The Virtuosi (Posts about credit cards)https://thephysicsvirtuosi.com/enContents © 2019 <a href="mailto:thephysicsvirtuosi@gmail.com">The Virtuosi</a> Thu, 24 Jan 2019 15:05:03 GMTNikola (getnikola.com)http://blogs.law.harvard.edu/tech/rss- The Magnetar Credit Card Swipehttps://thephysicsvirtuosi.com/posts/old/the-magnetar-credit-card-swipe/Corky<div><hr>
<p><a href="http://2.bp.blogspot.com/_fa6AZDCsHnY/TTumQ-3TLgI/AAAAAAAAAJE/BI366RfOTR0/s1600/nedflanderscredit.jpg"><img alt="image" src="http://2.bp.blogspot.com/_fa6AZDCsHnY/TTumQ-3TLgI/AAAAAAAAAJE/BI366RfOTR0/s320/nedflanderscredit.jpg"></a>
Ned Flanders' credit card doesn't satisfy the <a href="http://en.wikipedia.org/wiki/Luhn_algorithm">Luhn</a> checksum, but could probably still be erased by a magnetar.</p>
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<p>Hello, Internet! Today I'd like to talk about the Magnetar Credit Card
Swipe. Sounds like some sinister short on a derivatives deal, doesn't
it? Well, no need to worry, we don't deal with scary things like that
here. Instead, we are going to talk about a super-magnetized neutron
star speeding past Earth. A while ago I heard that a magnetar can erase
all the world's credit cards from half the distance to the moon. I did a
little research and it seems like this is the go-to "fun fact" about
magnetars. Almost every time they are brought up in a popular science
article, their credit card-erasing prowess is sure to get a mention. So
let's check it out! First things first, though. What the heck is a
magnetar [1]? Well, "magnetar" is just a spiffy name for a particular
flavor of neutron star. Now, neutron stars are already pretty extreme
objects. They've got a little more than the mass of the sun squished
down to the size of a big city with a central density over 10 trillion
times greater than lead.</p>
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<p><a href="http://4.bp.blogspot.com/_fa6AZDCsHnY/TUXMdXcrmoI/AAAAAAAAAJI/K2MVmanh5g4/s1600/earthmag.png"><img alt="image" src="http://4.bp.blogspot.com/_fa6AZDCsHnY/TUXMdXcrmoI/AAAAAAAAAJI/K2MVmanh5g4/s200/earthmag.png"></a>
Dipolar magnetic field</p>
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<p>What makes magnetars truly name-worthy (and more extreme than Doritos
and Mountain Dew at the X-Games) is the fact that they have strong
magnetic fields. And when I say strong, I mean <em>really</em> strong. Taking
the magnetic field to be dipolar (like the Earth's, see figure), the
field at the poles of a magnetar can be as high as 10^15 gauss [2]. For
comparison, the magnetic field of the Earth is about half a gauss and
the big magnets used in MRIs are about 10^4 gauss. So we're talking
about some big fields! Looking up the specs for a typical credit card,
it looks like most take about 1000 gauss to erase. And now we can get
started. I always forget the exact form of the magnetic field of a
dipole, but Jackson doesn't. He tells me that it is $$
\vec{B}\left(\vec{r}\right) =
\frac{3\vec{n}\left(\vec{n}\cdot\vec{m}\right) -
\vec{m}}{|\vec{r} |^3}$$ where m is the magnetic dipole moment, r is
the displacement vector to where we are measuring the field, and n is a
unit vector that points to where we are measuring. To find the magnitude
(which is what we really care about), we just take the dot product of
the vector with itself and take the square root. Working this out we get
$$ B\left(\vec{r}\right) = \frac{|\vec{m}|}{r^3}{\left[
3\cos^2{\theta}+1\right]}^{1/2} $$ where theta is the angle between
the magnetic moment vector and the direction vector n. But we are just
looking for a rough estimate here so let's just set the cosine term to
one. Now we have
$$ B\left(r\right) = \frac{2|\vec{m}|}{r^3}$$
We can now use the above to our advantage. Since we are assuming we know
the polar magnetic field strength, we can rearrange and solve for the
magnetic moment. At the pole of a star the distance is just going to be
the radius, so for radius R and field strength Bp, we get
$$ |\vec{m}|} = \frac{1}{2}B_{p}R^3 $$
Plugging this back in, we get a nice little formula for the strength of
the magnetic field in terms of the stellar radius and the polar field
strength
$$B(r) = B_{p}{\left(\frac{R}{r}\right)}^3 $$
And we are almost there! Rearranging now for r, we get $$r =
R{\left[\frac{B_p}{B(r)}\right]}^{1/3} $$ Huzzah! Now we just need
to plug in some values. Well, we'll use Bp = 10^15 gauss, B(r) = 1000
gauss (strong enough field to erase credit cards) and R = 10 km, which
gives... $$ r \approx 10km \times
{\left(\frac{10^{15}gauss}{10^{3}gauss}\right)}^{1/3} $$ so $$ r
\approx \mbox{100,000 km} $$ The moon is about 380,000 km away. So we
find that the magnetar will erase all credit cards up to a little over a
quarter the distance to the moon. Not bad! However, all this talk of
"half" and "quarter" is a bit misleading given that our best guesses
here will be order of magnitude. But, overall, we see that a magnetar at
roughly the Earth-Moon distance would have a good shot at erasing your
credit cards. Fun fact confirmed!
[1] For a really nice Scientific American article about magnetars, see
<a href="http://solomon.as.utexas.edu/~duncan/sciam.pdf">here</a>. For more
information than you could probably ever want, see
<a href="http://solomon.as.utexas.edu/~duncan/magnetar.html">here</a>.
[2] For those of you who prefer your magnetic field strengths given in
units named after someone played by David Bowie in a
<a href="http://en.wikipedia.org/wiki/The_Prestige_(film)#Cast">movie</a>, you may
note the conversion: 1 gauss = 10^-4 tesla.</p></div>credit cardsmagnetarneutron starscott bakulahttps://thephysicsvirtuosi.com/posts/old/the-magnetar-credit-card-swipe/Sun, 30 Jan 2011 17:22:00 GMT