What Are the Implications of Approaching the Speed of Light for Space Travel?

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SUMMARY

The discussion focuses on the implications of approaching the speed of light for space travel, emphasizing that as an object accelerates towards light speed, its relativistic mass increases, causing acceleration to taper off. The key equations derived include the primary relativistic solution for velocity as a function of time, v = c[sin(At/c)], and the time to reach a given speed, t = (c/A)[arcsin(v/c)]. The conversation also highlights the challenges of maintaining constant acceleration due to increasing mass and the practical feasibility of such missions, particularly for a robotic interstellar probe with an initial mass of 100 metric tons.

PREREQUISITES
  • Understanding of relativistic physics concepts, including mass-energy equivalence.
  • Familiarity with calculus, particularly integration techniques.
  • Knowledge of fundamental physics equations: F=ma and a=dv/dt.
  • Basic understanding of trigonometric functions and their applications in physics.
NEXT STEPS
  • Explore the use of hyperbolic functions in special relativity calculations.
  • Research the implications of relativistic mass increase on spacecraft design.
  • Learn about the practical challenges of constant acceleration in space travel.
  • Investigate advanced propulsion systems capable of achieving significant fractions of light speed.
USEFUL FOR

Physicists, aerospace engineers, and anyone interested in the theoretical and practical aspects of high-speed space travel and relativistic physics.

  • #31
Wow, a little relativity might be handy star drive. Corrections become significant at high energy levels. Yours, unfortunately, fail on more levels than I care to acknowledge.
 
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  • #32
For all newcomers to this thread, the two equations that I derived and posted previously are incorrect. Please accept my apology for those errors.

Thanks to jtbell, whom I think first had the insight into exactly what my problem was.

After additional study and some computer experiments, it is now clear that for relativistic mass computations, one cannot use m=M[T]. Instead they must use m=M[T]^3, where [T] is more conventionally known as “Gamma”.

I have already done an alternate derivation using m=M[T]^3 which resulted in an entirely different function solution that appears to be promising.

I tried to do something interesting and honestly thought that I had a computer supported confirmation that it was correct or I would never have posted the original piece.

A poor choice of terminology was to “prove”. Instead, confirm by computer integration of very small intervals, would have been a better way to say it.

When wrong, admit that and learn why. I see through a glass darkly. Perhaps someday I may see more clearly.

This is the end of this thread for me. The Forum Master is welcome to pull it down, as unproductive.

Thanks for your feedback.


Incorrect
<br /> v =c[\sin(At/c)]<br />

Incorrect
<br /> t =(c/A)[\arcsin(v/c)]<br />

Note: The following constraints apply;

Incorrect
<br /> [At/c]&lt;(\Pi/2)<br />

Incorrect
<br /> t &lt;(1/A) [(\Pi)(c)/2)]<br />
 
  • #33
I have already done an alternate derivation using m=M[T]^3 which resulted in an entirely different function solution that appears to be promising.
It should be what I've written in #30.
When wrong, admit that and learn why. I see through a glass darkly. Perhaps someday I may see more clearly.
Starting from #24, you can derive the result by using the chain rule.
This is the end of this thread for me. The Forum Master is welcome to pull it down, as unproductive.
This thread was as productive as it can get.
 
  • #34
Hi
I thought I was done, but now I have a followup question.

Let [T] = Gamma

Usually, I see this
t=t' [T]

But since for an accelerating ship along the axis of flight
m=M[T]^3

It seems intuitive that perhaps
t=t' [T]^3

I got some answers from my spreadsheet that suggest this may be true.

However, I've been manipulating the math very much and I can't seem to prove it.

Is this true?

Thanks
 
  • #35
Star Drive said:
Hi
I thought I was done, but now I have a followup question.

Let [T] = Gamma

Usually, I see this
t=t' [T]

But since for an accelerating ship along the axis of flight
m=M[T]^3

It seems intuitive that perhaps
t=t' [T]^3

No, it isn't true.
 
  • #36
starthaus said:
No, it isn't true.

Thanks for your feedback.

Please let me explain a bit more where I'm coming from.

Case 1.
As a ship is moving through space at say 0.9c, but is 'coasting' at steady velocity.
Here it seems clear that to me, that an Earthly observer would see.
t=t'[T]

Now assume the same scenario, but the ship is also under 1G acceleration.
m=m'[T}^3

I'd expect the observed time t to change for me on Earth. But I'm still unclear about what the increasing mass (cubed) does to what I would observe. (I haven't had calculus and D.E. in over 20 years, so please keep it basic. Make it verbal if you can.)

Thanks
 
  • #37
Star Drive said:
Thanks for your feedback.

Please let me explain a bit more where I'm coming from.

Case 1.
As a ship is moving through space at say 0.9c, but is 'coasting' at steady velocity.
Here it seems clear that to me, that an Earthly observer would see.
t=t'[T]

Now assume the same scenario, but the ship is also under 1G acceleration.
m=m'[T}^3

I'd expect the observed time t to change for me on Earth. But I'm still unclear about what the increasing mass (cubed) does to what I would observe. (I haven't had calculus and D.E. in over 20 years, so please keep it basic. Make it verbal if you can.)

Thanks

I can't make it verbal, I have shown you the complete correct solution, you should try to understand it.
 
  • #38
I'd expect the observed time t to change for me on Earth.
There's the http://en.wikipedia.org/wiki/Clock_hypothesis" that states that time dilation is independent of acceleration. Relativistic mass has nothing to do with it.
 
Last edited by a moderator:
  • #39
Ich said:
There's the http://en.wikipedia.org/wiki/Clock_hypothesis" that states that time dilation is independent of acceleration. Relativistic mass has nothing to do with it.



Thats some of the best help which I've been offered.

Thanks :~)
 
Last edited by a moderator:

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