relativistic rocket

[Rocket Man]

suppose for the moment, that we had a means of converting mass directly into harnessable energy.

if that energy were put out the back of a rocket as light, it will behave as though the propellant mass were moving at c relative to the rocket, but only if you look at it from a newtonian dynamics perspective.

 

if we factor in inertial dialation, at what speed will mass need to travel as propellant to muster the same force as that mass-energy from a laser?

[TriggerGrinn]

I wonder too what ratio between photon momentum and a given mass velocity..

 

I don't want to begin answering this, but maybe these equations would be related..

 

[math]p=\frac{h\nu}{c}=h\lambda[/math]

 

 

[swansont]

Let's look at a simple case of one impulse.

 

For a photon of energy E ejected, you have p=E/c = mv, so the final speed of the rocket is E/mc

 

For a rocket, the energy and momentum are divided up between the ejecta and the rocket, and their total mass is m

 

KE: m1v1^2 + m2v2^2 = 2E

p: m1v1 = m2v2

 

do some math and you get [math]v = \sqrt{(2E/{m_r(m_r/m_e +1)})}[/math]

(r=rocket, e=ejected mass)

 

if the mass of the rocket is big compared to the ejected mass, then m_r = m is a good approximation

 

[math]v = \sqrt{(2Em_e)/m}[/math] will be a good approximation

 

So which one is bigger? If we take the ratio,

 

[math]v_{photon}/v_{mass} = \sqrt{(E/2m_e)}/c[/math]

 

E/m is small, unless you are converting mass to energy and ejecting that same amount of mass (i.e. ejecting a proton for every proton completely annihilated) and c is really big, so the photon looks like it's a lot less efficient. The utility of a photon drive is that you don't have to lug the ejecta up outof your gravity well. As long as you can generate the energy, you can have propulsion.

[Janus]

When talking about the efficiency of a rocket, we are generally dealing with how much delta V we can get from our fuel/reaction mass.

 

For this we use the rocket equation

 

[math]\Delta V = V_e ln(MR)[/math]

 

where Ve is the exhaust velocity

and

MR is the mass ratio, (the mass of the fully fueled rocket divided by the mass of the unfueled rocket)

 

Note that for any given mass ratio, the dela v depends on the exhaust velocity.

 

The above equation works well for every day velocities, but for relativistic velocities you need to use the Relativistic rocket equation:

 

[math]\Delta V = c \tanh{\left (\frac{V_e}{c} \ln{(MR)}\right)}[/math]

 

Again we note that the delta v depends on the exhaust velocity. We get the best efficiency if we have an exhaust velocity of c. With such as "photonic" drive one could expect the following maximum delta v:

 

for a MR of 2, we can get .6c

for 3, .8c

4, .882c

5, .923c

6, .946c

7, .96c

...

100, .9998c

[Rocket Man]

i was thinking more about comparing the momentum of the two propellants.

assume you have a fixed amount of energy and the choice between blasting matter or light. what velocity does matter need to travel at to match the momentum of photons

[Janus]

i was thinking more about comparing the momentum of the two propellants.

assume you have a fixed amount of energy and the choice between blasting matter or light. what velocity does matter need to travel at to match the momentum of photons

 

It depends on how much matter you are talking about. For instance, if you had 9^16 joules to work with, you could have one electron moving at just slightly less than c, 1kg moving at .866c or 100 kg at .14c.

[Rocket Man]

It depends on how much matter you are talking about. For instance, if you had 9^16 joules to work with, you could have one electron moving at just slightly less than c, 1kg moving at .866c or 100 kg at .14c.

 

i've just done a little research,

the momentum of matter is equal to gammaMV

the energy equivalent of that same matter is gammaM(C^2)

 

the momentum of light is E/C

so if we use the same energy but use photons, we get

p = (gamma M(C^2))/C

 

which comes down to

P= gamma MC

where gamma is calculated by the velocity of the mass had we used it instead.

 

compared to matter, gamma MC is a lot larger than gamma MV because V has an upper limit of C and the gamma functions have the same input value.

edit: perhaps i should have rewritten the mass momentum to avoid confusion..

 

i'm sure i've missed something because i simply don't see how light can produce more momentum per joule than matter which has inertial dialation at high speeds.

[Janus]

i've just done a little research,

the momentum of matter is equal to gammaMV

the energy equivalent of that same matter is gammaM(C^2)

 

[math] \gamma mc^2[/math]

 

is the total energy equivalence of the matter, which includes the energy equivalence of the rest mass of the matter (The energy released if you convert this matter into energy) But you are not converting this matter into energy, you are just shootiing it out the back as reaction mass.

 

What you want here is the kinetic energy of the matter alone, which is found by

 

[math]c^2( \gamma m -m)[/math]

 

the momentum of light is E/C

so if we use the same energy but use photons, we get

 

p = (gamma M(C^2))/C

 

[math] p = \frac{c^2( \gamma m -m)}{c}[/math]

 

which comes down to

P= gamma MC

[math] p = c( \gamma m -m)[/math]

where gamma is calculated by the velocity of the mass had we used it instead.

 

compared to matter, gamma MC is a lot larger than gamma MV because V has an upper limit of C and the gamma functions have the same input value.

 

Now we have

[math] c( \gamma m -m)[/math]

compared to

[math]\gamma mv[/math]

 

let's assume that v = .866 c, which gives a gamma of 2. Then for photons:

[math]p= c( 2m -m) = cm[/math]

and and for matter

[math]p= 2 m (.866c) = 1.732cm [/math]

 

matter has more momentum at the same energy.

 

at .1c

 

we would get

for photons

[math]p = .005cm[/math]

and matter

[math]p= .1005cm[/math]

 

at .99c

 

we would get

for photons

[math]p = 6cm[/math]

and matter

[math]p= 6.93cm[/math]

 

 

i'm sure i've missed something because i simply don't see how light can produce more momentum per joule than matter which has inertial dialation at high speeds.

 

Actually, it has more to do with the relationship between energy and momentum. You get the same type of effect in non-relativistic physics.

 

consider

[math]p=mv[/math]

and

[math]E= \frac{mv^2}{2}[/math]

 

Now, lets take two masses m1 and m2, with equal kinetic energy. so that

 

[math]\frac{m_1 v_1^2}{2} = \frac{m_2 v_2^2}{2}[/math]

 

 

[math]m_1 v_1^2 = m_2 v_2^2[/math]

 

[math]\frac{m_1}{m_2}=\frac{v_2^2}{v_1^2}[/math]

 

[math]\frac{m_1}{m_2}=\left (\frac{v_2}{v_1}\right )^2[/math]

 

Thus the ratio of the masses is equal to the inverse of the square of the ratio of the velocities. If m1 is twice the mass of m2 then v1 is .707 that of v2

 

if v1 is 1 m/s and m1 = 1kg then

v2 is 1.414 m/s and m2 =.5kg

 

comparing momentums

 

[math]1kg(1\frac{m}{s}) = 1\frac{kgm}{s}[/math]

and

[math].5kg(1.414\frac{m}{s}) = .707\frac{kgm}{s}[/math]

 

The larger mass (m1) has a greater momentum than the smaller mass (m2) when both are at the same energy. No relativistic effects involved.

[Rocket Man]

no, i meant if you have a fixed energy to play with, including the mass-energy equivavlence as well as the kinetic energy and the choice to convert to matter or light from that same amount of energy.

 

so if you have fixed [math]E[/math], you can have a mass of [math]M=\frac{E}{\gamma C^2}[/math]

while [math]P=\gamma MV[/math]

 

put the two together you get [math]P=\gamma \left(\frac{E}{\gamma C^2}\right)V[/math]

leaving [math]P=\frac{EV}{C^2}[/math] for mass

and [math]P=\frac{E}{C}[/math] for light

 

it makes a little sense, [math]\Delta P[/math] must remain 0 which also means that the displacement of the centre of gravity must also be 0.

if after a short period, all motion were ceased, and all energy was converted to mass, the light would have travelled further so the rocket would have moved a proportionally equal amount further than it's mass-expelling counterpart.

[Janus]

no, i meant if you have a fixed energy to play with, including the mass-energy equivavlence as well as the kinetic energy and the choice to convert to matter or light from that same amount of energy.

 

so if you have fixed [math]E[/math], you can have a mass of [math]M=\frac{E}{\gamma C^2}[/math]

 

You mean that you have x amount of energy that you can either release as photons, or partially convert to matter while using the rest to provide the velocity?

 

This second choice seems a little silly, as it would be easier to store the mass that you are ejecting as matter rather than trying to store it as energy and then converting.

 

Also, in this case, the momentum you will get will depend on how much of the energy is converted to matter. From 0 if you converted all of it to matter (leaving none for kinetic energy and giving it 0 velocity) to approaching that of the photons as the amount of matter approaches 0 and its velocity approaches c.

[Rocket Man]

i'm not talking practicalities, this is entirely hypothetical, (i've been trying to describe this particular question since the start of the thread)

 

notice the "gamma C squared" in that term, gamma is a function of the velocity of the propellant as [math]E=\gamma mc^2[/math]

so for the same amount energy in total, the higher the velocity, the more the momentum. untill you reach the point where your ejecta is massless and travelling at C according to

[math]P=\frac{Ev}{c^2}[/math]

 

basically, i'd like someone to prove/disprove because i don't quite see how this is logical.

[Janus]

i've just done a little research,

the momentum of matter is equal to gammaMV

the energy equivalent of that same matter is gammaM(C^2)

 

the momentum of light is E/C

so if we use the same energy but use photons, we get

p = (gamma M(C^2))/C

 

which comes down to

P= gamma MC

where gamma is calculated by the velocity of the mass had we used it instead.

 

compared to matter, gamma MC is a lot larger than gamma MV because V has an upper limit of C and the gamma functions have the same input value.

edit: perhaps i should have rewritten the mass momentum to avoid confusion..

 

i'm sure i've missed something because i simply don't see how light can produce more momentum per joule than matter which has inertial dialation at high speeds.

 

 

Okay, now that I understand where you are coming from, let's attack this from a different direction.

 

Start with a a given mass (M) that can be converted to energy, either in whole or in part.

 

If we covert it all to energy in the form of photons this energy will be:

 

[math]E=Mc^2[/math] (The total energy equivalence of M)

 

and the momentum:

 

[math]P_p= \frac{E}{c} = \frac{Mc^2}{c} = Mc[/math]

 

If we convert all but a part (m) to energy and that energy is in the form of the kinetic energy of m, the total energy is:

 

[math]E= \gamma mc^2 [/math]

 

Since the total energy involved is conserved, it must stay the same both before and after conversion, thus.

 

[math]E=Mc^2[/math]

 

and

 

[math]E= \gamma mc^2 [/math]

 

are the same and so,

 

[math]Mc^2= \gamma mc^2 [/math]

[math]M= \gamma m [/math]

 

Solving for gamma:

 

[math] \gamma = \frac{M}{m} [/math]

 

The momentum will be

 

[math]P_m = \gamma mv[/math]

 

substituting from above:

 

[math]P_m =\frac{M}{m} mv[/math]

 

[math]P_m =Mv[/math]

 

Comparing Pm to Pp:

 

[math]\frac{P_p}{P_m} = \frac{Mc}{Mv} = \frac{c}{v}[/math]

 

Which is the same relationship you arrived at from a different direction.

 

Here, however we can see why the gamma factor doesn't come into play in the final analysis:

 

If we go back to the equation:

 

[math]M= \gamma m [/math]

 

and re-arrange:

 

[math]m= \frac{M}{\gamma} [/math]

 

We see that as you convert more and more of M to energy to increase the velocity of m(the rest mass of the remaining matter), m decreases such that that amount of mass that gamma has to work on in

 

[math]P_m = \gamma mv[/math]

 

decreases at the same rate as gamma increases.

 

As a result, the momentum increases in a steady fashion until it reaches a maximum when all the matter is converted to photons.