B Relative mass and inertia

1. Jan 29, 2017

Chris Miller

If a bullet were to strike the Earth at a velocity close enough to c that its relativistic mass approached that of the Earth, would the damage be equivalent to that of an object of similar invariant mass colliding at a non-relativistic velocity? (The reason I ask is that in Liu Cixin's sci-fi, The Dark Forest, a star is destroyed by hitting it with a small projectile traveling at near c.)

2. Jan 29, 2017

Ibix

I wouldn't use relativistic mass in this (it's largely a deprecated concept these days outside of pop sci, due to it engendering a lot of confusion), but yes, in theory. The kinetic energy ofa mass moving at speed $v$ is $(\gamma-1)mc^2$, where $\gamma =1/\sqrt {1-v^2/c^2}$. That number can be as large as you like, tending to infinity as you approach the speed of light.

Whether or not you can actually destroy a star with a kinetic strike, I don't know. Earth, yes.

3. Jan 30, 2017

Chris Miller

Thanks for the clarification, Ibix. Amazing, almost unbelievable, to me that something the size of a BB... a grain of sand... even a neutron? could theoretically destroy the earth (and why not a star then?) if its velocity were close enough to c. Hope those big particle accelerators have some safety features.

4. Jan 30, 2017

SiennaTheGr8

5. Jan 30, 2017

Staff: Mentor

They aren't needed, at least as far as accidentally letting a world-destroying particle loose. The only way of accelerating a particle to world-destroying energies is to supply that much energy to the particle in the first place - the particle never has more energy than what you put into it to accelerate it.

Now, it would be an interesting exercise to calculate the amount of energy required to accelerate a particle to a speed such that its relativistic mass is equal to the mass of the earth.... try it.

6. Jan 30, 2017

Mister T

The only kind of mass I ever talk about is the ordinary mass, so I'll restate your proposal in those terms. A bullet of mass $0.001$ kg collides with Earth, mass $6 \times 10^{24}$ kg. In a frame of reference where Earth is at rest, the bullet is moving so fast that it has an energy of $6 \times 10^{24}$ kg. (Note that I'm measuring energy in kilograms. To convert to joules you would multiply by $c^2 \approx 9 \times 10^{16}$ J/kg.)

Therefore we have $\gamma \approx \frac{6 \times10^{24}}{0.001} = 6 \times 10^{27}$.

(Note that when a particle's speed is so close to the speed of light that the difference is negligible, we speak of $\gamma$ rather than the speed because the former is more meaningful. This is analogous to speaking of the speed rather than of $\gamma$ when the speed is so small that the difference between $\gamma$ and $1$ is negligible.)

The LHC is the biggest particle accelerator. Its protons move at nearly the speed of light. The ratio of Earth's mass to the proton mass is $\frac{6 \times10^{24}}{2 \times10^{-27}} \approx 3 \times10^{51}$ but those protons are given only enough energy to make $\gamma \approx 7500$. I think we're safe because we'd need a $\gamma$ of $3 \times10^{52}$ to make that proton as dangerous as your bullet. We have nothing remotely capable of producing that much energy.

SLAC moves electrons at nearly the speed of light and achieves a $\gamma$ of about $98\ 000$. The ratio of Earth's mass to the electron mass is $\frac{6 \times10^{24}}{9 \times10^{-31}} \approx 7 \times10^{54}$. Again, we're safe.