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

Dale

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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

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Yet another thing that takes time to realise 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

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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

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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.

• sysprog and SiennaTheGr8

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).

vanhees71

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That's even worse! Don't confuse young people with wrong math!

They are confused enough by well-meaning "didactics". Right now, there's a big debate about the final highschool examps in math in Germany. The students complain they have been "too difficult". Today, the problems have been posted at some newspaper online pages, and I was shocked to see that they were just at the normal level you expect from a freshman entering our physics lectures. Obviously with all these didactical "achievements" following the socalled "Pisa shock" everything became worse and worse. Now there are all these videos pretending to explain something right even with formulae, which are simply utter nonsense. I've had to learn that there are "alternative facts" and "fake news" in all these "anti-social media", but "fake math" tops them all :-((.

Orodruin

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but I don't think that I would call those SI units any longer.
The meter and the second are still what they are defined as in the SI unit system and you can go out and measure a distance with the same meter ruler you would use if you are using the regular SI system. I can still say that the baseline of the DUNE experiment is 1300 km. The only difference is that that 1300 km is actually a 1/energy and the conversion factors (such as 1 m/s) are dimensionless rather than dimensionful. The meter is still the distance of 1/energy that light travels in 1/299792458 seconds. It is just a matter of whether you want to give your conversion factors physical dimension or not.

• vanhees71

Orodruin

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Obviously with all these didactical "achievements" following the socalled "Pisa shock" everything became worse and worse. Now there are all these videos pretending to explain something right even with formulae, which are simply utter nonsense. I've had to learn that there are "alternative facts" and "fake news" in all these "anti-social media", but "fake math" tops them all :-((.
I think the technical term, to quote the amazing Tom Lehrer, is "new math".

sysprog

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 $E=m$. 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 $E=m{c^2}$.

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PeterDonis

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In the geometrized system, distance and time are not dimensionless, but they both are set to 1, which makes speed dimensionless.
No, they are not both "set to 1". They are both measured in the same units, so their ratio (speed) is dimensionless.

When you lose dimension, and by fiat set $c=G=1$, you can as a consequence , instead of saying $E=mc^2$, say $E=m$.
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.

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

It's not problematic that both energy and mass are defined along [L] (length)
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.

When we say that $c = 1, \hbar = 1, G = 1$,
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.

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.
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 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' $1$).
This doesn't make sense; you can't "map" something with units to something that doesn't have units.

• Orodruin

sysprog

PeterDonis said:
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

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I would probably start into that branch-off with something like...
I have started a separate thread with a reference to this one:

sysprog

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

Thanks!
PeterDonis said:
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?

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Orodruin

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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

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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.

• weirdoguy

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.

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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

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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

lomidrevo

well, I might misunderstood the post, but this is really confusing then:
Each of the light pulses has energy E/2

Orodruin

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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

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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.

• lomidrevo

Orodruin

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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

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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 $p_i^2 = 0$.

• PeroK and lomidrevo

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

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