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Dear physics stack exchange,

I was looking over the darker non-mainstream physics wing of youtube when I came across a video with a person who gave a simple experiment to prove you could accelerate an electron to go faster than the speed of light. If you follow the link your see his apparatus and the key idea here being the use of a classical projectile problem. Given an apparatus with a distance $d$, electric field of strength $\textbf{E}_{o}$, charge $q$, mass $m$, and initial velocity $\textbf{v}$. Then,

$$ \textbf{v} = \langle v_{o} \cos(\theta) , v_{o} \sin(\theta) \rangle $$ $$ \textbf{E}_{o} = \langle 0 , E_{o} \rangle $$ $$ a_{net} = \frac{qE_{o}}{m}. $$

It would take $t = \frac{d}{v_{o} \cos{\theta}}$ to traverse the distance and acquire a final y-velocity of $v_{y}= at + v_{y_{o}} = \frac{qE_{o}d}{mv_{o} \cos(\theta)} + v_{o} \sin(\theta)$ so then the total end speed would be,

$$ |\textbf{v} | = \sqrt{(v_{o}\cos(\theta))^{2} + \left( \frac{qE_{o}d}{mv_{o} \cos(\theta)} + v_{o} \sin(\theta)\right)^{2}} .$$

From this equation he asserts you could find out what combination of initial angle or velocity could surpass the speed of light given you only need to come close to some large speed in the y-direction which if then combined with a non-zero x-velocity would yield a speed greater than $c$.

My question being what velocity would special relativity predict with a constant downward electric force or even better what would its trajectory be? I have looked over the wikipedia page on relativistic acceleration/forces but i'm always at a loss with determining stand alone equations in terms of coordinate/proper time for trajectories, especially ones that are not in a single direction. Any help or information is much appreciated.

Sincerely, A freshman going on sophomore year

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In relativistic mechanics the Lorentz force is defined: $$ \frac{d\mathbf{p}}{dt} = q\left(\mathbf{E} + \mathbf{v}\times\mathbf{B}\right). $$

This looks exactly like the non-relativistic expression. But while the notation is the same, there are some definitional differences between relativistic and non-relativistic mechanics.

The mistake is that the correct expression for momentum is $$ \mathbf{p} = \gamma m \mathbf{v}, \quad\quad\quad \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}, \quad\quad\quad v^2 = \mathbf{v}\cdot\mathbf{v} = {v_x}^2 + {v_y}^2 + {v_z}^2 $$

The important thing is that the relativistic Lorentz factor, $\gamma$, depends on the magnitude of velocity, $v$.

Using non-relativistic mechanics when the electric force is applied in the $\hat{y}$ direction, only the $y$-component of velocity changes. The non-relativistic momentum is just $m\mathbf{v}$, so the $y$-component of velocity changes independently of the $x$- and $z$-components.

Using the correct relativistic formulation, an electric force in the $\hat{y}$ direction affects all three components of velocity. Changing any one component of momentum will affect the magnitude of velocity, which mixes into the other components of the relativistic momentum. The $x$-component of velocity can change even if the $x$-component of momentum didn't.

In a reference frame where $\mathbf{B}=0$ and $\mathbf{E}=E_0 \hat{y}$, we have $$ \frac{d\mathbf{p}}{dt} = q\mathbf{E} $$

To find the final velocity we solve the Lorentz force equation: $$ \mathbf{p} = \mathbf{p}_0 + q \mathbf{E}\, \Delta t.$$

The $x$-component of momentum doesn't change: $$ p_x = p_{x,0},$$ but the $x$-component of velocity does! Any change in $v_y$ means $\gamma \ne \gamma_0$, so $v_x \ne v_{x,0}$. $$ \gamma m v_x = \gamma_0 m v_{x,0} $$

Looking at both the $x$- and $y$-components: \begin{align} p_{x} &= p_{x,0} & p_{y} &= p_{y,0} + q E_0 \Delta t \\ \gamma m v \cos\theta &= \gamma_0 m v_0 \cos\theta_0 & \gamma m v \sin\theta &= \gamma_0 m v_0 \sin\theta_0 + q E_0 \Delta t. \end{align}

The final Lorentz factor, $\gamma$, depends on the final velocity, $v$, which makes the algebra a bit of a mess. In principle we can solve the system of two equations for $v$ and $\theta$.

Combining the two equations we can write: \begin{align} \left( \gamma v \right)^2 &= \left( \gamma_0 v_0 \cos\theta_0 \right)^2 + \left( \gamma_0 v_0 \sin\theta_0 + \frac{q}{m} E_0 \Delta t \right)^2 \\ \frac{v^2}{1-\frac{v^2}{c^2}} &= {\gamma_0}^2 {v_0}^2 + \frac{q^2}{m^2} {E_0}^2 {\Delta t}^2 + 2 \frac{q}{m} E_0 \Delta t\, \gamma_0 v_0 \sin\theta_0 \end{align}

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  • $\begingroup$ Thank you so much for this answer, it's much appreciated. $\endgroup$ – The victorious truther Jun 1 '20 at 18:01
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It has always been very easy to prove speeds faster than $c$ when using Newtonian mechanics. The flaw in the logic is using $F = m a$ for relativistic velocities. I won't work the whole problem, but we have force only in the $y$ direction, so that the momentum in the $x$ direction is constant. Using $F=\frac{\partial P}{\partial t}$:

$\text{Px}(\text{t})=\frac{m\ {vx}(t)}{\sqrt{1-\frac{v(t)^2}{c^2}}}$

$\text{Py}(\text{t})=\frac{m\ {vy}(t)}{\sqrt{1-\frac{v(t)^2}{c^2}}}$

The force in the $x$ direction is $0$, so setting

$Px'(t) = 0$

and doing a little equation manipulation we get

$\text{vx}'(t)=-\frac{\text{vx}(t)\ \text{vy}(t)\ \text{vy}'(t)}{c^2-\text{vy}(t)^2}$

In the $y$ direction the momentum equation is

$Py'(t) = q\ E0$

which gets you a considerably more complicated equation for $vy'(t)$ which I will not reproduce here.

where

$vx(t)$ is the velocity in the $x$ direction

$vy(t)$ is the velocity in the $y$ direction

$Px(t)$ is the momentum in the $x$ direction.

$Py(t)$ is the momentum in the $y$ direction.

$vx'(t)$ is a negative quantity showing the velocity in the $x$ direction is decreasing even though the momentum is constant. So the premise that we are adding a constant $x$ velocity to an increasing $y$ velocity that approaches the speed of light is incorrect.

Addition

The two momentum equations are coupled differential equations that while very complicated, Wolfram Mathematica handled them. They are two long to put in here, but plugging in some numbers is doable. Using

$c = 1\ \ \ $ allows all velocities to be in terms of $c$

$v0 = c/100$

$\theta =30 {}^{\circ}$

$q = 1$

$E0 = 10$

$m = 1$

we get after solving the coupled differential equations and plugging in the values:

$\text{vx}(t)=\frac{\sqrt{\frac{3}{10}}}{2 \sqrt{99990 t^2+3 \sqrt{1111} t+1000}}$

$\text{vy}(t)=\frac{1}{2} \sqrt{\frac{3999600 \sqrt{1111} t^2+133320 t+\sqrt{1111}}{999900 \sqrt{1111} t^2+33330 t+10000 \sqrt{1111}}}$

$vx$ plot

enter image description here

$vy$ plot

enter image description here

$v=\sqrt{\text{vx}^2+\text{vy}^2}$ plot

enter image description here

The $x$ velocity goes to zero as the y velocity approaches $c$

Just for grins, computing the value at $t = 10000000$ with high precision we get

for $vy$

$0.99999999999999994999624962996312507482106$

and for $v$

$0.99999999999999995000000000500025376875156$

both close to but below $c$

Another simpler way to show that the total velocity is limited by the velocity of light is to use the fact that the momentum in the x direction is constant, which we will call $P0$. We have

$v=\sqrt{\text{vx}^2+\text{vy}^2}$

and

$\text{P0}=\frac{m\ \text{vx}}{\sqrt{1-\frac{\text{vx}^2+\text{vy}^2}{c^2}}}$

Solving for $vx$ we get

$\text{vx}=\frac{\text{P0} \sqrt{c^2-\text{vy}^2}}{\sqrt{c^2 m^2+\text{P0}^2}}$

Using that value to compute $v=\sqrt{\text{vx}^2+\text{vy}^2}$ we get

$v=c \sqrt{\frac{m^2 \text{vy}^2+\text{P0}^2}{c^2 m^2+\text{P0}^2}}$

From that expression we see that $\text{v}\to c$ as $\text{vy}\to c$ so that relativity is not violated.

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    $\begingroup$ It's a little confusing with $vx^{'}(t)$ by which you mean $v_{x}^{'}(t)$, right? $\endgroup$ – The victorious truther May 31 '20 at 21:57
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    $\begingroup$ Write v_x to get $v_x$. $\endgroup$ – G. Smith May 31 '20 at 23:16
  • $\begingroup$ I never use subscripts, but I have edited to remove an incorrect space for $vx$ in the $Px(t)$ equation. Sorry. $\endgroup$ – Bill Watts Jun 1 '20 at 3:09
  • $\begingroup$ That's a curious thing to not use... Also, shouldn't the force be defined as the total time derivative of the momentum? Is that a notation-related thing? $\endgroup$ – Philip Jun 1 '20 at 10:37
  • $\begingroup$ @Philip I avoid subscripts because I do my actual calculations with Mathematica and quite often using subscripts causes undesired results. You are right about the derivatives, which is a Mathematica translation issue, although in this case the partial derivative is the total derivative. $\endgroup$ – Bill Watts Jun 1 '20 at 17:53

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