Special theory of relativity question

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SUMMARY

The discussion centers on the behavior of light in a "light clock" setup as described by the special theory of relativity. When the light clock is at rest, the time for light to travel to a mirror and back is calculated as '2L/c'. However, when the clock moves with velocity 'v', observers see the light pulse travel at an angle due to the relative motion, which raises questions about the constancy of light's direction. Participants clarify that while the speed of light remains constant, the perceived direction changes for different observers, illustrating the relativity of simultaneity and the independence of observations in different inertial frames.

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  • Understanding of the special theory of relativity
  • Familiarity with the concept of inertial frames
  • Basic knowledge of light propagation and speed of light
  • Concept of simultaneity and its relativity
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  • #61
PAllen said:
If a light ray has angle A in a given frame, than in a frame moving at speed v in the +x direction, the angle in the new frame (A`) is given by:

cot(A`) = (cot(A) - (v/c) cosec(A)) gamma

This differs from Galilean aberration by factor of gamma. I don't know if this effect has been observed - the difference from the Galilean formula is *extremely* small for the Earth's motion.

While I don't know that the relativistic correction is big enough to see for effects of the Earth's changing velocity relative to astronomic sources, there is another context where the relativistic corrections is believed to be observed. This same equation explains (a portion of) relativistic beaming, where a larger solid angle of emitted light transforms to a smaller solid angle when the source is rapidly approaching; and the converse for a receding source. This is believed to explain why relativistically moving plasma jets from compact sources typically look similar if they are orthogonal to our line of view, but an approaching jet is *much* brighter than a receding jet.
 
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  • #62
Mentz114 said:
This is a good question so I'll have another go.

The answer is to use the wave model of light. When the light beam is turned on, imagine a circular wave front expanding from the emitter. After the moment of emission, the receiver moves a certain distance before the wavefront intersects with it. The light then appears to have traveled at an inclined path in the ground frame. For a collimated beam, a detailed analysis would show that interference only supports the path between the emitter and the detector it was aimed at in the rest frame.

So, the theory(s) required is wave optics, or QED as described by R. Feynman, with the assumption that speed of light is invariant.

Thanx Mentz. You have clearly resolved all doubts. Light in nature is produced as spherical wave by oscillating charged particles. For gettin a linear beam of light, the 3D spherical wave is passed thru an aperture in some opaque screen.
If we imagine the linear light beam source is composed of a point light source and a opaque screen with an aperture, we can see the light beam comin out of aperture as a light beam coming out at an angle. If there is some distance between the point light source and the aperture, there will be a delay for the light to move from the point source to the movin aperture. Thus only an angled light beam will come out of aperture as calculated by simple geometry.
Please post ur comment if u think my understandin is wrong or u have a better answer.
 
  • #63
lovetruth said:
Thanx Mentz. You have clearly resolved all doubts. Light in nature is produced as spherical wave by oscillating charged particles. For gettin a linear beam of light, the 3D spherical wave is passed thru an aperture in some opaque screen.
If we imagine the linear light beam source is composed of a point light source and a opaque screen with an aperture, we can see the light beam comin out of aperture as a light beam coming out at an angle. If there is some distance between the point light source and the aperture, there will be a delay for the light to move from the point source to the movin aperture. Thus only an angled light beam will come out of aperture as calculated by simple geometry.
Please post ur comment if u think my understandin is wrong or u have a better answer.

Yes, you are right about the aperture. It will move relative to the point source after the light is emitted and so produce an angled beam. No QED required.:smile:

I made a primitive movie of the situation with a point source. It's here

www.blatword.co.uk/space-time/wavemove.mpeg

size is only 560Kb.
 
Last edited:
  • #64
Mentz114 said:
Yes, you are right about the aperture. It will move relative to the point source after the light is emitted and so produce an angled beam. No QED required.:smile:
Exactly, that's also what I meant with "if it went straight up then it would go through the side of the laser!" :smile:
I made a primitive movie of the situation with a point source. It's here

www.blatword.co.uk/space-time/wavemove.mpeg

size is only 560Kb.

Looks good!

Cheers,
Harald
 
  • #65
Mentz114 said:
Yes, you are right about the aperture. It will move relative to the point source after the light is emitted and so produce an angled beam. No QED required.:smile:

I made a primitive movie of the situation with a point source. It's here

www.blatword.co.uk/space-time/wavemove.mpeg

size is only 560Kb.

Thx for the reply.
 

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