Relativistic Energy
The relativistic energy of an object with mass $ m $ moving at velocity $ v $ is given by $$ E = \gamma m c^2 = \frac{m c^2}{\sqrt{1 - \frac{v^2}{c^2}}} $$ where $ c $ is the speed of light. At low speeds, when $ v \ll c $, this expression can be approximated as $$ E \approx m c^2 + \tfrac{1}{2} m v^2 $$. More generally, at any velocity, energy can also be expressed in terms of momentum $ p $ as $$ E^2 = (p c)^2 + (m c^2)^2 $$.
In special relativity, energy and mass are not distinct quantities-they are two facets of the same physical reality.
Every object with a rest mass $ m $ possesses an intrinsic energy, called its rest energy, given by Einstein’s celebrated equation:
$$ E_0 = m c^2 $$
When the object moves, its total energy becomes the sum of its rest energy and the energy associated with motion, forming the total relativistic energy:
$$ E = \gamma m c^2 = \frac{m c^2}{\sqrt{1 - \frac{v^2}{c^2}}} $$
At low speeds ($ v \ll c $), this simplifies to the sum of the rest energy and the familiar classical kinetic energy $ \tfrac{1}{2} m v^2 $.
$$ E \approx m c^2 + \tfrac{1}{2} m v^2 $$
As velocity approaches the speed of light ($ v \to c $), the Lorentz factor $ \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} $ grows without bound, meaning that the energy required to further increase the speed tends toward infinity.
$$ \lim_{v \to c} E = \lim_{v \to c} \gamma mc^2 = \infty $$
Therefore, no object with nonzero mass ($ m>0 $) can ever reach the speed of light $ c $, since doing so would require infinite energy.
Yet we know that certain particles-such as photons-do in fact move at the speed of light.
Why can photons travel at the speed of light?
Massless particles ($ m=0 $), like photons, do not follow the formula $ E = \gamma m c^2 $. Instead, they obey the general relationship between energy and momentum:
$$ E^2 = (p c)^2 + (m c^2)^2 $$
For $ m = 0 $, this simplifies to $ E = p c $, which means that energy is directly proportional to momentum.
$$ E = pc $$
This explains why photons possess finite energy $ E $ and momentum $ p $ even though they have zero rest mass ($ m=0 $), and why they always travel at the speed of light $ c $.
Note. According to Planck’s law, the energy of a photon is finite and expressed as $ E = h \nu $, where $ h $ denotes Planck’s constant and $ \nu $ represents the photon’s frequency.
Explanation and Derivation
In classical mechanics, the total energy of a body is the sum of its potential energy $ U $, kinetic energy $ K = \tfrac{1}{2}mv^2 $, and possibly other contributions (such as elastic or gravitational energy).
In special relativity, however, this separation is no longer adequate. When a body moves at speeds comparable to that of light, the relationship between mass, energy, and velocity changes fundamentally.
Einstein showed that every body with mass $ m $ has a total energy given by:
$$ E = \gamma m c^2 = \frac{m c^2}{\sqrt{1 - \dfrac{v^2}{c^2}}} $$
where:
- $ c $ is the speed of light,
- $ v $ is the object’s velocity,
- $ \gamma = \dfrac{1}{\sqrt{1 - v^2/c^2}} $ is the Lorentz factor.
From this equation, we see that a body possesses an energy $ E_0 $ even when at rest ($ v = 0 $):
$$ E_0 = m c^2 $$
This is Einstein’s iconic result: mass itself is a form of energy.
In other words, mass and energy are interchangeable-mass can be converted into energy and vice versa.
Let’s now consider two regimes: one at very low speeds (the classical limit) and one approaching the speed of light (the relativistic regime).
The Low-Velocity Limit
When $ v $ is much smaller than $ c $-that is, $ v \ll c $-the Lorentz factor approaches unity: $ \gamma \approx 1 $.
$$ \lim_{v \to 0} \gamma = \lim_{v \to 0} \dfrac{1}{\sqrt{1 - v^2/c^2}} = 1 $$
Thus, the total energy $ E = \gamma mc^2 $ reduces to its rest energy $ E_0 = mc^2 $.
$$ E \approx E_0 = mc^2 $$
To explore this in more detail, we can expand the Lorentz factor as a Taylor series.
First, rewrite the Lorentz factor in exponential form:
$$ \gamma = \dfrac{1}{\sqrt{1 - v^2/c^2}} = (1 - \frac{v^2}{c^2})^{-1/2} $$
The Taylor expansion of a generalized binomial (power law) for any exponent $ \alpha $, valid when $ |x| < 1 $, is:
$$ (1 + x)^\alpha = 1 + \alpha x + \frac{\alpha(\alpha - 1)}{2!}x^2 + \frac{\alpha(\alpha - 1)(\alpha - 2)}{3!}x^3 + \cdots $$
In our case, $ \alpha = -\tfrac{1}{2} $ and $ x = -\dfrac{v^2}{c^2} $.
$$ (1 - \frac{v^2}{c^2})^{-1/2} = 1 + ( - \frac{1}{2} ) \cdot ( - \frac{v^2}{c^2} ) + \frac{1}{2!} \cdot ( - \frac{1}{2} ) \cdot ( - \frac{1}{2} - 1 ) \cdot ( - \frac{v^2}{c^2} )^2 + \cdots $$
$$ (1 - \frac{v^2}{c^2})^{-1/2} = 1 + \frac{1}{2}\frac{v^2}{c^2} + \frac{3}{8}\frac{v^4}{c^4} + \cdots $$
Expanding a function as a Taylor series means approximating it near a given point-in this case, around $ v = 0 $.
This shows that successive terms become progressively smaller:
$$ (\frac{v^2}{c^2} \ll 1), (\frac{v^4}{c^4} \ll \frac{v^2}{c^2}), ... $$
Therefore, keeping only the first few terms provides an excellent approximation in the classical limit:
$$ \gamma \approx 1 + \frac{1}{2}\frac{v^2}{c^2} $$
The higher-order terms are negligible when $ v \ll c $.
Substituting this approximation into the relativistic energy expression gives:
$$ E = \gamma mc^2 = (1 + \frac{1}{2}\frac{v^2}{c^2}) \cdot mc^2 $$
$$ E = mc^2 + \frac{1}{2} mv^2 $$
The first term, $ m c^2 $, is constant and represents rest energy; the second term, $ \tfrac{1}{2} m v^2 $, is the classical kinetic energy.
Thus, classical mechanics naturally emerges as the low-velocity limit of special relativity.
The High-Velocity Regime
As an object’s velocity $ v $ approaches the speed of light $ c $ ($ v \to c $), its kinetic energy increases dramatically with even the smallest change in speed.
For a body with rest mass $ m $, the total relativistic energy is expressed as:
$$ E = \gamma m c^2 = \frac{m c^2}{\sqrt{1 - \frac{v^2}{c^2}}} $$
This total energy combines two contributions: the rest energy $ E_0 $ and the kinetic energy $ K $ associated with motion, according to
$$ E = E_0 + K $$
By definition, the kinetic energy is simply the difference between the total energy $ E $ and the rest energy $ E_0 $:
$$ K = E - E_0 $$
Since $ E = \gamma mc^2 $, we obtain:
$$ K = \gamma mc^2 - E_0 $$
At rest ($ v = 0 $), the Lorentz factor equals one ($ \gamma = 1 $), so the rest energy is
$$ E_0 = \gamma mc^2 = 1 \cdot mc^2 = mc^2 $$
Substituting $ E_0 = mc^2 $ into the equation for $ K $ gives:
$$ K = \gamma mc^2 - mc^2 $$
$$ K = mc^2 (\gamma - 1) $$
This is the relativistic kinetic energy, defined as the difference between total and rest energy: $ K = E - E_0 $. Unlike the classical expression $ \tfrac{1}{2}mv^2 $, this formula remains valid at any velocity.
As $ v \to c $, the Lorentz factor grows without bound ($ \gamma \to \infty $):
$$ \lim_{v \to c} \gamma = \lim_{v \to c} \dfrac{1}{\sqrt{1 - v^2/c^2}} = \infty $$
Consequently, the energy required to further accelerate the object also diverges:
$$ \lim_{v \to c} E = \lim_{v \to c} \gamma mc^2 = mc^2 \cdot \lim_{v \to c} \gamma = \infty $$
Therefore, no object with nonzero mass can ever reach the speed of light, since it would require infinite energy ($ E \to \infty $).
$$ E = \gamma mc^2 = \frac{mc^2}{\sqrt{1 - v^2/c^2}} $$
Mathematically, this divergence arises because the numerator ($ mc^2 $) is positive for $ m>0 $, while the denominator approaches zero from above ($ 0^+ $), driving $ E $ to infinity.
But what happens if a particle has zero mass?
For a massless particle, such as a photon, the numerator $ mc^2 $ also vanishes ($ m = 0 $):
$$ \lim_{v \to c} E = \lim_{v \to c} \gamma mc^2 = mc^2 \cdot \lim_{v \to c} \gamma = \frac{0}{0} $$
This limit is indeterminate, meaning it doesn’t necessarily tend to infinity.
In this situation, the correct approach is to use the general energy-momentum relation:
$$ E^2 = (pc)^2 + (mc^2)^2 $$
where $ p = \gamma m v $ represents the relativistic momentum.
This equation holds for all particles, including those with zero mass, and thus provides a unified description of both cases:
- for $ m>0 $, it describes massive particles;
- for $ m=0 $, it describes massless particles.
Hence, it represents the most general form of the relativistic energy relationship.
For a massless particle ($ m=0 $), the equation simplifies to:
$$ E^2 = (pc)^2 $$
Taking the square root of both sides yields:
$$ \sqrt{E^2} = \sqrt{(pc)^2} $$
$$ E = pc $$
Thus, for a massless particle, its energy is equal to its momentum multiplied by the speed of light:
$$ E = p c $$
This is the general energy formula for massless particles, such as photons.
Now, let’s examine what happens to the momentum $ p $ as $ v \to c $:
$$ \lim_{v \to c} p = \lim_{v \to c} \gamma m v = \lim_{v \to c} \frac{m v}{\sqrt{1 - v^2/c^2}} = \frac{0}{0} $$
Again, when $ m = 0 $, the expression becomes an indeterminate form $ 0/0 $.
However, if the mass tends toward zero ($ m \to 0 $) at the same rate that the denominator tends toward zero ($ \sqrt{1 - v^2/c^2} \to 0 $), the limit remains finite:
$$ m \propto \sqrt{1 - v^2/c^2} \Rightarrow p = \text{finite}, \quad E = p c = \text{finite} $$
If the momentum $ p $ is finite, then the energy $ E = p c $ is also finite as $ v \to c $.
This describes exactly the behavior of photons: they possess finite energy $ E $ and momentum $ p $, yet have zero rest mass ($ m = 0 $).
Photons exist precisely at this limiting speed-they cannot exist at $ v < c $, but only at $ v = c $, where the formula $ E = \gamma mc^2 $ ceases to apply and is replaced by $ E = pc $.
Note. In classical mechanics, the total mechanical energy (the sum of kinetic and potential energy) is conserved. In relativity, however, mass itself represents a form of energy, $ E_0 = m c^2 $. Thus, the conservation principle extends beyond mechanical energy to include rest energy as well. When a system’s mass decreases, the energy released increases by an equivalent amount, and vice versa: $$ \Delta E = -\,\Delta m\, c^2 $$ This is the profound meaning of Einstein’s relation: mass and energy are interchangeable, yet their total sum remains constant.
And so forth.
