Express velocity

  • #1
Hello everyone.
This is not an homework but I'd like to express velocity ##v## in this formula :
$$t' = \frac{t}{\sqrt{1-\frac{v^2}{c^2}}} $$
I've found
$$ v^2 = -(\frac{t^2}{t'^2}+1)c^2 $$

I'm sure that is wrong but I don't manage to express ##v## and find the correct answer.
Thanks in advance.
 

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  • #2
berkeman
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The 2nd equation looks to be close, with just one sign error. Can you show the steps you are trying to isolate v in the first equation? :smile:
 
  • #3
The 2nd equation looks to be close, with just one sign error. Can you show the steps you are trying to isolate v in the first equation? :smile:
If it's a problem with a sign maybe this is correct :
QSpiUJU.png
 

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  • #5
Yes, that looks fixed now, good work! What equation editor are you using BTW? :smile:
Yes, it works ! So finally we have :
$$v = \sqrt{(-\frac{t^2}{t'^2}+1)c^2}$$
Othewise I use ShareLaTex https://fr.sharelatex.com/ . Sorry, I'm french so it's the french version.
 
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  • #7
Janus
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Yes, it works ! So finally we have :
$$v = \sqrt{(-\frac{t^2}{t'^2}+1)c^2}$$
Othewise I use ShareLaTex https://fr.sharelatex.com/ . Sorry, I'm french so it's the french version.
Okay. So now you have c2 as a factor under the radical sign. Do you see a way to reduce this down further? And do you see a way of making the other part under the radical look cleaner?
 
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  • #8
Okay. So now you have c2 as a factor under the radical sign. Do you see a way to reduce this down further? And do you see a way of making the other part under the radical look cleaner?
In the radical we multiply by c^2 so we can just multiply the square root by c :
$$v = c\sqrt{(-\frac{t^2}{t'^2}+1)}$$
Also, dividing by a number is like multiplying by the inverse. So, I think we can have something like that :
$$v = c\sqrt{-{t^2}{t'^2}+1}$$
 
  • #9
Ibix
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In the radical we multiply by c^2 so we can just multiply the square root by c :
Correct (although you'd normally phrase it as "take out a factor of c outside the square root", or something like that).
$$v = c^2\sqrt{(-\frac{t^2}{t'^2}+1)}$$
So this isn't right.
Also, dividing by a number is like multiply by the inverse. So, I think we can have something like that :
$$v = c^2\sqrt{-{t^2}{t'^2}+1}$$
Only if you define ##t'=1/t'##, which is a bit confusing. You could define something like ##f'=1/t'##, then use that. It's not a standard usage, though.
 
  • #10
Janus
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In the radical we multiply by c^2 so we can just multiply the square root by c :
$$v = c^2\sqrt{(-\frac{t^2}{t'^2}+1)}$$
Also, dividing by a number is like multiply by the inverse. So, I think we can have something like that :
$$v = c^2\sqrt{-{t^2}{t'^2}+1}$$
Ibex already pointed out that there is an error in your first equation.
For the second part, I was thinking more along the lines of what happens when you reverse the order of the terms.
 
  • #11
Ibex already pointed out that there is an error in your first equation.
For the second part, I was thinking more along the lines of what happens when you reverse the order of the terms.
Reverse the order of the terms ??? Maybe you mean put +1 before ## -\frac{t^2}{t'^2}##
So we have :
$$ v = c\sqrt{1-\frac{t^2}{t'^2}}$$
 
  • #12
Janus
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Reverse the order of the terms ??? Maybe you mean put +1 before ## -\frac{t^2}{t'^2}##
So we have :
$$ v = c\sqrt{1-\frac{t^2}{t'^2}}$$
Right, Now consider that at times it is more convenient to work with velocities as expressed as a fraction of c such that ## \beta = \frac{v}{c} ##
so how would you express this equation using ##\beta ## ?
 
  • #13
Right, Now consider that at times it is more convenient to work with velocities as expressed as a fraction of c such that ## \beta = \frac{v}{c} ##
so how would you express this equation using ##\beta ## ?
So we have :
$$ \beta = \sqrt{1-\frac{t^2}{t'^2}}$$
 
  • #14
Janus
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So does the right hand side of the equation remind you of anything?
 
  • #15
So does the right hand side of the equation remind you of anything?
It looks like what we have in the initial equation :
$$ \sqrt{1-\frac{v^2}{c^2}}$$

But we change ##v## and ##c ## .
 
  • #16
Janus
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It looks like what we have in the initial equation :
$$ \sqrt{1-\frac{v^2}{c^2}}$$

But we change ##v## and ##c ## .
Right, so if we take ## \frac{t^2}{t'^2} ## and express it as ##\left ( \frac{t}{t'} \right )^2##

We get
$$\beta = \sqrt{1- \left ( \frac{t}{t'} \right )^2} $$

and
$$ \frac{t}{t'} = \sqrt{1- \beta^2} $$
 
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