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I was sitting on the bus on my way home, and this popped into my mind.

As a car accelerates, the air pressure at the back of the car increases as air is pushed back due to inertia. This pressure will be constant, assuming a constant rate of acceleration and that the acceleration lasts long enough for an equilibrium to be reached.

The question is therefore: How high must the acceleration rate be for the air near the back window to undergo nuclear fusion?

Some further assumptions:

  • The air is 100 % nitrogen (or hydrogen, if that makes the approximation simpler)
  • When the car is not accelerating, the pressure is $10^5\ \mathrm{Pa}$ and the temperature is $273.15\ \mathrm{K}$ (STP).
  • The car is a square prism of dimensions $1 \mathrm{m}\times 1 \mathrm{m} \times 3 \mathrm{m}$ with the direction of travel perpendicular to the square faces. $3 \mathrm{m^3}$ is approximately the EPA definition of the interior volume of a mid-sized car. There is nothing else (seats etc.) in the car.
  • The car is an isolated system, so neither matter nor energy can enter or leave
  • The window cannot melt

I anticipate that this might lead to some minor problems with relativity...

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  • $\begingroup$ It is not clear. How the air is pushed back at the back of the car? What a car acceleration hs to do with that? $\endgroup$ – jaromrax May 10 '17 at 7:14
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    $\begingroup$ @jaromrax The air has inertia and tries to stay where the car was prior to the acceleration, which make it "move backwards" in the non-inertial reference frame of the accelerating car. The same effect is more obvious if someone leaves a book on the dashboard which falls to the floor when the car accelerates. A neat way to prove to yourself that the air is accelerating (in the non-inertial frame) is to get in a car with a helium balloon: the balloon's motion is determined by the pressure gradient in the air, so when you stomp on the accelerator pedal the balloon goes forward. $\endgroup$ – rob May 11 '17 at 19:26
  • $\begingroup$ Now I get it, it is a question on making such a pressure (and temperature) in the back seats of the car. The air density is 0.001, while Sun core density 150g/cm3, white dwarf 10^6. I think one should open a window to let some more air in... $\endgroup$ – jaromrax May 12 '17 at 7:40
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    $\begingroup$ I think a rough but useful estimate could be obtained by viewing the gas as having an initial gravitational potential energy in that acceleration; the gas will like get compressed to a thin film, thereby transforming all this potential energy to thermal energy. Average that to each proton, corrected by some binding energy and offering some to plasma electrons, one can work out the gravity required to initiate thermo fusion. I'm too lazy to crunch the numbers at this time... $\endgroup$ – Carl Lei Aug 7 '17 at 13:35
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OK, well the way to do this is to fill your car with a rather low-pressure mixture of Deuterium (${}^2H$) and Tritium, (${}^3H$) (yes, this does not meet your requirements: the size of my envelope does not allow for calculations any harder than this one, sorry). The pressure needs to be low enough such that the mean free path of the atoms is about the length of the inside of the car, which I'll take to be 2 metres.

Then you put your foot on the accelerator really hard, and some of the Tritium atoms will fall from the front of the car and hit Deuterium atoms close to the rear windscreen. Apparently the collision energy needs to be about $0.1\,\mathrm{MeV}$: this is the kinetic energy that the Tritium atom needs to have gained (it's better for the Tritium to do the falling towards the Deuterium as it's heavier so it does not need to fall so fast).

The first thing to do is to check whether this needs the speed to be relativistic, as there's no way I'm doing this calculation if it does. So, using $E=mv^2/2$ or equivalently $v = \sqrt{2E/m}$, we want $v \ll c$ The atomic mass of Tritium is about $5\times 10^{-27}\,\mathrm{kg}$ and $0.1\,\mathrm{MeV} \approx 1.6\times 10^{-14}\,\mathrm{J}$. And plugging this in we get

$$v \approx \sqrt{\frac{3.2\times 10^{-14}\,\mathrm{J}}{5\times 10^{-27}\,\mathrm{kg}}} \approx 2.5 \times 10^6\,\mathrm{ms^{-1}} \ll c $$

So we're fine.

So now we can just use the constant-acceleration formulae we learn in school: $v = u + at$ and $s = ut + at^2/2$, and solve for $a$ in terms of $s$ and $v$, both of which we know. And the formula is

$$ a = \frac{v^2}{2s} $$

So $$a \approx \frac{\left(2.5 \times 10^6\,\mathrm{ms^{-1}}\right)^2}{4\,\mathrm{m}} \approx 1.6\times 10^{12}\,\mathrm{ms^{-2}}$$

Or, in other words $a \approx 1.6\times 10^{11}\,\mathrm{g}$. As well as breathing apparatus you will need a g-suit.


As you can tell, I am trying to avoid work today.

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For fusion one needs order of MeV energies in the center of mass system of the fusing nuclei.

As all molecules are accelerated at the same time and direction, it is only fusion with the molecules accumulating on the glass back window that has meaning i.e. the question should be at what car acceleration the atoms of air reach MeV energies with respect to the back window glass. There, stopped hydrogen atoms may be hit by the accelerating ones and if the center of mass energy of the two hydrogens is of the order of hydrogen fusion, it will happen.

Of course the glass will have melted long before. Look at the temperatures for hydrogen fusion to happen at this link , paragraph :"fusion a closer look", about 10million Kelvin.

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