Haven't they also tested it with super sonic jets?Gonzolo said:Yes, its time must "slow down", not its speed. That happens with decaying particles.
If you mean that it flows according to the laws of physics, I agree. But the fact that c is invariant is practically a law of physics now.
Rates of flow of energy... I agree that all must always be consistent.
Not quite sure what you mean. When a rocket slows down near a planet, the planet couldn't care less. The equal and opposite effect is between the rocket and its fuel.
Well, until you can demonstrate that such a "force" and "frictive medium" exist, time dilatation is true, and has, on the other hand, been verified experimentally with very fast particles and very precise atomic clocks. "some sort of force", "seems", "think", and unstated "physical laws" won't do much to convince serious physicists. If you are confortable with classical mechanics (end of highschool level), I invite you to grab any book on special relativity and look for a single mistake in the mathematical reasoning.
Ohhhhh...LOL, wow i was being dumb, i get what he meant now.Gonzolo said:NoTime is referring to the fact that if you leave New York by jet at 7 am (sunrise), you will arrive in L.A. at about 10 am New York Time (it takes about 3-4 hours to go from New York to L.A.).
But during this time, the earth has rotated so that it is now sunrise in L.A., so perhaps 7 am local time.
Time zones... jet lag... GMT... Eastern time... Central time... "Tonight at 8/7 central..."
Hopefully, no one confuses this with relativity.
Wow :surprisedArmoSkater87 said:Ohhhhh...LOL, wow i was being dumb, i get what he meant now.
IMO this is becoming less and less true as technology progresses.Gonzolo said:Ok, first of all, I want to make it clear that whatever we learn in school (pre-university) is called classical physics (and electricity). For most people, it is all that they will ever need to know about physics, and even for most engineers (and I dare say most physicists), they will never have to use relativity
I believe so, with extremely precise atomic clocks.PatPwnt said:Haven't they also tested it with super sonic jets?
Because an observer on the ground sees the satellites in motion relative to them, Special Relativity predicts that we should see their clocks ticking more slowly (see Lecture 32). A straightforward calculation using special relativity predicts that the on-board atomic clocks on the satellites should fall behind clocks on the ground by about 7 microseconds per day because of the slower ticking rate.
Further, the satellites are in high orbits, where the curvature of spacetime due to the Earth's mass is less than it is at the Earth's surface. A prediction of General Relativity is that clocks closer to a massive object will seem to tick more slowly than those located further away (see Lecture 20 on Black Holes). As such, when viewed from the ground, the clocks on the satellites would appear to tick faster than identical clocks on the ground. A calculation using general relativity predicts that the clocks in each GPS satellite should get ahead of ground-based clocks by 45 microseconds per day.
The combination of these special and general relativitic effects means that, if not accounted for, the clocks on-board each satellite would tick faster than clocks on the ground by about 38 microseconds per day (45-7=38)! This sounds small, but the high-precision required of the GPS system requires nanosecond accuracy, and there are 1000 nanoseconds in a microsecond. If these effects were not taken into account, a navigational fix based on the GPS constellation would be false after only 2 minutes, and in general errors in global positions would accumulate at a rate of about 10 kilometers each day!
Application of Force means energy is expressed. Distance is a consequence of the mass's Acceleration. Acceleration of this mass is a consequence of a Force exerted upon it, which requires another mass with velocity and a collision of both masses. If Distance is zero, no Force exists which means no Energy is expressed. And, if no Energy is expressed, no Force can exist.Gonzolo said:Suppose you swing the masses at 1 revolution per second. Clearly, the 2nd one does a greater diameter : d2, than the 1st : d1. It therefore has a greater speed, d2/t > d1/t. This is totally classical, any engineer knows this. Same frequency of revolution, but different speed.
Application of a force doesn't mean using energy : remember W = fd. If d is 0, W is zero, and f and d must be in the same direction. Our swinging masses have d and f in perpendicular directions => no energy is used (f along string, d on circle diameter).
The speed (motion) is implied to be internal in the first clock, because the force upon the first clock must supply constant supply force for the second clock.Gonzolo said:The force is split in two. There is no "appear", the speeds are completely (classically) different as I said above.
When I think of static force, it means the movement is not as apparent to my vision and I have to use deduction to find it, where kinetic force means I notice it visually and is axiomatic. I don't see the first clock moving as much as I see the first clock. I know the first clock isn't loosing it mass where force occurs, so it looses motion outwardly to the second clock. This motion travels to the string and then to the second clock. It is a wave of energy.Gonzolo said:No, they are exactly the same types of force, both centripetal and static. There is no "collective speed", see what I said above.
When a constant supply of energy is supplied to this circumstance, and we isolated all outside forces upon this system (for now), we may see this in a linear way, because all things will settle as constant. So we may apply the what you are calling linear logic to it.Gonzolo said:Only a unique angular speed. But all I have said about relativity applies to linear speed (length per second), not angular (angle per second). I'm always talking about linear speed (that is what c is.).
But how is the balance of forces occuring here expressed? Perfect distribution of force accross the first clock and accross the second clock or is there a greater density of force, thus energy being expressed in one the clocks?Gonzolo said:No energy loss at all. No flow of energy. Only balanced forces. Let's get this classical thing right before going back to relativity.
This not how energy is defined in standard physics. When a rock is set on the ground, gravity exerts a force on the rock, the rock exerts a force on the earth, yet no energy is converted, because W = fd = f(0) = 0. For energy to be used, the rock has to move in the direction of the force. If you open a trap door under the rock, then there is energy conversion, because d will not be zero, the rock will move along the force (it will gain kinetic energy).omin said:...constant force is exerted upon the clocks and therefore is an expression of energy.
I don't see what you mean by "internal speed".omin said:The speed (motion) is implied to be internal in the first clock, because the force upon the first clock must supply constant supply force for the second clock.
Static forces means the forces are in equilibrium. Whereas Dynamic forces means their is movement along forces. You might call our system kinematic I suppose (not dynamic though), but in static and kinematic cases, there is no tranfer of energy. When something is already spinning (kinematic), centripetal force balances centrifugal force, but no force is along the motion.omin said:When I think of static force, it means the movement is not as apparent to my vision and I have to use deduction to find it, where kinetic force means I notice it visually and is axiomatic. I don't see the first clock moving as much as I see the first clock. I know the first clock isn't loosing it mass where force occurs, so it looses velocity outwardly to the second clock. This velocity travels to the string and then to the second clock. It is a wave of energy. I know this much.
There is no build up of energy, most was given in the beginning, whatever you add is to compensate for losses due to gravity (this should be ignored). There is more towards the outside though. A simpler system (without losses) is a spinning dumbell in space. You do not have to constanly supply energy to compensate losses in this case.omin said:What I don't know is if the wave of energy is building up toward the outside or if it's perfectly distributed.
If you do wish to supply constant energy (over what is need to compensate up and down oscillations), then it will be spinning faster and faster and faster and faster and this complicates things besides the point.omin said:When a constant supply of energy is supplied to this circumstance, and we isolated all outside forces upon this system (for now), we may see this in a linear way, because all things will settle as constant. So we may apply the what you are calling linear logic to it.
This is a contradiction in terms. We must not confuse things. The basis of the argument without this defined will make nothing intelligeable. I can't go through everything and do a logic operation without making it too time consuming. I'm not moving on untill we speak clearly on this issue.Gonzolo said:This not how energy is defined in standard physics. When a rock is set on the ground, gravity exerts a force on the rock, the rock exerts a force on the earth, yet no energy is converted, because W = fd = f(0) = 0.
Force always implies motion. Otherwise, potential energy would be matter of some sort that created velocity, rather than having it internalized in some way. Energy is niether created nor destroyed.Gonzolo said:No problem. You must first understand that a force, as defined in physics, (and as can be extended in all fields of life), does not necessarily imply motion, and where there is no motion, there can be forces (in equal and opposite directions, such as the net force is zero). The walls of a house exert a force on the roof all the time. Yet nothing moves.
It's been an ongoing assumption of mine for awhile. Here is the basis of my reasoning:Gonzolo said:"Force always implies motion."
Where did you learn that?
If it suits your needs, I have no problem with that. But until I learn alternative expressions for rotary motion (which I am 100% open to), I see no reason for abandoning centripetal and centrifulgal forces. The concepts have proven themselves useful beyond Newton's wildest dreams. Another example is that we couln't find the radius of the trajectory of an electron in a B-field in a simpler way.omin said:When I learned to count it began with the number one, not zero. Zero has never been taught to me as a value.
This assuption I'm introducing a new theory is smattering. That's what the physics books are implying and teaching, I've checked through the classic mechnanics of atleast four, and nothing is contradictory about what I've said about Net Force equaling zero and energy concepts.
This is the proof I have to offer: Force that equals zero only means that it is not sensed and implies momentum concepts, vs. energy which accounts for the motion that seems to dissapear or cancel.
Do you have proof that proves what I am saying is incorrect in representing things accurately?
And just because a book may not explain a nuance of classic physics doesn't mean it doesn't exist like: velocity is just average velocity, which means velocity never truely occurs in perfection, but acceleration does.
I found that quote funny.Otherwise, potential energy would be matter of some sort
No the coefficient of friction is the roughness of the surface, the force of friction is the coefficient times the normal force (vertical force).Istn't the vertical force implied in the coefficiant of friction?