2
$\begingroup$

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 .

$\endgroup$
3
$\begingroup$

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}

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Work as derivative in time of energy? You mean power? $\endgroup$ – pasaba por aqui May 10 at 7:04
  • $\begingroup$ Thanks, I edit it. I meant the rate of work done - or the power delivered - by the force. $\endgroup$ – Cinaed Simson May 23 at 19:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.