- #1
Anixx
- 81
- 12
Logically, it seems that if we arrive at the point of destination before we left the point of departure, we have traveled with greater-than-infinite speed.
So, I wonder whether anyone introduced faster than infinite speeds in a physical or mathematical theory.
Classically, it seems, for travel from point ##(x_0, t_0)## to ##(x_1,t_1)## along the ##x## spatial axis and ##t## temporal axis, the mean speed would be
$$s=(x_1-x_0){\mathcal {H}}\int_{t_1-t_0}^\infty \frac1{t^2} dt$$
But if we let the integral to be divergent without undertaking Hadamard regularization, the speed will be
$$s=(x_1-x_0)\int_{t_1-t_0}^\infty \frac1{t^2} dt=(x_1-x_0)\left((1-\operatorname{sgn}(t_1-t_0))\pi\delta(0)+ \frac{1}{t_1-t_0} \right)$$
Here ##\pi\delta(0)## formally represents the divergent integral ##\frac12 \int_{-\infty}^\infty \frac1{t^2} dt## in non-Hadamard-finite-part sense. In other words, intead of taking the Hadamard finite part integral and getting negative speed for negative travel time, we leave the integral divergent and get faster-than-infinite speed.
For instance, when a particle propagates in a direction, a virtual particle-antiparticle pair may arise ahead of it, with antiparticle moving in the direction to meet the original particle. Then the antiparticle may annihilate with the original particle, while the virtual particle becoming rea and travels in the original direction. This is how quantum tunneling works. For a side observer this may look like the distance between the pair birth and the annihilation points has been traveled by the particle faster than with infinite speed.
So, I wonder whether anyone introduced faster than infinite speeds in a physical or mathematical theory.
Classically, it seems, for travel from point ##(x_0, t_0)## to ##(x_1,t_1)## along the ##x## spatial axis and ##t## temporal axis, the mean speed would be
$$s=(x_1-x_0){\mathcal {H}}\int_{t_1-t_0}^\infty \frac1{t^2} dt$$
But if we let the integral to be divergent without undertaking Hadamard regularization, the speed will be
$$s=(x_1-x_0)\int_{t_1-t_0}^\infty \frac1{t^2} dt=(x_1-x_0)\left((1-\operatorname{sgn}(t_1-t_0))\pi\delta(0)+ \frac{1}{t_1-t_0} \right)$$
Here ##\pi\delta(0)## formally represents the divergent integral ##\frac12 \int_{-\infty}^\infty \frac1{t^2} dt## in non-Hadamard-finite-part sense. In other words, intead of taking the Hadamard finite part integral and getting negative speed for negative travel time, we leave the integral divergent and get faster-than-infinite speed.
For instance, when a particle propagates in a direction, a virtual particle-antiparticle pair may arise ahead of it, with antiparticle moving in the direction to meet the original particle. Then the antiparticle may annihilate with the original particle, while the virtual particle becoming rea and travels in the original direction. This is how quantum tunneling works. For a side observer this may look like the distance between the pair birth and the annihilation points has been traveled by the particle faster than with infinite speed.