# How much pressure do you need to make air go near lightspeed?

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If you had a ball made of a material that could withstand any pressure, how much pressure would you need in the ball to make it so when you popped it the air would go near lightspeed?

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Posted (edited)

You can do this by dimensional analysis. If the air is made up of just nitrogen,

$\left[P\right]=ML^{-1}T^{-2}$

(units of pressure)

And your fundamental constants are the mass of the nitrogen molecule, $$\hbar$$ and $$c$$ (the speed of light.)

$\left[\hbar\right]=ML^{2}T^{-1}$

$\left[m_{N_{2}}\right]=M$

$\left[c\right]=LT^{-1}$

$P=\left(m_{N_{2}}\right)^{j}\hbar^{k}c^{l}$

Gathering all together,

$M^{j}\left(ML^{2}T^{-1}\right)^{k}\left(LT^{-1}\right)^{l}=M^{j}M^{k}L^{2k}T^{-k}L^{l}T^{-l}=M^{j+k}L^{2k+l}T^{-k-l}=ML^{-1}T^{-2}$

So the power equations are,

$j+k=1$

$2k+l=-1$

$-k-l=-2$

whose solutions are,

$k=-3$

$l=5$

$j=4$

So your pressure would be the order of,

$P=\frac{\left(m_{N_{2}}\right)^{4}c^{5}}{\hbar^{3}}\simeq2.5\times10^{51}\:\textrm{Pa}\simeq2.5\times10^{45}\:\textrm{atm}$

That's like $$10^{29}$$ times the density at the centre of the Sun. I think you're gonna make a black hole. Don't do it at home!

Even though this is just dimensional analysis, if you take $$m_{\textrm{air}}=.7m_{N_{2}}+.2m_{O_{2}}+.1m_{H_{2}}$$, you get a better approximation for the average mass of the air molecules.

Edited by joigus
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Or one atmosphere if the temperature is high enough.
In normal circumstances, air molecules travel at about the speed of sound (that's not a coincidence)

If you want them to travel near the speed of light, that's about a million times faster (give or take)

That means you need to increase their energy a million million times (because the energy varies as the square of the velocity)

And the average energy is proportional to temperature.
So you need to heat them up from about 280K to about 280,000,000,000,000 K

I think that's about 10,000 times hotter than a supernova.

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1 m^3 of air at stp has easily sufficient mean particle velocity to escape its own gravity well, so providing it's not gravitationally bound to anything else, all you need do is wait a little while for dark energy to kick in. Not sure there's any current limit to the ultimate expansion velocities this could achieve.

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5 minutes ago, sethoflagos said:

1 m^3 of air at stp has easily sufficient mean particle velocity to escape its own gravity well

As long as you can avoid collisions, which is difficult. The mean free path of air molecules at room temperature is of order 100 nm.

And escape velocity is << c

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3 hours ago, swansont said:

As long as you can avoid collisions, which is difficult. The mean free path of air molecules at room temperature is of order 100 nm.

And escape velocity is << c

I'm assuming the process occurs in deep extragalactic space.

My understanding is that providing some divergent velocity is maintained, eventually there will be sufficient distance between particles for differential Hubble expansion to drive relative velocities to c and beyond.

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38 minutes ago, sethoflagos said:

I'm assuming the process occurs in deep extragalactic space.

The mean free path is still the same, at the point you pop the container.

Anyway, the relevant question is whether the thermal energy is comparable to the mass energy of one of the molecules, under the assumption that all of the thermal energy goes into one particle. If that's not the case, then there's not enough energy to get close to c.

k is 8.6 x 10^-5 eV/K

if T is 300K, then kT is ~ 0.026 eV

You need ~ 1 GeV for each nucleon, to give you a KE equal to the mass energy (which would imply a gamma of a tad more than 2, which is v = 0.866c)

So there will be a certain number of atoms where this will hold (around 10^11, so much, much less than a mole). You just then need the improbable circumstance where one particle has all the KE, and doesn't collide before escaping.

But that's not really what the question was - it wasn't one particle going near lightspeed. John Cuthber's treatment is the way to go.

And I'm still trying to figure out why escape velocity matters (or dark matter, for that matter. Space expansion doesn't give you a local velocity)

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7 minutes ago, swansont said:

Anyway, the relevant question is

It's not clear.

joigus' interpretation is entirely valid in one sense. However, the only thing leaving a black hole at light speed is what? .... Hawking radiation?

Similarly, John Cuthber's plasma fusion reactor is also quite valid. Though again, it isn't exactly going to be 'air' leaving that scenario at light speed.

My own post simply suggested a point of view which resulted in air molecules travelling at, and indeed exceeding, c.

Maybe not the most interesting point of view to some, but a view nonetheless.

wrt your last point, if the air can't escape it's own self-gravitation, it isn't going anywhere fast. Is it?

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22 minutes ago, sethoflagos said:

joigus' interpretation is entirely valid in one sense. However, the only thing leaving a black hole at light speed is what? .... Hawking radiation?

Hawking radiation won't leave at lightspeed, unless it's a photon.

JC didn't say anything about a fusion reactor, he just said that you'd need to get the gas bery hot to get to that speed.

Quote

My own post simply suggested a point of view which resulted in air molecules travelling at, and indeed exceeding, c.

No, it doesn't result in that.

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1 hour ago, swansont said:

Hawking radiation won't leave at lightspeed, unless it's a photon.

You may be correct. I don't know. Hence the question marks in my post.

1 hour ago, swansont said:

JC didn't say anything about a fusion reactor, he just said that you'd need to get the gas bery hot to get to that speed.

No he didn't. But the temperature JC quoted exceeded the minimum necessary to initiate fusion reactions.

1 hour ago, swansont said:

No, it doesn't result in that.

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1 minute ago, sethoflagos said:

Nothing has a speed that exceeds c.

Expansion is not a speed as we normally discuss speed.

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1 hour ago, sethoflagos said:

Including a particle that has receded beyond the Hubble horizon?

Its local speed is less than c

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2 hours ago, sethoflagos said:

Correct.

But relative to it's non-local origin... ?

Then space is being added between you and it. Independent of any motion of the object.

Here’s a better explanation

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13 hours ago, sethoflagos said:

No he didn't. But the temperature JC quoted exceeded the minimum necessary to initiate fusion reactions.

I suspect the temperature is so high that atomic nuclei get torn apart rather than fused.

It's completely beside the point.

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