Hi there,

This has been discussed elsewhere

(https://forum.image.sc/t/mean-gray-intensity-integrated-density-and-raw-integrated-density/4983/3

https://www.researchgate.net/post/Why_is_integrated_density_better_than_the_mean_to_measure_fluorescence_in_Image_J

https://www.researchgate.net/post/Is_integrated_density_or_mean_grey_value_better_to_quantify_fluorescence2)

But (From ImageJ manual):

1) Mean gray value (MGV) Average gray value within the selection. This is the sum of the gray values of all the pixels in the selection divided by the number of pixels. Reported in calibrated units (e.g., optical density) if Analyze▷Calibrate…↓ was used to calibrate the image. [...]

2) Integrated density (ID) The sum of the values of the pixels in the image or selection. This is equivalent to the product of Area and Mean Gray Value.

So

3) Mean Gray Value*Area = Integrated density

Indeed, there is no clear-cut answer to your question, it basically depends on what you are measuring and which biological significance have your results. Both are correct, but it depends on what you want to find.

To use a parallelism in physicochemical terms: To me, this has always been something like the difference between amount of a substance (Integrated Density in images) and concentration of a substance (Mean Gray Value in images) considering volumes (that would be areas in 2D images). You can get the same amount of any substance with different concentrations, or the same concentration with different amounts (playing around with the volumes)... right?

If you assume that the areas/sizes of the particles you are comparing have similar distributions, then you can use MGV.

If your treatment (or the difference between the groups you are comparing) involve changes in the size distribution of the particles you are analyzing (cells, ROIs,...) then it might be convenient to use ID or at least consider that some differences could be hidden in MGV and arise when using ID.

Moreover, if you are using a confocal system, then you will also have to take into account that you are working in a 3D space. So the so called area, actually comes from a maximal projection of a 3D space and then, the proper way to do this is considering the 3D space, not just the area (voxel vs pixel). However, even in this case, you'll have to consider that z resolution is not as good as xy resolution in your system.

If the later is the case, you may want to take a look at this:

(https://imagejdocu.tudor.lu/plugin/stacks/3d_ij_suite/start

https://imagejdocu.tudor.lu/plugin/analysis/3d_analysis/start)

Hope this helps,

Cheers,

J

Hi, as discussed above, the two are closely related quantities and the analogy of mass vs. concentration describes very well the difference.

If you are evaluating the whole images, then both provide the same information - the area in this case would be the area of the image, which is likely to be constant - let's say the field of view of your camera.

However if you want e.g. to compare the average fluorescence intensity of cells (between a control and a sample), then rather then comparing the whole images, you should look at the intensity within the cells only. A typical approach is to find an intensity threshold (Image/Adjust/Threshold...) to differentiate cells from the background. Divide the resulting binary image by 255 (cells will have value 1, background 0) and divide the original image by the binary (create 32-bit output). Cells then have the original intensity, while background is infinity.

In this case the mean intensity is much more meaningful than the integrated value since the area is no more a constant. It is the area occupied by the cells in each image, which can vary from image to image.

If on the other hand the intensity within the cells can be considered constant (it is the same for the control and the sample) then both the mean and the integrated intensity of the whole image can be considered a measure of the area occupied by the cells. E.g. if you want to compare cells density between sample and control.

Thank you@Radek Machan