# LOG#015. Time of flight.

Suppose we get a beam made of massive particles. The rest mass (the names invariant mass or proper mass are also popular) is $m$. The particle travels a distance L in its inertial frame. The particle has an energy E in that frame. Therefore, the so-called “time of flight” from two points at x=0 and x=L will be:

$T=\dfrac{L}{v}=\dfrac{m \gamma L}{m \gamma v}=\dfrac{ m \gamma L}{p}=\dfrac{m \gamma c^2 L}{pc^2}$

where we use the relativistic definition of momentum $p=m \gamma v$. Now, knowing that the relativistic energy is $E=m \gamma c^2$, using the time that a massless light beam uses to travel the proper distance L, easily calculated to be $\tau = L/c$, we will get:

$T=\dfrac{E L}{pc^2}=\dfrac{E \tau}{pc}=\dfrac{\tau}{pc/E}$

But the dispersion relation in Special Relativity is a function $E=E(p,m)$, and the squared of the relativistic momentum or momenergy yields the well known relationship

$p^2c^2 = E^2-m^2c^4$

i.e.,

$\dfrac{p^2c^2}{E^2}=1-\dfrac{m^2c^4}{E^2}$

and then,

$\dfrac{pc}{E}=\sqrt{1-\left( \dfrac{mc^2}{E}\right)^2}$

Thus, the time of flight is finally written as follows:

$\boxed{T=\dfrac{\tau}{\sqrt{1-\left( \dfrac{mc^2}{E}\right)^2}}}$

This equation is very important in practical applications. Specially in Astrophysics and baseline beam experiments, like those involving the neutrinos! Indeed, we usually calculate the difference between the photon (or any other massless) time of arrival and that of massive particles, e.g. the neutrinos. Several neutrino experiments can measure this difference using a well designed experimental set-up. The difference between those times of flight (the neutrino time of flight minus the photon time of flight) is:

$T-\tau=\dfrac{\tau}{\sqrt{1-\left( \dfrac{mc^2}{E}\right)^2}}-\tau$

or equivalently

$\Delta t=T-\tau=\tau \left( \dfrac{1}{\sqrt{1-\left( \dfrac{mc^2}{E}\right)^2}}-1\right)$

or as well

$\displaystyle{ \boxed{Q=\dfrac{\Delta t}{\tau}= \dfrac{1}{\sqrt{1-\left( \frac{mc^2}{E}\right)^2}}-1}}$

This last expression can also be expressed in terms of the speed of light and neutrinos, since $c=L/\tau$ and $v_\nu=L/T$, so

$\dfrac{ T/L-\tau/L}{\tau/L}=\dfrac{1/v_\nu-1/c}{1/c}=\dfrac{c-v_\nu}{v_\nu}$

Therefore,

$\boxed{Q=\dfrac{\Delta t}{\tau}=\dfrac{c-v_\nu}{v_\nu}}$

In the case of know light left-handed neutrinos, the rest mass is tiny (likely sub-eV and next to the meV scale), and then we can make a Taylor expansion for Q ( with $x=(mc/E)<<1$):

$Q\approx \dfrac{x^2}{2}+\mathcal{O}(x^4)$

Then, we would expect, accordingly to special relativity, of course, that

$\dfrac{c-v_\nu}{v_\nu}=\dfrac{T-\tau}{\tau}= \dfrac{1}{2}\left(\dfrac{m_\nu c^2}{E}\right)^2$

We can guess how large it is plugging “typical” values for the neutrino mass and energy. For instance, taking $m_\nu \sim meV$ and $E\sim GeV$, the Q value is about

$Q\sim \left( meV/GeV\right)^2= 10^{-24}$ !

Nowadays, we have no clock with this precision, so the neutrino mass measurement using this approach is impossible with current technology. However, it is clear that if we could make clocks with that precision, we would measure the neutrino mass with this “time of flight” procedure. It is a challenge. We can not do that in these times (circa 2012), and thus we don’t measure any time delay in baseline experiments. Then, neutrinos move with $v_\nu \approx c$ and since there is no observed delay (beyond the OPERA result, already corrected), neutrinos are, thus, ultra-relativistic particles, and for them $E=pc$ with great accuracy.

Moreover, from supernova SN1987A we do observe a delay due to the fact that stellar models predict that neutrinos are emitted before the star explodes! During the collapse, and without going deeper in technical details on the nuclear physics of dense matter, neutrinos scape almost instantaneously (in about some seconds) from the collapsing star but the photons find theirselves blocked during some hours due to the pressure of the dense matter. After some time passes, the block-out ends and photons are released. Historically, this picture was not the physical portrait of collapsing stars around the eighties of the past century, and the observations of supernovae entered, crucially, into scene in order to solve the puzzle of why neutrinos arrived before the photons (indeed, superluminal neutrinos were already tried and postulated to solve that paradox). However, SN1987A was helpful and useful due to the insight it offered about the nontrivial subject of collapsing stars and dense nuclear matter. Before SN1987A there were some assumptions about the nuclear processes inside collapsing stars that proved to be wrong after SN1987A. The puzzle was solved soon with the aid of SN1987A data taken by experimental set-ups like (Super)Kamiokande, and the interaction and collaboration between nuclear physicists, astrophysicists and high energy physicists. Generally speaking, that was the path towards the currently accepted models of nuclear physics in the inner shells of exploding stars.

In summary, photons find a several hours long block-out due to the dense collapsing material from some inner shells of the star, while neutrinos come out, and when the block-out ends, photons are shot into the sky (Bang! A supernova is born). The delayed time is consistent with the above expressions. The neutrinos arrive before the photons not because they are faster but due to the fact they are emitted before and they don’t find any resistence unlike photons; light is blocked by some time due the nuclear conditions of the matter inside the imploding star, but it ends after some hours. Furthermore, since neutrinos travel almost at speed of light, light generally is not able to reach them, even at astrophysical distances!

Remark (I): Generally, experimental physicists use a different formula in order to measure the time difference between the arrivals of neutrinos and photons. We will derive it easily:

$\boxed{\dfrac{c - v_\nu }{v_\nu}=\dfrac{\Delta t}{\tau}}$

This equation implies that

$\boxed{\dfrac{c - v_\nu }{c}=\dfrac{v_\nu \Delta t}{c\tau}=\dfrac{v_\nu T-v_\nu \tau}{c\tau}=\dfrac{L-L'}{L}}$

where L is the distance from CERN to Gran Sasso and L’ is the path length traveled by the neutrinos in the proper time ($\tau$).  Therefore, we write

$\dfrac{c- v_\nu }{c}=\dfrac{(T-\tau)}{\tau(c/v_\nu)}$

And now, since

$T=\dfrac{L}{v_\nu},\tau =\dfrac{L}{c}\longrightarrow \dfrac{T}{\tau}=\dfrac{c}{v_\nu}$

Thus,

$\dfrac{c - v_\nu}{c}=\dfrac{(T-\tau)}{\tau (T/\tau)}$

or

$\boxed{Q'=\dfrac{c - v_\nu }{c}=\dfrac{T-\tau}{T}=\dfrac{\Delta t}{T}}$

Obviously, it is an analogue of the previous formula Q. Q’ provides a measure of how much the propagation velocity of neutrinos differs from the speed of light. Experiments are consistent with a null result (the difference in velocity is measured to be zero). In conclusion, the neutrinos move almost at speed of light (with current experiments not being able to distinguish the tiny difference between both velocities, that of neutrino and the speed of light).

Remark (II): In the slides of OPERA-like experiments, you find the equation

$\boxed{\dfrac{v_\nu -c}{c}=\dfrac{\delta t}{TOF_c -\delta t}}$

and there, it is usually defined

$\delta t= TOF_c -TOF_\nu \longleftrightarrow TOF_\nu = TOF_c -\delta t$

There is a complete mathematical equivalence between my notation and those of experimental HEP neutrino talks handling the time of flight, if you note that:

$\delta t=-\Delta t \;\;\;\;\;\;\;\;\;\;\;\;\;TOF_c = \tau \;\;\;\;\;\;\;\;\;\;\;\;\; TOF_\nu=T$

Then, all the mathematical equations match.

Remark (III): An alternative way to calculate the relationship between neutrino speed and photon speed (in vacuum) is given by the next arguments. The total relativistic energy for the neutrino reads

$E_\nu=\dfrac{m_\nu c^2}{\sqrt{1-\dfrac{v_\nu^2}{c^2}}}$

Then, algebra allows us to write

$E^2=\dfrac{(m_\nu c^2)^2}{1-\dfrac{v_\nu^2}{c^2}}$

$E^2-(m_\nu c^2)^2=\dfrac{v_\nu^2 E^2}{c^2}$

$c^2E^2-c^2(m_\nu c^2)^2=v_\nu^2 E^2$

$v^2_\nu=c^2\left(\dfrac{c^2E^2-(m_\nu c^2)^2}{E^2}\right)=c^2\left( 1-\left(\dfrac{m_\nu c^2}{E}\right)^2\right)$

$v_\nu=\pm c\sqrt{1-\left(\dfrac{m_\nu c^2}{E}\right)^2}$

That is,

$\boxed{\dfrac{\vert v_\nu -c \vert}{c}=\bigg| \sqrt{1-\left(\dfrac{m_\nu c^2}{E}\right)^2}-1\bigg|}$

In the case of neutrinos, ultra-relativistic particles, and with high energy E compared with their rest mass, we get

$\dfrac{\vert v_\nu -c \vert}{c}\approx \dfrac{1}{2}\left(\dfrac{mc^2}{E}\right)^2$

as we obtained before.

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PS: This article was semifinalist in the 3quarksdaily 2012 online contest. Here the logo picture I got from the official semifinalist list http://www.3quarksdaily.com/3quarksdaily/2012/06/3qd-science-prize-semifinalists-2012.html

# 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 $(d-1)=(D-2)$.