1905 paper Does the Inertia of a Body Depend Upon its Energy-Content?

[edpell]

1905 paper "Does the Inertia of a Body Depend Upon its Energy-Content?

In his 1905 paper Einstein uses a Taylor expansion and keeps only the first term (an approximation) to derive E=mc^2 which is good if v<<c but if v is not <<c then we need more terms. If we keep the first three terms then mc^2 is divided by
(1 + 3/4(v/c)^2 + 15/8(v/c)^4)
have I done the Taylor series correctly? Do we all agree? This number of terms would be good as long as (v/c)^4 <<1 say up to v=0.4c. Is this correct?

 

[Meir Achuz]

Your expansion is wrong. The second term is (1/2)(v/c)^2 which lead to (1/2)mv^2.
Also, generally in an expansion like that if the second term isn't already small, there's not much point in continuing the expansion.

 

[edpell]

clem the Taylor expansion is
1 + (1/2)(v/c)^2 + (3/8)(v/c)^4 + (15/8)(v/c)^6 + ...
the 1 is canceled by the -1 in the previous equation
a factor of (1/2)(v/c)^2 is factored out so that Einstein gets
K0-K1 = (1/2)(E/c^2)v^2 by taking only the first term he then equates m=E/c^2
or E=mc^2
if we factor out (1/2)(v/c)^2 the Taylor expansion minus one is to four terms
1+3/4(v/c)^2+(15/8)(v/c)^4 which must appear below the mc^2 yes? no?

 

[Meir Achuz]

You mean you are finding corrections to (1/2)mv^2? That is an archaic, clumsy way to proceed, only of interest to historians.

 

[edpell]

In his paper Einstein says "Neglecting magnitudes of fourth and higher orders we may place..."
He seems to say that it is an approximation. I never knew that E = m$$c^2$$ is an approximation that only hold if $$(v/c)^2$$ <<1.

 

[Phrak]

In his 1905 paper Einstein uses a Taylor expansion and keeps only the first term (an approximation) to derive E=mc^2 which is good if v<<c but if v is not <<c then we need more terms. If we keep the first three terms then mc^2 is divided by
(1 + 3/4(v/c)^2 + 15/8(v/c)^4)
have I done the Taylor series correctly? Do we all agree? This number of terms would be good as long as (v/c)^4 <<1 say up to v=0.4c. Is this correct?

Einstein is not an easy read! One thing to keep in mind in reading Einstein is his humble respect for Newton. So he is careful not to say something like "I'm right, and Newton was wrong". You have to carefully read between the lines.

E=mc² is the accurate result. Einstein is claiming that the Newtonian result, E = E₀ + (1/2)mv² is an approximation and true when v<<c.

He claims that the expansion is accurate and the Newtonian result, E = E₀ + (1/2)mv² is an approximation.

E₀ doesn't really show-up in Newtonian energy because measurements of energy are always measurements of energy differences. For instance, when we calculate the energy in a battery we don't include the energy that is the mass of the battery. Since E₀ is the rest mass energy, it doesn't usually change except when elementry particles change into other elementry particles. E₀ drops-out of energy differerence equations except for these cases. So without atomic reactions it shows-up on both sides of an equation, and cancels:

$$E_0 + E_a = E_0 + E_b$$

$$E_a = E_b$$

 

[edpell]

So the politics of physics is getting in the way of a clear statement of the physics of physics (in Einstein's paper)? Of course he ended up with lifetime tenure at the Institute for Advanced Studies so I would have to say he made the smart choice.