DrStupid said:
No, I'm not. Please refer to my counter example above. With these modifications of the laws of motion force would be frame dependent without violating Galilean relativity.
The first law of motion (which is part of Galilean Relativity) says, in part, that a force applied to a body at rest will change that body's motion (ie. the body will experience a change in motion relative to its initial rest frame). And it says that all inertial reference frames are equivalent: the laws of motion are the same in all inertial reference frames.
So according to Galilean Relativity, the same physical interaction between two unit bodies will have the same result in all inertial frames of reference. Let's say that a standard interaction is the collision of two 1 kg steel balls, Ball A which is unconstrained and stationary, and Ball B which is moving at one unit of velocity in the rest frame of Ball A. Ball B is moving toward ball A at one unit of velocity immediately before it collides with Ball A. Galilean Relativity says that Ball A will experience the same change in motion whether the interaction takes place on a uniformly moving rail car or on a stopped rail car. If it did not, one would be able to differentiate between inertial reference frames, thereby destroying the basic postulate of Galilean Relativity.
The equivalence of inertial reference frames means that the graph of velocity vs time experienced by Ball A during the collision as measured in the frame of the rail car would be the same regardless of the (uniform) velocity of the rail car. Galilean Relativity also postulates that time and distance are the same in all inertial reference frames This means that the resulting change in velocity would be observed to be the same in all inertial reference frames. (This is just a matter of applying the Galilean transformation to the experiment).
I think we would both agree to this point. But if I am mistaken, let me know where we differ.
Now, let's involve the concept of force. By the First Law we know that Ball A experiences a force because it changes its motion. So let's attach a spring to Ball A and have Ball B move toward that spring at unit velocity and have a high speed camera record the interaction so we can measure the compression of the spring as a function of time.
Since the interaction is the same in all reference frames, the graph of spring compression vs. time will be the same regardless of how fast the rail car is uniformly moving. And since time and distance are the same for all inertial reference frames, the compression distance as a function of time will be the same for all inertial observers, regardless of how fast they are moving. So the area under that graph would be the same for all inertial observers. Since motion change when the spring compresses, we will say that the compression of the spring represents a force. We won't assume Hooke's law. We will just say that a force, F, at a given time is represented by the compression distance. The area under the graph is proportional to the impulse received by Ball A.
This shows that the impulse, ##I=\int Fdt## on Ball A is the same for all inertial observers. And since time is the same for all inertial observers, the time averaged force is the same. So let ##I = \int Fdt = F_{avg}\Delta t##.
Now repeat the experiment with two unit balls, Ball A and Ball A', each undergoing simultaneous collisions with unit ball traveling at unit velocity. The change in motion of Ball A will be identical to the change in motion of Ball A'. So the change in momentum (product of no. of unit balls and velocity) is double. And, however you want to define force, the average force is doubled. And since the mass has doubled with the same change in velocity, momentum change has doubled.
Since the change in velocity of Ball A, ##\Delta v##, is the same in all frames, this means that the average force, ##F_{avg}## multiplied by the time, is proportional to the change in momentum: ##F_{avg}\Delta t \propto m\Delta v \rightarrow F \propto \dot{p}##.
You could also do the same experiment using a series of collisions, with the successive collisions taking place with a Ball B traveling at unit velocity with respect to the rest frame of Ball A moving at ##v_i + \Delta v##. You can see in that case that doubling the duration of the same average force doubles the change in momentum. And since ##\Delta t## is doubled and the force is the same, ##2\Delta v/2\Delta t = \Delta v/\Delta t## is constant . Take any number of successive standard collisions and you can see that ##n\Delta v/n\Delta t = \Delta v/\Delta t## is constant. (ie. acceleration is constant)
Now if you take the limit as mass/time/unit velocity approach 0, you can see that a series of the same collisions, or any number of simultaneous collisions or any combination of these, has the same result: ##Fdt \propto mdv##. So if the force is constant over time, dp/dt is constant. And if mass increases proportionally with force, the change in velocity per unit of time is the same.
And that is just Newton's second law.
No, I would say that this wouldn't be an appropriate standard force unit.
So why don't you suggest a standard force unit. One does not have to assume Hooke's law, as the above shows.
It would have to be something that can be considered constant over time and space in some reference frame. Your definition of force as ##F=\dot{p}(1+v^2/c^2)## makes it impossible to have such a constant force. So a spring maintained at the same compression distance and applied to a free body would experience a different force depending on the speed of the inertial reference frame in which the spring is situated. I am not sure why you would want to presume that the force changes when nothing observable changes.
I will no longer spend any effort into the falsification of your claims because it is not my duty and it seems that it doesn't stop you from repeating them. You are responsible for proper justification. So please show me step by step how you get F=m·a from Galilean relativity and the first law.
See above.AM