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Gravity is not a force – free-fall parabolas are straight lines in spacetime
401–410 of 451 posts
Re: Gravity is not a force – free-fall parabolas are straight lines in spacetime
#402Ok, great opportunity for me to ask a dumb question that's been bothering me for a while, for practical reasons I won't go into. How is gravity like a force at all, even in Newtonian physics? It seems like mismatched units. Gravity is an acceleration, not a force. F=ma, right? So if gravity were a force, it would produce an acceleration that was dependent on the mass, and it doesn't, so it seems to me like the only s…
In Newtonian physics it's a force with a magnitude that is proportional to the mass of the body. For two different objects at the same point in the same gravitational field, the difference in gravitational force due to their differing mass exactly cancels out the difference in acceleration due to their differing mass, so they both accelerate at the same rate.
Sure, but the phrase "it's a force with a magnitude that is proportional to the mass" is like me asking "how heavy are you?" and you replying "75 liters", expecting me to know that humans are basically the density of water.
My point is that it really seems like "gravity" is not a force, it's an acceleration, but there is something you could call "force due to gravity" that you reverse-engineer from the known acceleration, and that means you need to multiply by the mass. Clearly different masses will just cancel, so the resulting acceleration is the same.
I'm fine with saying "force due to gravity" or even "gravitational force". Which is what you're describing in your first sentence, and I have no disagreement with that. "Force of gravity" starts to sound a little off, and I bet if I ask "what is gravity at Earth's surface?" I'll get back "9.8 m/s^2", which is an acceleration not a force.
Re: Gravity is not a force – free-fall parabolas are straight lines in spacetime
#403Physicist here, gravity is a force, just a different one. Also, like everything else in physics: it depends how you observe it. For instance, electromagnetism comes from the curvature of a U(1) bundle over space time, the (local) U(1) symmetry yields electromagnetic interactions. For gravity the symmetry is the (local) Pointcaré (SO(1,3) + translations) symmetry and curvature of spacetime itself. Also gravity on Eart…
Correct me if I'm wrong please, but the 'curved space' strategy describes the velocity of two objects in a 2 body scenario, but it doesn't describe the behavior when the two objects have zero velocity relative to each other, right?
The 'two spheres in a box' experiment for testing the gravitational constant has no relative velocity at all, so how could 'curved space' describe the force between them?
Re: Gravity is not a force – free-fall parabolas are straight lines in spacetime
#404Earlier quoted context omitted.
In Newtonian physics, a sphere and point mass are exactly interchangeable as long as you are not inside the sphere. If you are outside the sphere, the equivalence is exact, regardless of distance. Proving this is a classic problem in undergraduate physics.
Thanks! Shame I never took undergraduate physics but now that you've rung my bell I think we may have discussed this in high school. What an unintuitive result that being even a meter under ground breaks what is, up to that point, a fine model.
>What an unintuitive result that being even a meter under ground breaks what is, up to that point, a fine model.
It doesn't break it at all. The meter above you can be treated as a hollow shell, which surprisingly has zero net pull, and the solid sphere below can be treated as a point mass just as before.
Just remember these two facts, each provable with a simple integral calculation, usually done in high school physics or freshman college physics: a uniform sphere has the same gravitational pull on an object as a point mass at it's center, and the net gravitational pull on an object inside a spherical shell is zero.
This all works under perfect spheres, uniform (at the spherical shell level at least) density... There are other cases it works, but this simple case is the basic idea.
Re: Gravity is not a force – free-fall parabolas are straight lines in spacetime
#405Earlier quoted context omitted.
> Also if the light wasn't moving straight that would mean it's changing direction, which is the same as an acceleration, and a beam of light traveling thru a gravitational field feels no acceleration, because it's not accelerating. You could have both a deviation (i.e tangential acceleration) and a constant speed.
Any change in direction is an acceleration (by definition). Even an object moving in a perfect circle at constant radians per second is nonetheless undergoing a constant non-zero acceleration just due to change in direction. Acceleration is any change in a velocity vector, including simply a change in direction, and requires a force (if the object has mass)
Yes, obviously. But the message seemed to say that the constant speed of light implied that the derivative of the velocity with respect to time had to be zero.
Re: Gravity is not a force – free-fall parabolas are straight lines in spacetime
#406Earlier quoted context omitted.
We don’t know, because we don’t know if the past or future are objectively real, or if we ride the wave so to speak.
Eh, that view doesn't make sense since every particle has it's own light cone, hence it's own time cone. PBS spacetime did an episode about this recently.
Does it exist earlier or later in the time cone ‘already’?
Re: Gravity is not a force – free-fall parabolas are straight lines in spacetime
#407Earlier quoted context omitted.
Newton's first law provides a guide: "Every object persists in its state of rest or uniform motion in a straight line unless it is compelled to change that state by forces impressed on it." So, the definition of a force as something that causes an object to change its movement from a "straight line" comes to us from Newton's laws. > Why does the existence of a transformation that makes movement under a supposed force…
So what you are saying is that if we assumed that all particles, including light are magnetic, and everything that has a mass, emits a corresponding magnetic field with a strength relative to its mass, we could not form a similar theory of "general magnetic relativity" in which the frame of reference under magnetic fields would behave in a similar way it does for gravity? That seems kinda odd. What exactly would prov…
I think (and, again, I'm absolutely not an expert) that a fundamental difference between the two is that from our perspective gravity "effects" particles with 0 mass, while the electromagnetic field does not effect particles with 0 charge.
So a photon, which has 0 mass is still bent by gravity. Everything that we've observed that moves through spacetime is bent by gravity. That's why we say that gravity is a warping of spacetime itself, where electromagnetism isn't.
If you tried to build the "general electromagnetic theory of relativity", then 0 charged particles wouldn't follow a straight line on a geodesic of spacetime. With gravity, everything follows a straight-line on the geodesic of a curved straight line, regardless of its mass.
As to why such a difference exists between gravity and electromagnetism, that's well above my pay grade.
Re: Gravity is not a force – free-fall parabolas are straight lines in spacetime
#408Earlier quoted context omitted.
> is there any physical way to distinguish these fundamentally different situations? Are you asking if there is a way to distinguish a timelike object from a lightlike object? Of course there is. The fact that, for something that has a very, very small invariant mass, it might be practically difficult does not change the fundamental principle. Also note that the reason it was difficult, for example, to tell whether n…
It still doesn't sound physically distinct any more than distinguishing any continuous quantity as being zero or nonzero. If we measure something that looks like 0, we can't be sure if it's just below the sensitivity of our instruments. For neutrinos, even if we accelerated an rocket and somehow checked if a neutrino was at rest relative to it, we might find that it's not. That means we won't know if we need more spe…
> It still doesn't sound physically distinct
If you try what I described with a light ray, it will be moving away from you at c no matter how much you accelerate in its direction.
If you try it with a massive object, even a neutrino with a very, very tiny invariant mass, that will not be the case; its speed relative to you will decrease as you accelerate after it, eventually to zero.
There is no continuum between those two possibilities; they are distinct and discrete. The only continuum is in the latter case, where the final speed of the object relative to you will depend continuously on how long you accelerate.
> even if we accelerated an rocket and somehow checked if a neutrino was at rest relative to it, we might find that it's not. That means we won't know if we need more speed or if it's impossible
Yes, you will know, because you will know if the neutrino's speed relative to you has decreased or not. If it has, it's possible to bring it to rest relative to you. If it hasn't, it's not. See above.
> I suppose it's a bit easier than that because we only have to accelerate the rocket fast enough that the neutrino's speed becomes measurably less than c, rather than 0.
Exactly.
> But still, what if we can't even get it to go fast enough for that?
That's basically the position we are in now: we have no way of building a rocket or other device that can accelerate after a neutrino long enough to tell whether its speed relative to the rocket is measurably decreasing. So we have to resort to indirect measurements. But as I said before, that doesn't change the principle.
> even photons have a nonzero upper bound to their possible rest mass
Yes, because, as I said, practically speaking we can't run the obvious and straightforward experiment I described, to confirm that a photon moves away from you at c no matter how much you accelerate after it. So we have to resort to indirect measurements, like trying to measure its invariant mass by other means. But that doesn't change the principle.
Re: Gravity is not a force – free-fall parabolas are straight lines in spacetime
#409Earlier quoted context omitted.
> From the point of view of a photon, no time elapses between its the origin and destination endpoints. No, this is not correct. The correct statement is that the concept of "elapsed time" does not apply to a photon; it only applies to timelike worldlines, not null worldlines. To put it another way, if your statement were true, it would mean that the origin and destination events were the same point in spacetime. But…
(Shrug) You can go argue with Neil deGrasse Tyson, it's over my pay grade. https://www.youtube.com/watch?v=5ELA3ReWQJY
> You can go argue with Neil deGrasse Tyson
Show me an actual textbook or peer-reviewed paper Tyson has written where he makes this claim. Pop science videos don't count. (Tyson is by no means the only one; Brian Greene is notorious for the same thing.)
You won't be able to because there aren't any. No scientist who talks about a photon "experiencing zero time" in informal contexts will try it in a textbook or paper. That's because they know that if they did, other scientists would call them out on it, so they confine such claims to contexts where there are no other experts so there's nobody to call bullshit.
Another point is that if this concept were actually scientifically useful, somebody would be using it in a textbook or peer-reviewed paper. The fact that nobody is is a huge clue that the concept is not scientifically useful. It's only useful for selling pop science books or getting views of pop science videos, where, again, there are no other experts around.