An unknown celestial body of very large dimension (size of a galaxy) and
which mainly emits a clean monochromatic yellow light (600 nanometers), is
spotted in the deep sky.
Astrophysicists then begin to calculate its trajectory and its speed.
The speed of movement is 240000 km/s (0.8c).
It is an object that is fleeing the earth, at an angle of 60° with
respect to the line of sight.
We ask to calculate the observable wavelength of the object in the
terrestrial frame of reference.
Will it be observed in the visible spectrum, and if so, what will be its
observable color?

R.H.

Le 30/06/2022 à 16:17, Richard Hachel a écrit :
> An unknown celestial body of very large dimension (size of a galaxy) and which
> mainly emits a clean monochromatic yellow light (600 nanometers), is spotted in
> the deep sky.
> Astrophysicists then begin to calculate its trajectory and its speed.
> The speed of movement is 240000 km/s (0.8c).
> It is an object that is fleeing the earth, at an angle of 60° with respect to
> the line of sight.
>
> We ask to calculate the observable wavelength of the object in the terrestrial
> frame of reference.
>
> Will it be observed in the visible spectrum, and if so, what will be its
> observable color?
>
> R.H.

For now, I hear the sweet chirping of cicadas in the distance.

It's pleasant, France at night, in summer.

But the answer to my little problem... nay!

<http://news2.nemoweb.net/jntp?zLfqOGf6Pmg0zhZeyiTs3ZFXJSk@jntp/Data.Media:1>

R.H.

--
"Mais ne nous trompons pas.
Il n'y a pas que de la violence avec des armes : il y a des situations de
violence."
Abbé Pierre.
<http://news2.nemoweb.net/?DataID=zLfqOGf6Pmg0zhZeyiTs3ZFXJSk@jntp>

czwartek, 30 czerwca 2022 o 16:17:42 UTC+2 Richard Hachel napisał(a):
> An unknown celestial body of very large dimension (size of a galaxy) and
> which mainly emits a clean monochromatic yellow light (600 nanometers), is
> spotted in the deep sky.
> Astrophysicists then begin to calculate its trajectory and its speed.
> The speed of movement is 240000 km/s (0.8c).
> It is an object that is fleeing the earth, at an angle of 60° with
> respect to the line of sight.
>
> We ask to calculate the observable wavelength of the object in the
> terrestrial frame of reference.
>
> Will it be observed in the visible spectrum, and if so, what will be its
> observable color?
>
> R.H.

k * (1 + v/c cosf)
cos60 = 1/2; k = 5/3

fo/fs = 5/3 (1 + 0.4) = 5/3 * 1.4 = 2.333

600 nm / 2.333 = 260nm, probably UV.

niedziela, 3 lipca 2022 o 13:46:03 UTC+2 Alsor napisał(a):
> czwartek, 30 czerwca 2022 o 16:17:42 UTC+2 Richard Hachel napisał(a):
> > An unknown celestial body of very large dimension (size of a galaxy) and
> > which mainly emits a clean monochromatic yellow light (600 nanometers), is
> > spotted in the deep sky.
> > Astrophysicists then begin to calculate its trajectory and its speed.
> > The speed of movement is 240000 km/s (0.8c).
> > It is an object that is fleeing the earth, at an angle of 60° with
> > respect to the line of sight.
> >
> > We ask to calculate the observable wavelength of the object in the
> > terrestrial frame of reference.
> >
> > Will it be observed in the visible spectrum, and if so, what will be its
> > observable color?
> >
> > R.H.
> k * (1 + v/c cosf)
>
> cos60 = 1/2; k = 5/3
>
> fo/fs = 5/3 (1 + 0.4) = 5/3 * 1.4 = 2.333
>
> 600 nm / 2.333 = 260nm, probably UV.

rather:
k/(1 - v/c cosf) = 2.777

about 200 nm.

Le 03/07/2022 à 13:46, Alsor a écrit :
> czwartek, 30 czerwca 2022 o 16:17:42 UTC+2 Richard Hachel napisał(a):
>> An unknown celestial body of very large dimension (size of a galaxy) and
>> which mainly emits a clean monochromatic yellow light (600 nanometers), is
>> spotted in the deep sky.
>> Astrophysicists then begin to calculate its trajectory and its speed.
>> The speed of movement is 240000 km/s (0.8c).
>> It is an object that is fleeing the earth, at an angle of 60° with
>> respect to the line of sight.
>>
>> We ask to calculate the observable wavelength of the object in the
>> terrestrial frame of reference.
>>
>> Will it be observed in the visible spectrum, and if so, what will be its
>> observable color?
>>
>> R.H.
>
> k * (1 + v/c cosf)
>
> cos60 = 1/2; k = 5/3
>
> fo/fs = 5/3 (1 + 0.4) = 5/3 * 1.4 = 2.333
>
> 600 nm / 2.333 = 260nm, probably UV.

Thank you for your reply.

Are there any other reviews?

R.H.

niedziela, 3 lipca 2022 o 14:07:00 UTC+2 Richard Hachel napisał(a):
> Le 03/07/2022 à 13:46, Alsor a écrit :
> > czwartek, 30 czerwca 2022 o 16:17:42 UTC+2 Richard Hachel napisał(a):
> >> An unknown celestial body of very large dimension (size of a galaxy) and
> >> which mainly emits a clean monochromatic yellow light (600 nanometers), is
> >> spotted in the deep sky.
> >> Astrophysicists then begin to calculate its trajectory and its speed.
> >> The speed of movement is 240000 km/s (0.8c).
> >> It is an object that is fleeing the earth, at an angle of 60° with
> >> respect to the line of sight.
> >>
> >> We ask to calculate the observable wavelength of the object in the
> >> terrestrial frame of reference.
> >>
> >> Will it be observed in the visible spectrum, and if so, what will be its
> >> observable color?
> >>
> >> R.H.
> >
> > k * (1 + v/c cosf)
> >
> > cos60 = 1/2; k = 5/3
> >
> > fo/fs = 5/3 (1 + 0.4) = 5/3 * 1.4 = 2.333
> >
> > 600 nm / 2.333 = 260nm, probably UV.
> Thank you for your reply.
>
> Are there any other reviews?
>
> R.H.

OK.

1/k 1/(1 - v/c cosf) = 0.6 / 0.6 = 1

null Doppler shift: 600nm observed.

Le 03/07/2022 à 18:51, Alsor a écrit :
> niedziela, 3 lipca 2022 o 14:07:00 UTC+2 Richard Hachel napisał(a):
>> Le 03/07/2022 à 13:46, Alsor a écrit :
>> > czwartek, 30 czerwca 2022 o 16:17:42 UTC+2 Richard Hachel napisał(a):
>> >> An unknown celestial body of very large dimension (size of a galaxy) and
>> >> which mainly emits a clean monochromatic yellow light (600 nanometers), is
>> >> spotted in the deep sky.
>> >> Astrophysicists then begin to calculate its trajectory and its speed.
>> >> The speed of movement is 240000 km/s (0.8c).
>> >> It is an object that is fleeing the earth, at an angle of 60° with
>> >> respect to the line of sight.
>> >>
>> >> We ask to calculate the observable wavelength of the object in the
>> >> terrestrial frame of reference.
>> >>
>> >> Will it be observed in the visible spectrum, and if so, what will be its
>> >> observable color?
>> >>
>> >> R.H.
>> >
>> > k * (1 + v/c cosf)
>> >
>> > cos60 = 1/2; k = 5/3
>> >
>> > fo/fs = 5/3 (1 + 0.4) = 5/3 * 1.4 = 2.333
>> >
>> > 600 nm / 2.333 = 260nm, probably UV.
>> Thank you for your reply.
>>
>> Are there any other reviews?
>>
>> R.H.
>
> OK.
>
> 1/k 1/(1 - v/c cosf) = 0.6 / 0.6 = 1
>
> null Doppler shift: 600nm observed.

Correct.

R.H.

On Thursday, June 30, 2022 at 7:17:42 AM UTC-7, Richard Hachel wrote:
> A celestial body emits a monochromatic yellow light (600 nanometers)... The
> speed of movement is 240000 km/s (0.8c). It is fleeing the earth, at an angle
> of 60° with respect to the line of sight...what will be its observ[ed] color [on
> earth]?

The question is ambiguous, because you failed to specify whether the direction of movement (60 degrees from the line of sight) was expressed in terms of the inertial reference system of the celestial body (in which case the received wavelength on earth is 600 nanometers) or of the earth (in which case the received wavelength is 1400 nanometers).

Le 03/07/2022 à 20:13, Stan Fultoni a écrit :
> On Thursday, June 30, 2022 at 7:17:42 AM UTC-7, Richard Hachel wrote:
>> A celestial body emits a monochromatic yellow light (600 nanometers)... The
>> speed of movement is 240000 km/s (0.8c). It is fleeing the earth, at an angle
>> of 60° with respect to the line of sight...what will be its observ[ed] color
>> [on
>> earth]?
>
> The question is ambiguous, because you failed to specify whether the direction
> of movement (60 degrees from the line of sight) was expressed in terms of the
> inertial reference system of the celestial body (in which case the received
> wavelength on earth is 600 nanometers) or of the earth (in which case the received
> wavelength is 1400 nanometers).

The wavelength is given in the initial frame of reference of the emitting
object. It emits at 600 nanometers (from yellow).

The angle is given in the terrestrial reference, the object moves
according to an angle of 60° compared to the axis of sight.

<http://news2.nemoweb.net/jntp?z3G9Ild6hhyCP6h_ET8j4GdKapc@jntp/Data.Media:1>

The correct answer was given by Alsor in this thread.

R.H.

Le 04/07/2022 à 01:02, Richard Hachel a écrit :
> Le 03/07/2022 à 20:13, Stan Fultoni a écrit :
>> On Thursday, June 30, 2022 at 7:17:42 AM UTC-7, Richard Hachel wrote:
>>> A celestial body emits a monochromatic yellow light (600 nanometers)... The
>>> speed of movement is 240000 km/s (0.8c). It is fleeing the earth, at an angle
>>> of 60° with respect to the line of sight...what will be its observ[ed] color
>>> [on
>>> earth]?
>>
>> The question is ambiguous, because you failed to specify whether the direction
>> of movement (60 degrees from the line of sight) was expressed in terms of the
>> inertial reference system of the celestial body (in which case the received
>> wavelength on earth is 600 nanometers) or of the earth (in which case the received
>> wavelength is 1400 nanometers).
>
> The wavelength is given in the initial frame of reference of the emitting
> object. It emits at 600 nanometers (from yellow).
>
> The angle is given in the terrestrial reference, the object moves according to
> an angle of 60° compared to the axis of sight.
>
> <http://news2.nemoweb.net/jntp?z3G9Ild6hhyCP6h_ET8j4GdKapc@jntp/Data.Media:1>
>
> The correct answer was given by Alsor in this thread.
>
> R.H.

<http://news2.nemoweb.net/jntp?XUoLCWP5S0TM9Ogv_Ee8mMjr1cM@jntp/Data.Media:1>

<http://news2.nemoweb.net/?DataID=XUoLCWP5S0TM9Ogv_Ee8mMjr1cM@jntp>

R.H.

--
"Mais ne nous trompons pas. Il n'y a pas que de la violence
avec des armes : il y a des situations de violence."
Abbé Pierre.

On Sunday, July 3, 2022 at 4:02:56 PM UTC-7, Richard Hachel wrote:
> >> A celestial body emits a monochromatic yellow light (600 nanometers).... The
> >> speed of movement is 240000 km/s (0.8c). It is fleeing the earth, at an angle
> >> of 60° with respect to the line of sight...what will be its observ[ed] color
> >> [on
> >> earth]?
> >
> > The question is ambiguous, because you failed to specify whether the direction
> > of movement (60 degrees from the line of sight) was expressed in terms of the
> > inertial reference system of the celestial body (in which case the received
> > wavelength on earth is 600 nanometers) or of the earth (in which case the received
> > wavelength is 1400 nanometers).
> The wavelength is given in the initial frame of reference of the emitting
> object. It emits at 600 nanometers (from yellow).
>
> The angle is given in the terrestrial reference, the object moves
> according to an angle of 60° compared to the axis of sight.

Again, in that case the wavelength received on earth is 1400 nanometers. See above.

> The correct answer was given by Alsor in this thread.

No, the correct answer is that your question was ambiguous, because you failed to specify whether the direction of movement (60 degrees from the line of sight) was expressed in terms of the inertial reference system of the celestial body (in which case the received wavelength on earth is 600 nanometers) or of the earth (in which case the received wavelength is 1400 nanometers). You have now specified that the angle is in terms of the Earth's reference frame, so the answer is 1400 nanometers. Do you understand this? [Hint: Aberration.]

Le 04/07/2022 à 01:24, Stan Fultoni a écrit :
> On Sunday, July 3, 2022 at 4:02:56 PM UTC-7, Richard Hachel wrote:
>> >> A celestial body emits a monochromatic yellow light (600 nanometers)... The
>> >> speed of movement is 240000 km/s (0.8c). It is fleeing the earth, at an angle
>>
>> >> of 60° with respect to the line of sight...what will be its observ[ed] color
>>
>> >> [on
>> >> earth]?
>> >
>> > The question is ambiguous, because you failed to specify whether the direction
>>
>> > of movement (60 degrees from the line of sight) was expressed in terms of the
>> > inertial reference system of the celestial body (in which case the received
>> > wavelength on earth is 600 nanometers) or of the earth (in which case the
>> received
>> > wavelength is 1400 nanometers).
>> The wavelength is given in the initial frame of reference of the emitting
>> object. It emits at 600 nanometers (from yellow).
>>
>> The angle is given in the terrestrial reference, the object moves
>> according to an angle of 60° compared to the axis of sight.
>
> Again, in that case the wavelength received on earth is 1400 nanometers. See
> above.
>
>> The correct answer was given by Alsor in this thread.
>
> No, the correct answer is that your question was ambiguous, because you failed
> to specify whether the direction of movement (60 degrees from the line of sight)
> was expressed in terms of the inertial reference system of the celestial body (in
> which case the received wavelength on earth is 600 nanometers) or of the earth (in
> which case the received wavelength is 1400 nanometers). You have now specified
> that the angle is in terms of the Earth's reference frame, so the answer is 1400
> nanometers. Do you understand this? [Hint: Aberration.]

No.

<http://news2.nemoweb.net/jntp?dBhf5m21aaf5Eh80DnEs-gkiAT4@jntp/Data.Media:2>

The correct answer was given by Alsor.

The equation to use is this:

<http://news2.nemoweb.net/jntp?dBhf5m21aaf5Eh80DnEs-gkiAT4@jntp/Data.Media:1>

I don't understand your comments.

R.H.

--
"Mais ne nous trompons pas. Il n'y a pas que de la violence
avec des armes : il y a des situations de violence."
Abbé Pierre.
<http://news2.nemoweb.net/?DataID=dBhf5m21aaf5Eh80DnEs-gkiAT4@jntp>

On Sunday, July 3, 2022 at 4:54:19 PM UTC-7, Richard Hachel wrote:
> >> >> A celestial body emits a monochromatic yellow light (600 nanometers)... The
> >> >> speed of movement is 240000 km/s (0.8c). It is fleeing the earth, at an angle
> >> >> of 60° with respect to the line of sight...what will be its observ[ed] color
> >> >> [on earth]?
> >> >
> >> > The question is ambiguous, because you failed to specify whether the direction
> >> > of movement (60 degrees from the line of sight) was expressed in terms of the
> >> > inertial reference system of the celestial body (in which case the received
> >> > wavelength on earth is 600 nanometers) or of the earth (in which case the
> >> received wavelength is 1400 nanometers).
> >>
> >> The wavelength is given in the initial frame of reference of the emitting
> >> object. It emits at 600 nanometers (from yellow). The angle is given in the
> >> terrestrial reference, the object moves according to an angle of 60° compared
> >>to the axis of sight.
> >
> > Again, in that case the wavelength received on earth is 1400 nanometers.. See
> > above.
> >
> >> The correct answer was given by Alsor in this thread.
> >
> > No, the correct answer is that your question was ambiguous, because you failed
> > to specify whether the direction of movement (60 degrees from the line of sight)
> > was expressed in terms of the inertial reference system of the celestial body (in
> > which case the received wavelength on earth is 600 nanometers) or of the earth (in
> > which case the received wavelength is 1400 nanometers). You have now specified
> > that the angle is in terms of the Earth's reference frame, so the answer is 1400
> > nanometers. Do you understand this? [Hint: Aberration.]
>
> No.

Yes, you have specified that the angle of 60 degrees from the line of sight is in terms of the Earth's system of reference, so the received wavelength is 1400 nm.

What you are doing is computing the answer for the case where the angle is 60 degrees from the line of sight in terms of the celestial body's system of references, which means it would be moving at 120 degrees from the line of sight in terms of the Earth's system of reference (due to aberration), which would result in a wavelength of 600 nm. But this is not the question you have specified. You specifically said the 60 degrees is in terms of the Earth's reference system, so the answer is 1400 nm.

Do you understand this?

Le 04/07/2022 à 02:20, Stan Fultoni a écrit :
> Yes, you have specified that the angle of 60 degrees from the line of sight is
> in terms of the Earth's system of reference, so the received wavelength is 1400
> nm.
>
> What you are doing is computing the answer for the case where the angle is 60
> degrees from the line of sight in terms of the celestial body's system of
> references, which means it would be moving at 120 degrees from the line of sight
> in terms of the Earth's system of reference (due to aberration), which would
> result in a wavelength of 600 nm. But this is not the question you have
> specified. You specifically said the 60 degrees is in terms of the Earth's
> reference system, so the answer is 1400 nm.
>
> Do you understand this?

No.

Nothing.

R.H.

On Monday, July 4, 2022 at 4:16:17 AM UTC-7, Richard Hachel wrote:
> Le 04/07/2022 à 02:20, Stan Fultoni a écrit :
> > Yes, you have specified that the angle of 60 degrees from the line of sight is
> > in terms of the Earth's system of reference, so the received wavelength is 1400
> > nm.
> >
> > What you are doing is computing the answer for the case where the angle is 60
> > degrees from the line of sight in terms of the celestial body's system of
> > references, which means it would be moving at 120 degrees from the line of sight
> > in terms of the Earth's system of reference (due to aberration), which would
> > result in a wavelength of 600 nm. But this is not the question you have
> > specified. You specifically said the 60 degrees is in terms of the Earth's
> > reference system, so the answer is 1400 nm.
> >
> > Do you understand this?
>
> No.

It's been clearly explained several times. What part of it don't you understand? Are you interested in learning, or will you just run away?

Le 04/07/2022 à 16:23, Stan Fultoni a écrit :
> On Monday, July 4, 2022 at 4:16:17 AM UTC-7, Richard Hachel wrote:
>> Le 04/07/2022 à 02:20, Stan Fultoni a écrit :
>> > Yes, you have specified that the angle of 60 degrees from the line of sight is
>>
>> > in terms of the Earth's system of reference, so the received wavelength is
>> 1400
>> > nm.
>> >
>> > What you are doing is computing the answer for the case where the angle is 60
>> > degrees from the line of sight in terms of the celestial body's system of
>> > references, which means it would be moving at 120 degrees from the line of
>> sight
>> > in terms of the Earth's system of reference (due to aberration), which would
>> > result in a wavelength of 600 nm. But this is not the question you have
>> > specified. You specifically said the 60 degrees is in terms of the Earth's
>> > reference system, so the answer is 1400 nm.
>> >
>> > Do you understand this?
>>
>> No.
>
> It's been clearly explained several times. What part of it don't you
> understand? Are you interested in learning, or will you just run away?

I hope you are joking?

You don't know Richard Hachel.

The correct answer was given by a speaker.

I have nothing to add.

R.H.

On Monday, July 4, 2022 at 7:34:20 AM UTC-7, Richard Hachel wrote:
> >> > Yes, you have specified that the angle of 60 degrees from the
> >> > line of sight is in terms of the Earth's system of reference, so
> >> > the received wavelength is 1400 nm. Do you understand this?
> >>
> >> No.
> >
> > It's been clearly explained several times. What part of it don't you
> > understand? Are you interested in learning, or will you just run away?
>
> The correct answer was given by a speaker.

No, you gave the incorrect answer. The correct answer to the question you asked is 1400 nm.

The mistake you are making is that you are computing the answer for the case where the angle is 60 degrees from the line of sight in terms of the celestial body's system of references, which means it would be moving at 120 degrees from the line of sight in terms of the Earth's system of reference (due to aberration), which would result in a received wavelength of 600 nm on earth. But this is not the question you asked. You specifically said the 60 degrees is in terms of the terrestrial reference system, so the answer is 1400 nm.

What part of this do you not understand? Remember, all your beliefs amount to the claim that 1=0, which we proved previously, and you ran away. Now we've shown that you can't even correctly answer a simple high school homework question, and again you run away. Are you not interested in learning about this subject at all?

Le 04/07/2022 à 16:53, Stan Fultoni a écrit :
> No, you gave the incorrect answer. The correct answer to the question you asked
> is 1400 nm.
>
> The mistake you are making is that you are computing the answer for the case
> where the angle is 60 degrees from the line of sight in terms of the celestial
> body's system of references, which means it would be moving at 120 degrees from
> the line of sight in terms of the Earth's system of reference (due to aberration),
> which would result in a received wavelength of 600 nm on earth. But this is not
> the question you asked. You specifically said the 60 degrees is in terms of the
> terrestrial reference system, so the answer is 1400 nm.
>
> What part of this do you not understand? Remember, all your beliefs amount to
> the claim that 1=0, which we proved previously, and you ran away. Now we've shown
> that you can't even correctly answer a simple high school homework question, and
> again you run away. Are you not interested in learning about this subject at all?

I was very clear on the question, and I even made a small diagram.
The question was: if the initial wavelength of the object is 600
nanometers, what will be the wavelength perceived by the terrestrial frame
of reference.

<http://news2.nemoweb.net/jntp?5VqSOM_JknzM4Or25MKS567kwKI@jntp/Data.Media:1>

The object moves away from the earth (there will therefore be a
longitudinal Doppler effect to take into account) at an angle of 60° with
respect to the line of sight.

A speaker found the right answer without my prompting him.

In this case, there will be no variation of the wavelength which will
always be 600 nanometers (yellow)

I don't see how you manage to find such a large wavelength, and such a
displacement in the infrared.

And for a purely transverse displacement (µ=90°), you will find how
much? ? ?

R.H.

On Monday, July 4, 2022 at 11:22:16 AM UTC-7, Richard Hachel wrote:
> I was very clear on the question...

Well, you originally stated an ambiguous question, because you failed specify whether the 60 degree angle was in terms of the star's reference frame or the earth's reference frame. When I asked you to specify this, you clearly stated that the 60 degrees is in terms of the terrestrial reference frame, so the answer is 1400 nm.

> The question was: if the initial wavelength of the object is 600
> nanometers [in terms of the star's reference system, and it is moving
> at 0.8c at an angle of 60 degrees from the line of sight in terms of the
> terrestrial reference system], what will be the wavelength perceived by
> the terrestrial frame of reference.

Again, the answer to this question is 1400 nm, for the reason explained previously.

> The object moves away from the earth (there will therefore be a
> longitudinal Doppler effect to take into account) at an angle of 60° with
> respect to the line of sight.

Yes, but you must specify whether the 60 degree angle is in terms of the reference system of the star or of the earth. You specified (at my prompting) that it is defined in terms of the earth's reference system, so the answer is 1400 nm, because the start is moving away from the earth.

> A speaker found the right answer without my prompting him.

Someone posted several different answers; they obviously couldn't figure out which Doppler formula to use; and they obviously didn't know what they were doing because they didn't even ask you to specify whether the 60 degrees was defined in terms of the star's system or the earth's system. Without the information, the question is ambiguous. I pointed out the ambiguity in your question, and got you to specify the angle, and then provided you with the correct answer, which is 1400 nm. I also explained the error you were making in thinking that the answer is 600 nm, i.e., that's the answer for the case when the star is moving at 60 degrees from the line earth-to-star in terms of the star's reference system.
> In this case, there will be no variation of the wavelength which will
> always be 600 nanometers (yellow)

Again, you are mistaken. The answer to the question you asked, with the star moving at 0.8c in a direction that is 60 degrees from the earth-to-star line of slight is 1400 nm. See above.

> I don't see how you manage to find such a large wavelength, and such a
> displacement in the infrared.

This is Relativity 101. Since the star is moving away from earth at 0.8c at an angle of 60 degrees from the direct line of sight, it is significantly red-shifted. The first-order and the second-order Doppler terms both produce red shift in this case. On the other hand, if you specified 60 degrees in terms of the star's reference system, then it is 120 degrees in terms of the earth's reference system, which means it is actually getting closer to the earth, so the first-order Doppler cancels the second-order Doppler effect, to give no net shift in the frequency. But that isn't the question you asked.
> And for a purely transverse displacement (µ=90°), you will find how
> much? ? ?

Again, you must specify whether the angle is defined in terms of the terrestrial reference system or the star's reference system. If you specify 90 degrees in terms of the earth's reference system then the received wavelength is 1000 nm, whereas if you specify 90 degrees in terms of the star's reference system then the received wavelength is 360 nm.

Now do you understand?

Le 04/07/2022 à 21:08, Stan Fultoni a écrit :

>> And for a purely transverse displacement (µ=90°), you will find how
>> much? ? ?
>
> Again, you must specify whether the angle is defined in terms of the terrestrial
> reference system or the star's reference system. If you specify 90 degrees in
> terms of the earth's reference system then the received wavelength is 1000 nm,
> whereas if you specify 90 degrees in terms of the star's reference system then the
> received wavelength is 360 nm.

But no!

It's the opposite.

If the object passes transversely in the sky, for us, there is no
longitudinal relativistic effect to take into account.

The equation boils down to:
λ'=λ.sqrt(1-v²/c²)

Let λ'=600*0.6=360nm (blue shift)

R.H.

On Monday, July 4, 2022 at 12:52:34 PM UTC-7, Richard Hachel wrote:
> >> And for a purely transverse displacement (µ=90°), you will find how
> >> much? ? ?
> >
> > Again, you must specify whether the angle is defined in terms of the terrestrial
> > reference system or the star's reference system. If you specify 90 degrees in
> > terms of the earth's reference system then the received wavelength is 1000 nm,
> > whereas if you specify 90 degrees in terms of the star's reference system then the
> > received wavelength is 360 nm.
>
> But no! It's the opposite. If the object passes transversely in the sky, for us...
> the equation boils down to: λ'=λ.sqrt(1-v²/c²) (blue shift)

As always, you are mistaken. For an object moving transversely in terms of our system of reference, we will receive a red-shifted signal due to the time dilation. On the other hand, if the source object is moving transversely in terms of its own system of reference, we will receive a blue-shifted signal, as explained above.

Now do you understand?

On Monday, July 4, 2022 at 12:52:34 PM UTC-7, Richard Hachel wrote:
> >> And for a purely transverse displacement (µ=90°), you will find how much ???
> >
> > Again, you must specify whether the angle is defined in terms of the earth's
> > reference system or the star's reference system. If you specify 90 degrees in
> > terms of the earth's reference system then the received wavelength is 1000 nm,
> > whereas if you specify 90 degrees in terms of the star's reference system then the
> > received wavelength is 360 nm.
>
> But no! It's the opposite. If the object passes transversely in the sky, for us...
> the equation boils down to: λ'=λ.sqrt(1-v²/c²) (blue shift)

As always, you're mistaken. For an object moving transversely in terms of our system of reference, we will receive a red-shifted signal (1000 nm) due to the time dilation. On the other hand, if the source object is moving transversely in terms of its own system of reference, we will receive a blue-shifted signal (360 nm), as explained above. Now do you understand?