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

#### DaveC426913

Gold Member
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.
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.

I am unaware of any now-deleted content, so I may not grasp the target or scope of your post.

#### sysprog

As the OP, I can't help but wonder to how much of this thread you are referring, and how much of my content you deem nonsensical.

I am unaware of any now-deleted content, so I may not grasp the intent or target of your post.
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.

#### Nugatory

Mentor
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.
None of your content - if there were a problem with that you would have heard about it from one or more of the mentors.
I am unaware of any now-deleted content, so I may not grasp the target or scope of your post.
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.

#### vanhees71

Gold Member
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!
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)).

#### neilparker62

Homework Helper
Ask not where the 1/2 went to but whence it came! (see third term).

#### cmb

An incorrect formula doesn't get correct when changing the system of units.
But I don't see that E=m is incorrect.

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.

#### vanhees71

Gold Member
$E=m$ 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.

#### cmb

$E=m$ is correct for a particle at rest only.
The OP's question didn't go beyond that (E=mc^2). I am glad we are not disagreeing.

#### vanhees71

Gold Member
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).

#### weirdoguy

The OP's question didn't go beyond that (E=mc^2)
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.

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

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