B I need an equation for relativistic exhaust velocity (Ve)

1. Jul 6, 2017

BitWiz

Does anyone have an equation that -- given energy applied to a mass over a distance -- will give me an (ideal) final velocity of the mass?

If I direct a gigajoule to accelerate a gram over a distance of a meter, I obviously get into real trouble with c using conventional equations.

I've found equations that apply Lorentz -- but they seem to do it as an adjustment after a raw, conventional velocity has been obtained, using this final, conventional velocity to plug into Lorentz. That can't work in this case. I'm looking for a function Ve( joules, mass, distance ) that works for all non-negative parameter values, and takes relativity into account. Can you help?

Thank you very much!

2. Jul 6, 2017

Staff: Mentor

Take the ratio of total energy to rest energy: that gives you the relativistic $\gamma$ factor. Then invert the formula for $\gamma$ in terms of $v$ to get $v$.

3. Jul 6, 2017

Staff: Mentor

You should check your math first. Hint: what is the rest energy of one gram of mass, in gigajoules?

4. Jul 6, 2017

Staff: Mentor

The distance is actually irrelevant to the final velocity. It is only relevant to determining the acceleration required to achieve that velocity. The final velocity only depends on the total energy applied.

5. Jul 6, 2017

c2 / 1000 ?

6. Jul 6, 2017

Staff: Mentor

If you calculate this numerically and convert to gigajoules, yes. What do you get?

7. Jul 6, 2017

BitWiz

So, v = v' * rest_mass / (Rest_mass + Jo
~898756 Gj ?

8. Jul 6, 2017

Staff: Mentor

No. Try starting from what I said in post #2.

I think you're too high by an order of magnitude.

9. Jul 6, 2017

BitWiz

Whoops.
If a joule = 1 (kg) / c2, then 1kg = c2j, and a gram is 1000th of that?

If c is ~ 3e8, then c2 is 9e16j for a kg. For a gram, 9e13j. For a gigajoule, 9e4?

10. Jul 6, 2017

BitWiz

You're saying that gamma is 1 + ( mass-equivalent-of-energy-added / rest mass ) ?

Thanks!

11. Jul 6, 2017

Staff: Mentor

That's one way of putting it, yes. Or you could just do everything in energy units; it works out the same either way.