# How to tell the difference between acceleration and gravity

1. May 31, 2014

### randyu

I have a question.
I am in a frame. I construct a spring scale and weigh a ball to be 200 lbs on the floor. I feel no forces on my motions other than a gravitational field like being on the surface of the earth. I climb a very very tall ladder and weigh the ball again and find it to be 100 lbs. Wouldn't I say that I am in a gravitational field and not an accelerating field. And wouldn't I be able to calculate something about the nature of this gravitational field?
Conversely, if I climb the same very tall ladder and the ball still weighs 200 lbs. Wouldn't I then know I'm in an accelerating field and not gravitational?
Just confused, thanks for any help.
Randy

2. May 31, 2014

### D H

Staff Emeritus
That's a very tall ladder you write of. It would need to be over 2640 kilometers tall if you're on the surface of the Earth.

One of the consequences of the equivalence principle is that acceleration due to gravity and due to (say) a rocket are locally indistinguishable. Why "locally"? For precisely the reasoning you used in your opening post. Climbing a 2640 kilometer tall ladder is not a local experiment.

3. May 31, 2014

### randyu

Thanks DH,

I didn't really realize the principle was based on locality (guess I really must not have thought about it that much). Thought I might have an original thought; not really but.

So how does this nonlocality tie in with light being locally c, and the expansion of the universe being faster than c far enough away; and yet this doesn't violate the "local speed of light". I mean, it gets a little confusing to me. Impossible to wrap my mind around. How could a universe be bounded and yet have infinite acceleration. Seems like a flat U is the only possibility to me. And, of course, I can't understand that either. Any insight? I mean like what's the point of having a local requirement to light speed and the equivalence principle. Why not just generalize it all. Would that make it any easier to understand? Or am I talking about about general relativity, which I doubt I'll ever understand.

Randy

4. May 31, 2014

### Staff: Mentor

No, you would know that you were in a universe that obeyed different laws of physics from relativity. If the ladder stays the same length as it accelerates (along with the very long spaceship or whatever it is that all this is contained in), then the acceleration must vary with "height" up the ladder; it must get smaller as you go "higher", and larger as you go "lower".

Check out this portion of the Wikipedia page on Rindler coordinates (which are the natural coordinates for a family of accelerating observers in flat spacetime):

The difference between being in an accelerating field and being in a gravitational field is that in a gravitational field, there will be tidal gravity, which is *not* quite the same as simply the "acceleration due to gravity" varying with height, as the example we've been discussing shows. The correct way to detect tidal gravity is this: take two objects in free fall, that start out close together and at rest relative to each other. If they do not remain at rest relative to each other, tidal gravity is present. Note that in the "accelerating field" case, the objects *will* remain at rest relative to each other, like any other objects moving inertially that start out at rest relative to each other in special relativity.

5. May 31, 2014

### Staff: Mentor

Because the "speed" at which the universe expands is not a "speed" in the usual sense. Many popular accounts describe it that way, but only shows that you can't trust popular accounts if you actually want to understand the science. Objects far enough away from us "move faster than light" away from us only if you adopt a particular slicing of spacetime into space and time *and* interpret the coordinate velocity (i.e., the change in spatial coordinates with time) in that slicing as a "speed", even for objects that are spatially separated by large distances. But under that interpretation of "speed", things can travel faster than light; for example, light emitted by a galaxy that is receding "faster than light" from us will recede from us even faster than the galaxy itself does.

To the best of our knowledge, our universe is not (spatially) bounded. However, there are solutions to the Einstein Field Equations that describe a spatially bounded universe with accelerating expansion. Again, this is possible because the "speed" at which the universe expands is a different kind of "speed" from the one that's limited to the speed of light.

6. May 31, 2014

### randyu

Okay, I must really be confused. Are you saying that on earth if I drop 2 objects at the same time, they will NOT remain at rest relative to each other? I understand they will both move towards each other as they both accelerate to a common cg. Small effect as this is, I'm not talking about a different universe than ours. So I don't understand how I would know a universe that obeyed different laws of physics from relativity if the ball still weighs 200 lbs as I measure it on the ladder.

7. May 31, 2014

### randyu

So how should one interpret that change in spatial coordinates with time as something different than speed? I've heard the universe's spatial behavior labeled as "cosmic expansion", not using the word speed. But that begs my question, what is cosmic expansion if not coordinate velocity? For a local (out there in space) event, what is happening that causes the perception of faster than light travel? I mean I don't understand what particular slicing of spacetime we are adopting. Seems like we make an observation, and the measurements tell us the speed exceeds light.

How can the coordinate velocity be interpreted differently than speed?

8. May 31, 2014

### Staff: Mentor

Yes.

Not if they are spatially separated, no.

That's if they start out separated tangentially, yes. They will move towards each other even if they start out at rest relative to each other; moving towards each other is *not* staying at rest relative to each other.

If, OTOH, the objects start out separated radially, they will move away from each other; if you drop two objects at the same instant, one from a higher altitude than the other, the one that is lower will fall slightly faster, so it will move away from the one that is higher, even though they start out at rest relative to each other.

In the absence of gravity, neither of the above effects will occur. If you take two objects and release them at rest relative to each other far away from all gravitating bodies, they will remain at rest relative to each other.

Because relativity says that's impossible, even if the ladder is inside an accelerating spaceship. If the relative distances inside the spaceship remain the same (e.g., the ladder's length remains constant), the acceleration must vary with height along the ladder; it must be larger at the bottom than at the top. So an object that weighs 200 lbs at the bottom of the ladder must weigh less at the top.

9. May 31, 2014

### Staff: Mentor

You can interpret it as a kind of speed, as long as you understand that this kind of speed is not limited to the speed of light. In other words, the term "speed" can label different concepts, and you need to carefully distinguish which one you are talking about in a given context.

Also, it's important to realize that coordinates are abstractions; they are numbers we use to label events, and you can never assume that they have any particular physical interpretation. Any physical observable must be expressible as an invariant, i.e., a quantity that is the same regardless of what coordinates you choose. Coordinate velocity obviously is not such a quantity, since it changes when you change coordinates.

Ned Wright's cosmology FAQ and tutorial has some good entries on this:

http://www.astro.ucla.edu/~wright/cosmology_faq.html#FTL

http://www.astro.ucla.edu/~wright/cosmo_02.htm

Only if we interpret the measurements in a particular way. We don't actually *see* objects moving away from us faster than light. We use observational data to *infer* that they have a coordinate velocity faster than light in a particular set of coordinates. The first link to Ned Wright's FAQ above discusses this.

10. Jun 1, 2014

### randyu

In a spaceship frame at constant acceleration (no rotation) vertically upward, if I weigh the ball on the ground level, shouldn't I get the same weight at any other place vertically upward? I'm in the spaceship, I weigh the ball, I move up the ladder, the ball weighs the same. I conclude I'm in an accelerating frame. What do you mean the acceleration must vary with height? The acceleration, is the acceleration of the frame. Don't understand it varying. Do you mean as observed from the outside by someone else? The length of the ladder measured inside the frame certainly doesn't vary. Using the requirement of "local" measurements, I don't understand how constant acceleration must vary with height along the ladder.

11. Jun 1, 2014

### A.T.

Proper acceleration measured with an accelerometer (or a scale) is frame invariant.

If the length of the ladder doesn't vary in the rest frame of the ladder, then the ladder gets shorter in the initial inertial frame (length contraction). To get shorter the back must have more coordiante acceleration, than the front. This coordinate acceleration in the inertial frame is also the proper acceleration measured with accelerometers attached to the ladder.

If the proper accelerations were the same along the ladder, the ladder would be stretched and eventually break. See:

12. Jun 1, 2014

### m4r35n357

In talking about the need for the ladder to retain its shape I believe that PeterDonis is describing "Born rigid acceleration", this may be a new concept to you so try looking into that.
The second issue is just the difference between a "parallel" field and a "spherical" one. Two points travelling inwards along different radial lines will approach each other, that is all. The radially separated particles getting further apart is the other part of tidal gravity, but I think you understood that (the field gets weaker with altitude).

13. Jun 1, 2014

### Staff: Mentor

Once again, no. If the universe worked according to Newtonian physics, yes, you would. But the universe works according to relativity, and so you don't.

Please read the Wikipedia page I linked to on Rindler coordinates. Also please remember the key qualifier I gave: the acceleration must vary with height if the length of the accelerating spaceship remains constant.

Also take a look at the Bell Spaceship Paradox; this illustrates what happens, in relativity, if you do actually stipulate that the acceleration remains absolutely constant with "height".

What happens in this case is that the length of the spaceship does *not* remain constant; it stretches (gets longer) over time. (In the usual formulation of the "paradox", there are two spaceships connected by a string and the string's length is what is shown to increase with time; but the same logic applies to a single long spaceship.)

In other words, what these two scenarios (Rindler and Bell) are telling you is that, contrary to your intuition derived from Newtonian physics, it is impossible in relativity to have an object (or a "frame") in which objects that remain at rest relative to each other can all have exactly the same constant proper acceleration.

There is no such thing in relativity; that is, in the "frame" (a better term would be the Rindler coordinate chart) in which the accelerating spaceship is at rest, there is no single "acceleration"; it varies with spatial position (i.e., height).

Correct.

Last edited: Jun 1, 2014
14. Jun 1, 2014

### Staff: Mentor

Perhaps it will help if I point out another counterintuitive thing that relativity tells us: in this long spaceship with the ladder inside it that is accelerating, the "rate of time flow" also varies with height; time flows more slowly at the bottom of the latter (where the weight is larger) than at the top (where the weight is smaller). Einstein constructed a thought experiment to show this as part of an argument that showed that gravitational time dilation should exist; it's described in his book Relativity: The Special and the General Theory, which I recommend reading.

15. Jun 1, 2014

### randyu

wow, thanks guys,
thanks PeterDonis, I realize that it doesn't sound like I'm doing my reading, but I am. Thanks for your patience.
thanks D H, A.T and m4r35n357, your input has been valuable to me.

I say wow, because it's all starting to make some better sense to me now. So then, when referring to "local" events and experimental measurements, are we really inferring Newtonian physics interpretations?

Also, it seems like this all somehow ties into why light speed measurements here on earth always show isotropic behavior. At least that's the way I had understood it. But I'm confused by something I read Peter, that said: Ref: http://adsabs.harvard.edu/abs/1988AmJPh..56..811B

What does that mean, "any anisotropy in the speed of light is contrived and not physically significant". Why is it not physically significant? Seems like it SHOULD be. And how is it contrived? Again, I'm confused.

16. Jun 1, 2014

### Staff: Mentor

Not really, because even "local" measurements don't always match up with Newtonian predictions. For example, Einstein's thought experiment showing that there must be time dilation in an accelerating spaceship describes a measurement that, according to our modern definitions, is "local"; meaning, the result can be detected even over a region of spacetime small enough that tidal gravity effects are not detectable.

Btw, yes, that implies that the difference in acceleration with height that I've been referring to, since it is present even in flat spacetime, could also be detectable even over a region of spacetime small enough that tidal gravity effects are not detectable. Going back and reading D H's original response to you in this thread, I think this means that he was not quite correct. The difference between being in an accelerating spaceship and being in a gravitational field is that tidal effects, as I described them in an earlier post, are present in the latter case but not the former. The change in acceleration with height, in and of itself, is not a tidal effect; you have to look more closely at the behavior of freely falling objects that start out at rest relative to each other, as I described.

This probably deserves a separate thread (IIRC we had one on this general topic not too long ago, but I can't find it right now). The full paper is behind a paywall so I can't read it, and I'm hesitant to speculate on what is meant just based on the abstract. The most I can say is that it doesn't seem to me like what this abstract is talking about is related to the issues we're discussing in this thread.

(One thing in this abstract does puzzle me, and makes me more hesitant to speculate on what the paper means based on just the abstract. It says:

This doesn't make sense to me; AFAIK in standard relativity (special or general), the "invariant interval" along a timelike curve can *always* be identified with the lapse of proper time along the curve. So I don't understand what the abstract is talking about here. It's quite possible that this paper was speculative and not "mainstream" physics; not every paper that gets published in a physics journal is correct.)

17. Jun 1, 2014

### Staff: Mentor

One way to see this mathematically is to look at *how*, precisely, the acceleration varies with height in the two cases. In the case of the Earth's gravitational field, we have

$$a = \frac{GM}{r^2}$$

Whereas, in the case of the accelerating spaceship in flat spacetime, we have

$$a = \frac{c^2}{d}$$

where $d$ is the distance from the spaceship's "Rindler horizon" (note that for a 1 g acceleration, the Rindler horizon is about 1 light-year below the bottom of the spaceship). So in the former case, we have an inverse square dependence on height, but in the latter case, we have an inverse linear dependence on height. It's fairly straightforward to show that, for anything other than the inverse linear dependence on height, tidal effects will be present; but that they will not be with the inverse linear dependence.

18. Jun 1, 2014

### randyu

Thanks again. I think I get what you are describing about tidal effects, accelerating frame and gravity. But, I do think light speed fundamental behavior is tied together with the geometric formulation of gravity.

$G =\ 8πG/(c*4)\ \times\ T$ (sorry for the mess, I need a course in latex too)

If c were not physically isotropic would this still be correct. I mean, isn't light's characteristics fundamental to general relativity.

19. Jun 1, 2014

### Staff: Mentor

$c$ in this formula is a conversion factor for units--it converts between distance units and time units. It's not an actually measured quantity; it's a universal constant by definition, because unit conversions are universal constants. Whether or not the actually measured "speed of light" is isotropic is a separate question from what value of $c$ we use to do unit conversions.

Yes, but "speed" is not one of those characteristics. The fundamental characteristic of light for relativity is that it moves on null curves, i.e., the invariant "path length" along the spacetime curve that light follows is zero. This is a geometric property, not a "speed".

20. Jun 1, 2014

### WannabeNewton

The idea that gravity is a manifestation of dynamical curved space-time geometry does not require relativity, as Newtonian gravity can also be cast in the same geometric framework and Newtonian gravity does not know about relativity, so to speak. However it is certainly true that relativistic principles play a role in the myriad of properties of dynamical space-time in general relativity, the least of which (or perhaps greatest of which) is global causal structure.