#### The question was

Projection effects can trick you into thinking that an
astrophysical object is moving “superluminally”. Consider a quasar that ejects
gas with speed v at an angle θ with respect to the line-of-sight of the
observer. Projected onto the sky, the gas appears to travel perpendicular to
the line of sight with angular speed v

_{app}/D, where D is the distance to the quasar and v_{app}is the apparent speed. Derive an expression for v_{app}in terms of v and θ. Show that, for appropriate values of v and θ, v_{app}can be greater than 1.#### My Answer was

PSR J0108-1431, the closest known pulsar to the Earth, is 280 light years distant . If the ejected gas position was measured at times a year apart, the angular difference in position α would be 1/128 = 0.0078125. We remember that sin's and tan's of very small angles are very close to the angle.

I don't know why we have to bring relativity into this, so I haven't. Consider the diagram below, α is unrealistically big for visibility.

Our famous observer is at O and the pulsar is at P. The
ejected gas arrives at G after, say a year, and it's velocity v is represented
by PG. The apparent velocity v

_{app}isis represented by the line PG_{app}. We added the line PQ at right angles to OG_{app}and passing thru P. By simple geometry, we get angles in this order:
Starting at O: PGO = π - α - θ
then QGP = α + θ

Starting at G

_{app}: PG_{app}O = π/2 - α then QPG_{app}= α
calculating PQ in two different ways we get

PQ = v

_{app}cos α = v sin (α + θ)
Therefore

v

_{app}= v sin (α + θ) / cos α
This is not quite what was wanted,
but for maximum leverage we obviously want v = π/2 - α so v lies along
PQ. We then get

v

_{app}= v / cos α
Let's assume the
gas comes from PSR
J0108-1431 and that it's velocity is 0.99997, v

_{app}is 1.000000517. If v is 0.99996, v_{app}is 0.999990517. It's a close run thing.#### Embarassing

Sadly my answer was completely wrong as I discovered when I looked at the UCSB answer, which was very clear. They have a slightly different illustration (from Hartle). O is the Observer again and N is the pulsar. They also do not use relativity but they do take account of the fact that if the cloud is moving rapidly towards the observer, the time taken for the light to arrive will be shorter, making the apparent velocity larger. That was at least one of my mistakes.From Hartle via UCSB |

The formula they arrive at is

v

_{app}= v sin θ / (1 - v cos θ)
I have omitted a

*c*because, for us, it is always 1.
This gives much more exciting results. For example at
v=0.99, the apparent velocity can reach nearly 7 times the speed of light! Here
is a graph of v

_{app}vs θ for a more modest v = 0.95.v_{app} vs θ for v = 0.95 |

Regarding your comment "At the end Petra equates v sin θ / (1 - v cos θ) and v (sin θ + cos θ). I couldn't follow that and tested it for a few values. It's not true.", I was not equating the two values, but rather "equating" two inequalities, basically a/(1-b) > 1 iff a+b>1, assuming 1-b is positive.

ReplyDeleteYou say “do not use relativity”, but relativity is used in that the spread of light is assumed the same in both frames (the speed of the light for the observer is not increased by v cos \theta). So the calculation depends on the first principle that the speed of light is the same in different frames.

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