I Derivation of E=pc & E=MC2: Which Came First?

[imsmooth]

I have seen E=pc used to derive E=MC2. I have seen E=MC2 used to derive E=pc. This is circuitous. Which came first and how is E=pc derived?

 

[etotheipi]

It's just the relativistic dispersion relation with $m=0$

 

[PeterDonis]

I have seen E=pc used to derive E=MC2. I have seen E=MC2 used to derive E=pc.

Neither of these look correct. Can you give any references?

 

[PeterDonis]

how is E=pc derived?

It's obvious from the fact that the 4-momentum of a light ray is null.

 

[lomidrevo]

Both equations come from relativistic energy-momentum relation. See special cases, 1. and 3., in article:
https://en.m.wikipedia.org/wiki/Energy–momentum_relation

 

[imsmooth]

Both equations come from relativistic energy-momentum relation. See special cases, 1. and 3., in article:
https://en.m.wikipedia.org/wiki/Energy–momentum_relation

Thank you.

 

[Demystifier]

According to the book
https://www.amazon.com/dp/0393337685/?tag=pfamazon01-20
(which is a serious book, not a crackpot one), Einstein made 7 different proofs of $E=mc^2$ and all 7 proofs had a mistake.

 

[vanhees71]

What's wrong with the original proof in Einstein's addendum to his famous moving-body paper of 1905?

 

[dx]

The mass of a moving particle is

m = m₀/√(1 - v²)

All relativistic effects are contained in the idea that the mass increases with velocity, or the equivalence between mass and energy. Another way to say it is that the momentum is pc = Ev/c (the "flux of energy"), where E = mc². In terms of this m, the momentum is p = mv. If you substitute pc = Ev/c in the formula for the mass, you get

(m/E)²(E² - p²) = m₀²

Using E = mc², you get

E² - p² = m₀².

So essentially you get E = pc just by substituting v = c in the formula pc = Ev/c.

 

[etotheipi]

The mass of a moving particle is

m = m₀/√(1 - v²)

Therefore all relativistic effects are derivable from, or attributed to the increase of mass with velocity, or in other words, the equivalence between mass and energy. Another way to say it is that the momentum is pc = Ev/c (the "flux of energy"), where E = mc². In terms of this m, the momentum is p = mv. If you substitute pc = Ev/c in the formula for the mass, you get

(m/E)²(E² - p²) = m₀²

Using E = mc², you get E² - p² = m₀². So essentially you get E = pc just by substituting v = c in the formula pc = Ev/c.

Your units are all over the place, and you also lose points for bringing up relativistic mass.

 

[dx]

Your units are all over the place, and you also lose points for bringing up relativistic mass.

Consider the lagrangian L = -m√1 - v². Expanding this out to first order, we get

L = 1/2mv² - m

So m appears in the potential energy place of the lagrangian L = T - U. I find it difficult to understand such things without thinking about the relativistic mass. Maybe there is another way to understand such things without introducing such an idea. If anyone knows, it would be interesting to hear.

 

[PeterDonis]

All relativistic effects are contained in the idea that the mass increases with velocity

No, they are not. Physicists spend several decades in the early 20th century figuring that out. That's why "relativistic mass" is no longer considered a useful concept.

The rest of your post is gobbledegook, and you are hijacking someone else's thread. You have now been banned from further posting in this thread.

 

[robphy]

Consider the lagrangian L = -m√1 - v². Expanding this out to first order, we get

L = 1/2mv² - m

So m appears in the potential energy place of the lagrangian L = T - U. I find it difficult to understand such things without thinking about the relativistic mass. Maybe there is another way to understand such things without introducing such an idea. If anyone knows, it would be interesting to hear.

L=T-U is not always applicable.

It seems that the lagrangian of a free relativistic particle
does not have a term that
corresponds to the "relativistic kinetic energy" T_{rel}= \left(\frac{1}{\sqrt{1-(v/c)^2}}-1 \right)mc^2.

 

[weirdoguy]

All relativistic effects are contained in the idea that the mass increases with velocity

No, and if the OP needs to read more on that I recommend this insight:
https://www.physicsforums.com/insights/what-is-relativistic-mass-and-why-it-is-not-used-much/

 

[Herman Trivilino]

All relativistic effects are contained in the idea that the mass increases with velocity, or the equivalence between mass and energy.

The notion of relativistic mass increasing with velocity is a consequence of the way relativistic mass is defined, It doesn't follow from the postulates of special relativity. It's a definition, not a conclusion.

The equivalence of rest mass and rest energy does follow from the postulates, and it is one of Einstein's greatest discoveries.

There is no need for any kind of mass other than what some call the rest mass. Once you adopt that concept, there is no need to call it anything other than the mass.

 

[john t]

Neither of these look correct. Can you give any references?

The way I rationalize it requires recognition of relativistic mass and thus momentum and starts with E=gamma mc^2. My algebra is attached as a screenshot.

 

[Nugatory]

The way I rationalize it requires recognition of relativistic mass and thus momentum and starts with E=gamma mc^2.

This approach has the disadvantage that it relies on $m_0$ being non-zero, so can only be generalized to apply to massless particles by using dubious mathematical techniques: proof by emphatic assertion and derivation by energetic handwaving.

 

[PeterDonis]

My algebra is attached as a screenshot.

This is not allowed per the PF rules. Please use the PF LaTeX feature to enter equations directly in the post. A link to the LaTeX Guide is at the lower left of the post window.

 

[jartsa]

If a pulse of light has momentum 1 Ns, then its momentum can decrease at rate 1 Ns/s for 1 s, until it has no momentum left.

IOW, if a pulse of light has momentum 1 Ns, then it can exert a force 1 N for 1 s, until it has no momentum left.

The work that the aforementioned light pulse does when it exerts a force 1 N for 1 s is:

E=F*d = F*c*t = pc

(d is distance, p=F*t )So our light pulse has energy:

E=F*d=F*c*d=1 N * 300000000 m/s * 1 s = 300000000 Joules

 

[john t]

This approach has the disadvantage that it relies on $m_0$ being non-zero, so can only be generalized to apply to massless particles by using dubious mathematical techniques: proof by emphatic assertion and derivation by energetic handwaving.

Thanks, Nugatory. I would like to know what math or physics principle was broken. There is a derivation of E=gamma mc^2 based on thought experiments and that admittedly requires the assumption of non-zero mass. Is the objection that my approach goes from a specific to a general equation? In that case, would it be legal to go from the general equation to the special one using the reverse of the algebra and, if so, how can my approach be unjustified? Note I did not call it a derivation, rather I said I rationalized it that way. I take your view seriously and, as a non-physicist, would apprediate any further comments. I like the thought-experiment approach (based on several publications I read), because I can understand all the algebraic and calculus steps and do not have the background for Minkowski formalism, and I would like to be able to say to myself that I understand the general equation, even as a simple chemist.

 

[PeterDonis]

The way I rationalize it

First, please note that I was asking the OP for references; I wasn't asking you for a rationalization.

Second, I'm not sure how having a "rationalization" helps. If there is a rigorous derivation of the formula, you don't need a rationalization; and if there isn't a rigorous derivation of the formula, you don't know if it's valid, so a rationalization of it could be misleading you.

 

[PeterDonis]

I would like to know what math or physics principle was broken.

As @Nugatory has already pointed out, your approach, which starts with $E = \gamma m$ and $p = \gamma m v$ (I am using units in which $c = 1$ for simplicity), is only valid for nonzero $m$. So you can't use your reasoning to justify any formula whatever for the $m = 0$ case.

would it be legal to go from the general equation to the special one using the reverse of the algebra

Yes. And in fact that's the preferred approach. See below.

how can my approach be unjustified?

Because the general equation covers both the case of nonzero $m$ and the case of zero $m$. So you can start with that general equation, specialize to the case of nonzero $m$, and obtain $E = \gamma m$ and $p = \gamma m v$ (by reversing the algebra you did); or, you can start with the general equation, specialize to the case of zero $m$, and get $E = p$ (again, I am using units where $c = 1$). So this approach covers all cases with one general equation.

Your approach, by contrast, requires two separate derivations, one for the case of nonzero $m$ (the one you gave) and one for the case of zero $m$ (which you didn't give--you just claimed, without supporting argument, that the general equation you obtained somehow applied to the case of zero $m$).

I like the thought-experiment approach

What does the argument you gave have to do with any thought experiment?

 

[Nugatory]

There is a derivation of E=gamma mc^2 based on thought experiments and that admittedly requires the assumption of non-zero mass. Is the objection that my approach goes from a specific to a general equation?

It's more the exact way you move from the specific to the general.
We start with a formula ($m_R=\gamma m_0$) whose derivation critically depends on $m_0$ being non-zero, work through the algebra from there to derive $E^2=(m_0c^2)^2+(pc)^2$; this is valid as long as the initial $m_R=\gamma m_0$ is valid, which is to say if $m_0\ne 0$. Thus, setting $m_0$ to zero and watching $E=pc$ pop out of the $E^2$ equation is basically saying that our conclusion is valid even when the premises it depends on are invalid - and that's pretty strong hint that we've been arguing from poorly chosen premises.

I'm not just completely quibbling here. In this case we end up with the right answer, but consider the similar problem with the Lorentz transformations (and the time dilation and length contraction formulas that follow from them). These are derived under assumptions that are equivalent to $v<c$, just as your argument is derived under the assumption that $m_0\ne 0$ - but when we try to generalize by dumping that assumption and plugging in other values of $v$ the results are nonsensical in a way that has confused generations of physics students. Why should one procedure any less invalid than the other?

 

[vanhees71]

It's clear that the idea of "massless classical particles" is an idealization. There is no such thing in nature in a literal sense. Rather it occurs in terms of the "naive photon picture" or, more logically within classical physics, as the eikonal approximation for electromagnetic waves.

There are two ways to deal with "massless classical particles"

(a) You start with massive particles and derive the single-particle Lagrangian from Noether's theorem using the (proper orthochronous) Poincare group. Up to equivalence the Lagrangian (in the 1+3 formalism) reads
$$L=-m c^2 \sqrt{1-\vec{v}^2/c^2},$$
where $\vec{v}=\mathrm{d} \vec{x}/\mathrm{d} t$ and $m$ is a constant which turns out to be the invariant mass (I don't use other notions of mass than the invariant mass, which is the same as mass in the Newtonian approximation). Note that this leads to a Poincare-invariant action,
$$S=-m c^2 \int_{t_1}^{t_2} \mathrm{d} t \sqrt{1-\vec{v}^2/c^2}.$$
Energy and momentum are those conserved quantities which originate from translation invariance of space and time, leading to
$$E=c p^0=\frac{m c^2}{\sqrt{1-\vec{v}^2/c^2}} = m c^2 \gamma, \quad \vec{p}=\frac{m \vec{v}}{\sqrt{1-\vec{v}^2}}=m \gamma \vec{v}.$$
This of course does not work for "massless" particles, because the action becomes trivial, i.e., identical to 0 and the expressions for $E$ and $\vec{p}$ written in terms of $\vec{v}$ become either 0 or indefinite when using that $\vec{v}^2=c^2$ for massless particles.

One way out of this dilemma in trying to derive the massless limit is to use the Hamiltonian formulation, i.e., you use $\vec{p}$ instead of $\vec{v}$ to express the energy:
$$E=c \sqrt{m^2 c^2+\vec{p}^2} \rightarrow c |\vec{p}|.$$
Now expressing $\vec{v}$ in terms of $\vec{p}$ and $E$ for massive particles you can again take the limit $m \rightarrow 0$:
$$\vec{v}=\frac{c^2 \vec{p}}{E} \rightarrow c \frac{\vec{p}}{|\vec{p}|}.$$

(b) You use the naive photon picture, i.e., the Einstein-de Broglie relation between frequency and wave vector for electromagnetic field modes in a vacuum and energy and momentum of the corresponding quanta. Since
$$\Box \vec{E}=0, \quad \Box=\frac{1}{c^2} \partial_t^2-\Delta,$$
the dispersion relation for free-field plane-wave modes
$$\omega=c |\vec{k}|.$$
Using then the Einstein-de Broglie relation
$$E=\hbar \omega, \quad \vec{p}=\hbar \vec{k}$$
you get
$$E=c |\vec{p}|.$$
Of course one has to keep in mind that the naive photon picture is at best a heuristic argument (as Einstein put it in his famous light-quantum paper of 1905). One should not think about photons in terms of classical localized "particles". The only correct picture is to treat them as Fock states of the electromagnetic field. They don't even admit the definition of a position observable!

 

[john t]

It's more the exact way you move from the specific to the general.
We start with a formula ($m_R=\gamma m_0$) whose derivation critically depends on $m_0$ being non-zero, work through the algebra from there to derive $E^2=(m_0c^2)^2+(pc)^2$; this is valid as long as the initial $m_R=\gamma m_0$ is valid, which is to say if $m_0\ne 0$. Thus, setting $m_0$ to zero and watching $E=pc$ pop out of the $E^2$ equation is basically saying that our conclusion is valid even when the premises it depends on are invalid - and that's pretty strong hint that we've been arguing from poorly chosen premises.

I'm not just completely quibbling here. In this case we end up with the right answer, but consider the similar problem with the Lorentz transformations (and the time dilation and length contraction formulas that follow from them). These are derived under assumptions that are equivalent to $v<c$, just as your argument is derived under the assumption that $m_0\ne 0$ - but when we try to generalize by dumping that assumption and plugging in other values of $v$ the results are nonsensical in a way that has confused generations of physics students. Why should one procedure any less invalid than the other?

Thanks Nugatory. I understand, and I do not think you are quibbling, and your analogy with the Lorenz situation makes the point clear. Can one say that my logic shows the consistency of the specific with the general equation, given the acceptance (naieve?) of relativistic mass?

 

[Sagittarius A-Star]

Can one say that my logic shows the consistency of the specific with the general equation, given the acceptance (naieve?) of relativistic mass?

No. In you attachment to posting #16, you are using the same symbol $E$ for different things in the first and the last equation (total energy vs. rest energy). With $m$ you can't mean relativistic mass, because in the first equation you wrote:

$E=\gamma mc^2$

But according to Wikipedia, relativistic mass $m_R$ ist defined as total energy (devided by c²):

The relativistic mass is the sum total quantity of energy in a body or system (divided by c²). Thus, the mass in the formula
$E = m_R c^2$
is the relativistic mass.
...
A so-called massless particle (such as a photon, or a theoretical graviton) moves at the speed of light in every frame of reference. In this case there is no transformation that will bring the particle to rest. ... This property of having no rest mass is what causes these particles to be termed "massless". However, even massless particles have a relativistic mass, which varies with their observed energy in various frames of reference.

Source:
https://en.wikipedia.org/wiki/Mass_in_special_relativity#Relativistic_mass

The concept of "relativistic mass" is out-dated and needless, because it is redundant to "total energy". Therefore, it should be avoided.

 

[john t]

This is not allowed per the PF rules. Please use the PF LaTeX feature to enter equations directly in the post. A link to the LaTeX Guide is at the lower left of the post window.

Then this disqualifies me from using the forum. I am legally blind, and use of LaTex is beyond my ability. (I installed it on my mac, and see that it is beyond me.) The keystrokes are just too difficult. I am able to use MathType, but you guys do not recognize that apparently. Maybe I can find another forum which allows screenshots of written text I prepare on an ipad.

 

[PeterDonis]

Then this disqualifies me from using the forum. I am legally blind, and use of LaTex is beyond my ability. (I installed it on my mac, and see that it is beyond me.) The keystrokes are just too difficult. I am able to use MathType, but you guys do not recognize that apparently. Maybe I can find another forum which allows screenshots of written text I prepare on an ipad.

Just to clarify why we normally don't allow equations in images, it's because there is no way to quote the individual equations in responses. However, given your particular situation, we can do our best to work with whatever you can give us.

You mentioned MathType; is that what you are using on your ipad to prepare equations? It's possible that there is a way to translate that to LaTeX.

 

[robphy]

Off-topic but Possibly useful:

https://math.microsoft.com/en

Not perfect... but...

Produced from ocr on an image
https://www.physicsforums.com/threads/e-pc-derivation.997649/#lg=post-6435618&slide=0
saved to my phone. I did not edit the LaTeX produced.

\left. \begin{cases} { E = \gamma m { c }^{ 2 } } \\ { p = \gamma mv } \end{cases} \right.

$$
\left. \begin{cases} { E = \gamma m { c }^{ 2 } } \\ { p = \gamma mv } \end{cases} \right.
$$

From a screenshot of a desktop screen with my iphone camera,
I got [without editing the LaTeX produced]

\left. \begin{array} { l } { E = \gamma m c ^ { 2 } \text { and } p = \gamma mv } \\ { E ^ { 2 } = m ^ { 2 } c ^ { 4 } \frac { 1 } { 1 - \frac { v ^ { 2 } } { c ^ { 2 } } } = m ^ { 2 } c ^ { 4 } [ \frac { 1 - v ^ { 2 } / c ^ { 2 } + v ^ { 2 } / c ^ { 2 } } { 1 - \frac { v ^ { 2 } } { c ^ { 2 } } } ] = } \end{array} \right.

$$
\left. \begin{array} { l } { E = \gamma m c ^ { 2 } \text { and } p = \gamma mv } \\ { E ^ { 2 } = m ^ { 2 } c ^ { 4 } \frac { 1 } { 1 - \frac { v ^ { 2 } } { c ^ { 2 } } } = m ^ { 2 } c ^ { 4 } [ \frac { 1 - v ^ { 2 } / c ^ { 2 } + v ^ { 2 } / c ^ { 2 } } { 1 - \frac { v ^ { 2 } } { c ^ { 2 } } } ] = } \end{array} \right.
$$

UPDATE:
My version of MathType 6.9 [in Microsoft Word in Windows] can output LaTeX and MathML (under cut and copy properties).

A long time ago, I played around with https://www.inftyproject.org/en/software.html
(new? https://www.inftyreader.org/ )
but I have't tried recently.

 

[Sagittarius A-Star]

Can one say that my logic shows the consistency of the specific with the general equation

I think you could derive an equation, that is also valid for photons, by avoiding $\gamma$ as a factor, with $m$ being the "invariant mass", which is defined, but zero for photons:
$$mc^2 = E \sqrt {1-\frac{v^2} {c^2} } \ \ \ \ \ \ and \ \ \ \ \ \ \vec p = \frac {E} {c^2} \vec v$$
$$m^2c^4 = E^2 (1-\frac{v^2} {c^2} ) = E^2 - p^2c^2$$

Update: Repetition of my LaTex-formulas as text, so that a text-to-speech software (for example Windows 10 Narrator) can read it (symbol by symbol after cursor-movements):
mc²=E √(1-v²/c²) and p=E/c² v,
m²c⁴=E² (1-v²/c²)=E²-p²c²