# Conservation of components of relativistic momentum transverse to applied force

Suppose a particle, say a positron is moving with initial velocity $${\bf v} = (c/3,c/3,c/3)$$, where $$c$$ is the speed of light. Then the relativistic momentum is $${\bf p} = \gamma m{\bf v}$$. Now suppose we apply a force, perhaps due to an electric field E $$=(0,0,E)$$, so the force is F$$= (0,0,eE)$$. In Newtonian mechanics this force can only change the z component the momentum and the x and y components are conserved as no force acts in their direction.
In the relativistic case I don't understand how the components of the momentum will change as they cannot be independent of each other. Here is my frame of thought. An increase in the $$z$$ component of the velocity will increase the $$\gamma$$ factor. But if the momentum in the $$x$$ and $$y$$ direction is conserved then the $$x$$ and $$y$$ components of the velocity must decrease to compensate for an increasing $$\gamma$$ factor.
Is this reasoning the physical reality?

• why does relativity make the components of momentum dependent?
– JEB
May 15 '20 at 18:07
• never mind, I get it. It's a good question.
– JEB
May 15 '20 at 18:56

Note that 4-velocity can be written:

$$u^{\mu} =\gamma c (1, \vec{\beta})$$

so you can read off $$\gamma$$ and the 3-velocity as long as there is a "1" in time-slot (that is, $$\gamma c$$ is factored out).

So you have

$$u^{\mu} = c\sqrt{\frac 3 2}(1, \frac 1 3, \frac 1 3, \frac 1 3)$$

and

$$F^{\mu\nu}=\frac E c\left(\begin{array}{cccc}0&0&0&-1\\0&0&0&0\\0&0&0&0\\1&0&0&0\end{array}\right)$$

The Lorentz force law says:

$$m\frac{du^{\mu}}{d\tau}=\frac{dp^{\mu}}{d\tau}=qF^{\mu\nu}u_{\nu}$$

$$a^{\mu}\equiv\frac{du^{\mu}}{d\tau} = \frac{eE}m\sqrt{\frac 1 6}(1, 0, 0, 3)$$

First, note that 4-acceleration is orthogonal to 4-velocity (which is always the case for every force and every particle):

$$a^{\mu}u_{\mu}=[\frac{eE}m\sqrt{\frac 1 6}(1,0,0,3)][\sqrt{\frac 1 6} c\gamma(3,-1,-1,-1)]=0$$

so that $$||u^{\mu}|| = c$$ and this $$||p^{\mu}|| = mc^2$$ after the acceleration (as it must).

Moreover, the transverse component of the 4-velocity and momentum don't change:

$$\frac{d u_1}{d\tau} = \frac{d u_2}{d\tau}=0$$

$$\frac{d p_1}{d\tau} = \frac{d p_2}{d\tau}=0$$

But....

$$u_i = \gamma v_i, \ \ i\in(1,2)$$

and if $$\gamma$$ changes, the $$v_i$$ must change.

Let's look at a small change in 4-velocity:

$$u^{\mu} \rightarrow u'^{\mu} = u^{\mu} + a^{\mu} \delta\tau$$

Then:

$$u'^{\mu} = c\sqrt{\frac 3 2}(1+x, \frac 1 3, \frac 1 3, \frac 1 3+3x)$$

where $$x = (\delta\tau)\frac{eE}{cm}\frac 1{\sqrt 6}\sqrt{\frac 2 3}$$.

Doing some rearranging:

$$u'^{\mu} = c\sqrt{\frac 3 2}(1+x)(1, \frac 1 {3(1+x)}, \frac 1 {3(1+x)}, (\frac 1 3+3x)/(1+x))$$

What we see here is that:

$$\gamma \rightarrow \gamma \cdot (1+x)$$

and the transverse components of 3 velocity:

$$v_i \rightarrow v_i / (1+x), \ \ \ i\in (1,2)$$

which shows exactly what the OP suspected: the change in Lorentz factor accounts for a change in transverse 3 velocity while conserving transverse momentum.

• Excellent answer, +1. I don’t think this needs to be in the answer but thought it would be worth stating here that for extended bodies or systems of particles it is possible to have a four acceleration which is not orthogonal to the four velocity. In that case the force changes the mass of the object. For example a force that compresses a spring. Your answer is correct since you explicitly state “particle” which implies that the mass is fixed
– Dale
May 15 '20 at 20:19
• that's interesting...I said it keeps "m" on shell...but if m changes, then it has to be violated. Moreover: given that any uniformly accelerated object must experience internal stress (Bell's Spaceship Paradox)....idk, could be a could problem there.
– JEB
May 16 '20 at 0:08