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chroot

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Yes, the photon's frequency defines its color.

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what physical attribute(s) of the photon embody the Hubble velocity color shift?

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russ_watters

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Is that a homework question? Can you think of a way/reason a photon's color would change?

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selfAdjoint

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Nereid

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All observers measure the speed of light (in a vaccuum) to be the same ... c. This is a prediction from Einstein's theory of Relativity, and there are no (AFAIK) good experimental or observational results that are inconsistent with this theory.

While photons from distant galaxies in images taken by the Hubble Space Telescope are detected as 'red-shifted' (their wavelengths are greater than comparable photons emitted in a lab near you) - an effect caused by the expansion of the universe - you don't have to go so far to see red-shifted photons. For example, helioseismology relies on the (local) Doppler effect (this involves both redshifting and blueshifting).

Another example, this time the classic Pound and Rebka experiment ... the photons from the emitter were received by the detector (or not!) at a different frequency, but there was no relative motion! This is called 'classic' because it first demonstrated the 'gravitational redshift' predicted by Einstein.

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selfAdjoint

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From Einstein's paper,

t(B) - t(A) = r(AB) / (c-v)

which would seem to say that he considers that the light (photon) from the rest system is 'approaching' moving point B at less than c ???

t(B) - t(A) = r(AB) / (c-v)

which would seem to say that he considers that the light (photon) from the rest system is 'approaching' moving point B at less than c ???

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t(A) - t(B) = r(AB) / (c-v)

While we are this stage of Einstein's formulation, there is something that puzzles me.

When the photon reaches point B, it would seem like we might consider that the communication of the photon from the rest frame to the moving frame has taken place. As the equation above is written, the photon is approaching B at velocity c-v. If we regard the photon as having made the transition to this frame and now consider that this frame has the properties where A and B are at rest with respect to one another, at the instant the photon is reflected from B, it seems like the velocity must be c-v in the opposite direction.

so t'(A) - t(B) = r(AB) / (c-v)

Or, I don't see any problem with still characterizing this frame as moving at velocity v as Einstein apparently does. However, if at the time the photon arrives at B, we assume that B (and A) are still regarded as moving at velocity v, it seems like the equation for the reverse path must be written:

t'(A) - t(B) = r(AB) / (c - v - v(B) + v(A)) = r(AB) / (c-v)

It feels pretty satisfying that two different ways of characterizing the frame yield the same resulting equation.

However, when Einstein formulates the second equation, he has to be using some principle(s) that I'm not aware of.

In the case where the frame is considered now at rest when the photon reaches B, in order to get his equation, the reflection must add velocity 2v to the reflection:

t'(A) - t(B) = r(AB) / (c-v + v + v) = r(AB) / (c+v).

Or in the other case, where he considers the frame as still moving at v, the reflection must add velocity v to reflected velocity of the photon:

t'(A) - t(B) = r(AB) / (c-v + v + v(B)) = r(AB) / (c+v).

What I need help with is that I don't see how the reflection can add any velocity to the reflected motion of the photon much less two different amounts depending on whether you now regard the frame as at rest or still moving. ???

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Nereid

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Moving this to Special & General Relativity ...

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Making some logical assumptions about the coordinate systems being used (which should really be spelled out explicitly), the equation

t(B) - t(A) = r(AB) / (c-v)

does not say that the photon moves at a velocity c-v in any frame.

What it says that if a photon is approaching another moving object, the time it takes in the frame where the object is moving is given by solving

d = c*t +A (a light beam starting out at d=A at t=0)

d' = v*t+B (an object starting at d=B at t=0 with velocity v)

for the time when d=d'. This means c*t+A = v*t+B, or (c-v)*t = B-A

When you go to the frame of the moving object, you find (of course) theat the velocity of the photon in the moving frame is equal to c. You finding that in the moving frame, the light beam and the object do not start out at the same time 't'. that the distance is lorentz contracted, and that time is dilated.

[add]

I'm going to explain this more fully, in the hope that it will get through. (I'm guessing that this is the same person I talked to before that was confused about this issue, though of course it's a resaonably common confusion)

.

Anyway....

We have two frames of reference. One is stationary, one is moving at velocity v. The origins of the two frames coincide. We have an object, who is also moving at velocity v (the same velocity that the two frames are moving).

in frame 1, a light beam starts out at t=A, v=c

In frame 1, the equation of the light beam is

x = A + c*t

To convert this to frame 2, we use the Lorentz transform

x' = gamma*(x - v*t) t' = gamma*(t - v*x/c^2)

where gamma = 1/sqrt(1-v^2/c^2)

since x=A and t=0, we find

in frame 2, a light beam starts out at

x' = gamma*A, and t' = -gamma*v*A/c^2.

Since the velocity of light is a constant, 'c', in any frame, the equation for the motion of the light beam in frame 2 is

1) x = gamma*A + c*(t' + gamma*v*A/c^2)

(the only equation for an object moving at 'c' that passes through the given point).

Similarly, in frame 2, we find that an object starts out at

x' = gamma *B and t' = -gamma*v*B/c^2

The relativistic velocity equation will give a velocity for the object in frame 2 as zero, because I am making the assumption that our frame is comoving with this object (but has an origin that's displaced).

Thus the equation of motion for the moving object as a function of time in frame 2 is

2) x' = gamma*B

So, the two objects will meet when eq 1) equals eq 2)

gamma*A + gamma*A*v/c

gamma*A + c*(t' + gamma*v*A/c^2) = gamma*B

which boils down to gamma*A(1+v/c) + c*t' = gamma*B

or t' = gamma*(B-A)/c - gamma*A*v/c^2

which is quite consistent, it says that the distance is contracted by gamma, and you have a time offset of -gamma*A*v/c^2 due to the relativity of simultaneity.

t(B) - t(A) = r(AB) / (c-v)

does not say that the photon moves at a velocity c-v in any frame.

What it says that if a photon is approaching another moving object, the time it takes in the frame where the object is moving is given by solving

d = c*t +A (a light beam starting out at d=A at t=0)

d' = v*t+B (an object starting at d=B at t=0 with velocity v)

for the time when d=d'. This means c*t+A = v*t+B, or (c-v)*t = B-A

When you go to the frame of the moving object, you find (of course) theat the velocity of the photon in the moving frame is equal to c. You finding that in the moving frame, the light beam and the object do not start out at the same time 't'. that the distance is lorentz contracted, and that time is dilated.

[add]

I'm going to explain this more fully, in the hope that it will get through. (I'm guessing that this is the same person I talked to before that was confused about this issue, though of course it's a resaonably common confusion)

.

Anyway....

We have two frames of reference. One is stationary, one is moving at velocity v. The origins of the two frames coincide. We have an object, who is also moving at velocity v (the same velocity that the two frames are moving).

in frame 1, a light beam starts out at t=A, v=c

In frame 1, the equation of the light beam is

x = A + c*t

To convert this to frame 2, we use the Lorentz transform

x' = gamma*(x - v*t) t' = gamma*(t - v*x/c^2)

where gamma = 1/sqrt(1-v^2/c^2)

since x=A and t=0, we find

in frame 2, a light beam starts out at

x' = gamma*A, and t' = -gamma*v*A/c^2.

Since the velocity of light is a constant, 'c', in any frame, the equation for the motion of the light beam in frame 2 is

1) x = gamma*A + c*(t' + gamma*v*A/c^2)

(the only equation for an object moving at 'c' that passes through the given point).

Similarly, in frame 2, we find that an object starts out at

x' = gamma *B and t' = -gamma*v*B/c^2

The relativistic velocity equation will give a velocity for the object in frame 2 as zero, because I am making the assumption that our frame is comoving with this object (but has an origin that's displaced).

Thus the equation of motion for the moving object as a function of time in frame 2 is

2) x' = gamma*B

So, the two objects will meet when eq 1) equals eq 2)

gamma*A + gamma*A*v/c

gamma*A + c*(t' + gamma*v*A/c^2) = gamma*B

which boils down to gamma*A(1+v/c) + c*t' = gamma*B

or t' = gamma*(B-A)/c - gamma*A*v/c^2

which is quite consistent, it says that the distance is contracted by gamma, and you have a time offset of -gamma*A*v/c^2 due to the relativity of simultaneity.

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