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

[sysprog]

You only need to set $c = 1$ to do that. Setting $G = 1$ is more problematic and can't really be properly addressed in a "B" level thread. In any case, this thread is specifically about setting $c = 1$, so please limit discussion to that.

Not if we only set $c = 1$. Again, setting $G = 1$ to get "geometrized units" is more problematic and is really out of scope for this thread.

You can't; setting all three of those to 1 is inconsistent. At best you can set two of them to 1. That's one of the issues that can't really be properly addressed in a "B" level thread; if you want to discuss it further please start a separate thread.

As an alternative, could you perhaps 'split' this part off to another thread, as was done, e.g., in the Remove an Aluminum Tube thread to create the (split) Remove an Aluminum Tube thread?

I would probably start into that branch-off with something like:

It's my understanding that among systems of natural units, the Planck units system sets not only "all three of those to 1", but also does the same with the Coulomb and Boltzmann constants, leaving the elementary charge not similarly normalized, because trying to normalize that too, ...​

but that beginning would be pretty much straight out of Wikipedia, and the whole matter isn't something I would think to bring up in a standalone thread; I'd prefer to have the refer-back to this thread for context ...

Either way, per your advisory, I'll stay off further discussion of that matter in this thread.

 

[PeterDonis]

I would probably start into that branch-off with something like...

I have started a separate thread with a reference to this one:

https://www.physicsforums.com/threads/natural-systems-of-units.971522/

 

[sysprog]

I have started a separate thread with a reference to this one:

https://www.physicsforums.com/threads/natural-systems-of-units.971522/

Thanks!

No, they are not both "set to 1". They are both measured in the same units, so their ratio (speed) is dimensionless.

Yes. I should have said that even though neither distance nor time is dimensionless, when both are expressed in the same dimension (the [L] dimension), speed, i.e. their ratio, is dimensionless. Thanks for the correction.

You only need to set $c = 1$ to do that.

True.

If they're the same thing, this question is trivial. What you mean is, it tells us nothing about how much energy in conventional units, the units we are used to from everyday life, is equal to how much mass in conventional units. That's what SI units are for, to tell us that by specifying a numerical value for $c$ that gives a useful answer to that question.

Well, it's not unusual in GR to see energy and mass both expressed in units of MeV, but I've also seen $m= \frac {MeV} {c^2}$, and although I've seen $E=m$, I've never seen the simple tautology $MeV=MeV$, or the bizarre-looking $MeV= \frac {MeV} {c^2}$, but simple substitution would allow $E=m{c^2}$ to yield $MeV = \frac {MeV} {c^2} \times c^2$, which would simplify to the tautology.

It entails no such thing. The units of acceleration in "natural" units (where $c = 1$) are inverse time/length. That should be obvious if you consider that acceleration is the derivative of velocity with respect to time/length, and the units of the derivative are the inverse of the units of the thing you are taking the derivative with respect to.

I recognize that. I was trying to suggest that if, for example, $c$, taken as (the speed component of) a velocity, is expressed in $m/s$, and acceleration is expressed in $m/s^2$, then setting $c=c^2$ could allow setting $s=s^2$, which would make the expression $m/s$ seem to be the same as $m/s^2$, which is part of what leads to bewilderment when people unfamiliar with the territory encounter $E=m$.

This doesn't make sense; you can't "map" something with units to something that doesn't have units.

I was trying to find a simple substitute for the $=$ sign, which would allow for distinction between simple equality with 1 and assignment of $c$ to a dimensionless 1. What would you suggest for resolving that ambiguity, if anything other than seeing the difference from context might occur to you?

 

[Orodruin]

I was trying to suggest that if, for example, ccc, taken as (the speed component of) a velocity, is expressed in m/sm/sm/s, and acceleration is expressed in m/s2m/s2m/s^2, then setting c=c2c=c2c=c^2 could allow setting s=s^2, which would make the expression m/s seem to be the same as m/s2

This is just not true for the reason that $c = c^2 = 1$ is dimensionally consistent whereas m/s = m/s^2 is not.

What would you suggest for resolving that ambiguity, if anything other than seeing the difference from context might occur to you?

I do not think that there is an ambiguity because setting c=1 is perfectly consistent. Not doing so would be akin to measuring width in cm and depth in inches. This would introduce a lot of arbitrary conversion factors of 2.54 cm/inch in your formulas. You can always reinsert correct powers of c by dimensional analysis when needed for use with SI units.

Well, it's not unusual in GR to see energy and mass both expressed in units of MeV, but I've also seen m=MeVc2m=MeVc2m= \frac {MeV} {c^2}, and although I've seen E=ME=ME=M, I've never seen the simple tautology MeV=MeVMeV=MeVMeV=MeV, or the bizarre-looking MeV=MeVc2MeV=MeVc2MeV= \frac {MeV} {c^2}, but simple substitution would allow E=mc2E=mc2E=m{c^2} to yield MeV=MeVc2×c2MeV=MeVc2×c2MeV = \frac {MeV} {c^2} \times c^2, which would simplify to the tautology.

You seem to be saying that it is a tautology that two sides of an equality have the same physical dimension. This is a requirement, not a tautology.

You will only see MeV/c^2 used as a unit in cases where c has not been set to one.

 

[PeroK]

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.

It's good for what it is, but it raises more questions than it answers. If I were more creative I think I would do a series of "Just hang on a minute ..." videos.

 

[jartsa]

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

Let's say a system of two antiparallel but otherwise identical light pulses has rest mass m. That system has rest energy: $E=mc^2$.

Each of the light pulses has energy $E/2$.

Can we not write that $E/2$ as $\frac{mc^2}{2}$? I guess we can, if we can remember what that m there means.

The system has internal kinetic energy E.

Can we not write that E as $mc^2$? I guess we can, no problem there.

 

[lomidrevo]

Let's say a system of two antiparallel but otherwise identical light pulses has rest mass m

First, light is massless. Second, there is no frame of reference where light is at rest.

 

[Orodruin]

First, light is massless. Second, there is no frame of reference where light is at rest.

He is not referring to the mass of each pulse, he is referring to the system of two pulses as a whole. As long as the pulses are not parallel they certainly have an invariant mass and a rest frame where the total momentum of the system is zero.

 

[lomidrevo]

well, I might misunderstood the post, but this is really confusing then:

Each of the light pulses has energy E/2

 

[Orodruin]

In the rest frame of the system, each light pulse has the same energy, i.e., 1/2 of the system's total energy. If they did not, then they could not have equal and opposite momenta.

 

[lomidrevo]

In the rest frame of the system, each light pulse has the same energy, i.e., 1/2 of the system's total energy. If they did not, then they could not have equal and opposite momenta.

I understand that light pulses must have the same energy. But I don't understand why it is $E/2$, where E is the total rest energy of the system. I assume that the system consist of some apparatus as well. So not all energy can be attributed to the light pulses. Am I missing something?

 

[PeroK]

I understand that light pulses must have the same energy. But I don't understand why it is $E/2$, where E is the total rest energy of the system. I assume that the system consist of some apparatus as well. So not all energy can be attributed to the light pulses. Am I missing something?

The mass of an object includes all forms of energy within the object: the mass and kinetic energy of its constituent particles, binding energy in its atoms and molecules, and any electromagnetic radiation within the object etc. For example, a box filled with EM radiation but otherwise empty has more mass than an empty box.

What light does not have is a rest mass. But, the energy of light contributes to the mass of a system of which it is a part.

 

[Orodruin]

It is what the invariant mass of a system is - the total energy in the rest frame.

 

[lomidrevo]

Thanks @Orodruin and @PeroK for your posts, I think I'm getting to understand it now.

In case the pulses are parallel, the system would have zero rest mass in all frames of reference (like any light beam). The energy of the system would be $E = p_{all}c$ where $p_{all}$ is the sum of momenta of all pulses involved.
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. In case they are antiparallel, and of the opposite momenta (with the same magnitude $p$), all observers would agree on the invariant rest mass of the system $m$ as per following
$$E = mc^2 = 2pc$$
Am I correct?

 

[Orodruin]

the system would have zero rest mass in all frames of reference

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. However, in general for a system with non-zero invariant mass, the invariant mass is the mass in the rest frame, aka rest mass.

The general definition of a system's invariant mass is $M^2 = P^2$, i.e., the square of the system's total 4-momentum (in natural units). For two light pulses, whose individual 4-momenta are light-like and equal to $p_1$ and $p_2$, respectively, we obtain
$$
P = p_1 + p_2 \quad \Longrightarrow \quad M^2 = (p_1+p_2)^2 = p_1^2 + 2p_1 \cdot p_2 + p_2^2 = 2p_1 \cdot p_2,
$$
since $p_i^2 = 0$.

 

[lomidrevo]

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]

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).

 

[PeterDonis]

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 $s^2$.

 

[PeterDonis]

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]

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 $10^{-10}$.

 

[PeterDonis]

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]

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.

 

[sysprog]

I like to use a system of units where $\hbar, c, G, N_A$ and $\pi$ are all set equal to 1.

... but not the elementary charge, $e$ ...

When the 2019 redefinition of SI base units takes effect on 20 May 2019, its value will be exactly $1.602176634 \times 10^{−19}$ C by definition of the coulomb.

 

[sysprog]

@stevendaryl What will the unit circle look like in your brave new world?

 

[Orodruin]

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]

I like to use a system of units where $\hbar, c, G, N_A$ and $\pi$ are all set equal to 1.

$\pi$, the number!

 

[vanhees71]

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]

I like to use a system of units where $\hbar, c, G, N_A$ and $\pi$ are all set equal to 1.

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]

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 $\text{rad}=1$.

 

[stevendaryl]

$\pi$, the number!

Yes, everything is more convenient if we set it to 1.

 

[vanhees71]

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]

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.

 

[PeterDonis]

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

Ah, got it.

 

[stevendaryl]

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]

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.