|BAND||SHORTEST WAVELENGTH||HIGHEST FREQUENCY||ENERGY PER PHOTON|
|radio||0.1 cm||3 x 1011 Hz||1.98 x 10-15 ergs|
|infrared||10-4 cm||3 x 1014 Hz||1.98 x 10-12 ergs|
|visible||3 x 10-5 cm||1015 Hz||6.6 x 10-12 ergs|
|ultraviolet||10-6 cm||3 x 1016 Hz||1.98 x 10-10 ergs|
|X-rays||10-9 cm||3 x 1019 Hz||1.98 x 10-7 ergs|
Remember that Hz are sec-1 and E/s units are (gm cm2 sec-1) x (sec-1) = gm cm2 sec-2 = ergs, an energy unit (it is about the amount of energy a fly needs to do a push-up).
|infrared||0.1 cm||3 K|
|visible||10-4 cm||3,000 K|
|ultraviolet||3 x 10-5 cm||10,000 K|
|X-rays||10-6 cm||300,000 K|
|(gamma)-rays||10-9 cm||3 x 108 K|
Note that the units (nm) cancel, and z is a dimensionless number. Note also that this number is positive, so the velocity is 'positive,' meaning that this galaxy is moving away from us. Now we can get the velocity, rewriting z = v/c as:
We can use the simple form for relating v to z because z is very small. Now that we have a velocity, we could use Hubble's law to find the distance.
(I didn't write all the digits, because we can't be that accurate here anyway.) However a few hundred km/s is typical for the random motion of galaxies independent of the Hubble expansion, so the most we can say about this galaxy is that it isn't very far away. It would be in the local group, in fact.
|non-relativistic||1 + z = 1 + v/c = 1 + 0.6c/c = 1 + 0.6 = 1.6,|
|relativistic||1 + z = [(1 + v/c)/(1 - v/c)]1/2 = [(1 + 0.6)/(1 - 0.6)]1/2 = [1.6/0.4]1/2 = 41/2 = 2.|
Since these are so different, we can see that when the velocity is this high the approximate formula (nonrelativistic) doesn't work and we need to use the real formula. We now have 1 + z, and so we use the formula 1 + z = (lambda)obs/(lambda)em, and we know that this is the factor desired. Also, we can find z = 2 - 1 = 1.
Evans's 324 | Astronomy Department