Mass to Momentum: Exploring the Physics of Space Travel

[FayeKane]

A spaceship becomes lighter and easier to accelerate as it burns fuel. If it doesn't literally burn fuel, but it "magically" converts mass into to momentum with 100% efficiency, you will find that the limit of v as m approaches zero is c.

You may ask, "At that limit (when v=c) where did all the mass go?" It was turned into energy.

But where did the energy go?

?

My GUESS is that it still exists, as momentum without mass--which happens to be a description of the photon. Also, it shows why mass can never travel at c.

But not being a physics Jedi, I'm not sure.

-- faye kane, idiot savant

 

[Naty1]

I am not sure I understand the relationship between your example and your queston.

Why not just ask "In an accelerating spaceship where does the energy of the fuel go?"

energy is a scalar, momentum a vector, one hint that momentum is not energy...

In any case the energy of the fuel is converted to kinetic energy...the work done in accelerating the spaceship...

 

[Dale]

If you have a unit mass (in units where c=1) at rest then the four-momentum is (1,0,0,0), which, if completely anhilated, by conservation of the four-momentum can be split into (.5,-.5,0,0) and (.5,.5,0,0), each of which is individually massless. The energy is still there, it is the first component of each four-momentum.

 

[DrGreg]

A spaceship becomes lighter and easier to accelerate as it burns fuel. If it doesn't literally burn fuel, but it "magically" converts mass into to momentum with 100% efficiency, you will find that the limit of v as m approaches zero is c.

You may ask, "At that limit (when v=c) where did all the mass go?" It was turned into energy.

But where did the energy go?

?

My GUESS is that it still exists, as momentum without mass--which happens to be a description of the photon. Also, it shows why mass can never travel at c.

But not being a physics Jedi, I'm not sure.

-- faye kane, idiot savant

First of all, you can't convert mass into momentum. You can convert mass into energy, or, rather, mass is a form of energy, so you can convert mass-energy into some other form of energy.

As momentum is conserved, a spaceship can gain momentum only if something else loses an equal amount. Traditionally, the "something else" would be the exhaust gases, and those gases would take away energy, too, so that's where all the energy goes.

Also it doesn't make sense to ask what happens when v = c, because that limit is never actually reached; you get ever closer to it without ever getting there.

 

[FayeKane]

Why not just ask "In an accelerating spaceship where does the energy of the fuel go?"

Because I know where the energy of the fuel goes in a physically-realizable rocket ship.

I guess my error was asking the equivalent of "if x, which is impossible, were possible, then what would y be?", which may superficially sound plausible, but is actually meaningless.

energy is a scalar, momentum a vector, one hint that momentum is not energy...

Yeah, I was playing fast and loose with the language, sorry. If my question was worth pursuing, I'd need to rephrase it for it to make sense, but it isn't.

-flk

 

[FayeKane]

If you have a unit mass (in units where c=1) at rest then the four-momentum is (1,0,0,0), which, if completely anhilated, by conservation of the four-momentum can be split into (.5,-.5,0,0) and (.5,.5,0,0), each of which is individually massless. The energy is still there, it is the first component of each four-momentum.

I always like discovering that I'm ignorant but didn't know it, so I can become not ignorant.

It appears that I don't understand four-momentum.

Okay, in the expression "(1,0,0,0)" the first term (1) is the fraction of the total momentum which is in the time direction (i.e., the momentum through time of the invariant mass), and the other three are the fractions of the total momentum whose vectors point in spatial directions, is that correct?

In fact, I'm sure that is incorrect, because then (.5, a, b, c) would not be massless.

And my interpretation of "(.5,-.5,0,0) and (.5,.5,0,0)" would be that you split a stationary mass into two equal parts and send them shooting off in opposite directions, which I KNOW is wrong because there IS no mass left after it has been annihilated.

If you are inclined to, would you explain in what way my understanding of the notation representing four-momentum is incorrect?

The Wikipedia article doesn't help because I can't ask questions of it.

-- the apparently permanently confused faye

 

[Count Iblis]

The first component is te enrgy, the last three components are the three momentum components. The inner product is efined as:

a dot b = a0 b0 - [a1 b1 + a2 b2 + a3 b3]

This is invariant under Lorentz transformations.

If you take the inner product of a four-momentum vector with itself (we say, "square the vector"), you'll get a result that will be the same when evaluated in any frame. But in the rest frame the vector is
(m, 0,0,0,0), so the result must always be m^2. So, we have:

P^2 = E^2 - q^2 = m^2

where q is the ordinary momentum. This is the mass energy relation.

Now, consider a spacecraft that burns anti-matter. If P1 is the initial four-momentum of the spacecraft at rest and P2 is the final four momentum and Pf is the four-momentum of the emitted photons, we have:

P1 = P2 + Pf

This means that:

Pf = P1 - P2

Squaring both sides and using the mass-energy relation P^2 = m^2 gives:

0 = m1^2 + m2^2 - 2 m1 m2 gamma --------->

gamma = 1/2 [m1/m2+ m2/m1]

So, the gamma factor the spacecraft ends up at after it has burned some amount of its anti-matter is given by the mass ratio. If the final mass approaches zero, the gamma factor tends to infinity, so the speed approaches the speed of light.