# Understanding the equivalence principle

1. Apr 10, 2015

### Jimster41

Mentor Note: This thread has been separated out of a previous one to keep discussion of this particular topic open, since the previous thread has been closed.

Sorry, the opportunity to ask questions is just too hard to resist.

I'm still searching for the right understanding of equivalence. Or that is what has got me in trouble. I had been picturing it as "there is no difference in the reality experienced by an object traveling in a curved (accelerated) world-line in a flat space-time and that experienced by an object traveling in a straight world-line (at rest) in a curved space-time" - and so the same mechanism must be at work in both cases. I must have that simplification wrong in a fundamental way somehow.

This lead me to imagine that although no one of the travelling twin's accelerations, through which he's squeezing some time-savings out of flat $M4$ space-time by taking a longer path than sis, would be a significant "bending" of space-time, but that integrated over his path the result would have to be equal (due to equivalence) to the curvature from mass that would have been required to provide the same difference of geodesic length if he had descended? from his sister (in altitude) in a curved space around that mass the whole time. I thought that was Einstein's "virtual field" allusion. So this is where your correction is very much to my confusion.

Lot's to learn. But this has been a good set up for that chapter on curvature.

Last edited by a moderator: Apr 10, 2015
2. Apr 10, 2015

### Staff: Mentor

No, that's not correct. It's also not what the EP says, but let's first see why it's not correct.

The physical observable that corresponds to a curved worldline is proper acceleration, i.e., what is measured by an accelerometer. (Or, equivalently, you can think of it as "weight"--the force you feel.) So there is an obvious difference between a object traveling on a curved worldline in flat spacetime, and an object traveling on a straight worldline in curved spacetime: the first has a nonzero accelerometer reading, while the second has a zero accelerometer reading.

(Btw, a straight worldline does not have to be "at rest". Any worldline with zero proper acceleration, i.e., zero force felt by an object traveling on that worldline, is straight. In any coordinate chart, there will be many such worldlines that are moving, not at rest.)

Now, as to what the EP actually says: there are different ways of phrasing it, but here's one I like: locally, being accelerated in flat spacetime and feeling a given proper acceleration $g$ is equivalent to being at rest in a gravitational field in curved spacetime and feeling the same proper acceleration $g$. So, for example, if I'm standing in an enclosed room on the surface of the Earth, I could not tell, by measurements made just within the room (in other words, "locally"), that I was not inside a rocket far out in empty space whose engine was applying a 1 g acceleration.

Note that this is purely about curvature of the worldline--basically it says that any worldline with a given curvature (proper acceleration) is indistinguishable, locally, from any other worldline with the same curvature. But this says nothing about the curvature of the spacetime; spacetime curvature is irrelevant, because of the "locally". What "locally" means is that we only look at a small enough region of spacetime such that any effects of spacetime curvature are negligible. So worldlines with the same curvature are locally indistinguishable even if they are in spacetimes with widely differing curvatures.

3. Apr 10, 2015

### Jimster41

So in "Interstellar" the explanation for why Mathew McCaunaughey and Anne Hathaway aged more slowly when they went down to the planet close to the Black-Hole, then came back up to the mother ship in orbit after, was really about how much proper acceleration they had to experience to take that curved world-line rather rather than a straight word-line which would have meant free-falling into the black hole (in that curved space). The effect is no different than if the same proper acceleration had been applied to a twin scenario round trip (curved world-line) in flat space-time, away from the mother ship then back.

I do hope I got that at least fairly right.

Last edited: Apr 10, 2015
4. Apr 10, 2015

### Mentz114

Yes, there is an equivalence there.

But think you're attributing too much to proper acceleration. It is feasible to have your twins start at the same place in outer space and that each be given a nudge which sends them on different looping paths that meet again later. Neither has experienced any proper acceleration ( bar the nudge) but they caould still have different ages when they meet.

Elapsed clock times depend on the space-time path. That's all.

5. Apr 10, 2015

### A.T.

It's not quite the same. In curved space time, proper acceleration alone cannot even tell you who aged more. Gravitational time dilation is a function of gravitational potential difference rather than proper acceleration difference.

6. Apr 10, 2015

### Jimster41

I think I understand what you are saying. In the case of curved space-time it's a function of both the amount/shape of curvature (the potential) and the proper acceleration applied to create a curved path through it?

7. Apr 10, 2015

### Staff: Mentor

I think the best way is to start from the excellent summation by Mentz114:

That's the only general rule that works in all situations. Anything else, whether it be proper acceleration, gravitational potential, etc., is just a proxy for "depends on the spacetime path", and any such proxy will have limited applicability--you will always be able to find scenarios where it doesn't work (for example, Mentz114 gave an example where proper acceleration as a proxy doesn't work--two free-fall paths between the same events in curved spacetime can still have different lengths, even though they both have zero proper acceleration). The best thing is to just compare the spacetime paths.

In the "Interstellar" scenario, for example, what makes the difference is not that the astronauts had to accelerate to get down to the planet, and then accelerate again to get back up to the mother ship. (Note that both the mother ship and the planet are in free-fall orbits, so they have zero proper acceleration--yet another illustration of why proper acceleration is not a good proxy for "length of spacetime path" in this scenario.) What makes the difference is that the spacetime path they follow is shorter. In this case, gravitational time dilation works better as a proxy for "length of spacetime path", because the situation is stationary--the hole's rotation doesn't change, and the orbits of the planet and the mother ship don't change, so the only variable is altitude--the planet is deeper in the hole's gravity well than the mother ship, so there is more gravitational time dilation there. (This is what A.T. was saying.)

No, this is certainly not correct. Even in the twin paradox scenario in flat spacetime, although proper acceleration is what allows the traveling twin to come back and meet the stay-at-home twin again, the amount of proper acceleration is not what determines the difference in aging. What determines that is how fast the traveling twin moves, relative to the stay-at-home twin, and how far away he goes before he turns around. In other words, length of spacetime path.

8. Apr 10, 2015

### DrGreg

9. Apr 10, 2015

### Jimster41

Thanks guys. Clearly you have been down this road before...a few times.

And I appreciate the memorable rule of thumb: Elapsed clock time depends on length of space time path.

I think I am beginning to understand why the s.t. Diagrams are so important. I did a quick search. I have a hard time imagining one showing the case of curved geodesic in curved s.t. Is it correct to call a curved path in a curved s.t. A "curved geodesic" since a plain geodesic is a straight path in either flat s.t. or curved s.t.

I'm going to dive back into my GR textbook with renewed urgency, so last question,
Is it more correct to say that relative elapsed clock time of two observers is always dependent on relative length of s.t. path, since there is not any absolute (extrinsic?) metric for the length of s.t. path.

Is the point of the s.t. diagram to create a reference coordinate frame of some kind (from one or the other observer, or from some third observer's reference frame) that allows a single ruler to be used to establish the path length difference.

Which is to say I kindof get the idea that only length of s.t. path is telling. Now I have the question of how the difference in path length is established. Comparison of proper observables is all you need, or Do you have to define some sort of commom metric frame? Seems like proper observables is insufficient, or else the whole drama of the returning twin would be moot. "How'd you get so old when I've only been gone a few hours?"

Last edited: Apr 10, 2015
10. Apr 10, 2015

### Staff: Mentor

Not really; the word "relative" is not appropriate since the length of a particular spacetime path between two given events is an invariant--all observers agree on it.

Not really. The point of the spacetime diagram is to be a diagram of the geometry of spacetime and the particular paths of interest in it. Think of it as like a chart of the Earth's surface. Different reference frames correspond to different charts; they show the same spacetime geometry from different perspectives, just as different charts can show the same geometry of the Earth's surface from different perspectives.

Yes. The path length itself is a direct observable--it's just the elapsed time on a clock that follows that path through spacetime.

11. Apr 11, 2015

### Mentz114

As PeterDonis has said, a clock measures the 'proper interval' ( st path). The metric is what defines the proper time. In Minkowski spacetime the infinitessimal element of proper time $\tau$ is

$d\tau^2=dt^2 -dx^2 -dy^2-dz^2\ \rightarrow\ \tau=\int_{start}^{end} d\tau$ (setting $c^2=1$)

If we calculate $\tau$ from another frame where $dx=\gamma dx' + \gamma\beta dt',\ dt=\gamma\ dt' + \gamma\beta\ dx'$ by substituting into the metric, the metric remains unchanged. Try it.

So $\tau$ is the same from all frames connected by a Lorentz transformation.

It is more than 'a rule of thumb'.

12. Apr 11, 2015

### Jimster41

I didn't mean it was just a rule of thumb. I just meant it was something (a correcting rule) I can remember when just trying to contemplate what the heck space-time is.

I believe in the language of math. I think it's has special power. And I would love to know what it feels like for equations to serve the purpose of contemplative thought in their own right. I believe that for some they do. A handful do for me, and the ones above are beginning to. Reading Susskind, I realized, he likes to think in these terms. It's natural. Or at least that's the way it seems. For me at least, it's agonizing, all compelling thought just sort of grinds to a halt, connotative sensation evaporates, and I'm just trying to follow the symbols.