Understanding the Deflection of Starlight by the Sun Using Elementary Analysis

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SUMMARY

The forum discussion focuses on estimating the deflection of starlight by the Sun using elementary analysis, specifically through dimensional analysis. The effective time of fall is calculated as 4.67 seconds using the Sun's diameter of 1.4 x 109 meters and the speed of light at 3 x 108 meters/second. Participants emphasize the importance of gravitational acceleration (gsun = 275 m/s2) and the geometry involved in the light's path, noting that the diameter serves as a scale factor for gravitational force. The conversation also highlights the significance of integrating the path of light to accurately determine deflection.

PREREQUISITES
  • Understanding of gravitational acceleration, specifically gsun = 275 m/s2.
  • Familiarity with the speed of light, C = 3 x 108 meters/second.
  • Basic knowledge of dimensional analysis and its application in physics.
  • Proficiency in calculus, particularly integration techniques for determining deflection angles.
NEXT STEPS
  • Study the principles of gravitational lensing and its mathematical foundations.
  • Learn about the integration of functions involving trigonometric identities in physics problems.
  • Research Einstein's theory of general relativity and its implications for light deflection.
  • Explore historical experiments that validated the predictions of light deflection, particularly the 1919 solar eclipse observation.
USEFUL FOR

Students of physics, astrophysicists, and anyone interested in gravitational effects on light, particularly in the context of general relativity and astronomical observations.

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Homework Statement


Estimate the deflection of starlight by sun using an elementary analysis.

gsun = 275 meters / sec2

Diameter of sun = 1.4 * 109 meters.

In the following, assume that the light just grazes the surface of Sun in passing.

A) Determine an "effective time of fall" from the diameter of Sun and speed of light. From this time of fall deduce the net velocity of fall toward Sun produced by the end of the whole period of gravitational interaction.

Homework Equations



C = 3*108 meters/ second

The Attempt at a Solution



This is an odd problem so I have the correct answer.
A) By dimensional analysis and by what is given this is what I did (and got the right answer, but why is this the right answer?)

Diametersun / C = effective time fall = 4.67 seconds (which is correct).

So I imagine the light traveling horizantally to the right (+x direction), then once it grazes the sun it bends slightly towards the -y direction.

But how is the time for this to occur related to the Diameter of the sun??
 
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The length of the path where the gravitational acceleration is close to (or at least not far below) gsun depends on the diameter.
 
mfb said:
The length of the path where the gravitational acceleration is close to (or at least not far below) gsun depends on the diameter.

Can you elaborate more on this? Maybe some math will help
 
Actually the diameter of the sun only comes in here because of the 'grazing' : in your x-y picture, the distance of closest approach (to the center of gravity of the sun) is a scale factor for the vertical component of the gravitational force. The rest is simple geometry, with a theta to integrate over from 0 to pi. So what you work out here is the maximum possible deflection case.
 
Chemp_93 said:
Can you elaborate more on this? Maybe some math will help
Does this help?

attachment.php?attachmentid=65895&stc=1&d=1390342838.png
 

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Thanks I appreciate the picture.

Okay but shouldn't we be interested in using radius rather than diameter in order to calculate the effective time since the gravitational force depends on the distance from the center of mass (center of sun) to the edge of the sun (where particle grazes the sun).
 
It is a rough estimate, a factor of 2 does not matter. I guess the diameter is a bit better than the radius as you have deflection both before and after the point of closest approach, but that is beyond the accuracy needed for the problem. A proper calculation would need an integration.
 
Ha! The OP is a longer exercise and we are looking at the first step.

At the point of grazing, the vertical acceleration is at a maximum gsun. Photon is exposed to that only a short time.

Look at the picture by Mentor: even far away the 'negligible force' has a vertical component.
Better to speak of acceleration. Something with a sine in it.
The acceleration itself is gsun R2/r2.
So happens that R/r is the same sine !

Angle θ runs from 0 to -π when the photon travels from far left to far right.

In (a), they first want to know an effective time only. The time a particle has to hang around at the point of grazing to experience the same amount of acceleration towards the center of the sun.


They want us to integrate dx / c from -∞ to +∞ with a weight sin3Θ

(Hint: x = R/tanθ ; change to dθ ) My guess is that the integral gives 2R/c.

Contrary to what Mentor claims: the 2 is still important. Spectacularly so in 1919 when Einstein was proved right...
 
BvU said:
Contrary to what Mentor claims: the 2 is still important. Spectacularly so in 1919 when Einstein was proved right...
My nick is mfb, not Mentor.

Theories cannot be proven, only disproven. An experiment can be in agreement with the theory prediction, but that does not mean the theory has to be right.

The factor of 2 is important if you want a proper calculation. It is irrelevant if you are interested in the order of magnitude (as in "do I have a chance to see the effect at all?").
 
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