Relativistic Kinetic Energy
One of the fundamental laws of Newtonian mechanics is the Second Law describing the relationship between a vector of force F, inertial mass m and a vector of acceleration a
F = m·a
where inertial mass m is considered a constant characteristic of an object independent of time, space or motion.
Since a vector of acceleration a, by definition, is the first derivative of a vector of speed u by time t, the same law can be written as
F = m·du/dt = d(m·u)/dt
The product of mass and a vector of speed is a vector of momentum of motion p=m·u, so the same Second Law can be written as
F = dp/dt
As we know from the previous lecture, the relativistic momentum differs from the Newtonian one by Lorentz factor γ and, assuming the movement of an object of the rest mass m0 is one-dimensional along X-axis with, generally speaking, variable speed u(t), the relativistic momentum at any moment of time t can be expressed as
| p(t) = γ·m0·u(t) = |
|
| F(t) = |
|
|
During an infinitesimal time period from t to t+dt an object will move along X-axis by distance dx = u(t)·dt.
The force will perform work during this interval
dW(t) = F(t)·dx = F(t)·u(t)·dt
This work will increase an object's kinetic energy.
Let's assume that at time t=0 our object is at rest, that is u(0)=0.
During the period from t=0 to t=T the force F(t) is acting on this object, so its speed is changing from 0 to some ending value u(T).
The total kinetic energy K[0,T] gained by an object during a time interval from t=0 to t=T as a result of acting force F(t) can be obtained by integrating the work increment dW(t) by time t on an interval [0,T].
| K = ∫[ |
|
|
]·u(t)·dt |
Let's use the formula for integration by parts
∫0T f '·g = f·g|0T − ∫0T f·g'
where
| f = |
|
and g = u(t) |
Then the f·g is equal to
| f·g = |
|
· u(t) | |
|
= |
| = |
|
The ∫0T f·g' would be equal to
| ∫0T f·g' = ∫ |
|
·u'(t)·dt |
Since u'(t)·dt = du(t), we can use variable u as an independent variable and integrate by it on an interval from u=u(0)=0 to u=u(T)
Our latter integral equals then
| ∫0T f·g' = ∫0u(T) |
|
= |
| = ∫0u(T) |
|
= |
= −m0·c²·√1−u²(T)/c² + m0·c² =
| = |
|
+ m0·c² |
Now we can calculate the full expression
K = ∫0T f '·g = f·g|0T − ∫0T f·g' =
| = |
|
− |
| − |
|
− m0·c² |
Notice that two fractions in the above expression have a common denominator, so their numerators can be added together and significantly simplified with everything except m0·c² canceling out.
Therefore, relativistic kinetic energy accumulated by an object during the time T equals to
| K(T) = |
|
− m0·c² |
Since the time T can have any value, and u(T) is the final speed at this moment, we can drop T and consider kinetic energy at any moment in time as a function of speed u at this moment
| K(u) = |
|
− m0·c² = |
where
| γ = |
|
Obviously, it's interesting to see the relationship between relativistic kinetic energy and classical definition K=½m·u².
To establish this connection, we assume that the object's speed u is small compared to the speed of light c and can approximate Lorentz factor γ with the first few members of its Taylor series.
As is well known, Taylor series for 1/√1−x² is
1+x²/2+3x4/8+5x6/16+...
Using it with x=u/c and ignoring members with x4 and higher degrees, we will get approximate value of relativistic kinetic energy
K ≅ m0·c²·[1+(u/c)²/2−1] =
= ½m0·u²
which exactly corresponds to a classical definition of kinetic energy.
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