# LOG#014. Vectors in spacetime.

We are going to develop the mathematical framework of vectors in (Minkowski) spacetime. Vectors are familiar oriented lines in 3d calculus courses. However, mathematically are a more general abstract entity: you can add, substract and multiply vectors by some number. We will focus here on the 4D world of usual SR, but the discussion can be generalized to any other D-dimensional spacetime. I have included a picture of a vector OQ above these lines. It is a nice object, isn’t it?

First of all, remark that the conventional kinematical variables to describe the motion in classical mechanics are the displacement vector (or the position vector), and its first and second “derivatives” with respect to time. These magnitudes are called velocity (its magnitude is the speed or modulus of the velocity) and acceleration (in 3d):

$\mathbf{r}(t)=(x,y,z)=\begin{pmatrix}x \\ y \\ z\end{pmatrix}\equiv\mbox{POSITION VECTOR}$

$\mathbf{v}=\dfrac{d\mathbf{r}}{dt}=\left(\dfrac{dx}{dt},\dfrac{dy}{dt},\dfrac{dz}{dt}\right)=\begin{pmatrix}\dfrac{dx}{dt}\\ \; \\ \dfrac{dy}{dt}\\ \; \\ \dfrac{dz}{dt}\end{pmatrix}\equiv\mbox{VELOCITY VECTOR}$

$\mathbf{a}=\left(\dfrac{dv_x}{dt},\dfrac{dv_y}{dt},\dfrac{dv_z}{dt}\right)=\begin{pmatrix}\dfrac{dv_x}{dt}\\ \; \\ \dfrac{dv_y}{dt}\\ \; \\\dfrac{dv_z}{dt}\end{pmatrix}\equiv\mbox{ACCELERATION VECTOR}$

In Classical (nonrelativistic) Mechanics, you learn that velocity and acceleration are enough (in general) to fully describe the dynamics and motion of objects. Velocity can be thought as the rate of change of the displacement vector in a tiny amount of time. Acceleration is similarly the rate of change of the velocity vector in some small time interval. You can studied motion con those three vectors, and the notion of “force” is introduced since Newton times as the object that can change the state of “rest” or “uniform” motion (with constant velocity) of any material body. Indeed, Newton’s fundamental law of Dynamics ( it is also called Newton second law) says that the total force (also a vector additive quantity) IS the rate of change of the linear “momentum” vector with respect to time,

$\sum \mathbf{F}=\dfrac{d\mathbf{p}}{dt}$

where the momentum is simply defined as the vector

$\mathbf{p}=m\mathbf{v}\equiv \mbox{MOMENTUM}$

and m is the so called “inertial mass” a measure of the inertia of bodies saying how hard  to change its velocity is. Larger the mass is, larger the force necessary to change its velocity is. The total foce is denoted as a sum

$\sum \mathbf{F}=F_1+F_2+\ldots\equiv \mathbf{F}_t \longleftrightarrow \mbox{TOTAL FORCE}$

May the Force be with you! In the case that the mass is constant ( the most general case in Mechanics), Newton second law can also be written as:

$\sum \mathbf{F}=\dfrac{d(m\mathbf{v})}{dt}=m\dfrac{d\mathbf{v}}{dt}$

i.e.,

$\boxed{\mathbf{F}_t=m\mathbf{a}}$

The squared length of vectors is defined as the scalar (dot) product:

$\mathbf{V}\cdot{\mathbf{V}}=V^2=\vert \mathbf{V}\vert^2=V^2_x+V^2_y+V^2_z\equiv \mbox{squared length of any vector V}$

You can also define some interesting differential operators (some king of “gadgets” making “other” stuff form some stuff when they act on vectors somehow):

$\nabla=e^i \partial_i =(\partial_x,\partial_y,\partial_z)=\left(\dfrac{\partial}{\partial x},\dfrac{\partial}{\partial y},\dfrac{\partial}{\partial z}\right)\equiv \mbox{NABLA OPERATOR}$

$\nabla^2=\nabla \cdot \nabla=\partial^i \partial_i=\partial^2_x +\partial^2_y +\partial^2_z=\dfrac{\partial^2}{\partial x^2}+\dfrac{\partial^2}{\partial y^2}+\dfrac{\partial^2}{\partial z^2}\equiv \mbox{LAPLACIAN OPERATOR}$

OK, that is a good crash course on elementary Dynamics and vectors. Back to the future! Back to relativity! You can extend this formalism into the spacetime with care, since we do know time is not an invariant any more in relativistic Physics. Fortunately, we saw that there is an invariant quantity in special relativity: the proper time $\tau$! We remember the important result:

$\dfrac{dt}{d\tau}=\gamma$

So, we proceed to the spacetime generalization of the previous stuff. Firstly, events in spacetime are given by:

$\boxed{\mathbb{X}=X^\mu e_\mu=(ct,x,y,z)=\begin{pmatrix}ct \\ \; \\ x \\ \; \\ y \\ \; \\ z\end{pmatrix}\equiv \mbox{SPACETIME EVENT}}$

The “spacetime length” is got using the following dot product

$\boxed{S^2=-(ct)^2+x^2+y^2+z^2=\mathbb{X}\cdot \mathbb{X}=X^\mu X_\mu \equiv \mbox{SPACETIME INVARIANT}}$

so we have the so called contravariant components

$X^\mu=(ct,\mathbf{r})$

and the covariant components

$X_\mu=(-ct,\mathbf{r})$

The spacetime interval bewtween two arbitrary events, e.g., A and B, will be:

$\Delta \mathbb{X}_{AB}= \mathbb{X}_B - \mathbb{X}_A= c(t_B -t_A)+ (x_B-x_A)+ (y_B-y_A)+(z_B-z_A)$

The  squared spacetime “length” separation (an invariant) between those two events is:

$\Delta \mathbb{X}^2_{AB}=\Delta \mathbb{X}\cdot \Delta \mathbb{X}=\Delta X^\mu \Delta X_\mu$

or equivalenty

$\Delta \mathbb{X}^2=-c^2 (t_B -t_A)^2+ (x_B-x_A)^2+ (y_B-y_A)^2+ (z_B-z_A)^2$

You can also write the last two equations in a differential ( or infinitesimal) way:

$d \mathbb{X}^2=d \mathbb{X}\cdot d \mathbb{X}=dX^\mu d X_\mu$

$d \mathbb{X}^2=-c^2 d t^2+d x^2+d y^2+d z^2$

The spacetime velocity will be then:

$\boxed{\mathbb{U}=U^\mu e_\mu=\dfrac{d\mathbb{X}}{d\tau} \equiv \mbox{SPACETIME VELOCITY}}$

We now use the trick based on the chain rule of differential calculus:

$\dfrac{d\mathbb{X}}{d\tau}=\dfrac{d\mathbb{X}}{dt}\dfrac{dt}{d\tau}$

Therefore,

$\mathbb{U}=\left( c\dfrac{dt}{d\tau},\dfrac{d\mathbf{r}}{dt}\dfrac{dt}{d\tau}\right)= (\gamma c,\gamma\mathbf{v})$

So, we get

$\mathbb{U}=U^\mu e_\mu \leftrightarrow U^0=\gamma c \;\;\;\; U^i=\gamma v^i$

i.e.

$\boxed{\mathbb{U}=\dfrac{d\mathbb{X}}{d\tau}=(c\gamma,\gamma \mathbf{v})}$

The spacetime velocity has a nice property:

$\mathbb{U}^2=\mathbb{U}\cdot\mathbb{U}=-c^2\gamma^2+\gamma^2 v^2=-c^2\gamma^2(1-\frac{v^2}{c^2})=-c^2$

i.e.

$\boxed{\mathbb{U}^2=U^\mu U_\mu=-c^2}$

The spacetime momentum that we will call momenergy ( the reasons will be seen later) is defined as:

$\boxed{\mathbb{P}=P^\mu e_\mu=m\mathbb{U}=(mc\gamma,\gamma m\mathbf{v})\equiv\mbox{MOMENERGY}}$

The momenergy components are given by

$\boxed{P^0=mc\gamma \;\;\;\; P^i=m\gamma v^i}$

In the same way that time is the “fourth” coordinate in the spacetime, we can identify the zeroth component of the spacetime momentum, $P^0$, as the total relativistic energy in the following way:

$P^0\equiv \dfrac{E}{c}$

so

$P^0=mc\gamma=\dfrac{E}{c}\longleftrightarrow \boxed{E=P^0 c=m\gamma c^2}$

or

$\boxed{E=Mc^2}$

where the relativistic mass is defined as

$\boxed{M\equiv m\gamma=\dfrac{m}{\sqrt{1-\dfrac{v^2}{c^2}}}}$

Define the relativistic (spacelike) momentum as

$\mathbf{P}=\gamma \mathbf{p}$

Now we can understand why the spacetime momentum was called momenergy, since it can be rewritten as:

$\boxed{\mathbb{P}=\left(Mc, \mathbf{P}\right)=\left(\dfrac{E}{c},\mathbf{P}\right)=\left(\gamma mc, \gamma m\mathbf{v}\right)}$

We can make an invariant from the squared momenergy:

$\mathbb{P}^2=\mathbb{P}\cdot \mathbb{P}=P^\mu P_\mu=-M^2+\mathbf{P}^2=-m^2c^2\gamma^2\left( 1-\dfrac{v^2}{c^2}\right)=-m^2c^2$

i.e.

$\boxed{\mathbb{P}^2=\mathbb{P}\cdot \mathbb{P}=P^\mu P_\mu=-m^2c^2 \equiv (\mbox{MOMENERGY})^2 \; \mbox{INVARIANT}}$

We can use this result to write the most general relation between energy, momentum and mass in SR:

$\mathbb{P}^2=\mathbb{P}\cdot \mathbb{P}=P^\mu P_\mu=\mathbf{P}^2-\dfrac{E^2}{c^2}=-m^2c^2$

Then, we have

$\dfrac{E^2}{c^2}=\mathbf{P}^2+m^2c^2$

or equivalenty

$\boxed{E^2=\mathbf{P}^2c^2+m^2c^4 \longleftrightarrow E^2=(\mathbf{P}c)^2+(mc^2)^2}$

Setting units with c=1, we get

$\boxed{(\mbox{ENERGY})^2=(\mbox{MOMENTUM)}^2+(\mbox{MASS})^2}$

Wonderful! Momenergy allows us to unify the concepts of mass, momentum and energy in a 4D spacetime world!

What else? We can also calculate the spacetime acceleration, but it is a bit more complicated. Indeed, we need an auxiliary result from calculus:

$\dfrac{d\gamma}{dt}=\dfrac{d}{dt}\left(1-\dfrac{v^2}{c^2}\right)^{-1/2}=-\dfrac{1}{2}\left(1-\dfrac{v^2}{c^2}\right)^{-3/2}(-2)\dfrac{\mathbf{v}\cdot\mathbf{a}}{c^2}=\dfrac{\gamma^3}{c^2}\mathbf{v}\cdot \mathbf{a}$

that is

$\boxed{\dfrac{d\gamma}{dt}=\dfrac{\gamma^3}{c^2}\mathbf{v}\cdot \mathbf{a}}$

Now, we proceed to the calculations:

$\boxed{\mathbb{A}=A^\mu e_\mu=\dfrac{d\mathbb{U}}{d\tau}}$

and with the same trick we saw before

$\dfrac{d\mathbb{U}}{d\tau}=\dfrac{d\mathbb{U}}{dt}\dfrac{dt}{d\tau}=\gamma \dfrac{d\mathbb{U}}{dt}$

and

$\mathbb{A}=\gamma \left(\dfrac{d\gamma}{dt} (c, \mathbf{v}) +\gamma \dfrac{d}{dt}(c,\mathbf{v})\right)$

so

$\mathbb{A}=\gamma \left( \dfrac{\gamma^3 \mathbf{v}\cdot \mathbf{a}}{c^2}(c, \mathbf{v}) +\gamma (0,\mathbf{a})\right)$

i.e.

$\mathbb{A}=\left( \dfrac{\gamma^4 \mathbf{v}\cdot \mathbf{a}}{c^2}(c, \mathbf{v}) +\gamma^2 (0,\mathbf{a})\right)$

Therefore:

$\boxed{\mathbb{A}=(A^0,\mathbf{A})\equiv \begin{pmatrix}PROPER\\ ACCELERATION\end{pmatrix}\longleftrightarrow A^0=\dfrac{\gamma^4 \mathbf{v}\cdot \mathbf{a}}{c}\;\; A^i=\dfrac{\gamma^4 \mathbf{v}\cdot \mathbf{a}}{c^2}v^i+\gamma^2 a^i }$

We can also calculate the proper acceleration invariant, i.e., the squared of the spacetime acceleration:

$\mathbb{A}^2=\mathbb{A}\cdot \mathbb{A}=\dfrac{\gamma^8}{c^4}\left[ v^2(\mathbf{v}\cdot \mathbf{a})^2-c^2(\mathbf{v}\cdot \mathbf{a})^2\right]+2\dfrac{\gamma^6}{c^2}(\mathbf{v}\cdot \mathbf{a})^2+\gamma^4(\mathbf{a}\cdot \mathbf{a})$

i.e.,

$\boxed{ \mathbb{A}^2=\mathbb{A}\cdot \mathbb{A}=\dfrac{\gamma^6}{c^2}(\mathbf{v}\cdot \mathbf{a})^2+\gamma^4 a^2=\alpha^2\equiv \begin{pmatrix}\mbox{SQUARED PROPER} \\ \mbox{ACCELERATION}\end{pmatrix}}$

We can calculate some important particular cases of the last equation. If the acceleration is linear, $\mathbf{v}\parallel \mathbf{a}$, the last equation provides:

$\alpha^2=\gamma^6\beta^2a^2+\gamma^4 a^2=\gamma^4 a^2(\gamma^2\beta^2+1)=\gamma^6 a^2$

i.e. linear accelerations have proper acceleration $\alpha =\gamma^3 a$. In the case of circular motions, where $\mathbf{v}\perp\mathbf{a}$, we get proper acceleration $\alpha=\gamma^2 a$. Thus, the centripetal proper acceleration, e.g. in a storage ring, is $\gamma^2\dfrac{v^2}{r}\approx \gamma^2 \dfrac{c^2}{r}$.

Finally, we  calculate the so-called Power-Force vector, sometimes it is also referred as Minkowski force:

$\boxed{\mathbb{F}=\mathcal{F}^\mu e_\mu=(\mathcal{F}^0,\mathcal{F}^i)=\dfrac{d\mathbb{P}}{d\tau}=\gamma \left( \dfrac{1}{c}\dfrac{dE}{dt},\dfrac{d\mathbf{P}}{dt}\right)=\gamma \left( \dfrac{1}{c}\dfrac{dE}{dt},\mathbf{F}\right)\equiv \left(\mbox{POWER-FORCE}\right)}$

where

$\mathbf{F}=\dfrac{d\mathbf{P}}{dt}=\dfrac{d(M\mathbf{v})}{dt}=\dfrac{d(\gamma m\mathbf{v})}{dt}=\gamma m\mathbf{a}+\gamma^3\dfrac{m}{c^2}(\mathbf{v}\cdot \mathbf{a})\mathbf{v}$

It is called Power-Force due to the dimensions of physical magnitudes there

$\boxed{\mathcal{F}^0= \dfrac{ \gamma}{c}\dfrac{dE}{dt}}$ and $\boxed{\mathcal{F}^i=\gamma F^i}$

Recall the previously studied spacetime and momenergy vectors. Here, it is often defined the transversal mass as $M_T=\gamma m=M$ (mass-like coefficient for  force perpendicular to the motion) and also the so-called longitudinal mass $M_L=\gamma^3 m=\gamma^2 M$ (mass-like term for the force parallel to the direction of motion).

Moreover, we could obtain the acceleration from the force in the following way:

$\mathbf{F}\cdot \mathbf{v}=\gamma m \left[\mathbf{v}\cdot \mathbf{a}\right]\left(1+\dfrac{\gamma^2 v^2}{c^2}\right)=\gamma m \left[\mathbf{v}\cdot \mathbf{a}\right]\left(1+\dfrac{\beta^2}{1-\beta^2}\right)=\gamma m \left[\mathbf{v}\cdot \mathbf{a}\right]\left(\dfrac{1}{1-\beta^2}\right)$

i.e.

$\mathbf{F}\cdot \mathbf{v}=\gamma^3 m\left[\mathbf{v}\cdot \mathbf{a}\right]$

and thus

$\mathbf{F}=\gamma m \mathbf{a}+\dfrac{1}{c^2}\left(\mathbf{F}\cdot\mathbf{v}\right)\mathbf{v}$

and therefore

$\boxed{\mathbf{a}=\dfrac{1}{\gamma m}\left(\mathbf{F}-\dfrac{\mathbf{F}\cdot\mathbf{v}}{c^2}\mathbf{v}\right)=\dfrac{1}{M}\left(\mathbf{F}-\dfrac{\mathbf{F}\cdot\mathbf{v}}{c^2}\mathbf{v}\right)}$

We observe that Newton’s secon law is not generally valid in SR, since $\mathbf{F}\neq m\mathbf{a}$. However, we can “generalize” the Newton’s second law to be covariantly valid in SR in this neat form:

$\boxed{\mathbb{F}=m\mathbb{A}}$

The non-trivial expression for the spacetime acceleration and the mathematical consistency of the theory do the rest of the work for us.

Finally, we can make some nice invariant operator as well. We define the spacetime generalization of nabla as follows:

$\boxed{\square =e^\mu \partial_\mu =\left(\dfrac{1}{c}\partial_t, \partial_x,\partial_y,\partial_z\right)=\left(\dfrac{1}{c}\dfrac{\partial}{\partial t},\dfrac{\partial}{\partial x},\dfrac{\partial}{\partial y},\dfrac{\partial}{\partial z}\right)=\left(\dfrac{1}{c}\dfrac{\partial}{\partial t},\nabla \right)}$

The spacetime analogue of the laplacian operator is sometimes called D’Alembertian (or “wave” operator):

$\boxed{\square^2=\square \cdot \square = \partial^\mu \partial_\mu= -\dfrac{\partial^2_t}{c^2}+\partial^2_x +\partial^2_y +\partial^2_z=-\dfrac{1}{c^2}\dfrac{\partial^2}{\partial t^2}+\dfrac{\partial^2}{\partial x^2}+\dfrac{\partial^2}{\partial y^2}+\dfrac{\partial^2}{\partial z^2}}$

or equivalently, in a short hand notation

$\boxed{\square^2=-\dfrac{1}{c^2}\dfrac{\partial^2}{\partial t^2}+\nabla^2}$

Remark: Some books and authors define the D’Alembertian as $\square =\square \cdot \square$. While that notation is indeed valid/possible, and despite it can safe time sometimes, it is obviously a tricky and non-intuitive, somewhat unlucky, notation. We warn you about it, and we highly recommend transparent and powerful notations like the one introduced here, keeping in mind the physical concepts behind all this framework.

Remark(II): The mathematical framework using vectors in spacetime can be easily generalized to spacetime with an arbitrary number of dimensions, e.g., D=d+1, where d is the number of space-like dimensions or, even we could consider D=d+q=s+t, where we have an arbitrary number d (or s) of space-like coordinates and an arbitrary number q (or t) of time-like coordinates. However, it is not so easy to handle with these generalizations by different reasons (both mathematical and physical). A useful quantity in that case is the spacetime “signature”, defined as the the difference between the number of space-like and time-like dimensions, that is,

$\mbox{Signature of spacetime}=(s-t)=(d-q)$

The easiest ( the most studied so far) case with a higher dimensional spacetime is that of signature $2$, the SR case, or its most natural generalization with signature $(d-1)=(D-2)$.