Is the 'time'-component of 4-force always zero?

Question

The 4-velocity of a particle is defined as \begin{equation}\tag{1} u_\sigma = \frac{d x_\sigma}{d\tau} = \left(\frac{dx_1}{d\tau}, \frac{dx_2}{d\tau},\frac{dx_3}{d\tau},c\frac{dt}{d\tau}\right), \end{equation} where $d\tau$ is the proper time interval. Since \begin{equation}\tag{2} dt = \frac{d\tau}{\sqrt{1 - \beta^2}}, \end{equation} we have \begin{equation}\tag{3} u_\sigma = \frac{d x_\sigma}{d\tau} = \left(\frac{dx_1}{d\tau}, \frac{dx_2}{d\tau},\frac{dx_3}{d\tau},\frac{c}{\sqrt{1 - \beta^2}}\right). \end{equation} The 4-momentum is defined as $p_\sigma = mu_\sigma$, where $m$ is the invariant mass of the particle. Thus, \begin{equation}\tag{4} p_\sigma = \left(m\frac{dx_1}{d\tau}, m\frac{dx_2}{d\tau}, m\frac{dx_3}{d\tau}, \frac{mc}{\sqrt{1 - \beta^2}}\right). \end{equation} If we now define \begin{equation}\tag{5} f_\sigma = \frac{dp_\sigma}{d\tau} \end{equation} then the 4-th component of $f_\sigma$ will always be zero because \begin{equation}\tag{6} \frac{mc}{\sqrt{1 - \beta^2}} \end{equation} is a constant. Is this conclusion correct? It doesn't seem to be, yet I am unable to argue against it.

A few observations:

• I am able to define proper time of particle only because it is in an inertial frame of reference. That is, it is travelling with a uniform velocity. Therefore, not just the 4-th component, but all components will be zero.
• If this is correct, is it even possible to write the analogue of Newton's second law in special theory or relativity?

I consulted a few questions asked before, for example this one and this one. But I was unable to get a clear understanding of 4-force. Could someone kindly point out where I am going wrong?

Answer 1

$\gamma mc$ is not a constant. It is a function of $\tau$.

Remember that the gamma factor depends on $v$. It is only a constant when $v$ is constant. For non-straight line paths, gamma can vary in general

I am able to define proper time of particle only because it is in an inertial frame of reference. That is, it is travelling with a uniform velocity. Therefore, not just the 4-th component, but all components will be zero.

This is false. In general, proper time, along any worldline, is:

$$\int \sqrt{g_{\mu \nu} \frac{dx^{\mu}}{d\lambda} \frac{dx^{\nu}}{d\lambda}} d\lambda$$

$\lambda$ is some labeling parameter of the worldline.

In inertial frames, the metric $g$ is the Minkowski metric. So the proper time of any worldline will be calculated using the distance formula $dt^2-dx^2-dy^2-dz^2$.

If you choose $\lambda =t$ in the formula, and use the Minkowski metric, you get:

$$\tau=\int \sqrt {1-v^2(t) }dt$$

Or $$d \tau= \sqrt{1-v^2(t) } dt$$

Answer 2

The zeroth component of 4-momentum is energy, kinetic plus rest-mass energy. If a particle increases kinetic energy, the zeroth component changes.

For instance, under a constant proper acceleration in one dimension in relativity (the “relativistic rocket”) the rapidity (which is the linear measure of speed in relativity) should increase constantly with proper time, $$ \varphi = \alpha~\tau/c. $$ The $\gamma$ factor in rapidity-speak is given by the hyperbolic cosine $\gamma=\cosh\varphi$ and in its role as time dilation we have $$ \frac{\mathrm dt}{\mathrm d\tau}=\cosh\left({\alpha\tau\over c}\right),\\ t = \frac{c}{\alpha}\sinh\left({\alpha\tau\over c}\right).$$ So then we would find that under this force the zeroth component of momentum will vary with $t$ like, $$ \gamma~m~c = m~c~\sqrt{1+\left(\frac{\alpha t}{c}\right)^2} $$ and so is obviously not constant.