Can you tell your absolute speed in space?

Normally in relativity your speed can only be known relative to another object,

given that as one approaches light speed more energy is required to accelerate faster, based on the energy consumption profile you would be able to calculate at what % of the speed of light you were going?

apart from being a useful 'speedometer' i thought it interesting that one of the effects of the light speed barrier is you can determine your absolute speed relative to yourself, your proper speed i guess it would be called if this speed barrier did not exist then you would have no way to know your speed

now since you know your own speed, you also know the absolute speed of any other object you encounter from the relative difference

following on from this, if you knew your speed you could slow your ship down by the exact speed you knew yourself to be going and be truly at rest, with respect to the reference frame of the universe

my question is am i right? it seems counter intuitive to derive seeming absolutes in in a relativistic universe

• The answer is of course no - you cannot. When you say "as one approaches light speed more energy is required to accelerate faster", you failed to specify from which reference frame's point of view. – Rob Jeffries Aug 3 '15 at 15:42
• i am talking about the reference frame of someone in the ship. you know your ship's energy use, and it's accelleration from observing outside objects. if you know this then you know your absolute speed – martinc Aug 3 '15 at 15:44
• In which case, more energy (actually you mean force) is not required to provide the same acceleration. – Rob Jeffries Aug 3 '15 at 15:47
• so your saying with the same amount of energy i could attain light speed? – martinc Aug 3 '15 at 15:50
• Watching the universe from the point of view of a arbitrarily fueled rocket is one of the reasons that "as one approaches light speed more energy is required to accelerate faster" is a sub-optimal way to talk about relativity. In the rockets frame it is the rest of the universe that is (a) getting harder to add speed to even thought the rocket experiences exactly the same acceleration it would have when at relative rest and (b) that effect is asymmetric over finite velocity changes. The whole business is better understood in terms of velocity composition. – dmckee --- ex-moderator kitten Aug 3 '15 at 16:01

I'm afraid this won't work. If for example you have a rocket motor capable of producing 1g of thrust then it will still produce 1g of thrust no matter how long you accelerate for (assuming you don't run out of fuel).

From the perspective of the observer on Earth your acceleration will indeed slow down, but at the same time the Earth observer sees your time as dilated. For you on the rocket the two effects balance out and as far as you are concerned the thrust stays at a steady 1g.

If you're curious to find out more about what happens under steady acceleration I can highly recommend the relativistic rocket article on John Baez's web site.

• hi, i'm not interested in the perspective of the earth observer just the occupant of the ship. – martinc Aug 3 '15 at 15:42
• @martinc: as I said in my first paragraph, the occupant of the ship will find the acceleration is constant, no matter how long they accelerate for. That means they can't use changes in their acceleration to determine an absolute speed because there are no changes in acceleration. – John Rennie Aug 3 '15 at 15:44
• if your motor produces a maximum of X amount thrust. to attain light speed X would need to be infinity. so your ship will have a maximum % light speed it can go based on energy requirements. you would accelerate more slowly, by your logic you would reach light speed and exceed it? i'm sure you don't mean to imply this – martinc Aug 3 '15 at 15:48
• @martinc: You need to bear in mind that as far as you on the rocket are concerned you are at rest, and it's the rest of the universe that is accelerating away from you. So the question is whether you'd see the rest of the universe exceed the speed of light relative to you. The answer is that this doesn't happen, though the spacetime geometry you see actually gets quite complicated. If you're interested it's called the Rindler metric. Nevertheless, the acceleration you feel remains steady at 1g and you can't use this to for absolute speed. – John Rennie Aug 3 '15 at 15:53

That won't work, and here is why. It's subtle.

Say that you are on a ship, leaving the solar system with some technology that is able to thrust you in such a way that you experience a constant acceleration of 1g, as measured on the ship.
You can measure this by placing a 1kg weight on a scale.

From the point of view of the passengers of the ship (where the ship is at rest), the Sun will initially accelerate at 1g. You can measure this in principle by looking at the redshift of Sun's light.
Soon enough though, the Sun will accelerate slower and slower as it approaches $c$, the speed of light.

Now you could use your schema to determine at what fraction of the speed of light you are going, with respect to the Sun. You see, you can only measure your speed and acceleration with respect to some object.

• I'm not sure about that last line. In SR acceleration is absolute and as you say earlier in your answer you can measure it independently of anything outside the rocket just by weighing something. – John Rennie Aug 4 '15 at 5:25
• Yes, I recognise my answer is ambiguous. It's like there are two kinds of acceleration: the one you can measure locally as a force (like the thrust of the engine) and the one defined as rate of change of the velocity. In Newtownian Much they are the same, but in SR they are related but different. – Andrea Aug 4 '15 at 8:04
• Crucially for OP's question: no matter what the speed of the ship relative to the Sun (or the CMB for that matter), it will have to expend the same amount of energy to keep that 1g "local" acceleration going (assuming no change in fuel mass). That is, the amount of energy for an object to feel a certain acceleration is independent of its speed, so it cannot be used to measure and absolute velocity. – Andrea Aug 4 '15 at 8:09
• There we agree! – John Rennie Aug 4 '15 at 9:00

The problem with questions like this is that they include many misunderstandings of physics!

For example, you say "as one approaches light speed more energy is required to accelerate faster". What you may not be aware of is that in classical mechanics, it's also true that to an observer on the ground, the faster you are going, the more energy you need to accelerate! This is what $\mbox{Ke}=\frac{1}{2}mv^2$ does. You might also not know that energy, in classical mechanics and in special relativity, is a frame dependent concept. From one frame your kinetic energy might be $\frac{1}{2}mv^2$, but from another your kinetic energy might be $0$. So "kinetic energy" isn't an absolute observable that can be calculated without using a frame of reference.

To clarify, I'm asserting that your arguments would apply just as easily to classical Newtonian mechanics, and that therefore you can answer your question (involving the rate of change of kinetic energy) by better understanding energy in classical mechanics.

Your other issues stem from issues with relativistic velocity additon. If, in the frame of the earth you shot a ship off at half the speed of light, and if in the frame of that ship you shot a bullet at half the speed of light, the bullet would be travelling in the frame of the earth at $\frac{0.5+0.5}{1+0.5\times 0.5}=0.8$ times the speed of light. It's a not-trivial consequence of special relativity, which is best learned by reading books or reading more questions on here. This is why your comments such as, "you would accelerate more slowly, by your logic you would reach light speed and exceed it" and "so your saying with the same amount of energy i could attain light speed" are wrong.

It's difficult to imagine how such a velocity addition rule could hold, as it contradicts everything you know about position/time coordinate systems, but it does hold.

Can you tell your absolute speed in space?

Yes. You just look at the CMBR dipole anisotropy. This tells you how fast you're going relative to the universe, and that's as absolute as it gets.

"From the CMB data it is seen that our local group of galaxies (the galactic cluster that includes the Solar System's Milky Way Galaxy) appears to be moving at 627±22 km/s relative to the reference frame of the CMB (also called the CMB rest frame, or the frame of reference in which there is no motion through the CMB) in the direction of galactic longitude l = 276°±3° , b = 30°±3°"

Note that this doesn't constitute an absolute frame in the strict sense, in that you can't tell how fast you're going if you're in a box. But all you have to do is look out of the window, and you see a blueshift in the direction you're headed, and a redshift in the other direction. Have a look at this question and ghoppe's answer which refers to Douglas Scott who says this:

"However, the crucial assumption of Einstein's theory is not that there are no special frames, but that there are no special frames where the laws of physics are different. There clearly is a frame where the CMB is at rest, and so this is, in some sense, the rest frame of the Universe. "

Normally in relativity your speed can only be known relative to another object

Yes, the cosmic background radiation. Or if you prefer, the universe.

given that as one approaches light speed more energy is required to accelerate faster, based on the energy consumption profile you would be able to calculate at what % of the speed of light you were going?

If you were in something like a rocket in gravity-free space, you could calculate the increase in your speed based upon how much fuel you'd burned. But it isn't a great way to find out how fast you're going. You'd be better off looking out the window at the CMB.

now since you know your own speed, you also know the absolute speed of any other object you encounter from the relative difference following on from this, if you knew your speed you could slow your ship down by the exact speed you knew yourself to be going and be truly at rest, with respect to the reference frame of the universe

Yep. Slow down by about 627 km/s and there's no CMB dipole anisotropy. You are at rest with respect to the reference frame of the universe.

my question is am I right? it seems counter intuitive to derive seeming absolutes in in a relativistic universe

Yes, you're right. I say this as a relativity guy. Some people will tell you there's no such thing as absolute motion, only relative motion. But all we're talking about is motion relative to the universe. Which is as absolute as it gets.

• Once again, there is nothing privileged about the frame of the CMB. It's not special in any fundamental way. Saying it is "as absolute as it gets" is another way of say that it is not absolute at all. – dmckee --- ex-moderator kitten Aug 3 '15 at 21:51
• There are many interesting things in your answer, which makes it a little confusing as an answer to OP'S question. In short: it is true that one can measure in practice the velocity wrt the universe with the CMB. But OP is NOT right: OP's method is fundamentally flawed, because it relies on a misunderstanding of the SR model. It would be nice if someone were able to clearly explain why that would not work. – Andrea Aug 4 '15 at 8:12
• @dmckee : no, it is absolute, because the universe is as absolute as it gets. Follow my link to ghoppe's answer which refers to Douglas Scott who called it the rest frame of the universe. Andrea : yes, the OPs method of using energy consumption as a speedometer isn't great, but I presumed he was making allowance for relativistic effects. He didn't suggest that you ended up going faster than 299,792,458 m/s. – John Duffield Aug 4 '15 at 8:28
• ::sigh:: Identifiable is not the same as special. This is relativity, after all. Physics is no different in the CMB frame than in our current frame. Ergo there is nothing absolute about it. – dmckee --- ex-moderator kitten Aug 5 '15 at 0:11
• @dmckee : sigh yourself, nobody said the physics is different. And read my answer, where I said it wasn't an absolute frame in the strict sense. Also read this article by Frank Heile, and note the use of the word absolute. You may also care to do an arXiv fulltext search on absolute frame. There's 279 hits. – John Duffield Aug 5 '15 at 8:45

This is a good question. Especially if someone is new to relativity and hasn't gone through many relativistic calculations, yet.

The problem you will run into with trying to find the absolute reference frame is two fold:

1. Einstein Velocity Addition is working against you. Consider the following:

$$u=\frac{v+v'}{1+vu'/c^2}$$

A while back I had wondered a similar question: if we were in space, could we send 6 probes in each direction at half the speed of light and have them all send a signal when their clocks reach 60 seconds?

It's not terribly complicated to figure out but does require a lot of calculations(which I am not going to repeat here). Best to do it in a spreadsheet.

What I found is that Einstein Velocity Addition always results in being unable to determine the absolute reference frame. Even if you're going at half the speed of light in the z axis, and you send each probe at half the speed of light in opposite directions along the z axis, what you think is half the speed of light going forward is only an increase of 0.3c wrt(with respect to) the stationary reference frame. While the other probe you sent toward the stationary reference frame is seen as decreasing in velocity by 0.5c. But both probes appear from your ship as going at 0.5c in each direction.

And if that is the case, then it means you put the same amount of energy into increasing the one probe by 0.3c wrt the stationary reference frame and decreasing the other probe by 0.5c wrt the stationary reference frame. You can calculate this using the relativistic energy equation:

$$E=\sqrt{p^2c^2+m_0^2c^4}$$

Where $p$ is momentum, $c$ is the speed of light, $m_0$ is rest mass, and $E$ is energy. And you will find that both probes in the example above require the same energy to change velocity from all reference frames.

Which brings us to the second problem:

1. Your perspective of distance and time is working against you. Consider the following:

$$\gamma = \frac {1}{\sqrt{1-v^2/c^2}}$$

Length and time in your perspective need to be corrected wrt the stationary reference frame. When you look out your ship, both probes look like they are leaving in the opposite direction at the exact same speed. But from the stationary reference frame, one has changed speed more than the other. When you put it all together, each relative frame is unable to use any conventional means to determine their own velocity wrt the stationary reference frame.

There is no known experiment whereby one can know their relative velocity without seeing other stationary objects, and that is assuming those other objects aren't also moving, as well. But if you do figure out a way, please let me know!