Relationship between propellant mass and thrust at relativistic speeds

Question

When answering this question about relativity and delta-V I realized I don't know the relationship between the mass of propellants and how much delta-v they would produce in an engine at relativistic speeds in a real world scenario.

From a pure physics perspective my thinking is that as the mass of the propellants increases with the speed of light their exit speed will drop but will produce the same amount of delta-v. If delta-v increased as mass increased then it would just continue along the same curve as relativity, meaning there's no reason you couldn't reach the speed of light, which is impossible according to current understanding of physics. However, I'm not sure I have a complete grasp, is my thinking right?

Also, would the propellant mass changing require any changes to the engines on a relativistic ship? My understanding is that nothing changes on the ship's frame of reference, things don't get more massive, so the engines would work the same, however is this correct?

Answer 1

From a pure physics perspective my thinking is that as the mass of the propellants increases with the speed of light their exit speed will drop but will produce the same amount of delta-v.

The mass of the propellant doesn't increase in the frame of reference of the accelerating spaceship; it increases (with the increasing kinetic energy of the ship) in some external non-accelerating frame. The exit velocity and propellant flow rate and therefore thrust and local delta-v all remain unchanged.

Answer 2

1. The situation is significantly simplified by applying the Principle of Relativity:

The principle of relativity is the requirement that the equations describing the laws of physics have the same form in all admissible frames of reference.

It is the same in the Special Relativity, too.

2. In the case if we have only linear acceleration, the problem is easily solved by the concept of rapidity. Essentially, we can sum or negate the velocoties as usual in the Newtonian Mechanics, we need to apply the well-known $\frac{1}{\sqrt{1\pm\frac{v^2}{c^2}}}$ only to the result.

Thus, you don't need to count with propellant masses and so on. The spaceship has a $\Delta v$ reserve, what you can calculate as usual. An acceleration caused by an inifinitezimal small $dm$ in the reference frame of the ship, will be the same as in the Newtonian system.

Finally apply the relativistic formula to convert back the rapidity to velocity.

For example, if the $\Delta v$ reserve of the spaceship is $10c$, the resulting final speed what it arrives will be $\frac{10}{\sqrt{1+\frac{(10c)^2}{c^2}}} \approx 0.995c$ .

But (another example) if it uses $6c$ $\Delta v$ to accelerate and $2c$ to decelerate, then you can substitute $4c$ into the formula.

The situation becomes more problematic if the movements are not linear. Rapidity is an additive property only in 1D linear systems. There we need analysis to calculate the properties of the trajectory, it is already not so easy.

_{P.s. around at 0.1c, the $\approx 1 \frac{\rm{atom}}{cm^3}$ interstellar hydrogen causes $\approx 100 \frac{W}{m^2}$ hard proton radiation. Its luminosity increases cubically with the speed. The energy of the ionizing radiation to kill a human is lesser than to boil a cup of caffee. Thit makes highly relativistic spaceships yet more theoretical.}