[Eternal Student (OP)]

Hi.
 
   Not urgent, just curious and it's too late to grab a pencil and paper.   Someone might already know the answer for this anyway, so I'll just ask.

   How does a red shift (or blue shift) affect the radiation spectrum from a black body? 

Basic Question        Consider a Doppler shift (at relativistic speeds or just Newtonian if you prefer)
    A Black Body which is at rest and at temperature T₁  produces radiation with a known distribution of intensity at different wavelengths  ( a black body spectrum ).
    An observer is in a state of uniform motion with velocity  v≠0  relative to the black body.
    In the rest frame of the observer, is the radiation they receive from the black body still going to have the right distribution to be consistent with a Black body spectrum but just with a different temperature T₂? 

Best Wishes.

[Origin]

.
    In the rest frame of the observer, is the radiation they receive from the black body still going to have the right distribution to be consistent with a Black body spectrum but just with a different temperature T₂? 

The distribution curve of the radiation is different at different temperatures I believe, so I think the curve for say 800C would have the same shape it would just be shifted in the moving frame.

[Eternal Student (OP)]

Hi.

   Thanks.   It's late and I won't be checking or thinking about the replies for a while - but @Origin    was astonishingly fast.  Thanks.

[Halc]

I think I agree with Origin's statement about the shape, but is the T2 shape of a stationary object just a shifted version of an object with a different temperature?  If not, then a shifted (as seen by a moving observer) object at T2 would not be the same shape as a stationary object with a 'proper temperature' (if there is such a term) of T2. I don't know my black body radiation enough to say one way or the other.

In the rest frame of the observer, is the radiation they receive from the black body still going to have the right distribution to be consistent with a Black body spectrum but just with a different temperature T2?

As for the statement above, the temp T2 is dependent not just on the frame, but the position of the observer in that frame.  So if the black body is moving at v in the +x direction, the observer (stationary in that frame) might see a red or blue shift depending on his position in that frame.
You know this, but the OP didn't make that clear. Maybe he's off to the side and the thing is going by him without any Doppler shift, at least for a moment, just like an ambulance going by momentarily has an unshifted pitch to its siren.

[Bored chemist]

It is well known that the "shape" of a red shifted BB spectrum still looks like a BB spectrum.
That's where the CMBR comes from.

[evan_au]

An object putting out a black body spectrum is measured as such in all inertial frames of reference.
- Different Frames of reference will see the object as having a different effective temperature.

To take it to the next level: Black holes experience extreme gravitational red shift.
- According to Hawking, black holes "glow" with a black body spectrum.
- The effective temperature varies with the mass of the black hole
- A stellar mass black hole has a temperature measured in nanokelvins
- Which is why stellar-mass black holes take a long time to evaporate.

https://en.wikipedia.org/wiki/Hawking_radiation#Overview

[Eternal Student (OP)]

Hi.

    I've had some sleep and can think a bit.    Thanks to everyone who put in a reply.
Let's do the easier things to reply to first:

Maybe he (the observer) is off to the side (of the black body) and the thing is going by him without any Doppler shift, at least for a moment,

    Yes, motion could be directly towards or away from the black body but it could be entirely tangential to the black body or any shade in between.    For Newtonian Mechanics there's no Doppler shift at all if the observer is moving tangentially (I think they often say "transversely") to the Black Body.   For a relativistic treatment there is still some time dilation which has an effect.
    I was going to restrict motion to being directly towards or away from the Black body but this would have made the question longer and messy.   People can make that restriction if they wish, it's still complicated enough with that restriction.
 - - - - - - - - -

The distribution curve of the radiation is different at different temperatures I believe, so I think the curve for say 800C would have the same shape it would just be shifted in the moving frame.

   Yes to the first part (but please keep all temperatures in Kelvin or you'll confuse a simple person like myself, most of the formulae will only work in Kelvin etc.)
The shape of the spectrum is usually shown in a diagram of spectral radiance  vs. frequency (or wavelength) of the radiation.
Wikipedia has info and diagrams:  https://en.wikipedia.org/wiki/Black-body_radiation#Planck's_law_of_black-body_radiation



The second part of your reply is less clear (to me anyway).  You (Origin) might be happy with the use of the word "shift" and, indeed, a Doppler effect is often called a Doppler "shift" so I can completely see why the phrase was used.   However, the change in frequencies is really obtained by multiplication by some constant not by adding some constant.  The distinction is irrelevant if you're shifting only one frequency.  However we are working with graphs to describe the shape of the spectrum and for a whole graph with a whole range of frequencies it will matter a lot.  Simple people (like me) would consider a "shift" as a translation of the graph instead of a stretch along the frequency axis.
   The situation is made a little more complicated since shifting the frequency of a photon will not only change its frequency, it will change its energy ( E=h.v   for a photon).   This means the intensity at a given frequency can get stretched as well, or to say it another way the spectrum shown on the usual graph of spectral radiance vs. frequency can be stretched in y-axis as well as in the x-axis.
    I'm sure you can solve the problem or answer the original question just by examining how the graphs of the spectrum change but it was complicated enough that I just walked away last night.   The easier method seems to be just using some algebra and directly examining the spectral radiance formula (but I don't know - that's just what I went for in the end and when the new day came.  It may not be the best or fastest way).

Best Wishes.

[Eternal Student (OP)]

Hi again.

   Now let's just reply to the remaining people.  Basically they were right.

It is well known that the "shape" of a red shifted BB spectrum still looks like a BB spectrum.
That's where the CMBR comes from.

    Yes.  I had certainly heard this.  However, I had never seen the mathematical proof that the spectrum is actually preserved and remains exactly the right shape to be a BB spectrum.  I thought it might just be an approximation and wanted to check.

An object putting out a black body spectrum is measured as such in all inertial frames of reference.
- Different Frames of reference will see the object as having a different effective temperature.

    Yes, also absolutely correct.  Thank you.
 - - - - - - - -
      I've done the calculation now.  I'm not going to bore anyone with the mathematical proof but the result does hold.   The assurance from  evan_au and Bored Chemist is enough to make me quite happy to just state it as fact.

   For a simple red-shift or blue-shift  (which is any shift where radiation of frequency v → k.v  for some k that is independent of v)  a Black Body spectrum will be mapped precisely to another BB spectrum but just one with a different temperature (and often a different amplitude or magnitude).   Most frequency shifts, like changing frames of reference or gravitational redshifts are of this type.
   The change in magnitude of the intensity of radiation received is less important but it is there.  (For a change of frames in relativity one way to explain this is that there is a Compton headlight effect plus a length contraction effectively bringing the observer much closer to the Black Body - but the analysis of the spectral radiance formula shows these results without needing to worry about why they make sense or what interpretation is given by anyone in different frames etc.)

   Anyway, that's it.   Thanks for all the replies.

Best Wishes.

[Bored chemist]

I thought it might just be an approximation and wanted to check.

You should have checked the data.

 The assurance from  evan_au and Bored Chemist is enough to make me quite happy to just state it as fact.

The assurance isn't from me; it's from the universe.

[Eternal Student (OP)]

Hi.

The assurance isn't from me; it's from the universe.

   You can't trust the universe, it's a slippery thing.  It's never as re-assuring as the theory.

Best Wishes.