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

[pervect]

I can only guess what Ziag's background is. It appears to me he's struggling with the Lorentz transform, and has some issues understanding the relativity of simultaneity. Getting sidetracked into the more advanced methods needed to handle accelerating frames just isn't going to be productive for anyone who can't handle the much simpler case of special relativity in inertial non-accelerating frames.

I'm not sure if it well help Ziang, but a paper by Scherr , et al, <<link>>, entitled "The challenge of changing deeply held student beliefs about the relativity of simultaneity" at least illustrates the fact that he is not alone in having trouble with understanding the relativity of simultaneity. The Scherr et al paper is written for teachers and not students, still it might be helpful.

 

[Ziang]

To derive the mass-momentum relation, SR defined momentum $\mathbf{p} = \gamma m \mathbf{v}$.
After the relation derived, SR defined mass $(mc^2)^2 = E^2 - (pc)^2$.
Thus, the definitions are circular. The definition of mass is not clear.

 

[Ibix]

Not really. You define the four velocity, $U$, which is normalised to $c^2$, then you observe (from experiment) that there is an object-specific quantity that we'll call $k^{(i)}$ for the $i$th body such that the sum of $k^{(i)}U^{(i)}$ over all bodies is conserved in collisions. Then you do a Taylor series expansion and show that $k^{(i)}$ is the Newtonian mass of the $i$th body

Science is empty if you don't ground it in experiment.

 

[SiennaTheGr8]

@Ziang

I think you're confusing yourself by imagining that there's only one "path" through all of this material. Really, there are many possible starting points that will all lead to the same set of equations.

Yes, most people in the know today would probably agree that mass in SR is best defined as the magnitude of the four-momentum. But Einstein developed SR without recourse to the four-vector formalism at all! My point is that there's more than one way to skin a cat. Don't get hung up on seemingly contradictory approaches.

I suggest getting yourself a good book on special relativity and starting from the beginning. Get an understanding of the basic kinematics first. Then you'll be able to easily grasp the dynamics you're grappling with.

 

[Ziang]

You define the four velocity, $U$, which is normalised to $c^2$, then you observe (from experiment) that there is an object-specific quantity that we'll call $k^{(i)}$ for the $i$th body such that the sum of $k^{(i)}U^{(i)}$ over all bodies is conserved in collisions. Then you do a Taylor series expansion and show that $k^{(i)}$ is the Newtonian mass of the $i$th body

Science is empty if you don't ground it in experiment.

So what experiments could be to get the masses for the Sun, Earth, and Moon, based on definitions of mass in SR?

 

[Ibix]

So what experiments could be to get the masses for the Sun, Earth, and Moon, based on definitions of mass in SR?

Same as in Newton, except you can't use gravitational experiments - you need GR for that. You could collide them with something (probably a sequence of identical asteroid-sized bullets at a range of velocities). Or make them acquire a known charge and use electromagnetism.

For something like a planet none of the above is really practical. You really need GR, then you can study orbits. Or show that Newtonian gravity is the weak-field low-speed approximation and just use the masses implied by Newtonian physics - errors will be tiny.

 

[Herman Trivilino]

To derive the mass-momentum relation, SR defined momentum $\mathbf{p} = \gamma m \mathbf{v}$.

No, you can't both define and derive something. You can take an approach where you define momentum, or you can take an approach where you derive momentum based on definitions of other things. But however you do it, you will have to start with some definitions of things, things you define rather than derive.

And by the way, $\gamma m \vec{v}$ is a valid expression for the momentum of only massive, as opposed to massless, particles. $\left( \frac{E}{c^2} \right) \vec{v}$ is a more general expression that's valid for all particles.

So what experiments could be to get the masses for the Sun, Earth, and Moon, based on definitions of mass in SR?

You can't do experiments with these objects, we don't have the technology to manipulate them. What you can do is observe, and based on those observations compare their masses with each other. By determining the value of the gravitational constant $G$ you compare Earth's mass with the mass of the standard body. All measurements of mass are comparisons of one object's mass with another.

The norm of the energy-momentum four-vector equals what we measure when we measure mass. All we're doing is saying the momentum of these objects is zero, so their energy is the norm of their energy-momentum four-vector, which in turn is equivalent to their mass.

 

[Sorcerer]

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?

Here’s my favorite way. (Ibknow the discussion has moved on, but OP asked for a quick and easy derivation, so here is one way to do it that seems to work). Start with a spatial differential dx. Then divide by proper time (so that everyone agree with what you do going foward).

From there, use the usual kinemetics. Multiply by mass to get momentum, take the time derivative to get force, integrate over distance to use the work-energy theorem to get an expression for kinetic energy, then notice that kinetic energy = total energy minus potential energy, and you’re pretty much done.

 

[SiennaTheGr8]

Here’s my favorite way. (Ibknow the discussion has moved on, but OP asked for a quick and easy derivation, so here is one way to do it that seems to work).Start with a spatial differential dx. Then divide by proper time (so that everyone agree with what you do going foward).

From there, use the usual kinemetics. Multiply by mass to get momentum, take the time derivative to get force, integrate over distance to use the work-energy theorem to get an expression for kinetic energy, then notice that kinetic energy = total energy minus potential energy, and you’re pretty much done.

Not following you at the end there.

 

[Sorcerer]

Not following you at the end there.

You’ll get an expression for kinetic energy that has two terms separated by a minus sign.

T = E - U.

The U, the “potential energy” (that is, the “rest” energy that remains when speed u=0),” will end up being mc².I probably should have mentioned that.

 

[SiennaTheGr8]

Ah, that's not usually referred to as "potential energy." It's "rest energy" / "invariant energy" / "proper energy" / "mass-energy."

 

[Sorcerer]

Ah, that's not usually referred to as "potential energy." It's "rest energy" / "invariant energy" / "proper energy" / "mass-energy."

True. I used that phrasing more as an analogy since it works out to have the same form. Ambiguity aside, the derivation works like a charm. Set speed to zero and you get what OP wanted. *shrugs*

 

[Ziang]

The GR could lead to the relation too?

 

[Sorcerer]

The GR could lead to the relation too?

For small intervals of spacetime, spacetime is approximately flat. So if it's true in SR it's true locally in GR, if I understand it correctly.

 

[Ibix]

The GR could lead to the relation too?

GR is a generalisation of SR, so all results in SR can be derived using the tools of GR.

In this particular case, you'd probably start from the notion of a worldline and do a slightly more rigorous derivation of the four-velocity as the tangent to the worldline. The rest is the same.

 

[Ziang]

Could Newtonian mechanics concepts/absolute space and time lead to the relation as well?

 

[SiennaTheGr8]

Here’s my favorite way. (Ibknow the discussion has moved on, but OP asked for a quick and easy derivation, so here is one way to do it that seems to work).Start with a spatial differential dx. Then divide by proper time (so that everyone agree with what you do going foward).

From there, use the usual kinemetics. Multiply by mass to get momentum, take the time derivative to get force, integrate over distance to use the work-energy theorem to get an expression for kinetic energy, then notice that kinetic energy = total energy minus potential energy, and you’re pretty much done.

I'm sure I've seen this done for the rectilinear case (though I can't remember where), but I thought I'd give it a crack for the general case. The idea is that one has already obtained the equation for relativistic momentum $\vec p = \gamma m \vec v$ and plugs it into the "work" side of the work-energy theorem to see what pops out on the "energy" side. Using $\vec f = \dot{\vec p} = mc^2 \, d(\gamma \vec \beta)/d(ct)$:
$$\begin{split}
\int^{\vec r_f}_{\vec r_i} \vec f \cdot d \vec r &= mc^2 \int^{\vec r_f}_{\vec r_i} \dfrac{d (\gamma \vec \beta)}{c \, dt} \cdot d \vec r \\[3pt]
&= mc^2 \int^{\vec \beta_f}_{\vec \beta_i} \vec \beta \cdot \left( \gamma \, d \vec \beta + d \gamma \, \vec \beta \right),
\end{split}$$
where
$$\begin{split}
d \gamma &= d \left[ \left( 1 - \vec \beta \cdot \vec \beta \right)^{-1/2} \right] \\[3pt]
&= \dfrac{1}{2} \left(1 - \vec \beta \cdot \vec \beta \right)^{-3/2} \, d ( \vec \beta \cdot \vec \beta ) \\[3pt]
&= \gamma^3 ( \vec \beta \cdot d \vec \beta ) ,
\end{split}$$
so
$$\begin{split}
\int^{\vec r_f}_{\vec r_i} \vec f \cdot d \vec r &= mc^2 \int^{\vec \beta_f}_{\vec \beta_i} \vec \beta \cdot \left[ \gamma \, d \vec \beta + \gamma^3 ( \vec \beta \cdot d \vec \beta ) \vec \beta \right] \\[3pt]
&= mc^2 \int^{\vec \beta_f}_{\vec \beta_i} \Bigl[ \gamma \left( 1 + \gamma^2 \beta^2 \right) \Bigr] ( \vec \beta \cdot d \vec \beta ) \\[3pt]
&= mc^2 \int^{\vec \beta_f}_{\vec \beta_i} \gamma^3 ( \vec \beta \cdot d \vec \beta ) \\[3pt]
& = mc^2 \int^{\gamma_f}_{\gamma_i} d \gamma \\[3pt]
&= mc^2 \Delta \gamma \\[3pt]
&= mc^2 + mc^2 \Delta \left( \dfrac{1}{2}\,\beta^2 + \frac {3}{8} \, \beta^4 + \frac{5}{16} \, \beta^6 + \frac{35}{128} \, \beta^8 + \dots \right),
\end{split}$$
where the $\beta^2$ term suggests calling everything in the parentheses "kinetic energy" $E_k$ and the invariant first term "rest energy" $E_0$. Then it makes sense to define $\gamma mc^2 = \gamma E_0 \equiv E$ as the total energy, and:
$$E^2 - (pc)^2 = (\gamma E_0)^2 - (\gamma \beta E_0)^2 = E_0^2,$$
though I suppose one might like to know whether $E$ is indeed conserved before calling it "energy."

I think I prefer to "justify" the existence of rest energy by Einstein's own reasoning (from his original $E = mc^2$ paper): if a body at rest emits identical light waves in opposite directions, then its kinetic energy remains the same in this frame (zero), but it loses energy nevertheless.

 

[Sorcerer]

I'm sure I've seen this done for the rectilinear case (though I can't remember where), but I thought I'd give it a crack for the general case. The idea is that one has already obtained the equation for relativistic momentum $\vec p = \gamma m \vec v$ and plugs it into the "work" side of the work-energy theorem to see what pops out on the "energy" side. Using $\vec f = \dot{\vec p} = mc^2 \, d(\gamma \vec \beta)/d(ct)$:
$$\begin{split}
\int^{\vec r_f}_{\vec r_i} \vec f \cdot d \vec r &= mc^2 \int^{\vec r_f}_{\vec r_i} \dfrac{d (\gamma \vec \beta)}{c \, dt} \cdot d \vec r \\[3pt]
&= mc^2 \int^{\vec \beta_f}_{\vec \beta_i} \vec \beta \cdot \left( \gamma \, d \vec \beta + d \gamma \, \vec \beta \right),
\end{split}$$
where
$$\begin{split}
d \gamma &= d \left[ \left( 1 - \vec \beta \cdot \vec \beta \right)^{-1/2} \right] \\[3pt]
&= \dfrac{1}{2} \left(1 - \vec \beta \cdot \vec \beta \right)^{-3/2} \, d ( \vec \beta \cdot \vec \beta ) \\[3pt]
&= \gamma^3 ( \vec \beta \cdot d \vec \beta ) ,
\end{split}$$
so
$$\begin{split}
\int^{\vec r_f}_{\vec r_i} \vec f \cdot d \vec r &= mc^2 \int^{\vec \beta_f}_{\vec \beta_i} \vec \beta \cdot \left[ \gamma \, d \vec \beta + \gamma^3 ( \vec \beta \cdot d \vec \beta ) \vec \beta \right] \\[3pt]
&= mc^2 \int^{\vec \beta_f}_{\vec \beta_i} \Bigl[ \gamma \left( 1 + \gamma^2 \beta^2 \right) \Bigr] ( \vec \beta \cdot d \vec \beta ) \\[3pt]
&= mc^2 \int^{\vec \beta_f}_{\vec \beta_i} \gamma^3 ( \vec \beta \cdot d \vec \beta ) \\[3pt]
& = mc^2 \int^{\gamma_f}_{\gamma_i} d \gamma \\[3pt]
&= mc^2 \Delta \gamma \\[3pt]
&= mc^2 + mc^2 \Delta \left( \dfrac{1}{2}\,\beta^2 + \frac {3}{8} \, \beta^4 + \frac{5}{16} \, \beta^6 + \frac{35}{128} \, \beta^8 + \dots \right),
\end{split}$$
where the $\beta^2$ term suggests calling everything in the parentheses "kinetic energy" $E_k$ and the invariant first term "rest energy" $E_0$. Then it makes sense to define $\gamma mc^2 = \gamma E_0 \equiv E$ as the total energy, and:
$$E^2 - (pc)^2 = (\gamma E_0)^2 - (\gamma \beta E_0)^2 = E_0^2,$$
though I suppose one might like to know whether $E$ is indeed conserved before calling it "energy."

I think I prefer to "justify" the existence of rest energy by Einstein's own reasoning (from his original $E = mc^2$ paper): if a body at rest emits identical light waves in opposite directions, then its kinetic energy remains the same in this frame (zero), but it loses energy nevertheless.

I never noticed that you can do a change of variable to dγ. All the times I’ve done this I’ve had to do trig substitution. Such an obvious thing but I missed it.

 

[SiennaTheGr8]

I didn't find that obvious, but in hindsight I guess I should have (since I already knew the energy equations I was working toward). Anyway, fun exercise!

 

[Ziang]

Is there any way to derive the relation based on Newtonian mechanics clearly and convincedly?

 

[Nugatory]

Is there any way to derive the relation based on Newtonian mechanics clearly and convincedly?

No, because that relation does not hold in Newtonian mechanics.

 

[Herman Trivilino]

Is there any way to derive the relation based on Newtonian mechanics clearly and convincedly?

No. Starting with the Newtonian expressions for the relationships between force, momentum, and energy you can derive some approximate relations. Using the postulates of special relativity you can derive more general expressions that match very closely, when the speed is low, those approximate relations.

In other words, you can derive the Newtonian expressions from the relativistic expressions, but you can't derive the relativistic expressions from the Newtonian expressions.

 

[Ziang]

No, because that relation does not hold in Newtonian mechanics.

Is there any way to modify Newtonian mechanics so that the mass-energy equivalence can be derived from absolute space and time?

 

[SiennaTheGr8]

No.

 

[Sorcerer]

Is there any way to modify Newtonian mechanics so that the mass-energy equivalence can be derived from absolute space and time?

Yes. By changing the transformation rule from the Galileo to the Lorentz. That is, by turning it into special relativity ;) (but no, not from absolute space and time lol)

 

[Ziang]

I heard of a modified Newtonian mechanics which could yield mechanical expressions and the mass-energy equivalence as well from absolute space and time. The theory also gives better results than Newtonian mechanics does. May I write down some of its expressions here?

 

[Sorcerer]

I heard of a modified Newtonian mechanics which could yield mechanical expressions and the mass-energy equivalence as well from absolute space and time. The theory also gives better results than Newtonian mechanics does. May I write down some of its expressions here?

This forum only deals with established and accepted theories. If it's experimental and cutting edge, there is a sub-forum, and if it isn't peer reviewed stuff, it probably isn't welcomed here.

In any event, the only reason you get E = mc² is because of the very assumption that time doesn't tick the same for all clocks. That is built into the Lorentz transformation itself. I suppose there are/were different interpretations of the math, but since E = mc² was discovered, and even before it, no one really doubted the math (Lorentz himself accepted his own transformations, despite his slowness in warming to Einstein's interpretation, including his equation that leads to length contraction, but Lorentz attributed it to the moving matter being deformed).

 

[PeterDonis]

I heard of

Can you give a valid reference (textbook or peer-reviewed paper)?

 

[Ziang]

Can you give a valid reference (textbook or peer-reviewed paper)?

I found it on Amazon. A thin book titled "Beyond the world of relativity". Mathematics in the book is simple. I think scientists will get interested in the way it leads to the equivalence from absolute space and time.

 

[PeterDonis]

I found it on Amazon. A thin book titled "Beyond the world of relativity".

Can you give a link?

 

[Ibix]

Can you give a link?

Partial (I think) copy at Google Books edit: removed the link because I'm pretty sure it doesn't meet our standards - you can find it easily enough by googling for the title.

I haven't had a proper look - Google Books doesn't seem to get along with my phone's browser.

 

[Ibix]

I haven't had a proper look

Now I have (don't know why I can scroll around now - I couldn't before - such are the mysteries of computers). If that is the book that Ziang is talking about then I very much doubt it meets our acceptable reference standards. It apparently simply assumes that Newtonian gravity is correct without any supporting argument, then concludes that event horizons don't exist (around equation 9).

 

[Sorcerer]

Now I have (don't know why I can scroll around now - I couldn't before - such are the mysteries of computers). If that is the book that Ziang is talking about then I very much doubt it meets our acceptable reference standards. It apparently simply assumes that Newtonian gravity is correct without any supporting argument, then concludes that event horizons don't exist (around equation 9).

Seems the author’s motivation is religious:

“In 1987, Nguyen resettled in the United States. Once his new life became stable, he tried to find the answers to his questions in Buddhism, and he recognized and accepted the Buddhist beliefs of space and time are absolute.”Poor fellow. If only he would have realized that relativity says SPACETIME intervals are absolute.

 

[Ziang]

It apparently simply assumes that Newtonian gravity is correct without any supporting argument,

All theories have some assumptions or postulates. So I thought this assumption could be accepted. And because of this assumption, I called it a modified Newtonian mechanics.
Compare to SR, it gives more expressions as potential energy and gravitational red/blue shift.
Compare to Newtonian mechanics, it also gives more expressions as Doppler effect and mass-energy equivalence.
Anyway, it shows a way to derive the equivalence from fundamental definitions of natural concepts in Newtonian mechanics, in absolute space and time.

 

[Sorcerer]

All theories have some assumptions or postulates. So I thought this assumption could be accepted. And because of this assumption, I called it a modified Newtonian mechanics.
Compare to SR, it gives more expressions as potential energy and gravitational red/blue shift.
Compare to Newtonian mechanics, it also gives more expressions as Doppler effect and mass-energy equivalence.
Anyway, it shows a way to derive the equivalence from fundamental definitions of natural concepts in Newtonian mechanics, in absolute space and time.

What about relativistic kinetic energy? How does he derive that, which has the Lorentz factor?

 

[PeterDonis]

All theories have some assumptions or postulates. So I thought this assumption could be accepted. And because of this assumption, I called it a modified Newtonian mechanics.
Compare to SR, it gives more expressions as potential energy and gravitational red/blue shift.
Compare to Newtonian mechanics, it also gives more expressions as Doppler effect and mass-energy equivalence.
Anyway, it shows a way to derive the equivalence from fundamental definitions of natural concepts in Newtonian mechanics, in absolute space and time.

It appears that the reference @Ibix found is indeed the one you were referring to. That reference is not a valid source for PF discussion; it is not a textbook or peer-reviewed paper and appears to be the author's personal speculation.

Thread closed.