Why does weak equivalence principle say gravity is equivalent to acceleration?

Question

I am told that the weak equivalent principle, that $m_i=m_g$ (inertial and gravitational masses are equivalent) is equivalent to the statement that in a small system you can't tell whether you are in a uniform gravitational field, or in an equivalent accelerating frame. My question is about light: Special relativity doesn't say anything about how light falls in a gravitational field, so if you are in a rocket ship accelerating at $g$ why should it be true that when you shine a light in the rocket ship, you see the same thing happening as if you shined a light on earth? (to clarify, if you shot a bullet in the rocket ship, I understand why the trajectory of the bullet would look the same as if the bullet were shot on Earth, but this is because special relativity (or classical mechanics) tells us what the trajectory of a bullet looks like on Earth).

Answer 1

Special relativity does not say anything about gravitational fields at all, let alone how light behaves in them. But it does say things about non-inertial frames, and therefore can quite happily predict how light will behave in an accelerated frame.

It is a common confusion about SR that it somehow can't deal with acceleration at all: it can. SR is a theory which deals with spacetimes which are flat, and whether spacetime is flat or not is a property of spacetime, not of some observer, whether accelerated or not: all observers, whether or not they are accelerated, agree on the flatness of spacetime (if they do sufficiently careful measurements). So an accelerated observer merely needs to do the appropriate coordinate transformations to and from their non-inertial frame, to an inertial frame to find out how light moves: as other people have said it moves along null geodesics.

The equivalence principles which underly GR are not important here. However note that a 'uniform gravitational field' is almost an oxymoron: if a gravitational field is uniform then there is no curvature: it's just acceleration and it can be transformed away as above. Gravitational fields in the sense GR talks about them -- curvature -- are never uniform.

The way to think about this is that frames are essentially choices of coordinate systems. For all (sufficiently smooth) spacetimes you can choose, locally, some particularly nice coordinate systems where the laws of physics look very simple. For some special spacetimes you can choose globally particularly nice coordinate systems. These special spacetimes are ones which are flat, and they are the spacetimes which SR deals with. But, for any spacetime, flat or not, you can choose coordinate systems which are not nice in this sense. These coordinate systems correspond to the frames of non-inertial observers.

But the physics can't depend on the coordinate system: it can only depend on whether the actual spacetime is nice or not -- whether it is possible to choose these nice global coordinate systems: if it is, then SR works, if it is not, then it doesn't.

Answer 2

Your statement of the weak equivalence principle (WEP) is fine with me, but I think you're not applying it in the correct context.

Special relativity, as mentioned by @Jerry Schirmer, says that light travels along null geodesics. This meaning that it travels at $c$ since for a null geodesic the spacetime interval is null (zero). $$ds^2 = - c^2dt^2 + dx^2 = 0 \implies \frac{dx}{dt} =c $$

In a locally flat space we can use special relativity to say that light will travel in this way.

Hope this clears things up some!

Answer 3

Think of it this way:

Imagine you are in a rocket ship that has transparent portholes (very small ones) diametrically opposed on the body of the ship, and the ship is accelerating at g. At some instant during flight, a laser beam pulse enters one of the portholes, and as soon as that event occurs, it is detected and another laser apparatus emits a laser pulse that crosses the cabin of the rocket ship at the same instant, alongside the one that entered the porthole.

Is there any reason that the laser pulse emitted from inside the cabin should not be deflected downward any less than the one that came through the porthole? Make the rocket portholes be separated as far apart as it takes in order for the effect to be measurable.

This thought experiment even works if the laser was fired in some other direction (like from a spot on the ceiling of the rocket ship). Additionally, clocks inside the rocket ship also slow down exactly the same amount they would on the surface of the Earth.

Answer 4

If you are in a flat space(time), i.e. without any source of gravitation, in a spaceship, and you emit a ray of light across the spaceship, both the spaceship and the light will be in the same frame of reference. The frame will be inertial - not accelerated - and therefore the light will follow a straight path.

Yet if a boost is applied to the spaceship, it means only the spaceship will be acelerated but not the light itself. Therefore light will still be in an inertial frame, while the spaceship will be in a non-inertial frame. That's why it appears to bend.

Now, if the rocket is given acceleration of 1g, i.e. equal to that on Earth, you will see exactly "the same thing happening as if you shined a light on Earth".

Answer 5

One of the greatest insights ever was Einstein's realization that locally a gravitational field is equivalent to acceleration. There are two caveats.. one is locally, and the other of course is that quantum effects are not involved. The result was General Relativity - a strictly classical description of gravity. Over 100 years later it's still the best we have, but it will need major change. Or indeed total replacement. The world is not classical.