Is mass still a scalar in special relativity?

Question

In Euclidean space is the space of classical Mechanics,

A scalar is the same for all observers that are to say remain invariant under the change of coordinate systems.

A Vector $\mathbf{V}$ is a collection of three numbers which transform as the components of a vector in $\mathcal{V}^3(R)$:

$$V_i\rightarrow V'_i=\sum_j R_{ij}V_j$$ where $R$ is the usual $3\times 3$ rotation matrix.

In Minkowski space,

$$\eta =\Lambda^T\eta \Lambda$$ where $\eta$ Minkowski metric. The above describes the transformation $\Lambda$ between the two frames of reference.

$$dx_\mu\rightarrow dx'_\mu=\Lambda^\sigma_\mu dx_\sigma$$

---

The four coordinate or four-momentum vector follows these transformations.

The mass of the particle transforms from one coordinate to another. But not there aren't any four-vectors associated with it Like with energy. So Is there any four-vector I can associate with this transformation? Or maybe we can write it as a scalar product but the scalar product remains invariant.

Answer 1

The mass of the particle transforms from one coordinate to another.

You make reference to the old notion of the relativistic mass \begin{equation} m\boldsymbol{=}\dfrac{m_{\rm o}}{\sqrt{1\boldsymbol{-}\dfrac{v^2}{c^2}}}\boldsymbol{=}\gamma_v\,m_{\rm o} \tag{01}\label{01} \end{equation} where $\,m_{\rm o}\,$ the rest mass. The relativistic mass $\,m\,$ is a source of confusion and it's not in use in contemporary Physics.

The rest mass $\,m_{\rm o}\,$ is a Lorentz invariant scalar.

What about the other quantity like force? Is there any four-vector for them? Is it, in general, I can find four-vector for euclidian vectors in Minkowski space?

From a rest mass $\,m_{\rm o}\,$ preserving 3-vector force $\,\mathbf f\,$ (like the Lorentz force of Electrodynamics) applied on a particle moving with 3-velocity $\,\mathbf u\,$ we produce a Lorentz 4-vector force $\,\mathbf F\,$ as follows \begin{equation} \mathbf F\boldsymbol{=}\left(\gamma_{\mathrm u}\mathbf{f}, \gamma_{\mathrm u}\dfrac{\mathbf{f}\boldsymbol{\cdot}\mathbf{u}}{c}\right)\,, \qquad \gamma_{\mathrm u}\boldsymbol{=}\left(1\boldsymbol{-}\dfrac{\mathrm u^2}{c^2}\right)^{\boldsymbol{-}\tfrac{1}{2}} \tag{02}\label{02} \end{equation} Under a Lorentz boost with velocity $\,\boldsymbol{\upsilon}\,$ the rest mass $\,m_{\rm o}\,$ preserving 3-vector force $\,\mathbf f\,$ is transformed as follows \begin{equation} \mathbf f' = \dfrac{\mathbf f\boldsymbol{+}\dfrac{\gamma_v^2}{c^2 \left(\gamma_v\boldsymbol{+}1\right)}\left(\mathbf f\boldsymbol{\cdot} \boldsymbol{\upsilon}\right)\boldsymbol{\upsilon}-\gamma_v \boldsymbol{\upsilon}\left(\dfrac{\mathbf f\boldsymbol{\cdot}\mathbf u}{c^{2}}\right)}{\gamma_v \left(1-\dfrac{\boldsymbol{\upsilon}\boldsymbol{\cdot}\mathbf{u}}{c^{2}}\right)} \tag{03}\label{03} \end{equation}

Some details in my answer here Are magnetic fields just modified relativistic electric fields? would be useful.\

Not all 3-vectors, say $\,\mathbf h\,$, have a Lorentz 4-vector partner. But if so then you must $''$discover$''$ two scalars $\,\rm a_{\rm h},b_{\rm h}\,$ so that to build the Lorentz 4-vector \begin{equation} \mathbf H\boldsymbol{=}\left(\rm a_{\rm h}\,\mathbf h, \rm b_{\rm h}\right) \tag{04}\label{04} \end{equation} For example in case of the rest mass $\,m_{\rm o}\,$ preserving 3-vector force $\,\mathbf f\,$ in equation \eqref{02} \begin{equation} \mathrm a_{\rm f}\boldsymbol{=}\gamma_{\mathrm u}\,,\qquad b_{\rm f}\boldsymbol{=}\gamma_{\mathrm u}\dfrac{\mathbf f\boldsymbol{\cdot}\mathbf u}{c} \tag{05}\label{05} \end{equation}

An easy and safe method to build new Lorentz 4-vectors is to take a Lorentz invariant scalar multiple of a known Lorentz 4-vector. By this method we build the Lorentz 4-vectors :

$\boldsymbol{\S\,1.} \texttt{ The Velocity 4-vector } \mathbf U$

The velocity 4-vector $\,\mathbf U\,$ is built by differentiation of the space-time position Lorentz 4-vector $\,\mathbf X\boldsymbol{=}\left(\mathbf x, c\,t\right)\,$ with respect to the Lorentz invariant scalar proper time $\,\tau$. Note that these properties follow the differentials $\,\mathrm d\mathbf X\,$ and $\,\mathrm d\tau\,$ respectively. \begin{equation} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\mathbf U\boldsymbol{=}\dfrac{\mathrm d\mathbf X}{\mathrm d\tau}\boldsymbol{=}\left(\dfrac{\mathrm d\mathbf x}{\mathrm d\tau}, c\dfrac{\mathrm d t}{\mathrm d\tau}\right)\boldsymbol{=}\left(\dfrac{\mathrm d\mathbf x}{\mathrm dt}\dfrac{\mathrm d t}{\mathrm d\tau}, c\dfrac{\mathrm d t}{\mathrm d\tau}\right)\boldsymbol{=}\left( \gamma_{\mathrm u}\,\mathbf u, \gamma_{\mathrm u}\, c\right)\,, \:\: \gamma_{\mathrm u}\boldsymbol{=}\left(1\boldsymbol{-}\dfrac{\mathrm u^2}{c^2}\right)^{\boldsymbol{-}\tfrac{1}{2}} \tag{06}\label{06} \end{equation} Under a Lorentz boost with velocity $\,\boldsymbol{\upsilon}\,$ the velocity 3-vector $\,\mathbf u\,$ is transformed as follows \begin{equation} \mathbf u' = \dfrac{\mathbf u\boldsymbol{+}\dfrac{\gamma_v^2}{c^2 \left(\gamma_v\boldsymbol{+}1\right)}\left(\mathbf u\boldsymbol{\cdot} \boldsymbol{\upsilon}\right)\boldsymbol{\upsilon}-\gamma_v \boldsymbol{\upsilon}}{\gamma_v \left(1-\dfrac{\boldsymbol{\upsilon}\boldsymbol{\cdot}\mathbf{u}}{c^{2}}\right)} \tag{07}\label{07} \end{equation}

$\boldsymbol{\S\,2.} \texttt{ The Linear Momentum 4-vector } \mathbf P$

The linear momentum 4-vector $\,\mathbf P\,$ is built as the Lorentz invariant rest mass $\,m_{\rm o}\,$ scalar multiple of the velocity Lorentz 4-vector $\,\mathbf U$. \begin{equation} \mathbf P\boldsymbol{=}m_{\rm o}\,\mathbf U\boldsymbol{=}\left( \gamma_{\mathrm u}\,m_{\rm o}\,\mathbf u, \gamma_{\mathrm u}\,m_{\rm o}\, c\right)\boldsymbol{=} \left( \mathbf p, \rm E/c\right) \tag{08a}\label{08a} \end{equation} where \begin{align} \mathbf p & \boldsymbol{=}\gamma_{\mathrm u}\,m_{\rm o}\,\mathbf u\boldsymbol{=}\texttt{the linear momentum 3-vector} \tag{08b}\label{08b}\\ \mathrm E & \boldsymbol{=}\gamma_{\mathrm u}\,m_{\rm o}\,c^2\boldsymbol{=}\texttt{ energy of the particle} \tag{08c}\label{08c} \end{align}

$\boldsymbol{\S\,3.} \texttt{ The Acceleration 4-vector } \mathbf A$

The acceleration 4-vector $\,\mathbf A\,$ is built by differentiation of velocity 4-vector $\,\mathbf U\,$ with respect to the Lorentz invariant scalar proper time $\,\tau$. \begin{equation} \mathbf A\boldsymbol{=}\dfrac{\mathrm d\mathbf U}{\mathrm d\tau} \tag{09}\label{09} \end{equation}

$\boldsymbol{\S\,4.} \texttt{ The Force 4-vector } \mathbf F$

For a rest mass $\,m_{\rm o}\,$ preserving 3-vector force $\,\mathbf f\,$ the force 4-vector $\,\mathbf F\,$ is built by differentiation of Linear Momentum 4-vector $\,\mathbf P\,$ with respect to the Lorentz invariant scalar proper time $\,\tau$. \begin{equation} \mathbf F\boldsymbol{=}\dfrac{\mathrm d\mathbf P}{\mathrm d\tau}\boldsymbol{=}\dfrac{\mathrm d\left(m_{\rm o}\,\mathbf U\right)}{\mathrm d\tau}\boldsymbol{=}m_{\rm o}\,\dfrac{\mathrm d\mathbf U}{\mathrm d\tau}\boldsymbol{=}m_{\rm o}\,\mathbf A \tag{10}\label{10} \end{equation}

Answer 2

Mass, from the perspective of $\vec{F}=\tilde{m}\vec{a}$, is a tensor quantity. If an object was observed to be moving in the $\vec{x}$ direction, the mass tensor would be $$\tilde{m}=\gamma\ m_0\left[\begin{matrix}\gamma^2&0&0\\0&1&0\\0&0&1\\\end{matrix}\right]$$ where $m_0$ is the Lorentz invariant rest mass(which is a scalar). Mass in this tensor form isn't used nowadays as it tends to complicate things unnecessarily. As such, rest mass is used instead. Neither definition is wrong, it's just that Lorentz invariant scalars are easier to work with.