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#1
03-06-2000, 01:52 PM
 C K Dexter Haven Right Hand of the Master Charter Member Join Date: Feb 1999 Location: Chicago north suburb Posts: 16,078
You might want to check out this earlier topic on the same subject... http://boards.straightdope.com/ubb/F...ML/000137.html
#2
03-07-2000, 12:20 AM
 Joe_Cool BANNED Join Date: Jun 1999 Posts: 2,815
I have a comment that is surely a waste of somebody's time, about an answer IAN gave to a mailbag question...

He stated that a hammer & feather will always fall at the same rate in a perfect vacuum. Not exactly true, since the earth accelerates toward the hammer slightly more than towards the feather, giving it an imperceptibly slanted vector that I'm too tired to figure out.

If they're dropped separately, though....

Grav. constant= 6.673e-11 m^3 Kg^-1 s^-2
Mass(earth)= 5.976e24 Kg
Mass(hammer)= 2.000 Kg
Mass(feather)= 0.001 Kg (1 g)
Distance dropped= 2.000 m (height of some random guy who's doing the dropping)

I'm not ignoring the significant digits rule, but since I can't find the mass of the earth to 27 digits anywhere, and we need to carry it out to at least 25 decimal places to see the difference, let's use pretend objects: a pretend earth that is precisely 5.976e24 Kg, a pretend hammer that is precicely 2 Kg, etc. all to arbitrary precision...let's say the values are good to at least 30 decimal places. And we'll pretend the hammer and feather have no volumes, to avoid the hassle of which one has farther to fall.

Gravitational acceleration towards any body:

G*m
= ------ (mass 2 cancel out since we're
d^2 figuring acceleration not force)

Acceleration of any body towards our pretend earth at 2m above the equator=
(6.673*10^-11)(5.976*10^24)
---------------------------
6378002^2

= 9.803076945371995882725286955486 m/s^2
(yes, for the real earth it's 9.81something, but dropping 24 significant digits makes a significant difference. this is theory here. and not the real earth)

Acceleration towards the hammer=
2*(6.673*10^-11)
-----------------
6378002^2

= 0.000000000000000000000003280815 m/s^2

Acceleration towards the feather= 0.001*(6.673*10^-11)
----------------------
6378002^2

= 0.000000000000000000000000001640 m/s^2

So the total acceleration between the earth and the hammer (both bodies accelerate towards each other) is:

9.803076945371995882725290236301 m/s^2,

and the total between the earth and the feather is:

9.803076945371995882725286957126 m/s^2.

Plugging these numbers into the equation d=(1/2)(at^2) and solving for time, the feather lands after
0.638776293290439842764991676717 seconds,

and the hammer after only
0.638776293290439842764991569880 seconds:

a full 1.06837e-25 seconds faster! if I could just save that much time every day, I'd...well I guess I'd never know the difference.

Of course there's the integral you have to mess with, since the acceleration increases as the objects get closer together, the interference caused by the gravitational attraction of the guy dropping the hammer & feather, magnetic effects, etc, but we'll ignore those because they're petty.

Ok, so is this, but so what?
#3
03-07-2000, 12:31 AM
 Joe_Cool BANNED Join Date: Jun 1999 Posts: 2,815
OOPS. Forgot the link. here it is:
http://www.straightdope.com/mailbag/mgravity.html
#4
03-07-2000, 12:44 AM
 WillGolfForFood Registered User Join Date: Dec 1999 Location: USA (Pennsylvania) Posts: 788
Ahh, but in the classic test, the hammer and feather are dropped *at the same time*. Given that, the earth will accelerate towards the gravitational center of both objects combined. (Of course, most of the earth's acceleration will be because of the hammer, but that's just the cause, not the effect.)
#5
03-07-2000, 05:27 AM
 RM Mentock Guest Join Date: Mar 1999 Location: Durham, NC, USA Posts: 3,199
Quote:
 Originally posted by CKDextHavn: You might want to check out this earlier topic on the same subject...
Wow. What are the odds? Are they greater than 10^-25?

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rocks
#6
03-07-2000, 09:43 AM
 ZenBeam Guest Join Date: Oct 1999 Location: I'm right here! Posts: 8,858
Joe_Cool, I'm sure your math is impeccable (i.e. it sounds about right, and no way am I cranking through those numbers ), but there is a fatal flaw in your calculations. When you drop the feather, where is the hammer? It's laying on the ground, at your feet! In other words, you should add the hammer mass to the Earth mass when calculating how fast the feather falls. Similarly, the Hammer is falling to the Earth, feather combination.

As an admitted precision fiend (come on, you can't deny it after your OP), I'll leave it to you to work through the details. Assume the hammer's mass is evenly distributed throughout the Earth when the feather is dropped.

Speculation: you'll either get a difference in time about 2000 times smaller (ratio of the hammer and feather masses), 10^25 smaller (I'll put my money here), or exact cancellation.

Further speculation: if you really had the hammer laying on the ground, the proximity of its mass to the feather will make the feather hit the ground sooner than the hammer falling with the feather on the ground.

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It is too clear, and so it is hard to see.
#7
03-07-2000, 01:23 PM
 ZenBeam Guest Join Date: Oct 1999 Location: I'm right here! Posts: 8,858
Quote:
 An even more important factor is the shape of the objects.
I was going to reply that JoeCool specified a spherical Earth, but looking through his OP, I see that he didn't.

Regardless, I don't think that shape is important here. Whatever the Earth's field is, both the hammer and feather see the same one. He specified the feather and hammer had no volume, which I took to mean point objects. He was looking at the difference between the two cases, not so much what either actual result was. Certainly the effect of the shape of the Earth swamps the difference he came up with, but it should effect both cases the same, so the difference shouldn't change.

Not specifying where the hammer is when the feather is dropped, and vice-versa, means the problem itself is ambiguous at the precision he is working at. That's why I was careful to specify that he distribute the hammer mass evenly throughout the Earth when droppng the feather.

I've been waiting for the "Which way is down" update also.

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It is too clear, and so it is hard to see.
#8
03-07-2000, 09:52 PM
 Joe_Cool BANNED Join Date: Jun 1999 Posts: 2,815
True, I forgot to specify in so many words that my example earth is a perfect sphere. but I thought it was implied clearly enough, since "radius" is the only dimension I specified. the only shape I know whose only dimension is radius would be a sphere.

Ok, ok, I didn't specify three-dimensional space (four, since we're figuring time as well), but that's pretty well implied, since in two-dimensional space there is no such thing as mass, only area, so gravity would not be applicable.

As for ZenBeam's comment, I meant for the objects to be dropped separately. So separately, in fact, that neither one affects the other's drop. The objects experience a quantum state fluctuation, instantaneously bringing one (being dropped) into existence 2 meters above the surface of the earth and changing the other's (not being dropped) position to a point so far away that its effects are negligible. For argument's sake, let's define negligible this way: gravitational acceleration caused by the object at this distance is &lt; 10^-1,000,000,000 m/s^2.

Good enough? Why not? If we can imagine a perfectly sized and shaped sphere and arbitrarily precise masses, why not a gross misapplication of quantum mechanics? hehe
#9
03-07-2000, 10:00 PM
 Joe_Cool BANNED Join Date: Jun 1999 Posts: 2,815
By the way, ZenBeam, I didn't actually crank through the numbers. I wrote a bc script (under Linux). &lt;shameless plug&gt;bc is a great little program. You can do calculations to any precision _in theory._&lt;/shameless plug&gt;

All told it took about 5 minutes.
#10
03-08-2000, 12:02 AM
 RM Mentock Guest Join Date: Mar 1999 Location: Durham, NC, USA Posts: 3,199
ZenBeam

An even more important factor is the shape of the objects. Since gravity acts as if the mass were concentrated at the center only in approximations, and we're dealing with effects the size of 10^-25, I'm pretty sure that those effects are swamped by other considerations.

That was the point you were making in the shape of the earth topic, wasn't it? (Which, is still unresolved I guess. I thought the mailbag column was going to be updated.)

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rocks
#11
03-08-2000, 03:42 AM
 RM Mentock Guest Join Date: Mar 1999 Location: Durham, NC, USA Posts: 3,199
Quote:
 Originally posted by ZenBeam: Regardless, I don't think that shape is important here. Whatever the Earth's field is, both the hammer and feather see the same one. He specified the feather and hammer had no volume, which I took to mean point objects.
I missed that! I wasn't referring to the shape of the earth (although I mentioned it)--I was referring to the different shapes of the hammer and feather. The OP converted them to points to avoid the "distance dropped" ambiguity--but that makes them awful hard to nail down.

What about the general relativity corrections? At this level of accuracy, GMm/r² doesn't work anymore, does it?

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rocks
#12
03-08-2000, 07:50 AM
 ZenBeam Guest Join Date: Oct 1999 Location: I'm right here! Posts: 8,858
Quote:
 Good enough?
Yes, but I still want to know what happens with the case I described.

Quote:
 I wrote a bc script (under Linux).
Excellent. It shouldn't be too hard to rerun it with the hammer mass added to the Earth for the feather dropping, and the feather mass added to the Earth for the hammer. Just to answer my "speculation" question.

Quote:
 What about the general relativity corrections? At this level of accuracy, GMm/r² doesn't work anymore, does it?
I had been wondering that also, but I wasn't going to bring it up. You are volunteering to work it out, aren't you?

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It is too clear, and so it is hard to see.
#13
03-08-2000, 11:01 AM
 RM Mentock Guest Join Date: Mar 1999 Location: Durham, NC, USA Posts: 3,199
Quote:
 Originally posted by ZenBeam: Excellent. It shouldn't be too hard to rerun it with the hammer mass added to the Earth for the feather dropping, and the feather mass added to the Earth for the hammer. Just to answer my "speculation" question.
Speculation? Yeah, right. (sotto voce: ZenBeam doesn't speculate, ZenBeam calculates). In JoeCool's calculation, he moved the hammer away to nearly infinity.

Since the mass of the earth is only 3x10^24 times the mass of the hammer, the added mass of the hammer would change the answer by approx. 1/3x10^24, which is twice the difference that Joe_Cool found in the OP. Y'all both are extremely sneaky: ZenBeam for calculating, and Joe_Cool for throwing away "vital" mass.

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rocks
#14
03-08-2000, 01:44 PM
 Joe_Cool BANNED Join Date: Jun 1999 Posts: 2,815
Ok, ZenBeam, here ya go.

Dropping the feather with the hammer's mass distributed evenly throughout the earth, or vice versa, you were right, of course. either dropped object falls in
.638776293290439842764991569826 seconds. That's a difference on the order of 10^-29 for the hammer, and 10^-25 for the feather.

With the thing not being dropped at the dropper's feet (or, since we're ignoring the mass of the dropper, collinear with the dropped object and the center of mass of the earth), the difference for the hammer is on the order of 10^-11 for both. But the feather does fall faster, obviously, since all that matters is the attractive mass, not the falling mass.

By the way, I cheated a bit. For the last part (object at feet), I didn't calculate the acceleration towards the center of mass of the entire system, I figured the attractions for the two systems separately and added them.
#15
03-25-2000, 07:12 PM
 Chronos Charter Member Moderator Join Date: Jan 2000 Location: The Land of Cleves Posts: 78,093
Joe_Cool wrote:
Quote:
 By the way, I cheated a bit. For the last part (object at feet), I didn't calculate the acceleration towards the center of mass of the entire system, I figured the attractions for the two systems separately and added them.
Actually, that's exactly correct. Cheating would be to just figure on the center of mass, because, as RM Mentock pointed out, the
"all mass concentrated at the center of mass" approximation is only valid for spherically symmetric mass distributions, i.e., not a (theoretical) spherically symmetric Earth + (theoretical) point mass on its surface. And the GR effects do come in way sooner than 10^-24; I would be tempted to actually calculate them, but at that level of precision, you'd need a theory of quantum gravity, which, alas, has not yet been developed.

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"There are only two things that are infinite: The Universe, and human stupidity-- and I'm not sure about the Universe"
--A. Einstein
#16
03-27-2000, 11:44 AM
 ZenBeam Guest Join Date: Oct 1999 Location: I'm right here! Posts: 8,858
Quote:
 And the GR effects do come in way sooner than 10^-24
I find this hard to believe. Any errors due to using Newtonian physics ought to cancel out*. What GR effects are going to affect the hammer and feather differently?

* Unless you're considering both the feather and hammer as point masses, and you're dropping one micro-black-hole into another. But this hardly seems like a valid approximation for the original case.

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It is too clear, and so it is hard to see.
#17
03-27-2000, 01:12 PM
 RM Mentock Guest Join Date: Mar 1999 Location: Durham, NC, USA Posts: 3,199
Quote:
 Originally posted by ZenBeam: Unless you're considering both the feather and hammer as point masses, and you're dropping one micro-black-hole into another. But this hardly seems like a valid approximation for the original case.
Joe Cool bites again. The OP said the hammer and feather have no volumes. I guess that's the same as point-mass.

When do we actually get to do a real experiment?

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rocks
#18
03-27-2000, 07:02 PM
 ZenBeam Guest Join Date: Oct 1999 Location: I'm right here! Posts: 8,858
Doh!
#19
03-27-2000, 09:09 PM
 Chronos Charter Member Moderator Join Date: Jan 2000 Location: The Land of Cleves Posts: 78,093
You're right, ZenBeam, the relativistic effects should affect the hammer and feather (almost) identically... I was just thinking about the effect on either. The black holes raise their own problems... If I recall my hawking correctly, a 2-kg or 1-g black hole would probably explosively evaporate before it hit the ground.
#20
03-28-2000, 12:01 AM
 Chronos Charter Member Moderator Join Date: Jan 2000 Location: The Land of Cleves Posts: 78,093
Further information on the Hawking evaporation: According to some back-of-the-envelope calculations I was just doing, a hole of mass 2 kg would radiate power at about 10^34 watts, which would result in total evaporation within 10^-17 seconds (less, in fact, since the power increases with decreasing size), much less than the time it would take to hit the Earth, so I think that we have to say that they're not QUITE point masses. If anyone really wants, I can post the calculations.

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"There are only two things that are infinite: The Universe, and human stupidity-- and I'm not sure about the Universe"
--A. Einstein
#21
03-28-2000, 07:43 AM
 ZenBeam Guest Join Date: Oct 1999 Location: I'm right here! Posts: 8,858
Quote:
 If anyone really wants, I can post the calculations.
Sure, I'd like to see them.

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It is too clear, and so it is hard to see.
#22
03-28-2000, 03:22 PM
 Chronos Charter Member Moderator Join Date: Jan 2000 Location: The Land of Cleves Posts: 78,093
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 = A*sigma*T^4 = 4*pi*R^2*sigma*( ------ )^4
G*m

c^4*C3^4
P = 4*pi*sigma* ----------
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
4*pi*c^2*sigma*C3^4
(max)[/code]
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
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.

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"There are only two things that are infinite: The Universe, and human stupidity-- and I'm not sure about the Universe"
--A. Einstein
#23
03-28-2000, 03:26 PM
 Chronos Charter Member Moderator Join Date: Jan 2000 Location: The Land of Cleves Posts: 78,093
OK, 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.

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"There are only two things that are infinite: The Universe, and human stupidity-- and I'm not sure about the Universe"
--A. Einstein
#24
03-31-2000, 12:14 AM
 ZenBeam Guest Join Date: Oct 1999 Location: I'm right here! Posts: 8,858
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.

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It is too clear, and so it is hard to see.
#25
04-02-2000, 12:33 AM
 Opus1 Guest Join Date: Apr 2000 Location: Tucson, AZ, USA Posts: 1,242
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.
#26
04-02-2000, 11:14 AM
 ZenBeam Guest Join Date: Oct 1999 Location: I'm right here! Posts: 8,858
Quote:
 there is no way, even theoretically, to detect the difference between a hammer and a feather hitting the earth
...and the plot thickens...

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It is too clear, and so it is hard to see.
#27
04-02-2000, 10:44 PM
 RM Mentock Guest Join Date: Mar 1999 Location: Durham, NC, USA Posts: 3,199
Quote:
 Originally posted by Opus1: 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.
Now, why do we add the sig figs together?

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rocks
#28
04-02-2000, 11:49 PM
 Opus1 Guest Join Date: Apr 2000 Location: Tucson, AZ, USA Posts: 1,242
&gt;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).
#29
04-03-2000, 08:46 AM
 ZenBeam Guest Join Date: Oct 1999 Location: I'm right here! Posts: 8,858
Quote:
 >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
Which means you're adding the significant figures, like RMM said.

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.

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It is too clear, and so it is hard to see.
#30
04-03-2000, 01:53 PM
 Opus1 Guest Join Date: Apr 2000 Location: Tucson, AZ, USA Posts: 1,242
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 &gt;= 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 &gt;= hbar. m = 1, so it just cancels out of the equation. Also, we are interested in maximal accuracy, so the &gt;= 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.
#31
04-03-2000, 01:54 PM
 RM Mentock Guest Join Date: Mar 1999 Location: Durham, NC, USA Posts: 3,199
Quote:
 Originally posted by Opus1: >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.
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.

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rocks
#32
04-03-2000, 08:09 PM
 JLPicard Guest Join Date: Jun 1999 Posts: 100
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! :-)

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Bad spellers of the world... UNTIE
#33
04-04-2000, 11:41 AM
 ZenBeam Guest Join Date: Oct 1999 Location: I'm right here! Posts: 8,858
Quote:
 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.
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 T*dv 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 + T*dv / V
From the uncertainty principle, dx*dp &gt;= hbar/2, where p = momentum = v*m, and m = mass of feather. So dx*dv &gt;= hbar/(2*m) (where hbar = h / (2*pi)).
Using the minimum uncertainty case, dv = hbar/(2*m*dx), so
dt = (dx + hbar*T/(2*m*dx))/V
Taking the derivative of dt WRT dx, and setting this to zero to get the minimum dt gives
d(dt)/d(dx) = 1 - hbar*T/(2*m*dx^2) = 0, or dx^2 = hbar*T/(2*m)
so
dx = sqrt(hbar*T/(2*m)) and dv = sqrt(hbar/(2*m*T))
A little more algebra gives
dt = sqrt(2*hbar*T/(m*V^2))

Using hbar = 1.05*10^-27 erg-seconds, T = 0.639 seconds, m = 2 grams, V = 626 cm/s dt = sqrt(8.56*10^-34) = 2.93*10^-17 seconds. For the 2 kg hammer, dt = 9.27*10^-19 seconds.

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It is too clear, and so it is hard to see.
#34
04-04-2000, 08:23 PM
 RM Mentock Guest Join Date: Mar 1999 Location: Durham, NC, USA Posts: 3,199
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.

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rocks
#35
04-08-2000, 12:10 AM
 ZenBeam Guest Join Date: Oct 1999 Location: I'm right here! Posts: 8,858
Opus1 wrote;
Quote:
 Solving this equation for time, we get t = (-v +- Sqrt[v^2 + 2 a x])/a. [...] 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.
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

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It is too clear, and so it is hard to see.

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