Frequency, Wavelength, & Redshift

AngleWyrm wrote:

I accept this: Frequency, measured in cycles per second, is the inverse of
wavelength, measured in meters.

frequency = 1/waveLength
waveLength = 1/frequency

Those signs shouldn't be "equals", but "is proportional to". You're
missing the "constant of proportionality", which is here the velocity
of the wave: nu = v / lambda and lambda = v / nu. You can verify that
the units work out that way.

Question: What happens if the frequency is shifted? The length of the wave's
cycle is shifted as well, right?

Yes.

This is probably over-extending an analogy with soundwaves and transfer
medium, and it may be what is confusing me: Sound, leaving a moving source
(wrt the observer) seems compressed on the approach, and expanded on the
departure--the doppler effect. This is a shift in the frequency &
wavelength.

The analogy between light and sound is pretty good, except that with
sound you also have to take into account the velocity of source &
receiver WRT the medium: if a stationary observer is up- or down-wind
from a stationary source of sound, a slightly different pitch will be
heard than in still air.

My grasp of the sound analogy is this: On the approach, the pings-per-second
appear closer together (higher frequency, shorter wavelength) because the
physical distance between pings departing the source is actually less. The
first ping has travelled it's "normal velocity" towards the observer, but
the the source has also travelled in the same direction, reducing the gap
between pings.

Is this what happens when we observe redshift in celestial bodies?

Yes, more or less, except that in a redshift (as opposed to a
blueshift) the "pings" start farther and farther apart.

--
Odysseus

Odysseus wrote,

if a stationary observer is up- or
down-wind from a stationary source of
sound, a slightly different pitch will be
heard than in still air.

Ody, are you sure about that? It would seem more likely that as long as
source and receiver are stationary relative to each other, the receiver
would hear no pitch change for it would exactly cancel. Whereas if the
receiver were say, in a balloon drifting with the wind, a pitch change
would be heard. oc

Tie a 20ft rope to a fence wiggle it up and down,and count the
ups(crests) Now shorten the rope to 10ft you now have to have either
half as many crests(ups) or the same amount in half the time?. Your
answer is? Bert

"G=EMC^2 Glazier" wrote in message
...
Tie a 20ft rope to a fence wiggle it up and down,and count the
ups(crests) Now shorten the rope to 10ft you now have to have either
half as many crests(ups) or the same amount in half the time?. Your
answer is? Bert

Depends how fast I am wiggling it. What's your point?

BV.
www.iheartmypond.com

BV Good point,but will wiggling faster create more crests or higher
crests? Around and around we go. Faster also means less time(yes?) The
wave goes along the rope to the fence.(yes) Bert

Bill Sheppard wrote:

Odysseus wrote,

if a stationary observer is up- or
down-wind from a stationary source of
sound, a slightly different pitch will be
heard than in still air.

Ody, are you sure about that? It would seem more likely that as long as
source and receiver are stationary relative to each other, the receiver
would hear no pitch change for it would exactly cancel. Whereas if the
receiver were say, in a balloon drifting with the wind, a pitch change
would be heard. oc

Sorry, I got on the wrong track: in the example I gave the effects
indeed cancel out. What I was trying to get at (before I got
confused) was that for sound the motions of the source and the
receiver WRT the medium don't have symmetrical effects, while for EM
radiation only their relative velocity matters.

If a 1-kHz source is travelling towards a detector at Mach 0.1 (in
still air), it'll read 1000/(1-.1) = 1111 Hz, but if the detector
travels at the same speed towards the stationary source it'll receive
a 1000(1+.1) = 1100 Hz pitch instead. The higher the speeds, the
greater the asymmetry (at half the speed of sound the apparent
pitches would be 2000 Hz and 1500 Hz respectively). If the moving
object in each of these cases is being carried along by a Mach-0.1
gale the situations would be reversed, as the stationary object would
be moving WRT the air while the other wouldn't be.

--
Odysseus

"G=EMC^2 Glazier" wrote in message
...
Tie a 20ft rope to a fence wiggle it up and down,and count the
ups(crests) Now shorten the rope to 10ft you now have to have either
half as many crests(ups) or the same amount in half the time?. Your
answer is?

Try spinning the rope, and it'll make harmonics.

Angle The wave only looks like it is moving from you to the fence. That
is no different than dropping a rock in the middle of a pond. I think
such a wave has the name "residual" Waves in the ocean are residual
waves They take on the looks of a wave coming on to shore when they are
lifted,and friction takes place near the shore.(under them) It is a
very good illusion when you see the pond waves making bigger and bigger
circles,and smaller and smaller ups,and downs and end up on the pond's
shore line,and see them make their last ripple hitting the shore. Well
gravity comes into play,and the final energy at the shore ends up a
heat. The fact that water is non-compressible sets this
wave action started. Bert

"G=EMC^2 Glazier" wrote in message
...
Angle The wave only looks like it is moving from you to the fence. That
is no different than dropping a rock in the middle of a pond. I think
such a wave has the name "residual" Waves in the ocean are residual
waves They take on the looks of a wave coming on to shore when they are
lifted,and friction takes place near the shore.(under them) It is a
very good illusion when you see the pond waves making bigger and bigger
circles,and smaller and smaller ups,and downs and end up on the pond's
shore line,and see them make their last ripple hitting the shore. Well
gravity comes into play,and the final energy at the shore ends up a
heat. The fact that water is non-compressible sets this
wave action started. Bert

Hmm...this concept would seem to imply that the gravity waves are "flowing"
through some medium as the ripples are flowing through the water. Are you
suggesting that space has some intrinsic, unseen dark matter that allows
gravity to flow through it? If so, how does gravity appear to work
instantaneously at a distance?

BV.
www.iheartmypond.com

Hi BV Well maybe gravitons would like to move instantaniously through
space ,but my intinsic space field slows it to "c". That fits well with
Einstein. Have been giving a lot of thought to this theory,and coming
up with some far out thoughts. Trying to tie BB right up to our
universe's present space time,and even to its future time. Bert