Need Deflection of Light Clarification

  • Thread starter Thread starter Devin
  • Start date Start date
  • Tags Tags
    Deflection Light
Click For Summary
Both Newtonian gravitation and General Relativity predict that light bends around massive objects, but this can create confusion regarding the speed of light. In General Relativity, the speed of light remains constant (c) only in local measurements, while the average speed over larger distances can differ due to spacetime curvature. The concept of vector displacement illustrates the difference between deflected and undeflected light paths, but no actual travel occurs along this vector. The curvature of spacetime affects how light propagates, similar to how great circles represent the shortest distance on a sphere. Understanding these principles clarifies the apparent conflict with special relativity.
Devin
Messages
24
Reaction score
1
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?
 
Physics news on Phys.org
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.
 
  • Like
Likes 1 person
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.
 
  • Like
Likes 1 person
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.
 
  • Like
Likes 1 person
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.
 
In an inertial frame of reference (IFR), there are two fixed points, A and B, which share an entangled state $$ \frac{1}{\sqrt{2}}(|0>_A|1>_B+|1>_A|0>_B) $$ At point A, a measurement is made. The state then collapses to $$ |a>_A|b>_B, \{a,b\}=\{0,1\} $$ We assume that A has the state ##|a>_A## and B has ##|b>_B## simultaneously, i.e., when their synchronized clocks both read time T However, in other inertial frames, due to the relativity of simultaneity, the moment when B has ##|b>_B##...

Similar threads

  • · Replies 4 ·
Replies
4
Views
7K
  • · Replies 74 ·
3
Replies
74
Views
4K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 60 ·
3
Replies
60
Views
13K
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 15 ·
Replies
15
Views
5K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
30
Views
14K
  • · Replies 66 ·
3
Replies
66
Views
13K