Undergrad Solve Constant Acceleration Problem in Special Relativity

[Albertgauss]

TL;DR

Constant Acceleration Special Relativity

Basically I just want to work out a constant acceleration problem in relativity, of the same kind of introductory physics.

Vo= 0.9999c
Vf = 0
D= 50 Au
Accel, Earth frame?
Accel, Ship frame?
Time of transit, Earth frame?
Time of transit, ship frame?

Motion is 1-D. All origins line up at the beginning of the problem. Neglect orbit of Earth, etc, approximate all motion just a 1-D motion as possible.

Wikipedia has a ton of acceleration formulas, but I don't know which one I would use. I also don't know how to proceed with the time (of either frame) calculation. I'm sure this has been done on this website before, but since the words "relativity" and "acceleration" are so common in physics it was tons of pages I couldn't search through. I am happy to be pointed out to worked-out example, somewhere, if it exists.

https://en.wikipedia.org/wiki/Acceleration_(special_relativity)

 

[pervect]

Let's see if I understand the questio. You want to find the proper acceleration needed to come to a stop from V0 = .9999c in a distance D = 50 au answer your question? Or, conversely, the proper acceleration needed to reach a velocity of V0 in a distance D?

Perhaps you're not familiar with the term proper acceleration. That would be the acceleration in the ship frame.

The basic formula are worked out in the "relativistic rocket equations". See http://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html. I'll take a stab at it later when I have time, assuming I understand the question properly.

 

[Albertgauss]

Yes, you are correct. "V0" is in the initial velocity here, and Vf the final velocity.

I admit I am a bit confused about the term "proper acceleration". I was thinking of it in analogy to the "proper time"--the amount of time that passes on the ship's clocks during the journey. Thus, the "proper acceleration" would be, in this thinking, the "acceleration" inside the ship, since acceleration can be felt inside the ship.

I did notice that a bunch of threads on the subject popped up below and I will start to read through them. I suppose it was only after I posted the question I got relevant posts.

I also just now was able to respond to this post. I previewed your thread and will start reading through that. It looks very good, clear, and concise.

 

[pervect]

From the fifth equation in the Baez reference, the expression for gamma, we find directly that:

$$\frac{a\,d}{c^2} = \gamma - 1 \quad \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

or

$$a = \frac{c^2}{d} \left( \gamma - 1 \right)$$

In your example, $\gamma = 70.7$. $c^2 / d$ is about 12 000 m/s^2 with d=50 au, so the final answer is about 69.7 * 12 000 = 840 000 m/s^2

$\gamma - 1$ would be the kinetic energy / unit rest mass, so this formula could probalby be derived by an energy conservation formula, though I don't quite see exactly how.

a, the prope acceleration, i.e. the acceleration in the ship frame, is presumed constant. The coordinate acceleartion in the Earth frame would vary with time, basically it'd be divided by some power of gamma. I don't have a reference handy to check on the relation between the coordinate acceleration (which has little physical significance) and the proer acceleration.

 

[Ibix]

I admit I am a bit confused about the term "proper acceleration". I was thinking of it in analogy to the "proper time"--the amount of time that passes on the ship's clocks during the journey. Thus, the "proper acceleration" would be, in this thinking, the "acceleration" inside the ship, since acceleration can be felt inside the ship.

"Proper" in this context is related to "property", something that is one's own. So yes, proper acceleration is the acceleration measured by accelerometers aboard the ship. It's a direct measurable, even in a closed box, so is the thing with physical consequences. For example a proper acceleration of 20g would be detrimental to your health.

Coordinate acceleration is the other option. It's the rate of change of the ship's velocity with respect to some chosen reference. In Newtonian physics it's equal to the proper acceleration if the reference is inertial, so the distinction is frequently glossed over. In relativistic physics they are not equal and you need to keep the definitions separate in your mind.

 

[Albertgauss]

That page turned out to be an excellent resource. I am glad I was made aware of it. I also verified for myself through the calculation that I get the same acceleration that you did.

I did have one question about the formulas on the page, though. I attached a picture. For the equations that have a blue arrow next to them that I put in, how do you get from hyperbolic functions to square root expressions? I thought maybe at first the author did some kind of expansion but I couldn't be sure.

 

[Ibix]

For the equations that have a blue arrow next to them that I put in, how do you get from hyperbolic functions to square root expressions?

Just by eliminating between the expressions, I think. For example, using that $1=\cosh^2\theta-\sinh^2\theta$ you can substitute the expression for $d$ in terms of $\cosh$ into the expression for $t$ in terms of $\sinh$ to get the square root expression.

 

[PeterDonis]

For the equations that have a blue arrow next to them that I put in, how do you get from hyperbolic functions to square root expressions?

By substitution. Note that the square root expressions don't have all of the same variables in them.

For example, the first equation is

$$
t = \frac{c}{a} \sinh \frac{a T}{c}
$$

Multiplying the second equation by $a / c$ (and using the far right equality of that equation) gives

$$
\frac{a T}{c} = \cosh^{-1} \left( \frac{ad}{c^2} + 1 \right) =
$$

Therefore we have

$$
t = \frac{c}{a} \sinh \left[ \cosh^{-1} \left( \frac{ad}{c^2} + 1 \right) \right]
$$

Which, if you make use of the formulas for composition of hyperbolic trig functions, will give you the formula for $t$ with the square root in it.

 

[Albertgauss]

Yes, I got it. Just needed a little help. I could derive them from the hints given here. All good to go.

 

[pervect]

Yes, you are correct. "V0" is in the initial velocity here, and Vf the final velocity.

I admit I am a bit confused about the term "proper acceleration". I was thinking of it in analogy to the "proper time"--the amount of time that passes on the ship's clocks during the journey. Thus, the "proper acceleration" would be, in this thinking, the "acceleration" inside the ship, since acceleration can be felt inside the ship.

Yes, proper acceleration is basically the acceleration felt inside the ship. And it has close ties with proper time as well. So it sounds like we're on th same track.

 

[pervect]

On the topic of the relation between coordinate and proper acceleration.

I dug up a few references to check my fallible memory. https://en.wikipedia.org/wiki/Acceleration_(special_relativity)#Proper_acceleration and post 2 in https://en.wikipedia.org/wiki/Acceleration_(special_relativity)#Proper_acceleration (ignore post 1, it's wrong. I haven't tracked down my exact error there, though).

The bottom line is that when acceleration is in the same direction as the velocity, the coordinate acceleration of a moving object is equal to the proper acceleration of said object divided by $\gamma^3$, where $\gamma = 1 / \sqrt{1 - v^2/c^2}$.

The proper acceleration is independent of velocity and coordinates, making it easier to work with than the coordinate acceleration, which is not.

Note that the relationship between coordinate and proper acceleration is different when the acceleration is transverse to the direction of motion. This can mostly be attributed to length contraction. Also - it is possible to work back from the $\gamma^3$ relationship between proper and coordinate acceleration to find the trajectory of a relativistic rocket, with enough effort.

MTW has a good treatment of accelerated motion in "Gravitation", and it's the approach I personally use, but MTW uses 4-vector and tensor methods that may not be accessible to most readers.

I think Rindler also has a treatment. But I don't own a copy of his book.