Speed of light-can this be correct?

[rtharbaugh1]

A ship leaves Earth orbit under an acelleration of one gravity, about 9.8 meters per second per second. How long will it take the ship to reach the speed of light as seen from Earth? Assume the speed of light is 3 x 10^8 meters per second.

I calculate about one year. Can this be correct?

 

[DW]

The velocity of the ship $$v$$ is related to its rapidity $$\theta$$ by
$$v = ctanh(\theta )$$. For a constant proper acceleration of $$\alpha$$ and starting with $$v = 0$$, the rapidity will be given as a function of proper time $$\tau$$ by
$$\theta = \frac{\alpha \tau}{c}$$. This results in
$$v = ctanh(\frac{\alpha \tau}{c})$$. Due to the asymptotic behavior of the tanh function, $$v$$ never reaches c. As you suggested let's consider a proper acceleration of $$\alpha = 1\frac{Ly}{y^2} \sim 1g$$ and calculate v when $$\tau = 1y$$ for $$c = 1\frac{Ly}{y}$$. This results in
$$v = ctanh(1) \sim 0.76c$$.

 

[rtharbaugh1]

Well I guess this makes sense to someone. I suppose rapidity $$\theta$$ has to do with velocity expressed as an angle on time/space light cones. I'm not sure what proper time is, nor proper acceleration. I presume Ly is light year? And what is y in the denominator?

I have heard of this asymptotic behavior before, of course, and am not trying to be a smart alec. Are you saying that after one year Earth time at one g acceleration the ship is traveling at .76c?

 

[selfAdjoint]

RIght, LY is light years, and y is years, and he has taken your estimate of one year (Newtonian) to accelerate to c as denoting an acceleration rate c per year. Since c is one light year per year, that comes out 1 light year per year per year, a good L/T^2 unit for acceleration, which he abbreviates 1 LY/Y^2.

Proper time is the time measured on the ship, which is "Newtonian".

Rapidity is a technical term in relativity. As he says, it is the quantity of which the speed is the hyperbolic tangent. The Lorentz transformations are analogous to rotations, but where the rotation matrices have sines and cosines, the Lorentz matrices have hyperbolic sines and cosines, and this trick with rapidity sets you up to use that mathematical formalism. The hyperbolic functions are then functions of the rapidity, which is denoted by theta to indicate its analogy to a rotation angle.

So once he has the rapidity and proper time, he can do regular acceleration math. The results are then governed by the behavior of the hyperbolic tangent, which we have to use to get back from rapidity to speed.

 

[DW]

Right, and just to be clear, proper time is the wristwatch time of the traveller and proper acceleration is the acceleration felt by the traveler and yes y is years. I am saying that if the traveler is pushed such that he feels as if he weighs about the same as on Earth for one year of time according to his own watch that he will only have reached 0.76c. You can't hold his coordinate acceleration constant for long enough to get him up to c as the force on him required to do so approaches infinity as his speed approaches c. For this scenario his speed as a function of proper time was
$$v = ctanh(\frac{\alpha \tau}{c})$$. His speed as a function of coordinate time t, or Earth time, is
$$v = \frac{\alpha t}{\sqrt{1 + \frac{\alpha ^{2}t^{2}}{c^2}}}$$.

 

[rtharbaugh1]

Thank you, DW and Self Adjoint, for your help. Your dedication is astonishing.

I notice that the proper acceleration $$\alpha$$ is the same in both formulations. This strikes me curious.

I feel rather like a toothless infant trying to gum nourishment from a steak, especially since my knowledge of hyperbolic functions is scant. But I think Self Adjoint knows where I am trying to go with this.

Meanwhile, before the facts interfere too deeply with my pleasure in speculation, has anyone else found it interesting that the acceleration of gravity at the surface of Earth seems to be related to the period of Earth's orbit around the sun by the value of c? Or is this just another one of those astonishing coincidences?

Thanks again. You have both been most helpful. I suppose I should just wander over to the Astrophysics board for advice on this seeming coincidence, but I thought I'd run it by you first. I am most grateful.

Thanks for being here,

Richard