# Derive E=mc^2: Kinetic Energy, Photons, Lorentz-Transformations

• ehj
In summary: W=\gamma m c^2-mc^2...into...W=\gamma m_0 c^2 - m_0 c^2...which is just the usual formula for kinetic energy.

#### ehj

The kinetic energy of an object can be shown to be:
Ekin = m*c^2 - m0*c^2

Where m is the relativistic mass, m0 is the rest mass and c is the speed of light.

Is it acceptable to derive E = mc^2 from this equation by saying the kinetic energy of a photon is mc^2 because it has a rest mass of zero, and thereby making a relation between photon/light energy and mass. Or is this argumentation too weak? Does the "zero rest mass" photon actually come from the derived equation?
If this derivation isn't that strong I would like to ask for another derivation which uses the relativistic momentum, relativistic kinetic energy and/or Lorentz-transformations etc. I would like to avoid: derivations that use the photon momentum from theories concerning electromagnetism, and derivations that use Taylor expansion since I havn't learned that yet. Is this possible :)?

E=mc^2 is used in the formula you gave, i.e. its not sufficient as a proof.
Most derivations I'm familiar with are fairly complex, but i recall being shown a surprisingly simple "derivation" a while ago... let me see if i can find it ...

http://www.sciam.com/article.cfm?id=significance-e-mc-2-means
you should be able to follow that, good luck

The actual formula is

E^2 = m^2.c^4 + p^2.c^2

where p is the momentum...
now if the object has rest mass. we use E = mc^2 as at rest, momentum will be 0.

But for particles like electrons and photons, rest mass is 0 but they have momentum.
So the equation for them reduces to E = pc

spideyunlimit said:
The actual formula is

E^2 = m^2.c^4 + p^2.c^2

where p is the momentum...
now if the object has rest mass. we use E = mc^2 as at rest, momentum will be 0.

But for particles like electrons and photons, rest mass is 0 but they have momentum.
So the equation for them reduces to E = pc

Electrons have zero rest mass, what a profound statement...

Even ignoring that you don't know what an electron is, the energy of a photo in given by E=hf which is not equal to the product of its momentum and the speed of light, since pc = hv, and v is not equal to f.

Overall, even if you were correct, you helped the OP in no way at all.

Last edited:
I made a big error... it's not electron.. it's photon only... don't laugh mister.

E = pc
pc gives mc^2... dimensionally same.. and this equation is correct... I'm sure you've never come across it ;)

spideyunlimit said:
I made a big error... it's not electron.. it's photon only... don't laugh mister.

E = pc
pc gives mc^2... dimensionally same.. and this equation is correct... I'm sure you've never come across it ;)

Ahh even more godly revelations. E = pc because the units are equal. Why bother even proving E=mc^2, just check the units guys! I have never come across it true, but that this equation is correct is false.

I wonder what happens if we click this link, aye?

h**p://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html

write http
;)

And nowhere does it say pc = mc^2.

I didn't equate them parametrically, I was trying to explain to you that the equation is true, and not false as you believed.
Anyway go through it.
have to go now bye... and by the way can we add our signatures?

What kind of wish washy logic do people go through when they reason:

I define p = mv. We have E^2 = (mc^2)^2 + (pc)^2. Now if we let m=0, we are only left with E= pc. Let's ignore the fact p = mv = 0 here, in fact let's change the definition to something else right about now.

I see now that i was incorrect: E= pc = hf, if we take p = h/lambda. But that is still far besides the point - your first post gives no useful info on how to prove E=mc^2, as you assume a deeper result.

EDIT: P.S - There you go : https://www.physicsforums.com/profile.php?do=editsignature [Broken]

Last edited by a moderator:
Do you need to be sarcastic to prove your superior intellect or is it just in your nature?

Gib Z the derivation you linked to uses taylor expansion, which was one of the things I would like to avoid :).

As best I can tell, E = mc^2 comes from the assumption that the formula for kinetic energy (which is derived in the wikipedia article using pretty simple integral calculus):

Ek = ymc^2 - mc^2

represents the difference between _total_ particle energy (ymc^2) and _rest_ energy (mc^2). In other words, saying a particle's rest energy is E=mc^2 is an interpretation of the kinetic energy formula.

My favorite "derivation" is the one from Weinberg's QFT book. It goes something like this: A Hilbert space of one-particle states has the property (by definition) that every vector is an eigenvector of P2, and they all have the same eigenvalue, so we'll just write this eigenvalue as -m2 and call m the mass of the particle. Done-diddly-done. (If the eigenvalue is positive, we'll write the eigenvalue as n2 instead and call n the mass. In this case the particle is called a tachyon).

But this calculation is pretty cool too (even though the notation is kind of sloppy):

Choose units such that c=1. The work W performed accelerating a mass m from speed 0 to v is

$$W = \int F dx = \int \dot p dx=\int\frac{d}{dt}(\gamma m\dot x) \dot x dt = \int(\frac{d}{dt}(\gamma m \dot x^2)-\gamma m \dot x\ddot x)dt$$

Since $\dot\gamma=\gamma^3\dot x\ddot x$, we have

$$\gamma m\dot x\ddot x =\frac{m}{\gamma^2}\gamma^3\dot x\ddot x=\frac{m}{\gamma^2}\dot\gamma=-\frac{d}{dt}(\frac{m}{\gamma})$$

so

$$W=\int\frac{d}{dt}(\gamma m \dot x^2+\frac{m}{\gamma})dt =\gamma m v^2+\frac{m}{\gamma}-0-m =\frac{m}{\gamma}(\gamma^2 v^2+1)-m =\frac{m}{\gamma}\gamma^2-m =\gamma m -m$$

We can restore factors of c if we want:

$$W=\gamma m c^2-mc^2$$

The mass m is often written as m0 by people who prefer to call it the "rest mass". They define $m=\gamma m_0$ and call that the "relativistic mass". This notation turns the result into

$$W=mc^2-m_0c^2[/itex] We can always write $p^2=-m^2$, where p2 is the Minkowski space square of the four-momentum, so the equation really says that $-E^2+\vec p^2=-m^2$, or equivalently $E^2=\vec p^2+m^2$. With c included explicitly, it's $E^2=\vec p^2c^2+m^2c^4$. The m on the right-hand side is the rest mass, so if you prefer the m0 notation, you should write $E^2=\vec p^2c^2+m_0^2c^4$. Note that if we simply choose to write the left hand side as mc2, this m must have dimensions of mass, so we can do this if we take this to be the definition of "relativistic mass" m. This is one of the reasons why I find "relativistic mass" to be such a lame and useless concept. Hello Fredrik. I have used a similar method (although without having to do the "c=1" as you did) of deriving the relativistic kinetic energy formula - by using the force's work. The problem is not deriving the formula for kinetic energy as you did, because I've already done that. But how do you get from that, the relativistic kinetic energy, to E = mc^2 ? I'm not that familiar with minkowski space, four-momentum yet as I'm only in high school. :| Last edited: ehj said: Gib Z the derivation you linked to uses taylor expansion, which was one of the things I would like to avoid :). It only uses Taylor expansion to show that the relativistic equation for kinetic energy is approximately the Newtonian expression for v << c. To get to E=mc^2 it only requires some basic integration. EDIT: peter0302 said: Do you need to be sarcastic to prove your superior intellect or is it just in your nature? Both? Here you go: dE=Fdx dE=d(mv)/dt.dx = d(mv)/dt.dx/dt.dt = v d(mv) = v^2 dm + mv.dv m=m0/(rt (1-v^2/c^2)) [m0=rest mass] =>m.(rt(1-v2/c2)) = m0 On differentiating the above equation, we get c^2.dm=v^2.dm + mvdv Thus dE = v^2.dm + mvdv = c^2.dm Thus dE= c^2.dm On integrating, E=mc^2 that derivation doesn't work spideyunlimit. You forget the arbitrary constant upon integration, and as you're using the work done by the force, the kinetic energy, you can't just throw the constant away. ehj said: The problem is not deriving the formula for kinetic energy as you did, because I've already done that. But how do you get from that, the relativistic kinetic energy, to E = mc^2 ? First a reminder: Your physics books may not have made this point clear, but the easiest way to define energy in classical (i.e. not quantum) mechanics is to first define work, and then say that energy is anything that can be converted to work. The result we obtained says that mc2 is such a thing, so it is energy. Compare the result we obtained with the non-relativistic result W=mv2/2-mv02/2. (The derivation of this result is similar to the one I did above, but easier since we can replace $\gamma$ with 1 before we start messing with derivatives). This result is the reason why we can think of mv2/2 as a kind of energy, and name it "kinetic energy". The formula says that the work is equal to the change in kinetic energy. The relativistic version of that formula (i.e. our result) says that work is equal to a change in the quantity mc2 (where m is viewed as a function of velocity). This is the reason why we can think of mc2 as a form of energy. "Mass energy" would be an appropriate term for it. The formula you're looking for (E=mc2) is nothing more than an assignment of a letter of the alphabet to this quantity, just as when we assigned the letter T (or whatever your book uses) to the kinetic energy. One difference worth noting between the relativistic and non-relativistic results is that a particle at rest has zero kinetic energy, but a non-zero mass energy. ehj said: that derivation doesn't work spideyunlimit. You forget the arbitrary constant upon integration, and as you're using the work done by the force, the kinetic energy, you can't just throw the constant away. What he should have said instead of "integrating" is "integrating from m0 to m". This makes the constant irrelevant, but changes the end result to $E=mc^2-m_0c^2$. I prefer using the letter W for the work and E for the terms in the final result, $W=E-E_0$. So all you really need is a verbal argument to derive E = mc^2 from the relativistic kinetic energy formula.. You interpret from the formula Ekin = mc^2 - m0c^2 that the total energy of the object is mc^2 and that the rest energy is m0c^2 by using the argument that the kinetic energy must be the difference between the final energy and the initial energy? Gib Z said: Both? LOL. An honest answer. :) spideyunlimit said: ... On integrating, E=mc^2 The klunky derivation I'm familiar with goes something like this: [tex]F=ma=m\frac{\delta v}{\delta t}=m\frac{\delta x}{\delta t^2}$$

$$W=F \delta x=m\frac{\delta x^2}{\delta t^2}=mv^2$$

Regards,

Bill

## 1. What does E=mc^2 mean?

E=mc^2 is the famous equation proposed by Albert Einstein, which states that energy (E) is equal to the mass (m) of an object multiplied by the speed of light (c) squared. This equation is a fundamental concept in the theory of relativity and is often used to calculate the energy released in nuclear reactions.

## 2. How does E=mc^2 relate to kinetic energy?

E=mc^2 shows the relationship between mass and energy, where mass can be converted into energy. In the case of kinetic energy, this equation can be used to calculate the amount of energy an object has due to its motion, as kinetic energy is directly proportional to an object's mass and the square of its velocity.

## 3. What is the role of photons in E=mc^2?

Photons are the fundamental particles of light, and they play a crucial role in E=mc^2. According to the equation, energy and mass are interchangeable, and photons are a form of energy that can be converted into mass and vice versa. This concept is demonstrated in Einstein's famous equation, where a small amount of mass can be converted into a large amount of energy.

## 4. How do Lorentz transformations relate to E=mc^2?

Lorentz transformations are a set of equations used in Einstein's theory of relativity to describe how space and time change when an object is moving at high speeds. They are essential in understanding the relationship between energy and mass, as they show that an object's mass increases as its speed approaches the speed of light, making it more challenging to accelerate. This concept is reflected in E=mc^2, where the speed of light is squared and plays a crucial role in the equation.

## 5. Can E=mc^2 be applied to everyday situations?

Yes, E=mc^2 can be applied in many everyday situations, such as nuclear power plants, where nuclear reactions release a tremendous amount of energy by converting a small amount of mass into energy. It also plays a role in understanding the behavior of particles in particle accelerators, as well as in medical imaging techniques like PET scans, which use the conversion of mass into energy to create images of the body.