Rocket Motor Placement SR: Effects on Spaceship Acceleration

[pervect]

I recall reading that in the context of special relativity, one will notice differences in the acceleration of an idealized born-rigid spaceship when one compares two identical spaceships, one with the rocket motor mounted in front, the other with the rocket motor mounted in the rear.

I'm looking mainly for references that point out that this happens in the first place, though some discussion of "why" could be helpful. I'd like to use this to motivate the need for the stress-energy tensor in SR, but it might be necessary to assume the reader already know about the stress-energy tensor before one could derive this. Perhaps it's not needed, I seem to recall some arguments based on how fast the rocket motor consumes fuel. But I don't recall exactly what I read or where I read it.

 

[PeterDonis]

one will notice differences in the acceleration of an idealized born-rigid spaceship when one compares two identical spaceships, one with the rocket motor mounted in front, the other with the rocket motor mounted in the rear.

I would imagine the key difference would be that if the rocket is in the front, the ship will be under tension, while if it is in the rear, the ship will be under compression. So if we imagine two ships that are identical when their engines are off, and then turn the engines on in such a way as to produce the same proper acceleration, the ship with the front-mounted engine would end up with a longer length (in its instantaneous rest frame) than the ship with the rear-mounted length. However, I can see a lot of caveats to this (mainly around exactly how "the same proper acceleration" would be defined). I have not seen any discussion of this in textbooks or papers that I have read.

 

[PAllen]

Also, the purpose of Born rigid idealization is to rule out material realities like compression and tension. I would highly doubt that given assumption of born rigidity throughout, that it would make a difference whether the engine is in front or back. An inconsequential difference is that if some proper acceleration is a given for the front, versus a given for the back, the acceleration profile along the rocket is different. But this is silly IMO. Identical cases are achieved by specifying one proper acceleration for the front, versus a slightly larger given acceleration for the back. I guess another difference is that given a proper acceleration for the front, there is a maximum length rocket possible (back to the Rindler horizon), while given a proper acceleration for the back, there is no limit to rocket length. However, I don't view these as differences in the idealized rocket description. Instead they are differences in how you specify initial conditions for front versus back.

 

[PeterDonis]

I would highly doubt that given assumption of born rigidity throughout, that it would make a difference whether the engine is in front or back.

It doesn't in the sense that the same Born rigid congruence of worldlines can be realized by a rocket with the engine in front or one with the engine in back, yes.

What I was saying was that, if you start with a rocket of a given unstressed length (i.e., the same congruence of inertial worldlines, before the rocket starts up), then which Born rigid congruence will end up describing the equilibrium state of the rocket under thrust might depend on whether the rocket engine is in the front or the back of the rocket. (Heuristically, because the rocket engine in front will stretch the rocket while the rocket engine in back will compress it.)

 

[jartsa]

There is an energy current from the motor to the other parts of the spaceship. That energy current has momentum. The direction of the momentum depends on the direction of the current, which depends on the location of the motor.

https://en.wikisource.org/wiki/Tran..._Law_of_the_Lever_in_the_Theory_of_Relativity

 

[PAllen]

It doesn't in the sense that the same Born rigid congruence of worldlines can be realized by a rocket with the engine in front or one with the engine in back, yes.

What I was saying was that, if you start with a rocket of a given unstressed length (i.e., the same congruence of inertial worldlines, before the rocket starts up), then which Born rigid congruence will end up describing the equilibrium state of the rocket under thrust might depend on whether the rocket engine is in the front or the back of the rocket. (Heuristically, because the rocket engine in front will stretch the rocket while the rocket engine in back will compress it.)

Fine, but I interpreted the OP as stating Born rigidity throughout. That’s what the first sentence implies to me.

 

[PAllen]

There is an energy current from the motor to the other parts of the spaceship. That energy current has momentum. The direction of the momentum depends on the direction of the current, which depends on the location of the motor.

https://en.wikisource.org/wiki/Tran..._Law_of_the_Lever_in_the_Theory_of_Relativity

Born rigidity obviates that whole discussion. To amplify, given Herglotz-Noether’s theorem and corollaries, specifying proper acceleration profile for one point on a born rigid body fully determines the congruence for the whole body. Then, if one specifies instead, the consequent acceleration profile for some other point, you have the identical congruence.

 

[stevendaryl]

I recall reading that in the context of special relativity, one will notice differences in the acceleration of an idealized born-rigid spaceship when one compares two identical spaceships, one with the rocket motor mounted in front, the other with the rocket motor mounted in the rear.

I'm looking mainly for references that point out that this happens in the first place, though some discussion of "why" could be helpful. I'd like to use this to motivate the need for the stress-energy tensor in SR, but it might be necessary to assume the reader already know about the stress-energy tensor before one could derive this. Perhaps it's not needed, I seem to recall some arguments based on how fast the rocket motor consumes fuel. But I don't recall exactly what I read or where I read it.

Initially, it makes a difference: If you push from the rear of a rocket, then the rocket will compress slightly, and then spring back. If you pull the front of the rocket, it will stretch slightly, and then spring back. After the rocket has been traveling for a while at constant proper acceleration, it won't make any difference how you accelerated it, except for the peculiarity that:

Accelerating by pushing the rear at acceleration $g$ is equivalent to accelerate by pulling the front at acceleration $g'$ where the two are related by:

$\frac{1}{g'} = \frac{1}{g} + \frac{L}{c^2}$

where $L$ is the length of the rocket.

 

[jartsa]

Born rigidity obviates that whole discussion. To amplify, given Herglotz-Noether’s theorem and corollaries, specifying proper acceleration profile for one point on a born rigid body fully determines the congruence for the whole body. Then, if one specifies instead, the consequent acceleration profile for some other point, you have the identical congruence.

Let's say we attach two accelerometers on two identical rocket motors, and then we attach the motors on identical blocks of matter, but the attachment points are not identical.

Now we ask: Are the readings on the accelerometers same after the motors have been running for a long time?

Let's say the energy part of stress-energy in the blocks is the same in both cases. The stress part obviously is not. The two blocks might have different inertias, because of the different stress-energies. If the inertias are different, then the readings on the accelerometers are different.

Addition: The blocks tend to vibrate when the motors are turned on. So there is a small violation of strict rigidity requirement. But after the vibrations have died off the blocks behave like rigid objects. I don't see why those small vibrations at the beginning could not be just ignored.

 

[pervect]

I suppose that what's somewhat easy to motivate is to say that if we have two rocket motors connected by a light weight rigid rod, we known that the rocket motor in front must accelerate at a lower proper acceleration than the rocket motor in the rear for the tension / pressure in the rod to be zero. (In this context tension and pressure represent the same concept with opposite signs).

This implies that the tension in the rod will not be zero if we have two identical rocket motors connected by a lightweight rigid rod.

If we consider two identical rocket motors connected by a string, we have the bell spaceship paradox, where we know that the string breaks. We are basically just saying that if the string doesn't break, the rocket motor in front accelerates less (a lower proper acceleration) than than the one in the rear.

Unfortunately, it doesn't seem quite as easy as I'd hoped to go from this observation to motivating the stress-energy tensor.

 

[PeterDonis]

I interpreted the OP as stating Born rigidity throughout.

If that's really the case, then I agree with your statement in post #7.

 

[PAllen]

Let's say we attach two accelerometers on two identical rocket motors, and then we attach the motors on identical blocks of matter, but the attachment points are not identical.

Now we ask: Are the readings on the accelerometers same after the motors have been running for a long time?

Let's say the energy part of stress-energy in the blocks is the same in both cases. The stress part obviously is not. The two blocks might have different inertias, because of the different stress-energies. If the inertias are different, then the readings on the accelerometers are different.

Addition: The blocks tend to vibrate when the motors are turned on. So there is a small violation of strict rigidity requirement. But after the vibrations have died off the blocks behave like rigid objects. I don't see why those small vibrations at the beginning could not be just ignored.

None of this has anything to do with the question I answered. Born rigidity is an idealization that is not achievable, in principle, by any real body. But that is what was asked about, so that is what I answered. Born rigidity is a mathematical construct with rigorously proven properties.

 

[PeterDonis]

the rocket motor in front must accelerate at a lower proper acceleration than the rocket motor in the rear for the tension / pressure in the rod to be zero.

In Born rigid acceleration, the tension/pressure in the material won't be zero. It can't be, because of hydrostatic equilibrium; the material is under proper acceleration, which means it has weight, which means the "higher" parts are pushing down on the "lower" parts, which have to push back to maintain equilibrium. Born rigidity just means there is equilibrium: none of the parts are moving relative to the other parts. But that equilibrium cannot be a zero-pressure/tension equilibrium; that is only possible for an object in free fall.

This, btw, is where I think the concept of stress-energy tensor would come in (although in the simplest case of linear acceleration, you don't need the whole tensor, just energy density and pressure along the direction of motion is enough).

This implies that the tension in the rod will not be zero if we have two identical rocket motors connected by a lightweight rigid rod.

It implies more than that: it implies that the tension in the rod will be increasing. Whereas, in the Born rigid case, the tension/pressure in the rod will be constant. That's the key difference between the Born rigid case and the Bell spaceship paradox case.

 

[stevendaryl]

I don't know where you're getting this formula from, but it's wrong. Please do not post in this thread again since you clearly do not have an "A" level understanding of the subject matter.

I think he got it from me. The metric for Rindler coordinates is:

$c^2 d\tau^2 = R^2 d\theta^2 - dR^2$

where $R$ and $\theta$ are related to $x$ and $t$ through:

$x = R cosh(\theta)$
$c t = R sinh(\theta)$

An object that is at "rest" at a constant value of $R$ will satisfy:

$x = \sqrt{c^2 t^2 + R^2}$

we can expand the square-root for small $t$ to get:

$x = R \sqrt{1 + \frac{c^2 t^2}{R^2}} \approx R (1 + \frac{1}{2} \frac{c^2 t^2}{R^2}) +$ higher-order terms $= R + \frac{1}{2} \frac{c^2}{R} t^2$ + higher-order terms

Comparing that with the nonrelativistic approximation for constant acceleration, $x = x_0 + \frac{1}{2} g t^2$ shows us that

$g = \frac{c^2}{R}$

or $R = \frac{c^2}{g}$

Of course, this is nonrelativistic, so it is only valid when the rocket first launches, and is moving slowly compared to the speed of light. But the magnitude of the proper acceleration is constant, so that answer is valid even when the rocket is moving relativistically.

So the proper acceleration $g$ varies inversely with the Rindler coordinate $R$.

So if the rocket has length $h$, then:

$R_{front} = R_{rear} + h$

So

$\frac{c^2}{g_{front}} = \frac{c^2}{g_{rear}} + h = \frac{c^2 + h g_{rear}}{g_{rear}}$

$g_{front} =\frac{g_{rear}}{1 + \frac{h g_{rear}}{c^2}}$

 

[PeterDonis]

I think he got it from me.

Yes, probably. But it wasn't very clear.