# Time To Near Light Speed

1. Jun 21, 2010

### Star Drive

This solution is from my notes about twenty years ago and the integration by form was made by a friend. We worked on this interesting question together. Many others have surely gone down this same road. I am not a physicist, nor have I had University Physics since 1968.

It should be realized that for a constant applied force F, the acceleration magnitude will taper off along a steep curve, as the relativistic effects increase the mass of the object being accelerated. As the object approaches the speed of light in the limit, the acceleration tends to fall to zero in the limit. In such an extreme case, the object would become infinitely massive in the limit, as it approaches light speed. This is a key reason why the speed of light can’t be exceeded.

c =Velocity of Light, assumed constant at 2.998 x E8 m/s.
t =Time in Seconds.

\Pi= 3.141593
(Approximately)

Capital letters represent the initial or rest quantities as viewed by the observer. In example; A = Initial Acceleration, M = Initial Mass., F = Initial Force (Held constant throughout mission)

Lower case Letters represent the instantaneous variables, as viewed by the observer. In example; a = Instantaneous acceleration, m = Instantaneous mass.

Fundamental physics relations apply;
a=dv/dt
F=ma

Equivalent relativistic factors [T] follow;

[T]=1/\sqrt{1-v^2/c^2}

[T]=c/\sqrt{(-v)^2 + c^2}

Instantaneous acceleration;
a=dv/dt

Instantaneous (relativistic) mass;
m=M[T]

The following basic substitutions yield;

F=ma

F=M[T]dv/dt

Fdt =M[T]dv

(F/M)dt =[T]dv

\int(F/M)dt = \int[T]dv

Where; F, M and c are constants

(F/M) \int dt= c \int dv/ \sqrt{(-v)^2+c^2}

Integrate Equation by form with CRC integral tables to obtain the following result.

(F/M) t=[c/\sqrt{1}] [\arcsin(v\sqrt{1/c^2})]

Solve relationship F=MA for A and substitute. (All initial Values)

At=c[\arcsin(v/c)]

At/c=\arcsin(v/c)

\sin[At/c]=v/c

The following equation is the primary relativistic solution for velocity, as a function of time, where the units are angles (radians) and time (seconds).

v =c[\sin(At/c)]

Note: The following constraints apply;

[At/c]<(\Pi/2)

t <(1/A) [(\Pi)(c)/2)]

t< (1/A) [4.7092] (10^8)

The following equation is the derived solution for time to any arbitrary speed, which is less than light speed.

t =(c/A)[\arcsin(v/c)]

As an example solution, for 1.0 Earth gravity initial acceleration, the time to 99.9% of the velocity of light is about 1.4788 years. (The initial force F is held constant)

It was surprising for the solution to be inversely linear. In example, it requires about 14.788 Years for the same results at 0.1 G.

A personal computer spreadsheet was used to iterate the solution for time at some 10,000 points along the curve from rest to 99.9% of the velocity of light. That result agreed very well for the mathematical solutions provided above, with t = 1.48 years at 1.0G. This machine computation is correct since F=ma and v=at is true, for very small changes of velocity v.

By summing the resulting time, for each of the 10,000 points with the PC spreadsheet, it was possible to approximate the integration for time. Please do have some fun and setup a spread-sheet for your self.

DISCUSSION POINTS:

It seems that the math yields a false solution for c >= speed of light. As the math implies, does the spacecraft velocity actually ‘slow down’, with continued acceleration, ‘after’ the speed of light is ‘exceeded’? Discussion: What does this really mean? LOL

Is anyone else surprised by the appearance of trigonometric functions in the two solutions? Discussion?

Is anyone else surprised by the apparent inverse linearity of the solution for time with respect to acceleration? Discussion?

As a thought experiment, consider a one-way Interstellar probe, which must be robotic and maintenance free The probe’s initial mass should be 100 metric tons. The chosen power plant must provide an initial or starting acceleration of 0.1 Earth gravities. The power plant must provide a constant reaction force F to propel the vehicle and operate continuously over the entire mission. The probe must accelerate from rest, relative to the mission planner, to a maximum velocity of 99.5% of the speed of light. Such a velocity would cause a relativistic mass increase of about x10 and a time compression of about x(1/10). Would the rocket engineers comment on the practical aspects of building such a vessel? Possible or impossible? Discussion?

2. Jun 21, 2010

### starthaus

You have more mistakes than one can count in your derivation. For the correct derivation, check this file : AccelerationInSR in my blog.

3. Jun 21, 2010

### Janus

Staff Emeritus
You actually need to use hyperbolic functions to solve for this.

The equation for finding the time needed to reach any given speed (in ship time) is:

$$T = \frac{c}{a} \tanh^{-1} \left ( \frac{v}{c} \right )$$

This can converted to "stay at home" time by

$$t=\frac{c}{a} \sinh \left ( \frac{aT}{c}\right )$$

As far as practicality goes:

If you had a rocket with an exhaust speed of 10% of c (much greater than anything we have), then:

Your ship to fuel ratio would be something of the order of 1 to 1e13 to reach 99.5% of c. (for every kg of ship, you would need 1e13 kg of fuel.)

4. Jun 22, 2010

### Star Drive

Thanks for your feedback.

Perhaps the alternate equations that were provided by two others on the forum, all solve for a different set of operating conditions.

Do those equations require that the ship's acceleration remain constant throughout the entire ship's mission?

I have proposed a solution (if my derivation is correct), which requires that the rocket's reaction force (F) must remain constant throughout the entire mission. Moreover because mass increases toward infinity (as the ship approaches the speed of light), ship acceleration approaches zero.

The difficulty presented by keeping the acceleration constant against ever increasing mass is probably unworkable.

However, it may be more reasonable to establish an initial or starting acceleration and then accept the acceleration penalty to come, due to increasing mass. (This is the case, which I have attempted to provide a solution for.)

Thanks for your assistance.

PS
I understand that my solution is suspect and could well be incorrect.

5. Jun 22, 2010

### starthaus

Yes.

Your derivation is not correct.

6. Jun 23, 2010

### Ich

They require that the proper acceleration be constant. That's the acceleration as measured on board. It's proportional to the propulsion force if the rocket's mass is constant.
The acceleration as measured in the initial reference frame (the "stationary" frame) is constantly decreasing.

7. Jun 23, 2010

### Star Drive

Thanks for your feedback.

In your equations, the acceleration (A) is held constant. But in my example, the force (F) is held constant while acceleration is permitted to decline over a relativistic curve. I think that this could be a critical difference.

Two equations below are the focus of my interest.

Re; I have a spreadsheet that checks 9,999 data points (.01%c per step) along the velocity curve from start to 99.99% of the velocity of light. The resulting time t, for each sub-interval, is computed and the results are summed to compare with results from the equations below.

The following point on the curve illustrates how close spreadsheet versus my mathematical solutions are at v = 99.90% of the speed of light. I have arbitrarily established initial acceleration at A = 9.81 m/s^2 (1.0G).

Spreadsheet = 46670665.53 Seconds
Equation t =(c/A)[arcsin(v/c)] = 46637732.68 Seconds
Percent difference = 0.0706%

Spreadsheet results for all points along the computed time curve are extremely close to the results given by the equations below. I’m struggling with how these computer results could just be coincidence, if the equations are incorrect.

The following equations are repeated from my initial posting for convenience.

This is my attempted relativistic solution for velocity, as a function of time, where the units are as follows. Angles (radians), t = Time (seconds), c = Speed of light (2.998E8 m/s), A = Initial acceleration, v= (Final Velocity)

$$v =c[\sin(At/c)]$$

The following equation is my proposed solution for time to any arbitrary speed, which is less than light speed.

$$t =(c/A)[\arcsin(v/c)]$$

Note: The following constraints apply for the preceding equations;

$$[At/c]<(\Pi/2)$$

$$t <(1/A)[(\Pi)(c)/2)]$$

The following example illustrates the problem that I am working on.

Assume a 1 kg mass rocket, which has a miniature rocket power plant within. Assume no loss of mass due to expenditure of fuel. Assume that the power plant is capable of continuous propulsion and starts the rocket’s journey with an initial acceleration of one Earth gravity (1.0G). Since f=ma, the continuous applied force will be (1 kg)(9.81 m/s^2)= 9.81 kg-m/s^2. The power plant’s reaction force F will be held constant throughout the mission, at the figure obtained.

With rocket velocity approaching the speed of light, its mass will approach infinity in the limit. As a result, one can expect the rocket’s acceleration (a = f/m) to decline toward zero. At mission’s end, that wonderful power plant continues to output the same constant reaction force F, as it did from the start.

Relative to mission control (my time), what is the time in seconds from start to velocity v = 10%C, 50%C, 90%C and 99.9%C?

Thanks in advance for anyone’s assistance

8. Jun 23, 2010

### Janus

Staff Emeritus
Your example is exactly the situation that the equations I gave are used for. In your example, the engine produces a constant thrust which equals a constant proper acceleration for the occupants of the ship. In other words, the acceleration experienced by the occupants of the ship maintains a constant 1g, and does not fall off as the ship gets nearer to c (even though the "stay at home" observers see the acceleration as decreasing). It is this acceleration experienced by the ship that is used in the equations given.

9. Jun 26, 2010

### Star Drive

Thanks for your feedback.

Derivations that my friend and I provided have been revisited and thoroughly checked. The integration solution, by form from a CRC table, was differentiated and the result was then compared with the integral from which it was derived. As a result of this recheck, it appears to me that the previously submitted derivation is correct.

Starthaus directed me to a derivation that was posted on his PF blog. The following hyperbolic functions were received from Janus, a PF mentor, of which a partial quote follows. Thanks to both for the information provided.

The equation for finding the time needed to reach any given speed (in ship time) is:

$$T = \frac{c}{a} \tanh^{-1} \left ( \frac{v}{c} \right )$$

This can converted to "stay at home" time by

$$t=\frac{c}{a} \sinh \left ( \frac{aT}{c}\right )$$

(End quote)

Respectfully, the previous equations have not been verified with any of many attempted spreadsheet solutions. The closest computer solution obtained for time t at 99.9% of the speed of light is on the order of twice the hyperbolic function solution time for [T]. This represents a significant difference in figures of about 50%.

Computer spreadsheet solutions track the following inverse sine function over the entire velocity curve from start to 99.9% of light speed. The difference of both solutions is within 0.0234%, for all points on the velocity curve.

This following equation is the attempted relativistic solution for velocity, as a function of time, where the units are as follows. Angles (radians), t = Time (seconds), c = Speed of light (2.998E8 m/s), A = Initial acceleration, v= (Final Velocity)

$$v =c[\sin(At/c)]$$

The following equation is the proposed solution for time to any arbitrary speed, which is less than light speed.

$$t =(c/A)[\arcsin(v/c)]$$

Note: The following constraints apply for the preceding equations;

$$[At/c]<(\Pi/2)$$

$$t <(1/A)[(\Pi)(c)/2)]$$

The following example illustrates the problem that I am working on.

Assume a 1 kg mass rocket, which has a miniature rocket power plant within. Assume no loss of mass due to expenditure of fuel. Assume that the power plant is capable of continuous propulsion and starts the rocket’s journey with an initial acceleration of one Earth gravity (1.0G). Since f=ma, the continuous applied force will be (1 kg)(9.81 m/s^2)= 9.81 kg-m/s^2. The power plant’s reaction force F will be held constant throughout the mission, at the figure obtained.

With rocket velocity approaching the speed of light, its mass will approach infinity in the limit. As a result, one can expect the rocket’s acceleration (a = f/m) to decline toward zero. At mission’s end, that wonderful power plant continues to output the same constant reaction force F, as it did from the start.

Initially, the spreadsheet solved for about 10,000 points along the velocity curve. However, that spreadsheet has now expanded to solve for about 30, 000 points.

The velocity was changed in uniform steps of 10,000 m/s from rest to very near the speed of light. At each step, the time t for that particular interval was computed. The total time t of all steps or intervals was then summed.

The spreadsheet has been solved with three different approaches. All approaches yield exactly the same results. For all approaches, equations f=ma and v=at apply for each sub-interval. The starting force F=ma is held constant during the entire mission.

Approach I: Use the relativistic mass formula for change of mass, as a function of velocity.

Approach II: Use the relativistic acceleration formula for change of acceleration, as a function of velocity. (The initial or starting force F=ma is held constant.)

Approach III: Use the relativistic time formula for change of time, as a function of velocity.

The following partial table of solutions (selected) was obtained. Results are reported in days, of twenty four (24) hours each (86, 400 seconds). Velocity units are meters per second. (Sorry; table poorly formatted)

(% C) (Velocity) (Spreadsheet) (Inverse Sine Function)
(10.0) (29,980,000) (35.430) (35.430)
(50.0) (149,900,000) (185.204) (185.203)
(90.0) (269,820,000) (396.083) (396.075)
(99.9) (299,500,000) (539.910) (539.783)

Constants used:
Initial acceleration = 9.81 m/s (1.0G)
Spaceship mass = 1.0 kg
Velocity of light = 2.998 x 10^8 m/s

This would perhaps be a good high school physics/ math/ computer science project. I’d be interested in receiving anyone else’s figures.

Thanks very much

10. Jun 26, 2010

### starthaus

The above formula is wrong. We've (Janus and I) given you the correct formulas earlier in this thread.

11. Jun 26, 2010

### Star Drive

Thanks for your feedback. I have made a considerable effort to prove the formulas offered by both you and Janus. However, I can't replicate the computation results of those formulas with a computer. Basically, it should be a matter of integrating simple physics relationships over time. Sometimes a derivation can look impressive and elegant, as this one does on your blog. However in any science, experimentation (testing) should and must be used to verify theoretical results. Have you ever tested those formulas? Thanks very much

12. Jun 27, 2010

### Austin0

Hi Is there some simple way to explain the presense of the trigh functions in the maths?
Were they derived from the hyperbolic world lines of accelerating systems n Minkowski spacetime or were both derived from Lorentz maths on a deeper level?
Would it be possible to present the acceleration equations using the usual forms of the gamma math without the trigh functions??

Was the concept of proper acceleration as infinitesimal short accelerations relative to a series of ICMIF 's integral and neccessary to the derivation of the equations or is it just an interesting way to view and explain it??
Thanks

Last edited: Jun 27, 2010
13. Jun 27, 2010

### espen180

I guess the fact that $$x^2-t^2$$ (c=1) is invariant under Lorentz boosts are a hint towards hyperbolic trajectories through space-time under constant proper acceleration?

14. Jun 27, 2010

### starthaus

No, the formulas have not been tested. Neither have your formulas but yours have an internal inconsistence while the ones given by Janus and me , do not.
If you want to disprove the formulas we've shown you , you will need to find an error in their derivation. Since they are textboook formulas, no one has found an error and they are widely accepted. On the other hand, it is easy to find the error in your formulas. If you compared your derivation with the one in the blog, you should have no problem in finding it by yourself. it is a useful exercise.

15. Jun 27, 2010

### starthaus

No.

$$F=\frac{dp}{dt}=\frac{d}{dt}\frac{mv}{\sqrt{1-(v/c)^2}}=m\frac{d}{dt}\frac{v}{\sqrt{1-(v/c)^2}}$$

Rest of your derivation is wrong since you start with an error.

16. Jun 27, 2010

### Star Drive

Revised Derivation

The original posting was correct, however it needed some refinement. Notations and extra steps have been added for clarity. In the mean time, the integration by form from CRC tables has been proven. This was done by differentiating the function back to the original integral and it was exactly the same. There are also direct ways to integrate that function, which involve trigonometric substitutions. However, it is best to avoid the solution path which leads to logarithms in the result.

The original posting was improperly formatted for LaTex notation. A few inconsistencies of upper/ lower case notation have also been corrected. Hopefully, this posting will correct those issues.

Upper case (capital) letters represent the initial or rest quantities as viewed by an Earthly observer. In example; A = Initial Acceleration, M = Initial Mass., F = Initial Force. Quantities A, M and F are regarded as constants throughout this derivation.

Lower case Letters represent the instantaneous (changing) variables, as viewed by an Earthly observer. In example; a = Instantaneous acceleration, m = Relativistic mass “instantaneous”.

Two equivalent relativistic correction factors [T] follow;

$$[T]=1/\sqrt{1-v^2/c^2}$$

$$[T]=c/\sqrt{c^2-v^2}$$

Instantaneous acceleration;
a = dv/dt

Relativistic mass “instantaneous”;
m = M[T]

The following basic substitutions yield;
f = ma
f = M[T] dv/dt
f dt= M[T] dv
(f/M) dt = [T] dv

$$\int(f/M)dt = \int[T]dv$$

$$\int(f/M)dt = \int[c/\sqrt{c^2-v^2}]dv$$

$$(f/M)\int dt= c \int dv/ \sqrt{-v^2+c^2}$$

Integrate the previous equation by form with CRC integral tables to obtain the following result, or integrate it directly by one of several methods.

$$(f/M) t=[c/\sqrt{1}] [\arcsin(v\sqrt{1/c^2})]$$

Simplify the previous equation and substitute f = F because force is held constant.

As a result; Initial acceleration A = F/M = f/M

$$At=c[\arcsin(v/c)]$$

$$At/c=\arcsin(v/c)$$

$$\sin[At/c]=v/c$$

The following equation is the proposed relativistic solution for velocity, as a function of time, where the units are angles (radians) and time (seconds).

$$v =c[\sin(At/c)]$$

Note: The following constraints apply;
$$[At/c]<(\Pi/2)$$

$$t <(1/A) [(\Pi)(c)/2)]$$

The following equation is the derived solution for time to any arbitrary speed, which is less than light speed.

$$t =(c/A)[\arcsin(v/c)]$$

Units Follow:
Time: seconds
Velocity “v”: meters/second (m/s)
Speed of light “c”: 2.998 x 10^8 m/s
Mass “m”: kilograms
$$Pi = 3.141593$$

17. Jun 27, 2010

### Staff: Mentor

You cannot use this in F = ma when the force is parallel to the direction of motion. It works only when the force is perpendicular to the direction of motion. You need to use instead, $m = m_0 \gamma^3$ which is m = M[T]3 in your notation.

Some books and articles call these two versions of "relativistic mass", the transverse mass and the longitudinal mass respectively.

18. Jun 27, 2010

### starthaus

No, it is just as wronf as before, you aren't listening.

No, it doesn't. Not in relativity.

19. Jun 27, 2010

### Star Drive

After some digging in my old calculus book, I was able to directly integrate the desired function by trigonometric substitution. Consult the attached .jpg image for details. Please excuse the hand writing. Thanks

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20. Jun 27, 2010

### starthaus

You persist in integrating the wrong function. Can you please stop this nonsense?