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

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It would be more appropriate to talk about invariant mass (which is why I have tried to use it more or less consistently here) as the system then does not have a rest frame.
Yes, I should improve my terminology a little bit.

Thanks for the last equation! As the result is in a form of dot product of two 4-vectors, it is even more enlightening now :)
 

vanhees71

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In natural units (mechanically) ##E = \sqrt{p^2c^2 + m^2 c^4}##.

In the geometrized system, distance and time are not dimensionless, but they both are set to ##1##, which makes speed dimensionless. That makes the speed of light dimensionless, too, so you can set it to ##1##, too. When you lose dimension, and by fiat set ##c=G=1##, you can as a consequence , instead of saying ##E=mc^2##, say ##E=m##.

View attachment 243096

One problem with that is that the resulting 'equation', without units, tells us nothing about how much energy is equal to how much mass.

It's not problematic that both energy and mass are defined along ##[L]## (length); however, if we want to quantify, we need to define how many meters of energy is equal to how many meters of mass, not just by an equal number, but also in terms of some other fundamental factual quantity, such that ##E=m## for only 1 unique value on that reference scale.

Even though ##E## and ##m## both use the length dimension ##[L]## and are both set equal to ##1## in the geometrized unit system, the multiplication factor (for conversion between geometrized and SI) for ##E## is ##G c^{-4}##, while for ##m##, it's ##Gc^{-2}##.

When we say that ##c = 1, \hbar = 1, G = 1##, we should remember that these are not really equalities. In each of them, only the RHS is properly dimensionless; the LHS is not dimensionless. The fact that we can simplify some equations by using these pseudo-equalities to eliminate some terms does not properly make the SI versions of the units meaningless.

The loss of dimensional information that allows setting ##c=1## means, among other absurdities, that ##c=c^2##, which entails the nonsensical notion that velocity is the same as acceleration.

I think what is really meant by ##c=1## is not actual equality, because in that 'equation', the LHS still factually has dimension, and only the RHS is dimensionless, so it could instead, I think more properly, be written as ##c \mapsto 1## or ##c \to 1## (i.e. ##c## 'maps to', or more precisely, 'is substituted for by' ##1##).

When converting back to SI units, in order to be able to present meaningful numerical results, we have to 'unforget' the ##c^2## coeffficient, and so return from our temporary sojourn in which we used ##E=m##, back to the more familiar and realistic ##E=m{c^2}##.
Obviously you haven't understood, how a system of units works. In principle the new SI, becoming effective on May 20, is precisely defined by fixing all the natural constants, of coarse in the sense that these are the natural constants according to our contemporary best understanding of nature.

There's one qualification for practical reasons: The Newtonian gravitational constant has not been fixed (yet), because the accuracy with which we can realize it with real measurements is too unprecise. Thus there's still one "special constant" used, namely the frequency of the hyperfine transition of Cesium to define the base unit for time, second. Everything else is then fixed by defining the constants ##c## (speed of light in a vacuum) ##h## (Planck's constant of action), ##N_A## (Avogradro's constant), and ##k_{\text{B}}##.

Thus all these constants are now indeed mere conversion factors from the natural system of units to SI units. According to our understanding of the fundamental structure of the natural laws this concept of fixing the fundamental constants will perhaps be finalized by fixing also ##G## (Newton's gravitational constant) at a value. Then everything is independent of special substances (like Cs for defining the Second).
 
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setting ##c=c^2## could allow setting ##s=s^2##,
This doesn't make any sense. First, you can't set any quantity with units equal to its own square. Second, even if we leave that aside, you just said the units of speed are ##m / s##, not ##s##, so since ##c## is a speed, setting ##c = c^2## would mean setting ##m / s## equal to ##m^2 / s^2##, not ##s## equal to ##s^2##.
 
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this concept of fixing the fundamental constants will perhaps be finalized by fixing also ##G## (Newton's gravitational constant) at a value. Then everything is independent of special substances (like Cs for defining the Second).
How would fixing the value of ##G## remove the need to define the second in terms of the Cs hyperfine transition?
 

vanhees71

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Setting also ##G=1## measures everything just in numbers. The meaning is then of course that everything is measured in Planck units.

Of course in practice we'll keep the SI. As with the change coming into official effect on May 20th, which bases all the base units of the SI by fixing as many of the fundamental constants (##h##, ##c##, ##k_{\text{B}}##, ##e##, ##N_{\text{A}}##) to certain values (which then by definition are exact) except for one constant, which is this Cs hyperfine splitting for defining the second. Here I leave out the Cd, which is somewhat special being rather a physiological unit than a physical unit.

The reason, not to also substitute the Cs hyperfine splitting, which depends explicitly on a certain substance and thus also a certain isotope etc. by fixing also the value of ##G## is, that the measurement of the latter is very uncertain compared to the very accurate possibilities to measure time. It might be that the Cs standard will be substituted some time by an even more accurate definition (e.g., recently nuclear clocks get in the region to become more accurate than atomic clocks).

Of course you really to be able to use the units you also have to give measurement protocols realizing them. This is fixed in the drafts of the "Mise en pratique" for the various base units. E.g., to realize the Ampere accurately one makes use of the Josephson and quantum-Hall effects. For details, see

https://www.bipm.org/en/measurement-units/rev-si/

BTW, as far as I know, the most actual changes due to the revision are in the electrical units as the Volt (relative change by about ##10^{-7}##) and the Ohm (relative change by about ##10^{-8}##); the (pseudo-)fundamental constants ##\epsilon_0## and ##\mu_0## now get relative uncertainties of order ##10^{-10}##.
 
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Setting also ##G=1## measures everything just in numbers. The meaning is then of course that everything is measured in Planck units.
Yes, but Planck units are not "everything is just numbers". Mass/energy and length/time still have inverse physical dimensions to each other, and ##G## still has units of length squared/inverse mass squared. It's just that the Planck mass is set numerically to ##1##, i.e., we measure all masses in "Planck mass units", so ##G = 1 / m_P^2## is numerically equal to ##1##. But it's not the dimensionless number ##1## in the same sense that ##c## and ##\hbar## are.
 

stevendaryl

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Everything else is then fixed by defining the constants ##c## (speed of light in a vacuum) ##h## (Planck's constant of action), ##N_A## (Avogradro's constant), and ##k_{\text{B}}##.
I like to use a system of units where ##\hbar, c, G, N_A## and ##\pi## are all set equal to 1.:wink:
 
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@stevendaryl What will the unit circle look like in your brave new world?
 

Orodruin

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Yes, but Planck units are not "everything is just numbers". Mass/energy and length/time still have inverse physical dimensions to each other, and ##G## still has units of length squared/inverse mass squared. It's just that the Planck mass is set numerically to ##1##, i.e., we measure all masses in "Planck mass units", so ##G = 1 / m_P^2## is numerically equal to ##1##. But it's not the dimensionless number ##1## in the same sense that ##c## and ##\hbar## are.
But isn't this exactly @vanhees71 point? If you define G numerically you take away the need to rely on Cs for the definition of the second because you are fixing the basic unit using G instead. If you work in natural units with energy as the base dimension, then G has dimension -2 and time has dimension -1 so fixing G sets a base unit for time.

If you want to call it dimensionless or not depends on if you want to keep one or zero physical base dimensions.
 

martinbn

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vanhees71

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Yes, but Planck units are not "everything is just numbers". Mass/energy and length/time still have inverse physical dimensions to each other, and ##G## still has units of length squared/inverse mass squared. It's just that the Planck mass is set numerically to ##1##, i.e., we measure all masses in "Planck mass units", so ##G = 1 / m_P^2## is numerically equal to ##1##. But it's not the dimensionless number ##1## in the same sense that ##c## and ##\hbar## are.
No, as soon as you have fixed all the fundamental constants, ##\hbar##, ##c##, and ##G##, or using natural units by setting all of them to 1, there's no free unit left, and all quantities are given by dimensionless numbers.

In HEP physics the convention is to only set ##\hbar## and ##c## to 1. Then you have one arbitrary unit left, which is usually choosen to be GeV for masses, energies, momenta, and temperature or fm for times and lengths. The conversion between the two is provided by the relation ##\hbar c \simeq 0.197 \; \text{GeV} \, \text{fm}##, i.e., you can measure masses, energies etc. in terms of 1/fm or lengths and times in terms of 1/GeV.

Fixing then also this remaining freedom of the choice of units such that also ##G=1## fixes all conventional conversion constants, and all quantities are "measured" in dimensionless numbers.

Of course also in Planck units length/time have inverse dimensions to mass/energy/momentum since 1/1=1 ;-)).
 

vanhees71

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I like to use a system of units where ##\hbar, c, G, N_A## and ##\pi## are all set equal to 1.:wink:
It depends on, how you define ##\pi##. If it's defined as usual as the ration between the circumference and diameter of a circle in a Euclidean space, you are not free to set ##\pi## to an arbitrary value.

There's no principle problem to set ##N_{\text{A}}=1##, then defining 1 mole of a substance to consist of one fudamental building block of this substance (e.g., 1 mole water would then consist of 1 water molecule). That's clear: In principle you don't need an additional unit for the quantity "amount of a given substance" but you can simple give the number of fundamental building blocks of this substance.
 

vanhees71

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But isn't this exactly @vanhees71 point? If you define G numerically you take away the need to rely on Cs for the definition of the second because you are fixing the basic unit using G instead. If you work in natural units with energy as the base dimension, then G has dimension -2 and time has dimension -1 so fixing G sets a base unit for time.

If you want to call it dimensionless or not depends on if you want to keep one or zero physical base dimensions.
Sure, that's the same as to write ##\pi/2 \text{rad}## for an angle to make explicitly clear that you express it in terms of radians. Of course in fact ##\text{rad}=1##.
 

vanhees71

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But you can't set something to 1 which by definition isn't 1. Obviously there's a misunderstanding of what are arbitrarily chosen units and what is a mathematical definition. That's why some people, e.g., think that ##\epsilon_0## and ##\mu_0## are some very fundamental natural constants although they are just arbitrary choices to have convenient numbers in electrical engineering when using SI units. The choice in the natural system of Planck units is ##\epsilon_0=\mu_0=1##, implying also ##c=1/\sqrt{\epsilon_0 \mu_0}##.

On the other hand, a system of units which is self-contradictory is, of course, of no use in science and engineering. So nobody will ever use such non-sensical definitions to begin with.
 

Mister T

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However, as soon as the pulses are just slightly non-parallel, all observes will agree that the system has a non-zero invariant rest mass.
Note that in these cases there's a frame of reference in which the two light pulses move in opposite directions.
 

stevendaryl

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But you can't set something to 1 which by definition isn't 1. Obviously there's a misunderstanding of what are arbitrarily chosen units and what is a mathematical definition.
That was why I put a smiley-face there. It's a joke. You can't set ##\pi## or ##e## to 1, even if it would make life simpler.
 

cmb

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The answers above tend to deal with why E=mc^2 is as it is.

Look at it from the derivation of the kinetic energy formula way might also make good sense.

If you sum each little snippet of momentum as the speed of a body increases, you get the cumulative sum of all the momentums. If you integrate that changing momentum with respect to velocity you get the integral of mv, which is 1/2 mv^2.

E=mc^2 is not derived like that, it is not the integral sum of a variable.

Most 'energy' formula work out like that, take charge stored in a capacitor, Q, is CV, while the energy is the sum of all the charge over the variable V, i.e. it is E=1/2 CV^2. Or the flux in an inductor is LI while the energy is the sum of flux with respect to variable current and is therefore 1/2 LI^2. Magnetic fields are 1/2 uB^2 ... etc....

'c' isn't a variable, so the Einstein formula is not the integral of a product over the range of a variable.
 

cmb

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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).
I believe that it is in Lorentz–Heaviside units that c=1, so E=m would be correct?

Just pick your system of units to make E=m correct!
 

PVC

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In explaining to a curious member on a another forum what [tex]E=mc^2[/tex] 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 [tex]K=1/2mv^2[/tex] 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 [tex]1/2[/tex]? 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...
In explaining to a curious member on a another forum what [tex]E=mc^2[/tex] 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 [tex]K=1/2mv^2[/tex] 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 [tex]1/2[/tex]? 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...
Hi, I would to said the formula you consider is not perfectly corrected. When You use the newtonian kinetic energy his definition spring out to live forces theorem and it contain the factor 1/2 the mass and the speed squared.When we write mc^2 we consider the rest energy of the particle in relativity. More correct is to write
mc^2/sqrt[1-(v/c)^2].
that is the total energy of a body of mass m that for v<<c it is coinciding with the kinetic energy ~ 1/2mv^2 + neglectable other terms
 

vanhees71

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If there is anything that comes close to a "derivation" of the expression for energy and momentum (as well as the other very important conserved quantities angular momentum and the center-of-momentum theorem) is to start from the symmetry properties of Minkowski space with the proper orthochronous Poincare group as its symmetry group. The names of the corresponding Noether charges (conserved quantities) are then choosen the same as in Newtonian physics, i.e., energy, momentum (space-time translation invariance) and angular momentum (rotations) and center-of-momentum velocity (Lorentz boosts).

Applying this program to the mechanics of a point particle leads from applying only space-time translation invariance and rotation invariance, as in Newtonian physics to the conclusion that the Lagrangian must be of the form
$$L(\dot{x},x)=L_0(|\dot{\vec{x}}|).$$
Now one can use the general formalism for inifinitesimal one-parameter group transformations to the Lorentz boost in an arbitrary direction to derive the Lagrangian, but there's a short-cut.

All one needs is that the variation of the action is Poincare invariant. For a single particle we can easily form an action that is itself Poincare invariant. Since the only variable we have is ##|\dot{\vec{x}}|## the only solution is the space-time Minkowski line element itself, i.e.,
$$\mathrm{d} s^2=c^2 \mathrm{d} t^2 - \mathrm{d} \vec{x}^2.$$
For an arbitrary time-like trajectory ##\vec{x}(t)## thus the only possibility is the choice
$$L_0=A \sqrt{1-\dot{\vec{v}}^2/c^2},$$
where ##A## is some constant factor.

To determine this factor, we can look at the non-relativistic limit, which we get for ##\vec{v}^2/c^2=\vec{\beta}^2 \ll 1##. Indeed
$$L_0=A \left (1-\frac{\vec{\beta}^2}{2} \right).$$
Thus we get, up to an irrelevant additive constant the non-relativistic kinetic energy, if we set ##A=-m c^2##.

One should note here that ##m## is the same quantity which we call mass in non-relativistic physics, and it is a Lorentz scalar. This implies that the only consistent notion of mass in special relativity is this invariant mass. Finally we thus arrive at
$$L_0=-m c^2 \sqrt{1-\dot{\vec{x}}^2/c^2}.$$
The symmetry under time-translation invariance gives the Hamiltonian as the conserved quantity which we call, as in Newtonian mechanics, energy.

The canonical momenta are
$$\vec{p}=\frac{\partial L_0}{\partial \dot{\vec{x}}}=m \gamma \dot{\vec{x}}, \quad \gamma=1/\sqrt{1-\dot{\vec{x}}^2/c^2}.$$
This leads to the Hamiltonian
$$H=\dot{\vec{x}} \cdot \vec{p}-L=m \gamma c^2=c \sqrt{m^2 c^2+\vec{p}^2}.$$
It's easy to show that ##p=(H/c,\vec{p})## is a four-vector. Indeed ##p^2=m^2 c^2## is an invariant under Lorentz transformations.
 
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That post by @vanhees71 makes me wish there were something like a 'doubleplusgood' in the reaction options. Thanks to @PeroK for his appropriate denunciation as nonsense of a now-deleted nonsensical post that had only a few mites of intrigue, apparently insufficient in the eyes of the moderators to make the post despite its nonsensicality worthy of retention. I sometimes wonder about the enforcement of standards here on PF, especially when it's visited censoriously upon something I post; however, I gratefully accept that the staff conscientiously exercises its good judgement to continually keep the Physics Forums free of unworthy content, which good judgement I think is part of what makes PF a great place for people afflicted with an affection for scientific truth to visit and participate.
 
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I sometimes wonder about the enforcement of standards
The best thing you can do to help is to use the Report button if you think a post is violating the rules. That brings it to the attention of the moderators.
 

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