# Energy calculation in STR

1. Sep 15, 2008

### vibhuav

In the LHC, protons are accelerated to 99.9999991% of lightspeed. So gamma ~7500. At gamma of 7500, the energy content of proton is 7 TeV. So two colliding protons together should have a total mass-energy equivalentof 14 TeV.

Now, I want to calculate the total energy differently. The relative speed between the two protons is (using v = (u+v)/(1+uv/c^2)) 0.99999999999999995949.... Now, I assume one proton is at rest and the other is approching it at a speed of 0.99999999999999995949c. So the total energy is:

mp*c^2 + gamma(0.99999999999999995949)*mp*c^2

which turns out to be 104235 TeV, much greater than the 14TeV that was in the closed system of 2 protons.

The total energy content of the closed system was calculated to be 14TeV by an observer seeing two protons approaching each other at speeds of 0.999999991c. So it should have matched the calculation from the viewpoint of one of the protons (ie it is not moving and the other is coming towards it at 0.99999999999999995949c).

What am I doing wrong?

2. Sep 15, 2008

### Staff: Mentor

Energy is a frame-dependent quantity. This is true both in classical mechanics and in relativistic mechanics.

The energy is larger in the second situation because the center of mass of the system is moving, and you have a large amount of kinetic energy associated with that motion.

Last edited: Sep 15, 2008
3. Sep 15, 2008

### vibhuav

I understand that KE is frame dependent in classical mechanics (due to its dependence on velocity), but how can the total mass-energy content be frame dependent in a closed system? Should'nt the total energy be conserved, and therefore all frames should agree?

How can I calculate the KE associated with the motion of the center of mass? Can you please eloborate on this, or point me to some good book?

4. Sep 15, 2008

### JesseM

"Conserved" just means that in whatever frame you choose, the energy does not change with time in that frame; it doesn't mean the energy is the same from one frame to another. The formula for the total energy, both mass-energy and kinetic, is:

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

where m is the rest mass and p is the relativistic momentum $$p = mv / \sqrt{1 - v^2/c^2}$$. Naturally the rest mass m does not change from frame to frame, but p does because the velocity is different in different frames, so the energy is different too. A little algebra shows that the above formula is also equivalent to $$E = \gamma mc^2$$, where m is again the rest mass and $$\gamma = 1/\sqrt{1 - v^2/c^2$$, so you can see that this also tells you that energy changes with velocity (and since the rest mass is mc^2, this formula also tells you that kinetic energy is $$(\gamma - 1)mc^2$$)
The center-of-mass frame is the one where the total momentum of the system is zero. This page gives info on how to calculate it in SR.

One question I have: in accelerator experiments it's assumed that particles of a given mass are only created when the collision has the equivalent amount of energy (so that heavier particles require more energetic collisions), does this refer specifically to the energy in the center-of-mass frame?

5. Sep 15, 2008

### tiny-tim

Hi vibhuav!

Put u = c tanha, v = c tanhb …

then the relative speed is c tanh (a+b).

In the com-frame, a = b = x, say, and the energies are both m coshx … total 2m coshx.

In the rest-frame of one particle, a = 0, b = 2x, and the energies are m cosh0 = m and m cosh(2x) … total m (1 + cosh(2x)) = 2m cosh2x, which is the com-frame energy times the gamma-factor for the fact that in this frame the com is moving with speed c tanhx.

6. Sep 15, 2008

### Staff: Mentor

Yes. In the center-of-mass frame, if the incoming particles are annihilated (e.g. in an electron-positron collision), all their energy can go into producing the new particles. The minimum energy needed is the total rest-energy of the new particles, in which case the new particles are at rest in the c-o-m frame.

Now look at this same situation in a non-c-o-m frame. The new particles are moving in this frame and have nonzero net momentum and kinetic energy. The incoming particles must have enough energy to provide this outgoing kinetic energy in addition to the new particles' rest-energy.

7. Sep 15, 2008

### ZikZak

Basically, yes. You need to conserve both energy and momentum in all frames, and from this you can determine that only the amount of energy present in the CM frame is available to make mass from.

The conceptual way to look at it is that the CM frame is the one in which total energy is the least. In order to make a particle, there must be enough energy as measured in all frames; therefore, it's necessary to have that amount in the CM frame.

8. Sep 15, 2008

### Staff: Mentor

Hi vibhuav,

I really think you should look into the energy-momentum http://en.wikipedia.org/wiki/Four-vector" [Broken]. I think this would help you to clarify the difference between conserved and frame invariant quantities. For this problem let me demonstrate how it works.

The energy momentum four vector is (E/c,px,py,pz) where E is the total energy (rest energy plus kinetic energy) and px, py, and pz are the x y and z components of the momentum respectively. This is a four-vector so it http://en.wikipedia.org/wiki/Lorentz_transform#Matrix_form" as the normal four vector (ct,x,y,z). So if we take a proton at rest its energy-momentum four-vector (in units where c=1) is
(938 MeV, 0, 0, 0)

Boosting this to .999999991 c in the x direction we get
p1 = (6.99143922 TeV, 6.99143916 TeV, 0, 0)
with a Minkowski norm
|p1| = 938 MeV
which is the invariant mass of the proton

Boosting another proton to the same speed in the opposite direction we get
p2 = (6.99143922 TeV, -6.99143916 TeV, 0, 0)
with a Minkowski norm
|p2| = 938 MeV
which is the invariant mass of the other proton

So the total four momentum is
p = p1 + p2 = (14 TeV, 0, 0, 0)
with a Minkowski norm
|p| = 14 TeV
which is the invariant mass-energy of the collision

So, in the rest frame of p1, the primed frame, we can do the same calculation

The proton at rest has a four-momentum of
p1' = (938 MeV, 0, 0, 0)
with a Minkowski norm
|p1'| = 938 MeV
which is the invariant mass of the proton

"Double-boosting" the other proton we get a four-momentum of
p2' = (104.2222217532222243 PeV, 104.2222217532222201 PeV, 0, 0)
witn a Minkowski norm
|p2'| = 938 MeV
which is the invariant mass of the proton

So the total four-momentum is
p' = p1' + p2' = (104.2222227 PeV, 104.2222218 PeV, 0, 0)
with a Minkowski norm
|p'| = 14 TeV
which is the invariant mass-energy of the collision

So basically, the four-mometum is a nice book-keeping device that lets you keep track of energy and momentum in any reference frame and correctly determine how the energy transforms and how the invariant mass transforms.

PS Sorry about the significant figures, I had to take those out that far to show that the E component was actually different than the px component.

Last edited by a moderator: May 3, 2017