<p>I have a passband filter <span class="math-container">$b$</span> and a spectrum (SED) <span class="math-container">$f_{\lambda}(\lambda)$</span>. Often the formula I find to get the flux in that filter <span class="math-container">$F_b$</span> is
<span class="math-container">$$F_{b}=\dfrac{\int f_{\lambda}(\lambda) T_{b}(\lambda)d\lambda}{\int T_{b}(\lambda)d\lambda}\,,$$</span>
where <span class="math-container">$T_b$</span> is the filter response in band <span class="math-container">$b$</span>, which is similar to the filters shown in figure 2 of <a href="https://arxiv.org/pdf/1810.00001.pdf" rel="nofollow noreferrer">The PLAsTiCC team et al (2018)</a>.</p>

<p>However, when I try to compute a apparent magnitude, I get the formula
<span class="math-container">$$m_b=-2.5\log_{10}\left(\dfrac{\int f_{\lambda}(\lambda)\,\lambda\,T_{b}(\lambda)d\lambda}{\int T_{b}(\lambda)\,\lambda\,d\lambda}\right)\,.$$</span></p>

<p>I don't understand from where does the <span class="math-container">$\lambda$</span> in the second formula arises, since I thought that
<span class="math-container">$$m_b=-2.5\log_{10}\left(\dfrac{F_b}{F_{\mathrm{reference}}}\right)\,.$$</span></p>

<p>For example, <a href="https://arxiv.org/pdf/1407.6095.pdf" rel="nofollow noreferrer">Casagrande &amp; VandenBerg (2014)</a> show both expressions in equations 1 and 3. However <a href="https://arxiv.org/pdf/1112.2698.pdf" rel="nofollow noreferrer">Bessell &amp; Murphy (2012)</a> (eq. 2) and <a href="https://arxiv.org/pdf/astro-ph/0210394.pdf" rel="nofollow noreferrer">Hogg, Baldry, Blanton, &amp; Eisenstein (2002)</a> (eq. 4) only show the second approach. Additionally, <a href="https://arxiv.org/pdf/astro-ph/9905116.pdf" rel="nofollow noreferrer">Hogg (2000)</a> shows the K-correction without the added <span class="math-container">$\lambda$</span>'s:
<span class="math-container">$$K=-2.5\log_{10}\left(\dfrac{1}{(1+z)}\dfrac{L_{\lambda/(1+z)}}{L_{\lambda}}\right)\,.$$</span></p>

<p>I don't know how to reconcile these different views. Can you help me?</p>

CC BY-SA 4.0
2020-02-25T16:40:21.907Z
@usernumber, I have fixed the link. When I feed equation (1) into equation (3), the $\lambda$ factor does not appear. That is why I am confused.
Catarina Alves
CC BY-SA 4.0
2020-02-25T17:05:19.213Z
@usernumber there's an additional lambda in the denominator's integral
planetmaker
<spectroscopy><photometry>
How to calculate magnitudes in photometry?
Catarina Alves
# How to calculate magnitudes in photometry?

(Asked by: username_0 on 2020-02-24)

I have a passband filter $b$ and a spectrum (SED) $f\_{\lambda}(\lambda)$. Often the formula I find to get the flux in that filter $F\_b$ is
$$F\_{b}=\dfrac{\int f\_{\lambda}(\lambda) T\_{b}(\lambda)d\lambda}{\int T\_{b}(\lambda)d\lambda}\,,$$
where $T\_b$ is the filter response in band $b$, which is similar to the filters shown in figure 2 of [The PLAsTiCC team et al (2018)](https://arxiv.org/pdf/1810.00001.pdf).

However, when I try to compute a apparent magnitude, I get the formula
$$m\_b=-2.5\log\_{10}\left(\dfrac{\int f\_{\lambda}(\lambda)\,\lambda\,T\_{b}(\lambda)d\lambda}{\int T\_{b}(\lambda)\,\lambda\,d\lambda}\right)\,.$$

I don't understand from where does the $\lambda$ in the second formula arises, since I thought that
$$m\_b=-2.5\log\_{10}\left(\dfrac{F\_b}{F\_{\mathrm{reference}}}\right)\,.$$

For example, [Casagrande & VandenBerg (2014)](https://arxiv.org/pdf/1407.6095.pdf) show both expressions in equations 1 and 3. However [Bessell & Murphy (2012)](https://arxiv.org/pdf/1112.2698.pdf) (eq. 2) and [Hogg, Baldry, Blanton, & Eisenstein (2002)](https://arxiv.org/pdf/astro-ph/0210394.pdf) (eq. 4) only show the second approach. Additionally, [Hogg (2000)](https://arxiv.org/pdf/astro-ph/9905116.pdf) shows the K-correction without the added $\lambda$'s:
$$K=-2.5\log\_{10}\left(\dfrac{1}{(1+z)}\dfrac{L\_{\lambda/(1+z)}}{L\_{\lambda}}\right)\,.$$

I don't know how to reconcile these different views. Can you help me?

**Tags:** spectroscopy, photometry

> **Comments:**
> **username_0**: @usernumber, I have fixed the link. When I feed equation (1) into equation (3), the $\lambda$ factor does not appear. That is why I am confused.  
> **username_1**: @usernumber there's an additional lambda in the denominator's integral  

---

## 0 answers

