Solve the Mystery: Where is My Error in E=mc^2 Calculation?

[Alkatran]

I was trying to get to the equation E=mc^2, but I can't seem to.

I started with the example of a moving observer with a stationary observer with a beam of light.. (you know, c^2*t^2 = c^2*t'^2+t^2*v^2) and isolated v^2:

v^2 = c^2(t^2-t'^2)/t^2 = c^2(1-(t'/t)^2)

Then took the formula for kinetic energy: E(k) = mv^2/2 and replaced v^2:

E(k) = m*c^2*(1-(t'/t)^2)/2

Now.. I got the right equation except that for it to be correct (1-(t'/t)^2)/2 = 1 so 1-(t'/t)^2 = 2, (t'/t)^2 = -1, t'/t = sqr(-1)

...I think it's pretty obivous I made a mistake somewhere... but where?

 

[quartodeciman]

If you really intend for mv²/2 to equal mc², then you are in fact saying that v² = 2c², which makes v approximately equal to 1.4c. Oops! v is larger than c and outside the range of standard special relativity. Substitute v² = 2c² back into your original equation and you will get (t')² = -t² and your unfortunate conclusion.

I believe you are wrong to attempt pulling mc² out of mv²/2, tempting as it may look. In fact, mv²/2 comes out of mc² as a single term of an expansion. To be totally correct, the kinetic energy in special relativity is

mc²(1 - 1/SQRT(1 - v²/c²))

This expands in powers of v/c as

mc²(v²/2c² + 3v⁴/8c⁴ + 5v⁶/16c⁶ + ...)

and so on to more terms. Multiply mc² through this series to yield

mv²/2 + 3mv⁴/8c² + 5mv⁶/16c⁴ + ...

. If v is tiny enough compared to c, then only the first term really counts. Otherwise, it is just one term of a series for the kinetic energy.

---

I will (later) try to produce a demonstration of E=mc² here that is similar to your attempt, but uses another fact of special relativity.

 

[turin]

Alkatran,
I think your mistake is conceptual. E = mc² is not a statement of kinetic energy; it is either a statement of inherent energy (if m is the rest mass) or total energy (if m is the relativistic mass).

AFAIK, the famous relationship can be derived from conservation of 4-momentum, identification of the time-like part of the momentum with energy, and the idea that a contraction of two 4-vectors forms an invariant scalar.

 

[quartodeciman]

Get to E=mc².

Suppose we have a particle of mass m₀ standing still at time t=0. From that time on it gets accelerated continuously along the positive x line. We will represent its instantaneous speed as a variable v, function of time t.

Suppose the initial energy of the particle is E₀ and the instantaneous energy of the accelerated particle is represented by variable E, function of time t.

We define a variable m of time t as m₀/SQRT(1 - v²/c²). It has the dimension of a mass. It is usually called the relativistic mass. For those who don't like relativistic mass and say that the particle always has an invariant mass m₀, I assuage them by calling m just a defined variable. Its purpose is to allow us to represent the magnitude of the momentum of a particle in the way Newton did it: p = mv. This is then just p = m₀v/SQRT(1 - v²/c²), and I think all agree that this variable is the instantaneous magnitude of the momentum of our particle at time t.

In fact I will get a little more extra mileage from having this defined m variable around, as you will presently see.

Start with the following equation of motion:

dE/dt = vdp/dt

. This is close to the work definition of energy gained on the particle and I think it can be obtained by a change of integration variable.

Now p = mv, so

dE/dt = 1/m*(pdp/dt)

. pdp/dt = 1/2*d(p²)/dt

, so

2mdE/dt = d(p²)/dt

. This is our dynamic equation.

I think this equation is perfectly general for the Newtonian as well as the relativistic case. I haven't used the actual definition of the variable m yet. Suppose I replace my definition of m from the earlier paragraph to a simple m = m₀. Then the variable m does NOT depend on the speed v of the particle at any time t from 0 on. This means dm/dt = 0.

------
Newtonian case:
For m = m₀ (at all times), we can divide by nonzero 2m and get

dE/dt = 1/2m*d(p²)/dt = d(p²/2m₀)/dt

. Then

E - p²/2m₀ = constant-in-time

. Since at t=0 E=E₀ and p = 0,

E - p²/2m₀ = E₀ - 0²/2m₀

. lastly,

E = E₀ + p²/2m₀ = E₀ + 1/2*m₀v²

. This case gives us no clue what the initial energy E₀ is supposed to be.

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Sorry for the derail to the Newtonian m = m₀ case! I just felt compelled to show how the familiar kinetic energy term comes out of the dynamic equation.

------
relativistic case:
For m = m₀/SQRT(1 - v²/c²) at all times,

Square this.

m² = m₀²/(1 - v²/c²)

Multiply by that denominator and then by c² to get

m²c² - m²v² = m₀²c²

p² = m²v²

from the p = mv definition, so

m²c² - p² = m₀²c²

, and then

p² = m²c² - m₀²c²

.

Ok.

d(p²)/dt = d(m²c²)/dt - d(m₀²c²)/dt

d(p²)/dt = d(m²c²)/dt

, since dm₀/dt and dc/dt equal 0.

d(p²)/dt = d(m²)/dt*c² = 2m*dm/dt*c²

, since dc/dt is 0.

Remember the dynamic equation:

2mdE/dt = d(p²)/dt

. Substitute for d(p²)/dt:

2mdE/dt = 2m*dm/dt*c²

Divide by nonzero 2m

dE/dt = dm/dt*c²

dE/dt = d(mc²)/dt

, since dc/dt = 0. Then

E - mc² = constant-in-time

At t=0,

E₀ - m₀c² = constant-in-time

E - mc² = E₀ - m₀c²

E = E₀ + (m - m₀)c²

.

NOTE: If E₀ were to equal m₀c², then

E would equal mc² at each time t. That is a hint and NOT a proof, but very seductive.

In that particular case, the particle has an initial resting energy E₀ and hence a kinetic energy of

E - E₀ = m₀(1 - 1/SQRT(1 - v²/c²)).

 

[quartodeciman]

Just a bit of lily-gilding for my previous reply

I had earlier promised

I will (later) try to produce a demonstration of E=mc2 here that is similar to your attempt, but uses another fact of special relativity.

It may not be obvious that I did this in my last post. Let me try to show that now.

In the last post, I derived (for the relativistic case) this equation from the assumption m = m₀/SQRT(1 - v²/c²) for all times t:

p² = m²c² - m₀²c²

and went on to take the derivative to use in the dynamic equation. Now I will derive something straight away from this equation.

p² = m²v² from the definition of p, so

m²v² = m²c² - m₀²c²

. Rearrange,

m²c² = m₀²c² + m²v²

and

c²m² = c² m₀²+ v²m²

. Let m' be just another way to write m₀ (particle rest frame value for particle mass).

c²m² = c² m'²+ v²m²
.
This looks just like Alkatran's

c²t² = c² t'²+ v²t²

, except m's (mass variables) occur where t's (time variables) occurred.

------

SIDEBAR

I read somewhere that m/t = m'/t' is an invariant of interest that might be used to prove the dynamical relation

m = m⁰/SQRT(1 - v²/c²)

from the kinematical relation

t = t'/SQRT(1 - v²/c²)

. But I don't know what that mass/time is supposed to signify.

------

continuing ...

m²v² = m²c² - m₀²c²

m²v² = (m²- m₀²)c²

Factor difference of squares:

m²- m₀² = (m + m₀)(m - m₀)

So,

m²v² = (m + m₀)(m - m₀)c²

Divide by 2:

m²v²/2 = (m + m₀)/2*(m - m₀)c²

Divide by nonzero (m + m₀)/2:

(m²/((m + m₀)/2)* v²/2 = (m - m₀)c²

. The relativistic kinetic energy is

K = (m - m₀)c²

. So,

K = (m²/((m + m₀)/2)* v²/2

. Define a new mass variable M by m²/((m + m₀)/2.

Then

K = Mv²/2

. This looks like the classic formula but has a different mass element. Note that (m + m₀)/2 means just the average value of m between time 0 and the time t of interest.

Now

M = m*(m/average-m-value)

. When m₀ < m, then

m₀ < average-m-value < m

. So

1 < m/average-m-value < m/m₀

, all of these mass elements having strict positive values.

So, I have at long last explained Alkatran's puzzle by showing that the kinetic energy of a particle is

K = Mv²/2 = (m²/((m + m₀)/2)* v²/2

, and that is NOT the same thing as

mv²/2.

------

NOTE

If we let let v -> 0, then

m -> m₀

and

M -> m₀²/((m₀ + m₀)/2) = m₀

, which yields the classical kinetic energy as a limit value:

K -> m₀v²/2

.