I Weak principle of equivalence (Galileo, Newton)

AI Thread Summary
The discussion centers on the weak principle of equivalence and whether gravitational force equals inertial force during free fall. It highlights that gravitational mass and inertial mass are equivalent, leading to the conclusion that in free fall, gravitational acceleration equals inertial acceleration. The concept of inertia is clarified as not being a force but a property of mass that describes resistance to acceleration. The conversation emphasizes that Newton's second law (F=ma) applies strictly within inertial frames, while inertial forces arise only in non-inertial frames. Ultimately, the complexities of these concepts point to the need for a deeper understanding of gravity as explained by General Relativity.
qnt200
Messages
28
Reaction score
2
TL;DR Summary
Does the force of inertia act on the body during free fall, according to the weak principle of equivalence (Galileo, Newton)?
Many tutorials that explain the weak principle of equivalence (Galileo, Newton) do not clearly state whether the body is affected by the force of inertia during free fall as a result of the gravitational acceleration of the body. In other words, the question is whether, during the free fall of a body in a gravitational field, gravitational force equals inertial force (Fg=Fi).
Is this incompatible with the concept of two masses (gravity and inertial mass)? What is GR's opinion on it?
 
Physics news on Phys.org
There is mass, gravitational and inertial. Inertia is another name for inertial mass.

Inertia is not a force.
 
  • Like
Likes malawi_glenn and qnt200
Thank you for explaining. It raises a few more questions for me about the logic of bodies in free-fall:
It is known that gravitational mass = inertial mass (established with enormous precision). That is why we can talk about only one mass. Therefore m = mg = mi. In free fall, mg = ma, i.e., g = a. The product of ma acts as inertia on the body, slowing it down during free fall so that it does not instantly reach the speed of light.
I think it is irrelevant whether the product of mass and acceleration is called a force.
I don't know how correct this attitude is?
 
qnt200 said:
Many tutorials that explain the weak principle of equivalence (Galileo, Newton) do not clearly state whether the body is affected by the force of inertia during free fall as a result of the gravitational acceleration of the body
This doesn’t make sense. Inertial forces are not a result of acceleration. Inertial forces are a result of using a non-inertial reference frame. Sometimes non-inertial frames are called accelerating frames, but an accelerating frame is a different concept than acceleration.

If you want to ask about inertial forces then you must specify the non-inertial reference frame you are using.
 
  • Like
Likes protonsarecool and qnt200
qnt200 said:
The product of ma acts as inertia on the body, slowing it down during free fall so that it does not instantly reach the speed of light.
It is difficult to follow what you are saying here. It seems to indicate that you are understanding things incorrectly (or at least not in the standard way).

Apparently you have this idea that if we exert a force on an object then, were it not for the object's inertia, that object would fly away at the speed of light. That is a somewhat reasonable intuition.

To be picky, if we stay within the Newtonian framework, there is nothing special about the speed of light. A massless object subject to a finite force would be pushed away with an undefined speed. We might think of that speed as infinite, but from a mathematician's point of view, "undefined" is more correct. A mathematically inclined physicist would likely say that you cannot exert a finite force on a massless object.

But never mind that. Set the speed of light to one side.

You want to rescue us from having the object move off at an undefined velocity by conjuring up a counter-force called the "force of inertia". So we have this original real force f=ma pushing the object away and this magical counter-force pushing back. But hold on a minute... if we have equal and opposite forces then the object should not move at all. Eppur si muove.

We see real objects subject to real forces move all the time. So this way of thinking about the "force of inertia" cannot be entirely correct.

The standard way of thinking about this is to stick to inertial frames. You have an interaction force ##f## and an object of mass ##m## moves off with an acceleration ##a## when acted upon by that force. End of story.

We do not wave our hands and conjure up a force to explain the resulting motion. We already have the original force ##f## that explains the motion and a law of motion (##f=ma##) that quantifies the result. There is no need for anything more. The property that we call "inertia" is already encapsulated in Newton's second law.One can rescue the idea of an "inertial force" associated with an object's mass by adopting a non-inertial frame. But others such as @Dale are happily explaining that part.
 
  • Like
Likes qnt200, Dale and PeroK
Dale, jbriggs444
Thank you for your interesting answers and guidelines indicating to me the scope of classical physics, i.e., the inertial frame of reference. F=ma is only valid in inertial frames of reference. It is actually clear to me that the full answers, especially regarding gravity, are found within GR.In addition to other things, you pointed out a very interesting question that I have been thinking about for a long time:
jbriggs444 said:
So we have this original real force f=ma pushing the object away and this magical counter-force pushing back. But hold on a minute... if we have equal and opposite forces then the object should not move at all.
 
Last edited:
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Hello! Let's say I have a cavity resonant at 10 GHz with a Q factor of 1000. Given the Lorentzian shape of the cavity, I can also drive the cavity at, say 100 MHz. Of course the response will be very very weak, but non-zero given that the Loretzian shape never really reaches zero. I am trying to understand how are the magnetic and electric field distributions of the field at 100 MHz relative to the ones at 10 GHz? In particular, if inside the cavity I have some structure, such as 2 plates...
Back
Top