Radio source Hercules A and galaxy 3C 348

  • #1
Nathan Warford
23
1
TL;DR Summary
Lack of information about 3C 348, the galaxy that produces the radio source Hercules A.
243430
I have found plenty of information about the radio source Hercules A. Its large radio jets are 1.5 million light years long from end to end, the supermassive black hole that produces Hercules A is 3 to 4 billion solar masses, and the black hole exists in the center of galaxy 3C 348. However, I have not been able to find much information about the galaxy itself aside from the fact that it's a type E3 elliptical galaxy. I have not been able to find any sources that provide reasonable estimates for the mass or size of 3C 348.

One thing that I know is that the radius of an elliptical galaxy is defined as the "half-light radius", which is the radius within which half of the galaxy's luminosity is contained. Based on an image of radio source Hercules A overlaid with galaxy 3C 348 (which I have provided), and knowing that Hercules A is 1.5 million light years long, I came up with a VERY rough estimate for the galaxy's diameter at about 125,000 lights years (radius of about 62,500 lights years). However, with the limited technology that I have at my disposal, all that I can do is eyeball the image. Are my measurements accurate, or is there a more accurate size to be found for 3C 348?
 

Answers and Replies

  • #2
phyzguy
Science Advisor
5,092
2,097
I would suggest going to the Sloan Digital Sky Survey, downloading an image of the galaxy, and measuring the half-light radius yourself. I went to the following link:
http://skyserver.sdss.org/dr15/en/tools/chart/image.aspxTyped in the coordinates from this page on Wikipedia , and got the image below. This is just a jpeg image, but you can download a FITS file with the actual pixel values, from which calculating the half light radius is fairly simple. Then you will have not just a number, but an understanding of what is involved in obtaining it.

Another useful source is NED. Go to the site below, and type 3C348 into the 'object name' field, and you will get a large amount of data and links to sources.

https://ned.ipac.caltech.edu/forms/byname.html

Herc_A.jpeg
 
  • #3
Nathan Warford
23
1
Finding pixel values is not difficult. I can do that easily enough in Photoshop. Once I have the half-light radius in relation to the galaxy, then I can perform pixel measurements and compare it to the FOV legend to find an angular radius. What I don't know how to do is calculate the half-light radius based on the image. All that I can do is guesstimate where half of the galaxy's luminosity is contained.

NED was no help. It only provided information about the radio source and not the galaxy.
 
  • #4
phyzguy
Science Advisor
5,092
2,097
Finding pixel values is not difficult. I can do that easily enough in Photoshop. Once I have the half-light radius in relation to the galaxy, then I can perform pixel measurements and compare it to the FOV legend to find an angular radius. What I don't know how to do is calculate the half-light radius based on the image. All that I can do is guesstimate where half of the galaxy's luminosity is contained.
I don't understand. If you have the pixel values, then you add up all of the pixel values, out to some radius where the flux falls off to equal the sky background. This gives you the total flux. Then you add up pixel values out to a radius where you find half the flux. This gives you the half-light radius. Why doesn't this work for you?

NED was no help. It only provided information about the radio source and not the galaxy.
I'm surprised to hear this. When I went to the NED results and scrolled down to the tab titled," Quick-Look Angular and Physical Sizes", it returned the sizes in the screenshot below. Since they are in visible and infrared light, they only see the galaxy and not the radio source. Why don't these sizes answer your question?

3c348.png
 
  • #5
Nathan Warford
23
1
I don't understand. If you have the pixel values, then you add up all of the pixel values, out to some radius where the flux falls off to equal the sky background. This gives you the total flux. Then you add up pixel values out to a radius where you find half the flux. This gives you the half-light radius. Why doesn't this work for you?
Because I don't know how to precisely measure flux with the tools I have at my disposal.
I'm surprised to hear this. When I went to the NED results and scrolled down to the tab titled," Quick-Look Angular and Physical Sizes", it returned the sizes in the screenshot below. Since they are in visible and infrared light, they only see the galaxy and not the radio source. Why don't these sizes answer your question?

View attachment 244047
I was not observant enough to notice that. I saw the text "Hercules A" and so I ignored it. I think that has the information that I need. Thank you for pointing it out to me. You have been very helpful.
 

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