# Applied momentum

1. Apr 9, 2012

### grav-universe

Let's say that a ship makes small jumps in speed with a momentum measured by the ship observers as p each time. The first jump from the rest frame brings them to a speed of v. They continue to do this n times, so the the total momentum applied locally according to them is n * p, right? After n jumps, their new speed relative to the rest frame is u. What is the equation that the rest frame would use to determine what the total locally applied momentum n * p is according to the ship observers in terms of the speed u the rest frame now observes?

2. Apr 9, 2012

### Staff: Mentor

Observers in the ship will alway measure a momentum of 0 for the ship. You can define p as the momentum of the ship after a "jump", measured in the system before the jump. In that case, the observers in the ship will always have the same velocity change v, and you can use the formula for relativistic velocity addition for each jump:
$$v_n=\frac{v_{n-1}+v}{1+v_{n-1}v/c^2}$$
with v0=0. Maybe this can be expressed in a closed form, and maybe it helps to modify this to get a differential equation.

3. Apr 9, 2012

### grav-universe

Right, I used relativistic addition of speeds to find a relation, but the ship observers measuring zero momentum afterward is not what is mean, but that which is applied.

Let's try this question first. Let's say that the ship expends some amount of energy to get to a small speed of v. According to the rest frame, the energy of the ship is m v^2 / 2 classically, small enough to be non-relativistic at this point, and that should equal the energy expended by the ship, correct? Now the ship expends the same amount of energy again to reach a speed of approximately 2 v. According to the rest frame, then, the energy of the ship is now 4 times greater with m (2 v)^2 /2 , but the energy expended by the ship is only twice as great. So how do we relate the energy of the ship as observed by the rest frame in accordance with the energy expended by the ship?

Last edited: Apr 9, 2012
4. Apr 9, 2012

### Staff: Mentor

Only if the ship has something with infinite mass in the observer frame to push away from. A real ship with current physics would have to accelerate something in the opposite direction. Usually, this energy is much larger than the kinetic energy gained by the ship.

In that case, the energy has to be larger. Rocket drives get more efficient (in terms of gained kinetic energy per fuel and per reaction mass) if the rocket has a large velocity in the direction of the acceleration. But you have to carry all this fuel and reaction mass in the rocket, and a lot of its energy is lost in space.

5. Apr 9, 2012

### Bill_K

The problem you describe as "a series of small jumps in velocity" is the uniformly accelerating particle, a standard part of any relativity course.

The solution is a hyperbola, xμ = (A cosh Bτ, A sinh Bτ) where A = c2/a and B = a/c, and a is the proper acceleration.

6. Apr 9, 2012

### Staff: Mentor

This is an interesting question, but both answers so far would be for a ship whose mass is not changing during the acceleration, which is not how I would have understood the question at first glance.

grav-universe, can you clarify your intention? Are you thinking of a rocket ejecting fuel such that the change in momentum is equal each time in the rocket's frame and the mass is diminishing, or are you thinking of a ship being pushed by some external mechanism such that the change in momentum is equal each time in the external mechanism's frame and the ships mass is constant.

7. Apr 9, 2012

### Austin0

Would this hold true as instantaneous velocity reached significant relativistic levels??

When coordinate acceleration was dropping off in the reference frame??

8. Apr 10, 2012

### Staff: Mentor

Well... use a small δv, put it in the relativistic velocity addition, calculate the kinetic energy as (γ-1)*m and evaluate dγ/dv.

Using c=1:

$$\delta v_{obs}=\frac{v_{obs}+\delta v}{1+v_{obs} \cdot \delta v} - v_{obs} \approx (v_{obs} + \delta v)(1-v_{obs}\cdot \delta v) - v_{obs} \approx \delta v (1 - v_{obs}^2)$$
Written as derivative, $\frac{dv_{obs}}{dv}=1-v^2_{obs}$, but keep in mind that "v" as absolute value has no intuitive meaning.

$$\frac{d\gamma}{dv_{obs}}=v_{obs}(1-v_{obs}^2)^{-3/2}$$

$$\frac{d\gamma}{dv}=\frac{d\gamma}{dv_{obs}} \frac{dv_{obs}}{dv} = v_{obs}(1-v_{obs}^2)^{-1/2} = v_{obs}\gamma$$

This is increasing with the velocity of the rocket.
However, as rockets have to carry and accelerate their own reaction mass, their overall performance is usually bad compared to other ideas.

9. Apr 10, 2012

### grav-universe

Good responses, thank you. Okay, so the ship achieves a relative speed to the rest frame of v after one jump and v_n after n jumps. Working through the addition of relativistic speeds formula, the relation works out to

[(1 - v/c) / (1 + v/c)]^n = 2 / (1 + v_n/c) - 1

[(1 - v/c) / (1 + v/c)]^n = (1 - v_n/c) / (1 + v_n/c)

ln[(1 + v/c) / (1 - v/c)] n = ln[(1 + v_n/c) / (1 - v_n/c)]

ln[(1 + v/c) / (1 - v/c)] = ln[(1 + v_n/c) / (1 - v_n/c)] / n

This quantity, depending only upon the final speed and the number of jumps, remains constant since ln[(1 + v/c) / (1 - v/c)] on the left side of the equation is constant, and if we multiply the left side by m c / 2 with small v, that would be the momentum gained, that which would be gained to weak approximation to match Newtonian with non-relativistic speed.

Okay, so here's my first question. If this is the momentum gained for small v after one jump, and if the ship observers say this is the amount of momentum that has been applied to get them at that speed, then wouldn't they say n times that momentum has been applied after n jumps? I understand this applied momentum might be considered different from classical or even relativistic momentum, but it is just that amount that has been "applied". Is there a better term for this in physics?

In this manner of thinking about it, if (m c / 2) ln[(1 + v/c) / (1 - v/c)] is the momentum gained after one jump to small v, then (m c / 2) ln[(1 + v/c) / (1 - v/c)] n would be the "applied" momentum after n jumps. But this is then just equal to ( m c / 2) ln[(1 + v_n/c) / (1 - v_n/c)] for any n according to the above relation, so applies to all speeds, large or small.

Sorry to ramble on but this relation keeps popping up in my work and I'm wondering how I should think about it, how it fits into Relativity in a sensible way. A while back I found the identical relation using collisions of elastic bodies, which I then started a thread about but it was promptly removed because I had determined what I claimed to be new laws of conservation of momentum and energy based upon the results, mistakenly stating that Relativity did not uphold these laws, not realizing until asking about it afterward that Relativity maintains the laws using relativistic momentum and energy by heating up bodies when they collide, adding mass, until they lose the heat and regain their original masses.

The reason I mention this last bit about the conservation of energy of momentum and energy is because I also have questions about that which relate to what I have been working on as well. I thought about going back to the other thread I was asking about it, but here is just as well, where also a better idea can be gained about precisely what I am driving at. For instance, when the ship is constantly accelerating, why doesn't it heat up and gain mass in the same way as with collisions? I mean, the same relation for "applied" momentum is found in both cases, albeit without assuming mass gain or heat loss in deriving them. Secondly, if there is heat loss in collisions, why does Relativity include it beforehand in the original set of laws of conservation? That is, shouldn't heat gain or loss be considered after the fact, whereas the original laws assume no loss for the ideal situation, and heat loss would then subtract from the final kinetic energies of the bodies, as we would for classical, for instance? Finally, shouldn't the amount of heat gain or loss be determined by the nature of the interaction rather than according to the original laws of conservation under ideal conditions?

Again, just to be clear and safe, I am not claiming this to be the true form of relativistic momentum, at least not in this forum, but I am only asking questions about what I am referring to as "applied" momentum, perhaps similar to that of "effective" speed, v / sqrt(1 - (v/c)^2) in some respects. This is the best forum of Relativity experts and I would like to know more about what this is I keep running across. To demonstrate that this relation holds up, at least without considering mass or heat gain in the original example of a ship making n jumps, since the momentum gained is also equal to the proper acceleration applied times the proper time, p = m dv = a dt, jumping in infinitesimal intervals, proper in each case since it is the ship observers that measure this applied momentum, we gain,

[(1 + dv/c) / (1 - dv/c)]^n = [(1 + v_n/c) / (1 - v_n/c)], where n = t_n / dt

[(1 + dv/c) / (1 - dv/c)]^[(a t_n) / (a dt)] = [(1 + v_n/c) / (1 - v_n/c)]

[(1 + dv/c) / (1 - dv/c)]^[k1 / (a dt)] = [(1 + v_n/c) / (1 - v_n/c)]^(k1 / a t_n) = k2

where k1 and k2 are constants, k2 constant because the left side of the equation is constant for some value k2 after one jump and the right side equal to that value after any number of jumps n.

ln[(1 + v_n/c) / (1 - v_n/c)] (k1 / a t_n) = ln(k2)

ln[(1 + v_n/c) / (1 - v_n/c)] (k1 / a t_n) = ln(k2), k3 {constant} = ln(k2) / k1

ln[(1 + v_n/c) / (1 - v_n/c)] = k3 a t_n

Where for small v_n, the left side of the equation becomes 2 v_n/c and a t_n = v_n, so k3 = 2 / c.

ln[(1 + v_n/c) / (1 - v_n/c)] = 2 a t_n / c

t_n = (c / (2 a)) ln[(1 + v_n/c) / (1 - v_n/c)]

This is the equation for the proper time that has passed according to the ship observers' clock with proper acceleration a after gaining a relative speed to the rest frame of v_n. To put it in its more familiar form without v_n and using the time of acceleration t according to the rest frame, we have

v_n/c = (a t / c) / sqrt[1 + (a t / c)^2]

t_n = (c / (2 a)) ln[(1 + (a t / c) / sqrt[1 + (a t / c)^2]) / (1 - (a t / c) / sqrt[1 + (a t / c)^2])]

= (c / (2 a)) ln[(sqrt[1 + (a t / c)^2] + a t / c) / (sqrt[1 + (a t / c)^2] - a t / c)]

= (c / (2 a)) ln[(sqrt[1 + (a t / c)^2] + a t / c)^2 / ((1 + (a t / c)^2) - (a t / c)^2)]

= (c / (2 a)) ln[(sqrt[1 + (a t / c)^2] + a t / c)^2]

= (c / (2 a)) (2 ln[sqrt[1 + (a t / c)^2] + a t / c])

= (c / a) ln[a t / c + sqrt[1 + (a t / c)^2]]

It appears this relation for applied momentum can be legitimately applied to a constantly accelerating ship as well as a ship accelerating in jumps. But this is under ideal conditions, not considering heat and mass gain. So again, why would we use relativistic momentum for collisions but not for constant acceleration? Why wouldn't the ship heat up and gain mass also? Why doesn't the heat and mass gain in collisions depend upon the type of interaction? Why are each described by such different equations when essentially the same process is taking place to accelerate the ship as with collisions of bodies? Why don't the laws of conservation and energy apply in the same manner to the constantly accelerating ship, with heat and mass gain?

Last edited: Apr 10, 2012
10. Apr 10, 2012

### Staff: Mentor

Wow, that is a big post. Can you answer my questions above and condense your post to one or two key questions?

11. Apr 10, 2012

### yuiop

I am puzzled by your claim that relativistic collisions require heating up in order to conserve momentum. It is perfectly possible to model relativistic elastic collisions where no heating up or energy losses are incurred during collisions. It is only inelastic collisions that involve kinetic energy losses due to heat in classical and relativistic contexts.

12. Apr 11, 2012

### grav-universe

Well, we would consider that the ship is constantly accelerating by whatever means that would entail to derive the usual acceleration formulas for SR. Those equations don't include a decrease in the rest mass, so I suppose the mechanism would have to be external, perhaps with the local frame bombarding it with particles to get it up to v with each jump. With a very small momentum applied to the ship within each local frame with infinitesimal v gained each time, the proper acceleration of the ship would be steady.

My question(s) here is that as momentum is applied to the ship, the temperature of the ship and its mass is the same (different from losing mass by ejecting particles), while with relativistic momentum and the conservation laws there is gain in mass and heat. Why, when they are governed by the same general process? The same relation I demonstrated above would apply to both elastic collisions and constant acceleration, but under ideal conditions, without mass and heat gain, would give different conservation laws. How should the momentum be applied differently to conserve energy and momentum according to Relativity and how would that change the acceleration formulas?

13. Apr 11, 2012

### grav-universe

Here is the thread where I asked about it. Apparently it is inelastic collisions I asked about. I should have reviewed it first, sorry. Okay so, a constantly accelerating ship is considered to be accelerated by elastic collisions? Hmm, that complicates my questions. What if we bombard the ship with particles which attach to it but are then released? Would that tend to heat up the ship and increase its mass even after all of the particles are released? Also, why would an inelastic collision heat up the ship, while an elastic collision which rebounds off the ship does not heat up the ship even though the change in momentum is about twice as great?

Last edited: Apr 11, 2012
14. Apr 12, 2012

### Staff: Mentor

OK, thanks for the clarification. Constant proper acceleration is relatively easy to solve, and as was mentioned above the result is a hyperbolic path which is asymptotic to c. The usual reference frame used to analyze this kind of scenario is the Rindler coordinates.

I don't know what the mass and temperature have to do with anything in this scenario. However, the reason that the normal conservation laws don't apply is simply that the system is not isolated. The same thing happens in Newtonian mechanics.

15. Apr 12, 2012

### grav-universe

Thanks Dalespam. I suppose the simplest answer to most of my questions would be that the relativistic acceleration formulas aren't the end-all formulas, applied to any situation, so maybe I was over-generalizing them, but only as applied under ideal conditions, just that of constant acceleration stated as a given, which happens to require perfectly elastic collisions in order to not gain mass which would affect the acceleration, not inelastic (or ejection of particles which would lose mass of course). I will take a few days to think more about this and try to pinpoint better what I am trying to ask.

Thanks again everybody.

16. Apr 12, 2012

### Staff: Mentor

In principle you can have partially inelastic collisions which preserve the mass of one of the particles.