- #71
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!
Where did the 1/2 go in E=mc^2?
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In summary: It is still fair to conclude that the v^2 in both is due to the same property, although the details are a bit different.
Just pick your system of units to make E=m correct!
- #72
PVC
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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
- #73
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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.
- #74
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.
- #75
PeterDonis
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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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sysprog
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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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- #77
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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sysprog
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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:
- #79
Nugatory
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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.
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sysprog
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Thanks, @Nugatory; I think that the post by @vanhees71 was very excellently explanatory.
- #81
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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:
- #82
neilparker62
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Ask not where the 1/2 went to but whence it came! (see third term).
- #83
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.
- #85
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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weirdoguy
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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.
- #88
Dale
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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:
- #90
Dale
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I agree, it just wouldn’t be SI.
- #91
cmb
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I am not sure I understand the point of your post.Dale said:
'Energy' is not one of the base units in SI, so actually following the SI convention you would tend to expect there to be a conversion factor between energy and mass.
Energy is defined in SI as [kg⋅m^2/s^2 ] so the conversion factor is clearly in the form (m/s)^2. If you pick a system of units based on a 'fundamental' m/s = 1 then E=m.
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Indeed. If metrologists were able to measure "great G" much better than now, we could degrade all the base units of the SI to mere conversion factors for convenience of having handy numbers for quantities in everyday life (including engineering). This final realization of the complete definition of the base units (and with them all units within the SI) in terms of natural constants is not realized yet, because of the problem to measure Newton's constant of gravity way more accurately. That's why the definition of the base units still rests on one "material-dependent constant", namely the hyperfine transition energy of Cs-133. That's of course a very good basis, because there's nothing more accurately measurable than times, although I believe that this standard will be substituted in the not to far future by something even more accurate, i.e., based on some frequency normal in the optical regime (maybe by a nuclear rather than an atomic transition). At least the corresponding experiments seem to reach the accuracy limit right now. Maybe soon they will top the Cs-133 accuracy, and then after some careful research maybe the frequency normal will be redefined. I don't think that measuring ##G## has the chance to reach the needed accuracy in the foreseable future.Dale said:
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This discussion reminds me of a lab I had in particle physics as an undergrad. We were measuring the timing of photon pairs resulting from electron-positron annihilations. After measuring the timing difference for one setup, one of the detectors was moved 10 cm further away from the source and the timing difference was measured again (of course it was about 10 cm/##c## different from the original measurement).
Professor: So! What have you done now?! (very excited - expecting the answer "measured the speed of light")
Me: We have checked the calibration of your ruler. (probably with a smug expression on my face)
Professor: So! What have you done now?! (very excited - expecting the answer "measured the speed of light")
Me: We have checked the calibration of your ruler. (probably with a smug expression on my face)
- #94
Dale
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Yes, definitely. And between length and time, and between any other combination of base units. That is how the SI works.cmb said:
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Theorists have a hard time in both the introductory and advanced lab. For me the experience with those labs was to decide that I want to become a theoretician. I have two anectdotes in mind:Orodruin said:
(1) In one of the first advanced labs we had to explain Atwood's machine to demonstrate that we've adequately prepared for the experiment. Happily I explained the working of the machine using Hamilton's principle with constraints. It took about 5 minutes. The tutor was totally amazed, how quickly this can be done ;-)).
(2) In one of the advanced labs we had to investigate some em. transition in a nucleus, which was a quadrupole transition, and the dipole transition was forbidden by some selection rule. First question: How much spin can a photon carry. I answered that there's no spin in the strict sense, but that it's 1, 2, 3,... The tutor was of the opinion the only right answer is 1. Then I replied that then there'd not be the quadrupole transition, because it's ##J=2##, and that there's spin and orbital angular momentum, but this split doesn't make sense for a photon and you have to argue with total angular momentum of the em. field. This went back an forth for a while, until a professor came along and declared that of course my multipole expansion formula is right and that we now should start doing the experiment ;-)).
Of course there were endless fights with the theory part of our lab reports, concerning the use of Gaussian units...
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cmb
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That wasn't what I meant.Dale said:
Energy is a derivative unit from the base units. You would not tend to expect a conversion between base units, you'd have to invent new units to make that happen.
What's the conversion between moles and seconds, without inventing a 'new' base unit?
- #97
cmb
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Surely you are describing the recent 2019 base unit changes (came into force 4 days ago, timely enough!), where all 7 base units are defined (not necessarily directly) by 7 physical quantities now?vanhees71 said:
I don't see the problem with the Cs-133 as it is not subject to a 'sample control' issue and 'anyone' (so to speak) can measure it. Is your point that it is not a 'precisely perfect' value? Can that be said of all the other physical values?
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- #98
Dale
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No need for a new base unit. It would just be a factor with units of mol/s. I don’t think there is any universal constant with those units, but no new units are needed.cmb said:
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The point is that it is not a "universal natural constant" like the speed of light, Planck's ##h##, etc. It still refers to a specific atom, namely Cs-133. This is of course not a problem in practice as you explained since indeed all Cs-133 atoms and their atomic states, including the fine-structure transition used to define the second are indistinguishable (according to very fundamental and very well established quantum theory), i.e., everybody wherever in the unierse can just use Cs-133 atoms to very accurately realize the SI units of time, the second.cmb said:
However, there's no problem to use another fundamental frequency normal to redefine the second again if this becomes possible (I think in the near future) and is necessary for the accuracy needed in high-precision experiments.
Of course, the just enforced revision of the SI units is great progress. Particularly to get rid of a single artefact to define one of the base units, the "grande K", has been an overdue step forward; last but not least because particularly this one most important prototype of the kg significantly shifted its mass compared to all the secondary prototypes!
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I always found giving mol a fundamental dimension misguided. It is a unit of something that should be fundamentally dimensionless in my opinion, i.e., a number of things.Dale said:
This is pretty evident in the new definition, where it is the only unit that is not connected to the other units (and ultimately to the definition of the second based on Cs) by the definition of a fundamental constant.
- #101
Dale
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I agree. I have always wondered why they didn’t treat it as dimensionless. They treat the radian as a dimensionless unit, so why not the mol also? (Of course I have also seen arguments to make radians dimensionful!)Orodruin said:
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Honestly I would rather have angles have dimensions than numbers, but I agree that neither is naturally dimensionful.Dale said:
- #103
Dale
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Me too. Luckily I know that dimensions are conventional so whenever I feel like it I depart from the SI convention and use my own. I don’t publish anything that way, but it helps me make sure I have the formulas right.Orodruin said:
- #104
cmb
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Out of interest, what is the proof that the speed of light is a constant?vanhees71 said:
- #105
Dale
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The definition of the meter makes c constant in SI units.cmb said: