So let’s start off…… [LHS=Left Hand Side; RHS=Right Hand Side]
Consider a body with mass m moving.
Now, From Einstein’s Energy-Momentum Relation,
E2=m2c4+p2c2
where E=energy
m=mass
c=speed of light in a vacuum=3*10^8 m/s (approximately)(+lightspeed can be measured)
p = Momentum
Since our body is ONLY moving, we are considering that it has ONLY kinetic energy.
So.
The Kinetic energy is just the energy possessed by a body by virtue of it’s motion.
Since Work = Force * Displacement * cosΘ [dot product of force and displacement]
=> Θ=0° [it’s the angle between the force and the displacement (θ)]
Now, in Segment-I of Light, we discussed about trigonometry (a bit) and told about the sin of an angle (ratio of the side opposite to the angle θ to the hypotenuse)
Similarly cos of an angle is defined as the ratio of the side other than the side opposite to the angle θ (the base) to the hypotenuse.
SO, the diagram is,
So, Work is now given by
W=F*s
According to newtonian physics,
F=ma
=> W=mas
According to the third equation of motion,
s=(v²-u²)/2a
where v=final velocity
u=initial velocity=0 (if body starts from rest)
a=acceleration produced due to change in velocity=(v-u)/t
Therefore, work now becomes,
W=m*a*(v²-u²)/2a
Cancelling out a,
W=m(v²-u²)/2
IF the body starts from rest,
W=0.5*m*v²
Since the work done is just the result of kinetic energy utilisation (more like potential here because it’s the ‘potential’ of the object to do work),
Ek=0.5*m*v²
Now, multiply both sides with m (mass of body),
Ek*m=0.5*(mv)²
Now, also by Newtonian physics,
mv=p=momentum
So, multiplying both sides by 2 and inputting the value of momentum,
2*m*Ek=p²……….(i)
Inputting this into the Einstein’s Energy-Momentum Relation,
E2=m2c4+p2c2
=> E2=m2c4+2mEkc2
Since, Ek=0.5*m*v²,
The Equation now becomes,
E2=m2c4+m2c2v2
Taking (mc)2 common in the right hand side,
E2=(mc)2(c2+v2)
Taking the square root of both sides,
E=mc√(c2+v2)……….(ii)
Now, just to check the validity of what we’ve done so far, put v=0;
=> Body is at rest
=> From what Einstein said,
E=mc²……….(iii)
Now, Comparing results (ii) and (iii),
mc²=mc√(c2+v2)
Cancelling out mc from both sides,
c=√(c2+v2)
Squaring both sides,
c2=c2+v2
Cancelling out c2 from both sides,
0=v2
=> v=0
=>Body is at rest
Which matches with our assumptions
Thus, we are correct so far.
Now, Consider a body in motion, possessing only kinetic energy
So
Using result (ii),
E=mc√(c2+v2)
Squaring both sides,
E2=(mc)2(c2+v2)
Since the body has only kinetic energy,
E=Ek=0.5*m*v²…….(iii)
So,
Ek2=(mc)2(c2+v2)
Inputting result (iii),
(0.25)m2v4=m2c2(c2+v2)
cancelling out m2,
(1/4)v4=c2(c2+v2)
Multiplying both sides with 4,
v4=4c2(c2+v2)
Dividing both sides with c2 and further simplifying,
v4=4c4+4c2v2
Subtracting 4c2v2 from both sides,
v4-4c2v2=4c4
Taking v2 common from the left hand side,
v2(v2-4c2)=4c4
Dividing both sides with v2-4c2,
v2=4c4/(v2-4c2)
Taking the reciprocal of both sides,
1/v2=(v2-4c2)/4c4
Applying distributive property of division to the right hand side,
1/v2=v2/4c4-4c2/4c4
Further simplifying the RHS,
1/v2=v2/4c4-1/c2
Adding 1/c2 to both sides,
1/v2+1/c2=v2/4c4
Taking v2c2 as LCM (lowest common multiple) in LHS,
(v2+c2)/v2c2=v2/4c4
Cross multiplying,
4c4(v2+c2)=v2(v2c2)
Treating v2(v2c2) as LHS because it feels better,
v2(v2c2)=4c4(v2+c2)
Opening brackets in LHS,
v4c2=4c4(v2+c2)
Dividing both sides with c2,
v4=4c2(v2+c2)
Taking square root of both sides,
v2=2c√(v2+c2)…….(iv)
Since c=speed of light in vacuum=299792458 m/s
Inputting this value into result (iv),
v2=2*299792458 m/s*√[v2+(299792458 m/s)2]
v2=599584916 m/s*√[v2+89875517873681764 m2/s2]
Now, a real solution to this equation is given by,
v=-299792458√[2(1+√2)] m/s OR 299792458√[2(1+√2)] m/s
Since we now discard the negative solution,
v=299792458√[2(1+√2)] m/s
where 299792458 m/s=speed of light in a vacuum
v now gives,
v=658,754,421.88413 m/s (approx)
writing this in scientific notation,
v=6.58*10^8 m/s (approx)
BUT, isn’t that greater than the speed of light?
SO, it’s a tachyon.
BUT, that’s hypothetical!
So, our assumption is false
Therefore, a body cannot have only kinetic energy……………….
