Explosive Pressure in p-pbar annihilation

[floersh]

I was wondering if one were to construct a reaction camber that contains a high vaccum and were to annihilate fractions of grams of matter/antimatter (lets say p-pbar) would the resulting energy release produce any pressure on the reaction chamber's inner wall?

 

I know that matter antimatter reactions pretty much destroy all of their input matter (assuming equal ratio) and the resulting particles would have a very small (even insignificant) size relative to the volume of the chamber.

 

So I am going to assume that the pressure would be negligable if any at all?

 

Am I right or wrong?

[insane_alien]

well, there would be light pressure. but that would be tiny, since gamma rays tend to go through things rather than reflect off them.

[timo]

Approximation:

 

Assume R to be the rate of annihilation (mass per time). Then the energy created per time-unit is Rc². Let this energy escaping at c, making the energy density (and thus the pressure) at distance d (assuming a spherical container) roughly [math] \frac{R c^2 \, dt}{4 \pi d^2 \ c \, dt} = \frac{Rc}{4 \pi d^2}[/math].

[floersh]

So lets say we had .5 grams of antideuteron which we react with .5 grams of deuteron.

 

E = mc^2 = 89.875517 terajoules

 

Now lets say our reaction chamber is 2.5 meters in diameter (spherical) and the reaction occurs in the center roughly..

 

P = E/4*pi*r^2 = 4.577322538412 terapascals

 

I would say that is not insignificant!!!

 

I am also assuming the fact that it occured in a vaccum means nothing???

[Severian]

And where are you going to get 0.5g of antideuteron from?

 

You are not Dan Brown are you?

[swansont]

P = E/4*pi*r^2 = 4.577322538412 terapascals

 

I would say that is not insignificant!!!

 

I would say that the units aren't right. Pressure is Force/Area, not Energy/Area.

 

And as Severian implied, a gram of mass to convert to photons is not insignificant.

[CPL.Luke]

out of curiosity how would that pressure manifest itself? If the energy was being carried by gamma rays emanating from the source of the blast. and they came into contact with a piece of paper, would that paper react as if the number of newtons(p*A) givin by severians formula were pushing on it, or would it react as if a very small force were acting on it, coming only from the gamma rays that were absorbed by the paper.

[insane_alien]

I would say that the units aren't right. Pressure is Force/Area, not Energy/Area.

 

And as Severian implied, a gram of mass to convert to photons is not insignificant.

 

there is nothing wrong with the units. work it out for yourself.

 

it will come to kg·m^-1·s^-2

 

which is the base units for pascals.

 

http://en.wikipedia.org/wiki/SI_derived_unit

[swansont]

there is nothing wrong with the units. work it out for yourself.

 

it will come to kg·m^-1·s^-2

 

which is the base units for pascals.

 

http://en.wikipedia.org/wiki/SI_derived_unit

 

Energy and force have different units, so it can't work out to be the same

 

mc²/r² is going to be kg-s^-2

[swansont]

out of curiosity how would that pressure manifest itself? If the energy was being carried by gamma rays emanating from the source of the blast. and they came into contact with a piece of paper, would that paper react as if the number of newtons(p*A) givin by severians formula were pushing on it, or would it react as if a very small force were acting on it, coming only from the gamma rays that were absorbed by the paper.

 

The force would only be from the photons that interacted, and the momentum transfer would depend on the interaction — forward scattering will give you less than absorption, which will be less than backward scattering.

[insane_alien]

ahh, i was going by athiests formula which gives correct units.

 

[math] \frac{R c^2 \, dt}{4 \pi d^2 \ c \, dt} = \frac{Rc}{4 \pi d^2}[/math].

 

i never noticed that floersh had an extra c in there.

 

using floersh's numbers with the right equation we get 3819.7 Pa

 

that is pretty insignificant for the energy released.

[floersh]

Ahh thats much more reasonable

 

.5 gram antideuteron annilating .5 gram deuteron in a 2.5m diameter vaccum chamber would create 15.268 kilopascals of preasure..

 

Thats less than atmospheric preasure..

 

Would anybody happen to know how to calculate the heat energy this chamber would have to withstand?

[insane_alien]

how did you get 15 kPa? for 1 gram per second you get a value of 3.8kPa.

 

did you forget to change the mass flowrate into SI?

[floersh]

how did you get 15 kPa? for 1 gram per second you get a value of 3.8kPa.

 

did you forget to change the mass flowrate into SI?

 

Humm..

 

Mass = .5g + .5g = 1g / 1000 = .001kg

If we annihilate only once per second then R = .001kg/s

 

If we have a diameter of 2.5m then the radius would be 1.25m

 

Area = 4 * pi * r^2 = 19.6349540849 m^2

 

c = 299792458 m/s

 

P = R * c / A = .001 * 299792458 / 19.6349540849 = 15268.3 Pa

 

or 15.2683 kPa

 

Am I missing something?? I think I have all of the units right..

[insane_alien]

ahh, you used the radius instead of the diameter. the d in the equation means diameter not radius.

[timo]

ahh, you used the radius instead of the diameter. the d in the equation means diameter not radius.

It means "distance"; in the sense of distance from the point of creation. It's the radius.

[insane_alien]

oh yeah, you physicists don't follow the same conventions we chemical engineers use. i keep forgetting. sorry about that. in my field d is always diameter.

[floersh]

Would anybody happen to know how to calculate the heat energy this chamber would have to withstand?

[insane_alien]

just calculate then energy released. assuming the chamber absorbs all of the energy created then all that energy will be transformed into heat.

[floersh]

just calculate then energy released. assuming the chamber absorbs all of the energy created then all that energy will be transformed into heat.

 

Ok assuming that is correct (which it really isn't due to 50% nuetrinos which I couldn't absorb).. How might I convert that to temperature in say Kelvin? I assume it would be a function of the chamber's material heat capacitance?

[insane_alien]

i thought it was all gamma photons produced. luckily, i am not a particle physicist

 

if you know the mass, and the heat capacity of the material then dT = E/(m*C)

 

where dT is the change in temperature

E is the energy

m is the mass of the chamber

C is the heat capacity of the material.