Relativistic kinetic energy

See also: Mass in special relativity, Tests of relativistic energy and momentum
If a body's speed is a significant fraction of the speed of light, it is necessary to use relativistic mechanics to calculate its kinetic energy.

## Nonrelativistic kinetic energy

The work done in accelerating a particle with mass *m* during the infinitesimal time interval *dt* is given by the dot product of *force* **f** and the infinitesimal *displacement* **dx**

f

⋅
d

x

d
t

d
t

=

f

⋅

v

d
# t

d

p

d
t

⋅

v

d
# t

v

⋅
d

p

=

v

⋅
d
(
m

v

)

,

{\displaystyle \mathbf{f} \cdot d \mathbf{x} \frac{dt}{dt} = \mathbf{f} \cdot \mathbf{v} d t = \frac{d \mathbf{p}}{d t} \cdot \mathbf{v} d t = \mathbf{v} \cdot d \mathbf{p} = \mathbf{v} \cdot d (m \mathbf{v})\,,}

where we have assumed the relationship **p** = *m* **v** and the validity of Newton's Second Law.

Applying the product rule to dv2 we see that:

d
(

v

⋅

v

# )
(
d

v

)
⋅

v

v

⋅
(
d

v

# )
2
(

v

⋅
d

v

)
.

{\displaystyle d(\mathbf{v} \cdot \mathbf{v}) = (d \mathbf{v}) \cdot \mathbf{v} + \mathbf{v} \cdot (d \mathbf{v}) = 2(\mathbf{v} \cdot d\mathbf{v}).}

Therefore, (assuming constant mass so that *dm* = 0), we have,

v

⋅
d
(
m

v

# )

m
2

d
(

v

⋅

v

# )

m
2

d

v
2

=
d

(

m

v
2

2

)

.

{\displaystyle \mathbf{v} \cdot d (m \mathbf{v}) = \frac{m}{2} d (\mathbf{v} \cdot \mathbf{v}) = \frac{m}{2} d v^2 = d \left(\frac{m v^2}{2}\right). }

Since this is a total differential (that is, it only depends on the final state, not how the particle got there), we can integrate it and call the result kinetic energy. Assuming the object was at rest at time 0, we integrate from time 0 to time t because the work done by the force to bring the object from rest to velocity *v* is equal to the work necessary to do the reverse:

E
k

=

∫
0
t

f

⋅
d

x

=

∫
0
t

v

⋅
d
(
m

v

# )

∫
0
t

d

(

m

v
2

2

)

=

m

v
2

2

.

{\displaystyle E\_\text{k} = \int\_0^t \mathbf{f} \cdot d \mathbf{x} = \int\_0^t \mathbf{v} \cdot d (m \mathbf{v}) = \int\_0^t d \left(\frac{m v^2}{2}\right) = \frac{m v^2}{2}.}

This equation states that the kinetic energy (*E*k) is equal to the integral of the dot product of the velocity (**v**) of a body and the infinitesimal change of the body's momentum (**p**). It is assumed that the body starts with no kinetic energy when it is at rest (motionless).

Hence

d
E

d
x

=

d
E

d
p

d
p

d
t

d
t

d
x

=

d
E

d
p

f
v

=
v

f
v

{\displaystyle \frac{ d E}{ d x} = \frac{ d E}{ d p}\frac{ d p}{ d t} \frac{ d t}{ d x}
 = \frac{ d E}{ d p} \frac{f}{v} = v \frac{f}{v}}

because

d
E

d
p

=
v

{\displaystyle \frac{ d E}{ d p} = v }

therefore

# E
∫

v

⋅
d

p

{\displaystyle E = \int \mathbf{v} \cdot d \mathbf{p}}

Consider a unit force acting on a particle of unit mass. The particle is at rest at time=0.
To the left is a graph of of such a particle. The horizontal axis is force \* time. Multiplying the horizontal axis by mass gives the relativistic momentum.

The vertical axis is nonrelativistic velocity. Velocity \* dt = the distance traveled under the given force during the time interval dt. So the area under the curve is force \* distance = energy. This is true both classically (green) and relativistically (red).

## But converting mv^2/2 to relativity isnt as easy as it first sounds

Naïvely multiplying half the rest mass and the dot product of four-velocity with itself results in a constant:

m
0

2

⋅

[

γ
c

γ

v
x

γ

v
y

γ

v
z

]

⋅

[

γ
c

γ

v
x

γ

v
y

γ

v
z

]

=

m
0

2

γ
2

(

c
2

v
x
2

v
y
2

v
z
2

)

=

m
0

c
2

2

{\displaystyle \frac{m\_0}{2} \cdot \begin{bmatrix} \gamma c \\ \gamma v\_x \\ \gamma v\_y \\ \gamma v\_z \end{bmatrix} \cdot
\begin{bmatrix} \gamma c \\ \gamma v\_x \\ \gamma v\_y \\ \gamma v\_z \end{bmatrix} =
\frac{m\_0}{2} \gamma^2 \left( c^2 + v\_x^2 + v\_y^2 + v\_z^2 \right) =
\frac{m\_0 c^2}{2}
}

because four-velocity squared is a constant.

V

2

=

c
2

{\displaystyle \mathbf{V}^2 = c^2}

If we multiply relativistic momentum times ordinary velocity (not four-velocity)

# E

m
0

2

[

γ
c

γ

v
x

γ

v
y

γ

v
z

]

[

0

v
x

v
y

v
z

]

=

m
0

γ

2

c
2

(

β
x
2

β
y
2

β
z
2

# )

m
0

γ

2

c
2

(

β
2

)

{\displaystyle E = \frac{m\_0}{2} \begin{bmatrix}
{ \gamma c} \\
\gamma v\_x \\
\gamma v\_y \\
\gamma v\_z
\end{bmatrix} \begin{bmatrix}
0 \\
v\_x \\
v\_y \\
v\_z
\end{bmatrix} = \frac{m\_0 \gamma}{2} c^2 (
\beta\_x^2 +
\beta\_y^2 +
\beta\_z^2) = \frac{m\_0 \gamma}{2} c^2 (
\beta^2 )}

Still not quite right.

## Relativistic kinetic energy

In special relativity theory, the expression for linear momentum is modified.

With *m* being an object's rest mass, **v** and *v* its velocity and speed, and *c* the speed of light in vacuum, we use the expression for linear momentum

p

=
m
γ

v

{\displaystyle \mathbf{p}=m\gamma \mathbf{v}}

, where

# γ
1

/

1
−

v
2

/

c
2

{\displaystyle \gamma = 1/\sqrt{1-v^2/c^2}}

.

Integrating by parts yields

E
k

=
∫

v

⋅
d

p

=
∫

v

⋅
d
(
m
γ

v

)

=
m
γ

v

⋅

v

−
∫
m
γ

v

⋅
d

v

=
m
γ

v
2

−

m
2

∫
γ
d

(

v
2

)

{\displaystyle \begin{align} E\_\text{k} &=
 \int \mathbf{v} \cdot d \mathbf{p} \\ &=
 \int \mathbf{v} \cdot d (m \gamma \mathbf{v}) \\ &=
 m \gamma \mathbf{v} \cdot \mathbf{v} - \int m \gamma \mathbf{v} \cdot d \mathbf{v} \\ &=
 m \gamma v^2 - \frac{m}{2} \int \gamma d \left(v^2\right)
\end{align}
}

If the jump from step 2 to step 3 is giving you trouble then try drawing it out. Its actually trivial. Instead of finding the area under the curve as one would expect they are instead finding the area above the curve and then subtracting to find the area under the curve. See the image to the right.

The area above the curve is:

∫
m
γ

v

⋅
d

v

=
m

c
2

−

m

c
2

γ

{\displaystyle \int m \gamma \mathbf{v} \cdot d \mathbf{v} = mc^2 - \frac{mc^2}{\gamma}}

Since

# γ

(

1
−

v
2

/

c
2

)

−

1
2

{\displaystyle \gamma = \left(1 - v^2/c^2\right)^{-\frac{1}{2}}}

,

E
k

=
m
γ

v
2

−

−
m

c
2

2

∫
γ
d

(

1
−

v
2

c
2

)

=
m
γ

v
2

m

c
2

(

1
−

v
2

c
2

)

1
2

−

E
0

{\displaystyle \begin{align}
 E\_\text{k} &= m \gamma v^2 - \frac{- m c^2}{2} \int \gamma d \left(1 - \frac{v^2}{c^2}\right) \\
 &= m \gamma v^2 + m c^2 \left(1 - \frac{v^2}{c^2}\right)^\frac{1}{2} - E\_0
\end{align}}

E
0

{\displaystyle E\_0}

 is a constant of integration for the indefinite integral.

Simplifying the expression we obtain

E
k

=
m
γ

(

v
2

c
2

(

1
−

v
2

c
2

)

)

−

E
0

=
m
γ

(

v
2

c
2

−

v
2

)

−

E
0

=
m
γ

c
2

−

E
0

{\displaystyle \begin{align}
 E\_\text{k} &= m \gamma \left(v^2 + c^2 \left(1 - \frac{v^2}{c^2}\right)\right) - E\_0 \\
 &= m \gamma \left(v^2 + c^2 - v^2\right) - E\_0 \\
 &= m \gamma c^2 - E\_0
\end{align}}

E
0

{\displaystyle E\_0}

 is found by observing that when

v

=
0
,

# γ
1

{\displaystyle \mathbf{v} = 0,\ \gamma = 1}

 and

E
k

=
0

{\displaystyle E\_\text{k} = 0}

, giving

E
0

=
m

c
2

{\displaystyle E\_0 = m c^2 \,}

resulting in the formula

E
k

=
m
γ

c
2

−
m

c
2

=
m
⋅
(
γ
−
1
)

c
2

≈

m

v
2

2

c
2

c
2

=

m

v
2

2

{\displaystyle E\_\text{k} = m \gamma c^2 - m c^2 = m \cdot (\gamma - 1 ) c^2 \quad \approx \quad m \frac{v^2}{2 c^2} c^2 = \frac{m v^2}{2}}

This formula shows that the work expended accelerating an object from rest approaches infinity as the velocity approaches the speed of light. Thus it is impossible to accelerate an object across this boundary.

The mathematical by-product of this calculation is the mass-energy equivalence formula—the body at rest must have energy content

E
rest

=

E
0

=
m

c
2

{\displaystyle E\_\text{rest} = E\_0 = m c^2 \!}

At a low speed (*v* ≪ *c*), the relativistic kinetic energy is approximated well by the classical kinetic energy. This is done by binomial approximation or by taking the first two terms of the Taylor expansion for the reciprocal square root:

E
k

≈
m

c
2

(

1

1
2

v
2

c
2

)

−
m

c
2

=

1
2

m

v
2

{\displaystyle E\_\text{k} \approx m c^2 \left(1 + \frac{1}{2} \frac{v^2}{c^2}\right) - m c^2 = \frac{1}{2} m v^2}

So, the total energy

E
k

{\displaystyle E\_k}

 can be partitioned into the rest mass energy plus the Newtonian kinetic energy at low speeds.

When objects move at a speed much slower than light (e.g. in everyday phenomena on Earth), the first two terms of the series predominate. The next term in the Taylor series approximation

E
k

≈
m

c
2

(

1

1
2

v
2

c
2

3
8

v
4

c
4

)

−
m

c
2

=

1
2

m

v
2

3
8

m

v
4

c
2

{\displaystyle E\_\text{k} \approx m c^2 \left(1 + \frac{1}{2} \frac{v^2}{c^2} + \frac{3}{8} \frac{v^4}{c^4}\right) - m c^2 = \frac{1}{2} m v^2 + \frac{3}{8} m \frac{v^4}{c^2}}

is small for low speeds. For example, for a speed of 10 km/s (22,000 mph) the correction to the Newtonian kinetic energy is 0.0417 J/kg (on a Newtonian kinetic energy of 50 MJ/kg) and for a speed of 100 km/s it is 417 J/kg (on a Newtonian kinetic energy of 5 GJ/kg).

The relativistic relation between kinetic energy and momentum is given by

E
k

=

p
2

c
2

m
2

c
4

−
m

c
2

{\displaystyle E\_\text{k} = \sqrt{p^2 c^2 + m^2 c^4} - m c^2}

This can also be expanded as a Taylor series, the first term of which is the simple expression from Newtonian mechanics:

E
k

≈

p
2

2
m

−

p
4

8

m
3

c
2

.

{\displaystyle E\_\text{k} \approx \frac{p^2}{2 m} - \frac{p^4}{8 m^3 c^2}.}

This suggests that the formulae for energy and momentum are not special and axiomatic, but concepts emerging from the equivalence of mass and energy and the principles of relativity.

### General relativity

See also: Schwarzschild geodesics
Using the convention that

g

α
β

u

α

u

β

=

−

c
2

{\displaystyle g\_{\alpha \beta} \, u^{\alpha} \, u^{\beta} \, = \, - c^2 }

where the four-velocity of a particle is

u

α

=

d

x

α

d
τ

{\displaystyle u^{\alpha} \, = \, \frac{d x^{\alpha}}{d \tau} }

and

τ

{\displaystyle \tau \,}

 is the proper time of the particle, there is also an expression for the kinetic energy of the particle in general relativity.

If the particle has momentum

p

β

=

m

g

β
α

u

α

{\displaystyle p\_{\beta} \, = \, m \, g\_{\beta \alpha} \, u^{\alpha} }

as it passes by an observer with four-velocity *u*obs, then the expression for total energy of the particle as observed (measured in a local inertial frame) is

E

=

−

p

β

u

obs

β

{\displaystyle E \, = \, - \, p\_{\beta} \, u\_{\text{obs}}^{\beta} }

and the kinetic energy can be expressed as the total energy minus the rest energy:

E

k

=

−

p

β

u

obs

β

−

m

c
2

.

{\displaystyle E\_{k} \, = \, - \, p\_{\beta} \, u\_{\text{obs}}^{\beta} \, - \, m \, c^2 \, .}

Consider the case of a metric that is diagonal and spatially isotropic (*g**tt*, *g**ss*, *g**ss*, *g**ss*). Since

u

α

=

d

x

α

d
t

d
t

d
τ

=

v

α

u

t

{\displaystyle u^{\alpha} = \frac{d x^{\alpha}}{d t} \frac{d t}{d \tau} = v^{\alpha} u^{t} \,}

where *v*α is the ordinary velocity measured w.r.t. the coordinate system, we get

−

c
2

=

g

α
β

u

α

u

β

=

g

t
t

(

u

t

)

2

g

s
s

v
2

(

u

t

)

2

.

{\displaystyle -c^2 = g\_{\alpha\beta} u^{\alpha} u^{\beta} = g\_{tt} \left(u^{t}\right)^2 + g\_{ss} v^2 \left(u^{t}\right)^2 \,.}

Solving for *u*t gives

u

t

=
c

−
1

g

t
t

g

s
s

v
2

.

{\displaystyle u^{t} = c \sqrt{\frac{-1}{g\_{t t} + g\_{s s} v^2}} \,.}

Thus for a stationary observer (*v* = 0)

u

obs

t

=
c

−
1

g

t
t

{\displaystyle u\_{\text{obs}}^{t} = c \sqrt{\frac{-1}{g\_{t t}}} \,}

and thus the kinetic energy takes the form

E
k

=
−
m

g

t
t

u
t

u

obs

t

−
m

c
2

=
m

c
2

g

t
t

g

t
t

g

s
s

v
2

−
m

c
2

.

{\displaystyle E\_\text{k} = - m g\_{tt} u^t u\_{\text{obs}}^t - m c^2 = m c^2 \sqrt{\frac{g\_{tt}}{g\_{tt} + g\_{ss} v^2}} - m c^2\,.}

Factoring out the rest energy gives:

E
k

=
m

c
2

(

g

t
t

g

t
t

g

s
s

v
2

−
1

)

.

{\displaystyle E\_\text{k} = m c^2 \left( \sqrt{\frac{g\_{tt}}{g\_{tt} + g\_{ss} v^2}} - 1 \right) \,.}

This expression reduces to the special relativistic case for the flat-space metric where

g

t
t

=
−

c
2

g

s
s

=
1

.

{\displaystyle \begin{align}
 g\_{tt} &= -c^2 \, \\
 g\_{ss} &= 1 \,.
\end{align}}

In the Newtonian approximation to general relativity

g

t
t

=
−

(

c
2

2
Φ

)

g

s
s

=
1
−

2
Φ

c
2

{\displaystyle \begin{align}
 g\_{tt} &= -\left( c^2 + 2\Phi \right) \, \\
 g\_{ss} &= 1 - \frac{2\Phi}{c^2} \,
\end{align}}

where Φ is the Newtonian gravitational potential. This means clocks run slower and measuring rods are shorter near massive bodies.