Forces and Accelerations

In Newtonian physics, an applied contact force acting on a rigid object will accelerate the object in the same direction as the applied force. Does this behavior hold for applied contact forces in relativity physics (STR)? For example, if an applied contact force pushes on the same rigid object in the direction perpendicular to the direction of motion, will the resulting acceleration be in the direction of the applied contact force?

Answer

No. In the STR, all contact forces will produce an acceleration in a direction not parallel to the applied force! For example, a rigid sphere is moving along the plus x-direction of an inertial reference frame. Now let an applied contact force act in the plus y-direction to increase the speed of the sphere in the y-direction. What happens to the speed in the x-direction? The x-component of the speed decreases the object slows down in its original direction, corresponding to a negative acceleration!

To understand why the object slows in the x-direction when the applied contact force is in the y-direction, we begin with the space-time interval: (interval)² = c² Δt² – Δx² – Δy² – Δz². For real objects traveling at speeds less than c, the time term is much larger than the spatial terms, and the interval is called the proper time τ. The linear momentum p^x in the x-direction in Newtonian physics is defined as p^x = m dx/dt (for an object that is not changing its mass, i.e., excluding objects such as a leaking bucket of water). The correct STR expression simply substitutes proper time τ for Newtonian time t so that p^x = m dx/dτ. For a low-velocity object, dτ ~ dt. But the actual relationship between τ and t depends on the magnitude of the object’s total velocity, a vector quantity, not just the speed component in the x-direction. Therefore, as the object speeds up in the y-direction, its speed in the x-direction must decrease to maintain a constant total velocity magnitude, otherwise its x-component of linear momentum would change, forbidden by the law of conservation of linear momentum. The pertinent relationship is dt/dτ =  . Remember, the mass is fixed in value.

One could write down the relativistic momentum  and argue that since m is constant, the component of velocity in the original direction must decrease to keep the momentum component constant.