Length Contraction Question?

1. Sep 1, 2004

CollectiveRocker

Here's another one for ya'll. A spacecraft antenna is at an angle of 10 degrees relative to the axis of the spacecraft. If the spacecraft moves away from the earth at a speed of .70c, what is the angle of the antenna as seen from earth?

2. Sep 1, 2004

da_willem

Seems to me like a classic homework problem. Try making pictures, find the right formulae and solve. If you get stuck show us what you got, and we'll try to help you.

3. Sep 1, 2004

chronon

I don't like the as seen from earth bit. It's straightforward to calculate how the shape in the earth's frame differs from that in the spacecraft's frame. However, as seen from earth asks what someone on earth would see. They certainly don't just see the spacecraft shortened by the relevant factor, rather they see it rotated. I doubt whether the angle they see would agree with that from the calculation.
(Of course there is also the non-relativistic problem that the angle as seen from earth depends on the orientation of the spacecraft.)

4. Sep 1, 2004

da_willem

Although I am aware of the fact that what an observer will see is different from what is, or can be calculated (I think you are referring to Terrell Rotation?), I think this is not the intent of the question.

5. Sep 1, 2004

CollectiveRocker

What is the correct formula to use for this problem?

6. Sep 2, 2004

Chronos

The Lorentz contraction.

7. Sep 2, 2004

CollectiveRocker

What is the Lorentz contraction?

8. Sep 2, 2004

da_willem

An object moving with a speed v can be shown to be contracted in the direction of its motion. The new length in terms of its 'rest length' will be:

$$L=L_0/ \gamma$$ with $$\gamma=\frac{1}{\sqrt{1-v^2/c^2}}$$

This length contraction is called Lorentz contraction. How is it that you do relativistic problems without knowing Lorentz contraction?

9. Sep 2, 2004

CollectiveRocker

So how to I proceed with the problem? Essentially right now, I have two right triangles representing the angles and components. Yet therein lies the problem. How can you use trigonometry is you only know one thing?

10. Sep 2, 2004

Staff: Mentor

Hint: Length is contracted only in the direction of motion.

11. Sep 15, 2004

juju

Hi,

You have a right triangle. The hypotenuse is the antenna. The long side is the ship. The short side is distance from the ship to the antenna perpendicular to the ship. Apply the lorentz copntraction to the long side only. You will then have a new right triangle with the same short side but different long side and hypotenuse. From this you can recalculate the angle.

juju

12. Dec 26, 2004

Dathascome

I'm studying relativity on my own and am having a problem along these lines. The problem I'm doing is similar. I have a ruler that is
moving with a speed .8c relative to a frame S and is 1 m long. First question is no prob, they just ask the length if the ruler is traveling parallel to it's velocity.
Second they ask what is the length if the ruler is moving at an angle of 60 deg as seen from mthe rest frame of the ruler. This was ok too, I just put the ruler as the hypotentuse with 60 degree angle and recognized that only the component of length parallel to the velocity will contract and used the same perpendicular length (since it is unchanged) and thus found the length in this frame.
The third part I'm confused about. They ask what is the length when the ruler makes an angle of 60 degrees with the velocity as seen from S. Now my problem is I'm confused about what length is my proper length in this situation. I definitly don't think it's the intial 1m. It seems to me like it would be the length found in the previous part of the problem, since that is the rest frame of the ruler, and hence it's proper length. But didn't having that length also go hand in hand with a differet angle? So can it still be the proper length in S? What I'm thinking is that I just use this as my proper length, say that only the component of this length parallel to the velocity will contract and thus find the length as seen from S. This is an odd problem in the book so i have an answer, which is .832, where as my answer is .84, but I feel like I'm doing something wrong because for this sort of problem i feel like this is to much error, and that I should have soemthing closer.
Thanks for the help in advance.

13. Dec 26, 2004

juju

Hi,

I think the general idea that you need is this.

There is no contraction if the rulers length is measured by an observer in the rest frame of the ruler. (The ruler does not change length relative to its own frame.)

The length contraction is only measurable by the observer in the frame to which the ruler has a realative velocity.

juju

14. Dec 27, 2004

Staff: Mentor

The proper length of the ruler is its length as measured in its own rest frame: that length does not change. As measured from frame S, that ruler will be a different (shorter) length. Realize that the y-component is not contracted. Now apply some trig and the pythagorean theorem to solve for the measured length of the ruler. (The book's answer is correct.)