Where did the 1/2 go in E=mc^2?

[DaveC426913]

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...

 

[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]

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]

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 pushingA 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 = pcE_car = p*1/2 v

E_light = p* v

- because car slows down but light doesn't.

 

[Dale]

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

 

[ZapperZ]

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]

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]

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]

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]

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]

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]

Another option is to expand $E=\sqrt{p^2c^2+m^2c^4}$ in a Taylor series:
https://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Taylor_seriesLetting $N=mc^2$ 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]

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]

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:

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]

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]

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]

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:

https://en.wikipedia.org/wiki/Geometrized_unit_system
Even mass and energy are measured in units of length!

 

[Orodruin]

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]

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]

...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.

 

[Dale]

The other thing that can take time to realize is the idea that the dimensionality of a unit is as much a matter of convention as its size. I remember that it wasn't until I was exposed to cgs units that I became aware that different systems could differ in the dimensionality of a unit.

 

[Orodruin]

Yet another thing that takes time to realize is the difference between units and physical dimensions. You can have all of the SI units also in a situation where you only consider one basic physical dimension (such as energy) just as well as you have them in a situation where you consider 2, 3, or more basic physical dimensions. The difference is that if you consider length and time to have the same physical dimension (1/energy), then 1 m and 1 s are just different units for physical quantities with dimension 1/energy.

 

[Dale]

You can have all of the SI units also in a situation where you only consider one basic physical dimension

Hmm, I am not sure I buy that. I mean I can see how you could have length and time having dimensions of 1/energy so 1 m and 1 s are just different quantities of 1/energy, but I don't think that I would call those SI units any longer.

The SI definition of the second begins "The second, symbol s, is the SI unit of time." So to me that seems that they intend the dimensionality of the second to be time, not 1/energy.

Maybe nSI for non-Standard International units. Or SMy for Standard My units.

 

[vanhees71]

That Minute Physics video is pretty darn good!

It's wrong in all equations. How can you find this nonsense good?

Also Einstein never said simply $E=mc^2$ but got it right from the very first paper, of course.

The simple point is that you want to work with four-vectors, and thus instead of deriving with respect to time when writing down the equations of motion of a point particle you derive with respect to proper time $\mathrm{\tau}$, defined by
$$\mathrm{d} \tau^2=\frac{1}{c^2} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu} \eta_{\mu \nu}.$$
Now you define the four-momentum,
$$p^{\mu}=m \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}.$$
Of course, $m$ is the usual mass, i.e., the same quantity as in Newtonian physics and thus a scalar. Now you have
$$p_{\mu} p^{\mu}=m^2 c^2$$
and thus
$$p^0=\sqrt{m^2 c^2 + \vec{p}^2}.$$
What's the meaning of $p^0$? To see this, let's check the Newtonian limit, i.e., $\vec{p}^2 \ll m^2 c^2$. Then you get
$$p^0=m c \sqrt{1+\vec{p}^2/(m^2 c^2)} \simeq m c \left (1+\frac{p^2}{2 m^2 c^2} \right) = m c + \frac{p^2}{2m} \frac{1}{c}.$$
Thus up to an additive constant and a factor $1/c$ you have the kinetic energy of a particle. This leads to the conclusion that it is very convenient to keep the "rest energy" $E_0=m c^2$ as an additive constant and define
$$E=p^0 c.$$
The correct energy-momentum relation thus reads
$$E=c \sqrt{m^2 c^2+\vec{p}^2}.$$
You can also rewrite everything in the non-covariant usual velocity in some frame of reference, i.e.,
$$\vec{v}=\frac{\mathrm{d} \vec{x}}{\mathrm{d} t}=\frac{\mathrm{d} \vec{x}}{\mathrm{d} \tau} \frac{\mathrm{d} \tau}{\mathrm{d} t} =\frac{\vec{p}}{m} \sqrt{1-v^2/c^2}.$$
This implies
$$\vec{p}=m \gamma \vec{v} \quad \text{with} \quad \gamma=\frac{1}{\sqrt{1-\vec{v}^2/c^2}},$$
and this gives after some simple algebra
$$E=m c^2 \gamma.$$
As it turns out these definitions are very natural in the sense that for them the same conservation laws are valid as in Newtonian physics. Also from a more fundamental point of view, based on the symmetry principles of relativistic spacetime this holds true: Analysing the dynamics from this point of view in the sense of Noether's theorem the so defined energy and momentum are the generators for temporal and spatial translations, i.e., these quantities are precisely the conserved quantities of space-time translation invariance, as in Newtonian mechanics.

 

[SiennaTheGr8]

@vanhees71, I'm saying that the Minute Physics video is very good for what it is—i.e., an extremely short summary of the thought experiment that Einstein used in his first derivation of the mass–energy equivalence, aimed at people with maybe a high-school background in physics.

I maintain that the video gets pretty close to the heart of the thought experiment while providing about the right level of detail for its target audience. It conveys the rationale for the "rest energy" concept (if a body at rest can lose energy without moving, it must have had some to begin with), and it gives a rough outline of how Einstein concluded that mass and rest energy are the same quantity (relativistic Doppler effect, energy conservation, correspondence principle).