Proton Mass-Energy: E=mc^2 Explained

[Sakha]

After some learning on atomic physics, I came with the following question.
Proton mass is 1.67262158 × 10E-27 kilograms, and its velocity is proportional to its energy. So then, in E=mc^2, you only plug the m, and you will always get the same value. I don't quite understand that a proton approaching to c has about 7 TeV (I got that from an LHC article). So then, according to E=mc^2, a proton (and every other particle) has the same energy independently of its velocity?

 

[yuiop]

After some learning on atomic physics, I came with the following question.
Proton mass is 1.67262158 × 10E-27 kilograms, and its velocity is proportional to its energy. So then, in E=mc^2, you only plug the m, and you will always get the same value. I don't quite understand that a proton approaching to c has about 7 TeV (I got that from an LHC article). So then, according to E=mc^2, a proton (and every other particle) has the same energy independently of its velocity?

E=mc^2 is only half the story. The full equation for total energy is:

E = \sqrt{\left(\frac{m_o v c}{\sqrt{(1-v^2/c^2)}}\right)^2 + \left(m_0c^2\right)^2}

When the particle is at rest and v=0 the equation reduces to the familiar E=mc^2 which is the energy due to the rest mass (m_o) of the particle.

The full equation is difficult to get on a T shirt or incorporate into a catchy tune

 

[JesseM]

An easier-to-remember form of that equation is:

E^2 = m^2 c^4 + p^2 c^2

...with the understanding that m stands for rest mass and p stands for the relativistic momentum, which is like classical momentum multiplied by gamma (i.e. p = mv/\sqrt{1 - v^2/c^2})

edit: actually kev made a small error in typing the equation, that 1-v^2/c^2 in the denominator of the parentheses should be in a square root.

edit#2: by the way, another simple form of the equation for the energy of a moving object is E = \gamma mc^2, where the gamma factor is given by the usual \gamma = 1/\sqrt{1 - v^2/c^2}...a little algebraic manipulation will show it's equivalent to the first equation I quoted.

 

[Sakha]

Oh.. I had absolutely no clue about that. It's weird that in most cases they never show the real formula.

 

[JesseM]

Oh.. I had absolutely no clue about that. It's weird that in most cases they never show the real formula.

Well, E=mc^2 is the real formula for the energy of a particle at rest (you can see that if the momentum p = 0, then the formula I posted reduces to E = mc^2), it just doesn't give you the additional kinetic energy for a moving particle.

 

[yuiop]

edit: actually kev made a small error in typing the equation, that 1-v^2/c^2 in the denominator of the parentheses should be in a square root.

Oops, your right I have edited the post to correct the mistake


P.S. I gave the non simplified equation with v explicitly stated so that Sakha could insert a value for v and check the validity of the equation against the information he has gleaned from the LHC.

P.P.S. I am surprised that there is so little mention of the LHC in this forum or the Cosmology forum considering it was switched on a few days ago after many years in the making and billions of dollars. Is there that much lack of interest or is everyone just patiently waiting for solid results?

 

[Sakha]

Another little question..
I know that c = \lambda \nu for photons, but does v =\lambda \nu for everything?

 

[JesseM]

Another little question..
I know that c = \lambda \nu for photons, but does v =\lambda \nu for everything?

Yes, wavelength \lambda is the distance between peaks of the wave, frequency \nu is the rate at which successive peaks are passing a given position (number of peaks per second, say), which is just 1/(time between successive peaks). Naturally, the speed at which the peaks are moving is (distance between peaks)/(time between peaks), so that's wavelength*frequency.

 

[yuiop]

Another little question..
I know that c = \lambda \nu for photons, but does v =\lambda \nu for everything?

I asume you know about the de Broglie hypothesis that all particles (not just photons) have an intrinsic wavelength?

"Waves of molecules
Recent experiments even confirm the relations for molecules and even macromolecules, which are normally considered too large to undergo quantum mechanical effects. In 1999, a research team in Vienna demonstrated diffraction for molecules as large as fullerenes[8].

In general, the De Broglie hypothesis is expected to apply to any well isolated object."

http://en.wikipedia.org/wiki/De_Broglie_wavelength

 

[MeChaState]

The rest-mass energy of a resting proton is about 938 MeV, from resting Electron 0.511 MeV. If you set them in motion the Kinetic Energy will add into it as JesseM explained.
To compare, the energy of visible light is 2-3 eV.
In LHC (Large Hadron Collider) they smash protons with an relative ENERGY of 7 TeV to generate particles (relative to lab 2 x 3.5 TeV - to make it more efficient by making total momentum 0). "Theoretically" the 7 TeV would be enough to create 7000/0.938 = 7450 protons!
They try to detect Higgs particles with - as theory predicts - a mass energy of 500 GeV.
What they do is, annihilating protons to pure energy and hope that that energy will generates new particles (with mass), according to E <=> mc2 or m = (1/c2) E!

Take care
David