Aberrations from the relativistic aberration of light
Albertito
as opposed to a non-relativistic one, is like trying
to 'pink a pink'. How can we 'pink a pink'? It's like
to loop the loop, or like to take the logarithm twice,
y = ln(ln(x)), but claiming it has been taken once.
That's how SR can deal with relativistic aberration
of light.
Under Galilean relativity, aberration of light arises
as a natural phenomenon, when we add/substract
two velocities
c' = c - v,
where,
c is the velocity of light, and
v the velocity of an observer.
The aberration angle, alpha, is then the angle between
vector c' and v. By multiplying both sides of that equation
by v, we get inner products,
c' v = c v - v v
and, assuming the norms |c'| = |c|, it yields
|c| |v| cos(alpha) = |c| |v| cos(phi) - |v|^2,
cos(alpha) = cos(phi) - |v|/|c|,
where phi is the angle between c and v.
Now, here relativists come into action, trying to loop
the loop.They pretend applying Einstein's addition
of velocities to velocities c and v, to attain somewhat
odd relativistic aberration. How can we 'pink a pink'?
Easy. We take the projection of c onto -v , to attain
c' = c cos(phi), then we add their respective rapidities
to attain a virtual/apparent velocity of light, c'', which is
supposed will carry the relativistic aberration angle, as
compared with v. We can add rapidities because c' and
-v are in the same axis. So, we attain
cos(alpha) = (cos(phi) - |v|/|c|)/( 1 - |v|cos(phi)/|c|)
But, by taking the projection of c onto -v to attain c',
we see the sum c'' is in the same axis as v, so the angle
between them can only be zero. Actually, c'' does not carries
any aberration angle at all. OTOH, if you try to attain a rapidity
from the velocity c, you get arctanh(1) = oo, infinity! This means
the direct sum (arctanh(1) - arctanh(|v|/|c|)) always diverges. So,
under SR, you have to use the trick of the projection of c onto -v,
to attain a virtual speed c' whose norm would be less than that of
c. We clearly see that the velocity c' = c cos(phi) has a norm
|c|cos(phi) < |c|. It's a clever trick, indeed. But, by doing so, you
screw up the 2nd postulate of SR.
Brilliant! We have 'pinked the pink'. Great! Is anybody there
that realized a relativistic aberration is actually an ABERRATION?
In order to test with success relativistic aberrations of light we
would need accuracies within the third-order approximation
of v/c, or beyond. But, that's technically unachievable.
"You can't pretend you have taken the logarithm of x once,
if you actually have taken it twice, y = ln(ln(x))"
There are no relativistic corrections for aberration of light.
Aberration of light is galilean and euclidean, essentially.
If you look at it deeply, you will see aberration of light, as
James Bradley discovered it, has the lorentz factor, gamma,
encoded into it. And, that means:
"Stop looping the loop, don't pink the pink,
You can't pretend you have taken the logarithm of x once,
if you actually have taken it twice, y = ln(ln(x))"
shal...@gmail.com
1) The speed of light emitted by a source is c, regardless of its
motion. That is, the photon does not move at a velocity of c with
respect to some universal rest frame (how does one even begin to
define such a frame anyway?).
2) Where an observer is moving at a relative velocity v \approx 0.5c
directly away from the source, then a photon emitted directly toward
the observer would approach it at relative velocity of v \approx 0.5c,
increasing the observed wavelength by 2 (half the speed of light,
twice the absorption time, twice the wavelength, half the frequency).
This is where the 1 - cos(\phi) v/c term comes from. The remainder of
the formula is just the kinematic time dilation of the observer.
I take it you're not a fan of relativity.
Albertito
In aberration of light, the motion of the source
does not play any role at all, only the motion
of the observer with respect to the incoming
photons. Any analysis of aberration of light which
included the motion of a source is meaningless.
shal...@gmail.com
I'm pretty sure it's called relativity theory because motion is
relative. If I say the source is moving away from the observer, just
take it to mean that the observer is moving away from the source --
not that I said the source was moving in the first place, mind you.
- Shawn
Androcles
<shal...@gmail.com> wrote in message
news:74184524-b1b4-44ae...@59g2000hsb.googlegroups.com...
|I always thought it worked like this:
|
| 1) The speed of light emitted by a source is c, regardless of its
| motion. That is, the photon does not move at a velocity of c with
| respect to some universal rest frame (how does one even begin to
| define such a frame anyway?).
|
| 2) Where an observer is moving at a relative velocity v \approx 0.5c
| directly away from the source, then a photon emitted directly toward
| the observer would approach it at relative velocity of v \approx 0.5c,
| increasing the observed wavelength by 2 (half the speed of light,
| twice the absorption time, twice the wavelength, half the frequency).
| This is where the 1 - cos(\phi) v/c term comes from.
Nope. \phi is simply the angle of incidence. When the light (or sound)
source is coming straight at you \phi is zero and cos(\phi) = 1.
If the light (or sound) source passes you by then \phi changes as it
does so. This is fairly obviously observed with the change in shift
of the sound of a passing car. The term is strictly Doppler's original.
| The remainder of
| the formula is just the kinematic time dilation of the observer.
No such animal exists.
http://www.androcles01.pwp.blueyonder.co.uk/Wave.xls
Feel free to check the equations or change the values in the yellow boxes.
|
| I take it you're not a fan of relativity.
You can take it I'm not a fan of stupidity, I take it you are not
a fan of reality.
--
Why did Einstein say
the speed of light from A to B is c-v,
the speed of light from B to A is c+v,
the "time" each way is the same?
Androcles
shal...@gmail.com
> <shala...@gmail.com> wrote in message
Androcles,
Like I said in this example, where the observer is moving directly
AWAY from the source, and the photon is moving directly TOWARD the
observer, it's implied that their direction vectors are identical, and
so phi = 0, cos(phi) = 1, and thus 1 - cos(\phi) v/c = 0.5, resulting
in a doubling of the wavelength. Add in kinematic time dilation. The
fact that you were trying to correct an already correct example by
simply restating my correct solution is kind of funny. Thanks for the
refresher in linear algebra though.
- Shawn
Androcles
<shal...@gmail.com> wrote in message
news:57c79e16-770f-48fc...@m45g2000hsb.googlegroups.com...
Androcles,
1- cos(0) = 1-1 = 0 when I went to school. Maybe it has changed
since then.
shal...@gmail.com
Androcles, you forgot to include v/c in your calculation:
1 - cos(\phi) * v/c = 1 - 1 * 0.5 = 1 - 0.5 = 0.5
- Shawn
shal...@gmail.com
It also donned on me that this thread initially had nothing to do with
the relativistic Doppler effect. In that case, my apologies Albertito.
- Shawn
Androcles
<shal...@gmail.com> wrote in message
news:6ce22bb7-5d5e-4745...@d1g2000hsg.googlegroups.com...
I'm sitting in a grandstand watching cars race by.
As a car approaches, I hear the pitch of the engine raised.
As the car recedes, I hear the pitch lowered.
The angle changes as the car passes.
Somewhere in between I hear the normal pitch.
That is when the angle is 90 degrees.
f' = f * [(1 - cos(90 degrees).v/c] (All credit to Doppler)
Because cos(90 degrees) = 0, I do not need v/c.
f' = f * [(1 - 0.v/c]
f' = f * [(1 - 0]
f' = f * 1
= f
http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img107.gif
f' = f * [(1 - cos(90 degrees).v/c] / sqrt(1 - v^2/c^2) (Einstein)
= f/sqrt(1 - {something})
hence f' does not equal f at 90 degrees
Doppler's equation when divided by sqrt(1 - v^2/c^2) produces nonsense.
Feel free to change the values in the yellow boxes and examine the
equations.
http://www.androcles01.pwp.blueyonder.co.uk/Wave.xls
--
Why did Einstein say
the speed of light from A to B is c-v,
the speed of light from B to A is c+v,
the "time" each way is the same?
Answer: Because he was a loud-mouthed idiot.
Androcles
http://www.androcles01.pwp.blueyonder.co.uk/
Androcles
<shal...@gmail.com> wrote in message
news:0a50d645-5d8b-4966...@m73g2000hsh.googlegroups.com...
There is no such animal as "the relativistic Doppler effect" anywhere
in Nature. When that DAWNS on you, you might see the light.
shal...@gmail.com
Brilliant pun!
- Shawn
Albertito
Don't worry, Shawn, Doppler effect, aberration of light
and addition of velocities are all related. This is the
simple equation that relates them all:
- c ln(z +1) = v + w,
where,
v and w are velocities of source and observer
in a given frame of reference,
c is the velocity of light, and
z is the Doppler shift.
Notice that c is velocity of light, not a speed, so it is a vector.
The above formula is true because c depends on the velocity
of the source, it is not an invariant.
So, |c ln(z +1)| is the magnitude, norm, of the sum v + w.
This formula can't be found in any textbook. If you love
SR, then replace the binary operator + by Einstein addition
of velocities. This will make c be invariant in magnitude,
but not in direction wrt observer. If you love Galilean relativity
then retain + as strictly an euclidean vector addition.
Let's evaluate the above equation for some cases
where velocity of the source is v = 0. Then
- c ln(z +1) = w
1) If z = -1, then |c| = oo, which means the observer
approched the source at infinite speed.
2) if z = e - 1, then c = - w, which means the observer
is receding from the source at speed |c|.
3) if z = 1/e - 1, then c = w, which means the observer
is approaching the source at speed |c|.
4) if z = 0, then w = 0, which means the observer
is at rest wrt the source.
If you analyze experimental data, no Doppler blueshift z < 0
less than -1 will be found, as measured directly from the
observed frequency f' and the original one f. Likewise, you
will find out there can be Doppler redshifts, z >0, with no
apparent upper bound. What does it mean?. It means
the equation - c ln(z +1) = v + w is true, and SR is wrong.
Dono
> In aberration of light, the motion of the source
> does not play any role at all, only the motion
> of the observer with respect to the incoming
> photons. Any analysis of aberration of light which
> included the motion of a source is meaningless.-
Shito,
This is a "pearl" that nails you to the "Immortal Fumbles".
You obviously don't know that motion is relative, do you?
There is no such thing as "motion of the source" vs. "motion of the
observer". All that counts is the relative motion, retard.
shal...@gmail.com
Brilliant contribution. Oh wait, nevermind, you just copied my reply
from 2 days ago and converted it to "jerkish". Perhaps Androcles will
understand the argument, now that it has been converted to his native
language.
- Shawn
shal...@gmail.com
If you are talking about redshifts and z in regard to the apparent
motion of receding galaxies, then yes, SR does not apply. This is not
an effect of relative motion, but of the metric expansion of space.
The photons are actually shifting in frequency/wavelength as they
travel, unlike with the relativistic Doppler effect where the
frequency/wavelength doesn't actually change. Is this what you're
referring to?
- Shawn
Androcles
<shal...@gmail.com> wrote in message
news:3c5c2266-b6f5-4d15...@i76g2000hsf.googlegroups.com...
| If you are talking about redshifts and z in regard to the apparent
| motion of receding galaxies, then yes, SR does not apply. This is not
| an effect of relative motion, but of the metric expansion of space.
Prove it.
| The photons are actually shifting in frequency/wavelength as they
| travel,
Shifting in frequency? No way, Jose.
Look, any wave has both a wavelength and a frequency.
Since c = /lambda * /nu, what is the speed of this wave?
http://tinyurl.com/yt742o
Albertito
Well, I guess you mean the product frequency*wavelength
doesn't actually change, under SR assumptions. No, I was
referring to a more essential phenomenon. If you can find
Doppler redshifts of frequencies larger than 1, why can't you
find Doppler blueshifts of frequencies less than -1? It seems
there is an asymmetry, doesn't it? SR can't deal with that
asymmetry properly. There is something wrong in SR, and
I think that something is its 2nd postulate.
Dono
Shaenee-boy
I simply don't read the stuff you post, it is crap most of the time.
So, you got one right, so what? Most of the time you post
crackpottery.