# Components of four-force

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 .

• – Frobenius May 10 at 5:24
• The relation $\: \mathbf{f}\boldsymbol{=}m_0 \mathbf{a}\:$ is not valid. – Frobenius May 10 at 5:42

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

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

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?

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