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##\pi##, the number!stevendaryl said:I like to use a system of units where ##\hbar, c, G, N_A## and ##\pi## are all set equal to 1.
Where did the 1/2 go in E=mc^2?
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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.PeterDonis said:
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 ;-)).
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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.stevendaryl said:I like to use a system of units where ##\hbar, c, G, N_A## and ##\pi## are all set equal to 1.
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.
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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##.Orodruin said:
martinbn said:
Yes, everything is more convenient if we set it to 1.
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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.
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.
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Note that in these cases there's a frame of reference in which the two light pulses move in opposite directions.lomidrevo said:
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Orodruin said:
Ah, got it.
vanhees71 said:
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.
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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I believe that it is in Lorentz–Heaviside units that c=1, so E=m would be correct?DaveC426913 said:
Just pick your system of units to make E=m correct!
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DaveC426913 said:
DaveC426913 said:
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
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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.
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.
sysprog
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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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sysprog said:
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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I didnt mean it like that, @PeterDonis; I was simultaneously lamenting and lauding PF's enforcement of standards -- it stings when it bites on my fingertips, but I recognize that the enforcement of the standards is part of what makes PF such a great site.PeterDonis said:
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DaveC426913
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As the OP, I can't help but wonder about the scope of this thread that you are including, and how much of my content you deem nonsensical or unworthy.sysprog said:
I am unaware of any now-deleted content, so I may not grasp the target or scope of your post.
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I didn't mean to imprecate any of your content, @DaveC426913. I think you're a great contributor here, and if I were to disagree with you about something, I would try to make that disagreement quite specific and plain. Regarding content, I meant to refer only to some of my own contributions, and to a post which the moderators decided to delete.DaveC426913 said:
None of your content - if there were a problem with that you would have heard about it from one or more of the mentors.DaveC426913 said:
There was a problematic post that was up for a while before any of the mentors saw it - which is why @PeterDonis stressed above that problematic content should be reported. @sysprog saw it and one of the replies while it was still up, and that’s what’s he’s talking about.
Any further discussion in this fork of the thread belongs in a new thread in the “Feedback” section of the forum.
sysprog
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Thanks, @Nugatory; I think that the post by @vanhees71 was very excellently explanatory.
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An incorrect formula doesn't get correct when changing the system of units. The correct formula is ##E_0=m##, i.e., you choose the arbitrary additive constant of the single-particle energy as ##E_0=m##. The correct formula for a particle moving at velocity ##v## (a dimensionless quantitity in such units) still is ##E=m/\sqrt{1-v^2}##, where ##m## is the socalled "rest mass" (a better name is "invariant mass", because you can also extend the discussion to massless particles as a limit, and such a particle can never be at rest but always goes with a constant speed c (=1 in your natural units)).cmb said:
Ask not where the 1/2 went to but whence it came! (see third term).
cmb
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But I don't see that E=m is incorrect.vanhees71 said:
Energy and mass are convertible, surely the only question is your units used for the conversion?
I think it is the point of the thread that E=mc^2 is *not* an equation based on a variable 'c'. There is no variable in that equation, only a conversion ratio because c is a constant. Like inches = centimetres.k^2 , whatever k is.
For sure, the disclosure that the relative frames are related by c^2 by inclusion of that term in the equation makes that fact much more evident. But I don't think it is essential, with the right units into which that relationship is already embedded.
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##E=m## is correct for a particle at rest only. For a particle moving with momentum ##\vec{p}## you have ##E=\sqrt{m^2+\vec{p}^2}## (all written in natural units with ##c=1##), at least if you follow the modern definition of mass exclusively as "invariant mass". Everything else leads to confusion and doesn't reflect the physical meaning of the quantity "mass".
Of course ##c## is not variable but to the contrary just a mere conversion constant to convert the space-time-distance unit from seconds to meter and vice versa. It's just fixed to an exact constant value within the SI.
Of course ##c## is not variable but to the contrary just a mere conversion constant to convert the space-time-distance unit from seconds to meter and vice versa. It's just fixed to an exact constant value within the SI.
cmb
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The OP's question didn't go beyond that (E=mc^2). I am glad we are not disagreeing.vanhees71 said:
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Yes, but ##E=m c^2## is wrong, and Einstein didn't like to put his "most famous formula" this way. He always stressed that it's better not to use the socalled "relativistic mass", which is a misleading concept appearing in his famous paper of 1905 and unfortunately is perpetuated until today to confuse students. We should fight this misconception (along with promoting the fact that nowadays also temperature and chemical potential are Lorentz scalar quantities, and the phase-space distribution function of classical statistics is a scalar field either).
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cmb said:
To avoid amibguities one should use subscript: ##E_0=mc^2##, because it's the rest energy we are talking about, not energy in general.
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The SI system does specify that the base units are dimensionally independent. From the BIPM website:Orodruin said:
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That does not mean that the units do not exist in other systems. A meter does not stop being a meter because you use a system of units that has less base units, that would be absurd.Dale said:
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I agree, it just wouldn’t be SI.