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Author Topic: Can cow farts make the Earth rotate faster?  (Read 57141 times)

Offline Geezer

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Can cow farts make the Earth rotate faster?
« Reply #125 on: 03/08/2011 23:20:47 »
Geezer,
you say "During IFD the angular velocity was not uniform. It increased a bit, then it slowed back down so that the daily cycle time was reduced and that resulted in a phase shift of the Earth's rotational cycle relative to our atomic clock "day" (pretty hard to argue with that)."
Yes, the earth had an off day in terms of timekeeping.
But only one bad day in the whole of forever. On average, it didn't happen.

No argument. It depends on your interpretation of "make the Earth rotate faster". The "faster" bit did happen, just not for very long.

My greater objection was to the point that phase and frequency are independent. That goes against forty years of peering at oscilliscopes.

Understood about the ball and chain thing, but I'm not sure it's any more realistic than the cows. I think I'd prefer some some gigantic rockets. It should not be too difficult to determine how much thrust they would have to develop in order to accelerate the Earth's rotation by a measureable amount.

I'm clueless about how to determine how long the deceleration would take. I suspect the function would be exponential, but how would we calculate the atmospheric friction torque, even to a crude approximation? 
 

Offline Geezer

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« Reply #126 on: 03/08/2011 23:32:21 »
Wait a minute! BC, you said "On average, it didn't happen".

After the short day, the average absolutely will reveal a difference, unless you are going to count cycles that have not happened yet, so you can't say "it didn't happen", particularly when a permanent change in phase marks the time when it really did happen.
 

Offline Bored chemist

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Can cow farts make the Earth rotate faster?
« Reply #127 on: 04/08/2011 06:59:54 »
Did someone say that the Earth is a closed system?  Does that exclude, Light, gravity, cosmic rays, dark matter, asteroids, meteors, comets and UFOs?  Not to mention interplanetary dust.  In the cosmic context I do not think earth is a closed system, if it was we could not exist as the dinosaurs would still be running about..  :o

We are ignoring quite a lot of things to simplify the model.


Geezer, I'm averaging over an infinite past history. (which is one such simplifying assumption).
« Last Edit: 04/08/2011 07:04:41 by Bored chemist »
 

Offline Geezer

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Can cow farts make the Earth rotate faster?
« Reply #128 on: 04/08/2011 08:43:31 »
Geezer, I'm averaging over an infinite past history. (which is one such simplifying assumption).


Are you really sure you want to do that? If the Earth doubled its angular velocity tomorrow, would you say it wasn't happening because it had never happened before?
« Last Edit: 04/08/2011 08:56:29 by Geezer »
 

Offline Bored chemist

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Can cow farts make the Earth rotate faster?
« Reply #129 on: 04/08/2011 19:56:44 »
If it did it briefly enough I might.
 

Offline JP

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Can cow farts make the Earth rotate faster?
« Reply #130 on: 04/08/2011 20:36:21 »
If it only happens once, you can chalk it up to observational error and ignore it, right?  :)
 

Offline Geezer

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Can cow farts make the Earth rotate faster?
« Reply #131 on: 04/08/2011 21:23:42 »
If it only happens once, you can chalk it up to observational error and ignore it, right?  :)

That's what I usually do.

"What the bleep was that?"

"Beats me. Anyway, it doesn't fit the model, so it's obviously a fluke. Ignore it."

Off topic, that's a very common occurrence when testing just about anything with a computer in it. The test engineers see some weird behaviour, but because they can't reproduce the problem, the development engineers tell them they either screwed up or were hallucinating, and close out the problem report.

The weird behaviour usually reappears about two hours before the product is supposed to be released to the market.
« Last Edit: 04/08/2011 21:55:38 by Geezer »
 

Offline JP

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Can cow farts make the Earth rotate faster?
« Reply #132 on: 04/08/2011 22:36:02 »
It's like the old joke about a mathematician, physicist and engineer trying to prove that all odd numbers are primes:

The mathematician says "1 is prime, 3 is prime, 5 is prime, so by induction all odds are primes."

The physicist says "1 is prime, 3 is prime, 5 is prime, 7 is prime, 9 isn't prime (but that's experimental error), 11 is prime, so all odds are prime."

The engineer says "1 is prime, 3 is prime, 5 is prime, 7 is prime, 9 is prime, 11 is prime, so all odds are prime."
 

Offline JP

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« Reply #133 on: 04/08/2011 22:55:39 »
Ok, I just came up with this argument that seems to prove by contradiction that "fart day" generates extra frequencies (via the Fourier transform).  Let me know what you think.

First, you need to know that the Fourier transform has an inverse.  From a signal over time, you can uniquely get the frequency spectrum of that signal, and from a frequency spectrum you can uniquely recover the signal over time. 

Second, if the earth rotated unimpeded by farting cows, you could model it by a periodic sinusoid, s(t).  Maybe this sinusoid starts and stops and maybe it goes off to infinity.  It doesn't matter.  It generates a frequency spectrum, say S(f).  You can go back and forth from S(f) to s(t) by Fourier transforms and inverse Fourier transforms.  I could just as easily have told you that the frequency spectrum is S(f) and you could have recovered the signal over time, s(t).  There is no loss of information in the Fourier transform.

Let's assume BC is right and that the frequency spectrum with the cows farting is the same as without.  If that's the case, then it's also given by S(f).  By the properties of the Fourier transform, a frequency spectrum S(f) means that the earth's rotation is given by s(t), which we know is true from above.

But this is identical to the signal without the cows farting, and we know the signals cannot be identical (there's a phase shift).  So the frequency spectra cannot be identical.

QED?
« Last Edit: 04/08/2011 23:12:19 by JP »
 

Offline damocles

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Can cow farts make the Earth rotate faster?
« Reply #134 on: 05/08/2011 02:28:55 »
Ok, I just came up with this argument that seems to prove by contradiction that "fart day" generates extra frequencies (via the Fourier transform).  Let me know what you think.

First, you need to know that the Fourier transform has an inverse.  From a signal over time, you can uniquely get the frequency spectrum of that signal, and from a frequency spectrum you can uniquely recover the signal over time. 

Second, if the earth rotated unimpeded by farting cows, you could model it by a periodic sinusoid, s(t).  Maybe this sinusoid starts and stops and maybe it goes off to infinity.  It doesn't matter.  It generates a frequency spectrum, say S(f).  You can go back and forth from S(f) to s(t) by Fourier transforms and inverse Fourier transforms.  I could just as easily have told you that the frequency spectrum is S(f) and you could have recovered the signal over time, s(t).  There is no loss of information in the Fourier transform.

Let's assume BC is right and that the frequency spectrum with the cows farting is the same as without.  If that's the case, then it's also given by S(f).  By the properties of the Fourier transform, a frequency spectrum S(f) means that the earth's rotation is given by s(t), which we know is true from above.

But this is identical to the signal without the cows farting, and we know the signals cannot be identical (there's a phase shift).  So the frequency spectra cannot be identical.

QED?

I think that the problem with this analysis has precisely to do with the phase shift. A frequency spectrum S(f) does not uniquely define a signal s(t) unless we have a firm boundary condition, such as s(0) = 0, To take a simple example the signal function (sin qt + sin 2qt) has precisely the same frequency spectrum as (sin qt + cos 2qt) -- equal spikes at f = q and f = 2q -- but they are quite different functions.
 

Offline JP

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« Reply #135 on: 05/08/2011 12:55:03 »
To take a simple example the signal function (sin qt + sin 2qt) has precisely the same frequency spectrum as (sin qt + cos 2qt) -- equal spikes at f = q and f = 2q -- but they are quite different functions.

The Fourier spectra of these are actually quite different.   For both signals you get four spikes: at +/-q and +/-2q.  In the first case, they're weighted by i/2, -i/2, i/2, -i/2, respectively.  In the second case, they're weighted by i/2, -i/2, 1/2, 1/2, respectively.  You get spikes at the same frequencies, but the weights are different, so it's a change in Fourier spectrum. 

At some point, it comes down to the definition.  I would say the frequency in the above case has changed, since I'm using the Fourier definition of frequency.  The fact that you get (different) complex weights allows you to invert the transform to get the signals back. 
 

Offline Bored chemist

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« Reply #136 on: 05/08/2011 18:06:35 »
The FT is an integral (of sorts).

If I integrate something then differentiate it again I lose information because I don't know the "constant of integration".

I think that zero frequency information is lost in the same way.
 

Offline JP

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« Reply #137 on: 05/08/2011 19:18:46 »
Zero frequency information shouldn't be lost via the FT.  It's what we call a "DC term" in frequency analysis.  If I have a signal s(t)+Constant, then the FT of that is equal to the Fourier transform of s(t) plus a zero-frequency component of amplitude Constant. 

If you invert the Fourier transform, you get exactly the original signal back.

This is because the FT is a definite integral (with infinite limits), and the inverse FT is also a definite integral.  There is no derivative being taken.

The FT and it's inverse do run into some issues with some signals.  I know that figuring out the class of functions it fails on is very difficult, but what was taught to me in physics and optics was that for almost every physical signal, the FT works.  (I do think it has problems representing an instantaneous phase change in a sine wave, for example, but this isn't physical.)
 

Offline Geezer

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« Reply #138 on: 05/08/2011 19:58:01 »
I do think it has problems representing an instantaneous phase change in a sine wave, for example, but this isn't physical.

I would guess that's because there is "no time" involved, in which case it's no longer a continuous function. It's really a different wave, or a wave that is in two places at the same time, so everything goes haywire.

Presumably that does not happen as soon as you introduce any sort of slope into the function.
 

Offline JP

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« Reply #139 on: 05/08/2011 21:35:34 »
That's not exactly true.  You can Fourier transform back and forth from a step function or a "top hat" without losing information, even though they're discontinuous.  I could have just done the computation wrong for the sin/cos, which is possible, or there might be something special about it. 

By the way, I'm talking about doing the calculation analytically here.  Of course, trying to do a discrete FT of a step is going to have some issues, since you can't sample the step with perfect resolution.
 

Offline Geezer

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« Reply #140 on: 06/08/2011 01:54:17 »
since you can't sample the step with perfect resolution.

Ah, right! Doesn't that boil down to giving it a certain amount of slope that doesn't really exist, in which case a step function would really be a ramp?
 

Offline JP

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« Reply #141 on: 06/08/2011 12:43:59 »
since you can't sample the step with perfect resolution.

Ah, right! Doesn't that boil down to giving it a certain amount of slope that doesn't really exist, in which case a step function would really be a ramp?

Something like that, I think.  When you actually sample the function, you have to do a discrete Fourier transform, which isn't quite the same as the continuous integral.  If the function has sharp features that you can't resolve perfectly with your sampling, it throws off the result a little bit.
 

Offline Geezer

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« Reply #142 on: 06/08/2011 17:54:50 »
Something like that, I think.  When you actually sample the function, you have to do a discrete Fourier transform, which isn't quite the same as the continuous integral.  If the function has sharp features that you can't resolve perfectly with your sampling, it throws off the result a little bit.

Right, but what about a "vertical" section of the signal? If it really is vertical, there is no phase angle between the start and end of the vertical section.
 

Offline JP

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« Reply #143 on: 08/08/2011 14:40:16 »
Something like that, I think.  When you actually sample the function, you have to do a discrete Fourier transform, which isn't quite the same as the continuous integral.  If the function has sharp features that you can't resolve perfectly with your sampling, it throws off the result a little bit.

Right, but what about a "vertical" section of the signal? If it really is vertical, there is no phase angle between the start and end of the vertical section.

The way I'm picturing this, the signal is sin(t) to the left of the discontinuity and sin(t+phi) to the right.  The phase of the sine shifts by phi at the discontinuity.
 

Offline Geezer

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« Reply #144 on: 08/08/2011 18:06:15 »

The way I'm picturing this, the signal is sin(t) to the left of the discontinuity and sin(t+phi) to the right.  The phase of the sine shifts by phi at the discontinuity.


Ewe! You mean a DC section. I'm not sure that constitutes a phase shift exactly. I think you have to instantaneously "jump" from one sine wave to another sine wave of the same frequency but different phase, and that results in an "infinite" slope rather than no slope.
 

Offline JP

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« Reply #145 on: 08/08/2011 18:52:28 »
Maybe we're at odds with each other on terminology.  Usually when I see DC, I think of a constant added to the signal.  I would tell you that sin(t)+C has a DC component of C.

When I think of phase of a sine function, I think of whatever's in the parentheses.  For sin(t), the phase is t, which increases linearly and continuously with time.  If it jumps from sin(t) to sin(t+phi) suddenly, that's a discontinuous change in phase, which I would probably call a phase jump or phase discontinuity, although I assumed (perhaps wrongly) that's what we meant by the term phase shift.

At any rate, I think I get your point.  Your sampling density can only put a lower limit on the slope.  If you know it was sin(t) on one side and sin(t+phi) on the other, you don't know if it took the entire interval to climb or if it jumped in a fraction of the interval.  Your discrete Fourier transform will look identical whether it did either of those, you you've lost information.  (I believe the usual discrete Fourier transform assumes linear changes in phase across intervals.)
 

Offline Geezer

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« Reply #146 on: 08/08/2011 21:46:17 »
I'm pretty sure an instantaneous phase change constitutes a vertical "jump". If the signal had a horizontal section and you analysed only that section, there would be no frequency components at all, whereas the vertical jump tends towards infinite frequencies.

EDIT: Of course, if you do the "jump" at precisely the right time, the signal would just have a sharp bend in it.
« Last Edit: 08/08/2011 22:15:52 by Geezer »
 

Offline JP

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« Reply #147 on: 08/08/2011 22:51:18 »
I'm pretty sure an instantaneous phase change constitutes a vertical "jump". If the signal had a horizontal section and you analysed only that section, there would be no frequency components at all, whereas the vertical jump tends towards infinite frequencies.

EDIT: Of course, if you do the "jump" at precisely the right time, the signal would just have a sharp bend in it.

Yep.  I agree.  The DC term I was talking about is a flat, horizontal component of the signal, or a flat, horizontal component added to a changing signal, which just shifts it's amplitude globally by a certain amount.  A discontinuous jump in the signal would have infinite slope. 

But if you sampled the signal, you'd only be able to say that it jumped between samples by a certain amount.  You couldn't tell if that was an instantaneous jump with infinite slope, or a steep but linear rise.  For that reason, sampling and doing a discrete version of the Fourier transform can lose information if you aren't careful about sampling finely enough.  (And you can never sample vertical segments densely enough.) 

By the way, one of the basic rules of determining sampling density is to use Nyquist sampling:
http://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem ,
which essentially says that you have to sample your signal more finely if you want to accurately recover high frequency components.
 

Offline Geezer

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« Reply #148 on: 09/08/2011 03:45:06 »
Great! Well, I think this cow has been more than adequately flogged.
 

Offline JP

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« Reply #149 on: 09/08/2011 16:44:39 »
Great! Well, I think this cow has been more than adequately flogged.

We went from cow farts to Fourier transforms in just 6 pages of posts!
 

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Can cow farts make the Earth rotate faster?
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