Why Is It Impossible To Reach The Speed Of Light?

[diazona]

You're missing something in there somewhere, the units don't match up. Maybe should it be v²/c² in the numerator of the first fraction, instead of just v²?

Anyway, once you do that, it's just like adding any other fractions. Convert them to a common denominator and add the numerators.

 

[RK1992]

You're missing something in there somewhere, the units don't match up. Maybe should it be v²/c² in the numerator of the first fraction, instead of just v²?

You're right.. I'll give it a go now I've corrected it

Edit: yep it was simple, lol..it's an elegant way of showing the acceleration decreases, I like it.

 

[Fredrik]

I'm using units such that c=1. If you want to restore factors of c, just replace every v in my calculations with v/c.

Your equation says that v^2\gamma^3+\gamma=\gamma^3[/tex]. This is equivalent to v^2\gamma^2+1=\gamma^2, and we have<br /> <br /> v^2\gamma^2+1=\frac{v^2}{1-v^2}+\frac{1-v^2}{1-v^2}=\frac{1}{1-v^2}=\gamma^2<br /> <br /> By the way, I found the lack of parentheses in your expression confusing at first. You can get parentheses of different sizes with \big(, \Big(, \bigg(, \Bigg(, or \left(. The last one has to be closed with \right).<br /> <br /> Edit: By the way, the calculations I did <a href="https://www.physicsforums.com/showthread.php?p=2605539" class="link link--internal">here</a> might be interesting (as an exercise, if nothing else).