# Apparent position and light travel time

1. Nov 16, 2007

### gonegahgah

This is a thought of mine... or popping my 2 cents into the pot.

I think it is fine to have the speed of light as constant.

It does seem to me that we are missing one important facet when I consider it.

It has to do with position.

Where things are...

... relative to us.

I would suggest the following important distinction:

> where we observe objects to be is where they would be now (if they still exist)

> not where they were in relative terms to us when the light 1st began its journey...

This seems to be counterintuitive doesn't it?

Effectively what I am saying is that by my view we essentially see the stars about where they would be now; and not where they were relative to us when each light particle first began its journey.

I only mention this for two reasons:

1. it doesn't seem to have been mentioned.

2. because it seems to me that in general that nothing about the current relative position seems to be implicitly stated but it could be an underlying assumption that may be creating bias in understanding.

ie. We are assuming we are seeing things where they were; whereas I am saying that we see their position as being about where they would be now.

2. Nov 16, 2007

### Staff: Mentor

Not just counterintuitive, but wrong as well. Why would you think that?

3. Nov 16, 2007

### gonegahgah

Hi Doc

Well when they send probes to the planets in the Solar System I don't recall coming across them talking about working out how to go to a planet where it is now relative to where it was when the light first reflected back from them.

Do they?

Isn't the process:

1. Observe where it is now.
2. Calculate its distance and direction.
3. Calculate where it will be from that.

If the flight time of light were involved the process would be:

1. Observe where we see planet.
2. Calculate its distance and direction.
3. Work back to where it was when the light was sent.
4. Work out where it will be from there.

Light takes about 8 minutes to travel from the Sun to us on Earth.
We are on mean 92.9 million miles (149.6 km) from the Sun.
Pluto is on mean 3666 million milles (5900 km) from the Sun.

That's about 40 times as far so at best light will take 320 minutes to reach us or over 5 hours. Pluto would travel quite a distance through space in 5 hours wouldn't it?

The average orbital speed of Pluto is 4.7km/sec.
In 5 hours it would travel approximately 84600 km.

What do they do?

Also I have a whole series of diagrams that I believe show how it works which I have created. Can diagrams be incorporated in these posts? Is it permissable for me to show what I am meaning?

Last edited: Nov 16, 2007
4. Nov 16, 2007

### Staff: Mentor

All we can directly observe is the light that reaches us "now", which left the planet some time ago. So it seems a triviality that we are seeing the planet (or stars or galaxies) where they were when the light was emitted.

Sounds right to me.

All the more reason to think that your view (ignoring light travel time) is mistaken. If the Sun (or Pluto) exploded "now" how would we know until the light reached us? It could have exploded 8 minutes ago, yet we still see it shining "now".

Why don't you read this website http://www.cv.nrao.edu/~rfisher/Ephemerides/ephem_use.html" [Broken], which discusses figuring out the position of objects in our solar system. Here's a quote from a section titled "Time Delay and Apparent Direction":
"When we look at a planet we are actually seeing the planet where it was when its light left the planet. This could be minutes or even hours before the current time. The procedure for compensating for this time delay is to compute the distance to the planet at the time of observation. From this compute the light travel time, recompute the planet's position for current time minus light travel time, and use this earlier planet's position with the current observatory, moon, and earth-moon barycenter positions in the Planet(a,b,c) equation above." ​
I'm no astronomer, but this seems eminently reasonable.

You can certainly upload attachments and images. But I'm not sure it will help.

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5. Nov 17, 2007

### gonegahgah

Thanks for your help Doc. Thanks for that site.

Can I just correct something of mine? I wrote the following steps for working out where a planet will be to send a probe to it (if we see planets where they were when the light left them):

1. Observe where we see the planet. (ie. where it was)
2. Calculate its distance and direction.
3. Work out where it is now. (I had: Work back to where it was when light was sent).
4. Work out where it will be from there. (for probe to meet it)

If we are seeing planets where they were when the light was first reflected (or from direct radio emissions) don't we have to work out where they are now?

Last edited: Nov 17, 2007
6. Nov 17, 2007

### Staff: Mentor

Yes. If you want to know where they are now, you have to take into account light travel time and work it out.

7. Nov 17, 2007

### gonegahgah

Cool. That means that the planets should be a little further on in their orbit, than where we see them to be, under this principle, at a time that we see them, because we see them where they were not where they are now.

I'll just reexamine that section you grabbed from the JPL page. I've included the first sentence as well.

So the procedure explained is:

1. compute the distance to the planet at the time of observation
2. from this compute the light travel time
3. recompute the planet's position for the current time minus light travel time
4. use this earlier planet's position

Ummm, why are they wanting the earlier planet's position?
Don't we want the current planet's position?

Why are they working back in time (current time minus light travel time)?
Shouldn't they be advancing the position of the planet for the current time taking into account how far more it would have travelled in the time the light travelled to us?

What use is an 'earlier position' in the planets orbit?

Last edited: Nov 17, 2007
8. Nov 17, 2007

### Staff: Mentor

Right.
The goal of the emphemeris is to find the relative position of all bodies at the same time. So, to make use of an observation, one must take into account the light travel time.

It depends on what you are trying to do. If you want to take current observations and calculate the position of planets now, you must take into account light travel time and project where the planets would be now. If you don't account for light travel time, you will misinterpret your observational data.

[Since this discussion has nothing to do with the original thread topic, I will split it off into its own thread.]

9. Nov 17, 2007

### gonegahgah

Thanks that is true. I thought "where have our posts gone" for a second. Then I found them.

I understand what you are saying but this is not equivalent to the steps outlined at JPL is it? Rather than a final position their result is an earlier position?

Their 4 steps outlined seeks to reverse the time of the light travel and find an earlier position for the planet which seems odd to me?

ie. step 3 - "recompute the planet's position for the current time minus light travel time"
ie. step 4 - "use this earlier planet's position" (rather than the extrapolated position)

Aren't these steps, our example ones, and those on JPLs page, different?

ie.

1. "compute the distance to the planet at the time of observation" = observe where we see the planet.
2. "from this compute the light travel time" = work out the time light took to journey here.
3. "recompute the planet's position for the current time minus light travel time" <> multiply that time by calculated speed to add distance travelled to the current position.
4. "use the earlier planet's position" <> use the extrapolated planet's position.

I use "multiply time by speed" as a simplification.

Even if it is possibly not what they meant - though it would surprise me for JPL to have their site incorrect - I am quoting their steps correctly aren't I?

Last edited: Nov 17, 2007
10. Nov 18, 2007

### Staff: Mentor

If you want to predict where in the sky to look right now to see Mars, for example, you need the actual position of Mars not now, but where it was at "current time minus the light travel time" (which is in the past). That's what they were describing. (I admit that it could have been a bit clearer.)

Using the emphemeris (which is a computer model best fit to loads of empirical data and gravitational modeling of all the bodies in the solar system) you can figure out very precisely where anything is at any time, future or past. The point is that light travel time must be accounted for in order to make sense of observational data.

11. Nov 18, 2007

### gonegahgah

So in other words the process they are showing is how to work out where to point the telescope at a particular time to see the desired planet?

ie.

1. Choose the time that you will be observing the planet.
2. Use the ephemeris data to work out where the planet will physically be at that time.
3. From this calculate the distance to the planet.
4. Calculate the length of time light takes to travel that distance.
5. Work out using the emphemeris data where the planet was that amount of time ago.
6. When the observation time arrives have your telescope pointing in the direction worked out in 5.

And voila the planet should appear in the middle of the telescope.

The above steps correspond to JPLs text then as:

1. "at the time of observation" ie when you choose to observe
2&3. "compute the distance to the planet"
4. "from this compute the light travel time"
5. "recompute the planet's position for the current time minus light travel time"
6. "use the earlier planet's position"

With the help you've given me my reworded steps above now hopefully accurately represent the steps explained by JPL? Is that correct Doc?

12. Nov 19, 2007

### pervect

Staff Emeritus
I took a quick look at the Nasa website, and I don't see any mention of how or where they calculate the geodesic path that light takes (to take into account gravitational lensing). Which strikes me as a strange omission.

13. Nov 19, 2007

### Staff: Mentor

Looks like you've got it.

I see no mention of that either. But since this is a solar ephemeris, which uses the standard background of "fixed" stars as a coordinate system, why would it matter?

14. Nov 19, 2007

### pervect

Staff Emeritus
I suspect that any such effects would be small, but I don't know their magnitude offhand. Basically, I've been strongly influenced by the GR approach to the problem outlined in Precis of General Relativity.

This is basically a statement that we can assign coordinates to events in the solar system, and then use GR to compute both the orbits of the planets and the orbits of the light beams - that's all we need to explain observations. But the orbits of the light beams would not in general be a straight lines. Philosophically, the main point is simply that it is possible to assign coordinates to events in the solar system, and that all we need to do is to compute the orbits of light and the planets.

I'll note in passing that the GR issue of "what constitutes now" in the solar system is tricky, its closely related to the issue of actually defining a specific coordinate system for the solar system. It appears that the "now" used by the ephermis programs at NASA is equivalent to the theoretically defined but in practice apparently little used "TCB" defined by the IAU according to

http://aa.springer.de/papers/8336001/2300381.pdf

so that the "now" of the ephermis is the "now" of the IAU coordinate system.

I hope I haven't "hijacked" the thread, I'm just a little puzzled / concerned by the manner in which the computations are carried out, using a "pure GR" perspective.

Last edited: Nov 19, 2007
15. Nov 19, 2007

### gonegahgah

That's cool pervect.

The following site http://aa.usno.navy.mil/data/docs/geocentric.php" [Broken] mentions gravitational lensing in its notes under Apparent position.

Thanks for checking my steps Doc. Hopefully they are correct now.

The steps that interest me most are:

1. "at the time of observation" ie when you choose to observe
2&3. "compute the distance to the planet" ie to where it will be at that time
4. "from this compute the light travel time"

The reason I am interested is that they are measuring the distance to the present position and not the distance to where the planet was when it first emitted the light.

They then do steps 5 & 6.

This measuring to the present position (and the effect this has) is the distinction I am hoping to make. Can you check this and see if it is still okay.

From this I hope to then demonstrate what I am saying with diagrams.

Would you agree that what they are saying is that they are measuring the distance to the present ephemeris position of the planet and from this working out how long the light has been travelling to us? (By 'present' I mean at the time the observation will be made).

Last edited by a moderator: May 3, 2017
16. Nov 19, 2007

### Staff: Mentor

Now that you mention it, those steps aren't quite correct for just the reason you point out. (The website was a bit ambiguous as well.) To find the apparent position of a solar body now, you need it's actual position at the time it emitted the light. That is based on its distance then, not its distance now. Anything else would make no sense.

Not OK.

That does sound like what the website said, but I attribute that to sloppy writing. Obviously the light travel time depends on the distance the light had to travel to reach the observer, which is the distance to where it was when the light was emitted, not to where it is now.

Believe it or not, this wiki page seems to have a more accurate description of what's actually done: http://en.wikipedia.org/wiki/Light-time_correction" [Broken]. Here's a relevant quote:
"A rigorous calculation of light-time correction involves an iterative process. An approximate light-time is calculated by dividing the object's geometric distance from Earth by the speed of light. Then the object's velocity is multiplied by this approximate light-time to determine its approximate displacement through space during that time. Its previous position is used to calculate a more precise light-time. This process is repeated if necessary, but one iteration is usually sufficient given the slow movements of planets."​

Last edited by a moderator: May 3, 2017
17. Nov 20, 2007

### gonegahgah

You have done well with your researching Doc. You have been able to locate more information on this subject than I ever could.

This is a valuable exercise for me. I appreciate that you obviously have a great deal of interest in the fields and it would seem to me that in your 'teaching' (or elucidating) role that you develop a more rigorous insight of your own. I do like that it is a growing experience.

I must admit it is a little scary that Wikipedia should be more accurate than the JPL site.

Unfortunately Wikipedia only gives two references in relation to their page on "Light-time correction" that you found for me. ie:

P. Kenneth Seidelmann (Ed.), Explanatory Supplement to the Astronomical Almanac (Mill Valley, Calif., University Science Books, 1992), 23, 393.
Arthur Berry, A Short History of Astronomy (John Murray, 1898 – republished by Dover, 1961), 258-265.

They do make reference to observations of eclipses of the moons of Jupiter by Ole Rømer in 1675 as a reference to some experimental data. I have already had a good look at that experiment a while ago so I will have to find my notes on that and revisit it to see how it applies.

I am also taking what you have helped me to understand about what is said and am in the process of exploring it diagramatically. I'll get back with this.

18. Nov 20, 2007

### pervect

Staff Emeritus
I'd suggest looking at ray tracing as a way of determining the apparent position of objects. In this case, all you need is one level of ray-tracing, which wikipedia calls ray-casating.

The physical process of vision is that light originates from some source, reflects or refracts from various objects, and is then seen by the observer.

What you do in ray-tracing, or ray-casting, is to reverse time. Instead of tracing the rays forwards in time, you trace them backwards. So you start out at "now", and then trace the wavefront of light backwards in time.

The light beam that hits mars (the time evolution of both mars' orbit and the lightbeam is time-reversed, so they both go backwards in time) is the one that you see "now" in the apparent direction of mars.

Note that this is very similar to a general ballistics problem, where the goal is to hit some moving target, but you are shooting light beams, rather than bullets, and you have reversed time. Because the trajectory of the planets and the light are governed by differential equations, this is an example of a boundary value problem.

Ray tracing (or ray casting) has been used in standard 3-d graphics, in special relativity, and in GR. You'll find a number of references on the WWW, among them are:

http://www.anu.edu.au/physics/Savage/TEE/site/tee/learning/raytracing/raytracing.html [Broken]
http://www.anu.edu.au/Physics/Savage/RTR/

the latter will possibly run on your PC (depending on your graphic card).

you might also want to look at the thread

for more suggestions.

Last edited by a moderator: May 3, 2017
19. Nov 22, 2007

### gonegahgah

At the moment I just use Corel Draw 12 to visualise things. Ray tracing would be interesting. And those sites too.

Doc, I will be getting pressed for time heading up to Xmas but I would like to keep exploring this as I can with you.

In the meanwhile while I'm considering what we have discussed I would like to get to an understanding with you on some diagrams that I have done. Hopefully this will help as well.

The first one is attached.

It shows a planet circling a sun. The sunlight travels out from the sun and the planet passes through the sunlight - like a sunlight shower - while the planet journeys in its orbit.

It has to do again with working out how long the light has travelled.

The diagram shows one photon making its way out to and colliding with the planet as the planet passes through the path of the photon.

Looking at the diagram would you say that we would measure how long the photon has been travelling by measuring along the path it took from the sun to the Earth - that it eventually collides with it - using this diagram? What problems are there with this?

Forgetting things like gravitational lensing for the moment.

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20. Nov 22, 2007

### Staff: Mentor

I don't see any problems at this point. Seems reasonable to me. What do you plan to conclude from this?

21. Nov 23, 2007

### gonegahgah

Nothing yet Doc. I'm just taking it step by step to demonstrate how complex it can be to think about.

However saying that, the first diagram I have just presented is how I would tend to think that things would work.

As we proceed though it may be your conclusion that it has flaws just as you found for each of the steps I took (until the final information you provided me from Wikipedia) for our examination of how to measure the time the light has been travelling.

I am providing a second diagram now which is not much different to the first. However it looks more at the Wikipedia conclusion that we measure the distance the light has travelled relative to where the sender was; not where it is now.

Actually before I do that I would like to see if you agree with my current thinking from another diagram first. This shows the sun again with the Earth travelling around it in Earth's orbit. (The distances and shapes are exaggerated of course).

Would you agree that - as the diagram is depicting - that the sunlight moves out from the sun like a sun shower and that the Earth passes through this shower? So in the diagram, would the Earth see the sun as it appears via the yellow cone? I drew the cone because we see the sun as having width and not just as a point. Meaning of course that some light travels from the edges of the sun to us; instead of it all travelling directly outwards.

Would you agree that the grey cone would continue out into space? It would actually be more parallel of course for far distant observers.

All in all is this diagram currently acceptable for the moment?

Just adding some clarification of my diagram. The drawn Earth on the left is depicting the position of the Earth when the sunlight reaches it and the first circle on the right depicts the position of the Earth when the photon begins it journey. The same goes for the photons where the photon begins at the sun and 4 time intervals are shown out towards the Earth matching the 4 time interval positions of the Earth in its orbit.

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22. Nov 24, 2007

### gonegahgah

Doc, I'm not sure what my conclusion is yet but I am enjoying the journey. I do have some preconceptions which is why I am taking things step by step. I hope that you won't regard me as wasting your time as I am learning a lot out of it.

I've animated my diagram below - something I just learned - to hopefully give a clarified example of what I am representing rather than the static image.

It is meant to show in the diagram that we pass through light from a paticular face of the sun as that face moves out from the sun. This is depicted by the light yellow cone and also by one of the photons in that face.

The face that is facing where the Earth was when another photon began it journey is also shown and depicted as continuing out into space - unseen by the Earth - via the grey cone.

For this stage I am just wanting to confirm if that looks okay so far? Or do you see early flaws with it?

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23. Nov 24, 2007

### Staff: Mentor

Cool animation! Seems OK to me, but that depends on what you plan to conclude.

I'm happy to see that you seem to have abandoned your initial stance that ignored light travel time.

Last edited: Nov 24, 2007
24. Nov 25, 2007

### gonegahgah

Thank you Doc. I'm not sure what exactly my conclusion will be. You have made me think hard and I am still working on diagrams to help me to come to a clearer understanding of it.

My initial 2 cents thought was that we see things about in the position where they are now - if they continue in a straight line - though I did think that the picture was a picture from the past.

I did realise that it took time for the light to travel between emitter and receiver but I thought that the observed position moved with the emitter.

I'm not completely shaken off that just yet (give me time) though my tree is certainly looking more leafless and unhealthy after your efforts to help me out of it!

Actually, it seems to have been the main intent of Ole Rømer - from what I understand in his experiments on the moons of Jupiter (specifically Io) passing through the shadow of Jupiter - to prove that the speed of light wasn't instantaneous; which was one of the prevailing thoughts at the time. Ole Rømer of course is the bloke mentioned in the Wikipedia page that you provided me with.

Later, others used his data to determine a rough speed of light of about 200 000 km/sec.

I have done another diagram now. This is the diagram that I had planned to do up last time that I mentioned.

According to what we have, from the Wikipedia page that you provided me with, to work out the time the light has travelled we have to work out the distance the emitter was from the receiver at the time that the emission was sent.

Using the ephemeris data will only give us the current distance to the emitter (planet, etc.) and not the distance to the emitter at the time the emission was sent. So a sort of iteration process was used to get a more accurate figure of how far the emitter was away at the time the emission began. (Though it was noted that the planets don't move very fast in the sky; I guess in a similar way to how on a train the further trees appear to be moving slower).

Back to my diagrams.

In my previous animation I used the distance the photon travels out from the sun to the Earth at the time of collision to determine the distance the photon has travelled to the Earth. Is this wrong?

The photon began its journey out at a position not directly in line with Earth but just happens to collide with the Earth as it passes. If you swing the distance depicted by this around to where the Earth was - as I have done in the left diagram - then you can see that the distance - shown as d1 - falls short of the Earth.

So going on the Wikipedia principle - that the distance that must be used is the distance between emitter and receiver at the time of emission - then should we take the distance between where the photons emit from and where the Earth is when they emit do determine how far the light has travelled and then work back using this to determine how long the light has been travelling?

So is the amount of time the light takes to travel - outwards from the Sun to the Earth - equal to the speed of light divided by the distance between where the photon emitted from and where the Earth was at the time of emission (i.e. at an angle to the emission rather than in the direction of the emission)? This distance is shown as d2 in the right diagram.

I still haven't come to that conclusion myself just yet (still working on those diagrams) but it would seem to be in better agreement with the Wikipedia page. What do you think Doc?

Just adding this to hopefully make my question a little clearer: Is it correct to say that by Wikipedia's principle that to work out how long the light has travelled, along the path shown by d1 to where the Earth will be, we would have to divide d2 by the speed of light?

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25. Nov 26, 2007

### gonegahgah

Hopefully I can clear up what I am saying with the following diagram Doc if the previous is looking a little illegible.

The first diagram on the left shows what is represented by the Wikipedia page. It says that at time 5 when we wish to make our observation that the ephemeris data for this example will tell us that the planet is at position 5 but the light will have actually travelled from position 1 to us via the dark yellow path. In this respect it tells us that we must measure the distance of flight of the light along the path between us and position 1 via this dark yellow path and use this to determine how long the light has been travelling.

The second diagram in the middle is supposed to be the same as the first diagram except that we are moving the relation of the actors (the planet, the photon & the Earth) around the planet; instead of around the Earth as in the first diagram. If you check the relative distances of the actors for each time step you will see that they match exactly between the two pictures. i.e. at time 1 the relative distances are the same, at time 2 the relative distances are the same, etc.

You can see that the middle diagram looks more like our Earth orbiting the Sun example which is shown on the right.

For the moment Doc are you firstly in agreement that the left diagram and middle diagram spatially represent the same occurrence as each other?

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