Need Deflection of Light Clarification

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SUMMARY

The discussion clarifies the behavior of light in the context of Newtonian gravitation and General Relativity, emphasizing that while the speed of light is constant (c) in local measurements, it can differ in global coordinate systems due to spacetime curvature. The concept of vector displacement is introduced, highlighting that it represents the difference between deflected and undeflected light paths, rather than a physical travel path. The confusion regarding luminiferous ether is addressed, reinforcing that local measurements are more significant than average speeds over larger distances.

PREREQUISITES
  • Understanding of General Relativity principles
  • Familiarity with the concept of spacetime curvature
  • Knowledge of vector displacement in physics
  • Basic grasp of local versus global measurements in physics
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  • Study the implications of spacetime curvature on light propagation
  • Explore the concept of geodesics in General Relativity
  • Investigate the differences between local and coordinate speeds of light
  • Learn about the historical context and implications of luminiferous ether theory
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Physicists, students of theoretical physics, and anyone interested in the implications of General Relativity on light behavior and spacetime concepts.

Devin
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Both the Newtonian theory on gravitation and the General theory of Relativity predict a change in the direction of a beam of light as it propagates past a large mass. Regardless of the precision of the Δ velocity component (the component that can represent the vertical shift towards a mass), If you think in terms of vector displacement, then you have a resultant vector who's magnitude is greater than the speed of light. I apologize for the awful depiction. https://twitter.com/ValorAtmC/status/480107448385019904/photo/1 ... I understand that if this were true, a conflict with special relativity is created. With that being said, can somebody please point out what I'm missing?
 
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In general relativity, the statement "the speed of light is a constant c" is true only in a local sense. That is, if you could measure the speed of a pulse of light as it passes by you, using measurements that are "close enough" to your own location, you would get c, no matter where you are.

However, the "average" speed between two "distantly" separated points, computed as Δx/Δt, can be different from c, because of effects due to the curvature of spacetime.
 
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Devin said:
can somebody please point out what I'm missing?
Your vector displacement is the difference between a deflected path and an undeflected one. Nothing travels along this displacement vector.
 
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The speed of light is equal to "c" locally in any small region as measured by local clocks and local rules in any local inertial frame. However, the coordinate speed of light is not necessarily "c" in any given global coordinate system. In generalized coordinates, and in curved space-time, the coordinate speeds don't have any particular physical significance, the locally measured speeds are much more physically significant.

Some of the effects are due to the curvature of space-time, and are rather similar to how a great circle can be the shortest distance between two points on the Earth's surface, yet appear to be "curved" when plotted out on an chart. Like a great circle on the surface of the Earth, the path of light through space-time is a geodesic.
 
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jtbell said:
In general relativity, the statement "the speed of light is a constant c" is true only in a local sense. That is, if you could measure the speed of a pulse of light as it passes by you, using measurements that are "close enough" to your own location, you would get c, no matter where you are.

However, the "average" speed between two "distantly" separated points, computed as Δx/Δt, can be different from c, because of effects due to the curvature of spacetime.

Thank you for the response. I was just confused because I assumed a luminiferous ether.
 
Bill_K said:
Your vector displacement is the difference between a deflected path and an undeflected one. Nothing travels along this displacement vector.


Thank you for the response.
 

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