Special Relativity: Going from A to B on a helix

In summary, the conversation discusses the possibility of calculating the length of a helix from point A to point B, both non-relativistically and relativistically. It is suggested to use an inertial frame to analyze the motion and clock rate, with formulas provided for calculating the time elapsed on the object's clock. The concept of a spaceship traveling on a helix is also mentioned.
  • #1
Passionflower
1,543
0
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?
 
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  • #3
Passionflower said:
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?
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
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
JesseM said:
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.
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?

Fredrik said:
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?
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
Passionflower said:
So you do not think that rotations are the big problem here, these don't matter here? Basically we are talking about a 3D problem, right?
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.
Passionflower said:
I am glad to hear you folks think it is easy, what are the formulas?
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 [tex]\frac{L}{v} \sqrt{1 - v^2 /c^2 }[/tex]
 
  • #7
JesseM said:
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.
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.

George Jones said:
I'm not quite sure what you mean. Is

https://www.physicsforums.com/showthread.php?p=2245766#post2245766

at all relevant?
No, not a helix in spacetime but a helix in space. Thanks for the quote though, that is also interesting.
 
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  • #8
Passionflower said:
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?

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.
 
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  • #9
starthaus said:
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.
I would appreciate it.

Looks like I grossly overestimated this problem.
 
  • #10
Passionflower said:
So you are saying that a spaceship can travel on a helix inertially? How could that be?
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
JesseM said:
the helix rest frame
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
Passionflower said:
I would appreciate it.

Looks like I grossly overestimated this problem.

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 :-)
 
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  • #13
Passionflower said:
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?
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
Passionflower said:
No, not a helix in spacetime but a helix in space.
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
Passionflower said:
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?

This is an example of a world line of helix-shaped path: [itex]t\mapsto (t,\cos t,\sin t,t)[/itex]. 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:

[tex]\tau(x)=\int_a^b\sqrt{-g(x'(t),x'(t))}dt[/tex]

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
Sounds analogous to a helicopter rotor.
 
  • #17
DaleSpam said:
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.
Of course! :blushing:

Thanks for that aha-erlebnis!

DaleSpam said:
I have some Mathematica notebooks playing around with it if you want.
I use Matlab but I am seriously considering moving to Mathematica.
 
  • #18
Passionflower said:
Of course! :blushing:

Thanks for that aha-erlebnis!
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
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
Radius of the ball: 1
Pace ([itex]2c\pi[/itex]) 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
I have not looked at this thread in quantitative detail, but it is impossible to have a rigid ball.
 
  • #21
Passionflower said:
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
Radius of the ball: 1
Pace ([itex]2c\pi[/itex]) 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
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
 
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  • #22
Passionflower said:
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
Radius of the ball: 1
Pace ([itex]2c\pi[/itex]) 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

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
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
Radius of the ball: 0.3
Pace ([itex]2c\pi[/itex]) 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.
 
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  • #24
Passionflower said:
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
Radius of the ball: 0.3
Pace ([itex]2c\pi[/itex]) 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.

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.
 
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  • #25
yuiop said:
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.
Well but doesn't that seem to be in contradiction with this:

DaleSpam said:
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.
 
  • #26
Passionflower said:
Well but doesn't that seem to be in contradiction with this:

If you add the conditionals I have inserted in brackets below, to what Dalespam said, then I think we are on the same page:

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 (a space diagram in say frame A) is the same as a helix in a (spacetime diagram in say frame B) is the same as uniform circular motion (in a space diagram in frame B).

If that is not what Dalespam meant, then my apologies in advance for paraphrasing. At least that is what I think he meant, but not 100% sure.

Anyway, if we have uniform circular motion in space in one frame, then as long as the tangential velocity of the perimeter is less than c, then the path in another frame will be a helix in space, but the total resultant velocity (tangential velocity combined with parallel velocity) of a perimeter particle will always be less than c in any inertial reference frame.
 
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  • #27
yuiop said:
If you add the conditionals I have inserted in brackets below, to what Dalespam said, then I think we are on the same page:
That is very likely, I assume the problem is with my understanding.

yuiop said:
Anyway, if we have uniform circular motion in space in one frame, then as long as the tangential velocity of the perimeter is less than c, then the path in another frame will be a helix in space.
Right but then in no single frame should the velocity be greater than c right?
 
  • #28
Passionflower said:
Right but then in no single frame should the velocity be greater than c right?

I would say that is crucial.

(My lawyers said I should add the clause "as long as the frame in question is an Inertial RF which by definition does not have a velocity greater than c relative to any other IRF").
 
  • #29
yuiop said:
I would say that is crucial.
Yes but then I still do not understand the results.
 
  • #30
Passionflower said:
So something must still not be correct in the way I want to calculate this.
I see that there a bunch of comments already, so sorry if I am being repetitive.

What you are doing here is specifying the whole helix in one frame. When you do that you can easily get superluminal velocities even though the tangential and translational velocities are each subluminal. What I think you really want to do is to specify the tangential velocity in the "circular" frame, and then boost it to the "helical" frame. That will ensure subluminal speeds.
 
  • #31
yuiop said:
So a helix in (a space diagram in say frame A) is the same as a helix in a (spacetime diagram in say frame B) is the same as uniform circular motion (in a space diagram in frame B).

If that is not what Dalespam meant, then my apologies in advance for paraphrasing. At least that is what I think he meant, but not 100% sure.
That is what I meant. Thanks for clarifying it!
 
  • #32
Passionflower said:
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
Radius of the ball: 0.3
Pace ([itex]2c\pi[/itex]) 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
If the pace of rotation is one rotation every 2 seconds (i.e. 0.5 rotations per second) in the frame where it's moving at 0.6c, then in its own rest frame the ball rotates at 0.5/0.8 = 0.625 rotations per second, so the tangential velocity in the rest frame of the ball is (2*pi*0.3)*0.625 = 1.178 light-seconds per second, so it's still faster than light (yuiop gave a speed of 1.18c in post #24 above so I think we did the calculations the same way). If you instead wanted the pace to be 0.5 rotations per second in the rest frame of the ball, in that case the tangential velocity would be 0.94c in this frame, but in that case the pace in the frame where the ball is moving at 0.6c would only be 0.5*0.8 = 0.4 rotations per second, or one rotation every 2.5 seconds. So if the coordinate time is 20 seconds to get from A to B, there will be 8 full rotations. So, the height of each coil would be 12/8 = 1.5 light-seconds, so according to this page the length of each coil must be sqrt(1.5^2 + (2*pi*0.3)^2) = 2.409, so the total length of the helix is 8*2.409 = 19.27. And if the time in the rest frame of A and B is 20 seconds, the speed of the clock in this frame must be 19.27/20 = 0.96c.
 
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  • #33
Passionflower, the worldline of the clock on the end of the ball in the "circle" frame is given by:
[tex]\left( ct, 0, r \; cos(t\omega), r \; sin(t\omega) \right)[/tex]

Then Lorentz transforming into the "helix" frame gives:

[tex]\left( ct\gamma, vt\gamma, r \; cos(t\omega), r \; sin(t\omega) \right) = \left( ct', vt', r \; cos\left(\frac{t'\omega}{\gamma}\right), r \; sin\left(\frac{t'\omega}{\gamma}\right) \right)[/tex]
 
  • #34
JesseM said:
If the pace of rotation is one rotation every 2 seconds (i.e. 0.5 rotations per second) in the frame where it's moving at 0.6c, then in its own rest frame the ball rotates at 0.5/0.8 = 0.625 rotations per second, so the tangential velocity in the rest frame of the ball is (2*pi*0.3)*0.625 = 1.178 light-seconds per second, so it's still faster than light (yuiop gave a speed of 1.18c in post #24 above so I think we did the calculations the same way).
The pace of a helix is defined as the length of a coil it is not the rate of rotation.

JesseM said:
So, the height of each coil would be 12/8 = 1.5 light-seconds, so according to this page the length of each coil must be sqrt(1.5^2 + (2*pi*0.3)^2) = 2.409, so the total length of the helix is 8*2.409 = 19.27.
I get 3.48 for each coil and a total length of 27.85.

I took the radius and the height (pace) of a coil divided by 2pi. See for instance: http://mathworld.wolfram.com/Helix.html
 
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  • #35
Passionflower said:
The pace of a helix is defined as the length of a coil it is not the rate of rotation.
OK, it's not familiar terminology for me. Doing a google search for "pace" and "helix" doesn't turn up anyone else using "pace" in these sense in the first two pages of results, are you sure this is standard usage? I also didn't know why you put (2*pi*c) after the word "pace", although now that you quote the mathworld page I see it does give parametric equations for the helix in which z=ct, and where "r is the radius of the helix and 2*pi*c is a constant giving the vertical separation of the helix's loops" (note that the c here has nothing to do with the speed of light, so it was a bit confusing for you to write 2*pi*c in the context of a relativity discussion without explaining this!)

Anyway, if the length of a coil is 2 light-seconds in the frame where the speed of the sphere is 0.6c, then the time for a rotation in this frame must be 2/0.6 = 3.333... seconds. So, in the sphere's rest frame the rotation time must be 3.333...*0.8 = 2.666... seconds. That means the tangential speed in the sphere's rest frame is (2*pi*0.3)/2.666... = 0.70686c...where did you get a tangential speed of 0.57c?

And if the height of a coil is 2 light-seconds, then the length along a single coil must be sqrt(2^2 + (2*pi*0.3)^2) = 2.7483 light-seconds, for a total length of 6*2.7483 = 16.49 light-seconds. Thus if the time is 20 seconds in the frame where the sphere moves at 0.6c, the speed of the clock in this frame must be 16.49/20 = 0.82c.
Passionflower said:
I took the radius and the height (pace) of a coil divided by 2pi. See for instance: http://mathworld.wolfram.com/Helix.html
Where on that page do you see anything about the length of the helix? The page I quoted said the length of a single coil with height H and radius R is given by L = sqrt(H^2 + (2 pi R)^2), do you think this formula is incorrect? It's not hard to see that this formula is correct, since you can imagine the helix as a curve drawn on a cylinder, and then if you slice the surface of cylinder on an axis parallel to the central axis and "unwrap" it (which is possible since a cylinder has zero intrinsic curvature, just like a flat plane--see here), you'll have a rectangle with width 2*pi*R, and the vertical height of a single coil will be H.
 
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