Grasshopper
Okay, starting with Einstein's energy equation, then assuming some energy is added to the moving particle. It seems pretty obvious that energy is convertible into mass (because of the speed of light squared factor maybe? see below). But I have some questions. First, of course, does the following show that energy is interchangeable with mass:

E = mc^2/(1 - (v/c)^2)^(1/2)

Particle absorbs some energy E1:

E2 = (mc^2 + E1)/(1 - (v/c)^2)^(1/2)

Factor out c^2, which gives:

E2 = (m + E1/c^2)c^2 /(1 - (v/c)^2)^(1/2)

But an energy divided by a velocity squared leaves units of mass (kilogram meters^2/seconds^2, divided by meters^2/seconds^2 ).
Looking at the equation then, we could say that m + E1/c^2 equals a new mass, M, so that

m + E1/c^2 = M Which can be written in terms of Einstein's energy equation:

E2 = Mc^2/(1 - (v/c)^2)^(1/2)

Which seems to indicate that adding energy E1 to E yields an increase in mass.

Is that fairly decent so far? If so, here's my problem:

If the above is right, why can't we make the same argument for Newtonian physics? (or maybe, why didn't they?)

Starting with the kinetic energy equation:

E = (1/2)mv^2

E2 = (1/2)mv^2 + E1

factor out (1/2)v^2, which gives:

E2 = (m + E1/v^2) *(1/2)v^2

where E1/v^2 is again in units of mass, because an energy divided by a squared velocity leaves units of mass. This seems to me to indicate that even in pre-Einstein physics energy is interchangeable with mass. Clearly this is incorrect (and yes, it is purely academic, since obviously Newton's energy equations are less accurate than Einstein's).

So, can someone please explain to me what is wrong with all this? Was I correct in the first part (relativity part)?

Thanks, as always!
MjolnirPants
It looks like yer doin the math right, but...
I don't think yer accountin fer momentum.
m=sqrt E^2-(pc)^2/c^2 where p is the total momentum. (I hope I wrote that right, I'm not used to usin plain text fer math notations...)
Feel free to correct me if I'm wrong, as I don't have time to do the math, but I'm purty sure that'll express that extra energy in terms o momentum, not mass.
MjolnirPants
Jes expoundin on my last post, which I made in kinda a rush.
What I'm sayin is that if ya plug some values into yer equation an solve it, then plug those same values into the energy-momentum equation an solve it, you'll see yer extra mass bein added to p instead o to m.
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