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.
Re: Doppler and photoelectric effect Greetings ! 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.
Re: Re: Doppler and photoelectric effect So the energy equation uses what frequency, actual or observed?
Re: Doppler and photoelectric effect 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. 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).
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 ! 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. Live long and prosper.