# B E=mc^2: where did the 1/2 go?

#### DaveC426913

Gold Member
In explaining to a curious member on a another forum what $$E=mc^2$$ means, I finally came to understand it better myself.

The member wanted to know why c is squared. What sense does it make to square a velocity? I stated comparing it to the kinetic energy formula $$K=1/2mv^2$$ A 1 ton car moving at 40mph has four times as much energy as a 1 ton car moving at 20mph. That's why you square the velocity to calc the energy.

Einstein's formula is the same conversion, except that c has been substituted for v (since the resultant photons are all moving at c).

It's as simple as that. Take a mass, figure out what velocity it is moving at, square the velocity and you get the amount of energy.

So, assuming my thoughts are correct, what happened to the $$1/2$$? Einstein's formula doesn't contain it.

(I suspect it has something to do with the car transferring its energy to another mass a la Newtons Law, so you're only counting half the energy? But I'm not sure.)

D'OH! I just saw the 'Helpful Posts' at the bottom. This exact question has already been answered...

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#### lomidrevo

D'OH! I just saw the 'Helpful Posts' at the bottom. This exact question has already been answered...
So you have probably already found out, that it is not very meaningful to compare the two equations. The first one is for the rest energy of a particle of mass m (not for photons!) and it comes out of special relativity. The second one is classical Newtonian kinetic energy. So there are actually two kinds of difference. Rest vs Kinetic, Relativistic vs Newtonian.

#### Dale

Mentor
Einstein's formula is the same conversion, except that c has been substituted for v (since the resultant photons are all moving at c).
Not really. The formula is for a mass at rest, so v=0 rather than v=c. The units are the same, but that is all.

I prefer to take this approach: energy and momentum together form a four-vector. The norm of the four-momentum is $m^2 c^2=E^2/c^2-p^2$. This simplifies to $E=mc^2$ for $p=0$

#### DaveC426913

Gold Member
OK, when I said before that I came to understand it better, I lied.

#### jartsa

So, assuming my thoughts are correct, what happened to the $$1/2$$? Einstein's formula doesn't contain it.

A 1000 kg car moving 10m/s has momentum 1000*10 Ns. The car can push for a time 1000 seconds by a force 10 Newtons until it stops.

The pushed thing gets energy :

E = mv * 1/2 v = momentum * the average speed during the pushing

A light pulse that has the same momentum as that car can push for a time 1000 seconds by a force 10 Newtons until all momentum has been used.

The pushed thing gets energy:

E= momentum * the average speed during the pushing = pc

Ecar = p*1/2 v

Elight = p* v

- because car slows down but light doesn't.

#### Dale

Mentor
But $E=mc^2$ is the energy of a massive object at rest (v=0), not the energy of light.

#### ZapperZ

Staff Emeritus
2018 Award
Didn't Minute Physics showed a rather quick derivation a million years ago?

The biggest and most common mistake that people made here is thinking that this can be derived via Newtonian energy equations. The starting point of the derivation is completely different.

Zz.

#### Orodruin

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The big revelation in $E_0 = mc^2$ is that the rest energy $E_0$ of an object is actually its inertia in its rest frame. It is rather straightforward to find that $E_0^2 = E^2 - (pc)^2$, which for small velocities $v/c = pc/E$ gives
$$E_0^2 = E^2 (1 - v^2/c^2) \quad \Longrightarrow \quad E = \frac{E_0}{\sqrt{1-v^2/c^2}} \simeq E_0\left(1 + \frac{v^2}{2c^2}\right)$$
so the non-constant, i.e., kinetic, energy for the object is actually $E_0 v^2/(2c^2)$.

The identification of the rest energy with the inertial mass in the rest frame is what is truly the insight here.

#### SiennaTheGr8

That Minute Physics video is pretty darn good!

#### Orodruin

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That Minute Physics video is pretty darn good!
Well, there are some gaping holes, but for illustrative purposes and small velocities v ...

#### SiennaTheGr8

Well, there are some gaping holes, but for illustrative purposes and small velocities v ...
Well it is Minute Physics...

Really, though, I think the video gets impressively close to the heart of the thought experiment.

One thing it leaves out (as it must at that length) is how Einstein knew that light's frequency and energy transform in the same way. I've rarely seen this explained correctly.

An incorrect explanation (no, Einstein didn't use the Planck–Einstein relation): https://terrytao.wordpress.com/2012/10/02/einsteins-derivation-of-emc2-revisited/

A correct explanation: https://www.mathpages.com/home/kmath354/kmath354.htm

An almost-correct explanation for part of it (spot the error!): https://books.google.com/books?id=thXT19cY9jsC&pg=PA164

#### DaveC426913

Gold Member
OK, so it seems they are derived from completely different principles.

Is it still fair, however, to conclude that the v^2 in both is due to the same property?
i.e.
1] if a car moves at 10mph, you calculate its energy based on squaring the velocity. (tripling the velocity results in 9 times the energy).
2] If the products of annihilation are moving at c, you calculate its energy based on squaring its velocity (i.e. c^2)
?

#### ZapperZ

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OK, so it seems they are derived from completely different principles.

Is it still fair, however, to conclude that the v^2 in both is due to the same property?
i.e.
1] if a car moves at 10mph, you calculate its energy based on squaring the velocity. (tripling the velocity results in 9 times the energy).
2] If the products of annihilation are moving at c, you calculate its energy based on squaring its velocity (i.e. c^2)
?
No, because there is still an implicit assumption of the validity of galilean transformation here. You can't just "square velocity", especially at relativistic speeds. It is why the full relativistic energy equation is E2 = (pc)2 + (mc2)2.

Zz.

#### DaveC426913

Gold Member
Dang.

Well, my explanation for this guy - albeit poorly-executed - will have to do anyway.

He is so naive about math he thinks that c^2 "cancels out", and therefore E==m.
(And no, he's not in grade school).

#### TeethWhitener

Gold Member
Another option is to expand $E=\sqrt{p^2c^2+m^2c^4}$ in a Taylor series:
Letting $N=mc^2$ and $d=p^2c^2$ in the formula at the link above gives you
$$E =mc^2+\frac{p^2}{2m}+\cdots$$.
The first term is the rest energy and the second term is the Newtonian kinetic energy.

#### Pencilvester

He is so naive about math he thinks that c^2 "cancels out", and therefore E==m.
If you’re using geometric units (which makes the math in physics much cleaner), then $c =1$, so he’s not wrong.

#### DaveC426913

Gold Member
If you’re using geometric units (which makes the math in physics much cleaner), then $c =1$, so he’s not wrong.
That's what I've been trying to get across to him.
E is proportional to m. (Double the m, you get double the E).
But to be able to convert, you still need the constant so that the units work out (even if the value is 1).

#### sysprog

DaveC426913 said:
So, assuming my thoughts are correct, what happened to the $$1/2$$ ? Einstein's formula doesn't contain it.
For an object at rest, it doesn't need to, but when the object is moving, it gets brought back in; that's succinctly exemplified in the following Prof. Tao comment:
Terence Tao said:
anon said:
So is E = mc^2 not exactly true, but only up to order O(v^3)?
Yes. In fact, for moving objects the exact formula for the energy is $E = mc^2 / \sqrt{1-v^2/c^2} = mc^2 + \frac{1}{2} mv^2 + O( v^4 )$, thus one can view the $O(v^4)$ term as a relativistic correction to the Newtonian formula $\frac{1}{2} mv^2$ for the kinetic energy.

#### Pencilvester

But to be able to convert, you still need the constant so that the units work out
But that’s the beauty of geometric units: you don’t need to do any converting to make the units work out. $E$ is not only proportional to $m$ in this system of units— the number represented by $E$ is simply equal to the number represented by $m$.
If you understand this, then I don’t understand what problem you have with someone saying that $E = m$.

#### Orodruin

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But to be able to convert, you still need the constant so that the units work out (even if the value is 1).
There is no conversion. In natural units energy and mass have the same physical dimension. When we say that we use units such that c=1, we really mean it.

#### DaveC426913

Gold Member
There is no conversion. In natural units energy and mass have the same physical dimension. When we say that we use units such that c=1, we really mean it.
c=1 is not the same as c^2 canceling out. The units don't cancel out.

Simple E=m would result in $$kg·m^2/s^2=kg$$ which is wrong.

#### PeroK

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c=1 is not the same as c^2 canceling out. The units don't cancel out.

Simple E=m would result in $$kg·m^2/s^2=kg$$ which is wrong.
Try this:

Even mass and energy are measured in units of length!

#### Orodruin

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c=1 is not the same as c^2 canceling out.
Yes it is. In natural units c=1 and is dimensionless. You repeating the same thing over and over will not change this.

Simple E=m would result in

kg⋅m2/s2=kgkg·m2/s2=kg​

kg·m^2/s^2=kg which is wrong.
It is wrong in natural units but not for the reason you think. In natural units, m and s are units of the same physical dimension. c = 299 792 458 m/s is then a conversion factor similar to 2.54 cm/inch. Both these conversion factors are just unit conversions, both are equal to one and dimensionless.

#### PeterDonis

Mentor
Simple E=m would result in
...in not using SI units, which do not satisfy $c = 1$ in the first place.

If you want to think of $c = 1$ as having units, you can think of it as, for example, 1 light-second per second. But that just raises the question, what's the difference between a "light-second" and a "second"? And the answer that "natural units" systems give is, nothing--they're the same physical dimension, so 1 light-second per second is the same as 1 second per second, which is the same as the dimensionless number 1.

#### PeroK

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...in not using SI units, which do not satisfy $c = 1$ in the first place.

If you want to think of $c = 1$ as having units, you can think of it as, for example, 1 light-second per second. But that just raises the question, what's the difference between a "light-second" and a "second"? And the answer that "natural units" systems give is, nothing--they're the same physical dimension, so 1 light-second per second is the same as 1 second per second, which is the same as the dimensionless number 1.
In fairness to @DaveC426913 and although it's commonplace in GR texts, it is still quite profound in my opinion. Certainly when I first encountered the idea I could only digest it as a mathematical trick to make the formulas simpler. The full depth of the concept only sank in gradually.

"E=mc^2: where did the 1/2 go?"

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