# Homework Help: How much curvature are we talking about?

1. Oct 11, 2008

### KiyoEtAlice

1. The problem statement, all variables and given/known data
I've attached a picture which include the question -- I found it on the net, and it's exactly the same as my homework.

2. Relevant equations
I'm not sure which formular to use.

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2. Oct 12, 2008

### Redbelly98

Staff Emeritus

You must know something. Show what you have attempted, or think will solve the question.

3. Oct 12, 2008

### Almanzo

In space, a circle touching the bullet's track on the inside would have a much larger radius (hence less curvature) than a circle touching the ball's track on the inside. However, the problem is to be solved in spacetime. Now, the bullet takes less time to complete its track than the ball (which is evidenced by its rising and falling over less distance in the same gravity field to arrive on the same spot (in space) as the ball. The spacetime track of the ball is therefore much longer than the spacetime track of the bullet. If one uses the Euclidean metric, that is, but I think one must in this case.

The radius of a circle segment equals the square of the length of half its chord divided by its height. The length of the chord is nearly equal to the time difference times c, and this time difference can be calculated (approximated) by classical Newtonian means. The squares of the time differences should be in the same proportion as the heights, as both objects are rising and falling in the same gravity field. Therefore, the radii should be equal, and so must the curvatures be equal.

Last edited: Oct 12, 2008
4. Oct 12, 2008

### Redbelly98

Staff Emeritus

PF guidelines are that HW posters should show an attempt at solving the problem before giving help.

One implication of this is, HW answerers should not give help until the asker has made some attempt. Please let the asker show what they know first, before helping out.

5. Oct 12, 2008

### KiyoEtAlice

Sorry, that's because I've tried using simple geometry and I think it's worng, but I'm blank about what formula I should use.

I'm trying to understand this. I understand the last bit, but I'm just trying to really understand it in deep... nothing about your explaination, I'm just blank and tired. So if anyone can make it more clear, I would really appreciate it. And thanks very much too!

6. Oct 12, 2008

### Redbelly98

Staff Emeritus

Normal geometry is the way to go here. Draw a figure with the circular arc, and the circle's center (way below the arc)

There are two options for working this out:

Equation of a circle (in terms of radius and coordinates of the center)
or
Trigonometry

7. Oct 12, 2008

### KiyoEtAlice

Okay thank you. Although, I am a little stuck, what is the length the "chord" (the distance it traveled)?

8. Oct 12, 2008

### Redbelly98

Staff Emeritus

Not quite. It's the distance "straight across" the circle's arc.

9. Oct 12, 2008

### KiyoEtAlice

hmm... so in this case is it provided or do I have to work it out? I'm still a little confused about what's been given :(

10. Oct 12, 2008

### Redbelly98

Staff Emeritus

You'll have to work it out. See your figure ... it's not simply the "10 m" distance.

11. Oct 13, 2008

### KiyoEtAlice

Okay, now I'm stuck agian. I'm not sure which formula to use to calculate it, or how to calculate it :(

12. Oct 13, 2008

### Almanzo

I am unsure about the source of the confusion.

There are three things to be considered:

1. To calculate the time needed to fall from a certain height in Earth gravity.

2. To calculate the length of a line segment in spacetime.

3. To calculate the radius of a circle, if a chord and and height of a circle segment are known.

The first is classical mechanics: vertical speed increases from zero to a certain speed. The average vertical speed is half the vertical speed on arrival. The time of fall is the height divided by the average vertical speed. It is also final vertical speed divided by g, where g is local gravity (9.81 m/s2). Therefore height fallen is proportional to the square root of time elapsed.

The second is special relativity. The timelike component is c times the time elapsed. The spacelike component is the lenght of the trail in space. The spacelike component, here is tiny compared to the timelike component; therefore the size of the segment is nearly the size of its timelike component.

The third is elementary geometry. A rectangular triangle can be drawn, having half the chord as short side, radius minus height as long side, and radius as hypothenusa. A second such rectangle has height as the short side and half the chord as the long side. The two are similar; hence height/half chord = half chord/radius.

Note that the distance to the horizon (on an airless world) is the geometric mean of the height on which one is standing and the radius of the world.

13. Oct 13, 2008

### Redbelly98

Staff Emeritus
In the figure, the "10 m" length and the "meters of light travel time" form the legs of a right triangle. The hypotenuse of that triangle is the chord length you are looking for.

There is a well-known formula from geometry, for calculating the length of the hypotenuse in a right triangle.

14. Oct 13, 2008

### KiyoEtAlice

I'll try it! thanks!

15. Oct 13, 2008

### KiyoEtAlice

Okay I get it thanks!