# Looking out the window of a fast ship

1. May 3, 2007

### Taremos

Okay, for the moment, forget about time dilation.

Are the following descriptions accurate if you do not account for time dilation?

I've been going over and over this in my head and I can't figure out why any of these don't make sense.

Setting: A ship going .99c.

1
Attached to the rear of the ship there is a pole pointing parallel to the direction in which the ship is cruising. The pole is 5 light seconds long. Looking out a rear-facing window which is directly next to where the pole is fastened you see the pole like looking up a tall building from it's base. Even after correcting for the illusion of inflated height caused by standing so close to the base, the pole appears to be 500 light seconds long, not 5 light seconds long.

2
There is another pole 10 meters long attached perpendicular to the direction of the ship, to the side of the ship. You are able to view this pole from a side-facing window, 10 meters behind where the pole is mounted to the hull. Looking out at it the pole appears to be shorter than it actually is. Also, looking at the portion closest to the hull it looks normal, but as you look at the pole further away from the hull, it appears to be warped so that you see it as if you were looking along at an angle much smaller than the one you are.

3
Looking at the same pole as mention in 2, you are able to view it from another view port 10 meters away from it. This time the view port is closer to the front of the ship than where the pole is mounted. Looking back at the pole it appears to be much further away and much much smaller than it actually is and the angle at which you see it does not appear correct. It looks like the pole is much further away than 10 meters.

4
There is yet another pole 5 light seconds in length attached to the hull of the ship. This pole is faceted to the front of the ship, parallel to the direction in which the ship is traveling. Similar to the rear-facing pole, there is a view port right next to the base of this pole. Peering out this port gives you a few of the pole from it's base, but the pole appears much closer than it is and is distorted so that each part of the pole appears to be about half the distance of what it is. It still appears to be 5 light seconds in length, but looking at the tip, the tip it looks as if you are only about 2.5 light seconds from it.

5
You observe a planet from the ship which is 10 light minutes away and almost directly in front of the ship, the ship will miss it by only a few hundred kilometers. You look upon the planet and judge it to be 10 light minutes away. A minute later you look again, and it appears to be about 9 light minutes away. Another minute passes and it looks about 8 light minutes away.

2. May 3, 2007

### daniel_i_l

I didn't have time to read everything but you cant just "forget" about time-dilation. If you do than lots of paradoxs can arrise, and even before that, time dilation comes from the same equations as length contraction so how can you use one but not the other? Whatever conclusions you get to by forgeting about time dilation aren't physics, or anything else that i can think of for that matter.
And in (1), you say that the pole is 5ls long, in which frame?

3. May 3, 2007

### Hurkyl

Staff Emeritus
Here's where you went wrong: there's no such thing as an absolute velocity. Velocities can only be measured relative to nearby objects, or relative to a coordinate chart.

So, let's define some charts:
Let X be an inertial coordinate chart where the ship is going 0.99c forward.
Let S be an inertial coordinate chart where the ship is at rest.
Let W be an inertial coordinate chart where the ship is going 0.99c sideways. (in the direction of the second pole)

You subsequently state a lot of measurements -- but they all have the same problem: you don't specify in which coordinate chart they are measured. Normally, when one describes an object without qualification, they mean how it appears when at rest. I will assume that's what you meant throughout this reply.

No. S sees a 5 light-second long pole. W also sees a 5 light-second long pole. X sees a very, very short pole.

I.e. in terms of what you actually observe, you see a 5 light-second long pole.

No. S sees a straight 10 meter pole. X sees a straight 10 meter pole. W sees a straight, very short pole. The pole is straight in all three coordinate charts; but an observer in X will see a "warped" pole due to the difference in time it takes for the light from different parts of the pole to reach him. But when he corrects for that, he will find the pole is straight. Observers in S and W will see no warping.

I.e. in terms of what you actually observe, you see a straight, 10 meter pole.

Nope. Everything is the same as in 2. You see a straight, 10 meter pole. The pole is always perpendicular to the ship in X, W, and in S (though not in most charts). I don't know for sure, but it's plausible an observer in X might actually see a non-right angle due to light transit time. Observers in S and W will always see a perpendicular angle.

There is absolutely no difference between this pole and the rear pole.

I will assume you meant the planet appears 10 light-minutes away in S's chart. In that case, after k minutes, the planet will appear at a distance of
10 light-minutes - 0.99c * k minutes

So, if my assumption was right, you have this one correct.

4. May 3, 2007

### Staff: Mentor

Maybe I'm missing something (and frankly, I didn't read past the 2nd case...), but it sounds pretty straightforward to me: a pole attached to your ship isn't moving wrt you, so they all look like they are the "right" size. Time dilation and length contraction don't come into play.

5. May 3, 2007

### Taremos

[QUOTE="Hurkyl" about #1]No. S sees a 5 light-second long pole. W also sees a 5 light-second long pole. X sees a very, very short pole.

I.e. in terms of what you actually observe, you see a 5 light-second long pole.[/QUOTE]

Since there's plenty to talk about here, lets start with #1...

I don't understand why that is. Okay, light hits the pole and begins the trip to the view port. The light that is coming from the pole is going 1c, since light always goes 1c regardless of the velocity of the object it is emerging from.

So lets start with light coming from the tip of the pole, a distance of 5 light-seconds away from the ship (1 498 962 290 meters from the ship). The light is traveling at 1c. The ship is traveling in the same direction as the light that will eventually find its way into the view port at the rear at .99c.

If the ship was still (in relation to light), the light would get there 5 second after it left the tip of the pole. But because the ship is moving at .99c in the same direction as the light, the light is only moving at .01c relative to the observer in the ship, thus it takes 500 seconds to get to the observer. Since it took 500 seconds, the light has spread out that much more and it will appear that much further away... 500 light seconds away.

Am I missing something?

6. May 3, 2007

### matheinste

Doesn't light move at c relative to all observers.

Matheinste.

7. May 3, 2007

### JesseM

When you say "forget time dilation", are you also forgetting about length contraction? Do you just want to know how these things would look in a Newtonian universe? Or are you having doubts that in relativity the pole would appear the same length to observers on the ship regardless of the ship's motion? If you're asking about relativity, then you can't ignore time dilation and length contraction, they are essential to understanding why it looks the same.

Last edited: May 3, 2007
8. May 3, 2007

### paw

Yes.

Relative to you, the observer on board the ship, neither the ship nor the pole is moving. Therefor it takes light just 5 light seconds to reach you from the tip of the pole. This is what Hurkyl meant by frame S. The fact that the ship is moving in a different frame is immaterial to your question.

9. May 3, 2007

### JesseM

But it's still true that all frames must agree on their physical predictions, so you could interpret Taremos' question as a question about how the predictions of the two frames can be consistent. You're certainly free to calculate what the observer on the ship will see using a frame where the ship is moving at 0.99c rather than the ship's rest frame, although it's simpler if you use the rest frame.

10. May 3, 2007

### paw

Absolutely, but his question was specifically asked in the ship centered frame.

11. May 3, 2007

### JesseM

He was asking what the observer on the ship would see using light-signals, but the numbers he used seemed to suggest he wanted to calculate this from the point of view of the frame where the ship was moving at 0.99c.

12. May 3, 2007

### paw

Ok. I thought that was what I said.

13. May 3, 2007

### JesseM

What the observer on the ship sees with his eyes is a physical question which has nothing to do with what frame you use. You said that "his question was specifically asked in the ship centered frame", but I don't think Taremos said anything to indicate he was using the ship's frame.

14. May 3, 2007

### pervect

Staff Emeritus
I'm not sure what the Original Poster (OP) meant when he said "forget about time dilation".

Also, since he was interested in "what you see", it's possible he was asking about Terrell rotation. There's a FAQ about this and a lot of google hits, too:

http://math.ucr.edu/home/baez/physics/Relativity/SR/penrose.html

I'll quote a brief excerpt, if it's relevant one might want to read the entire article:

15. May 3, 2007

### MeJennifer

What is called time dilation and length contraction is simply an effect on light signals between objects that are in relative motion with each other. The proper lengths and times of the objects in question are not affected by relative motion. Only acceleration (and by equivalence spacetime curvature) affects the future rate of accumulation of proper time.

Last edited: May 3, 2007
16. May 3, 2007

### paw

He said "relative to the observer in the ship". I took that to mean in the frame of the ship. If that's not what he meant then all bets are off....

17. May 3, 2007

### daniel_i_l

No light signals need to be passed to see the effects of LC ans TD. Just give each observer a grid of rulers and clocks.

18. May 3, 2007

### MeJennifer

You are wrong Daniel, without light signals one cannot measure time dilation and length contraction effects due to objects being in relative motion with each other. But as I said before acceleration (and spacetime curvature) does change the future rate of proper time accumulation for an object.

Feel free to demonstrate I am wrong by showing length contraction or time dilation without light signals by for instance using "chalk marks" on your grid showing the length of the object in relative motion or by directly comparing the difference in elapsed time of two clocks by bringing them together and without any of them ever accelerating or going through curved spacetime. It won't be possible, simply because, what I said before, length contraction and time dilation are effects on the signals between objects in relative motion not effects on the proper values of these objects.

Last edited: May 3, 2007
19. May 3, 2007

### daniel_i_l

Ok,
Lets say you have one observer (A) who has set up a system of evenly spaced clocks, each with their own observers (who are at rest relative to the clocks), that are all in sync in his frame. Now you have another observer (B) who is moving relative to that system. Observer (B) sets of a firecracker next to clock #1. The observer at clock #1 writes down the time shown on his clock (ta1) and observer B's clock (tb1) when he noticed the explosion. Observer (B) continues on and sets of another firecracker next to clock #2. The observer at clock #2 writes down the time shown on his clock (ta2) and observer B's clock (tb2) when he noticed the explosion. The observers at each clock send their data over to their frames central database there they find that:
ta2-ta1 =/= tb2-tb1
No light signals where needed. And the time it takes for the clock observers to see the explosions is neglegible as the clocks can be placed arbitrarily close.

20. May 3, 2007

### JesseM

In the first post he was just talking about what the observer would see, but you're right, in the next post he said "But because the ship is moving at .99c in the same direction as the light, the light is only moving at .01c relative to the observer in the ship", I had missed that part. That statement would be wrong in relativity, since as you say, the light is still moving at c in his frame.