Is there a connection between energy derivation and Newton's Law of Motion?

[PhiPhenomenon]

Hey guys,

I was trying to reverse engineer Einstein's formula for energy, E=γmc^2 by re-engineering Newton's Law of motion, F=ma. I was talking with my physics prof about deriving energy from this because I got two different answers but it gets weird because the incorrectly derived formula works.

F = ma = dp/dt -> F dx = mv dv -> E = ∫ F dx = ∫ mv dv = .5mv^2 + C

Then I did this

F = dp/dt = v dp (dx/dx) -> F dx = v dp -> E = ∫ v dp = vp + C

My prof told me that my last integral, ∫ v dp, is an illegal operation and that v must be converted into p/m which makes sense because it then follows that E = .5mv^2 = p^2/2m.

I did some fiddling around though because I was curious and I was able to derive E = γmc^2 and the formula always works. What I derived from the above was:

E = vp + C = vp + mc^2/γ, p=γmv

I'm just curious if anyone can point out why it works.

Also, I know that energy for a photon is equal to |p|c. When m=0 then v=c and I find it interesting that the rest mass, m/γ, is introduced above given energy equivalence. So, bad math or is there something to this?

 

[robphy]

Your professor is correct in that "∫ v dp = vp + C " is incorrect because v is not a constant... it is a function of p... namely (p/m) as you were told.

There are some unclear/inconsistent variable uses here:

" F dx = v dp -> E = ∫ v dp "
implies that E is the [relativistic] kinetic energy (via the work-energy theorem)

" E = γmc^2 "
implies that E is the relativistic energy and m is the rest-mass

but then you say
"find it interesting that the rest mass, m/γ,"
then m in this sentence must be the so-called "relativistic mass"

Now, when you say "E = vp + C = vp + mc^2/γ, p=γmv"
then, making the substitution for p=γmv [where m must be the rest mass],
one gets
E = vp + C = v(γmv) + mc^2/γ = m (γv^2+ c^2/γ)
where the m is factored out to unravel the expression... which is not recognizable as anything meaningful.


So, I think you have to go back and fiddle around some more... but be consistent in the meaning of your variables and don't do any illegal mathematical operations.

 

[PhiPhenomenon]

Thanks for the reply.

I still can't figure out why it works. At first thought that since dv = 0 the first integral is legal but that makes F = 0... So I still have no idea.

I also messed up in my original response, M/γ is actually the rest mass divided by gamma, not the formulation for rest mass.

I went a little further with it and also found a formula for kinetic energy using this:

Ek = PV(γ + 1) + Mc^2 (γ^-1 - γ^2) = γmc^2 - mc^2, P=γmc^2

Again, the math works but is it significant to anything or just a silly way of saying (gamma)mc^2 - mc^2?

 

[PhiPhenomenon]

Now, when you say "E = vp + C = vp + mc^2/γ, p=γmv"
then, making the substitution for p=γmv [where m must be the rest mass],
one gets
E = vp + C = v(γmv) + mc^2/γ = m (γv^2+ c^2/γ)
where the m is factored out to unravel the expression... which is not recognizable as anything meaningful.

After reading that over that is exactly what I got but by my calculations:

E = vp + C = v(γmv) + mc^2/γ = m (γv^2+ c^2/γ) = γmc^2 when I tried plugging in a few values assuming a=0.

 

[PhiPhenomenon]

E.g.:

Particle with a mass of 1000 MeV/c^2 is traveling at 0.6c -> γ = 1.25

E = 1.25 * 1000 MeV/c^2 * c^2 = 1250 MeV

Alternatively:

E = 1000 MeV/c^2 (1.25 * (.6c)^2 + c^2 / (1.25)) = 1000 MeV/c^2 (.45 c^2 + 0.8 c^2) = 1000 MeV/c^2 (1.25 c^2) = 450 MeV + 800 MeV = 1250 MeV

 

[PhiPhenomenon]

Just figured it out, I inadvertently made an expansion.

(γc^2) = γv^2 + c^2/γ

E = m(γc^2) = m(γv^2 + c^2/γ) = PV + mc^2/γ