Is the Speed of Light Consistent Across Rotating Reference Frames?

  • #1
phymath7
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TL;DR Summary
Can we divide the speed of light into components?
Does it violate the postulate of special relativity in any sense?
 
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  • #2
What do you mean by "divide the speed of light into components"? Are you referring to vector components?
 
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  • #3
Yes, the velocity of light has components (x,y,z or whatever basis). The invariance of the speed of light applies to the magnitude of the velocity. The components adjust themselves accordingly.
 
  • #4
Ibix said:
What do you mean by "divide the speed of light into components"? Are you referring to vector components?
Yeah, I referred to vector components.
 
  • #5
jtbell said:
Yes, the velocity of light has components (x,y,z or whatever basis). The invariance of the speed of light applies to the magnitude of the velocity. The components adjust themselves accordingly.
Thanks. But a detailed work on this topic would be appreciated (from books or notes).
 
  • #6
phymath7 said:
Thanks. But a detailed work on this topic would be appreciated (from books or notes).
It's easy enough - it follows from the invariance of the interval. If you have two events on the same light pulse's worldline separated by ##\Delta t## in time and ##\Delta x##, ##\Delta y## and ##\Delta z## in space, then ##c^2\Delta t^2=\Delta x^2+\Delta y^2+\Delta z^2##. You can rearrange this to get ##0=c^2\Delta t^2-\Delta x^2-\Delta y^2-\Delta z^2## and recognise the right hand side as the interval, ##\Delta s^2##, which is invariant - hence ##0=c^2\Delta t'^2-\Delta x'^2-\Delta y'^2-\Delta z'^2##, which says that the speed of light is the same in the other frame. Or if you aren't familiar with the interval, use the Lorentz transforms to eliminate the unprimed quantities from the first equation to get the last. You can also use the Lorentz transforms to transform the ##\Delta## quantities explicitly if you want to see how the light's velocity vector transforms. The x component of its three velocity is just ##\Delta x/\Delta t## and similarly for the y and z components. Likewise in the primed frame - just remember to use the primed spatial and time deltas.
 
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  • #7
Ibix said:
It's easy enough - it follows from the invariance of the interval. If you have two events on the same light pulse's worldline separated by ##\Delta t## in time and ##\Delta x##, ##\Delta y## and ##\Delta z## in space, then ##c^2\Delta t^2=\Delta x^2+\Delta y^2+\Delta z^2##. You can rearrange this to get ##0=c^2\Delta t^2-\Delta x^2-\Delta y^2-\Delta z^2## and recognise the right hand side as the interval, ##\Delta s^2##, which is invariant - hence ##0=c^2\Delta t'^2-\Delta x'^2-\Delta y'^2-\Delta z'^2##, which says that the speed of light is the same in the other frame. Or if you aren't familiar with the interval, use the Lorentz transforms to eliminate the unprimed quantities from the first equation to get the last. You can also use the Lorentz transforms to transform the ##\Delta## quantities explicitly if you want to see how the light's velocity vector transforms. The x component of its three velocity is just ##\Delta x/\Delta t## and similarly for the y and z components. Likewise in the primed frame - just remember to use the primed spatial and time deltas.
Does the 2nd postulate of special relativity imply that if I shift the origin of a co-ordinate system to make another(without rotation),only then the speed of light is constant in both reference frame but so is not true when I do the shifting with rotation?
 
  • #8
phymath7 said:
Does the 2nd postulate of special relativity imply that if I shift the origin of a co-ordinate system to make another(without rotation),only then the speed of light is constant in both reference frame but so is not true when I do the shifting with rotation?
No. The speed of light is the same in all inertial reference frames. You can see that easily enough in the maths above. The components of the 3-velocity vector change under rotation or Lorentz boost, but not its magnitude.
 
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  • #9
Ibix said:
No. The speed of light is the same in all inertial reference frames. You can see that easily enough in the maths above. The components of the 3-velocity vector change under rotation or Lorentz boost, but not its magnitude.
Another way of saying this is that although all inertial frames agree on the speed of a beam of light, they don't all agree on the beam's direction (relative to the frame axes). This is due to aberration.
 
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  • #10
I don't know, what you define as a "velocity vector of light". The speed of light in Maxwell's equations (or in the somewhat hidden form of ##\mu_0 \epsilon_0=1/c^2## in the SI) is the phase velocity of plane-wave solutions.

The closest related vector quantity to it is the wave-fourvector, ##k=(\omega/c,\vec{k})## of an em. wave in vacuum, which is necessarily a null-vector, ##\omega^2/c^2-\vec{k}^2=0## or ##\omega=c |\vec{k}|##. It transforms as a four-vector, and of course ##\omega## and ##\vec{k}## both change under boosts, describing the Doppler effect (change of ##\omega##) and aberration (change of direction of ##\vec{k}##). The speed of light is the same, because a null vector stays a null vector, i.e., you also have ##\omega'=c|\vec{k}'|## in the new frame.
 
  • #11
phymath7 said:
Does the 2nd postulate of special relativity imply that if I shift the origin of a co-ordinate system to make another(without rotation),only then the speed of light is constant in both reference frame but so is not true when I do the shifting with rotation?
The spacetime interval is invariant under rotations, translations, time translations, and Lorentz boosts.
 
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