Derivation of E=mc2 from Four-Vector Definitions in Special Relativity

[Ziang]

In SR, fundamental concepts velocity, acceleration, force, momentum were defined as four-vectors.
https://en.wikipedia.org/wiki/Four-vector
Does someone show me a derivation of the equation E = mc² from those definition?

 

[Ibix]

In your link, see the last equation in the "four momentum" section. Consider the special case where velocity is zero and hence the three momentum $\mathbf p=0$.

 

[Ziang]

I mean how the equations E = γm_oc² and E² = c²pp + (m_oc²)² come form four vector definitions of fundamental concepts in SR?

 

[SiennaTheGr8]

If you take the four-momentum as a given, then you already have the second of your equations, because mass in special relativity is defined as the magnitude of the four-momentum:

$(mc^2)^2 = E^2 - (pc)^2$.

As for the relation $E = \gamma mc^2$, consider that the four-momentum is just the four-velocity scaled by mass:

$(E/c , \mathbf{p}) = \mathbf{P} = m \mathbf{V} = m (\gamma c, \gamma \mathbf{v} )$,

so $E = \gamma mc^2$ and $\mathbf{p} = \gamma m \mathbf{v}$.

 

[Ibix]

What didn't you understand about the derivation given in the wiki article? Apart from "the four position" which you shouldn't really regard as a four vector (the four velocity is fine) it seemed ok to me.

 

[Ziang]

If you take the four-momentum as a given, then you already have the second of your equations, because mass in special relativity is defined as the magnitude of the four-momentum:

$(mc^2)^2 = E^2 - (pc)^2$.

So the energy-momentum relation above is not a consequence but it is by definition.

 

[SiennaTheGr8]

So the energy-momentum relation above is not a consequence but it is by definition.

That depends on what your starting point is!

You asked how to derive those equations from the four-vectors given on the Wikipedia article you linked to. That's the question I answered.

If instead you had asked how those equations were originally derived in the early 20th century, the answer would be different.

 

[Ziang]

I would like to make my question clearer.
In Newtonian mechanics, they derived directly the kinetic energy from 3D-vector definitions of velocity v = ds/dt, acceleration a = dv/dt, F = ma, p = mv, W = F*x.

Now in SR, the fundamental concepts were defined as four-vectors.
Velocity U = dX/dτ
Acceleration A = dU/dτ
P = m_oU
F =dP/dτ
W = F*X
and γ = dt/dτ

My question is can we derive the mass-energy relation directly from 6 definitions above only?

 

[Ibix]

Yes. It's the modulus of P.

 

[Ziang]

Please show me.
Thanks,

 

[Ibix]

Please show me.

It's the last equation in the section on the four momentum in the wiki article you linked!

 

[Ibix]

A bit more explicitly, the modulus-squared of a four vector is its time like component squared minus the sum of its spatial components squared. This is invariant. Thus it must be equal to $m_0c^2$ from the definition of U in the rest frame.

 

[SiennaTheGr8]

I would like to make my question clearer.
In Newtonian mechanics, they derived directly the kinetic energy from 3D-vector definitions of velocity v = ds/dt, acceleration a = dv/dt, F = ma, p = mv, W = F*x.

Now in SR, the fundamental concepts were defined as four-vectors.
Velocity U = dX/dτ
Acceleration A = dU/dτ
P = m_oU
F =dP/dτ
W = F*X
and γ = dt/dτ

My question is can we derive the mass-energy relation directly from 6 definitions above only?

Are you familiar with the Minkowski dot product? If you have the four-vectors $\mathbf{Q} = (Q_t , \mathbf{q})$ and $\mathbf{W} = (W_t , \mathbf{w})$, then:

$\mathbf{Q} \cdot \mathbf{W} = Q_t W_t - (\mathbf{q} \cdot \mathbf{w})$.

Additionally, the relationship between the magnitude of a four-vector and its components is given by the Minkowski dot product of the four-vector with itself. For example:

$\mathbf{Q} \cdot \mathbf{Q} = Q_t^2 - (\mathbf{q} \cdot \mathbf{q}) = Q^2$

(where $Q$ is the magnitude of $\mathbf{Q}$).

Apply all that to the four-momentum $\mathbf{P} = (E, \mathbf{p}c)$, whose magnitude is $P = mc^2$:

$\mathbf{P} \cdot \mathbf{P} = E^2 - (\mathbf{p}c \cdot \mathbf{p}c) = P^2 = (mc^2)^2$.

 

[Ziang]

In the link, they write
P = (m_oγc, m_oγu)
then they write E = γm_oc² with no an explanation ?
I know it is Einstein's famous equation. But how the equation E = γm_oc² come from the 6 definitions that I mentioned.

 

[exmarine]

Quote the OP.

I am not as qualified to answer this as some others on this forum, but perhaps I can help, as I have wondered and studied the same question. E=mc^2 originally (I think) came from a second little (3 page) paper Einstein also wrote in 1905. I’ll try to attach the translation I have to this message – I assume that is ok to do. In this paper, he refers back to Section 8 of his big SR paper earlier in 1905 where he derived an expression for the transformation of energy of a “system of plane waves of light”. He then shows that the release of energy in the form of radiation from a body decreases it mass “exactly like the kinetic energy of the electron” (in Section 10 of his big SR paper). But you need to read and study his exact words rather than my paraphrase, etc.

Regards.

PS. I assume you have access to his big SR paper? I'll attach it too just in case.

 

[SiennaTheGr8]

In the link, they write
P = (m_oγc, m_oγu)
then they write E = γm_oc² with no an explanation ?
I know it is Einstein's famous equation. But how the equation E = γm_oc² come from the 6 definitions that I mentioned.

Perhaps this will satisfy you. (This is basically what Einstein did, but he wasn't using the four-vector formalism.)

Start with the time component of the four-momentum: $\gamma m c^2$.

Now do a binomial expansion of it (remember that $\gamma = (1 - \beta^2)^{-1/2}$, where $\beta = v / c$):

$mc^2 \left(1-\beta^2\right)^{-1/2} = mc^2 + mc^2 \dfrac{1}{2} \, \beta^2 + mc^2 \dfrac {3}{8} \, \beta^4 + mc^2 \dfrac{5}{16} \, \beta^6 + mc^2 \dfrac{35}{128} \, \beta^8 + \dots$

That second term is $.5 mc^2 \beta^2 = .5 mv^2$, which is just the Newtonian kinetic energy. All the speed-dependent terms to the right must be higher-order contributions to the kinetic energy (not easily measured at non-relativistic speeds), and the first term $mc^2$ must be an energy contribution that doesn't depend on speed at all (it's the invariant energy $E_0$). The invariant energy term and the kinetic energy terms must sum up to the total energy, which we label $E$.

So $E = \gamma mc^2$.

 

[Ibix]

Just to add to what @SiennaTheGr8 said, yes, Einstein is essentially asserting that $\gamma m_0c^2$ is energy on the basis that it's similar to the Newtonian expression for energy. You really need to do an experiment to show that it is actually relevant to the real world. Bertozzi did this and filmed it in a very 1950s style - it's easily findable on YouTube.

 

[Ziang]

Thank you all,
The binomial expression worked well. But it is possible for the exist of other expressions that have the term 0.5mv².
For example, the following expression has that term:
KE = 0.5mv²(β⁰ + β¹ + β² + β³ + ... )

 

[PeterDonis]

it is possible for the exist of other expressions that have the term 0.5mv2.

What does this have to do with the topic of this thread?

 

[Ibix]

Thank you all,
The binomial expression worked well. But it is possible for the exist of other expressions that have the term 0.5mv².
For example, the following expression has that term:
KE = 0.5mv²(β⁰ + β¹ + β² + β³ + ... )

You can make up infinitely many expressions that look like Newtonian kinetic energy to leading order. But they don't follow from Einstein's postulates or any equivalent system, and they won't in general match experiment even approximately (edit: outside the Newtonian regime, anyway). It would be basically numerology (algebrology?) rather than scientific reasoning.

 

[Ziang]

What does this have to do with the topic of this thread?

Sienna say Einstein's expression has the term 0.5mv². So my question is why it is the right one because other expressions have that term too.
I figured out the other one from four-vector definitions

 

[SiennaTheGr8]

Thank you all,
The binomial expression worked well. But it is possible for the exist of other expressions that have the term 0.5mv².
For example, the following expression has that term:
KE = 0.5mv²(β⁰ + β¹ + β² + β³ + ... )

Are you asking if the appearance of $.5mv^2$ in the binomial expansion of $\gamma mc^2$ might just be a coincidence, so that the other terms don't really have anything to do with "energy"?

The thing to keep in mind is that energy is mainly a useful quantity insofar as it's conserved. That $.5 mv^2$ appeared in the binomial expansion is extremely suggestive. But ultimately the "justification" for calling $\gamma mc^2$ "total energy" is experimental: is it a conserved quantity? Particle accelerators prove every day that it is.

 

[PeterDonis]

Sienna say Einstein's expression has the term 0.5mv2. So my question is why it is the right one because other expressions have that term too.

Because the 0.5mv^2 term is not the only one you have to get right. You have to get all of the terms right. Isn't that obvious?

 

[Ibix]

Sienna say Einstein's expression has the term 0.5mv². So my question is why it is the right one because other expressions have that term too.
I figured out the other one from four-vector definitions

That's not the point. The point is that the quantity $\gamma m_0c^2$ emerges from the maths, so is automatically interesting. And in the case where v<<c it looks similar to what Newton called energy. So it's at least plausible that it's a more broadly applicable expression that Newton merely approximated. Then you go to experiment to see of it predicts the energy/velocity relation correctly.

 

[Ziang]

I figured out the other one from four-vector definitions

Let me show you my work. Let me use I as a symbol of integrate
KE = I(FdX) = mI(AdX) = mI(UdU) = 0.5mU²
Replace U = γv,
I get KE = 0.5mv²(β⁰ + β¹ + β² + β³ + ... )

 

[SiennaTheGr8]

Let me show you my work. Let me use I as a symbol of integrate
KE = I(FdX) = mI(AdX) = mI(UdU) = 0.5mU²
Replace U = γv,
I get KE = 0.5mv2(β⁰ + β¹ + β² + β³ + ... )

First, are your X, A, and U supposed to be three-vectors or four-vectors?

Second, the relation $\mathbf{f} = m\mathbf{a}$ (three-vectors) does not hold in special relativity.

 

[Ziang]

First, are your X, A, and U supposed to be three-vectors or four-vectors?

Second, the relation $\mathbf{f} = m\mathbf{a}$ (three-vectors) does not hold in special relativity.

They are four-vectors F, A, U, X that I found on wiki.
https://en.wikipedia.org/wiki/Four-force

 

[SiennaTheGr8]

It looks like you're assuming that familiar equations from Newtonian mechanics still work in special relativity. Whether you're talking about three-vectors or four-vectors, that's a bad assumption.

Now, it turns out that the work–energy principle does still hold in special relativity:

$d E_k = \mathbf{f} \cdot d \mathbf{r}$

(where $\mathbf{f}$ is the force three-vector, and $d \mathbf{r}$ is the infinitesimal three-displacement).

However, there's no four-vector equivalent to this (that I'm aware of).

Also, the four-velocity $\mathbf{V}$ is not equivalent to $\gamma \mathbf{v}$. Where did you get that from? The quantity $\gamma \mathbf{v}$ is the celerity (or proper velocity, though I very much dislike that term).

 

[Ibix]

Also note that if the approach can be made to work it will yield the work done by the force, i.e. the kinetic energy $(\gamma-1)mc^2$, which does not include the rest mass energy.

 

[Ziang]

It looks like you're assuming that familiar equations from Newtonian mechanics still work in special relativity. Whether you're talking about three-vectors or four-vectors, that's a bad assumption.

Now, it turns out that the work–energy principle does still hold in special relativity:

$d E_k = \mathbf{f} \cdot d \mathbf{r}$

(where $\mathbf{f}$ is the force three-vector, and $d \mathbf{r}$ is the infinitesimal three-displacement).

However, there's no four-vector equivalent to this (that I'm aware of).

Also, the four-velocity $\mathbf{V}$ is not equivalent to $\gamma \mathbf{v}$. Where did you get that from? The quantity $\gamma \mathbf{v}$ is the celerity (or proper velocity, though I very much dislike that term).

1. In my work, I assume dE = F dX,
where F and X are four vectors. If my assumption is wrong, please show me the definition of work in four-vectors.

2. About U --> γv, I do the same way as they transfer P to p :)

 

[SiennaTheGr8]

There is no four-vector "version" of work.

Well, I suppose you could take the Minkowski product of the four-force and the four-displacement and think of it as the spacetime analogue of "work," but for the case of constant mass it will always equal zero (unless I've done something wrong):

Four-force (for constant mass): $\mathbf{F} = (\gamma \mathbf{f} \cdot \mathbf{v}/c , \gamma \mathbf{f})$

Infinitesimal four-displacement: $d\mathbf{R} = (c \, dt, d \mathbf{r})$

$\mathbf{F} \cdot d\mathbf{R} = (\gamma \mathbf{f} \cdot \mathbf{v}/c)(c \, dt) - \gamma \mathbf{f} \cdot d \mathbf{r} = \gamma \mathbf{f} \cdot d \mathbf{r} - \gamma \mathbf{f} \cdot d \mathbf{r} = 0$.

Not a very interesting quantity.

If there's a spacetime "analogue" of energy $E$, it would have to be the invariant energy $E_0$ (aka mass). Consider that the three-momentum (using $c=1$) is $\mathbf{p} = E \mathbf{v}$, and that the four-momentum is $\mathbf{P} = E_0 \mathbf{V}$.

If you're interested in learning more about relativistic mechanics, I suggest consulting a textbook!

 

[Ibix]

There is no four-vector "version" of work.

Well, I suppose you could take the Minkowski product of the four-force and the four-displacement and think of it as the spacetime analogue of "work," but for the case of constant mass it will always equal zero (unless I've done something wrong):

Four-force (for constant mass): $\mathbf{F} = (\gamma \mathbf{f} \cdot \mathbf{v}/c , \gamma \mathbf{f})$

Infinitesimal four-displacement: $d\mathbf{R} = (c \, dt, d \mathbf{r})$

$\mathbf{F} \cdot d\mathbf{R} = (\gamma \mathbf{f} \cdot \mathbf{v}/c)(c \, dt) - \gamma \mathbf{f} \cdot d \mathbf{r} = \gamma \mathbf{f} \cdot d \mathbf{r} - \gamma \mathbf{f} \cdot d \mathbf{r} = 0$.

Not a very interesting quantity.

But it does include a direct statement that the change in the zeroth component of the four momentum is the change in energy. Which is relevant here.

 

[Ziang]

If you're interested in learning more about relativistic mechanics, I suggest consulting a textbook!

In textbooks, they derive the mass-energy relation by figuring out the work
dE = fdx
where f = dp/dt and p = γmv.
My question is that if the force f > 0 "violate" the SR range, because SR deal with inertial moving objects only?

 

[SiennaTheGr8]

SR does not only deal with inertial motion. It can handle acceleration just fine.

 

[Ibix]

Force is just much harder to use outside Newtonian physics. For example, the 3-force and the 3-acceleration aren't necessarily parallel. Generally, people seem to use a Lagrangian or Hamiltonian approach instead.