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    *EDIT* It's a bit late now but I've just realised I've been using gammas instead of lambdas for the wavelength :doh:

    A probe is launched from Earth with velocity v=0.8c. A beacon emits a light with a wavelength of \gamma=500nm in it's rest frame. After many years, NASA locate the probe using a telescope and measure the light from the beacon to have the wavelength \gamma=500nm in their rest frame. Is this possible? What is the explanation for this observation.

    Essentially the problem states that v/v\prime=\gamma\prime/\gamma=1

    so using the doppler shift equation I can solve the relationship between \beta and \theta

    \frac{\gamma\prime}{\gamma} = \frac{\sqrt{1-\beta^2}}{1+\beta \cos\theta} From the little I know about redshifting I'd guess that the light should be redshifted as the probe is moving away from us.

    But in my notes it says "We know that there is a (small) redshift when \theta = \pi/2. So somewhere between \theta = \pi/2 and \pi we must have a situation where \gamma\prime/\gamma=1, that is where there is no shift at all between the observed and source wavelengths. Under the right conditions (particular relationship between \theta and \beta) they can exactly balance".

    I don't understand why this is the case though... Why do they balance or cancel out? My understanding was any object moving away from the observer will always be redshifted, if only a little bit. I don't get it...
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    (Original post by AishaGirl)
    A probe is launched from Earth with velocity v=0.8c. A beacon emits a light with a wavelength of \gamma=500nm in it's rest frame. After many years, NASA locate the probe using a telescope and measure the light from the beacon to have the wavelength \gamma=500nm in their rest frame. Is this possible? What is the explanation for this observation.

    Essentially the problem states that v/v\prime=\gamma\prime/\gamma=1

    so using the doppler shift equation I can solve the relationship between \beta and \theta

    \frac{\gamma\prime}{\gamma} = \frac{\sqrt{1-\beta^2}}{1+\beta \cos\theta} From the little I know about redshifting I'd guess that the light should be redshifted as the probe is moving away from us.

    But in my notes it says "We know that there is a (small) redshift when \theta = \pi/2. So somewhere between \theta = \pi/2 and \pi we must have a situation where \gamma\prime/\gamma=1, that is where there is no shift at all between the observed and source wavelengths. Under the right conditions (particular relationship between \theta and \beta) they can exactly balance".

    I don't understand why this is the case though... Why do they balance or cancel out? My understanding was any object moving away from the observer will always be redshifted, if only a little bit. I don't get it...
    I never studied this, but a quick look at the Transverse Doppler Effect on wikipedia shows what's going on.

    Basically, the earth is moving through space in an orbit around the sun, which means it isn't necessarily going to be in line with the axis of the original direction the beacon was released in. Hence the Doppler effect will be different.

    This picture may help (from the wikipedia article):



    The left hand image is the reference frame of the beacon. The right hand image is the frame of the observer, so this is the one that explains the question you're dealing with. What it shows is how the wave fronts are actually spherical, so if the observer is at an angle to the motion, the doppler shift will be different.

    In the image, the little black dot on the right is the observer. It's pretty obvious that as the beacon passes the observer, the compact wave fronts at the front of the beacon quickly become spread out at the back. It's the same basic concept as when a car zooms by and makes the classed 'nneeeyowwwmmm' noise. The car is constantly moving in that situation but there's still some point at which the doppler effect is zero, at the midpoint where it goes from 'neeee' to 'yooowwwwm'.

    So for any situation where the motions is not on the same axis, there's always going to be a variation in redshift as the angle between the observer and the object varies.

    So this is how the observation angle can change the amount of doppler shift, and explains why there is some angle where the shift = 1.
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    (Original post by Pessimisterious)
    I never studied this, but a quick look at the Transverse Doppler Effect on wikipedia shows what's going on.

    Basically, the earth is moving through space in an orbit around the sun, which means it isn't necessarily going to be in line with the axis of the original direction the beacon was released in. Hence the Doppler effect will be different.

    This picture may help (from the wikipedia article):



    The left hand image is the reference frame of the beacon. The right hand image is the frame of the observer, so this is the one that explains the question you're dealing with. What it shows is how the wave fronts are actually spherical, so if the observer is at an angle to the motion, the doppler shift will be different.

    In the image, the little black dot on the right is the observer. It's pretty obvious that as the beacon passes the observer, the compact wave fronts at the front of the beacon quickly become spread out at the back. It's the same basic concept as when a car zooms by and makes the classed 'nneeeyowwwmmm' noise. The car is constantly moving in that situation but there's still some point at which the doppler effect is zero, at the midpoint where it goes from 'neeee' to 'yooowwwwm'.

    So for any situation where the motions is not on the same axis, there's always going to be a variation in redshift as the angle between the observer and the object varies.

    So this is how the observation angle can change the amount of doppler shift, and explains why there is some angle where the shift = 1.
    But the probe is moving at 0.8c, even if Earth was moving toward the probe, the probe would still be moving away at a faster rate and therefore still be redshifted? I understand that Earth is in orbit but I don't see a time when Earths position cancels out the probe moving at 0.8c. If the prove was moving at the same velocity as Earth then sure.

    Does this mean that at some specific time a galaxy will not appear to be redshifted even though it is technically?
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    (Original post by AishaGirl)
    But the probe is moving at 0.8c, even if Earth was moving toward the probe, the probe would still be moving away at a faster rate and therefore still be redshifted? I understand that Earth is in orbit but I don't see a time when Earths position cancels out the probe moving at 0.8c. If the prove was moving at the same velocity as Earth then sure.

    Does this mean that at some specific time a galaxy will not appear to be redshifted even though it is technically?
    But it's not just about the earth moving towards or away from it. It's about the relative alignment of the two velocity vectors.

    Maybe the beacon also ends up orbiting something?

    I think the point of the question is just to somehow demonstrate the transverse doppler effect and how things can change with the angle. This is the only thing I can guess from it, given the fact that the equation you're working with deals with \theta, which only happens in that context.

    And yeah, I guess it's possible for a galaxy to not redshift even though it actually is moving. This is definitely something for astrophysics people though! Haven't studied a jot of that for about 4 years now, ha.
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    (Original post by Pessimisterious)
    But it's not just about the earth moving towards or away from it. It's about the relative alignment of the two velocity vectors.

    Maybe the beacon also ends up orbiting something?

    I think the point of the question is just to somehow demonstrate the transverse doppler effect and how things can change with the angle. This is the only thing I can guess from it, given the fact that the equation you're working with deals with \theta, which only happens in that context.

    And yeah, I guess it's possible for a galaxy to not redshift even though it actually is moving. This is definitely something for astrophysics people though! Haven't studied a jot of that for about 4 years now, ha.
    OK cheers
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    Does any of this apply to GCSE?
    please...
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    (Original post by AadamG)
    Does any of this apply to GCSE?
    please...
    No you don't learn this at GCSE...
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    (Original post by AadamG)
    Does any of this apply to GCSE?
    please...
    If you're with OCR, you may have to deal with out. Our old teacher made a habit of having us go through calculations and lots of details on relativistic redshift :-;


    Also, with regard to the question, the reason the redshift may be small is that at the angle pi/2 or 90 degrees, the beacon is moving perpendicular to our line of sight, thus the relativistic velocity we consider is to be thought of as negligible, at least for this case.

    The ships velocity isn't along the line of sight, thus any sort of beaming from the ship along the line of sight wont be redshifted by the factor that 0.8c seems to imply. That's my guess.

    Quick Edit : Gotta add this on.

    The reason they may balance out is that at some point, the line-of-sight velocity will cancel out with the velocity of the earth or whatever place you are at with NASA, and as such there is no relative velocity that should be considered for the calculations with regard to the frames that the beacon and earth are in, thus there shouldn't be any redshift.

    Also one last edit kicking in my head :

    The reason it may be travelling orthogonal to the field of sight is that nothing will travel purely straight through space unless the forces on it superpose to be zero. If it is shot off from earth and travels for some time as the question implies, much less at 0.8c, it's going to end up being attracted by other large bodies, etc.
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    (Original post by Callicious)
    If you're with OCR, you may have to deal with out. Our old teacher made a habit of having us go through calculations and lots of details on relativistic redshift :-;


    Also, with regard to the question, the reason the redshift may be small is that at the angle pi/2 or 90 degrees, the beacon is moving perpendicular to our line of sight, thus the relativistic velocity we consider is to be thought of as negligible, at least for this case.

    The ships velocity isn't along the line of sight, thus any sort of beaming from the ship along the line of sight wont be redshifted by the factor that 0.8c seems to imply. That's my guess.

    Quick Edit : Gotta add this on.

    The reason they may balance out is that at some point, the line-of-sight velocity will cancel out with the velocity of the earth or whatever place you are at with NASA, and as such there is no relative velocity that should be considered for the calculations with regard to the frames that the beacon and earth are in, thus there shouldn't be any redshift.
    Well let's work it through.

    Using the doppler shift equation we can say that \frac{\gamma\prime}{\gamma} = \frac{\sqrt{1-\beta^2}}{1+\beta \cos\theta} \implies \cos\theta=\frac{1}{\beta}\Big( \sqrt{1-b^2}-1\Big)

    It's clear that the square root term will always be less than or equal to 1 and so the right side will always be negative given that \cos\theta < 0 for \pi/2 < \theta \leq \pi

    Then using the taylor expansion it becomes \beta \rightarrow 0 \implies \cos\theta \approx \frac{1}{\beta}\Big( 1-\frac{1}{2}\beta^2-1\Big)=-\frac{\beta}{2}

    So you are right and there is a time when the probe travels nearly perpendicular to pi/2. If we just plug in \beta=4/5 we get \theta=120^\circ.

    All well and good but I still fail to see where the speed of the probe is taken into account... As I mentioned earlier even if the probe was moving directly away from earth with \theta=0 it should still be redshifted due to the sheer speed of the probe compared to Earths orbit.
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    (Original post by AishaGirl)
    Well let's work it through.

    Using the doppler shift equation we can say that \frac{\gamma\prime}{\gamma} = \frac{\sqrt{1-\beta^2}}{1+\beta \cos\theta} \implies \cos\theta=\frac{1}{\beta}\Big( \sqrt{1-b^2}-1\Big)

    It's clear that the square root term will always be less than or equal to 1 and so the right side will always be negative given that \cos\theta < 0 for \pi/2 < \theta \leq \pi

    Then using the taylor expansion it becomes \beta \rightarrow 0 \implies \cos\theta \approx \frac{1}{\beta}\Big( 1-\frac{1}{2}\beta^2-1\Big)=-\frac{\beta}{2}

    So you are right and there is a time when the probe travels nearly perpendicular to pi/2. If we just plug in \beta=4/5 we get \theta=120^\circ.

    All well and good but I still fail to see where the speed of the probe is taken into account... As I mentioned earlier even if the probe was moving directly away from earth with \theta=0 it should still be redshifted due to the sheer speed of the probe compared to Earths orbit.
    Ya, to be honest that part has me in a loop too.

    My best guess would be that the folks on Earth are accelerating, at some point they will be accelerating relative to the probe, thus this acceleration may cancel out the relativistic doppler shift. That's true to some extent but on the scale of 0.8c, no idea.

    My only notion would be as I mentioned earlier with the probe being at 90 degrees to the line of sight, but ignoring any external influence, i.e. just considering the probe heading straight away and the folks on earth not orbiting just sitting there, I don't know.
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    (Original post by Callicious)
    Ya, to be honest that part has me in a loop too.

    My best guess would be that the folks on Earth are accelerating, at some point they will be accelerating relative to the probe, thus this acceleration may cancel out the relativistic doppler shift. That's true to some extent but on the scale of 0.8c, no idea.

    My only notion would be as I mentioned earlier with the probe being at 90 degrees to the line of sight, but ignoring any external influence, i.e. just considering the probe heading straight away and the folks on earth not orbiting just sitting there, I don't know.
    Yeah I dunno, going to ask the physics teacher at school see if he can explain it better .
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    (Original post by AishaGirl)
    Yeah I dunno, going to ask the physics teacher at school see if he can explain it better .
    phaha...
    i didnt mean to spark of a whole lecture
    thanks anyways tho
 
 
 
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