The principle of equivalence

by SinghRP
Tags: equivalence, principle
 P: 73 What’s your understanding of the principle of equivalence? In the literature, I find two meanings: (1) Gravitational mass is numerically equal to inertial “mass.” (This is a postulate.) (2) A mass at rest in a frame is equivalent to being in a “gravitational field” in an accelerated frame. (This one is a pseudogravity.) General relativity does not seem to explain either!
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 Quote by SinghRP What’s your understanding of the principle of equivalence? In the literature, I find two meanings: (1) Gravitational mass is numerically equal to inertial “mass.” (This is a postulate.) (2) A mass at rest in a frame is equivalent to being in a “gravitational field” in an accelerated frame. (This one is a pseudogravity.) General relativity does not seem to explain either!
There is experimental evidence for (1) See http://en.wikipedia.org/wiki/E%C3%B6...B6s_experiment
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 Quote by SinghRP (1) Gravitational mass is numerically equal to inertial “mass.” (This is a postulate.)
It's a postulate as far as the theory is concerned, but it's also an observed fact, at least to as high an accuracy as we have tested. See the link Mentz114 gave.

The key thing about this version of the principle is that it enables us to view gravity as due to spacetime curvature; if the ratio of gravitational mass to inertial mass were different for different objects, we could not do that.

 Quote by SinghRP (2) A mass at rest in a frame is equivalent to being in a “gravitational field” in an accelerated frame. (This one is a pseudogravity.)
This is not quite correct as a statement of this version of the principle. The correct statement is: Being at rest in an accelerated frame (for example, inside a rocket whose engines are firing), and feeling a certain force (say 1 g), is equivalent to being at rest in a gravitational field and feeling the same force (for example, being at rest on the surface of the Earth and feeling a 1 g force).

 Quote by SinghRP General relativity does not seem to explain either!
What makes you think that?

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The principle of equivalence

 Quote by SinghRP What’s your understanding of the principle of equivalence? In the literature, I find two meanings: (1) Gravitational mass is numerically equal to inertial “mass.” (This is a postulate.) (2) A mass at rest in a frame is equivalent to being in a “gravitational field” in an accelerated frame. (This one is a pseudogravity.) General relativity does not seem to explain either!
The way I would put it is a variation of number (2):

(2') An object in freefall in a gravitational field is locally equivalent to an object travelling inertially in gravity-free space.

In other words, the local effects of gravity can be made to vanish by choosing freefall coordinates.

For example, in gravity-free space, a tiny charged particle will obey an equation of motion given by:

$m \dfrac{dU^\mu}{d\tau} = q F^\mu_\nu U^\nu$

where $U^\mu$ is 4-velocity, $\tau$ is proper time, $q$ is the charge, and $F^\mu_\nu$ is the electromagnetic field strength tensor (which incorporates both the electric and magnetic fields).

The exact same equation will hold locally in the presence of a gravitational field, provided that you choose "freefall" coordinates.
 P: 284 I believe there are a couple of equivalence principles known as weak (WEP) and strong (EEP). The weak is the one about the equivalence of masses, the strong is really a statement about co-ordinate independence, gauge, diffeomorphisms etc.
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 Quote by stevendaryl (2') An object in freefall in a gravitational field is locally equivalent to an object travelling inertially in gravity-free space.
I realize this is largely a matter of definition, but to me this is part of (1), not (2). The reason we can make the local effects of gravity vanish is that inertial and gravitational mass are equal; if they weren't we couldn't do that. This is another way of saying that we can view gravity as a manifestation of spacetime curvature because inertial and gravitational mass are equal.

(2) is talking about something different: what happens when objects are *not* in free fall. The key point about (2) is that equal proper acceleration is what defines "equivalent" states of motion. The free fall case, zero proper acceleration, can be viewed as a special case of this; but that special case alone is not enough. We need the full principle of equivalence, covering *all* possible values of proper acceleration (not just zero), to justify the full machinery of GR for dealing with all kinds of motion in curved spacetime, not just inertial motion.
 P: 73 To PeterDonis: Answer to "What makes you think that? I get lost in mathematical jungle. I am a physicist, and I like to think in terms of physical models. I recall Feynman also: The glory of mathematics is that you don't have to say what you are talking about. The genesis of POE is in: (Gravitational mass) (Gravitational field intensity) = (Inertial mass) (Acceleration). (I agree with your "corrected" stateent.) But my statement is just an interpretation the above. Another expression: (Electical charge) (Electrical field intensity) = (Inertial mass) (Acceleration). I may interpret it as: an electrical charge at rest in a frame is equivalent to being in an "electrical field" in an acclerating frame.
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 Quote by SinghRP Another expression: (Electical charge) (Electrical field intensity) = (Inertial mass) (Acceleration). I may interpret it as: an electrical charge at rest in a frame is equivalent to being in an "electrical field" in an acclerating frame.
You cannot say that, because the resistance to motion of the charge is supplied by the matter, and the mass does not cancel from the acceleration equation as it does with gravity.

See post#10 below.
 P: 1,412  You're completely right, what I wrote was not as general as I had intended. What I meant to say was that physics in a gravitational field when described by freefall coordinates is equivalent to physics in a gravity-free space when described by inertial Cartesian coordinates. That's not what I said.
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 Quote by SinghRP To PeterDonis: Answer to "What makes you think that? I get lost in mathematical jungle. I am a physicist, and I like to think in terms of physical models. I recall Feynman also: The glory of mathematics is that you don't have to say what you are talking about. The genesis of POE is in: (Gravitational mass) (Gravitational field intensity) = (Inertial mass) (Acceleration). (I agree with your "corrected" stateent.) But my statement is just an interpretation the above. Another expression: (Electical charge) (Electrical field intensity) = (Inertial mass) (Acceleration). I may interpret it as: an electrical charge at rest in a frame is equivalent to being in an "electrical field" in an acclerating frame.
Not quite. Mathematically,

$M_{inertial}\ \vec{A} = M_{grav}\ \vec{g}_{grav}$

where $\vec{g}_{grav}$ is the gravitational field.

Since $M_{grav} = M_{inertial}$, you can divide through by $M_{grav}$ to get:

$\vec{A} = \vec{g}_{grav}$

Which means that acceleration due to gravity is universal, the same for all objects, regardless of their mass, or what they are made out of, or whatever. That's the principle character of "fictitious" or "inertial" forces such as the "g" forces that arise in an accelerating rocket. This means that gravity can be understand as locally equivalent to a fictitious or inertial force.

In contrast, if you start with the force due to electric fields, you have:

$M_{inertial}\ \vec{A} = Q \vec{E}$

where $Q$ is the electric charge, and $\vec{E}$ is the electric field.

If you do the same trick of dividing through by $M_{inertial}$, you find:
$\vec{A} = \dfrac{Q}{M_{inertial}} \vec{E}$

So the acceleration due to electrical forces is not universal; the acceleration depends on the charge-to-mass ratio. So electrical forces can't be interpreted as inertial, or fictitious forces, since they accelerate different objects in different ways.
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 Quote by PeterDonis I realize this is largely a matter of definition, but to me this is part of (1), not (2). The reason we can make the local effects of gravity vanish is that inertial and gravitational mass are equal; if they weren't we couldn't do that. This is another way of saying that we can view gravity as a manifestation of spacetime curvature because inertial and gravitational mass are equal. (2) is talking about something different: what happens when objects are *not* in free fall. The key point about (2) is that equal proper acceleration is what defines "equivalent" states of motion. The free fall case, zero proper acceleration, can be viewed as a special case of this; but that special case alone is not enough. We need the full principle of equivalence, covering *all* possible values of proper acceleration (not just zero), to justify the full machinery of GR for dealing with all kinds of motion in curved spacetime, not just inertial motion.
Doh! I misread my own post. You're exactly right, and I was wrong. What I meant to say was that physics in a gravitational field described using freefall coordinates is locally equivalent to physics in gravity-free space described by inertial coordinates.
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[Edit: Looks like our posts crossed in the mail, so to speak. The following may be superfluous, but I'll leave it in case there is any further comment.]

 Quote by stevendaryl I didn't say what is being described is in free fall, I said that you're using freefall coordinates to describe it. You can use inertial coordinates to describe the motion of an object that is accelerating, and you can similarly use freefall coordinates to describe the motion of an object that is not in freefall.
As long as inertial and gravitational mass are equal. If they're not, then trying to set up freefall coordinates in the presence of gravity will fail. That's why I said I view the use of freefall coordinates as an aspect of (1). But it is, as I said, a matter of definition; what's really important is the physics and how we can usefully model the physics, and I agree that local inertial frames in curved spacetime are the key to doing that.

 Quote by stevendaryl If you know how the physics works in one coordinate system (freefall coordinates), you can certainly do a coordinate transformation to find out what the equations look like in another coordinate system (such as a coordinate system in which an object on the surface of the Earth is considered at rest). It's just a coordinate change, and that's just mathematics, not physics.
Agreed. But there still remains the question of what the result will actually look like when you do this, which is what the (2) part of the principle as the OP stated it is talking about, IMO. See below.

 Quote by stevendaryl The physical content is exhausted by saying that freefall coordinates are the same (locally, which basically means ignoring the variation of the metric tensor with location) as inertial coordinates.
No, this isn't enough by itself, because this doesn't tell you what, specifically, being at rest in a gravitational field looks like in freefall coordinates. That's what the (2) part of the principle is for: it tells you that being at rest in a gravitational field is equivalent to following a particular kind of accelerated worldline in a local inertial frame.

Remember that Einstein originally enunciated the equivalence principle before he had derived the Field Equation. Today we would first solve the Field Equation to derive the Schwarzschild solution, and then observe that an object at rest in Schwarzschild coordinates follows an accelerated hyperbola in a local inertial frame; but Einstein couldn't do that yet. So one way of looking at (2) is that it was Einstein trying to guess what a particular solution would look like to a field equation he hadn't yet derived.
 PF Patron Sci Advisor P: 4,473 Well, if a free fall frame has the standard physics of SR locally, then it follows that any other frame will behave like an accelerated frame in SR. This is why modern authors use different (but essentially similar) EP's to Einstein. There are many categorizations used. The following is a common one, from: : http://relativity.livingreviews.org/.../fulltext.html Weak Equivalence Principle: "the trajectory of a freely falling “test” body (one not acted upon by such forces as electromagnetism and too small to be affected by tidal gravitational forces) is independent of its internal structure and composition. " Einstein Equivalence Principle: "1) WEP is valid. 2) The outcome of any local non-gravitational experiment is independent of the velocity of the freely-falling reference frame in which it is performed. 3) The outcome of any local non-gravitational experiment is independent of where and when in the universe it is performed." Strong Equivalence Principle: "1) WEP is valid for self-gravitating bodies as well as for test bodies. 2) The outcome of any local test experiment is independent of the velocity of the (freely falling) apparatus. 3) The outcome of any local test experiment is independent of where and when in the universe it is performed."
 P: 73 To PeterDonis - After rereading my statement (2) and then reading yours, I think I found the problem. I restate (2) as follows: A mass at rest in a frame is equivalent to being in a "gravitational field" when the frame is acclerating. To PAllen: I will read your post and come back to you after I understand it. To all: I will work out a text to explain without invoking POE: the bending of light beam, the speed of time, and the length of a rod in an acclerating rocket far from any mass; centrifugal force, and weightlessness. I will post it tomorrow. The clearest exposition of POE is given by George Gamow in -- guess what -- one of his popularization-series book: Gravity, ch 9, Dover Pubs, 2002. Thank all you.
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