Does Special Relativity Predict Zero Acceleration in Free Fall?

[georgir]

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.

 

[PeterDonis]

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.

 

[stevendaryl]

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]

 

[georgir]

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.

 

[PeterDonis]

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.

 

[stevendaryl]

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).

 

[stevendaryl]

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.

 

[PeterDonis]

It's really just like Newtonian physics

But this assumes that "Newtonian physics" of this sort can be consistently modeled in SR.

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?

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.

 

[Mentz114]

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.

 

[PeterDonis]

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.

 

[Mentz114]

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.

There is only the force of the floor pushing on your feet, and causing your worldline to be non-geodesic.

My post is withdrawn.

 

[Dale]

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.

 

[stevendaryl]

But this assumes that "Newtonian physics" of this sort can be consistently modeled in SR.

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?

Well, you've touched on an area that I've never seen explored, really. In practice, people use Newtonian physics for things like building bridges, and analyzing forces in structures, and use SR when analyzing elementary particles. I always assumed that that was just because Newtonian physics is simpler to use than SR, but that if you wanted to, you could do everything in SR. SR is in a sense a replacement for Newton's laws of motion.

In the same way that Newton's laws of motion don't tell us what the forces are, they only tell us how particles respond to forces, SR similarly doesn't tell us what the forces are, but only gives constraints on them (invariance under changes of rest frame) and tells how particles respond.

As to why the floor should be pushing up on my feet: I'm assuming that at some level, the floor pushes up on my feet because of electromagnetic forces. The actual electrons of my feet repel the electrons of the floor. It probably would be very difficult to give a first-principles model of that interaction, though, because probably quantum mechanics would have to come into play. Quantum mechanics is needed to explain the stability of atoms and molecules, after all.

But people don't try to apply Newtonian physics from first principles starting with elementary particles. Instead, they assume that objects in contact have a contact force. I don't see why SR would require a more first-principles analysis than Newtonian physics does.

 

[Dale]

Bingo. This is the key to it... tidal effects can always be neglected in a small enough locality, and if we do neglect them, we all agree... zero reading.

Yes. The corollary is that tidal effects cannot be neglected for the Langevin scenario. SR has no way to handle tidal gravity and thus predicts an erroneous reading for the accelerometer.

 

[PeterDonis]

SR is in a sense a replacement for Newton's laws of motion.

Note that you said "laws of motion", *not* "law of gravity". Yes, SR is a more accurate version of the laws of motion, which takes into account the finite speed of light and its consequences. But it's *not* a replacement for the law of gravity.

This doesn't necessarily mean that none of the scenarios you mention can be analyzed using SR. For example, if I'm calculating the stress on a beam, I can do the calculation in a local inertial frame, which means I can use SR if for some reason I need that level of accuracy. But note that in doing this, I am not adding a "force of gravity" to SR. I am using the equivalence principle to mimic the local effects of "gravity" with acceleration: I am basically modeling the beam as being inside an accelerating rocket and assuming that the results will apply.

As to why the floor should be pushing up on my feet: I'm assuming that at some level, the floor pushes up on my feet because of electromagnetic forces...

But people don't try to apply Newtonian physics from first principles starting with elementary particles. Instead, they assume that objects in contact have a contact force. I don't see why SR would require a more first-principles analysis than Newtonian physics does.

I agree with all this, but it's not the question I was asking. Consider: suppose I put you in an accelerating rocket. The bottom of the rocket pushes on your feet; this causes your body to compress, which you feel as weight. But the reason it causes your body to compress is that, at the start, only your feet feel a force, so only their worldline is curved; the worldlines of the other parts of your body are straight. It's the difference in path curvature between adjacent worldlines that gives rise to the internal forces between the parts.

In the case of me standing on Earth, if we try to analyze it with SR, we have a pair of worldlines that are both straight, so there is nothing to explain how internal forces can arise. There is no difference in path curvature, so SR predicts that there should be no internal forces.

 

[Dale]

Instead, your formula for what according to SR an accelerometer in free fall will read was not supported by your references and as for me, I will need some time to search more specific references myself. I won't respond anymore to such personal attacks but discuss such references and basic derivations.

How is pointing out your failure to provide a reference a "personal attack"? I am attacking your argument as being speculative, not you personally. You shouldn't take that as a personal attack.

My formula was supported by the references.

The references on the four-acceleration gave the formula in terms of the four-velocity. The references on the four-velocity gave the four-velocity in terms of the worldline. To obtain my formula, simply substitute the four-velocity into the four-acceleration. I can provide a reference on substitution too if needed.

The references on the four-acceleration also explained the relationship to proper acceleration. In the other thread I provided references showing that the proper acceleration is the acceleration measured by an accelerometer. I didnt feel the need to repeat those here.

Bottom line, my formula is correct, and well supported by references. Yours is neither. You can choose to ignore the facts, but that doesn't change them.

Meanwhile I think that the participants to this thread are not a bad sample of "mainstream" opinion, and the opinions are divided.

No, a poll on PF might constitute mainstream public opinion, but does not qualify as a mainstream scientific reference. Particularly not in terms of the PF rule against overly-speculative posts.

 

[Dale]

This thread was to discuss the several times repeated claim by Dalespam that according to SR an accelerometer in free fall will have a large reading.

My claim, as I stated in post 2 and repeated multiple times is: "That question cannot even be addressed by SR since SR does not handle gravitation. You cannot generally establish a self-consistent SR inertial frame in the presence of gravity. This question is outside the domain of applicability of SR."

As I have mentioned several times, all you can do in SR is to neglect gravity. When you do neglect gravity then you introduce errors in your analysis. Whether or not the use of SR is justified depends on the magnitude of those errors.

The formula used to give the SR prediction has been provided. Applying that formula to the Langevin scenario, those errors are large because the SR prediction is a high proper acceleration and the real accelerometer would measure 0.

Any time I mentioned a high accelerometer reading for free-fall it was specifically in reference to this, neglecting gravity for the Langevin scenario. If you treat the Langevin scenario using SR then you use the referenced formula for accelerometers and get an erroneously high prediction.

 

[Samshorn]

Any time I mentioned a high accelerometer reading for free-fall it was specifically in reference to this, neglecting gravity for the Langevin scenario. If you treat the Langevin scenario using SR then you use the referenced formula for accelerometers and get an erroneously high prediction.

What is the "Langevin scenario"? Are you referring to a twin scenario where the turn-around is achieved by looping around a distant star under the effect of the star's gravitation? If so, Langevin never mentioned any such thing. (Granted, another poster claimed the gravitational turn-around scenario was in Langevin's 1911 paper, but when I asked him to point to where in that paper the gravitational turn-around was discussed, he admitted there was no such thing in that paper, so it was just a mis-attribution.)

 

[PeterDonis]

What is the "Langevin scenario"? Are you referring to a twin scenario where the turn-around is achieved by looping around a distant star under the effect of the star's gravitation? If so, Langevin never mentioned any such thing.

Yes, that is what we're referring to. Even if Langevin didn't actually come up with it, it's named after him now (at least for this thread and its relatives).

 

[georgir]

This thread is a mess...

DaleSpam, you still have not addressed my last edit. So I will repeat

your quote again:

[..] SR predicts a very large accelerometer reading during the turnaround, and real free falling accelerometers read 0.

If there were noticeable tidal forces, then a real free falling accelerometer would also not read 0.

You have two separate cases:
a) tidal gravity is negligible, SR predicts 0 and real experiments show 0;
b) tidal gravity is significant, SR predicts >0, irrelevant how accurate, and real experiments also show >0
So your quote is still wrong no matter how you look at it.

PeterDonis, I just don't know where to start with your posts, I'll leave it for another day.

 

[Dale]

You have two separate cases:
a) tidal gravity is negligible, SR predicts 0 and real experiments show 0;
b) tidal gravity is significant, SR predicts >0, irrelevant how accurate, and real experiments also show >0

This is a false dichotomy and, in particular, it is not true in the case under consideration. Real accelerometers would still show 0 in the gravitational twin scenario, but the SR prediction is >0.

 

[PeterDonis]

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.

Really? Think carefully about the Bell Spaceship Paradox before you answer.

 

[georgir]

This is a false dichotomy and, in particular, it is not true in the case under consideration. Real accelerometers would still show 0 in the gravitational twin scenario, but the SR prediction is >0.

Again, no. SR will not predict >0. It predicts 0 without tidal effects. Whatever formula you think you have for an accelerometer's reading in SR is clearly wrong. There can be no change in the distance between any components of an accelerometer in its proper frame according to SR if the acting forces are uniform, that means zero reading.

 

[Dale]

Whatever formula you think you have for an accelerometer's reading in SR is clearly wrong.

The formula is the correct one for SR. If you disagree then please post the equation you believe to be the correct one for SR and provide your references.

 

[stevendaryl]

Note that you said "laws of motion", *not* "law of gravity". Yes, SR is a more accurate version of the laws of motion, which takes into account the finite speed of light and its consequences. But it's *not* a replacement for the law of gravity.

This doesn't necessarily mean that none of the scenarios you mention can be analyzed using SR. For example, if I'm calculating the stress on a beam, I can do the calculation in a local inertial frame, which means I can use SR if for some reason I need that level of accuracy. But note that in doing this, I am not adding a "force of gravity" to SR. I am using the equivalence principle to mimic the local effects of "gravity" with acceleration: I am basically modeling the beam as being inside an accelerating rocket and assuming that the results will apply.

Well, the exact equations of motion for a test particle in curved spacetime in an arbitrary coordinate system are:

$m (\dfrac{d^2 x^\mu}{d \tau^2}+ \Gamma^\mu_{\nu \lambda} \dfrac{d x^\nu}{d \tau} \dfrac{d x^\lambda}{d \tau}) = F^\mu$

Treating gravity like a force simply amounts to moving the $\Gamma^\mu_{\nu \lambda}$ term to the right side:

$F_{eff}^\mu = F^\mu - m \Gamma^\mu_{\nu \lambda} \dfrac{d x^\nu}{d \tau} \dfrac{d x^\lambda}{d \tau}$

Then it looks just like SR with an effective force $F_{eff}^\mu$

$m \dfrac{d^2 x^\mu}{d \tau^2} = F_{eff}^\mu$

The problem with this approach is that $F_{eff}^\mu$ doesn't obey the third law; there is no equal and opposite force. That's why I said that the problem with considering gravity a force is in correctly handling the evolution of the forces. If you assume that the particle has a negligible effect on the field, then I don't think there's any problem.

 

[stevendaryl]

Really? Think carefully about the Bell Spaceship Paradox before you answer.

You're certainly right about that, but that's a higher-order effect. It is not noticeable until the spaceship reaches relativistic speeds.

Anyway, the point holds, that if both ends are accelerated, then the amount of stretching is not an accurate measure of how much you are accelerating. To get zero stretching, I suppose you would have to have the electric field get weaker with height in just the right way.

 

[PeterDonis]

Well, the exact equations of motion for a test particle in curved spacetime in an arbitrary coordinate system are:

Remember I said we're using SR; we're working in a local inertial frame. In such a frame the Christoffel symbols are zero; there is no "gravitational field".

If you want to use non-inertial coordinates instead, with the Christoffel symbols present, you can, sure; for example, you could use Rindler coordinates for the accelerating rocket, which is basically equivalent to using Schwarzschild coordinates on the Earth's surface (provided you are working with a small enough range of altitudes that tidal effects can be ignored). I'm not sure this makes the analysis any easier, but you could do it.

The problem with this approach is that $F_{eff}^\mu$ doesn't obey the third law; there is no equal and opposite force. That's why I said that the problem with considering gravity a force is in correctly handling the evolution of the forces. If you assume that the particle has a negligible effect on the field, then I don't think there's any problem.

I don't follow this. The problem with $F_{eff}^\mu$ is not a problem about the particle affecting the field; it's about the "force" on the particle not obeying Newton's third law. That problem is independent of the particle's effect as a source of the field.

 

[stevendaryl]

Remember I said we're using SR; we're working in a local inertial frame. In such a frame the Christoffel symbols are zero; there is no "gravitational field".

I'm saying that treating the gravitational field as a force, and using SR is equivalent to the exact GR treatment, if you don't worry about the evolution/propagation of the force. In other words, I'm saying that one can use SR with an effective force and get good agreement with the exact GR equations.

 

[stevendaryl]

I don't follow this. The problem with $F_{eff}^\mu$ is not a problem about the particle affecting the field; it's about the "force" on the particle not obeying Newton's third law. That problem is independent of the particle's effect as a source of the field.

In SR, there is no action at a distance; objects are only affected by fields, not by other objects, and they only affect fields, not other objects. So the equivalent of the third law is that when a field affects an object, the object affects the field, as well, so as to conserve momentum, when you include the momentum of the field as well as particles.

The problem (or I should say, one of the problems) with regarding fictitious forces as real forces is that they act on particles, but are not acted upon by the particles.

 

[PeterDonis]

I'm saying that treating the gravitational field as a force, and using SR is equivalent to the exact GR treatment, if you don't worry about the evolution/propagation of the force. In other words, I'm saying that one can use SR with an effective force and get good agreement with the exact GR equations.

If by this you mean that you can use curvilinear coordinates in flat spacetime (e.g., Rindler coordinates), and then interpret the Christoffel symbols in those curvilinear coordinates as the "gravitational field", then yes, I agree.

But if you mean that you can *add* a "force of gravity" to SR as something extra, over and above just using curvilinear coordinates, then no, I don't agree. Doing that destroys the physical predictiveness of the theory.

For example, what does proper acceleration--path curvature of a worldline--mean in this "SR + gravity" theory? I know we've gone back and forth about what different types of accelerometers would read, but in standard SR and GR, a key element of the physical interpretation of the theory is that proper acceleration, path curvature of a worldline, is a direct physical observable; there is *some* device, call it an "ideal accelerometer", that measures it. A given worldline in the presence of gravity has *different* path curvature according to "SR + gravity" than it does according to GR, because "SR + gravity" still uses flat spacetime; so "SR + gravity" can't possibly get good agreement with the physical predictions of GR with regard to path curvature.