Components of four-force

Question

Temporal component of the four-accelaration is:

$$\mathbf{A}_t = \gamma_u^4\frac{\mathbf{a}\cdot\mathbf{u}}{c}$$

that, multipliying by the rest mass, should give a value of the temporal component of the four-force of:

$$\mathbf{F}_t = m_0 \gamma_u^4\frac{\mathbf{a}\cdot\mathbf{u}}{c} = \gamma_u^4\frac{\mathbf{f}\cdot\mathbf{u}}{c}$$

where I replaced $ m_0 \mathbf{a} = \mathbf{f} $.

I reach same value if I take derivate respect proper time of the temporal component of the four-momentum $m_0 \gamma_u c$:

$ { d \gamma_u \over dt } = {d \over dt} \frac{1}{\sqrt{ 1 - \frac{\mathbf{v} \cdot \mathbf{v}}{c^2} }} = \frac{1}{\left( 1 - \frac{\mathbf{v} \cdot \mathbf{v}}{c^2} \right)^{3/2}} \, \, \frac{\mathbf{v}}{c^2} \cdot \, \frac{d \mathbf{v}}{dt} \, = \, \frac{\mathbf{a \cdot u}}{c^2} \frac{1}{\left(1-\frac{u^2}{c^2}\right)^{3/2}} \, = \, \frac{\mathbf{a \cdot u}}{c^2} \, \gamma_u^3 $

$ { d \gamma_u \over d\tau } = { d \gamma_u \over dt }{ dt \over d\tau } = \frac{\mathbf{a \cdot u}}{c^2} \, \gamma_u^3 \, \gamma_u $

$ \mathbf{F}_t = { d \mathbf{P}_t \over d\tau } = m_0 c { d \gamma_u \over d\tau } = m_0 \frac{\mathbf{a \cdot u}}{c} \, \gamma_u^4 =\frac{\mathbf{f \cdot u}}{c} \, \gamma_u^4 $

However, wikipedia gives as correct value:

$ \mathbf{F}_t = \frac{\mathbf{f \cdot u}}{c} \, \gamma_u $

I'm making an error of $\gamma_u^3$ and I can not find where it is .

Answer 1

The relation $\:\mathbf f \boldsymbol{=} m_0 \mathbf a\:$ is not valid. Instead of it use this \begin{equation} \mathbf f \boldsymbol{=}\gamma_u m_0 \mathbf a\boldsymbol{+}\gamma^3_u m_0 \dfrac{\left(\mathbf a \boldsymbol{\cdot}\mathbf u\right)}{c^2}\mathbf u \quad \boldsymbol{\Longrightarrow} \quad \boxed{\:\:\mathbf f\boldsymbol{\cdot}\mathbf u \boldsymbol{=}\gamma^3_u m_0 \left(\mathbf a \boldsymbol{\cdot}\mathbf u\right)\vphantom{\dfrac{a}{b}}\:\:} \tag{A-01}\label{A-01} \end{equation} To reach that combine \begin{equation} \mathbf f \boldsymbol{=}\dfrac{\mathrm d\mathbf p}{\mathrm d t} \boldsymbol{=}\dfrac{\mathrm d\left(\gamma_u m_0 \mathbf u\right)}{\mathrm d t} \boldsymbol{=}\cdots \tag{A-02}\label{A-02} \end{equation} with yours \begin{equation} \dfrac{\mathrm d\gamma_u}{\mathrm d t} \boldsymbol{=}\cdots \tag{A-03}\label{A-03} \end{equation}

Answer 2

In your definiton $\vec{f} = m\vec{a}$ but the wikipedia defines $\vec{f} = \frac{d}{dt}(\gamma m \vec{u})$

Answer 3

The first calculation starting by writing the $4$ vector $A$ in terms of $\mathbf{u}$ and $\mathbf{a}$ was correct.

To obtain the $4$ force, presumably one would multiply $A$ by $m$.

The second calculation starting from the definition of force $dP/d\tau$ had an error.

It appears the error was to assume that both calculations would produce identical results - while ignoring the different starting points.

Here is corrected version for the second calculation:

\begin{align*} F= &\;\frac{dP}{d\tau}\\ = & \;(\frac{1}{c}\frac{dE}{d\tau},\frac{d\mathbf{p}}{d\tau})\\ =& \;\gamma\frac{d}{dt}(E/c,\mathbf{p})\\ =& \;\gamma(W/{c},\mathbf{f}) \end{align*}

where $\frac{d}{d\tau}=\gamma \frac{d}{dt}$,$\;\mathbf{f}=\frac{d\mathbf{p}}{dt}$, and $\frac{dE}{dt}=W$ is the rate of work done by the force $\mathbf{f}\cdot \mathbf{u}.$

Note the Wikepedia links you provided have both results as being correct.

But what is the work done by the $F$, i.e., $F\cdot U$ - which is a Lorentz invariant? Does the rest mass remain constant in either calculation?