# A The information about the curvature on the rest

Tags:
1. Oct 20, 2016

### Abel Cavaşi

What happens with the information about the curvature of the trajectory of a body when the body stops? We know that to assess the curvature of a path we must calculate the value of a fraction of which the denominator is speed module (cubed). But if speed module is canceled, this fraction can not be assessed. So, it preserves information about the curvature in that interactions that stop the body?

2. Oct 20, 2016

Staff Emeritus
I have no idea what you are talking about.

3. Oct 20, 2016

### Abel Cavaşi

Thanks for the reply! But, about the curvature of the trajectory, you have an idea? You know what is the curvature? You know how to calculate it? If yes, then what value does it for a body with zero speed?

4. Oct 20, 2016

### Staff: Mentor

Write down the formula for calculating the curvature. What happens to the formula when $v=0$

5. Oct 20, 2016

### Abel Cavaşi

Indeed, that's what I asked: "What happens to the formula when v=0"?

6. Oct 20, 2016

### jbriggs444

A good first step would be to heed @Dale's advice and

7. Oct 20, 2016

### Abel Cavaşi

I was addressed those who already know the formula. I did not open this topic to make education here. The formula can be found, for example, in the Wikipedia. $$\kappa=\frac{|\gamma'\times\gamma''|}{|\gamma'|^3}$$

8. Oct 20, 2016

### nasu

To ask about curvature you should first see to what "object" this curvature applies to.
So what do you think is the trajectory of a body at rest?

9. Oct 20, 2016

### Abel Cavaşi

Until you answer me the questions, you can not understand my logic.

10. Oct 20, 2016

### jbriggs444

Take the trajectory in given by the parameterization:

$$x(t)=t^3$$
$$y(t)=0$$
$$z(t)==0$$

Take the trajectory given by the parameterization:

$$x(t)=t$$
$$y(t) = 0$$
$$z(t) = 0$$

Do these two define the same path? Does this path include the point (0,0,0)? What is the curvature of the path at (0,0,0)?

11. Oct 20, 2016

### Staff: Mentor

Suppose, instead of the projectile coming to rest (as reckoned by one observer), its motion is reckoned by another observer who is traveling in an inertial frame of reference at constant velocity relative to the first observer. According to this observer, once the projectile hits the ground, it does not come to rest. Instead, it continues in the same direction at constant velocity tangent to the ground. So, according to this observer, the curvature of the projectile path has undergone a discontinuous change. How much of the previous curvature is preserved by the straight line path that the projectile is then experiencing after making contact with the ground?

12. Oct 20, 2016

### Abel Cavaşi

I expect an answer to the question that I raised it in this topic:
"what happens with the information about the curvature when the body stops?"

13. Oct 20, 2016

### Staff: Mentor

What is the difference between stopping and being at rest?

14. Oct 20, 2016

### Abel Cavaşi

To my knowledge, neither.

15. Oct 20, 2016

### Staff: Mentor

My answer is the answer to the question about what happens when the body stops. It is just the answer as reckoned by two different observers.

16. Oct 20, 2016

### Staff: Mentor

Then what is your problem with the concept of "at rest."

17. Oct 20, 2016

### Abel Cavaşi

But the information is objective. It can not disappear to any observer in the universe.

18. Oct 20, 2016

### Staff: Mentor

Is there anyone out there who has the slightest idea what this guy is talking about (besides him). OP, please do not respond?

19. Oct 20, 2016

### Abel Cavaşi

The problem with the rest is exactly that I raised in this topic.

20. Oct 20, 2016

### jbriggs444

Here, for instance is a coherent version of what you may have meant to ask.

Can we determine the curvature at a particular point on the path traced out by a point particle if we are given only information about the position, velocity and acceleration of the particle at a time when it was at rest at that point?