<p>If the object you're observing is something good and hot, like a star, it's pretty easy to identify a set of <a href="https://en.wikipedia.org/wiki/Balmer_series" rel="nofollow noreferrer">Ballmer lines</a>   by the relative spacing between the lines.  Then compare the absolute wavelengths to the known stationary values and <em>viola</em> you've got the Doppler shift value. </p>

<p>So this would "technically" be using the blackbody radiation from the <a href="https://en.wikipedia.org/wiki/Photosphere" rel="nofollow noreferrer">photosphere</a> of a star to illuminate its cooler atmosphere above, making its dark absorption lines visible.</p>

CC BY-SA 4.0
2019-01-16T06:06:44.010Z
I've added a sentence to make sure this addresses the OP's question about "...using Black-body radiation?"
uhoh
CC BY-SA 4.0
2019-01-15T20:19:24.650Z
2019-01-16T12:57:53.293Z
2019-01-16T12:57:53.293Z
Carl Witthoft
<p>Without any other information, you cannot distinguish between the two effects.  </p>

<p><span class="math-container">$$ T = T_0 (1 + z) $$</span></p>

<p>A blackbody spectrum of temperature <span class="math-container">$T$</span> is identical to a blackbody spectrum of temperature <span class="math-container">$T_0$</span> with redshift <span class="math-container">$z$</span>.</p>

<p>For stellar/galactic radiation, we can use the fact that the radiation is not a perfect blackbody.  For the CMB, we can use the fact that atomic combination happens only below a particular temperature.</p>

CC BY-SA 4.0
2019-01-15T00:58:57.137Z
2019-01-15T00:58:57.137Z
BowlOfRed
<blockquote>
  <p>How is it possible to distinguish between temperature and doppler
  effect?</p>
</blockquote>

<p>The blackbody curve varies in both peak <em>and</em> shape:</p>

<p><a href="https://i.sstatic.net/8mRWO.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/8mRWO.png" alt="enter image description here"></a></p>

<p>As you can see, even if you slide the curves left or right, the 5000k curve is still different than the 3000k one. The peak is at a different point relative to the curve as a whole, and especially the extinction point in the ultraviolet (around 0.2 um in this case).</p>

<p>At a minimum, you could compare the position of the peak to the extinction point and produce a temperature. In this example, blue has a span of about 0.3, while red is about 0.5.</p>

<p>So using the numbers in this example, if you were given a spectrum and measured a span of 0.3, you know it's blue. So if the peak of the curve is at 0.8 instead of 0.6 you would ascribe that to 0.2 um of doppler shifting (or other shifting effects).</p>

CC BY-SA 4.0
2019-09-18T14:30:08.937Z
the question isn't well written, but it seems that it's asking about how to differentiate a change in temperature from a change in velocity. For example, if the peak wavelength of the [Planck distribution](https://en.wikipedia.org/wiki/Planck%27s_law) increases by 1% we can't tell if the object became 1% cooler, or started to move at 1% the speed of light away from us.
uhoh
CC BY-SA 4.0
2019-09-18T15:04:01.457Z
This isn't correct. A red-shifted blackbody curve just looks like a cooler blackbody. e.g. the CMB.
ProfRob
CC BY-SA 4.0
2019-09-18T10:06:22.843Z
2019-09-18T10:06:22.843Z
Maury Markowitz
<p>I understand that the radiation of a body can be described using the curve for black-body radiation. In the sense that a hotter body will be blue shifted and a cooler body will be red shifted.</p>

<p>The doppler effect gives similar results in the sense that a body is approaching us is blue shifted. While an object that is travelling away from us is red shifted.</p>

<p><strong>How is it possible to distinguish between temperature and doppler effect?</strong></p>

CC BY-SA 4.0
2019-01-14T23:00:33.860Z
I think you misunderstand something, or the way you describe is not correct. A blackbody has nothing to do with blue/red shifting. Or, if you are talking about a hotter blackbody has its peak at bluer wavelength, that is you are describing the peak, not the blue/red shift.
Kornpob Bhirombhakdi
CC BY-SA 4.0
2019-01-15T20:16:36.533Z
@KornpobBhirombhakdi I don't think that's the point of this question. He wants to know if a red-shifted BB curve has any shape difference from a non-shifted BBcurve at a lower temperature.
Carl Witthoft
<star><temperature><radiation><doppler-effect><black-body>
How to Distinguish between Temperature and Doppler effect using Black-body radiation?
Marcelo Fonseca
# How to Distinguish between Temperature and Doppler effect using Black-body radiation?

(Asked by: username_0 on 2019-01-14)

I understand that the radiation of a body can be described using the curve for black-body radiation. In the sense that a hotter body will be blue shifted and a cooler body will be red shifted.

The doppler effect gives similar results in the sense that a body is approaching us is blue shifted. While an object that is travelling away from us is red shifted.

**How is it possible to distinguish between temperature and doppler effect?**

**Tags:** star, temperature, radiation, doppler-effect, black-body

> **Comments:**
> **username_1**: I think you misunderstand something, or the way you describe is not correct. A blackbody has nothing to do with blue/red shifting. Or, if you are talking about a hotter blackbody has its peak at bluer wavelength, that is you are describing the peak, not the blue/red shift.  
> **username_2**: @KornpobBhirombhakdi I don't think that's the point of this question. He wants to know if a red-shifted BB curve has any shape difference from a non-shifted BBcurve at a lower temperature.  

---

## 3 answers

### Answer by username_2 on 2019-01-15. Score: 3 | Accepted answer

If the object you're observing is something good and hot, like a star, it's pretty easy to identify a set of [Ballmer lines](https://en.wikipedia.org/wiki/Balmer_series) by the relative spacing between the lines. Then compare the absolute wavelengths to the known stationary values and *viola* you've got the Doppler shift value.

So this would "technically" be using the blackbody radiation from the [photosphere](https://en.wikipedia.org/wiki/Photosphere) of a star to illuminate its cooler atmosphere above, making its dark absorption lines visible.

> **Comments:**
> **username_3**: I've added a sentence to make sure this addresses the OP's question about "...using Black-body radiation?"  

---

### Answer by username_4 on 2019-01-15. Score: 7

Without any other information, you cannot distinguish between the two effects.

$$ T = T\_0 (1 + z) $$

A blackbody spectrum of temperature $T$ is identical to a blackbody spectrum of temperature $T\_0$ with redshift $z$.

For stellar/galactic radiation, we can use the fact that the radiation is not a perfect blackbody. For the CMB, we can use the fact that atomic combination happens only below a particular temperature.


---

### Answer by username_5 on 2019-09-18. Score: -1

> How is it possible to distinguish between temperature and doppler
> effect?

The blackbody curve varies in both peak *and* shape:

[![enter image description here](https://i.sstatic.net/8mRWO.png)](https://i.sstatic.net/8mRWO.png)

As you can see, even if you slide the curves left or right, the 5000k curve is still different than the 3000k one. The peak is at a different point relative to the curve as a whole, and especially the extinction point in the ultraviolet (around 0.2 um in this case).

At a minimum, you could compare the position of the peak to the extinction point and produce a temperature. In this example, blue has a span of about 0.3, while red is about 0.5.

So using the numbers in this example, if you were given a spectrum and measured a span of 0.3, you know it's blue. So if the peak of the curve is at 0.8 instead of 0.6 you would ascribe that to 0.2 um of doppler shifting (or other shifting effects).

> **Comments:**
> **username_3**: the question isn't well written, but it seems that it's asking about how to differentiate a change in temperature from a change in velocity. For example, if the peak wavelength of the [Planck distribution](https://en.wikipedia.org/wiki/Planck%27s_law) increases by 1% we can't tell if the object became 1% cooler, or started to move at 1% the speed of light away from us.  
> **username_6**: This isn't correct. A red-shifted blackbody curve just looks like a cooler blackbody. e.g. the CMB.  

---

