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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.

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    $\begingroup$ Mass is one of the Casimirs of ISO(d-1,1) $\endgroup$
    – nwolijin
    May 5 at 14:44
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    $\begingroup$ The mass of a particle is related to the square of the four momentum $m^2 = p^\mu p_\mu$ which makes it clear that $m$ is Lorentz invariant. $\endgroup$
    – Prahar
    May 6 at 10:01
  • $\begingroup$ @Prahar Mitra : But the 4-momentum $\,p_{\mu}\,$ is defined in dependence to the a priori considered invariant rest mass $\,m_{\rm o}$, see equation \eqref{08a} in my answer. So I don't think that this is a proof, unless there exists a different definition of $\,p_{\mu}\,$ independent of the rest mass $\,m_{\rm o},$ which I miss. But then from this different definition must be proved first that $\,p_{\mu}\,$ is a Lorentz 4-vector in order for your proof to be valid. $\endgroup$
    – Frobenius
    May 6 at 13:02
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    $\begingroup$ @Frobenius - I suppose it depends on how one defines the 4-vector $p^\mu$. The definition that one is presented with when starting to learn special relativity does involve $m_0$ in which case my argument would be circular and since this is the context that is relevant to OP, I'll have to agree with you. $\endgroup$
    – Prahar
    May 6 at 13:06
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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}

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    $\begingroup$ Can the community start phasing out the term "relativistic mass"? Is it even in use these days?, I have the feeling there is always such questions again and again. The term in use in the literature are essentially mass (meaning always rest mass) and perhaps effective mass in the context of modified dispersion in mediums. $\endgroup$
    – ohneVal
    May 6 at 8:29
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    $\begingroup$ @ohneVal : ...The relativistic mass m is a source of confusion and it's not in use in modern Physics. $\endgroup$
    – Frobenius
    May 6 at 8:38
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    $\begingroup$ Yes that is why I upvoted this, I support your cause bro, haha, this is not addressed to you but the community. $\endgroup$
    – ohneVal
    May 6 at 9:07
  • $\begingroup$ @ohneVal : I disagree with you. I think that the community suggests in many places what I mention in my answer and previous comment. $\endgroup$
    – Frobenius
    May 6 at 9:14
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    $\begingroup$ @Jerry Schirmer : Thanks for editing and comment. $\endgroup$
    – Frobenius
    May 6 at 16:22
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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.

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