A mysterious yet apparently true ~0.2 mag offset between ATLAS & ZTF Forced-Photometry

[Codeprey]

Hello,

I'm a transient follower and an honest user of ATLAS Forced-Photometry. Thanks for your great online service, it always provides vital information!

Recently I am studying two sources.

For the first source, I built the differential lightcurves using the results of both ATLAS and ZTF forced-photometry, trying to study the early bump. To improve the S/N I combined ATLAS result into 1-day bins. Compared with the ZTF result, the ATLAS lightcurve seems to be fainter at both the peak and the late-time plateau. This difference is also noticeable in the top right corner of the inset.

I tried to fix this difference by multiplying a factor 1.2 (which is an arbitrary attempt). It worked like a charm, although weird.

However, it unreliably changed the luminosity of the bump. So I tried to find if this difference is repeatable.

Then I found the same difference in the other source. Before this 'correction' the MW extinction corrected light curves look like this:

After the 'correction':

Looks like there should be an offset between c,o- and g,r- band results. I realize that could be due to the different wavelength range, so I check the effective wavelength for these bands. For ATLAS bands, lambda_c ~ 5300Å, lambda_o ~ 6700Å; For ZTF bands, lambda_g ~ 4800Å, lambda_r ~ 6400Å. For a red source, m_g>~m_c>m_r>~m_o.

For these two sources:
First one, from g,r we can deduct vLv ~ const, so (vLv)_r ~ (vLv)_o, since they are closed to each other. Therefore, an offset is existing, since they are clearly not equal to each other;
Second one, from m_g-m_r>0 we can deduct the source is red, there's no way to have m_g ~ m_o. Therefore, an offset is existing.

I cannot find out the reason, since both forced-photometry services use the similar PSF differential photometry, which means the results should be similar. It could also be a problem of the ZTF side, I will also keep tracking. If I need to provide more data or tests, please tell me.

Thanks in advance,
Peter