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## Relativistic Energy, Physics tutorial

Relativistic Work and Energy:Work is said to be done on the object when the forces displaces it in its (force's) direction. Consider the element of work dW done on particle when the force F

^{→}moves it through element of displacement ds^{→}. We can write this as:dW = F

^{→}.ds^{→}= FdsIn relativistic range, force is given as:

F = d/dt(m

_{0}v/√(1 - v^{2}/c^{2})Using product rule you can write this as:

F = m

_{0}/(√1 - v^{2}/c^{2})dv/dt + vd/dt(m_{0}/(√1 - v^{2}/c^{2}))= (m

_{0}(1 - v^{2}/c^{2})^{3/2})dv/dtPutting this result in work equation, we get

W =

_{0}∫^{s}Fds =_{0}∫^{s}(m_{0}/(1 - v^{2}/c^{2})^{3/2})(dv/dt)dsNow after applying chain rule we get

W =

_{0}∫^{v}(m_{0}v/(1 - v^{2}/c^{2})^{3/2})dvIntegrating by parts, we have

W = [m

_{0}c^{2}/√(1 - v^{2}/c^{2})] - m_{0}c^{2}W = mc

^{2}- m_{0}c^{2}This equation is equation of relativistic work. It tells us that we can't accelerate an object to speeds up to or greater than speed of light. If v→c, then m→∞. Amount of work required to get this status becomes infinite and impossible to furnish.

If, we ignore dissipative (frictional) forces and suppose constant potential energy for body, then work-energy theorem (form conservation of energy principle) tells us that work done on body seems as its kinetic energy. Therefore, if we write T for kinetic energy, we get

T = mc

^{2}= m_{0}c^{2}As always, you must demand that equation be consistent with definition of kinetic energy as we all know it at ordinary velocities. Therefore, in limit as v/c << 1, we can use series expansion formula

1/√1 -x = 1 + 1/2x + 3/8x

^{2}+....Here x = v

^{2}/c^{2}. Relativistic mass formula could then be approximated asm = m

_{o}/√(√(1 - v^{2}/c^{2}) ≈ m_{0}[1 + 1/2(v/c)^{2}] = m_{0}+ 1/2m_{0}(v/c)^{2}On solving this approximate value we get

T = [m

_{0}+ 1/2m_{0}(v/c)^{2}]c^{2}- m_{0}c^{2}[1 + 1/2(v/c)^{2}- 1] = 1/2m_{0}v^{2}for v/c<<1Thus, equation correctly represents kinetic energy at all velocities.

Mass-Energy Equivalence:T = mc

^{2}- m_{0}c^{2}= [m_{0}c2/√(1 - v^{2}/c^{2})] - m_{0}c^{2}We easily observe that kinetic energy T is function of velocity. Also, as T is energy, other two terms on right hand side of equation are also energy terms and we can write:

T(v) = E(v) - E(0)

or E(v) = T(v) + E(0)

Where E(v) = m

_{0}c^{2}/√(1 - v^{2}/c^{2}) = mc^{2}and E(0) = m_{0}c^{2}which value of E(v) for v = 0E(v) is total energy of particle moving at relativistic speed v, E(0) is energy of particle when it is at rest i.e. v = 0. Further more if we consider each term individually, we write:

E(0) = m

_{0}c^{2}E(v) = mc

^{2}These are famous Einstein's mass-energy equations. It is clear from the equations that mass and energy are related by factor c

^{2}and are equivalent. It means that principle of conservation of mass or energy no longer makes sense as conservation of mass-energy does.E(v) - E(0) = mc

^{2}- m_{0}c^{2}or ΔE = Δmc^{2}where Δm = m - m_{0}This equation defines that change in energy leads to corresponding change in mass. More usually, it says that when mass is destroyed, it seems as energy and also if energy disappears, it emerges as mass.

If particle has potential energy V, we could write more general equation of form

mc

^{2}= T + V + m_{0}c^{2}Now, we use that fact that E = mc

^{2}, E0 = m_{0}c^{2}and p = mvE

^{2}= (pc)^{2}+ E_{0}^{2}This Equation is relationship between momentum and energy of the particle. It describes why, in relativistic theory, we should replace conservation of total energy. We can then state that, as viewed from the specified frame of reference, total relativistic energy of isolated system remains constant.

Transformation of Momentum and Energy:Consider two observers o and o' in inertial frames of reference S and S' which are in relative motion along -axis at a relativistic velocity. For observer, components of momentum and relativistic energy of particle of rest m

_{0}with velocity V along positive -axis arep

_{x}= m_{0}v/(√1 - V^{2}/c^{2})P

_{y}= 0P

_{z}= 0and E = m

_{0}c^{2}/(√1 - v^{2}/c^{2})]Observer O' assigns to this particle components of momentum and relativistic energy as

P'

_{x}= m_{0}V'/(√1 - v'^{2}/c^{2})P'

_{y}= 0P'

_{z}= 0 and E' = m_{0}c^{2}/√(1 - V'^{2}/c^{2})Where V' is velocity of particle along positive -axis as measured by this observer. Notice that O' assigns to particle same rest mass m

_{0}.We have to find primed quantities in terms of unprimed ones. First of all transform velocity terms. That is, calculate quantities

V'/√1 - V'

^{2}/C^{2}and c^{2}/√(1 - V'^{2}/C^{2}) in terms of V using velocity transformation equation,i.e.

V' = (V - v)/ (1 - (v/c

^{2})V)1 - V'

^{2}/c^{2}= (1 - 2v^{2}/c^{2 }+ V^{2}v^{2}/c^{4})/(1 - 2Vv/c^{2}+ V^{2}v^{2}/c^{4})Get relativistic energy equation as assigned by observer O'. We get:

m

_{0}c^{2}/√(1 - V'^{2}/C^{2}) = (m_{0}c^{2}- m^{0}Vv)/(√1 - v^{2}/c^{2})(√1 - V^{2}/c^{2})E' = E - vP

_{x}/√(1 - v^{2}/c^{2})Also, momentum assigned by O' is

P'

_{x}= (Px - vE/c^{2})/√(1 - v^{2}/c^{2})Momentum in directions that are perpendicular to direction of motion is not changed. With this in mind, we collect transformation equations for relativistic energy and components of momentum together to get set given below:

P'

_{x}= (P_{x}- vE/c^{2})/√(1 - v^{2}/c^{2})P'

_{y}= P_{y}P'

_{z}= P_{z}E' = (E - vP

_{x})/(√(1 - v^{2}/c^{2})Relativistic energy and momentum equations obtained are of same form as Lorentz transformation equations. Therefore, we conclude that momentum and energy transform exactly as space-time quantities x, y, z and t.

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