
#1
Nov203, 02:23 AM

P: 51

what is the actual equation of e=mc^2? this is only the simplified equation, and i have forgotten the actual one already...




#2
Nov203, 04:44 AM

Emeritus
Sci Advisor
PF Gold
P: 10,424

You probably mean
E^{2} = p^{2}c^{2} + m^{2}c^{4}  Warren 



#3
Nov203, 07:09 PM

P: 21

what does the P stand for? and is that equation homogenous?




#4
Nov203, 07:26 PM

Emeritus
Sci Advisor
PF Gold
P: 10,424

what is the actual equation of e=mc^2?
It's a lowercase p, and it stands for (linear) momentum. I don't know what you mean by "homogenous."
 Warren 



#5
Nov303, 01:55 PM

P: 383

It all depends on whether you are a massist or an energist.
 A massist is willing to attribute mass values to anything, in any state of motion. For a massist, this m is really m_{0}, a mass attributed to something in its rest frame of reference. For a massist E^{2} = p^{2}c^{2} + m_{0}^{2}c^{4} p = mv are always true in any inertial frame. For light quanta, E = pc p = mc , because m_{0} = 0 for light quanta. But m = p/c = E/c^{2}, a mass value dependent upon total energy of a quantum. So E = mc^{2} is true for a light quantum as well as a particle with a nonzero rest mass.  An energist is willing to attribute energy values to anything, in any state of motion. For an energist, m can only be attributed to something in its rest frame, so the subscript 0 is never needed. For an energist, p^{2} = E^{2}/c^{2}  m^{2}c^{2} is always true in any inertial frame. The energy E must come from other physics. For light quanta, p = E/c is a given, so p^{2} = p^{2}  m^{2}c^{2} , so m^{2}c^{2} = 0 . Since c > 0, m = 0 for a light quantum. So, E = mc^{2}/(1  v^{2}/c^{2})^{1/2} only in the case of a particle with nonzero rest mass.  Most modern day physicists, especially highenergy physicists, tend to be energists rather than massists. 



#6
Nov403, 12:17 PM

P: 383

I said:
So, E = mc^{2}/(1  v^{2}/c^{2})^{1/2} only in the case of a particle with nonzero rest mass. I should have said: So, E = mc^{2}/(1  v^{2}/c^{2})^{1/2} only in the case of a particle with nonzero mass. 


#7
Nov403, 05:22 PM

P: n/a

http://www.geocities.com/physics_wor...ergy_equiv.htm If the particle is a tardyon (i.e. a particle which travels at speeds less than light) then m = m_{0}/sqrt[1(v/c)^{2}] Multiply both sides by c^{2} mc^{2} = m_{0}c^{2}/sqrt[1(v/c)^{2}] Substitute in E = mc^{2} to get E = m_{0}c^{2}/sqrt[1(v/c)^{2}] This equation can be rewritten as E^{2}  (pc)^{2} = (m_{0}c^{2})^{2} Pete 


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