Kinetic Energy and Rest Energy

[GeeKay]

Query: if two objects containing equal quantities of matter and antimatter were made to collide, would it be necessary to add the kinetic energy produced by the impact velocity to the energy release resulting from these two pieces of matter/antimatter coming together in the first place? If so, I'm left wondering how one can add still further to a full 100% of rest energy into radiant energy. Confused!

[studiot]

Yes, the KE should be part of the equations, but have you any figures comparing the relative magnitudes of the KE and anihilation energies?

 

You might be better off considering momentum conservation.

Edited February 4, 2014 by studiot

[Endercreeper01]

At relativistic speeds, the kinetic energy will play an important factor. In relativity, the energy contained in an object is given by [latex]E=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}[/latex].

The change in energy from being at rest and being in motion is the kinetic energy, i.e kinetic energy is [latex]\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}-mc^2[/latex]. At non-relativistic speeds, kinetic energy makes up a very small amount of the total energy in the object, and so it is only necessary when speeds are relativistic.

[Enthalpy]

An example would be a positron emitted by radioactivity. Its kinetic energy is significant.

The kinetic energy adds to everything else. It will increase the energy of the emitted photons.

[GeeKay]

The concept that prompted this 'thought experiment' (inspired by a magazine article) had ten tonnes each of matter and antimatter undergoing a head-on collision, with an impact speed of some 30,000 kps. The total energy release turns out to be about 400,000 megatons - but 'total' only if one leaves out the kinetic energies of the impact itself. This left me wondering whether antimatter, as distinct from standard matter, is a special case when it came to kinetic energies.

 

Also, my maths isn't to be trusted. According to my back-of-the-envelope calculations, two such ten-tonne objects impacting at 30,000 kps produces, in kinetic terms, an energy release far in excess of that produced by the matter/antimatter merger. Indeed, at 1.21e 26 joules, it appears to exceed that of the Sun itself. . . which is clearly preposterous!

[ajb]

You might be better off thinking at first about an electron positron collision analysed in the center of momentum frame.

[swansont]

At v = 0.1c the KE is only a tiny fraction of the rest energy — just 0.5%. And as ajb recommends, if you do that in the CoM frame, it's even smaller. So if you calculated the KE to be greater than the mass energy, at least one of those calculations is incorrect.

[imatfaal]

The concept that prompted this 'thought experiment' (inspired by a magazine article) had ten tonnes each of matter and antimatter undergoing a head-on collision, with an impact speed of some 30,000 kps. The total energy release turns out to be about 400,000 megatons - but 'total' only if one leaves out the kinetic energies of the impact itself. This left me wondering whether antimatter, as distinct from standard matter, is a special case when it came to kinetic energies.

 

Also, my maths isn't to be trusted. According to my back-of-the-envelope calculations, two such ten-tonne objects impacting at 30,000 kps produces, in kinetic terms, an energy release far in excess of that produced by the matter/antimatter merger. Indeed, at 1.21e 26 joules, it appears to exceed that of the Sun itself. . . which is clearly preposterous!

 

Just off the top - and without any relativistic corrections - kinetic energy is one half mass times velocity squared, rest mass converted completely to energy is mass times velocity of light squared. Velocity of light will always be bigger than than velocity of material.

[studiot]

Just to reinforce all that you have been told

 

V_collision = 30,000 km/sc = 3x10⁷m/s

 

c = Velocity of light is about 3 x10⁸ m/s

 

So v/c is about 0.1, which is the figure swansont mentioned

 

So go check your envelope.

 

[Sensei]

Also, my maths isn't to be trusted. According to my back-of-the-envelope calculations, two such ten-tonne objects impacting at 30,000 kps produces, in kinetic terms, an energy release far in excess of that produced by the matter/antimatter merger. Indeed, at 1.21e 26 joules, it appears to exceed that of the Sun itself. . . which is clearly preposterous!

 

Your calculations are complete nonsense.

 

Let's for instance 1.0078 grams of protons and 1.0078 of antiprotons (6.022141*10^23 * 2 = 1.2*10^24 of both particles)

 

E=m*c^2 = 2 * 0.0010078 kg * 299,792,458^2 = 1.8115*10^14 J

 

E.K. of one of them = 1/2*m*v^2= 1/2* 0.0010078*(299,792.458*0.1)^2= 4.53*10^11 J

 

Total E.K is 1/200 of total energy released.

Edited February 5, 2014 by Sensei

[imatfaal]

 

Your calculations are complete nonsense.

 

Let's for instance 1.0078 grams of protons and 1.0078 of antiprotons (6.022141*10^23 * 2 = 1.2*10^24 of both particles)

 

E=m*c^2 = 2 * 0.0010078 kg * 299,792,458^2 = 1.8115*10^14 J

 

E.K. of one of them = 1/2*m*v^2= 1/2* 0.0010078*(299,792.458*0.1)^2= 4.53*10^11 J

 

Total E.K is 1/200 of total energy released.

 

And your calculation are just a bit silly - it is completely spurious, lowers accuracy and heightens confusion to proceed as above. The quantity in the equation is mass - not number, so use 1 gram and make life so much easier, or use figures given, or better still just take m out of the equations as it is common.

[Endercreeper01]

The total mass is 20,000 kilograms (not including relativistic effects). Because the velocity is 0.1c, [latex]\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/latex] is 1.00503781526. Solving for the total energy gives us

[latex]E=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}=1.8065659 \times 10^{21} J[/latex]

Working out the kinetic energy, we find that

[latex]\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}-mc^2=1.8065659 \times 10^{21} - 20,000c^2=9.05552510888 \times 10^{18} J[/latex]

This means that the kinetic energy makes up only 0.5% of the total energy.

The energy from the collision is nowhere near the energy of the sun, as that is 1.78771393 × 10⁴⁷ joules.

[GeeKay]

Yes, I understand now. Soon after I'd made my howler (viz the above KE release 'rivalling that of the Sun') I revised it down to 9.0e x 18 joules, or a tad more than 2,000 megatons, but remained in doubt about it. So many thanks for all the corrections, and in particular the explanations concerning the calculations - the finer points of E = mc2 included!