# Equivalence principle

1. Dec 7, 2006

### nikolatesla20

Sorry for so many questions.

In Einsteins equivalence principle, it states that an observer in a close room at 1G accel would not know the difference between that or whether he was standing on the Earth. This makes fine sense to me. But it also states as well that it is independent of velocity.

The problem I'm wondering about is, the artificial force is created by the room's acceleration. Once the room stops accelerating, it will no longer generate this force. So, would you not eventually reach a speed limit? You would have to keep going faster and faster to keep up the force.

1. Imagine a main inside the room. The room is accelerating. The man feels a downward force.

2. Now, the man jumps up. The room is still accelerating, so the room comes up to meet him in the air. To the man, it appears he has fallen back down.

3. Now, the room stops accelerating, and just continues to move in a continuous velocity. The man jumps, but the room does not come to meet him, because now the man, and the room, are moving at the same velocity. In fact, the man is now moving at a higher velocity, since he jumped. He would move towards the ceiling of the room.

This is the problem I have with that illustration, anyway. The room would have to accelerate forever.

So if this is true, how can "warped spacetime" cause "constant acceleration"...so we are, in a sense, pushed into the Earth.

-niko

2. Dec 7, 2006

### wxrocks

Think of the movie 2001 -- or any time you see the G-Force simulator -- they are moving in a circullar motion. Because you are continually changing direction, you are always accelerating in a different direction.

3. Dec 7, 2006

### nikolatesla20

Yes, but in this case I am not referring to moving in a circle. The room moves in a straight line.

I'm just pointing out that the illustration, while helping one to understand the ideas, it flawed in that it cannot work "forever". Also to go along with that, it does not help understand why a body "wants" to continue "accelerating".

I mean if a body is following a free-fall inertial frame of spacetime, but the Earth keeps them from doing so (so that body feels the force of the Earth against them) - why doesn't the body just "stop". Just like a body at rest tends to stay at rest. I don't understand the connection here. Why does a body want to keep trying to move thru the Earth instead of just stopping, and then of course feeling no force at all?

-niko

4. Dec 7, 2006

### pervect

Staff Emeritus
You can accelerate at a constant rate (say 1G for definitness) for as long as you like, and you will never quite reach the speed of light.

This is ultimately a consequence of the way velocities add in special relativity.

vtotal = (v1 + v2) / (1 + v1*v2/c^2)

For the details of your velocity vs time for a constant acceleration, see the webpage on the "relativistic rocket equation".

http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]

Last edited by a moderator: May 2, 2017
5. Dec 8, 2006

### nikolatesla20

Of course you can accelerate for as long as you like, and you never can get to the speed of light.

But the analogy falls apart is my point. You have to keep going faster and faster to keep up the gravitational force, because it does not work unless you are accelerating. And, the faster you go, the more your mass increases, and the more you weigh. In fact you have 2 curves:

1. In one curve, you have to go faster. As you go faster you weigh more (according the relativity theory)

2. Since you weigh more, you need less acceleration over time, to feel the same effects.

At some point, these two curves pass over each other. After that point, you will weigh so much that you will be barely accelerating at all.

Isn't it considered somewhat strange that since spacetime is curved (I know that of course relativity is always somewhat non-intuitive) - but what I mean to say is, if free-fall and gravitation are the same, then there is one connection I am missing. When a object is in free fall in space, it will not move unless acted upon. According to relativity theory, an object under "gravity" is really moving thru spacetime, which is curved. The object is in free-fall, so it wants to continue moving in the shortest path, which is in this case the geodisc of spacetime.

However what I don't get yet, is why does the object "want" to keep moving. Once it hits the earth, its inertial motion would be stopped. For example, if we have a free-moving object in inertial motion in space, it feels not gravity. If something pushes against it with an acceleration, it feels a force because of inertia. If that acceleration stops, the object continues until it hits something. When it hits that something, it feels a force again until its momentum is spent. Then it would simply be stopped again, feeling no force.

So why, if we are following curved space because of our inertia, do we continue to accelerate and keep banging into the earth. It seems it needs to be more than just our own inertia. Somehow our body wants to be in a lower energy state or something. Otherwise why would it keep attempting to accelerate along the curvature.

-niko

6. Dec 8, 2006

### HallsofIvy

Yes, that's true. In order to emulate gravity "forever", the room would have to accelerate "forever".

Now, I'm completely lost. What does this have to do with the previous scenario? The "principal of equivalence" says gravitational force and and constant acceleration have, locally, the same effect. It doesn't say the ARE the same!

7. Dec 8, 2006

### nikolatesla20

Well the whole reason Einstein (the given reason anyway) came up with the idea of equivalence principle relating to gravity is because he noticed the effects of inertia could be similar, so you could think of gravity in "reverse" - instead of gravity pulling us down, it's really our own inertia along a spacetime curve that forces us in. At least I thought it was talking about inertia. It sure seemed like he was thinking of them as being the same.

Perhaps it should have rather been called "The similarity principle" ?

-niko

8. Dec 8, 2006

### JesseM

Mathematically, I believe the Equivalence Principle is always described the other way around, i.e. in terms of the equivalence between an observer moving inertially in flat spacetime and an observer in freefall in a gravitational field. In the limit of a smaller and smaller patch of spacetime, the laws of physics measured by a freefalling observer passing through this patch will get arbitrarily close to those measured by an inertial observer in flat spacetime, so that GR reduces to SR "locally". This could be stated in some precise way in terms of the equations of GR.

But you can derive the equivalence between an accelerating observer in flat spacetime and an observer at rest in a gravitational field from this more fundamental version of the equivalence principle. Just imagine the freefalling observer passing by a small box which is at rest in the gravitational field (sitting on a platform on top of a pole planted in the ground, say, with the ground outside of the patch of spacetime that the freefalling observer is making their observations in), and which contains tiny experimenters doing their own experiments--the freefalling observer will see this box moving upwards in exactly the same way that an inertial observer in flat spacetime would see a similar box that was accelerating "upwards" at a constant rate (with this box sitting on a platform on top of a pole which is connected to the nose of an accelerating rocket, with the rocket outside of the patch of spacetime that the inertial observer is making their observations in). So, because of the equivalence principle between the inertial observer and the freefalling observer, they both should see exactly the same results for any experiments that are being done by the mini-experimenters in the box.

Last edited: Dec 9, 2006
9. Dec 8, 2006

### rbj

"the faster you go" relative to whom? relative to yourself, you're not faster and your mass is no different in time if you are experiencing a constant acceleration.

from the point of view of some "stationary" observer (let's say unaccelerated, which is an absolute property), you appear to be increasing velocity in the direction of motion (but that might be perceived as decreasing velocity if, from this person's frame of reference you were originally already speeding away from him/her and your acceleration was in this observer's direction. then your relative velocity is decreasing as is also the relativistic mass.)

but in your own frame of reference, if an energy and propellant source was no problem, you could accelerate forever at 1 G and it would feel just like gravity (but eventually the nearby stars might start looking funny).

10. Dec 9, 2006

### nikolatesla20

Agreed exactly.

Just like I said it is kinda neat - that as you go faster, your mass increases, so it takes less acceleration to feel the same effect, so eventually you will be hardly accelerating at all, just before you hit the speed of light but you will never get there (the limit).

Now, I was just wondering from that analogy, if that is the state our atoms are in..cause I don't feel like I'm accelerating into the earth more and more..

So i was just observing that the by the analogy's standpoint you have to keep accelerating forever to keep up the force, but eventually, it will be as though (or very close to) not accelerating at all.

:P

-niko

Last edited: Dec 9, 2006
11. Dec 9, 2006

### pervect

Staff Emeritus
You might want to read http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html

which I will quote in part.

Note that the "relativistic mass" you are using is not independent of the observer. If you have a mass of 1kg, and are sitting still, someone whizzing by you at .999 c will say that you have a much higher relativistic mass. So you can't really say what your relativistic mass is unless you specify an observer. You seem to be tending towards the idea that relativistic mass is some measure of an "absolute velocity". (At least that's the impression I get). This is unfortunatealy a common idea, but it's totally incorrect.

Space-time in the rocket is not really curved, in the sense that the Riemann curvature tensor of the space-tiem of the accelerating rocket is zero.

The sense in which gravity in the rocket is the same as on a planet is related to the Christoffel symbols of the space-time rather than the curvature tensor. However, there doesn't seem to be be any clear way to describe the difference between these two different mathematical objects without using technical language which unfortunately a lot of people don't understand.

What this means in practice is this:

The gravity in the accelerating rocket is the same at a point, but the tidal forces (the rate of change of the gravitational forces) are not the same, even at a point.

I'm not sure I follow this question, but bodies do not naturally accelerate. They will accelerate only if a force is applied to them. Your accelerating rocket, for instance, is not following a geodesic in space-time. It is departing from a geodesic from the force caused on the rocket by the rocket exhaust.

12. Dec 10, 2006

### marlon

Hey Niko,

Yes, the acceleration of the room needs to last forever. So if the acceleration stops, so does the artifical force.

This is a classic question. The answer is that velocities in special relativity (SR) do not follow the classical addition laws. I mean, one cannot just add two velocities they way you do it in classical physics. I suggest you give some special attention to the velocity transformation laws in SR. You will notice that "adding" velocities in SR will not allow you to exceed the speed of light (as many people initially think)

greets

marlon

13. Dec 11, 2006

### dupont

14. Dec 11, 2006

### JesseM

This is a confused objection. The equivalence principle is about the equivalence between freefall and inertial movement in flat spacetime--when you land, it's because a force is acting on you, so of course you're no longer in freefall! Also, you must remember that the equivalence principle is only defined locally, ie in a small region of space over a small interval of time. It says that if a freefall observer does some experiments in this patch of spacetime, he won't see any different results than if an observer moving inertially in flat spacetime did the same experiments. That is all the equivalence principle is saying, it's true that over an extended period of time the freefall observer could figure out he was in freefall due to the landing (and also due to tidal forces), but that does not contradict the equivalence principle.
What's your point? If there are no gravitating masses, then spacetime is flat, and "free" movement is just inertial movement. It would be silly to point out that there is an equivalence between this and inertial movement in flat spacetime, since it already is inertial movement in flat spacetime.
I don't understand what you're saying--are you arguing the equivalence principle predicts they can't be synchronized, when in fact they can be synchronized? If so, why do you think this would be a prediction of the equivalence principle?

15. Dec 12, 2006

### dupont

dear JesseM

The equivalence principle (ep) has two formulations, either the free fall-inertial frame one or the standing still-accelerated frame one.
1) If the principle has any meaning, I must be able to transpose the results from one situation to the other. Spacetime changes then must be predicted by the ep to form the basis for a theory of gravity. So, I have to look at how clocks strike in an accelerated frame in flat space-time. There, I find that clocks at a different height (i.e. at a distance along the direction of acceleration) strike differently, in particular, one finds blue and red shifts. There, we have agreement between this prediction and measurements in gravity at rest.
If I do the same thing perpendicular to the direction of acceleration, I get disagreement. So, the ep can only be used for completely local measurements.
2) talking of local measurements
In an inertial frame without gravity, I never encounter any graviating masses, a no-brainer, of course.
In free fall, I need gravitating masses, so I can encounter them and measure them locally. This does not stop my free fall necessarily but it constitutes a local measurement that distinguishes free fall and inertial motion. This contradicts the ep.

16. Dec 12, 2006

### JesseM

Whenever I look it up in textbooks, the precise mathematical form of the equivalence principle is always stated in terms of freefall being equivalent to inertial movement locally (i.e., GR reduces to SR locally). Of course it isn't too hard to show that the standing still in gravity/accelerating in empty space equivalence follows as a necessary consequence, see the argument I made in my first post on this thread.
I still don't understand your argument. If you have two clocks a short distance apart in your accelerating ship, perpendicular to your direction of acceleration, you can indeed "synchronize" them in the sense that if they are sending signals to each other with each tick, each clock will itself tick at the same rate that it receives signals from the other clock, so they don't get out-of-sync over time. But exactly the same is true of two clocks arranged horizontally (i.e. the same height above the ground) a short distance apart in a gravitational field! Why do you think it wouldn't be?
The equivalence is in the way the laws of physics work locally, not in terms of what you see passing by you over time (anyway, since you are only making local measurements, over a short time-interval and in a small region you can only see small amounts of mass passing you, not enough to appreciably curve spacetime). You might as well argue that since a person driving his car on earth would see something different when looking out his window than a person driving his car on Pluto, that proves that the laws of physics don't possess translation invariance (the property of remaining unchanged at different points in space).

Last edited: Dec 12, 2006