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Is E=mγc2 correct? 
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#1
Nov612, 03:59 PM

P: 23

I've just saw this other equation E2=m2c4+p2c2 but what's going on with that? I thought E=mγc2 was the equation including relativistic mass. What is going on?



#2
Nov612, 04:15 PM

P: 378

Just algebra. If you substitute the expression for relativistic 3momentum (p=γmv) into E^{2}=m^{2}c^{4}+p^{2}c^{2}, you'll end up at E=γmc^{2}.



#3
Nov612, 04:27 PM

P: 23

I get it now, but, if you we're traveling near the speed of light and needed the Lorentz factor, would you make it E2=(mγ)2c4+(mvγ)c2 since you need to put in for relativistic mass in both places?



#4
Nov612, 04:41 PM

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Is E=mγc2 correct?



#5
Nov612, 04:46 PM

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Thanks
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One of the advantages of this formulation is that it works for photons and other massless particles as well. 


#6
Nov612, 05:08 PM

P: 23

That makes sense, but in the simplified thing E=mγc2, where does the γ come from then?



#7
Nov612, 05:28 PM

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PF Gold
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KE = (1/2) mv^2 ? 


#8
Nov612, 06:21 PM

P: 23

What I'm asking is what is the equation E2=m2c4+p2c2 used for if E=mγc2 is used for total energy? And is the γ in the just there for relativistic mass or what is that there for?



#9
Nov612, 06:22 PM

P: 378




#10
Nov612, 06:29 PM

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#11
Nov612, 06:48 PM

P: 378

The gamma isn't "for" anything. Some people do like to define a "relativistic mass" and make the relativistic equations look like the Newtonian ones. Consensus around here is that this is not helpful  for example you need two different relativistic masses to deal with force, and the acceleration is not parallel to the force (in general) anyway. Better to acknowledge that from the beginning. The gamma is a consequence of being in a universe with a Minkowski spacetime. It is not a modification to Newton. If you wish to recover the low velocity limit, do a binomial expansion of gamma and neglect terms of order v^{4} and higher. But you can't get from Newton to Einstein that way. Does that make some kind of sense? 


#12
Nov612, 07:30 PM

P: 23

Do E2=m2c4+p2c2 and E=mγc2 give you the same answer, because, it doesn't seem like they do?



#13
Nov612, 08:01 PM

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#14
Nov612, 08:01 PM

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#15
Nov612, 08:25 PM

P: 23

But E2=m2c4p2c2 has v velocity in p so where does that go?



#16
Nov612, 10:58 PM

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Did you try the algebra?
Here are a few stages along the way. Try to fill in the gaps. E^2 = (mc^2)^2 + m^2γ^2v^2c^2 = (mc^2)^2 ( 1 + v^2/(c^2  v^2) ) = (mc^2)^2 (c^2 / (c^2  v^2) ) then E = mγc^2 


#17
Nov712, 01:40 PM

P: 23

I still can't figure out how to factor or whatever your doing to get to that!



#18
Nov712, 02:49 PM

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