Split Thread The validity of classical physics (split from: DWFTTW)

[mender]

The link speaks only of relative velocity, Mender. Not relevant.

Until you understand that it is relative velocity that is the issue, your comments are irrelevant.

How can you not see that? Did you put the boat velocity to zero and change the river speed?

 

[mender]

The only known objects that come close are those that can be dragged by the water surface itself.

Dragged by the water surface? How does that differ from just below the surface or near the bottom?

Because that external force is the water itself, and the fluid ahead of the object is viscous, work must be done to accelerate that object. This means that the object cannot reach waterspeed.

Accelerate the object? Why would we do that?? We want it at rest with respect to the water, not moving through it. When it is at rest wrt the water, it is at waterspeed (figured I should make this more noticable so you understand that it is important). Moving it away from that condition will require force to counteract the drag that the water resists that movement with.

Is that so hard for you to understand?

 

[spork]

Humber however sees that the river is moving at 5 mph...

His reasoning has been consistent (and wrong)...

You are either giving humber too much credit, or have done what none of the rest of us have had the energy to do - understand the true (consistent?) nature of humber's confusion.

 

[spork]

Incidentally, it occurs to me that I must come across as selfish or lazy for not spending the necessary energy to straighten humber out. This is no excuse, but I can tell you I'd be willing to expend FAR more energy on this effort if humber was asking for help, or even admitting the very slightest possibility that he doesn't understand physics perfectly well.

 

[sol invictus]

Just for the pleasure of watching more squirming and flipflipping:

http://en.wikipedia.org/wiki/Hot_air_ballooning

Hot air ballooning is the activity of flying hot air balloons. Attractive aspects of ballooning include the exceptional quiet (except when the propane burners are firing), the lack of a feeling of movement, and the bird's-eye view. Since the balloon moves with the wind, the passengers feel absolutely no wind, except for brief periods during the flight when the balloon climbs or descends into air currents of different direction or speed.

http://www.ipy.org/index.php?/ipy/detail/weather_balloons_in_antarctica/

We can track the balloon throughout its flight using Global Positioning System satellites (GPS) and then convert the GPS track into information about the wind speed and direction at different heights in the atmosphere.

From a book called "Meteorological Measurement Systems:

http://books.google.com/books?id=Zj...X&oi=book_result&resnum=9&ct=result#PPA219,M1

 

[spacediver]

No, that's not correct. Momentum is conserved in that collision because it has no place to go - if all the mass is accounted for, there's nothing to carry it away. Kinetic energy is not conserved, because some gets converted into heat.

But momentum is a function of velocity and mass. If KE can dissipate due to heat, then surely momentum will also dissipate (according to my not-yet-getting-it thinking).

Here's an example that I can't get my head around: Suppose a metal sphere weighing 4 kg is flying through space at 10 m/s. Its momentum is 40 kg m/s. Its kinetic energy is 200 Newtons.

Suppose it briefly passes through a magnetic field, slowing it down to 5 m/s. Its momentum is 20 kg m/s. Its kinetic energy is 50 Newtons.

How is momentum conserved here?

My guess would be that the magnet itself has a mass and its velocity would change due to the interaction with the sphere? But if this is correct, then why, in post 205, did you (Sol Invictus) tell me that the following is incorrect?

If I had to guess, I'd say that the answer is that the loss of kinetic energy necessarily involves the change in the mass or velocity of some other particle.

The mass cannot change unless the collision is so energetic a nuclear reaction takes place. Lets assume that doesn't happen. Then the state of each object is characterized by its velocity, which is a vector with three components. So we have six numbers initially before the collision (three and three) and six after. Conservation of momentum is an equation that tells us something about (some combinations of) three of those numbers, and conservation of energy tells us about one.

Ok so if I understand correctly, the six numbers are a scalar and direction component for each of three axis. And when we specify momentum we need to talk about all three pairs (axes) to fully define the momentum of a particular object. But when we talk about KE, we're only talking about the KE along one axis (hence one pair)?

Is this correct?

The quickest answer is, "because". If you take Newton's laws, you'll see that mv is conserved, but mv³ is not.

Ok fair enough - it's an empirical fact about the universe which has been observed. I wrongly assumed that if m*v was conserved, then any combination of m's and v's would also be conserved, but I see how that's a flawed assumption.

 

[spacediver]

I forgot to mention one very important point...

If you want to gain an understanding of such physical principles, read and think about most any of these explanations. Get your head around the ones that seem to make more sense to you. Even just make your own random guesses if need be. But for GOD'S SAKE don't use humber's posts for anything other than entertainment. I know it may seem like I'm intentionally being silly here, but I'm not. The worst thing someone trying to understand physics could possibly do is to study horribly misconceieved and completely innaccurate descriptions. If humber's explanations were sold over the counter they would absolutely require a warning label "To be used for entertainment purposes only"

worry not, I've already confirmed for myself that Humber is incapable of expanding or correcting her mental models of the physical world. As such, I don't trust a thing she says.

 

[John Freestone]

This is absurd. It implies
(1) All objects regardless of drag, ( all values of c(u)>0) reach waterspeed.
(2) There is no force F_drag=0 at waterspeed.

That's what we said in English. Is it so absurd that you can use the formula to express the same things?

So, a half-filled oil drum, will travel as fast as a sleek canoe, and will require no work to keep is at waterspeed. ?

Yes - if you ignore the drag from the air, or any differences in waterspeed these two craft experience due to their different shapes. Just going on their sleekness, it's true. A dead starfish will reach the same waterspeed that a dead goldfish will. They'll take different amounts of time to get there. The draggier object will be off downstream ahead of the sleeker one. They will therefore be going different speeds to start with. They will, however, both approach the same speed (still separated by some distance) - waterspeed. At that point, drag will be zero. No force. Water not moving past objects. That's why no drag. No water passing. Not passing the water. Water, object, same speed. See? Speed, same. Same. speed.

The ideas is false because your mathematics is bad. What you have done is demonstrated that F_drag = c(u) u², or 0 = 0.

Zero does equal zero in our universe.

Really. I say put-up or shut-up is for those who make extraordinary claims and the don't support them. I have three times asked you for evidence that objects flow downstream at the speed of the water (or other medium), as you claim. I have looked. I find none. I say you are avoiding saying you are wrong because that would undermine the concept of the treadmill.

Don't try your lame and borrowed 'flip-flop' idea. I have made my position clear.

The score is at least 4-0*. You've made your position, er, reasonably clear, for you. It is you who has not provided supporting documentation.

*ETA: I missed four new ones. Eight - Nil.

 

[sol invictus]

But momentum is a function of velocity and mass. If KE can dissipate due to heat, then surely momentum will also dissipate (according to my not-yet-getting-it thinking).

Nope.

Here's an example that I can't get my head around: Suppose a metal sphere weighing 4 kg is flying through space at 10 m/s. Its momentum is 40 kg m/s. Its kinetic energy is 200 Newtons.

Suppose it briefly passes through a magnetic field, slowing it down to 5 m/s. Its momentum is 20 kg m/s. Its kinetic energy is 50 Newtons.

How is momentum conserved here?

You have to keep track of the momentum transferred to the eddy currents in the sphere, and their associated magnetic fields.

My guess would be that the magnet itself has a mass and its velocity would change due to the interaction with the sphere?

Right - and moreover, electromagnetic fields themselves store and carry momentum.

A simpler example is just the one we started with - two objects colliding. Imagine two objects of the same mass flying directly towards each other and the same speed. Total initial momentum equals zero. In an inelastic collision they stick together and the resulting object will be at rest. Zero momentum (and zero KE - the KE has changed, but the momentum hasn't). In an elastic collision they will bounce off each other, but they will always fly apart with equal velocities and in opposite directions. Total momentum equals zero, but this time KE hasn't changed (if the collision is perfectly elastic).

Ok so if I understand correctly, the six numbers are a scalar and direction component for each of three axis.

No - just two momentum vectors (one for each object). A vector has three components (three numbers).

But when we talk about KE, we're only talking about the KE along one axis (hence one pair)?

KE isn't a vector, it's a scalar (i.e. one number).

 

[spork]

But momentum is a function of velocity and mass. If KE can dissipate due to heat, then surely momentum will also dissipate (according to my not-yet-getting-it thinking).

You're thinking is not as far off as you may suspect.

Momentum IS conserved so long as nothing interferes with it. That's not the most precise or accurate way of saying it, but let's go further.

In a closed system with no external forces, momentum is conserved. If we have a bullet screaming through the atmosphere it will clearly slow down. Thus it seems it's losing momentum - and IT is. But we either have to consider the whole system (i.e. bullet and every molecule of atmosphere it hits) OR we can consider the bullet alone and treat the atmospheric interaction as an external force.

Here's an example that I can't get my head around: Suppose a metal sphere weighing 4 kg is flying through space at 10 m/s. Its momentum is 40 kg m/s. Its kinetic energy is 200 Newtons.

Suppose it briefly passes through a magnetic field, slowing it down to 5 m/s. Its momentum is 20 kg m/s. Its kinetic energy is 50 Newtons.

Yes, the metal sphere slows down - and the magnet speeds up. The magnetic field is just a way for some other object to reach out and touch the sphere. The two objects (sphere and magnet) interect. They are both part of the closed system. And the momentum within that closed system will be conserved.

My guess would be that the magnet itself has a mass and its velocity would change due to the interaction with the sphere?

D'OH!!! I should read before I write. You are getting this just fine.

Ok so if I understand correctly, the six numbers are a scalar and direction component for each of three axis. And when we specify momentum we need to talk about all three pairs (axes) to fully define the momentum of a particular object. But when we talk about KE, we're only talking about the KE along one axis (hence one pair)?

Not exactly. For a rigid object we can consider the three axes (x, y, z). The body has a translational velocity component in each axis and a "rotational velocity" component about each axis. Any of these numbers may be positive, negative, or zero.

EDIT: I see that Sol answered this now. I assumed the 6 numbers you were refering to were the state vector for a single rigid body. It looks like you're talking about the state vectors for two rigid bodies with no angular momentum (from Sol's answer).


Sol, I would think the momentum carried by the EM field would be negligible in the example given. No?

 

[spacediver]

Right - and moreover, electromagnetic fields themselves store and carry momentum.

I re-edited my post, but you were too quick on the draw, so I'll ask it again

If this is true, then why did you earlier tell me that my guess was incorrect? Here is what my guess was:

If I had to guess, I'd say that the answer is that the loss of kinetic energy necessarily involves the change in the mass or velocity of some other particle.

I know I was asking about KE and not momentum, but here's my thinking:

If an object loses kinetic energy due to heat, without changing its mass, then its velocity must be changed. Hence, the momentum of this object has decreased. Surely, if conservation of momentum is true, then there must be a corresponding increase in momentum in some other part of the system.

But if the loss of KE doesn't involve the change in mass or velocity of another set of particles (as you implied when you said my guess was incorrect), then there is no necessary corresponding increase in momentum in some other part of the system.

Do you see where I'm confused?

A simpler example is just the one we started with - two objects colliding. Imagine two objects of the same mass flying directly towards each other and the same speed. Total initial momentum equals zero. In an inelastic collision they stick together and the resulting object will be at rest. Zero momentum (and zero KE - the KE has changed, but the momentum hasn't). In an elastic collision they will bounce off each other, but they will always fly apart with equal velocities and in opposite directions. Total momentum equals zero, but this time KE hasn't changed (if the collision is perfectly elastic).

Ok I get this now. It's a useful scenario and I can clearly see how momentum is conserved and KE is not. It will be a useful example to meditate on since it makes intuitive sense and it jives well with my own mental model. I just need to square this example with the concern I have listed in the first part of this post.

KE isn't a vector, it's a scalar (i.e. one number).

But KE is a function of velocity, and velocity is a vector. So how can KE be a scalar?

 

[John Freestone]

(1) The water molecules already have the required velocity. That is what defines waterspeed.

Then they won't be accelerated. That won't be true if I just pour a cup of water into a river. The molecules will begin with no lateral speed. So, you think after a reasonable amount of time, the water I added to the river will be going significantly slower than the rest of it. Of course you do.

(2) Hot air balloons can use their heat engine to gain altitude. This can be used to gain lateral velocity over and above the wind.

I'm sure that's right, if you mean they can accelerate to one windspeed and then descend or ascend into a slower airflow. They will take a little time to slow down to windspeed, just as things take a little time to speed up to windspeed. You're moving goalposts again. Still, their 'heat engine' only gives them the ability to rise in the vertical component of their direction. The heat engine does not directly cause them to move sideways. You know that as well, but won't admit it.

(3) The River boat link is one of countless examples of relative velocity between observers. Not about fluid dynamics, or object speed in media.

Pants on fire.

The given velocities are nominal. If you deduce from that, no wonder you are in a muddle.

Oh yes, I should just deduce from measured velocities. I should get in a boat and measure the velocities with a spoon.

(4) One correct. Waves are indeed not the same.

as drowning.

I would cut down on the waffles.

I have been putting a bit on over xmas, it's true, but I didn't know my webcam was live.

 

[Subduction Zone]

magnetic fields can store momentum??? I don't think so, at least not an appreciable amount. spacediver you will find that momentum is transferred to whatever is holding the generator of the magnetic field. If that is attached to the ground it is being transmitted to the Earth. Now it takes an awful lot of momentum transfer to notice any change in the momentum of the Earth. That is why studies in changes of momentum are usually done with much smaller bodies.

 

[sol invictus]

Sol, I would think the momentum carried by the EM field would be negligible in the example given. No?

Not in all cases - I'm pretty sure I can arrange a thought experiment where all the momentum is carried away by EM radiation, for example. But it would be a bit contrived (and almost certainly not what the OP had in mind, now that I re-read it).

If this is true, then why did you earlier tell me that my guess was incorrect?

There may have been some confusion - what I was telling you is that loss of total KE does not necessarily imply change in total momentum.

If an object loses kinetic energy due to heat, without changing its mass, then its velocity must be changed. Hence, the momentum of this object has decreased. Surely, if conservation of momentum is true, then there must be a corresponding increase in momentum in some other part of the system.

Almost - except that you need to remember that momentum is a vector, so "increase" and "decreased" in that paragraph aren't quite right (think about the example of the two bits of clay colliding and you'll see why).

But KE is a function of velocity, and velocity is a vector. So how can KE be a scalar?

Does the term "dot product" mean anything to you? If not, just think of KE as the magnitude of the velocity (which is a scalar) squared, times the mass over 2. It's like the difference between speed and velocity.

magnetic fields can store momentum??? I don't think so, at least not an appreciable amount.

Then you've never seen what happens when the current to a big electromagnet is suddenly turned off.... to store momentum one needs both a magnetic and an electric field, but a change of reference frame takes care of that regardless .

 

[69dodge]

But KE is a function of velocity, and velocity is a vector. So how can KE be a scalar?

The kinetic energy of an object doesn't depend on the direction that the object is moving, just on how fast it's moving.

Consider two objects of equal mass moving at the same speed, but in opposite directions. The momentum of the pair is zero, because the directions cancel, but the kinetic energy of the pair is twice that of each object, because kinetic energy doesn't care about direction.

An object's momentum might be x kg m/s north or y kg m/s south, but its kinetic energy is just z joules.

(Zero momentum is special, in that it has no direction, or, if you like, any direction. But any other momentum needs a direction to be specified.)

 

[spacediver]

thanks - i think i've been given enough here to work with. Will post back if I'm still confused

 

[Subduction Zone]

Then you've never seen what happens when the current to a big electromagnet is suddenly turned off.... to store momentum one needs both a magnetic and an electric field, but a change of reference frame takes care of that regardless .

If you are talking about inductance and the tendency of the current to keep flowing that is another matter. I would not call that momentum, at leas it is not the momentum of a moving ball for example. If you slowed a metal ball with a magnetic field the momentum would go into whatever the coil was attached to. If it was attached to the Earth the momentum of the ball would be transferred to it, just as the thrower of the ball transferred an opposite momentum to the Earth when he threw it.

 

[sol invictus]

If you are talking about inductance and the tendency of the current to keep flowing that is another matter. I would not call that momentum, at leas it is not the momentum of a moving ball for example.

I'm talking about momentum. EM fields can also store energy and angular momentum.

http://www.jdis.com/wos/027.html

The linear momentum density in a configuration of EM fields is [latex]$\vec E \times \vec B/c$[/latex]. The energy density is [latex]$\vec E^2/2 + \vec B^2/2$[/latex] (times some constants that depend on your units).

Haven't you ever heard of a light sail?

 

[humber]

Dragged by the water surface? How does that differ from just below the surface or near the bottom?

Very small objects can be carried by the surface tension, they are an exception.
I am afraid that I am going to have to ask you to show where I mentioned anything other than relative motion to the water. I told Sol_invictus this three times.
To make that clear, waterspeed is when u=0, as defined by Raleigh's formula. That is alsao Sol_invictus' definition. Argue with him about that.

Accelerate the object? Why would we do that?? We want it at rest with respect to the water, not moving through it. When it is at rest wrt the water, it is at waterspeed (figured I should make this more noticable so you understand that it is important). Moving it away from that condition will require force to counteract the drag that the water resists that movement with.

When released into the water, the object will be accelerated to its final velocity. Again, if you disagree, this is also accepted by Sol_invictus. Argue with him about that.

Is that so hard for you to understand?

No, but you are trying to make it look like it is. We have the same definitions, and that is that.

ETA:
Still no evidence.

 

[Perpetual Student]

I'm talking about momentum. EM fields can also store energy and angular momentum.

http://www.jdis.com/wos/027.html

The linear momentum density in a configuration of EM fields is [latex]$\vec E \times \vec B/c$[/latex]. The energy density is [latex]$\vec E^2/2 + \vec B^2/2$[/latex] (times some constants that depend on your units).

Haven't you ever heard of a light sail?

S. I.:

How do you write mathematical expressions here (like the ones in the above quote)? I am able to use a system called aurora in a word document, but I cannot cut and paste the math to this forum. Do you have a way to write the math directly in your messages here?