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Old Apr 14th 2008, 10:02 PM   #1
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Electron Velocity

Is there a way to calculate this in a Non-Calculus way?

I saw one of my friends try to use a relativistic view, specifically:

$\displaystyle v = \sqrt{\frac{E^2 - m_0^2c^4}{m^2c^2}}$

I know that this can be simplified even more, but he chose to leave it as such... I don't even know if he was right or not to use this equation... That leaves me to my second question...

If you cannot use a non-calc method to do this, then what is the BEST way to do it?
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Old Apr 15th 2008, 05:03 AM   #2
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Originally Posted by Aryth View Post
Is there a way to calculate this in a Non-Calculus way?

I saw one of my friends try to use a relativistic view, specifically:

$\displaystyle v = \sqrt{\frac{E^2 - m_0^2c^4}{m^2c^2}}$

I know that this can be simplified even more, but he chose to leave it as such... I don't even know if he was right or not to use this equation... That leaves me to my second question...

If you cannot use a non-calc method to do this, then what is the BEST way to do it?
Ummmm....What problem are you trying to solve? It looks like you are trying to find the speed of the moving frame such that the electron has a specific energy?

-Dan
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Old Apr 15th 2008, 08:40 AM   #3
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Well, it deals with Linear Accelerators.

We're dealing with energies between 6 and 25 MeV.

I just want to know the best way to find the range of electron velocities for the given energy range.
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Old Apr 15th 2008, 09:09 AM   #4
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Originally Posted by Aryth View Post
Well, it deals with Linear Accelerators.

We're dealing with energies between 6 and 25 MeV.

I just want to know the best way to find the range of electron velocities for the given energy range.
Oh okay. The energy given is going to be the kinetic energy of the electrons, so
$\displaystyle K = (\gamma - 1)m_0c^2$
which leads to

$\displaystyle v = \left ( 1 - \frac{m^2c^4}{(K + mc^2)^2} \right ) c$

-Dan
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Old Apr 15th 2008, 05:06 PM   #5
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If I may ask... What is the $\displaystyle \gamma$ for?
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Old Apr 16th 2008, 07:58 PM   #6
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Would the gamma be the Lorentz factor?

$\displaystyle \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$
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Old Apr 17th 2008, 05:03 AM   #7
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Originally Posted by Aryth View Post
Would the gamma be the Lorentz factor?

$\displaystyle \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$
Yes, sorry I didn't specify that earlier.

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