Thermodynamics and the JFK Assassination

**Andrew Mason** · Feb1-05, 07:54 AM

There has been an ongoing debate about the physics of the fatal shot that struck President Kennedy in the head.

Although there is very strong evidence that the fatal shot came from the right rear (where the sixth floor sniper's nest in the Texas School Book Depository was located) JFK's head and body move backward and to the left with significant speed immediately after the head shot. Conspiracy theorists point to this rearward kick of the body as evidence of a shot from the grassy knoll which was located ahead and to the right of the President.

There is no question that matter exited the President's head with significant speed. A simple kinematic analysis shows that the forward ejection of matter from the head can carry more rearward momentum than the incoming bullet possessed, due to the sudden release of pressure built up in the skull as the bullet passed through. This was discussed in a 1976 Physics Today article by Nobel physicist Luis Alvarez. But Alvarez could not explain the physics of this pressure build up saying that one would "have to ask someone more knowledgeable in the theory of fluid mechanics than I am".

Experts have suggested that it is possible to generate the kind of pressure sufficient to burst the skull if the bullet generates a shock wave (ie. it passes through the head at a speed greater than the speed of a compression wave in the brain). The problem is that the speed of a compression wave in water (which the brain consists largely of) is about 5000 fps or 2.5 times the speed of a Mannlicher-Carcano bullet.

I think that it is simply a matter of the first law of thermodynamics:

$LaTeX Code: \\Delta KE = \\Delta Q = \\Delta U + \\Delta W = \\Delta (PV)$

where $LaTeX Code: \\Delta W$ is the work done by the system

Assuming that the head is filled with a liquid in a confined volume, if the bullet slows down it has to cause compression or produce heat. If the volume is confined, no work is done by the system (until the skull ruptures), so we have:

$LaTeX Code: \\Delta KE = V\\Delta P$

All the bullet has to do is deform sufficiently in entering the back of the skull to create enough drag force as it passes through the brain to slow down. By the first law, it will convert that loss of kinetic energy to pressure. So:

$LaTeX Code: \\frac{1}{2}m_{bullet}(v_i^2-v_f^2) = V\\Delta P$

In order to duplicate the result using the same ammunition used by Oswald (copper jacketed 6.5 mm. 160 grain (10 gram) bullets fired from an Mannlicher Carcano rifle at about 2000 feet/sec.) one just has to make the bullet slow down enough in passing through the target. Assuming the volume was about 2 litres (.002 m^2)(average weight of human head is 4.5-5 kg), the initial speed of the bullet was 600 m/s and the exit speed 150 m/s, (mass .010 kg):

the loss of KE is about 1700 J., which means that:

$LaTeX Code: \\Delta P = 1700/.002 = 850000 Pa$ or about 8.5 atmospheres.

Can anyone find a material flaw in this analysis?

AM

**arildno** · Feb1-05, 08:07 AM

Apart from the sheer ghoulishness of the subject matter, not as such initially, anyway..

**Bystander** · Feb1-05, 11:46 AM

Originally Posted by Andrew Mason

(snip)... human head is 4.5-5 kg), the initial speed of the bullet was 600 m/s and the exit speed 150 m/s, (mass .010 kg):

the loss of KE is about 1700 J., which means that:

$LaTeX Code: \\Delta P = 1700/.002 = 850000 Pa$ or about 8500 atmospheres.

AM

8.5 atm; 3/4 velocity loss? Little high. The "movement" occurs after the bullet strike (collision event occurs over 100s of μs vs. the 40 ms/frame for Zapruder film) and is more easily attributed to relative motion between an inert mass (JFK) and an accelerated mass (the limo).

**Andrew Mason** · Feb1-05, 01:17 PM

Originally Posted by Bystander

8.5 atm; 3/4 velocity loss? Little high. The "movement" occurs after the bullet strike (collision event occurs over 100s of μs vs. the 40 ms/frame for Zapruder film) and is more easily attributed to relative motion between an inert mass (JFK) and an accelerated mass (the limo).

It is difficult to say when the movement begins, because the Zapruder film is not continuous. The frames are 55 ms. apart and each frame captures only about 25 ms (30 ms. is for film motion). So we do not know when exactly between frame 312 and 313 that the bullet struck. It could have struck during the exposure of frame 313. Furthermore, it takes some time for the rearward recoil to reverse the forward motion of the head/body caused by the bullet impact from the rear. So there will be recoil occurring before we actually see reverse movement.

The beginning of the rearward motion is evident in frame z314 which is 110 ms after frame 312 was exposed, which could be as little as 55 ms after the impact.

Besides, the only person to move backward was JFK. There is no indication that the car suddenly sped up at that point. Driver William Greer said he stepped on the accelerator immediately, but he was turned to the rear at the moment of the impact and did not turn forward until a few frames later.

AM

P.S. I have corrected the conversion from 850000 Pa to 8.5 Atm. - thanks for pointing that out.

**Janitor** · Feb1-05, 01:23 PM

Another explanation that I have heard is that the disruption of the brain caused all sorts of spurious motor nerve signals that caused muscles to clench in a way that resulted in the neck movement.

**Andrew Mason** · Feb1-05, 02:04 PM

Originally Posted by Janitor

Another explanation that I have heard is that the disruption of the brain caused all sorts of spurious motor nerve signals that caused muscles to clench in a way that resulted in the neck movement.

That is a theory. But there is no evidence that this can or did occur. The rearward motion is quite uniform.

On the other hand we know that there was a forceful ejection of matter. We know from Newton's laws that the ejection of matter forward and to the right must have generated a recoil impulse to the head rearward and to the left, which is exactly what is seen.

AM

**kanato** · Feb3-05, 12:07 AM

where does your $LaTeX Code: V \\Delta P$ come from? If the bullet is causing compression, then there should be a change in volume of the fluid, and the work done on the system wouldn't be zero.

But it seems we're also assuming that the system can be treated with equilibrium thermodynamics, but I think with as fast as things happen in this event that it would be unreasonable to assume the the fluid is in any kind of hydrostatic equilibrium.

**Andrew Mason** · Feb3-05, 01:45 PM

Originally Posted by kanato

where does your $LaTeX Code: V \\Delta P$ come from? If the bullet is causing compression, then there should be a change in volume of the fluid, and the work done on the system wouldn't be zero.

I am assuming that the contents of the heat would be, like water, almost incompressible. In any event, all we are concerned about is whether there is any work done BY the system ie. an increase in volume. Until the head ruptures, there is no work done by the system. The net work done on the system, which has to be equal to the loss in kinetic energy of the bullet, is the increase in internal energy (PV) less the work done BY the system.

But it seems we're also assuming that the system can be treated with equilibrium thermodynamics, but I think with as fast as things happen in this event that it would be unreasonable to assume the the fluid is in any kind of hydrostatic equilibrium.

That is true. The pressure will likely be greatest in front of the bullet and the least behind the bullet. The thermodynamic analysis simply tells us that the pressure has to build up in the head if the bullet slows while passing through it. It tells us what the average pressure would be but does not tell us what the maximum would be at any location.

AM

**kanato** · Feb3-05, 09:01 PM

You are assuming that the internal energy of the system is PV? It's true for a non-interacting system that the internal energy will be proportional to PV, as in an ideal gas, $LaTeX Code: E = \\frac{3}{2} PV$ but that's not true in general.

**Andrew Mason** · Feb3-05, 10:16 PM

Originally Posted by kanato

You are assuming that the internal energy of the system is PV? It's true for a non-interacting system that the internal energy will be proportional to PV, as in an ideal gas, $LaTeX Code: E = \\frac{3}{2} PV$ but that's not true in general.

It has nothing to do with an ideal gas. It is just a recognition that pressure is a way of storing energy. Pressure IS energy density (E/V = F*distance/Volume = F/A = P) . So PV = energy. It is potential energy and can be converted back to kinetic energy by releasing that pressure (which is what is seen in frame 313).

AM

**kanato** · Feb3-05, 10:41 PM

Originally Posted by Andrew Mason

It has nothing to do with an ideal gas. It is just a recognition that pressure is a way of storing energy. Pressure IS energy density (E/V = F*distance/Volume = F/A = P) . So PV = energy. It is potential energy and can be converted back to kinetic energy by releasing that pressure (which is what is seen in frame 313).

AM

ok, I see what you are saying. But I don't think the pressure of the entire system would all increase right off the bat, just around the bullet, and then there would be pressure waves which would eventually dissipate into thermal energy. But "eventually" is probably far too long for this situation.

**Andrew Mason** · Feb4-05, 12:03 AM

Originally Posted by kanato

ok, I see what you are saying. But I don't think the pressure of the entire system would all increase right off the bat, just around the bullet, and then there would be pressure waves which would eventually dissipate into thermal energy. But "eventually" is probably far too long for this situation.

Yes. I think one can safely assume that in time frame of the bullet passing through (less than 1/1000th of a second) the kinetic energy that is lost is transferred to the contents of the skull and does not have enough time to dissipate significantly to the external environment.

AM

**timnich** · Nov30-09, 01:10 AM

Andrew, I have been doing research on the JFK assassination for 11 years. Your post I feel has answered a very puzzling problem with JFK's head wound. I feel that a different type of bullet was responsible for the head wound. The bullet came from the rear slightly left to right at about 3 to 4 degrees. I believe the bullet disintegrated into dozens of fragments less than 1 mm in length and in the process converted about 90% of its kinetic energy into enough pressure to cause a skull fracture with an internal force between 7000 and 15000 Newtons at the site of the wound. I have quoted your post in my research.

**cabraham** · Nov30-09, 06:59 AM

Originally Posted by timnich

Andrew, I have been doing research on the JFK assassination for 11 years. Your post I feel has answered a very puzzling problem with JFK's head wound. I feel that a different type of bullet was responsible for the head wound. The bullet came from the rear slightly left to right at about 3 to 4 degrees. I believe the bullet disintegrated into dozens of fragments less than 1 mm in length and in the process converted about 90% of its kinetic energy into enough pressure to cause a skull fracture with an internal force between 7000 and 15000 Newtons at the site of the wound. I have quoted your post in my research.

I've been a JFK assassination student for 46 yrs. Your info is very good and well researched. Maybe we can talk on the phone.

Claude