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## Free fall acceleration in SR

 Quote by stevendaryl The second approach is to treat gravity as an external force and use SR's equations of motion for such a force: $m \dfrac{d^2 x^\mu}{d \tau^2} = F^\mu$ That's ambiguous
No, it's worse than that. It's incorrect and inconsistent. There's no way to formulate a consistent theory of "SR + gravity" along these lines, and even if it were, the equation you give above obviously gives incorrect predictions (for example, it predicts that astronauts orbiting in the International Space Station will feel weight).

 Quote by PeterDonis No, it's worse than that. It's incorrect and inconsistent. There's no way to formulate a consistent theory of "SR + gravity" along these lines, and even if it were, the equation you give above obviously gives incorrect predictions (for example, it predicts that astronauts orbiting in the International Space Station will feel weight).
Well, that's an interesting result, itself. I'm a little surprised, though.

There is a sense in which it doesn't really matter, because today SR is considered a limiting case of GR, so there really is no good reason for worrying about what SR would predict in the absence of GR. But if we're trying to get into the frame of mind of a physicist living in the decade between the development of SR and the development of GR, then presumably we would have some strategy for dealing with gravity. It's interesting to speculate how someone might approach it.

When it comes to using SR (or Newtonian physics, for that matter) to describe forces, there are (or can be) two different aspects: (1) Describing how "test particles" are affected by the force, and (2) describing how the force itself evolves with time.

The way I understood the incompatibility of SR and gravity was that it was number (2) that caused problems. If you try to model the propagation of the gravitational field along the lines of the electromagnetic field, the resulting theory makes the prediction that "radiation" or a fluctuation in the field carries negative energy. You can't (or at least, I don't know how) have a sensible theory of dynamics if a system can "radiate" negative energy.

But you could make the approximation (which is what people generally do in applying SR or GR to problems such as orbital dynamics) that the gravitational field is approximately static, so we don't need to consider gravitational radiation. In other words, if we just worry about step (1)--the effect of gravity on the motion of "test particles".

I don't see how this approach would predict that people in orbit would feel a weight. The feeling of weight is really about things pressing against other things, such as the floor pressing against your feet. If both your foot and the floor were being acted upon by a force that is proportional to mass, then there would be no pressure of your foot against the floor, and so you wouldn't feel weight.

 Quote by PeterDonis for example, it predicts that astronauts orbiting in the International Space Station will feel weight
How so? It predicts that they will have a proper acceleration, but not that they can feel it or detect it with any device at all.

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 Quote by georgir How so? It predicts that they will have a proper acceleration, but not that they can feel it or detect it with any device at all.
To say that SR predicts a proper acceleration but doesn't predict that it's detectable is nonsense; a prediction is a prediction of an observation. SR predicts that astronauts inside the ISS would feel weight, would be able to stand on the "floor" of the station, etc., just as they would inside a rocket with its engine firing. That's what proper acceleration *means*, physically. The mathematical expression is not proper acceleration; it's just how proper acceleration, the physical, detectable phenomenon, is represented in the math.

Not to mention that the proper acceleration would be easy to detect even without looking at the reading on a scale: SR predicts that the astronaut would be able to *stand* on a scale in the ISS, just as it predicts that you would be able to stand on a scale inside the moving chamber in a giant centrifuge floating freely in flat spacetime.

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 Quote by stevendaryl There is a sense in which it doesn't really matter, because today SR is considered a limiting case of GR, so there really is no good reason for worrying about what SR would predict in the absence of GR.
True; but you can still, purely as a theoretical exercise, use SR to make (wrong) predictions in situations where gravity is not negligible. See further comments below.

 Quote by stevendaryl But if we're trying to get into the frame of mind of a physicist living in the decade between the development of SR and the development of GR, then presumably we would have some strategy for dealing with gravity. It's interesting to speculate how someone might approach it.
Interesting as history, perhaps. But not as physics.

 Quote by stevendaryl The way I understood the incompatibility of SR and gravity was that it was number (2) that caused problems.
No, both (1) and (2) cause problems. SR assumes that there are global inertial frames in which the worldlines of freely falling test objects are straight lines. Try to describe the worldline of a test object orbiting the Earth in an inertial frame; *any* inertial frame. It won't be a straight line. Thus SR predicts that such a worldline is not freely falling. See further comments below.

 Quote by stevendaryl I don't see how this approach would predict that people in orbit would feel a weight. The feeling of weight is really about things pressing against other things, such as the floor pressing against your feet. If both your foot and the floor were being acted upon by a force that is proportional to mass, then there would be no pressure of your foot against the floor, and so you wouldn't feel weight.
This is how we understand gravity from either a Newtonian viewpoint, or a GR viewpoint, yes. (In the case of GR, this is how we understand its Newtonian approximation, where we try to "translate" GR's statements about curved spacetime into intuitively more palatable statements about gravity as a "force".)

But again, this won't work within the framework of SR, because, once again, the worldline of such an object in an inertial frame is not a straight line. And since we are talking about an astronaut standing on the "floor" of the station, the astronaut and the "floor" will be at slightly different distances from the center (the floor will be a bit further from the center), so the curvature of their paths will be slightly different. That means they will push on each other, i.e., the astronaut will feel weight. Once again, it's the same as if the astronaut were inside the moving chamber of a giant centrifuge floating freely in flat spacetime; do you dispute that SR predicts that such an astronaut will feel weight?

Note that what I am invoking here is *not* tidal gravity. We do not have to assume any difference in the acceleration produced by the "force" between the astronaut and the floor. All that is necessary is the slight difference in distance from the center.

 Quote by PeterDonis To say that SR predicts a proper acceleration but doesn't predict that it's detectable is nonsense; a prediction is a prediction of an observation. SR predicts that astronauts inside the ISS would feel weight, would be able to stand on the "floor" of the station, etc., just as they would inside a rocket with its engine firing. That's what proper acceleration *means*, physically. The mathematical expression is not proper acceleration; it's just how proper acceleration, the physical, detectable phenomenon, is represented in the math. Not to mention that the proper acceleration would be easy to detect even without looking at the reading on a scale: SR predicts that the astronaut would be able to *stand* on a scale in the ISS, just as it predicts that you would be able to stand on a scale inside the moving chamber in a giant centrifuge floating freely in flat spacetime.
The "floor" is also getting accelerated in exactly the same way as the astronauts are, so the astronauts would not be able to stand on it. The situation is exactly analogous to the Newtonian model.

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 Quote by georgir The "floor" is also getting accelerated in exactly the same way as the astronauths are, so the astronauths would not be able to stand on it.
See my response to stevendaryl.

 Quote by PeterDonis To say that SR predicts a proper acceleration but doesn't predict that it's detectable is nonsense; a prediction is a prediction of an observation.
That doesn't seem right to me. As I said in another post, what you "feel" is not acceleration directly, but relative acceleration of the parts of a body.

Take the simple model of an accelerometer, a cubic box with a ball suspended in the center by identical springs connected to each of the 6 sides. If you put the contraption on an accelerating rocket ship, then the position of the ball relative to the walls will shift, and that shift indicates that the box is accelerating.

But now imagine that the box is charged, and so is the ball, and the charge/mass ratio is the same for box and ball. Then use a uniform electric field to accelerate the box. Then to first order, there is no shift in the position of the ball relative to the walls. The accelerometer will not detect acceleration.[/QUOTE]

 Quote by PeterDonis Note that what I am invoking here is *not* tidal gravity. We do not have to assume any difference in the acceleration produced by the "force" between the astronaut and the floor. All that is necessary is the slight difference in distance from the center.
Slight difference in distance means slight difference in acceleration. Exactly tidal forces.
Your analogy with centrifuge is flawed because there the "scale" is fixed to the centrifuge. If it were freefalling together with the astronaut, there would be no weight again.

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 Quote by stevendaryl The feeling of weight is really about things pressing against other things, such as the floor pressing against your feet. If both your foot and the floor were being acted upon by a force that is proportional to mass, then there would be no pressure of your foot against the floor, and so you wouldn't feel weight.
I should also note that, on the "SR + a force of gravity" viewpoint you are considering, this argument proves too much: it proves that I should not be able to feel weight on the surface of the Earth. Consider: in an inertial frame centered on the Earth, I and the ground just below me are both motionless. According to SR, that means we should both be freely falling, and should not push on each other at all, so I should not be able to stand on the ground.

 Quote by PeterDonis But again, this won't work within the framework of SR, because, once again, the worldline of such an object in an inertial frame is not a straight line. And since we are talking about an astronaut standing on the "floor" of the station, the astronaut and the "floor" will be at slightly different distances from the center (the floor will be a bit further from the center), so the curvature of their paths will be slightly different. That means they will push on each other, i.e., the astronaut will feel weight. Once again, it's the same as if the astronaut were inside the moving chamber of a giant centrifuge floating freely in flat spacetime; do you dispute that SR predicts that such an astronaut will feel weight?
I think it would be hard to model the feeling of weight realistically, but let's take a drastically simplified model of "feeling acceleration", which is: you have two masses connected by a spring. The spring has an equilibrium length. In this simplified model, "feeling weight" means that the spring is compressed or stretched.

If you put one of the masses on the floor of an accelerating rocket, with the spring holding the other mass up, then the spring will compress. Alternatively, if you attach one mass to the front of the rocket, and let the other mass dangle from the spring, the spring will stretch. So compression or stretching indicates acceleration.

But now, suppose instead of just accelerating one of the masses, you're accelerating both of them: For example, suppose they are both charged, with the same charge/mass ratio, and you are accelerating them upward by using a uniform electric field. Then the spring will be neither stretched nor compressed. So with this simple model of "feeling acceleration", you don't feel any acceleration when both masses are accelerated together.

So I don't know why you are saying that an astronaut on board a satellite will feel acceleration (or that SR predicts that he will).

 Quote by PeterDonis I should also note that, on the "SR + a force of gravity" viewpoint you are considering, this argument proves too much: it proves that I should not be able to feel weight on the surface of the Earth. Consider: in an inertial frame centered on the Earth, I and the ground just below me are both motionless. According to SR, that means we should both be freely falling, and should not push on each other at all, so I should not be able to stand on the ground.
I don't know why you say that. It's really just like Newtonian physics: there is a force diagram. The floor has an upward force due to the rock underneath it. It has a downward force due to gravity, and a second downward force due to my feet pressing down on it. I have an upward force due to the floor pressing on my feet, and a downward force due to gravity. The total force on me is zero. The total force on the floor is zero.

But even though the total force on me is zero, the downward gravitational force and the upward force due to the floor are not applied at the same point. The upward force applies only to my feet, while the gravitational force applies to all parts of my body. So my feet should accelerate upward relative to the rest of me, which tends to compress me. Of course, my bones act like springs, and provide a resistance force to being compressed, but I will compress a little bit. That little bit of compression is what it feels like to have weight.

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 Quote by stevendaryl It's really just like Newtonian physics
But this assumes that "Newtonian physics" of this sort can be consistently modeled in SR.

 Quote by stevendaryl The floor has an upward force due to the rock underneath it.
Why? I understand why it does in Newtonian physics. But why does it in SR? How do we get this prediction from SR? Both the floor and me are moving in a straight line in an inertial frame. Why should we be pushing on each other at all?

 Quote by stevendaryl But even though the total force on me is zero, the downward gravitational force and the upward force due to the floor are not applied at the same point. The upward force applies only to my feet, while the gravitational force applies to all parts of my body. So my feet should accelerate upward relative to the rest of me, which tends to compress me. Of course, my bones act like springs, and provide a resistance force to being compressed, but I will compress a little bit. That little bit of compression is what it feels like to have weight.
Is this supposed to be an argument within Newtonian physics? SR? GR?

With respect to Newtonian physics, I don't dispute it, although I haven't tried to think it through in detail.

With respect to GR, I don't think the argument is right (except maybe as a sort of "translation" into the Newtonian approximation), because in a local inertial frame, there is no "force of gravity" at all. There is only the force of the floor pushing on your feet, and causing your worldline to be non-geodesic.

With respect to SR, I can see how this argument could be applied to the astronauts in the ISS--who do not actually feel weight--but I don't see how it can be applied to me standing on the surface of the Earth, because, as I said before, I and the ground below me are both moving on straight lines in an inertial frame, so there's no reason for either of us to push on each other at all.

In other words, SR gets both predictions wrong: it predicts that astronauts in the ISS will feel weight, and your argument helps to explain why--but the prediction is wrong because they actually don't. But SR also predicts that the ground should not push on me, here on the surface of the Earth, because our worldlines are both straight lines in an inertial frame and are therefore freely falling--but the prediction is again wrong because they actually aren't.

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 Quote by PeterDonis I and the ground below me are both moving on straight lines in an inertial frame
Are you sure ? A geodesic in the Schwarzschild spacetime is falling freely radially. The ground is stopping you from following a geodesic and the accelerometer shows g=9.8 m/s2.

I think you probably mean this in some Newtonian way - in which case please disregard it.

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 Quote by Mentz114 Are you sure ? A geodesic in the Schwarzschild spacetime is falling freely radially.
Here I'm talking about what SR predicts, not what's actually true. You might need to read back through the discussion for context.

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 Quote by PeterDonis Here I'm talking about what SR predicts, not what's actually true. You might need to read back through the discussion for context.
Actually you say it here.

 Quote by PeterDonis There is only the force of the floor pushing on your feet, and causing your worldline to be non-geodesic.
My post is withdrawn.

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 Quote by georgir DaleSpam, are you perhaps confusing "SR predicts a very large accelerometer reading during the turnaround" with "SR predicts a very large acceleration of the accelerometer during the turnaround" ? The two are not the same. The accelerometer may be accelerating and read 0, because all of its components are accelerating completely uniformly. Anyway... enough with me posting without the full context. I'll be silent until I get more info. [edit: posted before seeing your last reply, feel free to disregard]
No, I am not confusing those two. That is the difference between coordinate acceleration and proper acceleration. It was discussed extensively in the other thread.

 Tags acceleration, free fall, special relativity