frac{\sqrt{r}\ dr}{(-a)\sqrt{r-2a}} = \int(coshF - 1)dF\ = sinhF - F $

But below that we have (Note: I was able to confirm the values of all three fractional exponents in the equations below as being $3/2$ after consulting the photograph of Herrick and Rennie that just happens to include this portion of the blackboard):^[6]

Figure 8

$ Approx: \ M \approx \displaystyle\int \frac{\sqrt{r} \ dr}{(-a)\sqrt{r-2a}} \approx \frac{1}{(-a)^{3/2}} \int \sqrt{\frac{r}{2}}\ dr = \frac{(2r)^{3/2}}{6(-a)^{3/2}}$

Note: I added approximately equal ($\approx$) signs to my transcription to make it clear that neither $M$ nor what follows it are exactly equal to the rest of the equation. After carrying out a number of calculations, I am sure the term with the square root symbol over $r/2$ was written too hastily (too short). It only makes sense if that is $\sqrt{r/2}$, so I've corrected my transcription to show that. Thus, the middle of the equation computes exactly:

$\displaystyle \approx\ \frac{1}{(-a)^{3/2}} \int \sqrt{\frac{r}{2}}\ dr\ =\ \frac{\sqrt{2}}{3\ (-a)^{3/2}}\ r^{3/2} =\ \frac{(2r)^{3/2}}{6(-a)^{3/2}}$.

The usual answer for the first part of the equation would be:

$\displaystyle \int\frac{\sqrt{r} \ dr}{(-a)\sqrt{r-2a}} = -2\ sinh^{-1}\left(\sqrt{\frac{r-2a}{2a}}\right) - \frac{\sqrt{r-(r-2a)}}{a} $

Which is also equal to (from further above):

$ = \displaystyle \int(coshF - 1)dF\ = sinhF - F $

Moving over to the right side of the board we have both the largest and overall the most difficult area of the blackboard to read.

Quite frankly, there are a number of places on the board where what was written was done so in a rather sloppy manner. And apart from various terms being repeated or otherwise having known values, it would be very difficult to transcribe accurately; especially on this much more concentrated half of the board. Many of the factorial symbols ($!$) in the equations below appear more like colons (:) than exclamation marks in the series expansions which follow these three equations:

Figure 9

$\therefore M = sinhF - F + m \displaystyle \left\{-\frac{r^{7/2}}{\sqrt{r-2a}} + \frac{7}{6}\ r^{5/2}\ \sqrt{r-2a} - \frac{35(-a)}{12}\ r^{5/2}\ \sqrt{r-2a} + \frac{35(-a)^3}{4} (sinhF - F)\right\} + \dots$

Where did the “$+\ m \left\{ ... \right\} $” terms in the equation above come from?

The following two right-hand sides of the equation make use of the identities found in Figure 6:

$= sinhF - F + m(-a)^3 \displaystyle \left\{-\frac{(coshF - 1)^{7/2}}{(coshF + 1)^{1/2}} + \frac{7}{6}(coshF - 1)^{5/2}(coshF+1)^{1/2} - \frac{35}{12}(coshF-1)^{1/2}(coshF+1)^{1/2} + \frac{35}{4}(sinhF-F)\right\}+ \dots$

$= sinhF - F + m(-a)^3 \displaystyle \left\{sinhF\left[-\frac{(coshF-1)^3}{coshF+1} + \frac{7}{6}(coshF - 1)^2 - \frac{35}{12}(coshF-1)\right] + \frac{35}{4}(sinhF-F)\right\}+ \dots$

Those equations are followed by the simplest definitions of the Taylor (Maclaurin) series for both the hyperbolic^[7] sine ($sinh)$ and cosine $(cosh)$ of $F$ which were used in the equations above to expand them:

Figure 10
$sinhF = F + \displaystyle \frac{F^3}{3!} + \frac{F^5}{5!} + \frac{F^7}{7!}+\ ...$

$coshF-1 = \displaystyle \frac{F^2}{2!} + \frac{F^4}{4!} + \frac{F^6}{6!}+\ ...$

And to the right of each of those we find these expansions;^[7] which show how the last one was derived:

$coshF + 1 = 2 +\ ...$ $(coshF-1)^3 = \displaystyle \frac{F^6}{8} +\ ...$

$(coshF-1)^2 = \displaystyle \frac{F^4}{4} + \frac{F^6}{4!}+\ ...$ $\displaystyle \frac{(coshF-1)^3}{coshF+1} = \frac{F^6}{16} +\ ...$

Figure 11

By substituting those expansion approximations into the last right-hand side of the equation in Figure 9, we proceeded as follows:

$= sinhF - F + m(-a)^3 \displaystyle \left\{\left(F + \frac{F^3}{3!} + \frac{F^5}{5!} + \frac{F^7}{7!}\right) \cdot \left[-\frac{F^6}{16} + \frac{7}{6}\left(\frac{F^4}{4} + \frac{F^6}{4!}\right)- \frac{35}{12}\left( \frac{F^2}{2!} + \frac{F^4}{4!} + \frac{F^6}{6!} \right)\right] + \frac{35}{4}\left( \frac{F^3}{3!} + \frac{F^5}{5!} + \frac{F^7}{7!} \right)\right\}+ \dots$
Or:
$= sinhF - F + m(-a)^3 \left\{\left(F + \frac{F^3}{3!} + \frac{F^5}{5!} + \frac{F^7}{7!}\right) \cdot \left[-\frac{F^6}{16} + \frac{7F^4}{24} + \frac{7F^6}{6\cdot4!} - \frac{35F^2}{24} - \frac{35F^4}{12\cdot4!} - \frac{35F^6}{12\cdot6!}\right] + \frac{35}{4}\left( \frac{F^3}{3!} + \frac{F^5}{5!} + \frac{F^7}{7!} \right) \right\}+ \dots$ which reduces to:

$= sinhF - F + m(-a)^3 \left\{\left(F + \frac{F^3}{3!} + \frac{F^5}{5!} + \frac{F^7}{7!}\right) \cdot \left[- \frac{35F^2}{4!} + \frac{49F^4}{12\cdot4!} -\frac{31F^6}{2^6\cdot3^3}\right] + \frac{35}{4}\left( \frac{F^3}{3!} + \frac{F^5}{5!} + \frac{F^7}{7!} \right) \right\}+ \dots$ But factoring out ($35/4$), gives us:

$= sinhF - F + m(-a)^3 \left\{\left(F + \frac{F^3}{3!} + \frac{F^5}{5!} + \frac{F^7}{7!}\right) \cdot\ \colorbox{lightgreen}{$\frac{35}{4}$}\ \cdot\ \left[- \frac{F^2}{3!} + \frac{7F^4}{3\cdot5!} -\frac{31F^6}{3\cdot7!}\right] + \colorbox{lightgreen}{$\frac{35}{4}$}\ \left( \frac{F^3}{3!} + \frac{F^5}{5!} + \frac{F^7}{7!} \right) \right\}+ \dots$

The substitutions, reductions and factoring of the equations above then provides the following equation for $M$ in Figure 12:

$ \therefore M = sinhF - F + \displaystyle\colorbox{lightgreen}{$\ \frac{35}{4}$}\ m (-a)^3 \left\{\left(F+\frac{F^3}{3!}+\frac{F^5}{5!}+\ ...\right)\left[-\frac{F^2}{3!}+ \frac{7F^4}{3\cdot5!} - \frac{31F^6}{3\cdot7!}\right] + \frac{F^3}{3!}+\frac{F^5}{5!}+\frac{F^7}{7!}+\ ...\right\}+\ ...$

Note that the initial parenthesis was not included on the board. Due to the length of the equation above, I split this image into two lines:

Figure 12

Note: I believe there should have been a "+ ..." at the end of the three terms inside the bracketed "$[\ ]$" area. However, in all of the equations which follow, an assumption must have been made that carrying out any calculations further than only the first or up to 3 terms for any series expansions was to be considered negligible! So the absence of the "+ ..." doesn't really matter. I decided to carry out the multiplication of all those factors to see whether I could get to the last equation by doing so. It must also be noted that the professor purposely truncated the terms within the parentheses in Figure 12 by dropping the $(F^7/7!)$ term; highly likely on purpose, in order to get a string of terms that summed to zero! I carried out a set of calculations which included it, and that resulted in a few terms (an $ F^9$, $ F^{11}$ and $ F^{13}$) that did not sum to zero.

Proceeding by multiplying the terms in parentheses by those in the brackets, we get:

$\displaystyle \left(F + \frac{F^3}{3!} + \frac{F^5}{5!}\right)\ \cdot\ \left[- \frac{F^2}{3!} + \frac{7F^4}{3\cdot5!} -\frac{31F^6}{3\cdot7!}\right]\ = $

$\displaystyle - \frac{F^3}{3!} + \frac{7F^5}{3\cdot5!} -\frac{31F^7}{3\cdot7!} -\frac{F^5}{3!\cdot3!} + \frac{7F^7}{3\cdot3!\cdot5!} -\frac{31F^9}{3\cdot3!\cdot7!}\ - \frac{F^7}{3!\cdot5!} + \frac{7F^9}{3\cdot5!\cdot5!} -\frac{31F^{11}}{3\cdot5!\cdot7!} = $

$\displaystyle - \frac{F^3}{3!} + \frac{7F^5}{3\cdot5!} -\frac{F^5}{3!\cdot3!} -\frac{31F^7}{3\cdot7!} + \frac{7F^7}{3\cdot3!\cdot5!} - \frac{F^7}{3!\cdot5!} -\frac{31F^9}{3\cdot3!\cdot7!}\ + \frac{7F^9}{3\cdot5!\cdot5!} -\frac{31F^{11}}{3\cdot5!\cdot7!} = $

And after adding in the remaining terms from within the braces ({ }) we have:

$\left\{\displaystyle \colorbox{pink}{$- \frac{F^3}{3!} + \frac{F^3}{3!}$} + \frac{F^5}{5!} + \frac{7F^5}{3\cdot5!} -\frac{F^5}{3!\cdot3!} +\frac{F^7}{7!} -\frac{31F^7}{3\cdot7!} + \frac{7F^7}{3\cdot3!\cdot5!} - \frac{F^7}{3!\cdot5!} -\frac{31F^9}{3\cdot3!\cdot7!}\ + \frac{7F^9}{3\cdot5!\cdot5!} -\frac{31F^{11}}{3\cdot5!\cdot7!}\right\} =$

We could have proceeded with immediately dropping the $ F^3$ terms; which cancel out, and then reducing the $ F^5$ and $ F^7$ terms. However, the professor decided to factor out those $ F^n$ variables from their fractions for his next step, but, he dropped the other higher order terms without even mentioning them! So, the following easily becomes what you see in Figure 13 without those additional terms:

$\left\{\displaystyle F^3 \left(- \frac{1}{3!} + \frac{1}{3!}\right) + F^5 \left(\frac{1}{5!} + \frac{7}{3\cdot5!} -\frac{1}{3!\cdot3!}\right) + F^7 \left(\frac{1}{7!} -\frac{31}{3\cdot7!} + \frac{7}{3\cdot3!\cdot5!} - \frac{1}{3!\cdot5!}\right) -\frac{31F^9}{3\cdot3!\cdot7!}\ + \frac{7F^9}{3\cdot5!\cdot5!} -\frac{31F^{11}}{3\cdot5!\cdot7!}\right\} =$

The equation on the blackboard continued as follows:

$= sinhF - F + \displaystyle \frac{35}{4} m (-a)^3 \left\{F^3\left(-\frac{1}{3!}+\frac{1}{3!}\right) + F^5\left(-\frac{1}{6\cdot3!}+\frac{7}{3\cdot5!}+\frac{1}{5!}\right) + F^7\left(-\frac{1}{6!} +\frac{7}{3\cdot6!} -\frac{31}{3\cdot7!} +\frac{1}{7!}\right)\right\}$

This image shows only the $F^5$ and $F^7$ terms:

Figure 13

Note that both "$3!\cdot5!$" and $6!$ equal 720, so that clears up what we see in the denominators of the first two $F^7$ terms.

In summary: We already know the $F^3$ terms are equal to zero. What about the $F^5$ terms? Well, upon turning the denominators into values you can work with, they become: -(1/36) + (7/360) + (1/120), and after converting to an LCM of 360 we have: -(10/360) + (7/360) + (3/360). So, the $F^5$ terms are also equal to zero! Likewise, even the $F^7$ terms reduce to zero! However, keep in mind that the equation on the blackboard did not include any of the $F^9$ or $F^{11}$ terms. Noting that, you should be able to conclude that someone wanted to emphasize that all of the terms which are displayed on the board become zero!

So, the final equation becomes nothing more than: $\colorbox{yellow}{$ M = sinhF - F $}$^[8] (which we already saw in Figure 7)!!

Apart from the fact that we know Samuel Herrick was hired to develop an equation for the film, I have no evidence (at this time) connecting him directly to what was written on the board. However, no one else comes to mind when asking, 'Who could have made all the changes we discovered in the equations above to produce these results?'

Figure 14

For the majority of those seeing the blackboard for the first time, its most visually striking aspect is what you see here!

But who wrote this?
And why?

Cursory observers without spending any time examining the equations, almost always comment on these exclamation marks as being factorial symbols; which would make no sense, since taking multiple factorials of zero would always result in 1 (0!=1 and 1!=1). Nor does it explanin the "?" mark at the end. In light of what we discovered above when evaluating the terms for the equations, that very much lends credence to those who believe the diagonal (") marks next to the zeros are equal signs pointig to each of the three $F^n$ terms and pointing out that they all become zero. I am currently attempting to gather more information concerning the question, 'Who wrote these bold symbols on the board?" and "Why?" For those who wish to see this as the professor becoming surprised after having evaluated each term in succssion; first one "!", then "!!" but then "!!!!" (four instead of only 3 exclamation marks), it becomes difficult to follow through with the plausibility of such a proposal; especially when it comes to having four "!"s followed by a "?". If the professor was still questioning whether the $F^7$ terms would become zero, then why already set it equal to zero; no less, following that by four "!"s of astonishment? Many see these large, bold symbols as simply being added by the production crew for a dramatic visual effect when an actor places check marks next to each of them. Do you have any specific documentation that would help answer these questions?

In that scene where the character "Klaatu" examines the blackboard, then puts a number of checkmarks next to varous equations, dramatically approving their correctness, he tells the boy that the professor merely needs a little help with his math. But where? He didn't point out with the chalk anything questionable! However, he then proceeds to write the following on the blackboard:

Try differentiating

$ sinh F - F = M - \frac{3}{8}m(-a)^3 M^3 $

Note: This is almost exactly the same as Equation [06] that Herrick wrote here; except the $(-a)$ shown here cancels out the minus sign in front of the last term, so it becomes: $sinhF - F = M\colorbox{lightgreen}{$ + $}\frac{3}{8}m\ \colorbox{lightgreen}{$a^3$}\ M^3$.

So what would the result be after doing that? And differentiate both sides with respect to what other variable? We have already seen $sinhF$ a number of times in the equations above. And in Figure 7 and our Summary, we even found $M=sinhF-F$; which according to the series expansion in Figure 10, gives us: $M=(F^3/3!)+(F^5/5!)+(F^7/7!)+\ ...$ or simply $F^3/6$ for only the first term (the same as seen in Herrick's Equation [11] here). One could also solve for $M$ in terms of $m$ and $a$, but that isn't the point here! The professor most likely needs solutions either $m$ or $a$, or both.

The Math symbols and equations displayed here use LaTeX under MathJaX.

Footnotes

¹[Return to Text] Kirby, David A., Lab Coats in Hollywood: Science, Scientists, and Cinema (2011, Cambridge, Massachusetts: MIT Press), pp. 40-41.

²[Return to Text] Kirby, David A., Lab Coats in Hollywood: Science, Scientists, and Cinema (2011, Cambridge, Mass.: MIT Press), p. 80. Which can be found at archive.org (since July 23, 2020). Figure 4.4 on page 81 also includes a photograph of Samuel Herrick and actor Michael Rennie standing in front of the blackboard. Kirby wrote that the signs “Don’t Erase” and “Don’t Touch” were added, because "Once Herrick left that set there was no one who could put it back together..." (p. 80).

³[Return to Text] See: "Appendix 3: Astrodynamical Terminology, Notation and Usage," pp. 477 ff. in Astrodynamics: Orbit Determination, Space Navigation, Celestial Mechanics, Volume 1 (1971, Van Nostrand Reinhold Company) by Samuel Herrick. There's a copy here at archive.org.

⁴[Return to Text] Images of various parts of the blackboard on this page are being used solely for academic reasons and to show that the equations have been accurately transcribed. They come from my observations of many different DVDs, online presentations and Youtube videos of the film; which is Copyrighted by 20th Century Fox studios, in order to arrive at the best possible transcription of the equations on the blackboard. If you believe there are any errors in the transcriptions; after examining all possible angles of the board yourself, then please contact me.

⁵[Return to Text] The term $(t-\tau)$ appears in a number of orbital equations and astrodynamics. One form of Kepler's Mean Anomaly equation is: $M = n(t-\tau)$ where M is the Mean Anomaly, which is the angular distance a hypothetical body would have traveled from periapsis if it were moving in a circular orbit at a constant angular velocity (mean motion).
n is the mean motion, the average angular velocity of the orbiting body around the central body.
$t$ is the current time. And $\tau $ is the time of periapsis passage, i.e., the time when the orbiting body was at its closest point to the central body (periapsis or perigee for Earth orbits).

⁶[Return to Text] This is the same photo referenced by David A. Kirby (see Footnote 2 above) as "Figure 4.4" on page 81 of his book. We are citing it here as: Samuel Herrick Papers (Ms1978-002), Special Collections and University Archives, University Libraries, Virginia Polytechnic Institute and State University, B02F02, photo of Samuel Herrick and Michael Rennie in front of the blackboard. You can also view that for yourself in the following video at timestamp 56:36 which I have cued this link to: Archival Adventures - Episode 81: Samuel Herrick Papers (Ms1978-002).

⁷[Return to Text] The blackboard shows only the begining terms of each of the series expansions which are often expressed using these summation symbols:

$\colorbox{lightgreen}{$sinh(x) = \displaystyle x+\frac{x^{3}}{3!}+\frac{x^{5}}{5!}+\frac{x^{7}}{7!}+ \dots\ $} = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}$ $\colorbox{lightgreen}{$cosh(x) = \displaystyle 1+\frac{x^{2}}{2!}+\frac{x^{4}}{4!}+\frac{x^{6}}{6!}+ \dots\ $} = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}$

Note: The identity on the board of $coshF+1$ as "$=2+ \dots$" might only be meaningful to someone familiar with these expansions. It can be defined as:

$cosh(x) + 1 = \displaystyle \colorbox{lightgreen}{$ 2 + $} \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \dots\ = \left(\sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}\right) + 1$ Then we come to $(coshF-1)^3$ with only a single term shown; the reason becomes clear once you see:

$(cosh(x)-1)^2 = \displaystyle \left(\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+\dots\right)^2 = \colorbox{lightgreen}{$ \frac{x^4}{2!}+\frac{x^6}{4!}$} +\frac{x^8}{320}+\dots$ And then to get the cube, you need to multiply that again by $cosh(x)-1$:

$\displaystyle \left(\frac{x^4}{2}+\frac{x^6}{24}+\frac{x^8}{320}+\dots\right) \cdot \left(\frac{x^2}{2}+\frac{x^4}{24}+\frac{x^6}{720}+\dots\right) = \colorbox{lightgreen}{$ \frac{x^6}{8}$} +\frac{x^8}{32}+\frac{7x^{10}}{1920}+\dots$ for only the first three terms!

So, it's understandable as to why these expansions were limited to only their first terms when substituting them into the equations which follow! And as shown on the blackbord in Figure 11, it made it quite easy to see that $[(coshF-1)^3]/(coshF+1) = F^6 / 2 \cdot 8 +\ ... =\colorbox{lightgreen}{$ F^6 / 16 +\ ... $}$

⁸[Return to Text] For those who might be interested: This happens to be the simplest form of the Hyperbolic Kepler equation where $M$ is called the "Mean Anomaly" and $F$ is called the "hyperbolic eccentric anomaly", but in these equations $e$ (the ecentricity) has been set to 1, so the orbit becomes parabolic rather than hyperbolic.

First Posted on: 13 July 2025 (2025.07.13).
Updated on: 14 JUL 2025 (2025.07.14); added a few more images, made some corrections and clarifications and more footnote data.
15 JUL 2025 (2025.07.15); More clarifications, added a few more images and completely rearranged where one section of the board was introduced for a much more logical presentation!
18 JUL 2025 (2025.07.15); More clarifications, typo fix, rearranged location of some figures and gave all of them a label, and had to correct a huge error on my part!
20 JUL 2025 (2025.07.20); Typo fixes, arranged the data better, and made more clarifications.
22 JUL 2025 (2025.07.22); Added link to page on Herrick's Math Notes, etc.
24 JUL 2025 (2025.07.24); Added Intro about variables, corrected an ambiguous square root symbol on the board in transcription.

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Exactly What was On the Blackboard in the1951 Film, The Day the Earth Stood Still ?And What Do the Equations Mean?

The Blackboard

Exactly What was On the Blackboard in the
1951 Film, The Day the Earth Stood Still ?
And What Do the Equations Mean?