7 April 2005 Astronomy 241, Spring 2005 1 Today in Astronomy 241: pulsars Today’s reading: Carroll and Ostlie pp. 608-626, on pulsars and the pulsar mechanism. Artist’s conception of the formation of the double pulsar PSR J0737-3039 (John Rowe Animations and U. Manchester)
Slide 1Today in Astronomy 241: pulsars
Today’s reading: Carroll and Ostlie pp. 608-626, on pulsars and the
pulsar mechanism.
Artist’s conception of the formation of the double pulsar PSR
J0737-3039 (John Rowe Animations and U. Manchester)
Pulsars
Discovered serendipitously by Jocelyn Bell and Anthony Hewish in
1967.
Compact, pulsed radio sources. Hundreds are now known. Most pulsars
have periods near 1 s; a relatively recently- identified class of
millisecond pulsars shows periods in the 1-10 ms range. The periods
of most pulsars increase extremely gradually (dP/dt ~ 10-15
typically); they are vey precise clocks.
Very soon after they were discovered, they were identified as
rotating neutron stars.
7 April 2005 Astronomy 241, Spring 2005 3
Pulsars (continued)
Pulsars have extremely strong magnetic fields (typically 1012
gauss), arising from the original star’s magnetic field and
conservation of magnetic flux.
This gives rise to the pulsar radiation mechanism: synchrotron
radiation from electrons in the neutron star’s magnetic field,
beamed preferentially along the magnetic poles. If the magnetic and
rotation axes are not the same, stellar rotation swings this
“flashlight beam” around, and a distant observer sees pulses when
the beam cross his/her view. Structural changes within each star
give rise to occasional glitches in pulsar period. (Starquakes?
Phase transitions?)
7 April 2005 Astronomy 241, Spring 2005 4
Why is pulsar radiation beamed?
This is explained in PHY 218. 1. Hot electrons and ions at the
neutron-star surface find
themselves loaded onto magnetic field lines. They’re hot enough
that the electron speeds are relativistic.
2. The Lorentz force keeps them all tied to given field lines, so
as the star rotates, they are flung outwards along the lines of B,
like beads on a swinging wire. At the poles, where the field is
strongest, this means vertically outwards.
3. Accelerated charges radiate light. Relativistic accelerating
charges radiate lots of light, and radiate it mostly in the
direction of their velocity. So lots of light gets beamed outward
along the magnetic poles .
See here for the details.
Some pulsars
PSR 1937+214: the first (and still the fastest) millisecond pulsar,
and one with a particularly small period derivative.
PSR 1845-19: the slowest pulsar yet observed, with P = 4.308 s. PSR
0531-21: the pulsar in the Crab nebula (P = 0.0333 s), and thus the
remnant of Supernova 1054. Pulses seen in radio and in visible
light. Its energy loss from rotational slowing is equal to the
radiation output of the Crab.
P
P
7 April 2005 Astronomy 241, Spring 2005 6
Some pulsars (continued) Geminga: the closest pulsar to Earth (20
pc), with P = 0.237 s, it shows pulses at X-ray and γ-ray
wavelengths but not in the radio. PSR 1913+16: the first binary
pulsar observed.
The period derivative is too large to be due just to magnetic
dipole radiation; this is taken to be the first direct observation
of gravitational radiation. (Nobel Prize to Hulse and Taylor,
1993.) More on this on in a couple of weeks.
15 -1
1 2
2.47583(1) 10 Hz s 1.4410(5) 1.3874(5) 0.6171308(4) 27906.9808968 s
(7.75 hours)orb
M M M M e P
ω
ω −
Minimum rotation period
If a star is spun faster than this it will break up (i.e. gravity
would not be enough to supply centripetal acceleration):
3 min 2 RP
Today’s in-class problems
1. Problem 15.14. 2. Problem 15.15. 3. Problem 15.16.
Note: the Sun’s rotational period is 26 days, and its average
surface magnetic field is about 2 gauss. The average density of a 1
M white dwarf and a 1.4 M neutron star are
6 -3
14 -3
NS
ρ
ρ
Today’s in-class problems (continued)
Answers and/or secrets to problems done last class: 1. As in two
problems done last week, use
2. This simply requires straightforward differentiation, as long as
you remember that
0
. Fp
ρ = =∫
Today’s in-class problems (continued)
3. Expand the binomial and the arcsinh(x) in Taylor series and
multiply it out. It turns out that all terms in x and cancel out,
so one must keep fifth order terms in the expansions:
to obtain
3 5 7
x x x O x
x x x x O x
+ = + − +
= − + +
( ) ( ) ( ) ( )
1 11 2 3 2 8 1 3 6 40
f x x x x O x x
x x x O x
= + − + −
+ − + +
Today’s in-class problems (continued)
or
Thus
5 6
5 1
3 3 92 3 3 4 2 8 2 40
15 91 40 40
8 . 5x
x x x x xf x x x x x O x
x O x
5 5 P a x a C
b ρ ρ ≅ = =
Pulsars
Some pulsars
