Well, assuming in the other universe there are intelligent beings, and those intelligent beings develop the mathematical concepts of real numbers and of Euclidean space, and for some reason they ask for the ration of circumference and diameter of a circle in Euclidean space, they'll find the same value.
Of course a completely different question is, assuming there exists an universe which is not locally euclidean, and which contains intelligent beings, will they develop the (alien to them) concept of Euclidean space?
> OTOH, the red shift due to the quasar speed just gives all observed energies a constant factor, without changing the ratio of energy differences
This is an amazingly bad assuption.
No, it's amazingly simple math.
Every frequency is scaled with a common factor f, which does not depend on the frequency. This is true both for the relativistic Doppler effect and the non-relativistic (medium-relative) Doppler effect (Ok, the cosmological red shift isn't exactly a Doppler effect, but it's a common factor nevertheless).
The frequency of the emitted light is, according to quantum mechanics, given by the difference of the energy levels; that is, if you compare the ratio of the differences of the energy levels, what you actually do is to compare the ratio of frequencies.
Thus if the measured light frequencies are nu1 and nu2, the red shift changes the ratio nu1/nu2 to (f*nu1)/(f*nu2). Thus the red shift cancels out.
Well, take conventional sphere as an example (that certainly is a Riemannian surface). The spherical excess E (i.e. the difference of the sum of the angles of a spherical triangle to pi) is given by
with s = (a + b + c)/2, and a, b, c the sides of the triangle (divided by sphere radius if that is not 1).
Now you can see that if a, b, c go to zero, so does E. That is, you can make the spherical excess arbitrary small (i.e. making the sum of the angles arbitrary close to pi) by just making the triangle small enough. Yes, it will never be exactly pi, but the difference can certainly be made small enough that it will not show up in the first 100 digits (of course, if the sphere has earth radius, then your triangle will have to have sides far smaller than Planck length to get the first 100 digits right (s/r has to be below about 10^-50, but the planck length is just about 10^-35 meters).
Now, the curvature radius of space around earth is IIRC about one light year (roughly 10^16 meters), so as long as you keep below about 10^-36 meters, you should be OK in the first 100 digits of pi (at least as far as the sum of angles in a triangle is concerned). Unfortunately, this is still below planck length, so you just won't get your 100 digits in the real world anyway:-)
Ok, let's look at circumference over diameter instead. Let's embed a sphere of radius R into 3D euclidean space (because that simplifies the calculations). Now let's assume we have a circle with radius r on the sphere. then that circle has the radius R sin(r/R) in the embedding euclidean space, thus its circumference is 2 pi R sin(r/R), which gives a "spherical pi" of pi R/r sin(r/R), which for small r can be Taylor-expanded to pi (1 - 1/6 (r/R)^2 + O((r/R)^4)). Again, we need r/R to be below about 10^-50 to get pi right in the first 100 digits.
Also note that in all considerations above, no deformation of the sphere is done. The topology (which is a global property of the sphere) doesn't enter into the picture (after all, we are making the triangle small, thus probing only local properties of the sphere).
The molecular hydrogen spectrum is composed from different effects: The electronic states, for which the proton mass is almost negligible (because it's about the electron movement, and the protons are just so much heavier; just like you can neglect the movement of the sun when looking at planetary orbits), and the rotational and vibrational states, which depend strongly on the proton mass (because it's about the movement of the protons themselves). So any change in the ratio of electron and proton mass will show up in the ratio of energy differences between electronic and rotational/vibrational states. OTOH, the red shift due to the quasar speed just gives all observed energies a constant factor, without changing the ratio of energy differences (because that common factor will just cancel out).
that I face the risk that the first 100 digits of PI that I have memorized could change, and the knowledge becomes useless?
Well, that was true since Einstein: The value of pi as you learned it is only valid in Euklidean (flat) space, and our space is Riemannian (curved). However, to your relieve, the Riemannian space is locally Euclidean, so if you restrict yourself to a small enough volume, your 100 digits are accurate again. Unless you get into trouble with quantum physics (I'm now too lazy to calculate if you could get 100 digits of pi right on Earth without getting close to the Planck length).
What you miss is the damage done by a false alarm. If a fire brigade comes to a building which doesn't burn, well, that's just the cost of the fire brigade going to that building. Not a big deal. However, look at the effect this false bomb alarm had. A false alarm with such drastic consequences several times in a row will surely cause much damage in itself. Add to that that true bomb alarms are much more rare than true fire alarms, and you have the big risk of the alarm being ignored, based on the expected damage of not ignoring it being higher than the expected damage of ignoring it.
purposely display an image of a dangerous item where none exists, inciting a scare like the one witnessed Wednesday, disrupting thousands of lives and paralyzing a major terminal
More importantly: After enough false alarms, the screeners will more likely not react should a real bomb appear. "Oh well, surely just another software fault, just like the three we've had earlier this week. We better don't scare our passengers again..."
But did you also expect Microsoft to admit it? The headline doesn't read "Microsoft hides flaw details", but "Microsoft admits to hiding flaw details".
It's easy: The more the big, influential companies get hurt by patents, the more likely we are to get rid of them (the patents, that is). So you can be anti-patent and at the same time hope that MS will get hurt by patent violations as much as possible, without contradicting yourself.
Of course the TV stations could respond by e.g. leaving that flag set for some time after the ads, so you'd lose some of the following part of the movie.
What about pulling the plug from the wall? I guess they'll do a law that all players must have sealed plugs which may only be put in/released by professional electricians who need a special license to be allowed to unplug media players. Of course, every plugging and unplugging will be recorded and sent to the corresponding companies/organizations. And of course you will have to prove that the unplugging was really necessary.
Now that this problem seems to have been solved, 3DR have expanded their team considerably, from 22 to 31 members, marking what many hope to be the final stage of the development cycle.
What happens again if you add more people to a late project?:-)
Consider the simple fact, drawn from the official temperature records of the Climate Research Unit at the University of East Anglia, that for the years 1998-2005 global average temperature did not increase (there was actually a slight decrease, though not at a rate that differs significantly from zero).
If average temperature did not increase since 1998, it means the first year to not have a higher temerature than the one before is 1999. Note that Napster was first released in 1999. So it's quite obvious that those two events are related. Thus we now have the final proof that piracy indeed counteracts global warming!
Unless I should find out that my friend in reality comes from space and researches for the next edition of an electronic book which has the words "don't panic" written on it in large, friendly letters...
Whoa whoa whoa, slow down. First I learn there's a "nice" command, and now you're telling me there's a "man" command too? This is way too much information for one day.
Obviously it is. Otherwise you would have noted that he told you about the mount command as well.:-)
Slashdot Mirror
Yes, but does it run Linux? ...
Well, imagine a Beowulf cluster of them
And price-performance = performance/price, and not price/performance?
I guess you'd also take the job with the lower monthly wage, because a smaller time/money ratio means more money in the same time, right?
Let's see ... 2019-07-17 :-)
take digit sums -> 12,7,8
multiply -> 672
subtract digit sum of year -> 666
Ok, now we know what will happen on that date!
Well, assuming in the other universe there are intelligent beings, and those intelligent beings develop the mathematical concepts of real numbers and of Euclidean space, and for some reason they ask for the ration of circumference and diameter of a circle in Euclidean space, they'll find the same value.
Of course a completely different question is, assuming there exists an universe which is not locally euclidean, and which contains intelligent beings, will they develop the (alien to them) concept of Euclidean space?
No, it's amazingly simple math.
Every frequency is scaled with a common factor f, which does not depend on the frequency. This is true both for the relativistic Doppler effect and the non-relativistic (medium-relative) Doppler effect (Ok, the cosmological red shift isn't exactly a Doppler effect, but it's a common factor nevertheless).
The frequency of the emitted light is, according to quantum mechanics, given by the difference of the energy levels; that is, if you compare the ratio of the differences of the energy levels, what you actually do is to compare the ratio of frequencies.
Thus if the measured light frequencies are nu1 and nu2, the red shift changes the ratio nu1/nu2 to (f*nu1)/(f*nu2). Thus the red shift cancels out.
Well, take conventional sphere as an example (that certainly is a Riemannian surface). The spherical excess E (i.e. the difference of the sum of the angles of a spherical triangle to pi) is given by
/ 2))
:-)
tan(E/4) = sqrt(tan(s/2)*tan((s-a)/2)*tan((s-b)/2)*tan((s-c)
with s = (a + b + c)/2, and a, b, c the sides of the triangle (divided by sphere radius if that is not 1).
Now you can see that if a, b, c go to zero, so does E. That is, you can make the spherical excess arbitrary small (i.e. making the sum of the angles arbitrary close to pi) by just making the triangle small enough. Yes, it will never be exactly pi, but the difference can certainly be made small enough that it will not show up in the first 100 digits (of course, if the sphere has earth radius, then your triangle will have to have sides far smaller than Planck length to get the first 100 digits right (s/r has to be below about 10^-50, but the planck length is just about 10^-35 meters).
Now, the curvature radius of space around earth is IIRC about one light year (roughly 10^16 meters), so as long as you keep below about 10^-36 meters, you should be OK in the first 100 digits of pi (at least as far as the sum of angles in a triangle is concerned). Unfortunately, this is still below planck length, so you just won't get your 100 digits in the real world anyway
Ok, let's look at circumference over diameter instead. Let's embed a sphere of radius R into 3D euclidean space (because that simplifies the calculations). Now let's assume we have a circle with radius r on the sphere. then that circle has the radius R sin(r/R) in the embedding euclidean space, thus its circumference is 2 pi R sin(r/R), which gives a "spherical pi" of pi R/r sin(r/R), which for small r can be Taylor-expanded to pi (1 - 1/6 (r/R)^2 + O((r/R)^4)). Again, we need r/R to be below about 10^-50 to get pi right in the first 100 digits.
Also note that in all considerations above, no deformation of the sphere is done. The topology (which is a global property of the sphere) doesn't enter into the picture (after all, we are making the triangle small, thus probing only local properties of the sphere).
The molecular hydrogen spectrum is composed from different effects: The electronic states, for which the proton mass is almost negligible (because it's about the electron movement, and the protons are just so much heavier; just like you can neglect the movement of the sun when looking at planetary orbits), and the rotational and vibrational states, which depend strongly on the proton mass (because it's about the movement of the protons themselves). So any change in the ratio of electron and proton mass will show up in the ratio of energy differences between electronic and rotational/vibrational states. OTOH, the red shift due to the quasar speed just gives all observed energies a constant factor, without changing the ratio of energy differences (because that common factor will just cancel out).
Well, that was true since Einstein: The value of pi as you learned it is only valid in Euklidean (flat) space, and our space is Riemannian (curved). However, to your relieve, the Riemannian space is locally Euclidean, so if you restrict yourself to a small enough volume, your 100 digits are accurate again. Unless you get into trouble with quantum physics (I'm now too lazy to calculate if you could get 100 digits of pi right on Earth without getting close to the Planck length).
What you miss is the damage done by a false alarm. If a fire brigade comes to a building which doesn't burn, well, that's just the cost of the fire brigade going to that building. Not a big deal. However, look at the effect this false bomb alarm had. A false alarm with such drastic consequences several times in a row will surely cause much damage in itself. Add to that that true bomb alarms are much more rare than true fire alarms, and you have the big risk of the alarm being ignored, based on the expected damage of not ignoring it being higher than the expected damage of ignoring it.
More importantly: After enough false alarms, the screeners will more likely not react should a real bomb appear. "Oh well, surely just another software fault, just like the three we've had earlier this week. We better don't scare our passengers again
But did you also expect Microsoft to admit it?
The headline doesn't read "Microsoft hides flaw details", but "Microsoft admits to hiding flaw details".
True, that ISNOT consistent ... oops! :-)
It's easy: The more the big, influential companies get hurt by patents, the more likely we are to get rid of them (the patents, that is).
So you can be anti-patent and at the same time hope that MS will get hurt by patent violations as much as possible, without contradicting yourself.
Rejected: Too much prior art.
Better read them some of the poems written by Paula Nancy Millstone Jennings. :-)
Of course the TV stations could respond by e.g. leaving that flag set for some time after the ads, so you'd lose some of the following part of the movie.
And all rooms get automatic doors which lock whenever a commercial is sent, so I cannot just go to the toilet during the ads?
What about pulling the plug from the wall? I guess they'll do a law that all players must have sealed plugs which may only be put in/released by professional electricians who need a special license to be allowed to unplug media players. Of course, every plugging and unplugging will be recorded and sent to the corresponding companies/organizations. And of course you will have to prove that the unplugging was really necessary.
What happens again if you add more people to a late project?
If average temperature did not increase since 1998, it means the first year to not have a higher temerature than the one before is 1999. Note that Napster was first released in 1999. So it's quite obvious that those two events are related. Thus we now have the final proof that piracy indeed counteracts global warming!
Unless I should find out that my friend in reality comes from space and researches for the next edition of an electronic book which has the words "don't panic" written on it in large, friendly letters ...
Obviously it is. Otherwise you would have noted that he told you about the mount command as well.