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max8166
2010-Apr-09, 12:23 PM
A simple question on relativity?

If an astronomer points a telescope at Mars and simultaniously fires a laser at the point which he is observing, does the laser light:
a) travel at light speed so time stops and the laser hit's the spot.
b) travel at 300,000 km/s so the planet continues on its orbital speed of about 24 km/s and so misses the spot by 24km for every light second mars is away from Earth.
c) miss the spot by about 48km for every light second Mars is away from Earth as the light from the planet the astronomer is viewing has had to spend light seconds to arrive at Earth.
d) miss the spot by 24 km for every light second just because the planet had already moved that much while the light was travelling to Earth.

(This is not a question about tangents and other relative tradectories and I am well aware Earth is moving too, this is a question about the nature of light in a simple example)

Hungry4info
2010-Apr-09, 12:52 PM
I'm pretty sure the answer is B.

For A, our reference frame is what matters, not the photon's.
For C, once the photons arrive at Mars, they bounce back and come toward Earth along with the rest of the photons from Mars, so the amount that the laser missed is not exaggerated by the photon's return trip.
For D, that requires the photon to be at Mars and on its return trip immediately being fired.

Hornblower
2010-Apr-09, 01:17 PM
If we are aiming at the apparent position at the instant of firing, I would say C. For every light-second of distance away, the target has moved 24km by the time its light reaches us along the line of sight. It will move another 24km before our light gets there.

Jeff Root
2010-Apr-09, 01:24 PM
C, see? Si!

-- Jeff, in Minneapolis

neilzero
2010-Apr-09, 02:12 PM
48 kilometers per second = 172,800 kilometers per hour = 4147200 kilometers per day = 1,513,655,000 kilometers per Earth year, so the number in B maybe the correct frame of reference. Neil

max8166
2010-Apr-09, 02:23 PM
48 kilometers per second = 172,800 kilometers per hour = 4147200 kilometers per day = 1,513,655,000 kilometers per Earth year, so the number in B maybe the correct frame of reference. Neil

Sorry? I'm not following you. Are you saying B is correct but the orbital speed of Mars is wrong? I took the orbital speed of Mars from the wiki link here (http://en.wikipedia.org/wiki/Mars)

Ken G
2010-Apr-09, 03:52 PM
The answer is E, but the spirit of C is the correct spirit. The reason C is wrong is that it is not the orbital speed of Mars that counts, but its tangential speed relative to Earth, which depends on where Earth and Mars happen to be. You also have to use the current distance between them. But I see what you are asking-- you have two lags to worry about, the lag of the light you see from Mars, and the lag of the laser on its way to Mars. (I won't mention the aberration due to Earth's rotation, that muddies the water even more!)

max8166
2010-Apr-09, 06:34 PM
But doesn't time stop for the photon that travels to Mars at light speed? The experience of time from the earth would continue but our reference frame has changed to that of the Photon now, we need to determine the distance from the target in terms of 24 km or 48 km (2X) times the distance in light seconds.
What does "time stops for the photon" mean if around the photon planets are merrily carrying on their passage whilst the photon wings it's way to the red planet?
I think we can agree the astronomer is viewing the planet as it was several light seconds previously so we know the photon cannot go back in time to when the target was observed, however from the point of view of the laser photon can we really say the planet will not only move the distance it should move in the time it takes a photon to travel there, but twice that distance in (from the photons perspective) zero time?

Ken G
2010-Apr-09, 07:40 PM
But doesn't time stop for the photon that travels to Mars at light speed?Not in an absolute way, only in its frame. It's not a very good frame to use-- stay in the Earth frame instead. The answer can't depend on the frame, but there are easy frames and hard frames to use to get the answer.

Jeff Root
2010-Apr-10, 01:27 AM
But doesn't time stop for the photon that travels to Mars at light speed?
Photons do not experience time. Because of that, photons do not experience
anything at all. A photon comes into existence and goes out of existence
instantaneously in its view. It would consider itself to have a lifetime of zero
length. Any observer will see the photon come into existence in one place,
travel at the speed of light to another place, and then go out of existence,
after a finite time. The only value a photon can ever measure for anything
is "zero". So what a photon would experience, if it could experience
anything, is perfectly useless and irrelevant to any calculation.

The experience of time from the earth would continue but our reference
frame has changed to that of the Photon now, we need to determine the
distance from the target in terms of 24 km or 48 km (2X) times the distance
in light seconds.
In the frame of a photon, every distance is zero. In the frame of a photon,
every time is zero. In the frame of a photon, there is no such thing as
"light seconds", or any other units of any quantity at all.

What does "time stops for the photon" mean if around the photon planets
are merrily carrying on their passage whilst the photon wings it's way to
the red planet?
It means the same kind of thing as time dilation in general. Let's look at
a case where we have significant time dilation, but not a total stopping of
the clock.

I put you in a box and ship you via GalactEx to Alpha Centauri. When you
arrive, the addressee refuses to take the box, so it is immediately shipped
back to Earth. Because of the great speed of the GalactEx spacecraft, I
measure only 8.5 years went by from the time I handed the box over to
GalactEx and the time they returned it to me unopened. Because of the
effect of travelling at very high speed relative to Earth, it seems to you that
you have only been in the box for a couple of months. And indeed, the
clock you had with you inside the box shows that you were only in the box
for a couple of months.

Now I'll try something else. I use my matter transducer to transform you
into a photon. The output of the transducer is carefully aimed at Alpha
Centauri. You, in the form of a photon, are shot out of the transducer at
the speed of light, headed for where Alpha Centauri will be in 4.2 years.
When you get to Alpha Centauri, you hit a strategically-located and very
carefully-oriented mirror and are reflected back toward where Earth will
be in 4.2 years. You, in the form of a photon, eventually hit the photon
detector of my matter transducer, which transforms you back into a living
human being, or close enough.

To me, 8.4 years have gone by. To you, no time has gone by at all. You
are now standing at the output of my matter transducer instead of at the
input, but no time has passed for you since I started to transform you into
a photon.

I think we can agree the astronomer is viewing the planet as it was several
light seconds previously so we know the photon cannot go back in time to
when the target was observed, however from the point of view of the laser
photon can we really say the planet will not only move the distance it should
move in the time it takes a photon to travel there, but twice that distance in
(from the photons perspective) zero time?
The planet had already moved half the total distance at the time the photon
was launched. An astronomer on Earth sees Mars where it was about five
to twenty minutes earlier, depending on how far apart the two planets are.
Shooting the photon directly at the location viewed in the telescope sends it
to a location that is ten or forty minutes or so out of date according to your
clock, and according to the clock of someone on Mars. (Clocks on Earth and
Mars running at practically identical rates because the relativistic effects are
miniscule.) The photon doesn't know or care what is going on. It not only
doesn't know where Mars is, it isn't even aware of its own existence, just
as you were not aware of your trip to Alpha Centauri and back as a photon.

-- Jeff, in Minneapolis

George
2010-Apr-10, 04:46 AM
I think we can agree the astronomer is viewing the planet as it was several light seconds previously so we know the photon cannot go back in time to when the target was observed, however from the point of view of the laser photon can we really say the planet will not only move the distance it should move in the time it takes a photon to travel there, but twice that distance in (from the photons perspective) zero time? You ask an interesting question.

If the laser is aimed by the correct amount of angle ahead of Mars in its orbit, as a shotgun is aimed in bird-hunting, then both the observer who shot the laser and the photons themselves will experience the photons impacting Mars. Since the photons traveled to Mars in zero amount of time yet Mars had to move ahead in its orbit, which takes time the photons don't have, how could both the observer and photons agree that Mars was hit? [I think I'm restating your question, right?] The answer is that you can't mix the two frames together and think that what the photon sees is what the observer sees. For the photons, Mars never moves, yet it is directly in its path. For the observer, Mars has to move into the pathway of the laser pulse. The photons can't ever see what the observer sees as if they conducted chit-chat with the observer about where Mars is before leaving, else they would be in the observer's frame, which is never allowed for photons. [Hopefully, Ken or others will explain this better as I am still trying to get my head around SRl.]

max8166
2010-Apr-10, 09:28 AM
We can simplify this by using the moon as our target, this will make the observer relatively stationary and make the distance a measurable quantity.
(I'm going to round the numbers for easy maths!)

Moon orbits earth once a month so travels at 2 x pi x 400,000 = (well it works out about) 1000 meters per second.
Exact numbers : http://en.wikipedia.org/wiki/Moon
And so our intrepid astronomer mounts his laser gun on his most powerful telescope, lines them up exactly but remembers there can be a discrepany of 2.6 x 1300 meters, so includes this in his telescopes field of view.

Now he expects to see a laser lit point on the surface after 2.6 seconds as this is the time it takes for the light to travel to the moon and be reflected back but only expects the moon to move by 1.3 x 1000 meters during the time the light travels to it.

So what does our astronomer see?

George
2010-Apr-10, 04:47 PM
Now he expects to see a laser lit point on the surface after 2.6 seconds as this is the time it takes for the light to travel to the moon and be reflected back but only expects the moon to move by 1.3 x 1000 meters during the time the light travels to it.

So what does our astronomer see? What you've stated looks correct. Why is this a question?

[Added: Perhaps this will help... The laser gun had to be aimed 2.6 km in front of the Moon -- 1.3 km due to the time lag for light to travel from the Moon to the observer and 1.3 km to allow for the motion of the Moon during the 1.3 seconds it takes the beam to hit the Moon.]

noncryptic
2010-Apr-11, 02:14 PM
What does "time stops for the photon" mean?

In practice: nothing.
It's fun to consider the Frame of Reference of a photon, and it quickly leads to the conclusion that it is useless, because you can't do useful calculations with zeros en infinities.

In Relativity it is absolutely crucial to not mix Frames of Reference: any observation is only valid in the FoR of the observer making the observation.

So if you consider de FoR of a photon, its observation of time and space is valid only in de FoR of the photon, and not in the FoR of the observer who fires the laser beam.

The tendency to mix FoRs comes from the fact that in every day life the relative velocities involved are a tiny fraction of the speed of light so that the differences in observations due to difference in velocities are not noticeable.
So in every day life there is no need to consider different FoRs; in effect there is only one FoR, and the notion that observations do in fact depend on the relative velocities of the observers is counter-intuitive to that.