If we derive the momentum in Special Relativity we get :

$$ \dfrac{\mathrm{d}\boldsymbol{p}}{\mathrm{d}t} = \dfrac{\mathrm{d}m \gamma \boldsymbol{v}}{\mathrm{d}t} = m\dot{\gamma}\boldsymbol{v}+m\gamma\dot{\boldsymbol{v}} = \boldsymbol{f} $$

It's the relation between the 3-force and 3-acceleration in SR we can find on Wikipedia on other textbook. We can see that if it exist a "work" done by a force and the motion is not in the direction of that force (velocity is not longitudinal to the force) then, the "object" feel an acceleration in the transverse direction.

For example the object is a bullet with :

$$ \boldsymbol{v} = (v_x,v_y) \quad \boldsymbol{f}=(f_x,0) $$

$\boldsymbol{f}$ can be due to an uniform electric field or gravity. The work is not 0, and the velocity in $y$ direction is not 0 too. Even if there is no force in the $y$ direction we have :

$$ \dfrac{\dot{\gamma}v_y}{\gamma}=\color{red}{-}\dot{v_y} $$

And so, the acceleration is not parallel to the force like in Newtonian/Classical physics, where $\gamma$ does not exist. Here, the change of energy due to the force along $x$ imply an acceleration in $y$ direction.

I read several text where authors spoke about longitudinal and transverse mass, but it seems it's an old concept which is don't use anymore.

This phenomenon can be expressed with simple word ? I don't find an explanation in my own brain apart "The derivation chain rule"...

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    $\begingroup$ BTW: \dot{v_y} and \dot{v}_y typeset differently ($\dot{v_y}$ and $\dot{v}_y$ respectively). People (like me) who are hung up on typesetting generally prefer the latter. $\endgroup$ Jun 3, 2019 at 19:10
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    $\begingroup$ BTW, that issue of longitudinal and transverse mass is one of the reasons why relativistic mass is not used in modern treatments of special relativity. See physics.stackexchange.com/q/133376/123208 $\endgroup$
    – PM 2Ring
    Jun 4, 2019 at 3:25
  • $\begingroup$ Yes it's a point of view I understand. Btw for me it's simple to attach some characteristics to objects in their own ref' (like charge, mass ...) and after thinking about cinematic effects. In addition, you can point out that if we change the frame in order $v_y^\prime=0$, then $f_y^\prime \neq 0$, due to Lorentz transform on the 4-force vector $\boldsymbol{F} = \gamma(\boldsymbol{f} \cdot \boldsymbol{v},\boldsymbol{f})$ $\endgroup$
    – WilliamFr
    Jun 4, 2019 at 13:04
  • $\begingroup$ Pretty sure that should be $\dot{\gamma} v_y/\gamma = \color{red}{-} \dot{v}_y$, by the way. $\endgroup$ Jun 4, 2019 at 14:58
  • $\begingroup$ Oh yes, sorry ! I correct it immediately $\endgroup$
    – WilliamFr
    Jun 4, 2019 at 15:06

2 Answers 2


If a positively charged particle A moves to the north, and a negatively charged particle B also moves to the north, and if both particles enter an electric field that points straight down in our frame, then the system consisting of the two particles starts gaining more rest-mass, and losing northwards speed. The system's northwards momentum stays constant.

In other words the moving system scoops up mass-energy that is originally not moving , and that causes the slowdown.

The system's kinetic energy decreases and the system's internal kinetic energy increases. So "inelastic collision" is a good term to describe this phenomenon.

So both particles "scatter".

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    $\begingroup$ In this mind experiment, are the positive and negative charge at the same place initially ? Did you take into account the attraction force between the two particles ? $\endgroup$
    – WilliamFr
    Jun 5, 2019 at 10:24
  • $\begingroup$ Those particles were far apart, for the sake of simplicity. But let's now say that a system's name is "hamburgerl" and it consists of billion particles and it flies into a microwave field while spinning around. What happens to A) its rest mass B) its speed C) its rotational inertia D) its angular velocity $\endgroup$
    – stuffu
    Jun 6, 2019 at 9:35

"Increase of speed in the transverse direction causes a decrease of speed in the longitudinal direction" means the same as: "Motion causes time dilation". That can be also expressed as "The faster a clock moves, the slower the clock-hand of that clock moves".

If we have a particle moving to the north-east direction, then we can say the east-wards component of the motion is a clock's clock hand moving, and the north-wards component of the motion is the whole clock moving, or the other way around.

I'll add something about transverse and longitudinal masses:

After a car has braked down from 0.99c to 0.9 c, its brakes have tons of mass-energy which is still moving quite fast which means that the brakes have huge amount of momentum. This is one way how a relativistic car can have a large longitudinal inertia.

If the driver of said car locks the brakes at 0.99 c, then the car starts losing mass-energy quite fast, in this case the mass energy and its momentum go to the road, which means that even a slippery road feels a large force when a relativistic object is sliding on it.

After a car has braked down from 0.99c to 0.9 c, its brakes have tons of mass-energy - now that extra mass has an effect on steering forces. These steering forces change the velocity of the transverse mass. The car has the same mass-energy and the same transverse inertia before and after the braking.

  • $\begingroup$ If I understand correctly your answer you said : The work done by the longitudinal force modify the longitudinal speed and thus the $\gamma$ factor. This change imply time dilatation phenomena which is "feel" by the particle in that way : its transverse speed is redefine/rescale with a new $\gamma$ factor and so varies in time, which is the definition of acceleration. That's correct ? Eventually, this acceleration is a direct consequence of time dilatation, like a "transverse Doppler Effect" $\endgroup$
    – WilliamFr
    Jun 4, 2019 at 15:04
  • $\begingroup$ I really don't know. :) My point was very simple : How about if you try to think a single moving particle as a mechanical clock. My second answer is better anyway. $\endgroup$
    – stuffu
    Jun 4, 2019 at 23:24
  • $\begingroup$ Finally I think it's the point. If you study the 4-velocity vector $\boldsymbol{U}=(\gamma c, \gamma \boldsymbol{v})$, it's the derivation of the 4-position vector with respect to the proper time. Thus the $\gamma$-factor come from the time dilatation. If $\gamma$ varies (due to a work done by a force along $x$ for example, the longitudinal axis) then all the component of the four-vector varie in time. The fact particle changes its inertial frame imply variation of transverse 4-velocity. If we give energy to a system ($\gamma$ varies) but we don't apply force, we introduce an acceleration $\endgroup$
    – WilliamFr
    Jun 5, 2019 at 11:13

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