# Time To Near Light Speed

#### Star Drive

I'm not saying there are any transverse forces! What I'm saying is that you're using the transverse mass m = M[T] (in normal notation, $m = m_0 \gamma$) instead of the longitudinal mass m = M[T]^3 (in normal notation, $m = m_0 \gamma^3$). Your force is along the direction of motion so you should be using the longitudinal mass.
I took the existing spreadsheet and changed the relativistic mass multiplier from [T] to [T]^3. (third-power). With that computation made on some 30,000 intervals for t, the total time sum jumps to a very large value of 7,967.1 days (86,400 seconds/ day). The arc sine function solution is on the order of 539.8 days. Before [T] was changed, the spreadsheet matched the arc sine solution. Both solutions are made at 99.9% of c.

Last edited:

#### starthaus

I took the existing spreadsheet and changed the relativistic mass multiplier from [T] to [T]^3. (third-power). With that computation made on some 30,000 intervals for t, the total time sum jumps to a very large value of 7,967.1 days (86,400 seconds/ day). The arc sine function solution is on the order of 539.8 days. Before [T] was changed, the spreadsheet matched the arc sine solution. Both solutions are made at 99.9% of c.
There is no "arcsine" in the correct derivation.

#### Star Drive

There is no "arcsine" in the correct derivation.
I got that point long ago, as you have so often pointed out.

I'm searching for a computer validation of "any" function that works. Its not about who is right or wrong.

If the function, which you offered is proven, I'll admit that and perhaps discuss the reason(s) why it turned out to be the "correct" one.

I've been reading and learning that the subject is far more complex than I initially thought.

So give me a break here because I may yet give you some satisfaction, in knowing that "I" suceeded in proving you to be correct.

If you'd like to help, offer me some basic physics relationships which ought to work for any small sub-interval (10,000 m/s steps) of velocity from "rest" to 99.9%c. Solve me something for t, as a function of velocity, with initial force f held constant. I'll program it and see how it stacks up.

#### starthaus

I got that point long ago, as you have so often pointed out.

I'm searching for a computer validation of "any" function that works. Its not about who is right or wrong.
The function that "works" is the solution of the equation of motion. I've given you that information already.

If the function, which you offered is proven, I'll admit that and perhaps discuss the reason(s) why it turned out to be the "correct" one.
What do you mean by "proven" ? Do you understand formal proofs? Do you understand that what you've been writing is incorrect?

So give me a break here because I may yet give you some satisfaction, in knowing that "I" suceeded in proving you to be correct.
It would be much more productive if you understood your errors. The solution we (several of us) have given you is known to be correct.

If you'd like to help, offer me some basic physics relationships which ought to work for any small sub-interval (10,000 m/s steps) of velocity from "rest" to 99.9%c. Solve me something for t, as a function of velocity, with initial force f held constant. I'll program it and see how it stacks up.
You have been given the correct formulas, just write the program for evaluating the functions.

#### Ich

Solve me something for t, as a function of velocity, with initial force f held constant. I'll program it and see how it stacks up.
You already did the calculation. You just have to compare the result with Janus' formulas($T=\gamma v/a$).

#### Chronos

Gold Member
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.

#### Star Drive

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.

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

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

Note: The following constraints apply;

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

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

#### Ich

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.

#### Star Drive

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

#### starthaus

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.

#### Star Drive

No, it isn't true.

Please let me explain a bit more where I'm comming 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

#### starthaus

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

#### Ich

I'd expect the observed time t to change for me on Earth.
There's the http://en.wikipedia.org/wiki/Clock_hypothesis" [Broken] that states that time dilation is independent of acceleration. Relativistic mass has nothing to do with it.

Last edited by a moderator:

#### Star Drive

There's the http://en.wikipedia.org/wiki/Clock_hypothesis" [Broken] 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: