Originally Posted by **ChipB** The picture I attached shows constant velocity of all particles starting immediately after the explosion of the bomb. There is no acceleration shown or implied, and no force needed to keep it going. Do you agree that this is how an ideal bomb would work? |

No, I don't think it does. Once a bomb has exploded all the ejected shrapnel should obey Newton's first law. One idealisation I made was that all the pieces of shrapnel were of identical size which looks similar to your scenario anyway. If all particles are being ejected with the same velocity then they cover the same distance in the same unit of time. However what you have is a linear scaling which means that particles further away from the origin travel further in the same unit of time just because they were further away from the origin. I don't know of any ideal bomb which works this way.

To be clear, what you have is something like:

- a particle 1cm from the bomb moves 2cm in a unit time
- a particle 5cm from the bomb moves 10cm in a unit time
- a particle 10m from the bomb moves 20m in a unit time

etc.

If the particles were travelling at say 10m/s you would have

- a particle 1cm from the bomb moves 10.01m in a second
- a particle 5cm from the bomb moves 10.05m in second
- a particle 10m from the bomb moves 20m in a second

This is not a linear scaling but a translation.

So I can only assume you are presuming the further away particles to start with higher initial velocities but I can't see what kind of motivation you might have for making such an assumption as it seems very arbitrary.

Also the rubber sheet analogy, which might produce something like this, isn't really like an idealised bomb either (the forces come from tension forces at the edges and stretching the manifold) so I am at a loss to know how you derived these results.

Edit:

I just did some calculations on ChipB's expansion model.

Basically his model comes down to v=kx where v is the velocity at position x offset from the big bang and k is some suitable constant. This equation ensures that all particle distance ratios remain constant as they expand which was indicated in ChipB's particle expansion diagram.

Constant ratio expansion requires $\displaystyle v=\frac{dx}{dt} = kx$

integrating as an ODE...

$\displaystyle \int_{x_0}^x \frac{dx}{kx} = \int_{t_0}^t dt$

yields

$\displaystyle \frac{1}{k}(ln(x-x_0)) = t-t_0$

results in an expansion of $\displaystyle x = e^{k(t-t_0)}+x_0$ therefore the ChipB's linear expansion model is actually an exponential expansion model in practice !! If you differentiate that expression twice you will get a non vanishing acceleration.