# B Measuring relativistic effects in a single frame

1. Feb 5, 2017

### BitWiz

Sorry if this is a dupe:

An astronaut is in a windowless, sensorless starship that is drifting in an unknown location in space. There is an accelerometer aboard which reads zero. The astronaut turns on his ideal linear propulsion engine for one second, losing a negligible amount of propellant mass, and reads the maximum value on his accelerometer before it returns to zero.

The astronaut then goes to sleep for an arbitrary amount of time. Any unknown forces or fields may affect his vehicle's speed and position on any axis during that time, any of which may or may not affect his accelerometer, but he does not experience any of them while asleep.

When he wakes up, he sees that his accelerometer reads zero. He repeats the experiment. Disregarding non-ideal effects such as component wear, or physical alterations to his ship, will he always see the same reading on his accelerometer?

Thanks!

2. Feb 5, 2017

### PeroK

Is there a purpose to this question? Both experiments are carried out in an IRF, so the results should be the same each time. There is no effect of an unknown "absolute" velocity if that is what you're asking.

3. Feb 5, 2017

### Ibix

The mass of his rocket reduces due to fuel expended, so he'll accelerate harder each time. But that's all, unless he ends up close enough to a source of gravity that tidal effects become important, or his ship becomes magnetised or charged and interacts with an external field during acceleration or something.

4. Feb 5, 2017

### Staff: Mentor

The answer you've been given ("No effects whatsoever") is a direct result of Einstein's first postulate, and also solidly confirmed by many experiments. In fact, you do something similar everytime you go to sleep -- when you awaken the surface of the earth is moving in a completely different direction in space, but if you couldn't see the sun, stars, and planets, you'd never know.

5. Feb 6, 2017

### BitWiz

Thanks. Then the relativistic effect of increased mass at higher speeds is strictly the observation of an external observer? If the rocket and its passenger were traveling at 1% or 99% of c, the reading on the accelerometer would remain the same. To the passenger, the rocket expended the same energy to produce the same result, I.e. Relativistic effects were unmeasurable. Is that correct?

6. Feb 6, 2017

### Mister T

Moreover it's based an an entirely unnecessary re-definition of what we mean by mass. The old-fashioned use of relativistic mass has diminished, especially during the last few decades. It was introduced as a way of explaining things, but it was quickly realized that it's not a necessary part of the explanation.

Note that the phrase "traveling at 1% or 99% of c" makes sense only if you specify the reference frame in which those speeds are measured. In that reference frame there will definitely be differences in the way the rocket behaves when the rocket engines are applied, even though the accelerometer readings on board the ship may be the same.

7. Feb 6, 2017

### Ibix

As Mr T noted, relativistic mass is largely a deprecated concept these days due to the endless confusion it causes. But yes, relativistic effects always happen to other people. You can always consider yourself at rest, so the Newtonian approximation is always a good one from your own perspective.

There's no absolute sense of velocity, so this is a meaningless statement without some external reference point to regard as stationary...

...but this is true anyway. The acceleration measured by your accelerometer is called the proper acceleration. Proper in the sense of "property", something that is your own, rather than "right".

More or less. The passenger can always build a particle accelerator inside the rocket and study relativity that way. Or do Fizeau's experiments on light speed. But I suspect that's cheating in the sense you mean it.

8. Feb 6, 2017

### BitWiz

Thank you Nugatory, Ibix, and Mister T for your very helpful answers. They clear up a lot of confusion.

I'd like to extend this argument a little further: The astronaut wakes up, his accelerometer says zero. He resets his proper clock to zero. He then lights his rocket engine and experiences a constant, measurable acceleration. (For simplicity, please ignore mass changes due to fuel and propellant usage, and energy-mass equivalence.) Using his proper clock and accelerometer, he calculates when he has achieved 1%c based on v = ta. Can he say he is now traveling at 1%c with respect to his initial state?

The astronaut makes a note of the energy he used to achieve 1%c. He then repeats the above 98 more times. Can he say that he is traveling at 99%c with respect to his original initial state? Will he see that each of his 1% efforts consumed the same amount of energy from his power supply?

I'm okay leaving this at 99% c to keep from turning this into a different discussion. ;-)

Thanks!

9. Feb 6, 2017

### Staff: Mentor

No, because he used the wrong equation to determine his final velocity relative to that initial state. (We are assuming he's in flat spacetime so velocity relative to a distant location has an invariant meaning; in curved spacetime it doesn't.) He should have used the relativistic rocket equation:

http://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html

The correct equation for $v$ as a function of $a$ and $T$ (where I've used a capital T because that's what the article uses to denote the proper time of the traveler) is, as given in that article:

$$v = c \tanh \left( \frac{aT}{c} \right)$$

The error if $aT / c$ is 0.01 is not large (about 1 part in 10,000), but it would be measurable using today's technology.

No. See above.

Not if "1%" is defined by his change in speed. But if it's defined as his change in rapidity, yes. Rapidity is just the argument of the $\tanh$ function above, i.e., it's $aT / c$. In other words, each increment of the traveler's clock time will consume the same amount of energy (which makes sense since he is feeling constant thrust from his rocket). But it won't translate into the same increment of speed relative to his starting point--that will get smaller and smaller (as you can see by looking at a graph of the $\tanh$ function).

10. Feb 7, 2017

### BitWiz

Thanks, Peter,

Say the rocket is able to mark an external, stationary wall along the route with a daub of paint each time the engine fires. Using the conversion of rapidity to velocity above, the daubs will be closer together in the wall's frame with each successive rocket motor firing. Correct?

The astronaut, seeing that his accumulated rapidity is 99% c, calculates that his true velocity with respect to his origin has increased step by step to about 0.757 c (c tanh(%c)), and the gamma has similarly increased to about 1.53. With length compression effects, are the daubs really getting closer together in the wall's frame?

Thanks!

11. Feb 7, 2017

### Staff: Mentor

If the motor fires at constant intervals of time by the ship's clock, yes.

Length contraction is irrelevant here, because the daubs aren't measuring the rocket's length; they're measuring the distance it travels in a fixed interval of time by the rocket's clock.

12. Feb 9, 2017

### BitWiz

Hi, Peter,

I thought ALL of space was contracted along the axis of motion according to the astronaut. At a gamma of 2, the astronaut can't observe himself traveling two light years in one year, correct? Therefore, both the astronaut-observed distance -- and the physical distance -- is contracted to compensate, correct?

13. Feb 9, 2017

### Staff: Mentor

That's irrelevant because you asked about the distance between the daubs in the wall's frame, not the astronaut's frame.

14. Feb 9, 2017

### Mister T

As Peter said, the distance between the daubs is not Lorentz contracted in the rest frame of the wall. In that frame, the rocket ship's length is Lorentz contracted.

In the rest frame of the rocket the distance between the daubs is Lorentz contracted. But the rocket's length is not.

15. Feb 10, 2017

### BitWiz

Thank you for the info gentlemen. Very much appreciated.

Let me steer the conversation back toward my original intent:

Using the original example of an astronaut observer in an enclosed rocket in a vast tract of empty space with only his accelerometer (a) and a clock (T) for company, I think it's established that he reads the same maximum acceleration during each interval that he fires his rocket, no matter how long or short that proper interval is, or how many times he fires his rocket -- with the restriction for now that his calculation of T * a < 99% c. His accelerometer always returns to zero when the engine is cut off. He observes that his rocket motor uses the same energy per second to achieve the same acceleration at all times. External curves in spacetime may speed him up or slow him down with respect to external observers, but he and his accelerometer don't notice or care as long as tidal effects aren't noticeable. Are these statements correct?

Thanks!

16. Feb 10, 2017

### Ibix

No such restriction is needed. $aT$ is only a valid estimate of the final speed of the rocket (a) in its original rest frame and (b) when $aT <<c$. If $aT\simeq c$ then it's an invalid approximation and we don't care what it says.

Yes.

With the stipulation that the fuel used never adds up to a detectable fraction of the rocket's mass, yes. This is implausible in practice, but fine as a simplification.

That's a trickier statement than you might think, but good enough in many cases.

Yes.

17. Feb 10, 2017

### BitWiz

Thank you, Ibix.

An engineer says: If the rocket is at rest with respect to an external observer, and fires its engine for time t, the rocket will attain velocity v, use Energy E, and accumulate kinetic energy Ek. If the rocket fires its engines again for time t , the engineer says it will NOT attain a velocity of 2v, because it's kinetic energy would have to be 4Ek, but only 2E will have been expended. The presumption is that the kinetic energy must come from the on-board fuel.

I understand this for a car on the road. If a car accelerates at the same rate for 2t, it will use 4E of fuel to obtain 2v and 4Ek. But a car must have the Earth (and thus an independent Earth observer) in its rest frame. In the case of the rocket, is it only an observer in the initial rest frame who sees the rocket accumulating 4Ek? Doesn't the astronaut see a linear relationship between the energy his rocket expends and his presumed proper velocity (subject to tanh(%c)), simply because there is nothing else in his frame except himself and his rocket?

Is the excess Ek that the Earth observer sees real? If so, must the energy come from on-board fuel? If "no" to either question, how do I convince my engineer that his physics is wrong?

Thanks!

18. Feb 10, 2017

### Ibix

You aren't accounting for the kinetic energy of the rocket exhaust. If you do that (remember the conservation of momentum and that the initial velocity is not zero for the second acceleration) you will find that the energy difference before and after each acceleration phase is the same. This is true whether you use Newtonian or relativistic expressions for energy and momentum.

19. Feb 10, 2017

### Umaxo

what this (half of the) sentence is supposed to mean? I cannot make sense of it.

20. Feb 10, 2017

### Ibix

I took it to mean gravity. If he didn't mean that, I need to revisit my last answer.

21. Feb 10, 2017

### Umaxo

So you interpret "external curves of spacetime" in similar way as curves are interpreted in "woman with beautiful curves", i.e. as curvature (of spacetime)?

22. Feb 10, 2017

### Ibix

I'm not sure that's an appropriate comparison, on many levels. But gravity is spacetime curvature, so yes, that's what I said.

23. Feb 10, 2017

### BitWiz

Hi, Ibix,

Is it the Earthbound *engineer* who isn't allowing for the kinetic energy of the rocket exhaust? Is that the mistake?

The astronaut sees his energy usage E is linear with respect to time T.

The engineer sees the rocket body's kinetic energy increase along a power curve.

At 3T, and assuming the rocket mass = 1, is the Exhaust-ek = 3E - (3v)2/2? In other words, is exhaust energy a constant over time to the astronaut, but a power curve to the origin observer?

24. Feb 10, 2017

### BitWiz

Hi, Umaxo,

As a five-dimensional being, I see gravitational effects as a beautiful set of curves. ;-)

25. Feb 11, 2017

### Ibix

I think you have it, yes. Basically you note that the rocket has an initial velocity U and an initial mass m+M. The rocket fires in a short pulse to avoid calculus. Then the exhaust of mass m has velocity U-u and the rocket has mass M and velocity U+v. You conserve momentum to find the relationship between u and v, and you'll find that U is not involved. You can then find the difference between the initial and final energies and again you'll find that U is not involved.

That's all Newtonian physics. Relativity is the same with different formulae for velocity addition, momentum, and energy.