Special Relativity: Going from A to B on a helix

1. Sep 15, 2010

Passionflower

Is this doable, a calculation to go from A to B on a helical path?

To simplify the problem we could consider A and B being parallel planes so that the end location of the helix is not of prime importance, it could end anywhere on the plane at B.

It is easy to determine the length of a helix from A to B non relativistically. but how do we approach the problem relativistically?

2. Sep 15, 2010

George Jones

Staff Emeritus
3. Sep 15, 2010

JesseM

Well, if you assume that in the frame where the helix is at rest, the speed along the helical path is constant, then the problem is easy since constant speed in some frame implies constant rate of time dilation relative to coordinate time in that frame, regardless of the specific path.

4. Sep 15, 2010

Fredrik

Staff Emeritus
What do you want to calculate? The proper time of the curve? Why not just write down the formula that defines the helix and use the definition of proper time?

5. Sep 15, 2010

Passionflower

So you do not think that rotations are the big problem here, these don't matter here? Basically we are talking about a problem with rotations in two dimensions, right?

Same answers here, since this is a problem with rotations, it seems a very hard problem. No?

I am glad to hear you folks think it is easy, what are the formulas?

6. Sep 15, 2010

JesseM

Rotations of what? Are you asking about how the problem would be analyzed in some non-inertial rest frame of the object traversing the helix? I was just saying the problem would be easy to solve in the inertial frame where the helix (and its endpoints A and B) was at rest, this frame obviously isn't rotating.
If L=length along helical path from A to B in frame where helix is at rest (not sure the formula for this but it's a purely geometrical problem, nothing to do with relativity), and v=constant speed of object moving along this path in the helix rest frame, then coordinate time to get from A to B in this frame will just be L/v, and the time elapsed on the object's clock will be $$\frac{L}{v} \sqrt{1 - v^2 /c^2 }$$

7. Sep 15, 2010

Passionflower

So you are saying that a spaceship can travel on a helix inertially? How could that be?

Think of a corkscrew going from A to B.

No, not a helix in spacetime but a helix in space. Thanks for the quote though, that is also interesting.

Last edited: Sep 15, 2010
8. Sep 15, 2010

starthaus

A. The length of the helix in the frame of the helix is simply the integral of the arc element from start to end. You can use any parametrization you wish. So, this part has nothing to do with relativity.

B. The length of the helix as measured by an observer that moves along the helix at relativistic speeds is simply the Lorentz contracted value calculated at point A. I did someplace the calculation for Lorentz contraction for circular motion, I can dig it out, the helical motion is just a slightly more complicated case.
See the case of arbitrary motion in https://www.physicsforums.com/blog.php?b=1959 [Broken], I can rederive the case of circular motion in a few steps.

Last edited by a moderator: May 4, 2017
9. Sep 15, 2010

Passionflower

I would appreciate it.

Looks like I grossly overestimated this problem.

10. Sep 15, 2010

JesseM

No, I'm saying you can use an inertial frame to analyze its motion and clock rate (see the edit to my last post giving some simple formulas). You understand that we don't need to use non-inertial frames to analyze non-inertial clocks, right?

11. Sep 15, 2010

Passionflower

I think we have some disconnect here, what rest frame?

Can you imagine a spaceship going on a "corkscrew" path from planet A to planet B?

12. Sep 15, 2010

starthaus

Yes :-). I added a pointer to my blog that shows how it's done. In effect, it is a piecewise linear approximation using https://www.physicsforums.com/blog.php?b=1959 [Broken]. The method seals with BOTH length contraction AND time dilation, so you are covered :-)

Last edited by a moderator: May 4, 2017
13. Sep 15, 2010

JesseM

In that case it would be the rest frame of A and B (again assuming the ship travels at constant speed in this frame). I was imagining for simplicity that the object was traveling along an actual physical helix, but of course the set of spatial positions the ship travels through in the A and B rest frame can be used to define the coordinates of an imaginary helix at rest in that frame.

14. Sep 15, 2010

Staff: Mentor

If you have uniform circular motion in some frame then you have a helix in spacetime. In some frames the axis of the helix will be the time axis, and in other frames the axis of the helix will be at some angle to the time axis. In those frames you will have a helix in space. So a helix in space is the same as a helix in spacetime is the same as uniform circular motion.

I have some Mathematica notebooks playing around with it if you want.

15. Sep 15, 2010

Fredrik

Staff Emeritus
This is an example of a world line of helix-shaped path: $t\mapsto (t,\cos t,\sin t,t)$. Let's call that curve x. If you want to calculate the proper time of x, you just calculate x'(t)=(1,-sin t,cos t,1) and insert the result in the definition of proper time:

$$\tau(x)=\int_a^b\sqrt{-g(x'(t),x'(t))}dt$$

where g is defined by g(u,v)=-u0v0+u1v1+u2v2+u3v3. I'm just not sure if that's what you wanted to calculate.

16. Sep 15, 2010

Naty1

Sounds analogous to a helicopter rotor.

17. Sep 16, 2010

Passionflower

Of course!

Thanks for that aha-erlebnis!

I use Matlab but I am seriously considering moving to Mathematica.

18. Sep 16, 2010

Staff: Mentor

You are welcome. Also, it is a good geometric principle in general to understand that a worldline feature which is in time in one frame is in space in other frames. This can help with other kinds of motion:

If a particle is undergoing simple harmonic motion then it is moving in a sinusoidal pattern in time in that frame, but in other frames it will move in a sin wave in space also.

In a variation in the barn-pole paradox the pole is moving parallel to the wall with the barn door and is then impulsively accelerated into the opening. This causes a bend in the worldsheet in time in one frame, and a bend in space in other frames, allowing the pole to enter.

Etc.

19. Sep 16, 2010

Passionflower

Ok, so I am still confused.

Let's assume we have a rotating test ball of radius r moving from A to B (in the direction of the axis of rotation), there is one clock in the center and one on the edge of the ball so that it circles around the axis of rotation over the center of the ball.

Distance A-B: 12
Velocity of the ball: 0.6
Pace ($2c\pi$) of the helix: 0.1

So we got:
Coordinate time A-B: 20
Proper time ball (clock at the center): 16
Number of full turns between A and B: 120
Length of the helix: 754.08

So now how do we get the velocity and coordinate time of the rotating clock?

Obviously length/time does not work because: 754.08 / 20 > 1

20. Sep 16, 2010

George Jones

Staff Emeritus
I have not looked at this thread in quantitative detail, but it is impossible to have a rigid ball.

21. Sep 16, 2010

JesseM

Well, the problem here is that you've made the ball rotate faster than light! If it has a rotation rate of 10 rotations per second in the frame where it's moving at 0.6c where the time dilation factor is 0.8, then in the rest frame of the center it must have a rotation rate of 10/0.8 = 12.5 rotations per second, and it has a radius of 1 light-second so the tangential speed of a point on a surface must be (2*pi*1 light-second)*(12.5/second) = 78.54 light-seconds/second

Last edited: Sep 16, 2010
22. Sep 17, 2010

yuiop

Try it with one full turn every 8 seconds in the rest frame (S) of the ball's centre. In this frame the rim speed is 2*pi/8 = 0.7854c. In frame S' where the ball is moving at 0.6c parallel to its rotation axis, the ball only rotates at a rate of one revolution every 10 seconds and the component of the velocity orthogonal to the rotation axis is 2*pi/10 = 0.62832c. The resultant velocity of the rim in S' is sqrt(0.62832^2+0.6^2) = 0.86878c. The spiral path length in S' is 17.376 ls. The proper elapsed time of the clock on the equator of the ball is 20*sqrt(1-0.86878^2) = 9.904 seconds. The proper time can also be calculated in the rest frame of the balls centre as 12*sqrt(1-0.7854^2) = 9.904 seconds.

23. Sep 17, 2010

Passionflower

Ok, that makes sense.

However I am still not out of the woods:

Let's assume we have a rotating test ball of radius r moving from A to B (in the direction of the axis of rotation), there is one clock in the center and one on the edge of the ball so that it circles around the axis of rotation over the center of the ball.

Distance A-B: 12
Velocity of the ball: 0.6
Pace ($2c\pi$) of the helix: 2

So we got:
Coordinate time A-B: 20
Proper time ball (clock at the center): 16
Number of full turns between A and B: 6
Length of the helix: 16.48
Tangential speed:0.57
Helix velocity: 0.82
Gamma: 1.77
Proper time: 11.32

Assuming this is the correct method I still have a problem with something:

Assume we make the radius 0.5

Then:

Length of the helix: 22.34
Tangential speed:0.94
Helix velocity: 1.12 !

So the tangential speed is high but below c, however the helix velocity is above c.

So something must still not be correct in the way I want to calculate this.

Last edited: Sep 17, 2010
24. Sep 17, 2010

yuiop

Not checked your figures, but offhand even though the tangential speed in S' is below c, the total resultant speed, sqrt((tangential speed)^2+(linear speed)^2) = sqrt(0.94^2+0.6^2) = 1.12c in S' in your example and the tangential speed in the rest frame (S) of the ball's centre is 2*pi*0.5/(16*6) = 1.18c if I have done it right.

When resolving a resultant velocity vector into its component parts, the individual components might appear to be less than c, but it is the total resultant velocity that counts. The tangential speed in S' is just one component. The "helix velocity" is the total resultant velocity in this case.

Last edited: Sep 17, 2010
25. Sep 17, 2010

Passionflower

Well but doesn't that seem to be in contradiction with this: