What is Hubble's Luminosity Law?

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SUMMARY

The discussion centers on Hubble's Luminosity Law, specifically the equation 5 log(R) = -m + k. Participants clarify that the variable 'k' represents the sensitivity of the detector and can be ignored when focusing on the overall shape of the graph. Setting 'k' to zero allows for a clearer understanding of the relationship between the apparent magnitude of stars and the angular size of reflection nebulae. The conversation emphasizes the importance of understanding how distance affects light reflection and magnitude perception.

PREREQUISITES
  • Understanding of Hubble's Luminosity Law
  • Familiarity with the concepts of apparent magnitude and angular size
  • Basic knowledge of graphing equations
  • Awareness of light reflection principles in astronomy
NEXT STEPS
  • Research the implications of setting 'k' to zero in astronomical equations
  • Explore the relationship between angular size and apparent magnitude in reflection nebulae
  • Study Hubble's original paper on luminosity and its applications
  • Learn about the sensitivity of astronomical detectors and their impact on data interpretation
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Astronomy students, educators, and anyone interested in understanding the mathematical relationships in astrophysics, particularly those studying Hubble's Luminosity Law and its applications in analyzing celestial objects.

angela6884
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Problem Statement: Please help me understand the variables for the equation, 5 log(R) = -m + k.
Relevant Equations: 5 log(R) = -m + k

Here is the link to where I found the equation. I know it's on wikipedia but I checked the Hubble's paper and it seems to be credible. I'm trying to make theoretical graphs by testing different inputs for the equation. Would it be dumb to get rid of the "k" in the equation? Sorry I'm in high school so I'm still learning the ropes of how to make an analysis paper.😖 Also I'm a new member so sorry if I posted on the wrong page!
 
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angela6884 said:
Would it be dumb to get rid of the "k" in the equation?

No, as 'k' just represents the sensitivity of your detector. All it does if you graph the equation is move the entire graph up or down. If you're just interested in the overall shape of the graph and what it means, you can ignore the 'k'.

Setting k=0 and plotting the magnitude on the y-axis shows you that the apparent magnitude of the star increases (gets more negative) as the angular size of the nebula increases. Or, rather, that if reflection nebula are indeed illuminated by reflecting light from a nearby star, their angular size should be proportional to the magnitude of the star.

The reason for this is a bit unclear for me. I'm not sure if it has something to do with the way that angular size and magnitude change as distance from Earth changes, or if it is because of how parts of a nebula further away from a star will reflect less light, since they lie further from the star and the light is more spread out by the time it reaches that part of the nebula and reflects. I'm leaning towards the latter.
 
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Drakkith said:
No, as 'k' just represents the sensitivity of your detector. All it does if you graph the equation is move the entire graph up or down. If you're just interested in the overall shape of the graph and what it means, you can ignore the 'k'.

Setting k=0 and plotting the magnitude on the y-axis shows you that the apparent magnitude of the star increases (gets more negative) as the angular size of the nebula increases. Or, rather, that if reflection nebula are indeed illuminated by reflecting light from a nearby star, their angular size should be proportional to the magnitude of the star.

The reason for this is a bit unclear for me. I'm not sure if it has something to do with the way that angular size and magnitude change as distance from Earth changes, or if it is because of how parts of a nebula further away from a star will reflect less light, since they lie further from the star and the light is more spread out by the time it reaches that part of the nebula and reflects. I'm leaning towards the latter.
Oh okay that makes sense! And yes I think you're latter inference makes sense but, Ill try to read Hubble's paper and get you your answer! Thank you for your assistance!
 

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