3
$\begingroup$

If we begin with the equation for time dilation:

$$ \Delta t^\prime = \frac{\Delta t}{\sqrt{1-\frac{v^2}{c^2}}} $$

Now, for a particle moving in space with no potential energy,

$$ E = \frac{1}{2} mv^2 $$

So

$$ v^2 = \frac{2E}{m} $$

Then

$$ \Delta t^\prime = \frac{\Delta t}{\sqrt{1-\frac{2E}{mc^2}}} $$

$$ = \frac{\Delta t}{\sqrt{1-2}} $$

$$ = \frac{\Delta t}{\sqrt{-1}} $$

Explain please?

Improve this question
$\endgroup$
3
  • 9
    $\begingroup$ This $E = \frac{1}{2}mv^2$ is only correct in the limit of $v \ll c$ (the Newtonian or non-relativistic limit). The generally correct expression is $E = \sqrt{(mc^2)^2 + (pc)^2}$. It follows that rest of your work is also incorrect except in that limit. Never mix relativistically correct and Newtonian math. Unless you really know what you are doing, of course. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Aug 5, 2013 at 2:52
  • $\begingroup$ @dmckee that could probably be an answer $\endgroup$
    – David Z
    Commented Aug 5, 2013 at 2:55
  • 2
    $\begingroup$ After enough comment edits I'll even get the units right. ::sigh:: That's why particle physicists work in $c = 1$ units: because we're too dumb to handle SI units. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Aug 5, 2013 at 2:56

2 Answers 2

Reset to default
7
$\begingroup$

This $E = \frac{1}{2}mv^2$ is only correct in the limit of $v \ll c$ (the Newtonian or non-relativistic limit). The generally correct expression is $$E = \sqrt{(mc^2)^2 + (pc)^2}\,.$$ It follows that rest of your work is also incorrect.

Never mix relativistically correct and Newtonian math. Unless you really know what you are doing, of course.

Improve this answer
$\endgroup$
1
1
$\begingroup$

Generally, $E = \gamma mc^2$ which becomes $E = \frac 1 2 mv^2 + mc^2$ for $v$ << $c$.

You've gone wrong by equating $E$ to just kinetic energy, and then using $E$ equal to just the rest energy $mc^2$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ You could taylor expand if you want to not use the slow-motion approximation, (contd $\endgroup$
    – Abhimanyu Pallavi Sudhir
    Commented Aug 5, 2013 at 12:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.