Acceleration in special relativity

[timetraveller123]

Homework Statement

Homework Equations

The Attempt at a Solution

it is known that
$a_0$is the proper acceleration
$a= \frac{a_0}{\gamma^3}$
hence integrating it gets
$v = \frac{a_0t}{\sqrt{1 + (\frac{a_0 t}{c})^2}}$
but this is in terms of t how to make it in terms of t'

 

[Orodruin]

How is the proper time of any worldline related to the coordinate time?

 

[timetraveller123]

i don't really get your question it is too complicated for me i only know the basics could please simplify it for me thanks

 

[Orodruin]

How does $dt'$ relate to $dt$? Hint: Time dilation.

 

[timetraveller123]

ooh that
are you asking for this
$dt = \gamma(dt' + \frac{xu}{c^2})$
or
$dt = \frac{dt'}{\gamma}$

 

[Orodruin]

Which one describes time dilation?

 

[timetraveller123]

the second one

 

[Orodruin]

the second one

So how can you use that to find $t$ as a function of $t'$?

 

[timetraveller123]

the t in time dilation is the time of event measured by each observer but in my case t is the length of the time that has passes from the start

 

[timetraveller123]

oh okay so considering infinitesimal times is it
$dt' = \frac{dt}{\sqrt{1 - \frac{v}{c}^2}}\\$
substituting for v(t)
$dt' = \frac{\sqrt{c^2 + (a_0 t)^2}}{c} dt$
is it that ?

 

[Orodruin]

You have put $dt'$ on both sides in your post ...

 

[timetraveller123]

oh sorry now?

 

[timetraveller123]

but still this is the answer to the second part of the problem how to get answer to first without doing second part

 

[Orodruin]

I did not notice before, but you have mixed up the relation between $dt'$ and $dt$. The proper time elapsed should be smaller than the coordinate time and so the $\gamma$ has to go on the other side, i.e., $dt' = dt/\gamma$.

but still this is the answer to the second part of the problem how to get answer to first without doing second part

The way you are doing it you will get the answer to (b) first. Why is that a problem? It is possible to do it in a different way, but you should not let getting the result to a later part as a byproduct when solving an earlier part confuse you and think that it is impossible to do it that way just because you get a result that is asked for later on the way.