Undergrad What is the Principle of Equivalence and how was it determined?

[stevendaryl]

I am not sure what stevendaryl means above, but a good discussion of the EP can be found in chapter 4 here:
http://www.preposterousuniverse.com/grnotes/

If you have equations of motion that are valid in flat spacetime (say, for charged particles moving in an electromagnetic field), then the EP basically says that if we replace partial derivatives by the appropriate covariant derivatives (I know that prescription is a little ambiguous, because of operator ordering problems, but it is a good heuristic) then we get equations of motion that are valid in the presence of gravity (at least to the extent that we can treat the gravity as approximately unaffected by the particles and fields under consideration; so we can use the prescription to describe the motion of rocks and light near the Earth, but not to describe the motion of the Moon, which is large enough to have a significant gravitational effect).

 

[atyy]

What I'm after here is for someone to mathematically justify that the EP serves as the basis for such claim(not entering into the claim itself). Considering not only the metric and it first derivatives, but also the second derivatives necessary to define the Riemann curvature. I just don't see how the EP says anything about the second derivatives of the metric.

There are several versions of the EP.

The heuristic one is says something about the local laws of physics, and fails for the "nonlocal" laws of physics, eg. http://arxiv.org/abs/0806.0464.

The non-heuristic one is called (universal) minimal coupling, and it is essentially exact for the known laws of physics, eg. http://arxiv.org/abs/0707.2748.

Although I don't understand it, there is an argument that universal minimal coupling can be derived eg. http://arxiv.org/abs/1007.0435v3 (section 2.2.2).

 

[PeterDonis]

Point 1, how gravity affects matter, pretty much only involves the first derivatives of the metric tensor

This is only true within a local inertial frame. Once you go beyond a local inertial frame, tidal gravity affects the relative motion of neighboring geodesics, and tidal gravity involves second derivatives of the metric tensor.

Curvature does not (directly) influence the motion of particles.

Yes, it does. See above.

What curvature does not affect directly is covariant derivatives at a particular event, as you point out elsewhere. Those are only affected by the connection coefficients, i.e., by first derivatives of the metric. But when you start trying to parallel transport vectors and tensors from one event to another, knowledge of the covariant derivative at one event is not sufficient. You need to know the curvature tensor, because parallel transport is path dependent, and the curvature tensor tells you the path dependence.

The reason the EP doesn't involve curvature is that it only talks about a single event, or more precisely a sufficiently small region of spacetime around a single event. But that doesn't mean curvature doesn't affect the motion of particles at all. It only means it doesn't affect them (to the accuracy of measurement) within a sufficiently small region of spacetime around a single event.

 

[PeterDonis]

If you want to compute the path of a particle under the influence of gravity, then you use the geodesic equation.

But this requires a choice of coordinate chart (and not just within a single local inertial frame--see below). I don't think that is sufficient to support the broader claim that "curvature doesn't affect the motion of particles".

The EP basically says that test particles move on geodesics.

I think this is too broad. The EP only talks about what happens within a local inertial frame.

 

[stevendaryl]

This is only true within a local inertial frame. Once you go beyond a local inertial frame, tidal gravity affects the relative motion of neighboring geodesics, and tidal gravity involves second derivatives of the metric tensor.

If you assume that test particles follow geodesics, then tidal effects follow from that assumption. So you don't need to assume any kind of coupling to second (or higher-order) derivatives of the metric. That's why I inserted the word "directly" in the statement about the effect of gravity on the equations of motion.

What curvature does not affect directly is covariant derivatives at a particular event, as you point out elsewhere. Those are only affected by the connection coefficients, i.e., by first derivatives of the metric. But when you start trying to parallel transport vectors and tensors from one event to another, knowledge of the covariant derivative at one event is not sufficient. You need to know the curvature tensor, because parallel transport is path dependent, and the curvature tensor tells you the path dependence.

What I'm saying is that if you know, for a particular coordinate system in a particular small region of spacetime, what the connection coefficients are at each point in that region, that's all you need to know to predict the motion of test particles or the evolution of small-amplitude fields. Curvature is computable from that knowledge, so it's not an additional piece of information.

 

[stevendaryl]

I think this is too broad. The EP only talks about what happens within a local inertial frame.

Well, if you know what happens in every local inertial frame, then doesn't that imply what happens globally? Under the assumptions that:

1. We're talking about test particles and weak fields whose effect on gravity is negligible, and
2. There are no direct couplings of the equations of motion to curvature or higher-order derivatives of the metric.

Nonminimal coupling can never be ruled out except experimentally, but I would say that if there are nonminimal couplings, that to me means that the EP does not hold for situations in which nonminimal coupling is relevant. Or to put it another way, to me, the impact of the EP is the claim that there is minimal coupling of matter and fields to gravity.

 

[loislane]

Let's take Newtonian gravity, for a simpler example than GR.

If you let \Phi be the gravitational potential, then gravity affects matter through the equation:

m \dfrac{d^2 x^j}{dt^2} = - m \dfrac{\partial \Phi}{\partial x^j}

Note: this involves the first derivative of \Phi

Matter affects gravity through the equation:

\sum_j \dfrac{\partial^2 \Phi}{(\partial x^j)^2} = 4 \pi G \rho

where \rho is the mass density. That involves the second derivative of \rho.

GR modifies both of these equations: Instead of the first equation, we have:

m \dfrac{d^2 x^\mu}{d \tau^2} = - m \Gamma^\mu_{\nu \sigma} \dfrac{dx^\nu}{d\tau} \dfrac{dx^\sigma}{d\tau}

where \Gamma^\mu_{\nu \sigma} is constructed from the first derivatives of the metric tensor.

Instead of the second equation, we have:

G_{\mu \nu} = 4 \pi T_{\mu \nu}

where G_{\mu \nu} is constructed from the second derivatives of the metric tensor, and T_{\mu \nu} is the energy-momentum tensor.

So what you're saying is mathematically unsustainable is a feature of both GR and Newtonian gravity. Which is a kind way of saying that you're completely wrong about this.

You build your point on an identification of the Newtonian gravitational potential with the metric tensor in GR. Without proof this is what I call hand-waving in the mathematical sense.
And certainly I don't pretend to know physics, that's why I am asking in this forum. There's simply some mathematical facts I happen to be acquainted with. But you certainly seem to pretend you do understand everything in this particular issue. If only that were true how fortunate we'd all be. Experience tells me the chances of that are scarce.

 

[stevendaryl]

This is only true within a local inertial frame. Once you go beyond a local inertial frame, tidal gravity affects the relative motion of neighboring geodesics, and tidal gravity involves second derivatives of the metric tensor.

I think that there is a very close analogy with Newtonian gravity. In Newtonian gravity, you have a gravitational potential \Phi, and the path of a test particle is given by:

m \dfrac{d^2 x^j}{dt^2} = -m \partial_j \Phi

The motion of the particle only depends on the first derivative of \Phi

Newtonian gravity certainly has tidal effects, which involve the second derivatives of \Phi. But the existence of tidal effects follows from the above equation of motion, except in the special case in which \nabla \Phi is a constant vector.

 

[stevendaryl]

You build your point on an identification of the Newtonian gravitational potential with the metric tensor in GR. Without proof this is what I call hand-waving in the mathematical sense.

Look, it's either one or the other. Either you actually pick up a GR textbook, and learn that subject, or else you have to settle for hand-wavy arguments. If you're not willing to learn the math, then handwavy is the best you can get. There is no notion of "proof" outside of rigorous reasoning, which you can only learn from actually studying the subject.

And certainly I don't pretend to know physics, that's why I am asking in this forum.

This forum is not appropriate for learning a technical subject. Use a textbook for that.

 

[stevendaryl]

But you certainly seem to pretend you do understand everything in this particular issue.

I don't claim to understand everything, or even most things about GR, but the issues that you are stumbling over are not advanced topics in GR, they are the basics that you learn in the first course on GR.

 

[loislane]

Well, if you know what happens in every local inertial frame, then doesn't that imply what happens globally? Under the assumptions that:

1. We're talking about test particles and weak fields whose effect on gravity is negligible, and
2. There are no direct couplings of the equations of motion to curvature or higher-order derivatives of the metric.

Nonminimal coupling can never be ruled out except experimentally, but I would say that if there are nonminimal couplings, that to me means that the EP does not hold for situations in which nonminimal coupling is relevant. Or to put it another way, to me, the impact of the EP is the claim that there is minimal coupling of matter and fields to gravity.

If you assume minimal couplig from the start you are assuming the EP in exact(not just approximate) form. The problem with that assumption is that as commented by PeterDonis it leads you to a completely coordinate dependent formulation of the geodesic equation.

 

[stevendaryl]

If you assume minimal couplig from the start you are assuming the EP in exact(not just approximate) form.

Exactly. To me, the EP is the claim that gravity effects particles via minimal coupling.

 

[loislane]

I think that there is a very close analogy with Newtonian gravity. In Newtonian gravity, you have a gravitational potential \Phi, and the path of a test particle is given by:

m \dfrac{d^2 x^j}{dt^2} = -m \partial_j \Phi

The motion of the particle only depends on the first derivative of \Phi

Newtonian gravity certainly has tidal effects, which involve the second derivatives of \Phi. But the existence of tidal effects follows from the above equation of motion, except in the special case in which \nabla \Phi is a constant vector.

I think you are relying too much on Newtonian gravity, you do know is not exactly correct, don't you?

 

[PeterDonis]

To me, the EP is the claim that gravity effects particles via minimal coupling.

I agree with this formulation (the only potential quibble I would have would be to say "spacetime geometry" instead of "gravity"--"how spacetime tells matter how to move", so to speak). My previous comments weren't really about the physics but about ordinary language terminology. I agree with everything you have said about the physics.

The problem with that assumption is that as commented by PeterDonis it leads you to a completely coordinate dependent formulation of the geodesic equation.

Christoffel symbols are coordinate dependent, but covariant derivatives are not; they are proper tensorial objects. So the geodesic equation expressed in terms of covariant derivatives is properly covariant.

As I said above, I was not really commenting about the physics; I was commenting about terminology. I thought the phrase "curvature does not affect the motion of particles" might be misleading. But the "minimal coupling" formulation is saying the same thing, just in different words. The physics is the same either way.

 

[stevendaryl]

I think you are relying too much on Newtonian gravity, you do know is not exactly correct, don't you?

The particular points being discussed are true of both Newtonian gravity and General Relativity.

 

[PeterDonis]

I think you are relying too much on Newtonian gravity

No, he's not. He is saying that, if you want more than what has already been said in this thread, you are basically asking us to provide you a textbook on GR in the limited space of a PF thread. That's not going to happen. If you want more details, please consult a textbook.

 

[PeterDonis]

At this point the OP's question has been more than thoroughly answered. Thread closed.