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Posted
  On 11/21/2014 at 12:48 PM, Strange said:

 

Same as anything else:

\displaystyle m_o c^2 \left(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} - 1 \right)

Do you think scientists measure neutrino KE by Joules?

Posted (edited)
  On 11/21/2014 at 2:18 PM, DimaMazin said:

Do you think scientists measure neutrino KE by Joules?

 

:confused:

 

I don't expect they measure the mass in kg, either. But you can multiply by the appropriate factor to convert to whatever units you want; it doesn't change the equation.

Edited by Strange
Posted
  On 11/21/2014 at 2:18 PM, DimaMazin said:

Do you think scientists measure neutrino KE by Joules?

 

eV is more likely because it's the appropriate scale for atomic/nuclear/particle systems, but ultimately it doesn't matter since the conversion is trivial.

Posted
  On 11/21/2014 at 2:49 PM, Strange said:

 

:confused:

 

I don't expect they measure the mass in kg, either. But you can multiply by the appropriate factor to convert to whatever units you want; it doesn't change the equation.

I only have wrongly called energy of rest mass by gravitational energy :doh: . Therefore you think I use another equation. :lol::rolleyes:

Posted
  On 11/22/2014 at 3:45 PM, DimaMazin said:

I only have wrongly called energy of rest mass by gravitational energy :doh: . Therefore you think I use another equation. :lol::rolleyes:

 

No, you were just talking nonsense. Not particularly funny.

Posted
  On 11/22/2014 at 4:01 PM, Strange said:

 

No, you were just talking nonsense. Not particularly funny.

EK(eV units)=(gamma-1)E0(eV units) * 6.24150934*1018 :P

E0 - rest energy

Posted
  On 11/23/2014 at 6:48 AM, DimaMazin said:

EK(eV units)=(gamma-1)E0(eV units) * 6.24150934*1018 :P

E0 - rest energy

 

That won't get you energy in Joules, it will get you an answer in eV2/J. However, this doesn't affect the physics of the problem. So what's the point?

Posted
  On 11/23/2014 at 10:22 AM, swansont said:

 

That won't get you energy in Joules, it will get you an answer in eV2/J. However, this doesn't affect the physics of the problem. So what's the point?

Yes.

E=gamma*E0

EK=E-E0

EK(eV units)=gamma*E0(eV units) - E0(eV units)=(gamma-1)E0(eV units)

Posted
  On 11/23/2014 at 2:52 PM, DimaMazin said:

Yes.

E=gamma*E0

EK=E-E0

EK(eV units)=gamma*E0(eV units) - E0(eV units)=(gamma-1)E0(eV units)

 

Is there a point to these rather mangled equations?

Posted
  On 11/23/2014 at 2:52 PM, DimaMazin said:

Yes.

E=gamma*E0

EK=E-E0

EK(eV units)=gamma*E0(eV units) - E0(eV units)=(gamma-1)E0(eV units)

 

What is the point?

 

While unit consistency is important (e.g. everything here has units of energy) the actual unit system used is a secondary consideration — it does not need to be specified in the equation.

Posted
  On 11/23/2014 at 3:04 PM, Strange said:

 

Is there a point to these rather mangled equations?

Let's check.

gamma=2

E0=1eV

then Etotal=2eV

and EK=1eV

E0=EK it's correct when gamma=2 :P

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