Dppler and photoelectric effect

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Will

Doppler and photoelectric effect

When using the doppler effect equations to determine the speed of celestial objects, what happens to the energy of the photons? If a certain device required a 600nm wavelenth of light and frequency 5E14 Hz to induce current, would photons from a celestial body moving away from earth and emitting this wavelength have sufficient energy to induce current, since observed f is less than 5E14 Hz and E=hf ?
I was curious as to how astrophysicists can distinguish between one lightsource traveling at v1 with frequency f1 and another with v2,f2 if the observed f is the same for both.
 
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drag

Science Advisor
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Re: Doppler and photoelectric effect

Greetings !
Originally posted by Will
I was curious as to how astrophysicists can
distinguish between one lightsource traveling
at v1 with frequency f1 and another with v2,f2
if the observed f is the same for both.
There are two types of shifts for light.
One is normal doppler and the other is
relativistic. The relativistic can only
be distinguished by comparing the
frequency to the one we would expect in
this case (based on expected materials,
compared atomic/molecular emmision lines
and so on). The normal doppler effect
is distinguished by comparing the energy
of the photon according to E = h * frequency
(the signs/smilies menu doesn't work for me,
sorry) to the actual reception frequency.

Live long and prosper.
 

Will

Re: Re: Doppler and photoelectric effect

Originally posted by drag
Greetings !

The normal doppler effect
is distinguished by comparing the energy
of the photon according to E = h * frequency
(the signs/smilies menu doesn't work for me,
sorry) to the actual reception frequency.

Live long and prosper.

So the energy equation uses what frequency, actual or observed?
 

Janus

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Re: Re: Re: Doppler and photoelectric effect

Originally posted by Will
So the energy equation uses what frequency, actual or observed?
The actual frequency is the observed frequency.
 

Alexander

Re: Doppler and photoelectric effect

Originally posted by Will
When using the doppler effect equations to determine the speed of celestial objects, what happens to the energy of the photons? If a certain device required a 600nm wavelenth of light and frequency 5E14 Hz to induce current, would photons from a celestial body moving away from earth and emitting this wavelength have sufficient energy to induce current, since observed f is less than 5E14 Hz and E=hf ?
They would not. Energy depends on reference system.

Say, water will turn turbine falling from 100 m elevation, but it's energy vanishes if you place turbine at zero elevation and becomes negative if you move turbine higher. Bullet energy becomes zero in the co-moving with the bullet reference system.



I was curious as to how astrophysicists can distinguish between one lightsource traveling at v1 with frequency f1 and another with v2,f2 if the observed f is the same for both.
They can't. There is no way of figuring out "original" frequency of passing photon. Thus, they have to use additional information (say, about nature of source). For example, scientists are not exactly sure what was original temperature of CMBR when it was emitted. All they know is that at temperatures about T<(3-5)x1000 K hydrogen becomes transparent to radiation (with temperature T), so thery assume that this was the temperature of radiation (thus, of universe) at time of decoupling of CMBR from matter (hydrogen).
 
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drag

Science Advisor
1,055
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Greetings !

Will, I'm terribly sorry !
I was somewhat confused myself and now
I apparently transfered my confusion to you.
But, as long as I admit my mistake and
fix it now, all is not lost ! :smile:

Anyway, The normal Doppler effect for light
is the same as for sound. As the source
moves towards/away from you the frequency of
the photons (EM waves) increases/decreases.

Now, when you "add" relativity you must
also use the Lorentz transformation.
The energy of a photon from such a source
will be higher for you than it is
for the source (in terms of the Lorentz
transformation at least).

Now, the Lorenz transformation only really
comes into play when the source is moving at
relativistic velocities. So, the normal
doppler shift is more significant before
you get to velocities close to the speed
of light (0.8 c +). What's important is that
the relativistic energy and hence frequency
increase of the photon occurs in any case -
no matter whether the source moves towards or
away from you.

The doppler shift is v/c.
The Lorentz transformation is 1/ sqrt(1 - sqr(v / c)).

How is the actual frequency discovered ?

Like I mentioned above it is done by comparing
spectral lines and looking for matches for the
type of materials and energy of reactions you
expect to be present at the source.
Fopr example, one well known frequency is
1.420 Gigahertz (hydrogen - which is abundant
in space - 90% of all normal matter) and it's
frequency shift from a the source can be used
to set the whole spectrum "right".

Another issue worth mentioning is the Universal
expansion. Basicly if you consider that
the Universe expands by about 65 km per Mega Parsec
per second then the photons travelling from the
source to you will also expand according to
the amount of time it took'em to reach you
(their frequency will decrease).
So, you have to account for that in your
calculations and spectral analyses of distant,
in astronomical terms, sources. Alternatively, this
can also be used to make a general estimate of
distance to the source.

Hope I helped this time. :wink:

Live long and prosper.
 

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