# Squared in e=mc^2?

1. Feb 25, 2015

### paulo84

Hi,

Sorry to bring this up again but I have a new question. I understand the e=mc part, you've only got to look at a nuclear explosion, but I don't really get the 'squared'. Given that a nuclear explosion is literally a 3 dimensional shape (let's leave exploring possible hidden dimensions for now), shouldn't it be e=mc^3? Or am I just looking at it too literally?

If someone could just lay out the maths as to why this isn't the case...should be able to get my head around it.

2. Feb 25, 2015

### Staff: Mentor

3. Feb 25, 2015

### paulo84

Ok thanks. And do I just take k.e. = 1/2mv^2 as true because it's observable, or is there something else I can know about it to help me understand it? (Hope I'm not pushing the thread too off topic).

4. Feb 25, 2015

### Staff: Mentor

the kinetic energy formula comes initially from Newtonian Mechanics

There is a derivation halfway into the Wikipedia article

http://en.m.wikipedia.org/wiki/Kinetic_energy

5. Feb 25, 2015

### Staff: Mentor

It is a definition.

6. Feb 25, 2015

### Staff: Mentor

Well, that has a derivation too, that relates kinetic energy to work:
http://www.wikihow.com/Derive-Kinetic-Energy-Formula
In short, work is force times distance and kinetic energy is what you get when you apply a constant force to an object, accelerating it at a constant rate.

7. Feb 25, 2015

### paulo84

Aha! That makes perfect sense now. For some reason I don't remember Force ever being described as a constant acceleration in school (and I never figured it out because I wasn't looking at this stuff from the right angle then). This will help me with other bits given that I know some calculus relative to acceleration, speed, distance...thanks a lot guys! :)

8. Feb 26, 2015

### HallsofIvy

I don't think anyone else has pointed out that energy cannot be equal to "mc" or "mc^3" because the units would be wrong.

Energy has units to "Joules" (in the mks system) which can be reduced to "kg m^2/sec^2" (You can see that from two more basic formula- "work" energy, the energy expended to move an object a given distance against a given force, is "force times distance". Force has units of "Newtons" which are "kg m/sec/^2" so force times distance is kg m^2/sec^2. "Kinetic energy", the energy an object of a given mass has, moving at a given speed is "(1/2)mv^2- one half the mass of the object times the square of its speed. Energy must have units, in the mks system, of "kg m^2/sec^2". So obviously, if some energy is a function of mass and speed, it must be some constant times mass times speed squared.

9. Feb 26, 2015

### ChrisVer

How could someone connect the $mc^2$ with a 2-dimensional object, so that he would try to bring it up to "3dimensions"?
Most of the cases the masses are considered to be those of point-like particles.

Apart from this, Hallsoflvy's post gives the best description, and it's the one that helps most when you want to go from natural units (where you drop all c's of SR by setting $c=1$ or [length]=[time]) to SI units let's say. In natural units you don't have to carry all those confusing c's, and write just $E=m$ for particles at rest (so energies and masses -as well as momenta- are measured in the same units, that's why you can see that the mass of electron is given in "eV" units and not in "kg")...

Last edited: Feb 26, 2015
10. Feb 26, 2015

### paulo84

Ok trying to get my head around that...for a particle at rest E=m, or E=mc^2 where c=1, so to get the mass of an electron would it not be m=E/c^2, or m=E for the particle at rest. eV I just looked up is an electronvolt, so is an electronvolt a type of natural unit?

I wasn't aware the mass would be referring to something like an electron only, I was thinking in terms of speed squared = distance squared/time squared, and as distance squared would give you an area rather than a volume, well yeah that was it basically. Somewhat confused thinking...

I wonder if there's still something to it if you consider the mass to be a unit of distance, given that mass is basically occupied space which is kind of what distance is too...then you'd have (mass * distance * distance) which would give you a volume. Except if it's the mass of a point then it's basically not even distance, it's zero dimensional rather than one dimensional. I think this is just even more confused thinking...

11. Feb 26, 2015

### ChrisVer

eV is a unit for energy. It has some certain definition (as all the units), but this doesn't change that it is a unit for energy.
Well it's better to forget what I wrote for E=m for now. Maybe you will come across it later in your studies.

Would you consider the same thing for the classical k.e. you've seen written above: ~mv^2 ?
It's in fact irrelevant since you don't have distance^2 alone, but you have time^2 as well.

There is no need for mass to be "occupied space". There are several definitions of mass, eg the inertia mass which exists in Newton's second law F=ma , which is just a constant of proportionality that relates the force acted on an object to its acceleration (so giving what is said in schools, how "well" does an object react to an applied force).
The invariant mass of Special relativity, which is the one in E=mc^2, is defined as the energy the body has in its rest frame [the frame in which the body is at rest] - or more generally as the 4-momentum squared.
No "occupied space" in each of these definitions. What connects space with mass are the densities.

In general, dimensional analysis can only be helpful as a check or getting some "idea" of the units you are working with (like making characteristic units)

Last edited: Feb 26, 2015
12. Feb 26, 2015

### paulo84

I understand that eV is a unit for energy, it just didn't seem to be an S.I. unit. Yeah time squared is there as well, it's just time squared is a difficult concept to get my head around, which is probably why I didn't take it into account. Apologies for that.

mass=volume * density, or volume=mass/density. If you looked small enough subatomically such that the density were constant (let's assume for a moment that at infinitesimal levels you do get constant density), then there would be a direct link between volume and mass, wouldn't there? Yeah I know I'm speculating here...distance isn't necessarily occupied space, can be unoccupied, but an occupied distance is measured in the same units as an unoccupied distance.

Anyway I'm curious as to the 4-momentum squared?

Sorry if I'm coming across like a muppet...

13. Feb 26, 2015

### HallsofIvy

Yes, you seem to be very confused about this. "$E= mc^2$" is true for the intrinsic energy of any object of mass m, not just an electron. And it is not for a moving object, but one that is motionless in the given frame of reference. I do not know what you mean by "consider the mass to be a unit of distance"- different objects, with different densities can have the same volume with very differing volumes. You cannot "consider the mass to be a unit of distance'.

14. Feb 26, 2015

### paulo84

Ok, firstly - 'Most of the cases the masses are considered to be those of point-like particles'. - ChrisVer. I took this the wrong way. I do actually understand that it can be any object of mass m. 'And it is not for a moving object, but one that is motionless in the given frame of reference' - yes this confuses me. As for considering mass to be a unit of distance or volume, well that's just speculation on my part. It's probably best if it's just ignored.

EDIT: ok looking back at jedishrfu's link, it seems e=mc^2 applies to stationary bodies, while for non-stationary bodies you need the full form of the equation, which I'm not about to type out. I understand that an object of mass m, if converted into energy, unleashes a huge amount of energy (the speed of light squared being a big number). That is the basic way in which I think I understand this. I don't understand concepts like time squared. I would like to understand this stuff better.

Last edited: Feb 26, 2015
15. Feb 26, 2015

### ChrisVer

That's why I mentioned above the square of the 4-momentum. The complete relationship would be:
$E^2= m^2 c^4 + |p|^2 c^2$
With $|p|^2$ the (known) 3-momentum squared of your mass :$\vec{p} \cdot \vec{p} =p_x^2 + p_y^2 + p_z^2$. If you go to a frame in which the object is not moving (is at rest) $|p|^2=0$ and the above reduces to $E^2=m^2 c^4 \Rightarrow E= mc^2$.

The 4-momentum is a 4-vector (1 time-component, 3-spatial components) which has the energy as the time-component, and the spatial momentum in its spatial components. It's Minkowski product with itself (its square) is a Lorentz-invariant (does not depend on the reference frame) which is called the invariant-mass or rest-mass. You should have a look in SR for more information (because I don't know your maths/physics background) ...

16. Feb 26, 2015

### paulo84

Yeah, I'll have a look. A lot of terms there I'm not familiar with, like Minkowski and Lorentz. 'Energy as time-component' makes sense to me because I was wondering if there was a relationship between time and energy (time being basically change and energy also seeming to be change) and obviously I get the basic maths that |p|^2=0 leads you to E=mc^2, I'm just not sure where the E^2, m^2, c^4 etc. are coming from in the first place.

OK, another bit. I'm going to try and relate this to acceleration to see if it helps me understand. Speed squared=distance squared/time squared=area (of a square shape)/unit of time/unit of time.

So therefore see speed squared as being like an accelerating square shape??

EDIT: or no, sorry. It's like something accelerating through 2 dimensions rather than just through one. And accelerating through those 2 dimensions at an equal rate.

17. Feb 27, 2015

### ChrisVer

Then you should start from the basics of the Special Relativity, before pursuing any further. Because no matter what kind of questions you will ask, you won't be able to understand the answers or you might end up in making wrong interpretations of them.
"time-component" is just the 0-th component of a four-vector. It doesn't get transformed by spatial rotations, but is subject to boosts (change of velocity).
To understand why E appears again I'd suggest going through an SR book.

No that is not correct. Are you familiar with derivatives?
$\frac{dx}{dt} \frac{dx}{dt}$ is having units m^2/s^2.
$\frac{d^2 (x \cdot x) }{dt^2}$ is also having units m^2/s^2. However they are not equal.
The first is the velocity squared and by taking the sides of a square to be measured in m/s units it can be seen (it's not) as the area of a square (no-change) of side dx/dt, the 2nd can be seen (is not) as the acceleration of the "change" of the area of a square with side $x$.

$\frac{d^2 (x \cdot x) }{dt^2}= 2 \frac{dx}{dt} \frac{dx}{dt} + 2 \frac{d^2x}{dt^2} x$

Last edited: Feb 27, 2015
18. Feb 28, 2015

### PWiz

Sorry to butt in, but isn't the total energy of an object in any non-inertial frame given by $E=mc^2=\gamma m_0c^2$? (m=relativistic mass)

19. Feb 28, 2015

### ChrisVer

I am sorry, I don't understand your question...I don't understand why you put the "non-inertial frame", it's the energy of a mass m0 particle in some Lorentz frame (boosted by the gamma factor).
But for the formula you are correct (for massive particles). For the rest frame ($\gamma =1$), this formula becomes $E=m_0 c^2$ (where m0 is the rest mass).
However, because the relativistic mass is a bad choice, you can use instead $E=\sqrt{ m^2_0 c^4 + p^2 c^2}$ to take into account a boosted frame (not-at rest).
That is because the four momentum is $(c=1)$, $p^\mu = (E , \vec{p} )$. If you take its minkowski product with itself, gives you $p^\mu p_\mu \equiv \sum_{\mu,\nu=0}^3 n_{\mu \nu} p^\mu p^\nu = E^2 - |\vec{p}|^2$ which is a lorentz invariant; a Lorentz transformation doesn't change its value (nice exercise for starting SR). So you can choose a particular frame, where the particle is at rest (has no momentum and no kinetic energy), and so the energy will be just its rest-mass $E=m_0$. In that frame $p^\mu p_\mu = m_0^2$, and this holds to any reference frame. So:
$E^2 - |\vec{p}|^2 = m_0^2 \Rightarrow E = \sqrt{ m_0^2 + |\vec{p}|^2}$
or inserting the c's:
$E = \sqrt{ m_0^2 c^4+ |\vec{p}|^2c^2}$

For massless particles, there is no rest frame. However in their cases, even by treating them as classical waves, you can find that the energy is $E=|\vec{p}|$ and the $p^\mu p_\mu =0$.

Last edited: Feb 28, 2015
20. Feb 28, 2015

### PWiz

I used the term "non-inertial frame" to be on the safer side because I've only tried learning SR, where problems are exclusively limited to non-accelerating frames. I don't want to comment on anything I'm unsure of. Anyway, what I actually don't understand is why in some posts in this thread the rest energy of an object is being given by $E=mc^2$ when this equation gives the total energy (rest+kinetic) instead.

21. Feb 28, 2015

### ChrisVer

It depends on what you mean by $m$. Because in general the relativistic mass is not used (nobody uses it), m stands for the rest mass (no kinetic energy)

22. Feb 28, 2015

### PWiz

Ah, I really fell for that one. Thanks.