mass lightning question

[dragon]

I have a question on mass lighting. If the Starship is done accelerating and is at a constant velocity and the mass lighteners are turned off what happens.

[Bounty]

If the Starship is done accelerating and is at a constant velocity and the mass lighteners are turned off what happens.

Starships that lose engine power just come to a halt, which might be caused by the mass lightening field shutting down.

[dragon]

If the Starship is done accelerating and is at a constant velocity and the mass lighteners are turned off what happens.

Starships that lose engine power just come to a halt, which might be caused by the mass lightening field shutting down.

Hum that shouldn't work after all there has to be a force exerted in order for decelleration to occur. And with no power their inertail dampeners will also be down, which means they would get squished if they just stop.

[Starglider]

Conservation of momentum suggests that the ship's velocity vector would change to be the vector it would have had if the mass lightening system had not been operating (ignoring the effects of gravity, which complicate the analysis). However this violates conservation of energy, unless you have the mass lightening system pumping in and removing energy in the right quantities to match the change in KE. You could directly conserve KE by slowing the ship so that its effective KE (relative to when the mass lightening field was turned on) remains constant whether the field is on or off, but that breaks conservation of momentum. Personally I'd rather have conservation of energy than conservation of momentum, but it depends on the exact physical realisation of the system. The only major sci-fi setting I designed had a hybrid STL/FTL drive system that worked as an effective combined mass lightener/antigravity system/warp drive strictly limited by conservation of energy (extra energy had to be pumped in or removed through the drive to balance things, particularly when traversing gravitational PE gradients). I tried to set this up as carefully as possible to give the tactical environment I wanted (and no easy relativistic planet smashing) while being as physically consistent as possible (though with completely invented fantasy physics).

[Ted C]

I have a question on mass lighting. If the Starship is done accelerating and is at a constant velocity and the mass lighteners are turned off what happens.

Presumably, the ship's velocity would drop in proportion to its mass increase in order to conserve momentum. Any imbalance in the ship's kinetic energy would have to be converted to heat, raising or lowering the ship's temperature accordingly.

[Batman]

If the Starship is done accelerating and is at a constant velocity and the mass lighteners are turned off what happens.

Starships that lose engine power just come to a halt, which might be caused by the mass lightening field shutting down.

Hum that shouldn't work after all there has to be a force exerted in order for decelleration to occur.

The force needed to accelerate them was never applied in the first place under mass lightening. The ship would simply revert to the status it should be in if real world physics had been in force.

And with no power their inertial dampeners will also be down, which means they would get squished if they just stop.

How about no? No engine power /= no power, period. Notice how outside TUC which involved a DELIBERATE attack on that subsystem they NEVER lose artificial gravity?

[Drooling Iguana]

They also lost artifical gravity during the pre-credits teaser of an episode of Enterprise, though only for a minute or two.

[Sriad]

Conservation of momentum suggests that the ship's velocity vector would change to be the vector it would have had if the mass lightening system had not been operating (ignoring the effects of gravity, which complicate the analysis). However this violates conservation of energy, unless you have the mass lightening system pumping in and removing energy in the right quantities to match the change in KE. You could directly conserve KE by slowing the ship so that its effective KE (relative to when the mass lightening field was turned on) remains constant whether the field is on or off, but that breaks conservation of momentum. Personally I'd rather have conservation of energy than conservation of momentum, but it depends on the exact physical realisation of the system. The only major sci-fi setting I designed had a hybrid STL/FTL drive system that worked as an effective combined mass lightener/antigravity system/warp drive strictly limited by conservation of energy (extra energy had to be pumped in or removed through the drive to balance things, particularly when traversing gravitational PE gradients). I tried to set this up as carefully as possible to give the tactical environment I wanted (and no easy relativistic planet smashing) while being as physically consistent as possible (though with completely invented fantasy physics).

Isn't conservation of momentum just another way of saying conservation of KE and direction?

[Cyborg Stan]

No. In particular, Conservation of Momentum appears often in discussions of collisions, which are divided into two types - Elastic (KE conserving) and Inelastic (KE not conserved). For example :

We fire a 1g bullet at 1000 m/s into a 100g wooden block at rest. Total KE is (.5*m*v^2) (.5 * .001 kg * 1000^2 (m^2/s^2)) + (.5 * .1 kg * 0^2) = 500 J. Total Momentum would be m*v, or (.001 kg * 1000 m/s) + (.1g * 0 m/s) = 1 kg*m/s. Bullet gets embedded into the block on impact, losing KE to deform it. Momentum is still 1 kg*m/s, but the KE is now .5 * .101kg * (1/.101)^2, or a little under 5 J.

God, I'm bored.

[Connor MacLeod]

Can you really even "pick" between CoE and CoM? I'd imagine the two are pretty closely inter-related with other important aspects of science.

[Kuroneko]

At first I thought it adjust the velocity as well, but it simply doesn't work.

Conservation of momentum suggests that the ship's velocity vector would change to be the vector it would have had if the mass lightening system had not been operating (ignoring the effects of gravity, which complicate the analysis). However this violates conservation of energy, unless you have the mass lightening system pumping in and removing energy in the right quantities to match the change in KE. You could directly conserve KE by slowing the ship so that its effective KE (relative to when the mass lightening field was turned on) remains constant whether the field is on or off, but that breaks conservation of momentum.

It's not an either-or proposition. One will always fail to conserve momentum. In the instantaneously comoving inertial frame before the mass operation, there must be no momentum change. However, the momentum is already zero in that frame, so the change in velocity must also be zero, and therefore zero in every other inertial frame. Thus, there always exist frames in which CoM fails (in fact, almost every frame). On the other hand, CoE fails as well, as increasing the mass amounts to conjuring energy. Perhaps one can hand-wave that away by blaming subspace--the standard Trek pseudo-answer.

[Wyrm]

At first I thought it adjust the velocity as well, but it simply doesn't work.

Conservation of momentum suggests that the ship's velocity vector would change to be the vector it would have had if the mass lightening system had not been operating (ignoring the effects of gravity, which complicate the analysis). However this violates conservation of energy, unless you have the mass lightening system pumping in and removing energy in the right quantities to match the change in KE. You could directly conserve KE by slowing the ship so that its effective KE (relative to when the mass lightening field was turned on) remains constant whether the field is on or off, but that breaks conservation of momentum.

It's not an either-or proposition. One will always fail to conserve momentum. In the instantaneously comoving inertial frame before the mass operation, there must be no momentum change. However, the momentum is already zero in that frame, so the change in velocity must also be zero, and therefore zero in every other inertial frame. Thus, there always exist frames in which CoM fails (in fact, almost every frame). On the other hand, CoE fails as well, as increasing the mass amounts to conjuring energy. Perhaps one can hand-wave that away by blaming subspace--the standard Trek pseudo-answer.

I've been thinking about this, and I've realized that, if the mass lightening field actually works as its name suggests —that is, in the comoving frame, the ship appears to gain mass and not change velocity— then in all other frames, the action of turning off the field must result in the velocity in that frame not changing if both CoE and CoM apply. Here's why:

In the comoving frame, we see the turning off of the field as adding internal energy to the ship (commesurate with the ship's mass energy), bringing it back up to the original mass. However, we can view the ship gaining this mass as it accumulating a four-momentum μmU to add to its own four-momentum m'U just before turning off the field. Thus, μmU + m'U = mU, where U is the four-velocity of the ship in this frame.

What happens in another frame, a frame where the ship is moving? Well, what happens is that the components of the four-momentum μmU change in the new coordinate system. Specifically, there is a Lorentz boost that projects part of the energy component of the four-momentum as spatial linear momentum. Therefore, in a frame where the ship is moving when the field is turned off, we see the ship not only gain internal energy (mass) back, but also see it get a kick of momentum in the direction of motion — just enough momentum is added, in fact, to keep the ship puttering along at the same speed it had when the field was on.

Furthermore, I need not have invoked the above paragraph. In the paragraph preceeding I said, "U is the four-velocity of the ship in this frame." Actually, I could've left off the "in this frame" part completely, because the four-velocity is a geometrical object; it is defined and has an algebra independent of the coordinate system used. Therefore, the equation μmU + m'U = mU holds in all frames of references, which after all are only different coordinate systems, if it holds in one.

So, to answer dragon's question, what would happen is that the ship's velocity would not change in all frames of reference. In comoving frames, the ship would stay still, and in frames where the ship moves, the ship would receive just enough momentum to keep it going at the same speed with the increased mass.

The problem with mass lightening fields in general is more subtle than merely CoE or CoM problems. Both conserve fine. The problem is when you throw angular momentum into the mix. Not so much a definite violation of angular momentum conservation, but rather the requirement that angular momentum be conserved renders mass lightening impotent, at least in the form we have now.

I submit the following example (I did not originate it, just ripped it off... er, adapted it from this webpage: http://www.physicsguy.com/subphys/Subsp ... ics.html#7 — Jason Hinson here suspends disbelief andμ talks about subspace fields as if they were real, except when talking about angular momentum conservation here):

Frame A: Frame A is initially comoving with the ship, and in which the following events in coordinate time of A.

Event 0: A ship at our origin event has internal energy E_0.

x_{ship,0} = (ct,x,y,z)_A = (0,0,0,0)_A
p_{ship,0} = (E/c,p_x,p_y,p_z)_A = (E_0/c,0,0,0)_A

where (ct,x,y,z)_A is a four-position vector with components t, x, y, and z within coordinate system (reference frame) A, and similarly with (E/c,p_x,p_y,p_z)_A a four-momentum vector with components in coordinate system A. Notice that for the four-position, t is multiplied by c — this puts all of the components of the vector in the same units.

Event 1: The ship has turned on its mass lightening field (the webpage example explicitly calls it a subspace field, but this argument applies to any way to temporarily hide mass from the universe in an attempt to fool it). This field hides μE_0 of the total energy of the ship from observers outside the field, thus making the ship appear to have mass of only (1-μ)E_0/c² = (1-μ)m.

The ship then uses its internal energy to create a photon (γ1) of momentum with magnitude |p| traveling in the negative-y direction. At this event, the photon and the ship are still in the same position momentarily, but with the photon moving with a momentum p_y = -p and the ship gaining the opposite momentum p_y = p, they will soon move apart.

x_{ship,1} = (ct_1,0,0,0)_A
p_{ship,1} = (E_1/c,0,p,0)_A

x_{γ1,1} = (ct_1,0,0,0)_A
p_{γ1,1} = (p,0,-p,0)_A

Event 2: The ship has reached its destination and uses more of its internal energy to emit another photon (γ2) in the positive-y direction. Meanwhile, the first photon, γ1, has conitinued moving from the origin at the speed of light.

x_{ship,2} = (ct_2,0,y,0)_A
p_{ship,2} = (E_2/c,0,0,0)_A

x_{γ1,2} = (ct_2,0,-c(t_2-t_1),0)_A
p_{γ1,2} = (p,0,-p,0)_A

x_{γ2,2} = (ct_2,0,y,0)_A
p_{γ2,2} = (p,0,p,0)_A

Event 3: The ship has turned off its mass lightening field and its lost mass has been restored to it. However, it does not have as much internal energy as it had when it began this exercise, as some has been used to create photons. Therefore, its internal energy is now E_3. Meanwhile, the two photons continue on their merry way.

x_{ship,3} = (ct_3,0,y,0)_A
p_{ship,3} = (E_3/c,0,0,0)_A

x_{γ1,3} = (ct_3,0,-c(t_3-t_1),0)_A
p_{γ1,3} = (p,0,-p,0)_A

x_{γ2,3} = (ct_3,0,y+c(t_3-t_2),0)_A
p_{γ2,3} = (p,0,p,0)_A

Adding up all the p_i's, we get

At Event 0: ∑ p_{i,0} = (E_0/c,0,0,0)_A
At Event 3: ∑ p_{i,3} = (E_3/c + 2p,0,0,0)_A

For conservation of energy to apply, E_0 = E_3 + 2pc, or E_0 - E_3 = 2pc — the difference in energy of the ship is equal to the energies of the two photons. If this is true, then p_{ship,0} = p_{ship,3} + p_{γ1,3} + p_{γ2,3}.

Frame B: Frame B is initially moving with velocity v on the x axis relative to A, all of the four-position and four-momentum vectors gain the same Lorentz boost. In this frame, the four-position and -momentum vectors at events 0 and 3 have the following form:

Event 0:
x_{ship,0} = (ct',x',y',z')_B = (0,0,0,0)_B
p_{ship,0} = (E'/c,p'_x,p'_y,p'_z)_B = (γE_0/c,-γβE_0/c,0,0)_B

where (ct',x',y',z')_B is four-position vector with components t, x, y, and z within coordinate system (reference frame) B, and similarly with (E'/c,p'_x,p'_y,p'_z)_B, a four-momentum vector. The components are primed because each coordinate is the result of a Lorentz transformation from frame A to frame B. β = v/c, where v is the velocity of B in A, and γ = 1/(√{1-β²}).

Event 3:
x_{ship,3} = (γct_3,-γβct_3,y,0)_B
p_{ship,3} = (γE_3/c,-γβE_3/c,0,0)_B

x_{γ1,3} = (γct_3,-γβct_3,-c(t_3-t_1),0)_B
p_{γ1,3} = (γp,-γβp,-p,0)_B

x_{γ2,3} = (γct_3,-γβct_3,y+c(t_3-t_2),0)_A
p_{γ2,3} = (γp,-γβp,p,0)_A

At Event 0: ∑ p_{i,0} = (γE_0/c,-γβE_0/c,0,0)_B
At Event 3: ∑ p_{i,3} = (γ[E_3/c + 2p],-γβ[E_3/c + 2p],0,0)_B

Conservation of energy requires E_0 = E_3 + 2pc, which we have already enforced. Therefore CoE applies. Of course, I could have sidestepped the above calculations by stating that, since the four-momenta are geometical objects, their algebra is frame-invariant. Therefore, if CoM and CoE apply in one frame, they apply to all frames.

Now we consider angular momentum. Since all motion is restricted to the xy plane, we need only consider angular momentum along the z axis, L_z(x,p) = x p_y - y p_x. (Note that because the angular momentum is a three-vector, we cannot expect it to have the nice properties under the Lorentz boost as we expect from the four-vectors we have been considering.)

Frame A:
Event 0:
x_{ship,0} = (0,0,0,0)_A
p_{ship,0} = (E_0/c,0,0,0)_A
L_z(<ship,0>_A) = 0 0 - 0 0 = 0

where L_z(<c,e>_A) is the angular momentum of component c at event e in frame A.

Event 3:
x_{ship,3} = (ct_3,0,y,0)_A
p_{ship,3} = (E_3/c,0,0,0)_A
L_z(<ship,3>_A) = 0 0 - y 0 = 0

x_{γ1,3} = (ct_3,0,-c(t_3-t_1),0)_A
p_{γ1,3} = (p,0,-p,0)_A
L_z(<γ1,3>_A) = 0 (-p) - (-c(t_3-t_1)) 0 = 0

x_{γ2,3} = (ct_3,0,y+c(t_3-t_2),0)_A
p_{γ2,3} = (p,0,p,0)_A
L_z(<γ2,3>_A) = 0 p - (y+c(t_3-t_2)) 0 = 0

At Event 0: ∑ L_z(<i,1>_A) = 0
At Event 3: ∑ L_z(<i,3>_A) = 0

Angular momentum is easily conserved in this frame.

Frame B:
Event 0:
x_{ship,0} = (0,0,0,0)_B
p_{ship,0} = (γE_0/c,-γβE_0/c,0,0)_B
L_z(<ship,0>_A) = 0 0 - 0 (-γβE_0/c) = 0

Event 3:
x_{ship,3} = (γct_3,-γβct_3,y,0)_B
p_{ship,3} = (γE_3/c,-γβE_3/c,0,0)_B
L_z(<ship,3>_B) = (-γβct_3) 0 - y (-γβE_3/c) = yγβE_3/c

x_{γ1,3} = (γct_3,-γβct_3,-c(t_3-t_1),0)_B
p_{γ1,3} = (γp,-γβp,-p,0)_B
L_z(<γ1,3>_B) = (-γβct_3)(-p) - (-c(t_3-t_1))(-γβp) = γβpct_1

x_{γ2,3} = (γct_3,-γβct_3,y+c(t_3-t_2),0)_A
p_{γ2,3} = (γp,-γβp,p,0)_A
L_z(<γ2,3>_B) = (-γβct_3)p - (y+c(t_3-t_2))(-γβp) = γβp(y-ct_2)

At Event 0: ∑ L_z(<i,1>_B) = 0
At Event 3: ∑ L_z(<i,3>_B) = yγβE_3/c + γβpct_1 + γβp(y-ct_2)
= γβ[yE_3/c + pct_1 + p(y-ct_2)]
= γβ[yE_3/c + pct_1 + py - pct_2]
= γβ[y(E_3/c + p) - pc(t_2-t_1)]

These two angular momenta must be equal, so
y(E_3/c + p) - pc(t_2-t_1) = 0
or
y(E_3/c + p) = pc(t_2-t_1)

From our requirement of CoE, E_0/c = E_3/c + 2p implies E_0/c - p = E_3/c + p, so we use this to simplify the equation:

yE_0/c - yp = pc(t_2-t_1)
yE_0/c = p[y + c(t_2-t_1)]
E_0 = pc[1 + c(t_2-t_1)/y]
pc = E_0/[1 + c(t_2-t_1)/y]

So the energy of the photons depend only on the beginning internal energy (rest mass) of the ship, the time elapsed between the emissions of the photons, and the distance traveled. Nowhere in this equation does E_1 or E_2 show up, much less μ. Nothing has been gained by using the mass-lightening field.

Is this it for mass lightening fields? Well, no. Not quite. Between events 0 and 1, a proportion μ of the ship's four-momentum disappears in all frames. Simiarly, between events 2 and 4, the ship's four-momentum reappears. Although in the long run four-momentum was conserved, from somewhere between events 0 and 3, the universe was missing four-momentum of μmU. If we were demand four-momentum conservation at all times, then where did this four-momentum μmU go? Where did the four-momentum come from to restore the ship's mass?

We propose that spacetime is filled with a field that acts as a reservoir of four-momentum (I call it the "mass reservoir", 'mass' because "four-momentum reservoir" is unwieldy), and the transfer of four-momentum is controlled by the mass lightening field — though it probably should be called a mass shunting field. The μmU the ship lost when the field turned on was absorbed by this mass reservoir. The four-momentum reasborbed by the ship when the field turned off came from this reservoir. However, if turning off the field in all situations leaves the mass reservoir completely unchanged, then we are left with the same problem with angular momentum conservation. However, we may demand that the changes wrought by the shunt field to be local. This solves our problem, because if we want savings on energy moving things around with mass ligthening fields, the mass reservoir must be asked to absorb some angular momentum without changing its total four-momentum. This is only possible if the four-momentum is rearranged in the mass reservoir as a function of space.

Let us suppose that the mass reservoir is a four-momentum field where each point in space has a defined four-momentum density, π(x). We may integrate this over a region of space to get a four-momentum for that region, p(R), from which we can extract a mass and a four-velocity for that region of the mass reservoir. The region occupied by the mass shunt field forms such a region, and the shunt field would operate in such a way to increase the mass reservoir within the region it occupies by μ times the four-momentum it encloses in real space. If we demand that this phenomenon conserves four-momentum, then this four-momentum must come from the four-momentum the region encloses in real space, so that four-momentum reduces to 1-μ its former value.

This means that when the field is turned on prior to event 1, there is at that event a lump of four-momentum sitting in the mass reservoir a region about that point above that of the four-momentum the reservoir carried at event 0. After event 2, the ship reabsorbes four-momentum from the mass reservoir at its new coordinates to bring its mass back up. If this lump created before event 1 were carried with the ship to be reabsorbed by the ship, then the overall distribution of four-momentum in the mass reservoir does not change, and so it cannot absorb any angular momentum (and we're right back where we started). Therefore, we must leave the four-momentum we left behind in event 1, and create a dip in the four-momentum at our new location. This redistributes the four-momentum of the mass reservoir without changing the total four-momentum of the reservoir.

Going back to our example, let's take into account the redistribution of four-momentum in the mass reservoir.

Frame A:

Event 0: As with the previous example, but we add the following statements about the condition of the mass reservoir, both at the origin of our journey and our (currently projected) destination.

x(R1)_0 = (0,0,0,0)_A
∆p(R1)_0 = (0,0,0,0)_A
L_z(<∆,R1,0>_A) = 0 0 - 0 0 = 0

x(R2)_0 = (0,0,y,0)_A
∆p(R1)_0 = (0,0,0,0)_A
L_z(<∆,R2,0>_A) = 0 0 - y 0 = 0

We express the four-momentum of the reservoir at each event as a delta from a base, undisturbed four-momentum distribution that may vary with space and time. Fortunately, the cross-product distributes across vector addition, so L_z(x,p_1 + p_2) = L_z(x,p_1) + L_z(x,p_2).

Event 0.5: Between event 0 and event 1, the shunt field activates, transfering four-momentum of μp_{ship,0} from the ship to R1 of the mass reservoir. The remainder of the momentum is kept by the ship. There is no change to the four-momentum of R2, except the changes wrought by forces within the mass reservoir itself.

x_{ship,0.5} = (ct_{0.5},0,0,0)_A
p_{ship,0.5} = ((1-μ)E_0/c,0,0,0)_A

x(R1)_{0.5} = (ct_{0.5},0,0,0)_A
∆p(R1)_{0.5} = (μE_0/c,0,0,0)_A

Event 2.5: Between event 2 and 3, the shunt field shuts down, transfering four-momentum of μp_{ship,0} from R2 of the mass reservoir to the ship. The transfer conserves four-momentum. There is no change to the four-momentum of R1, except the changes wrought by forces within the mass reservoir itself. We ignore the two photons for the moment

x_{ship,2} = (ct_{2.5},0,y,0)_A
p_{ship,2} = (E_3/c,0,0,0)_A

x(R2)_{2.5} = (ct_{2.5},0,y,0)_A
∆p(R2)_{2.5} = (-μE_0/c,0,0,0)_A

Event 3: When it's over, the ship and photons are as they are described in the original Event 3. In addition, we have two regions of the mass reservoir that have changed four-momentum and need to have their angular momentum accounted for:

x(R1)_3 = (ct_3,0,0,0)_A
∆p(R1)_3 = (μE_0/c,0,0,0)_A
L_z(<∆,R1,3>_A) = 0 0 - 0 0 = 0

x(R2)_3 = (ct_3,0,y,0)_A
∆p(R2)_3 = (-μE_0/c,0,0,0)_A
L_z(<∆,R1,3>_A) = 0 0 - y 0 = 0

Note that ∆p(R1)_3 + ∆p(R2)_3 = 0 and L_z(<∆,R1,3>_A) + L_z(<∆,R2,3>_A) = 0, so the fact that the ship's method of travel conserves energy, and linear and angular momentum does not change.

Frame B:
Event 0:
x_{ship,0} = (0,0,0,0)_B
p_{ship,0} = (γE_0/c,-γβE_0/c,0,0)_B
L_z(<ship,0>_A) = 0 0 - 0 (-γβE_0/c) = 0

x(R1)_0 = (0,0,0,0)_B
∆p(R1)_0 = (0,0,0,0)_B
L_z(<∆,R1,0>_B) = 0 0 - 0 0 = 0

x(R2)_0 = (0,0,y,0)_B
∆p(R1)_0 = (0,0,0,0)_B
L_z(<∆,R2,0>_B) = 0 0 - y 0 = 0

Event 3:
x_{ship,3} = (γct_3,-γβct_3,y,0)_B
p_{ship,3} = (γE_3/c,-γβE_3/c,0,0)_B
L_z(<ship,3>_B) = (-γβct_3) 0 - y (-γβE_3/c) = yγβE_3/c

x_{γ1,3} = (γct_3,-γβct_3,-c(t_3-t_1),0)_B
p_{γ1,3} = (γp,-γβp,-p,0)_B
L_z(<γ1,3>_B) = (-γβct_3)(-p) - (-c(t_3-t_1))(-γβp) = γβpct_1

x_{γ2,3} = (γct_3,-γβct_3,y+c(t_3-t_2),0)_A
p_{γ2,3} = (γp,-γβp,p,0)_A
L_z(<γ2,3>_B) = (-γβct_3)p - (y+c(t_3-t_2))(-γβp) = γβp(y-ct_2)

x(R1)_3 = (γct_3,-γβct_3,0,0)_A
∆p(R1)_3 = (γμE_0/c,-γβμE_0/c,0,0)_A
L_z(<∆,R1,3>_A) = (-γβct_3) 0 - 0 (-γβμE_0/c) = 0

x(R2)_3 = (γct_3,-γβct_3,y,0)_A
∆p(R2)_3 = (-γμE_0/c,γβμE_0/c,0,0)_A
L_z(<∆,R1,3>_A) = (-γβct_3) 0 - y (γβμE_0/c) = -yγβμE_0/c

At Event 0: ∑ L_z(<i,1>_B) = 0
At Event 3: ∑ L_z(<i,3>_B) = yγβE_3/c + γβpct_1 + γβp(y-ct_2) + 0 + (-yγβμE_0/c)
= γβ[y(E_3/c + p) - pc(t_2-t_1) - yμE_0/c]
= γβ[y(E_3/c + p - μE_0/c) - pc(t_2-t_1)]

These two angular momenta must be equal, so
y(E_3/c + p - μE_0/c) - pc(t_2-t_1) = 0
or
y(E_3/c + p - μE_0/c) = pc(t_2-t_1)

From our requirement of CoE, E_0/c = E_3/c + 2p implies E_0/c - p = E_3/c + p, so we use this to simplify the equation:

yE_0/c - yp - μE_0/c = pc(t_2-t_1)
y(E_0-μE_0)/c = p[y + c(t_2-t_1)]
(1-μ)E_0 = pc[1 + c(t_2-t_1)/y]
pc = (1-μ)E_0/[1 + c(t_2-t_1)/y]

Here, at last, is the gold! We have an energy dependence on μ, and furthermore, the savings are in proportion to (1-μ). It's like, instead of dragging the whole mass of the ship m around, you drop your μm backpack off at your start point and pick up an identical μm backpack at your destination, so you only have to drag your (1-μ)m self about (the larger μ is, the better).

This, of course, assumes that the ship would regain what it had lost, which assumes the mass shunt field "remembers" the four-momentum it gave to the mass reservoir, like a receipt. Things change when it does not. However, this seems a good place to stop for now.

[bz249]

Sorry why the conservation of energy is so critical question? In this case we only talk about the conservation of kinetic energy, and as far as I know there is no conservation law for that (oh sorry there is a weak one, that conservative fields conserve the total mechanical energy). So the simpliest answer to the question that the mass lightening conserve the linear momentum, but not the kinetic energy. It uses some external energy source (to pump in the required energy) during accelaration and the excess energy simply dissipates during a deadstop.

[Wyrm]

Sorry why the conservation of energy is so critical question?

Because without it you cannot guarantee that the universe will work the same way tomorrow as it did today. It's called "Noether's theorem" — for every continuous symmetry (time translation being one), you can derive one conservation law. From time symmetry (the guarantee that you can perform a physics experiment at any time and come up with the same result) you get the conservation of energy. Wihtout conservation of energy, if you turn on a light today and illuminate the room today, you can't guarantee that turning on the same light tomorrow would not cover the entire room in cocoa butter.

In this case we only talk about the conservation of kinetic energy, and as far as I know there is no conservation law for that (oh sorry there is a weak one, that conservative fields conserve the total mechanical energy).

Kinetic energy is just the excess in internal energy resulting from projecting the moving object's four-momentum onto your own time basis vector. It's a trick of coordinate systems, so it shouldn't be surprising that it isn't conserved. Four-momentum, on the other hand, is conserved. Always.

So the simpliest answer to the question that the mass lightening conserve the linear momentum, but not the kinetic energy. It uses some external energy source (to pump in the required energy) during accelaration and the excess energy simply dissipates during a deadstop.

Dead stop relative to whom, bz249? If in one reference frame, the ship goes from traveling at a certain velocity v to a dead stop, then I can always find a frame such that, in this frame, the ship goes from dead stop to traveling at an equal and opposite velocity, -v. Whence did this kinetic energy come from, dearheart? Futhermore, you say that the ship goes to a dead stop and dissipates the kinetic energy (presumably as heat) which is radiated. In the frame where the ship goes from dead stop to -v, the ship also radiates energy. Where did this energy come from?

I need not accept answers that are clearly broken, no matter how simple they appear.

[Kuroneko]

I've been thinking about this, and I've realized that, if the mass lightening field actually works as its name suggests —that is, in the comoving frame, the ship appears to gain mass and not change velocity— then in all other frames, the action of turning off the field must result in the velocity in that frame not changing if both CoE and CoM apply.

Yes, that's exactly what I said. Although you don't actually need both CoE and CoM for the conclusion that the change in velocity is zero in every inertial frame.

In comoving frames, the ship would stay still, and in frames where the ship moves, the ship would receive just enough momentum to keep it going at the same speed with the increased mass. ... The problem with mass lightening fields in general is more subtle than merely CoE or CoM problems. Both conserve fine.

Those statements are contradictory... perhaps you mean to ignore the intermediate steps of the subsequent scenario (as you do later), but at this point your meaning is rather unclear. There is a violation of momentum conservation at several steps in the experiment (e.g., between events 2 and 3); your invocation of an external field is necessary not to just conserve angular momentum, but also momentum and energy.

We propose that spacetime is filled with a field that acts as a reservoir of four-momentum (I call it the "mass reservoir", 'mass' because "four-momentum reservoir" is unwieldy), and the transfer of four-momentum is controlled by the mass lightening field ...

Well, it seems to work on a formal level, but it also makes mass lightening completely unnecessary. One could just as easily extract an arbitrary amount of four-momentum and put it back later, according to this scheme.

Let us suppose that the mass reservoir is a four-momentum field where each point in space has a defined four-momentum density, π(x). We may integrate this over a region of space to get a four-momentum for that region, p(R), from which we can extract a mass and a four-velocity for that region of the mass reservoir. ...

Although not on a curved spacetime.

It is more than a little bit disconcerting to have a reservoir of `stuff' that has energy and momentum but that either does not move itself at all or one that costs no energy to accelerate. In the end, what's being "shunted" is not the mass or four-momentum but the all conservation problems in the physical universe.

[metavac]

We propose that spacetime is filled with a field that acts as a reservoir of four-momentum (I call it the "mass reservoir", 'mass' because "four-momentum reservoir" is unwieldy), and the transfer of four-momentum is controlled by the mass lightening field ...

Well, it seems to work on a formal level, but it also makes mass lightening completely unnecessary. One could just as easily extract an arbitrary amount of four-momentum and put it back later, according to this scheme.

Just dropping in, but if I understand this correctly, Wyrrn is just describing a rocket and a net. That's how the math works out for me, at least.

[Wyrm]

Yes, that's exactly what I said. Although you don't actually need both CoE and CoM for the conclusion that the change in velocity is zero in every inertial frame.

Yes, I see that now.

Those statements are contradictory... perhaps you mean to ignore the intermediate steps of the subsequent scenario (as you do later), but at this point your meaning is rather unclear. There is a violation of momentum conservation at several steps in the experiment (e.g., between events 2 and 3); your invocation of an external field is necessary not to just conserve angular momentum, but also momentum and energy.

I admit to some clunky wording. However, I protest that there were "several" violations of CoM. There were only two, when the field was turned on and when the field was turned off. And the second was an undoing of the first violation; when the field turned off, all of the missing momentum should be accounted for.

Well, it seems to work on a formal level, but it also makes mass lightening completely unnecessary. One could just as easily extract an arbitrary amount of four-momentum and put it back later, according to this scheme.

The ship's lighter with the field on, right?

I would say that making the ship mass that is lost actually disappear completely from our reckoning is beside the point and a completely unnecessary requirement. The "mass lightening field" is being used for its practical effect — that the ship's mass is reduced for as long as the field is turned on. I say, I don't care where the mass goes in the universe, so long as it isn't in the ship to bog it down when we accelerate, and we can restore the mass on demand by turning the field off.

You must admit that a field that makes everything inside appear less massive is fanciful enough already, even if the mass defect has to go somewhere else for a time. Ignoring the parts when the mass was vanished away and was restored allowed me to gloss over the process, and ask what was needed to make an energy savings possible.

Let us suppose that the mass reservoir is a four-momentum field where each point in space has a defined four-momentum density, π(x). We may integrate this over a region of space to get a four-momentum for that region, p(R), from which we can extract a mass and a four-velocity for that region of the mass reservoir. ...

Although not on a curved spacetime.

'Ey, if I wanted to bring curved spacetime into the discussion, I would've at least mentioned the metric.

It is more than a little bit disconcerting to have a reservoir of `stuff' that has energy and momentum but that either does not move itself at all or one that costs no energy to accelerate.

Like with the first calculation involving the angular momentum not taking account of the angular momentum of the mass reservoir, I'm glossing over the dynamics of the reservoir itself. I wouldn't be surprised if the mass reservoir extracted an energy premium for being torqued. (Just as long as it's not too big.)

In the end, what's being "shunted" is not the mass or four-momentum but the all conservation problems in the physical universe.

True, but the neutrino was proposed for similar reasons, wasn't it? Physicists didn't want to give up on CoM and CoE for beta decay? The assumption of CoM, CoE, and conservation of angular momentum, plus an assumption of energy savings as a function of the proportion of mass hidden leads us to this conclusion.

Just dropping in, but if I understand this correctly, Wyrrn is just describing a rocket and a net. That's how the math works out for me, at least.

Not quite. It's more like leaving a large ballast behind at your start point, then picking up an identical ballast at your destination to restore your original mass.

And the username's Wyrm, dammit. A 'wyrrn' is a huge fuck-off space bug that lays its eggs in chief engineers. (Alien stole its moves from Doctor Who.)

[metavac]

Just dropping in, but if I understand this correctly, Wyrrn is just describing a rocket and a net. That's how the math works out for me, at least.

Not quite. It's more like leaving a large ballast behind at your start point, then picking up an identical ballast at your destination to restore your original mass.

I don't think that dog hunts. A ballast analogy places a massive body in a massive scalar field coupled to a repulsive electrostatic one. An action in which ballast is release at t = t_0, r = r_0 and is recovered some t = T, r = R later will result in the body returning to r = r_0. You want to recover less mass at r = R so that r'(t) = 0 relative to the field at r = R.

And the username's Wyrm, dammit. A 'wyrrn' is a huge fuck-off space bug that lays its eggs in chief engineers. (Alien stole its moves from Doctor Who.)[/quote]

Sorry about that. Wyrm as in dragon. Got it.

[Wyrm]

I don't think that dog hunts. A ballast analogy places a massive body in a massive scalar field coupled to a repulsive electrostatic one. An action in which ballast is release at t = t_0, r = r_0 and is recovered some t = T, r = R later will result in the body returning to r = r_0. You want to recover less mass at r = R so that r'(t) = 0 relative to the field at r = R.

You do know that analogies fail at some point, do you not? And that's why you shouldn't reason with them and only use as a tool to intuitive understanding.

Anyway, you're even using the wrong analogy. There are two distinct but identical ballasts, one at A and another at B. You drop one ballast at A, move to B, and pick up the one sitting at B. Clear?

[Kuroneko]

I admit to some clunky wording. However, I protest that there were "several" violations of CoM. There were only two, when the field was turned on and when the field was turned off.

Mea culpa--yes, there are exactly two in that scenario.

The ship's lighter with the field on, right? I would say that making the ship mass that is lost actually disappear completely from our reckoning is beside the point and a completely unnecessary requirement.

Yes, but that's not at all what I was alluding to. How about simply extracting four-momentum from the field and putting it back when one is done? Such an operation would be for most practical purposes equivalent to the above scenario, just in opposite order.

You must admit that a field that makes everything inside appear less massive is fanciful enough already, ...

Yes. That's why I was proposing to simply skip that step. The name would be nothing but a fanciful analogy.

True, but the neutrino was proposed for similar reasons, wasn't it? Physicists didn't want to give up on CoM and CoE for beta decay?

In this case, the gap is too large, and has other problems as well. For example, the most obvious application of mass-lightening technology is to use on a gas reservoir, thus decreasing its temperature and pressure. If turning off the field restores original energy and momentum of the gas particles, this allows us to simply destroy entropy, since there we do not need to do any work to compress a gas. If the operation is affected by volume, we can keep the same volume and still use the lightened gas as a heat sink, still violating the second law.

[metavac]

I don't think that dog hunts. A ballast analogy places a massive body in a massive scalar field coupled to a repulsive electrostatic one. An action in which ballast is release at t = t_0, r = r_0 and is recovered some t = T, r = R later will result in the body returning to r = r_0. You want to recover less mass at r = R so that r'(t) = 0 relative to the field at r = R.

You do know that analogies fail at some point, do you not? And that's why you shouldn't reason with them and only use as a tool to intuitive understanding.

Yes, I do, and I suspect the reason you went with that analogy is to get the reader to think of a spaceship shedding its mass imparted with negligible momentum with respect to hull. Problem is that there's another piece of the system in play, that electrostatic field, that will now do work on the ship. By the way, the ballast analogy works best with a solar or laser sail.

Anyway, you're even using the wrong analogy. There are two distinct but identical ballasts, one at A and another at B. You drop one ballast at A, move to B, and pick up the one sitting at B. Clear?

Which, like I pointed out, would lead to the ship circuiting B right back to A, provided the analogy holds. We can modify it by eliminating the coupled potential gravitational and electrostatic fields, but now you're just hurling mass out (a rocket) and hoping to run into another body with the right momentum (a net) to stop. That is, of course, unless we simply accept the violation in energy-momentum conservation that you and kuroneko pointed out.

[skotos]

We can modify it by eliminating the coupled potential gravitational and electrostatic fields, but now you're just hurling mass out (a rocket) and hoping to run into another body with the right momentum (a net) to stop.

That is more or less how Star Trek treats subspace. It even offers the beginning of a an explanation to the myriad "sub-space anaomalies" that crop up in Star Trek. Let's suppose that one can usually hurl mass out to accelerate and find the corresponding mass to stop at one's destination. Occasionally one would come across an area of space where the corresponding subspace did not allow you to "pick up your ballast". In those situations the ship would move faster of slower or sideways, or would heat up or cool down, or suffer numerous other effects.

To summarize, I suspect that "ballast analogy" can be used to not only explain mass lightening, but also to explain some or most of the various subspace anomalies that we see in Star Trek. But I will leave it to you and Wrym and Kuroneko to hash it out or show that I am an idiot.

[Wyrm]

Yes, but that's not at all what I was alluding to. How about simply extracting four-momentum from the field and putting it back when one is done? Such an operation would be for most practical purposes equivalent to the above scenario, just in opposite order.

Um, first, please go into more detail on how such a strategy is to work. How does borrowing energy from the reservoir reduce the amount of energy total that the ship has to spend to get from A to B?

Second, even if there exists a theoretical field configuration that allows such schenannighans, that doesn't mean that the appropriate field is necessarily physical.

In this case, the gap is too large, and has other problems as well. For example, the most obvious application of mass-lightening technology is to use on a gas reservoir, thus decreasing its temperature and pressure. If turning off the field restores original energy and momentum of the gas particles, this allows us to simply destroy entropy, since there we do not need to do any work to compress a gas. If the operation is affected by volume, we can keep the same volume and still use the lightened gas as a heat sink, still violating the second law.

And if all gasses are affected similarly, then the crew would be surrounded by supercooled air and freeze to death, if not experience a temperature drop in their own bodies. Either way, they're crewcicles. This makes the field useless for its intended purpose, carting ship and crew around a solar system cheeply.

Since this doesn't happen for subspace fields (and let's stop pretending that we're not talking about them), either this isn't how subspace fields work, or something more subtle is going on. One of the things I dutifully noted in my long-ass post was that the mass of the ship when viewed from outside the field has its apparent mass (internal energy) reduced. The above requirement of not turning the crew into crewcicles (and since a lot of particle physics depends on individual particle mass, the requirement that our crew remains alive) seems to dictate that the physics that takes place wholey within the field remains the same.

If the physics performed inside the field remain the same, then you cannot cool the gas by putting in a subspace field, so the person outside sees a gas that appears to be at a lower internal energy, but sees it behave thermodynamically like a gas at higher temperature. Then again, he's veiwing the experiment through the subspace field which he can also measure.

You do know that analogies fail at some point, do you not? And that's why you shouldn't reason with them and only use as a tool to intuitive understanding.

Yes, I do, and I suspect the reason you went with that analogy is to get the reader to think of a spaceship shedding its mass imparted with negligible momentum with respect to hull. Problem is that there's another piece of the system in play, that electrostatic field, that will now do work on the ship.

I don't think we're talking about the same kind of ballast, dearheart. You seem to be talking about ballast as used in electrical engineering, especially in flourecent lights where the "ballast" is indeed an induction coil. I'm talking about "ballast" as used in the shipboard sense.

bal·last (blst)
n.

1. Heavy material that is placed in the hold of a ship or the gondola of a balloon to enhance stability.
2. 1. Coarse gravel or crushed rock laid to form a bed for roads or railroads.
   2. The gravel ingredient of concrete.
3. Something that gives stability, especially in character.

Ie, I when I say "ballast," I mean a chunk of lead, or a stone. And if you think a real spaceship wouldn't need ballast, think again. Ballast on a spaceship would keep the center of mass right smack where it should be, along the axis of thrust. Otherwise, your ship will spin about as a lever arm forms.

Is that clearer?

We can modify it by eliminating the coupled potential gravitational and electrostatic fields, but now you're just hurling mass out (a rocket) and hoping to run into another body with the right momentum (a net) to stop.

No. Read the example again. We don't "hurl" mass into the reservoir or "run into" it. In the comoving frame, shunting mass to and from the field gives no acceleration, and therefore change in speed, as it would for a net or rocket.

[metavac]

Yes, I do, and I suspect the reason you went with that analogy is to get the reader to think of a spaceship shedding its mass imparted with negligible momentum with respect to hull. Problem is that there's another piece of the system in play, that electrostatic field, that will now do work on the ship.

I don't think we're talking about the same kind of ballast, dearheart. You seem to be talking about ballast as used in electrical engineering, especially in flourecent lights where the "ballast" is indeed an induction coil. I'm talking about "ballast" as used in the shipboard sense.

Ie, I when I say "ballast," I mean a chunk of lead, or a stone. And if you think a real spaceship wouldn't need ballast, think again. Ballast on a spaceship would keep the center of mass right smack where it should be, along the axis of thrust. Otherwise, your ship will spin about as a lever arm forms.

Is that clearer?

Yes, and we were talking about similar things, although I was talking about ballast to control buoyancy and I have no idea why anyone would think a sailing ballast analogy has much of anything to do with changes in motion along a desired axis.

Here's the problem as I understand it. You're looking a ship whose captain wants to change velocity relative to the comoving frame you're in. You can do this by adding or shedding mass-energy, but to conserve the 4-current, the energy-momentum of the shed mass-energy must sum with the energy momentum of the ship after the field is turned on to give you what you measured before you turned the field on. That, my friend, is exactly what a rocket does.

Now as I understand it, kuroneko (and Hinson) for that matter, argue that global spacetime is foliated by a bundle such that the product of any tangent vector at a given point with a vector in the corresponding space is non-zero. That's to say, there exists at every point in spacetime a space of vectors with non-zero momentum.

Is that clearer?

[Wyrm]

Yes, and we were talking about similar things, although I was talking about ballast to control buoyancy and I have no idea why anyone would think a sailing ballast analogy has much of anything to do with changes in motion along a desired axis.

It doesn't, except that dropping the ballast makes your ship easier to move. That's the point.

Here's the problem as I understand it. You're looking a ship whose captain wants to change velocity relative to the comoving frame you're in. You can do this by adding or shedding mass-energy, but to conserve the 4-current, the energy-momentum of the shed mass-energy must sum with the energy momentum of the ship after the field is turned on to give you what you measured before you turned the field on. That, my friend, is exactly what a rocket does.

Except the field doesn't actually provide acceleration, as would be required for a rocket. The shedding of mass/4-momentum by way of the field and the acceleration that moves the ship are accomplied with two different mechanisms. Activating and deactivating the field here yields no change in velocity in any frame.

Shedding 4-momentum does not a rocket make. For real rockets, modifying the 4-momentum current is a necessary condition, but not a sufficient condition. Fundamentally, rockets cause acceleration, whch cause modification of the 4-velocity, and it's not hard to show that all accelerations on an object in a frame are orthogonal to the 4-velocity. This doesn't happen with the turning on and off of the field. The field only scales the 4-momentum of the ship (which is itself is a scaling of the 4-velocity by the ship's rest mass) and obviously has no component orthoginal to it.

Now as I understand it, kuroneko (and Hinson) for that matter, argue that global spacetime is foliated by a bundle such that the product of any tangent vector at a given point with a vector in the corresponding space is non-zero. That's to say, there exists at every point in spacetime a space of vectors with non-zero momentum.

Irrelevant, because the operation of the field does nothing to the ship but scale its 4-momentum. It therefore does not cause acceleration in the ship and therefore the analogy with the field operation with a rocket fails hard.