List of relativistic equations

The following notations are used very often in special relativity:

Lorentz factor

${\displaystyle \gamma ={\frac {1}{\sqrt {1-\beta ^{2}}}}}$

where β = ${\displaystyle {\frac {v}{c}}}$ and v is the relative velocity between two inertial frames.

For two frames at rest, γ = 1, and increases with relative velocity between the two inertial frames. As the relative velocity approaches the speed of light, γ → ∞.

Time dilation (different times t and t' at the same position x in same inertial frame)

${\displaystyle t'=\gamma t}$

Derivation of time dilation

Applying the above postulates, consider the inside of any vehicle (usually exemplified by a train) moving with a velocity v with respect to someone standing on the ground as the vehicle passes. Inside, a light is shone upwards to a mirror on the ceiling, where the light reflects back down. If the height of the mirror is h, and the speed of light c, then the time it takes for the light to go up and come back down is: ${\displaystyle t={\frac {2h}{c}}}$ However, to the observer on the ground, the situation is very different. Since the train is moving by the observer on the ground, the light beam appears to move diagonally instead of straight up and down. To visualize this, picture the light being emitted at one point, then having the vehicle move until the light hits the mirror at the top of the vehicle, and then having the train move still more until the light beam returns to the bottom of the vehicle. The light beam will have appeared to have moved diagonally upward with the train, and then diagonally downward. This path will help form two-right sided triangles, with the height as one of the sides, and the two straight parts of the path being the respective hypotenuses: ${\displaystyle c^{2}\left({\frac {t'}{2}}\right)^{2}=h^{2}+v^{2}\left({\frac {t'}{2}}\right)^{2}}$ Rearranging to get ${\displaystyle t'}$: ${\displaystyle \left({\frac {t'}{2}}\right)^{2}={\frac {h^{2}}{c^{2}-v^{2}}}}$ ${\displaystyle {\frac {t'}{2}}={\frac {h}{\sqrt {c^{2}-v^{2}}}}}$ ${\displaystyle t'={\frac {2h}{\sqrt {c^{2}-v^{2}}}}}$ Taking out a factor of c, and then plugging in for t, one finds: ${\displaystyle t'={\frac {2h}{c}}{\frac {1}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}={\frac {t}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}}$ This is the formula for time dilation: ${\displaystyle t'=\gamma t}$

In this example the time measured in the frame on the vehicle, t, is known as the proper time. The proper time between two events - such as the event of light being emitted on the vehicle and the event of light being received on the vehicle - is the time between the two events in a frame where the events occur at the same location. So, above, the emission and reception of the light both took place in the vehicle's frame, making the time that an observer in the vehicle's frame would measure the proper time.

Length contraction (different positions x and x' at the same instant t in the same inertial frame)

${\displaystyle \ell '={\frac {\ell }{\gamma }}}$

Derivation of length contraction

Consider a long train, moving with velocity v with respect to the ground, and one observer on the train and one on the ground, standing next to a post. The observer on the train sees the front of the train pass the post, and then, some time t′ later, sees the end of the train pass the same post. He then calculates the train's length as follows: ${\displaystyle \ell =vt'\,}$ However, the observer on the ground, making the same measurement, comes to a different conclusion. This observer finds that time t passed between the front of the train passing the post, and the back of the train passing the post. Because the two events - the passing of each end of the train by the post - occurred in the same place in the ground observer's frame, the time this observer measured is the proper time. So: ${\displaystyle \ell '=vt=v\left({\frac {t'}{\gamma }}\right)={\frac {\ell }{\gamma }}}$

This is the formula for length contraction. As there existed a proper time for time dilation, there exists a proper length for length contraction, which in this case is ℓ. The proper length of an object is the length of the object in the frame in which the object is at rest. Also, this contraction only affects the dimensions of the object which are parallel to the relative velocity between the object and observer. Thus, lengths perpendicular to the direction of motion are unaffected by length contraction.

Lorentz transformation

${\displaystyle x'=\gamma \left(x-vt\right)}$

${\displaystyle y'=y\,}$

${\displaystyle z'=z\,}$

${\displaystyle t'=\gamma \left(t-{\frac {vx}{c^{2}}}\right)}$

Derivation of Lorentz transformation using time dilation and length contraction

Now substituting the length contraction result into the Galilean transformation (i.e. x = ℓ), we have: ${\displaystyle {\frac {x'}{\gamma }}=x-vt}$ that is: ${\displaystyle x'=\gamma \left(x-vt\right)}$ and going from the primed frame to the unprimed frame: ${\displaystyle x=\gamma \left(x'+vt'\right)}$ Going from the primed frame to the unprimed frame was accomplished by making v in the first equation negative, and then exchanging primed variables for unprimed ones, and vice versa. Also, as length contraction does not affect the perpendicular dimensions of an object, the following remain the same as in the Galilean transformation: ${\displaystyle y'=y\,}$ ${\displaystyle z'=z\,}$ Finally, to determine how t and t′ transform, substituting the x↔x′ transformation into its inverse: ${\displaystyle x=\gamma \left(\gamma \left(x-vt\right)+vt'\right)}$ ${\displaystyle x=\gamma \left(\gamma x-\gamma vt+vt'\right)}$ ${\displaystyle x=\gamma ^{2}x-\gamma ^{2}vt+\gamma vt'\,}$ ${\displaystyle \gamma vt'=\gamma ^{2}vt-\gamma ^{2}x+x\,}$ ${\displaystyle \gamma vt'=\gamma ^{2}vt+x\left(1-\gamma ^{2}\right)}$ Plugging in the value for γ: ${\displaystyle \gamma vt'=\gamma ^{2}vt+x\left(1-{\frac {1}{1-\beta ^{2}}}\right)}$ ${\displaystyle \gamma vt'=\gamma ^{2}vt+x\left({\frac {1-\beta ^{2}}{1-\beta ^{2}}}-{\frac {1}{1-\beta ^{2}}}\right)}$ ${\displaystyle \gamma vt'=\gamma ^{2}vt-x\left({\frac {\beta ^{2}}{1-\beta ^{2}}}\right)}$ ${\displaystyle \gamma vt'=\gamma ^{2}vt-\gamma ^{2}\beta ^{2}x\,}$ Finally, dividing through by γv: ${\displaystyle t'=\gamma \left(t-\beta {\frac {x}{c}}\right)}$ Or more commonly: ${\displaystyle t'=\gamma \left(t-{\frac {vx}{c^{2}}}\right)}$ And the converse can again be gotten by changing the sign of v, and exchanging the unprimed variables for their primed variables, and vice versa. These transformations together are the Lorentz transformation: ${\displaystyle x'=\gamma \left(x-vt\right)}$ ${\displaystyle y'=y\,}$ ${\displaystyle z'=z\,}$ ${\displaystyle t'=\gamma \left(t-{\frac {vx}{c^{2}}}\right)}$

Velocity addition

${\displaystyle V'_{x}={\frac {V_{x}-v}{1-{\frac {V_{x}v}{c^{2}}}}}}$

${\displaystyle V'_{y}={\frac {V_{y}}{\gamma \left(1-{\frac {V_{x}v}{c^{2}}}\right)}}}$

${\displaystyle V'_{z}={\frac {V_{z}}{\gamma \left(1-{\frac {V_{x}v}{c^{2}}}\right)}}}$

Derivation of velocity addition

The Lorentz transformations also apply to differentials, so: ${\displaystyle dx'=\gamma \left(dx-vdt\right)}$ ${\displaystyle dy'=dy\,}$ ${\displaystyle dz'=dz\,}$ ${\displaystyle dt'=\gamma \left(dt-{\frac {vdx}{c^{2}}}\right)}$ The velocity is dx/dt, so ${\displaystyle {\frac {dx'}{dt'}}={\frac {\gamma \left(dx-vdt\right)}{\gamma \left(dt-{\frac {vdx}{c^{2}}}\right)}}}$ ${\displaystyle {\frac {dx'}{dt'}}={\frac {dx-vdt}{dt-{\frac {vdx}{c^{2}}}}}}$ ${\displaystyle {\frac {dx'}{dt'}}={\frac {{\frac {dx}{dt}}-v}{1-{\frac {dx}{dt}}{\frac {v}{c^{2}}}}}}$ Now substituting: ${\displaystyle V_{x}={\frac {dx}{dt}}\,\quad V'_{x}={\frac {dx'}{dt'}}}$ gives the velocity addition (actually below is subtraction, addition is just reversing the signs of Vx, Vy, and Vz around): ${\displaystyle V'_{x}={\frac {V_{x}-v}{1-{\frac {V_{x}v}{c^{2}}}}}}$ ${\displaystyle V'_{y}={\frac {V_{y}}{\gamma \left(1-{\frac {V_{x}v}{c^{2}}}\right)}}}$ ${\displaystyle V'_{z}={\frac {V_{z}}{\gamma \left(1-{\frac {V_{x}v}{c^{2}}}\right)}}}$ Also, the velocities in the directions perpendicular to the frame changes are affected, as shown above. This is due to time dilation, as encapsulated in the dt/dt′ transformation. The V′y and V′z equations were both derived by dividing the appropriate space differential (e.g. dy′ or dz′) by the time differential.

It is possible to express the above coordinate transformation via a matrix. To simplify things, it can be best to replace t, t′, dt, and dt′ with ct, ct', cdt, and cdt′, which has the dimensions of distance. So:

${\displaystyle x'=\gamma x-\gamma \beta ct\,}$

${\displaystyle y'=y\,}$

${\displaystyle z'=z\,}$

${\displaystyle ct'=\gamma ct-\gamma \beta x\,}$

then in matrix form:

${\displaystyle {\begin{pmatrix}ct'\\x'\\y'\\z'\end{pmatrix}}={\begin{pmatrix}\gamma &-\gamma \beta &0&0\\-\gamma \beta &\gamma &0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}}{\begin{pmatrix}ct\\x\\y\\z\end{pmatrix}}}$

The vectors in the above transformation equation are known as four-vectors, in this case they are specifically the position four-vectors. In general, in special relativity, four-vectors can be transformed from one reference frame to another as follows:

${\displaystyle {\boldsymbol {\mathsf {a}}}'=\Lambda {\boldsymbol {\mathsf {a}}}}$

In the above, ${\displaystyle {\boldsymbol {\mathsf {a}}}'}$ and ${\displaystyle {\boldsymbol {\mathsf {a}}}}$ are the four-vector and the transformed four-vector, respectively, and Λ is the transformation matrix, which, for a given transformation is the same for all four-vectors one might want to transform. So ${\displaystyle {\boldsymbol {\mathsf {a}}}'}$ can be a four-vector representing position, velocity, or momentum, and the same Λ can be used when transforming between the same two frames. The most general Lorentz transformation includes boosts and rotations; the components are complicated and the transformation requires spinors.

Invariance and unification of physical quantities both arise from four-vectors.[1] The inner product of a 4-vector with itself is equal to a scalar (by definition of the inner product), and since the 4-vectors are physical quantities their magnitudes correspond to physical quantities also.

Property/effect	3-vector	4-vector	Invariant result
Space-time events	3-position: r = (x1, x2, x3) ${\displaystyle \mathbf {r} \cdot \mathbf {r} \equiv r^{2}\equiv x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\,\!}$	4-position: X = (ct, x1, x2, x3)	${\displaystyle {\boldsymbol {\mathsf {X}}}\cdot {\boldsymbol {\mathsf {X}}}=\left(c\tau \right)^{2}\,\!}$ ${\displaystyle {\begin{aligned}&\left(ct\right)^{2}-\left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\right)\\&=\left(ct\right)^{2}-r^{2}\\&=-\chi ^{2}=\left(c\tau \right)^{2}\end{aligned}}\,\!}$ τ = proper time χ = proper distance
Momentum-energy invariance	${\displaystyle \mathbf {p} =\gamma m\mathbf {u} \,\!}$ 3-momentum: p = (p1, p2, p3) ${\displaystyle \mathbf {p} \cdot \mathbf {p} \equiv p^{2}\equiv p_{1}^{2}+p_{2}^{2}+p_{3}^{2}\,\!}$	4-momentum: P = (E/c, p1, p2, p3) ${\displaystyle {\boldsymbol {\mathsf {P}}}=m{\boldsymbol {\mathsf {U}}}\,\!}$	${\displaystyle {\boldsymbol {\mathsf {P}}}\cdot {\boldsymbol {\mathsf {P}}}=\left(mc\right)^{2}\,\!}$ ${\displaystyle {\begin{aligned}&\left({\frac {E}{c}}\right)^{2}-\left(p_{1}^{2}+p_{2}^{2}+p_{3}^{2}\right)\\&=\left({\frac {E}{c}}\right)^{2}-p^{2}\\&=\left(mc\right)^{2}\end{aligned}}\,\!}$ which leads to: ${\displaystyle E^{2}=\left(pc\right)^{2}+\left(mc^{2}\right)^{2}\,\!}$ E = total energy m = invariant mass
Velocity	3-velocity: u = (u1, u2, u3) ${\displaystyle \mathbf {u} ={\frac {\mathrm {d} \mathbf {r} }{\mathrm {d} t}}\,\!}$	4-velocity: U = (U0, U1, U2, U3) ${\displaystyle {\boldsymbol {\mathsf {U}}}={\frac {\mathrm {d} {\boldsymbol {\mathsf {X}}}}{\mathrm {d} \tau }}=\gamma \left(c,\mathbf {u} \right)}$	${\displaystyle {\boldsymbol {\mathsf {U}}}\cdot {\boldsymbol {\mathsf {U}}}=c^{2}\,\!}$
Acceleration	3-acceleration: a = (a1, a2, a3) ${\displaystyle \mathbf {a} ={\frac {\mathrm {d} \mathbf {u} }{\mathrm {d} t}}\,\!}$	4-acceleration: A = (A0, A1, A2, A3) ${\displaystyle {\boldsymbol {\mathsf {A}}}={\frac {\mathrm {d} {\boldsymbol {\mathsf {U}}}}{\mathrm {d} \tau }}=\gamma \left(c{\frac {\mathrm {d} \gamma }{\mathrm {d} t}},{\frac {\mathrm {d} \gamma }{\mathrm {d} t}}\mathbf {u} +\gamma \mathbf {a} \right)}$	${\displaystyle {\boldsymbol {\mathsf {A}}}\cdot {\boldsymbol {\mathsf {U}}}=0\,\!}$
Force	3-force: f = (f1, f2, f3) ${\displaystyle \mathbf {f} ={\frac {\mathrm {d} \mathbf {p} }{\mathrm {d} t}}\,\!}$	4-force: F = (F0, F1, F2, F3) ${\displaystyle {\boldsymbol {\mathsf {F}}}={\frac {\mathrm {d} {\boldsymbol {\mathsf {P}}}}{\mathrm {d} \tau }}=\gamma m\left(c{\frac {\mathrm {d} \gamma }{\mathrm {d} t}},{\frac {\mathrm {d} \gamma }{\mathrm {d} t}}\mathbf {u} +\gamma \mathbf {a} \right)}$	${\displaystyle {\boldsymbol {\mathsf {F}}}\cdot {\boldsymbol {\mathsf {U}}}=0\,\!}$