Momentum to Energy Conversion: Learn How it is Done

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Momentum for accelerated particles is expressed in MeV/c, and to convert this to energy, one can multiply the momentum by the speed of light (c). The relationship between momentum, energy, and mass is defined by the equation E² = (pc)² + (mc²)². This means that if the momentum is 0.79 MeV/c, multiplying by c gives an energy of 0.79 MeV. Understanding this conversion is crucial for analyzing particle physics and energy dynamics.
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Hello.

I'm reading some accelerator physics paper showing the momentum unit for the accelerated particles as MeV/c.

I'm not interested in the momentum but energy thus I really want to convert this to energy.

How it is done? Simply multiplying c? if so, there must be some reason for justify the action.

Please help me to go further.
 
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Momentum, energy, and mass are related by ##m^2 c^2=E^2/c^2-p^2##
 
Or, the way I like to remember it: ##E^2 = (pc)^2 + (mc^2)^2##.

E, pc and mc2 all have units of energy (MeV or joules or whatever)
 
Oh thanks people!

Thus..since momentum in the paper is, for example, 0.79 MeV/c so pc should be simply the momentum multiplied by c, 0.79 MeV right?
 
Right.
 
In an inertial frame of reference (IFR), there are two fixed points, A and B, which share an entangled state $$ \frac{1}{\sqrt{2}}(|0>_A|1>_B+|1>_A|0>_B) $$ At point A, a measurement is made. The state then collapses to $$ |a>_A|b>_B, \{a,b\}=\{0,1\} $$ We assume that A has the state ##|a>_A## and B has ##|b>_B## simultaneously, i.e., when their synchronized clocks both read time T However, in other inertial frames, due to the relativity of simultaneity, the moment when B has ##|b>_B##...

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