4 ms·
I find the coverage of this drive simply bizarre. There is nothing impossible about it, in principle. We know E=mc^2 (mass and energy are equivalent). We know
by PeterWhittaker 11y ago
I find the coverage of this drive simply bizarre. There is nothing impossible about it, in principle.
We know E=mc^2 (mass and energy are equivalent).
We know that rockets move by ejecting mass at high speed in a direction counter to the desired direction of movement. The rocket shifts forward an equivalent amount (m1v1=m2v2).
We know that it is possible to focus EM radiation emissions.
So far so good.
The "magic" here (used in the sense of sufficiently advanced tech, of course), is that the EM emission is entirely unidirectional.
So instead of throwing mass out the nozzle, we're throwing energy. And since E=mc^2, we get the same effect: RmRv=E/c (rocket mass times rocket speed equals Energy over c (since m2v2 = (E/c^2)c).
Very, very, very cool.
- kragen 11y agoBut the energy isn't going out a nozzle. You're describing a lightsail or photonic laser thruster, not the EM drive. If the EM drive works, it violates linear conservation of momentum (and therefore, if relativity holds, conservation of energy). This makes it very unlikely indeed that it works. I hope the explanation isn't something terribly boring.
- wai1234 11y agoThat C^2 factor is a bit of a problem. Calculate the mass equivalent of any reasonable amount of energy you intend to "throw" and you will be disappointed. In the other direction, that c^2 term works pretty well (for atomic bombs anyway). For the relationship of the mass-energy equation to the force exerted by radiation. see Wikipedia: https://en.wikipedia.org/wiki/Radiation_pressure https://en.wikipedia.org/wiki/Radiation_pressure
- Geee 11y agoEM drive has a closed cone, there's no nozzle. They tested identical machine with open nozzle, and it didn't generate the same force.