Sure, I’d like to see them.
It is too clear, and so it is hard to see.
I should have known someone would ask… Ok, here goes (not for the faint of heart):
According to Hawking, black holes radiate energy as if they were black bodies, with a peak wavelength approximately equal to the Schwartzchild radius (I think there’s probably a 2pi in there, too, but can’t find a quick reference on it). A discussion of the why and how of this radiation probably belongs in a different thread, or you can read about it in A Brief History of Time.
OK, math starts<BLOCKQUOTE><font size=“1” face=“Verdana, Arial”>code:</font><HR><pre>here:
G*m
R = —
c^2
lambda = R (aproximately)
C3 C3*c^2
T = ------ = ------
lambda G*m
C3*c^2
P = AsigmaT^4 = 4piR^2sigma( ------ )^4
G*m
c^4*C3^4
P = 4pisigma* ----------
G^2*m^2
dm/dt = (dE/dt)/c^2 = P/c^2
4*pi*sigma*c^2*C3^4
dm/dt = ---------------------
G^2*m^2
G^2*m^3
lifetime = m/(dm/dt) = -----------------
4pic^2sigmaC3^4
(max)
Plugging in the values of the various constants, and taking 2 kg as the mass, will (hopefully) give the results in my previous post.
Explanation of symbols:
lambda = peak wavelength
R = Schwartzchild radius
G = Gravitational constant
m = mass (2kg)
c = speed of light
T = effective temperature
C3 = Wien's constant (not to be confused with c)
P = power = dE/dt
A = surface area
sigma = Boltzman's constant
E = energy
The weak point in this argument is the "approximately" in the lambda = R; if I'm off by 2*pi there, it would result in my final answer being off by a factor of 1.56*10^3; however, the point is that this still gives a lifespan much shorter than the time taken to hit the Earth.
------------------
"There are only two things that are infinite: The Universe, and human stupidity-- and I'm not sure about the Universe"
--A. EinsteinOK, does anyone know how to get equations to line up right in here? If you cut and paste that into Notepad, and put extra linespaces between the equations, it should look right.
“There are only two things that are infinite: The Universe, and human stupidity-- and I’m not sure about the Universe”
–A. Einstein
Thanks Chronos. I’d never seen a calculation for that before.
On a related note, I just today came across an article at newscientist.com which I thought readers of this thread might be interested in. It suggests microscopic black holes don’t evaporate all the way, but rather stabilize at around a Planck mass.
It is too clear, and so it is hard to see.
Very nice calculations, Joe Cool, but you unfortunately did not take into account the Heisenberg Uncertainty Principle. It prevents you from knowing both the momentum and location of an individual object to more than a certain degree of accuracy (~10^-34 J*s). Since you require 24 sig figs for both mass and distance to the earth in your calculations, this puts you on the order of 10^-48, which is impossible. So there is no way, even theoretically, to detect the difference between a hammer and a feather hitting the earth, even if we could construct a perfectly round earth, infinitely small point objects, etc.
…and the plot thickens…
It is too clear, and so it is hard to see.
Now, why do we add the sig figs together?
rocks
>Now, why do we add the sig figs together?
We don’t. We multiply them. 10^-24 * 10^-24 = 10^-48. Technically, we’re not multiplying the sig figs but rather the accuracy with which we can measure each quantity. For example, something with 3 numbers after the decimal point (2.000 meters, e.g.) can actually be anywhere from 1.9995 to 2.0005. Thus the 3 numbers after the decimal point give a range of 10^-3 meters for the measurement. So 3 sig figs after the decimal point implies an uncertainty to the order of -3. 2.000 actually has 4 sig figs (since there’s one before the decimal point), so we’re only up to 10^-46, not 10^-48, but my off-the-cuff estimation is still well beyond the bounds of what Heisenberg allows (10^-34).
Which means you’re adding the significant figures, like RMM said. :rolleyes:
I don’t understand Opus1’s explanation of why you multiply the accuracy, but I did a crude estimate of the minimum quantum mechanical error in the time for the feather to drop. I got around 10^-17 seconds, far larger than the 10^-25 second difference joe_Cool calculated.
It is too clear, and so it is hard to see.
That seems to be a contradiction: you multiply them, but technically you don’t?
There also seems to be a mis-application of the use of significant figures. If you start with two quantities both of which have three significant figures, their product still has only three significant figures.
rocks
Sigh. I knew I’d actually have to do REAL calculations some time! Joe Cool said something to the effect that we need to carry it out 25 decimal places to see the difference. Another spot he said 30, somewhere else he implied 24 or 27. You can read the message itself. The problem is, you cannot determine both the position and initial velocity of the hammer (or feather) to arbitrary precesion. The limit is ~10^-34. If you have 24 decimal places, this implies an uncertainty of 10^-24. Multiply the uncertainty for both the initial velocity and the initial position, and you get 10^-48, which is less than 10^-34, and therefore impossible due to the Heisenberg Uncertainty Principle (HUP). Thus Joe Cool’s choice of “arbitrary precision” is impossible. That’s all I was really doing with multiplying the sig figs.
Now, let’s do the calculations to see how much precision we CAN obtain for the time element. According to Newton, x = x0 + v0 t + .5 a t^2. We’ll choose our starting point to be 0, so x0 = 0. Our x should be 2.0000… to as much precision as possible, and our v0 should be 0.0000… to maximal accuracy as well.
Solving this equation for time, we get t = (-v ± Sqrt[v^2 + 2 a x])/a. We are not interested in the time it takes to fall, however. Instead, we are interested in the DIFFERENCE in time it takes between successive trials of the same object, that is to say, the difference in times you can obtain from performing the same experiment. To do this, we replace t, v, and x with delta t, delta v, and delta x. Since I’m lazy, I’ll just leave the equation as is, but stating that from now on, all t’s, x’s and v’s in equations are actually deltas.
We also know from HUP that delta p * delta x >= h bar (hbar ~= 10^-34). The mass of the hammer is stated to be 2 kg, but we’ll make it one to simplify the calculations. Since p = momentum = m*v, we can rewrite HUP as m * delta v * delta x >= hbar. m = 1, so it just cancels out of the equation. Also, we are interested in maximal accuracy, so the >= sign becomes a simple = sign. Rewriting the equation and solving for delta x gives: delta x = hbar/delta v. Plugging this into our Newtonian equation yields:
t = (-v ± Sqrt[(2a hbar/v)+ v^2])/a
We now wish to find a value of delta v that mimimizes t. To do this, we take the derivative of t w/ respect to v, set it equal to zero, and obtain:
v = a^(1/3) * hbar^(1/3) / 2^(2/3) ~= 6.3*10^-12
Since delta x = hbar/delta v, it equals 1.6*10^-23.
These are the precisions which we should measure for our momentum and location in order to minimize our delta t. So, what is delta t? Plugging our v and x values back into the equation, we get t ~= 2.6*10^-12. This is a far greater number than the 10^-25 second difference between the hammer and the feather.
Thus, if we were to drop a hammer to the earth 100 separate times, we could not state how long it will take to hit the ground any more precisely than a picosecond, way more than the difference in time between a hammer and a feather. In other words, on any one trial, the hammer or the feather could hit first–there’s no way of knowing.
Zenbeam said he got 10^-17 seconds for the quantum mechanical accuracy. One of us messed up our calculations somewhere. I posted mine for the whole world to scrutinize, so maybe he can look and see where one of us went wrong.
Do you guys ever worry that your heads are going to explode? Cause I don’t think anyone should be thinking this hard and not getting paid for it! 
Bad spellers of the world… UNTIE
I haven’t had time to look over Opsu1’s derivation to figure out where it differs from mine. Being an MSU alumnus, I had an important meeting starting at 9:18 pm last night (Woohoo! National champs!! ). Here is my derivation. One caveat is that this is in CGS units, and I haven’t used CGS for 17 years, although I’m pretty sure I got it correct.
From the OP, the time to fall, T =0.639 seconds.
Using g = 980 cm/s, the final velocity V=626 cm/s.
Call the initial uncertainty in position and velocity dx and dv. The velocity uncertainty dv will give a position uncertainty Tdv when the feather is about to hit the Earth in T seconds. This is in addition to dx. Since the feather is falling with velocity V when it hits the ground, the uncertainty in position translates into a total uncertainty in time of
dt = dx / V + Tdv / V
From the uncertainty principle, dxdp >= hbar/2, where p = momentum = vm, and m = mass of feather. So dxdv >= hbar/(2m) (where hbar = h / (2pi)).
Using the minimum uncertainty case, dv = hbar/(2mdx), so
dt = (dx + hbarT/(2mdx))/V
Taking the derivative of dt WRT dx, and setting this to zero to get the minimum dt gives
d(dt)/d(dx) = 1 - hbarT/(2mdx^2) = 0, or dx^2 = hbarT/(2m)
so
dx = sqrt(hbarT/(2m)) and dv = sqrt(hbar/(2mT))
A little more algebra gives
dt = sqrt(2hbarT/(mV^2))
Using hbar = 1.0510^-27 erg-seconds, T = 0.639 seconds, m = 2 grams, V = 626 cm/s dt = sqrt(8.5610^-34) = 2.9310^-17 seconds. For the 2 kg hammer, dt = 9.2710^-19 seconds.
It is too clear, and so it is hard to see.
Do you guys see my last post as posted in the minute after Opus1’s, but it is threaded before? Now, that’s a heisenberg simul-post.
rocks
Opus1 wrote;
This is were the mistake is. To get the delta time, you can’t just replace t with dt, etc. You have to be more careful. Using your notation:
(T+t) = (-(V0+v) ± Sqrt[(V0+v)^2 + 2 a (X0-X+x)])/a
where X0 = V0 = 0, X = 2 meters, and T is the time to drop X if v and x were zero.
T+t = (-v ± Sqrt(v^2 + 2 a x - 2 a X))/a
Or, to first order in v and x,
T+t = -v/a ± Sqrt(-2 X / a) * (1 - x/X) = -v /a ± Sqrt(-2 X / a) ± Sqrt(-2 x^2/(a X))
With x = v = 0, T = ± Sqrt(-2 X / a), so you are left with
t = -v/a ± x * Sqrt(-2 /(a X))
which looks pretty similar to my equation, dt = dx / V + T*dv / V
It is too clear, and so it is hard to see.