Andrew Mason said:
Not yet, as far as I can tell. But there would be an inherent detection problem. Consider a charged massless particle passing through an electric field. What would happen? It would not experience a detectable force, F = ma, since m = 0 and, since it always travels at the same speed, the speed of light, a = dv/dt = 0.
AM
Having constant speed v = |\mathbf{v}| does not imply zero acceleration \mathbf{a} = d \mathbf{v}/d t. All you can say with certainty is that the dot product:
<br />
v^{2} = \mathbf{v} \cdot \mathbf{v}<br />
<br />
2 \, v \dot{v} = 2 (\mathbf{v} \cdot \dot{\mathbf{v}})<br />
<br />
(\mathbf{v} \cdot \mathbf{a}) = 0<br />
is zero.
Furthermore, 2nd Newton's Law has to be modified in relativistic limit:
<br />
\mathbf{F} = d \mathbf{p}/ d t, \; \mathbf{p} = m \, \gamma \, \mathbf{v}, \gamma = (1 - \beta^{2})^{-\frac{1}{2}}, \beta = v / c<br />
Using the chain rule:
<br />
\dot{\mathbf{p}} = m \, \left(\gamma \, \dot{\mathbf{v}} + \dot{\gamma} \, \mathbf{v}\right)<br />
<br />
\dot{\gamma} = \frac{d \gamma}{d v} \, \dot{v} = \left(-\frac{1}{2}\right) \, \gamma^{3} \, \left(-\frac{2 \, v \, \dot{v}}{c^{2}}\right) = \frac{\beta \, \gamma^{3} \, \dot{v}}{c}<br />
Unfortunately, when m = 0, we have an indeterminate expression 0 \cdot \infty. That is why we eliminate \mathbf{v} in terms of momentum \mathbf{p}:
<br />
p = m \, c \, \gamma \, \beta \Rightarrow \gamma^{-2} = \left(\frac{p}{m \, c \, \beta}\right)^{-2}<br />
<br />
1 - \beta^{2} = \left(\frac{m \, c}{p}\right)^{2} \, \beta^{2} \Rightarrow \beta^{2} = \frac{p^{2}}{p^{2} + (m \, c)^{2}}, \; 1 - \beta^{2} = \frac{(m \, c)^{2}}{p^{2} + (m \, c)^{2}} \Rightarrow \gamma = \frac{\sqrt{p^{2} + (m \, c)^{2}}}{m \, c}<br />
Substituting everything and simplifying, we get:
<br />
\mathbf{F} = m \, \frac{\sqrt{p^{2} + (m \, c)^{2}}}{m \, c} \, \left( \mathbf{a} + \dot{v} \, \frac{p \, \mathbf{p}}{m \, c} \right)<br />
Taking m \rightarrow 0, \dot{v} \rightarrow 0 and we still have an indeterminate form 0/0. Let us evaluate \dot{v} in terms of \dot{\mathbf{p}}:
<br />
\dot{v} = \frac{d v}{d p} \, \dot{p}<br />
<br />
\frac{d v}{d p} = c \, \frac{d \beta}{d p} = c \, \frac{1 \cdot \sqrt{p^{2} + (m \, c)^{2}} - p \frac{p}{\sqrt{p^{2} + (m \, c)^{2}}}}{p^{2} + (m \, c)^{2}} = \frac{m^{2} \, c^{3}}{\left(p^{2} + (m \, c)^{2}\right)^{3/2}}<br />
So,
<br />
\frac{\dot{v}}{m} = \frac{m \, c^{3} \, \dot{p}}{\left(p^{2} + (m \, c)^{2}\right)^{3/2}} \rightarrow 0, \; m \rightarrow 0<br />
the second term in the parentheses in the expression for the relativistic force becomes zero in the massless limit and we have:
<br />
\mathbf{F} = \frac{p \, \mathbf{a}}{c}, \; m = 0<br />
Thus, a finite force can give rise to a finite acceleration of a massless particle in the same direction as the force. But, a non-zero acceleration must be perpendicular to the velocity. Thus, the force must be perpendicular to the velocity of the particle. Lorentz force always fulfills this condition, but an electrostatic field might not.
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