## Monday, September 18, 2006 ... //

### Frank Wilczek: podcast

Some people might be interested in

although I have not yet heard it because of time limitations. It you can't play the m4b (=mp4) file but you want to, download iTunes. Also, Lisa Randall's Warped Passages are now available in paperback.

#### snail feedback (1) :

Lubos:

Emotionally I refuse to buy into any conspiracy theory of 9/11. But as far as physics is concerned I find your model of the twin tower falling ridiculous wrong, even in the updated version you show at the time of this comment. Let me show you why you are wrong.

First, the observed video shows all floors roughly reach the ground at about the same time and about the same speed. Of course that is some simplification, I already said "roughly". Let's also simplify so it begins to fall from the top of the floors, not from the middle. Another simplification is if you track the top floor, the falling is roughly a uniform acceleration, although the acceleration should be some what less than the g of free fall.

We can write:

h = 1/2 * a * T^2

or
h = 0.5 * Vf^2/a

and
T = 2*h/Vf

with a <= g and unknown.

Assuming the gravity potential convert into dynamic energy 100% without any waste. Consider that the floors fall an average of half the height of the tower:

m*g*(0.5*h) = 0.5 M * Vf^2

with Vf the final velocity.

thus
Vf = sqrt(gh)

thus
T = 2*h/Vf = 2*sqrt(h/g)

That is the same number as you got in your "elastic" model, right?

BUT, the above is NOT what happened, because we have made conflicting assumptions. The floors falls to the ground at the same time and same speed. That means the the collisions are COMPLETELY INELASTIC, because the pieces acquired the same speed after multiple collisions. So there are energy losses. You can not assume all gravity potential convert to dymanic energy when there are energy loss in the inelastic collision.

Assume half of the energy is lost, for completely plastic collision, then the final Vf should be reduced by a factor of sqrt(2). This will increase the time T by the factor sqrt(2). Thus the T is:

T = 2*sqrt(2*h/g)

But even that result is too small.

Let's consider the model not from the point of view of energy, but from the point of view of impulse, momentum.

If the falling takes time T, the debrises, started from rest and fall to the ground in T seconds, the impulse obtained from gravity is:

Pg = m*g*T

BUT, gravity is not the only thing that provides impulse during the whole T seconds. The ground is still supporting at least the portion of the building that has not started fall yet. This provides upward, and hence negative impulse.

Again assuming uniform acceleration, during the whole T seconds, the ground is supporting an average of 2/3 of the body of the building (you do the exercise to figure out 2/3). So the supporting impulse is:

Ps = (2/3)M*g*T

Combine the two impulse:

P = 1/3 * M * g *T

Impulse equals final momentum:

P = M*Vf

Vf = 1/3 * g* T

but we also have, for uniform acceleration:

T = 2*h/Vf

T = sqrt(6*h/g)

With a building height of 410m (417m minus the antenna):

T = sqrt(6*410/9.8) = 16 seconds.

Even the above estimate is some what too small. We are assuming that at any moment, the ground is supporting only the portion of building that has not begin to fall yet. But we also need to consider that as the falling floors hit down on the building with tremendous force. This force is propagated down by the supporting stell columns. Thus the ground not only supports the floors that has not begin to fall yet, the ground is also taking the hit of that crumbing forces.

That is, the impulse provided by the ground should be:

Ps >> 2/3 * M * g * T

This will result in a T
T >> sqrt(6*h/g)

The 9/11 video shows the building collapsing much faster. I don't know how to make it out of it. But the physics doesn't look right unless it is a demolition.

Before you start jumping to conclusion, I want to remind you that the supporting force from the ground is VERY IMPORTANT to consider, because whether there is ground support or not is the whole difference between a demolition fall, and none-demolition fall.