J.C.MacSwell

Other than simply hacking away at an object with a weapon or tossing it off a building, there are two primary ways to have an object damaged. The first is to simply use it in the way it was intended, and the object will become damaged naturally. This process is called wear and tear.

 

The second is to simply let it sit, and the object will become damaged simply by not using it. A good example of this is an old, abandoned house. It will have cobwebs and unstable floors galore, simply due to years of not being lived in. I'm sure this phenomenon has a name, but I don't know what it is. Maybe one of you guys can tell me what it is.

 

My question is: If all else is equal, what will damage an object more: Using it, or not using it?

1. abuse (hacking etc,)

2. use (wear and tear)

3. neglect (lack of maintenance over time)

I think that is a common term you might be looking for. Possibly decay or weathering as well.

Here, V must increase to keep P and n constant. n is the mole of the air inside the balloon, it cannot be changed as hot air doesn't leak out. With the constant n and larger V, its density is smaller, lighter.

I'm pretty sure GDG had it right for a hot air balloon. Your right that it doesn't or shouldn't leak out, but it certainly is free to get out, as it is open at the bottom.

http://www.inhabitat.com/wp-content/uploads/hotairbal-ed01.jpg

It would appear that you are assuming 100% packing efficiency. That would make for some very uncomfortable hippos...

101%!

I think we are assuming the majority are liquid hippos.

2nd and 3rd paragraphs:

So...how does that differentiate time dilation due to expansion displacement vs movement displacement?

Not disagreeing or agreeing there's a difference, but I don't see it in those paragraphs.

Someone here shared this video on time dilation with me recently and I thought it was one of the better ones. Check it out.

If you could view Earth in real time after accelerating out of Earth's frame, you would see people age quickly. They would see you age slowly in your frame if viewing in real time was possible.

Ouch.

A few corrections.

 

 

 

Actually, that's what it is at 4 C, the temperature at which it is densest at standard pressure. However, it is usually very close to 1 g/cm³ at other temperatures.

 

 

 

Actually, moist air is less dense than dry air (probably a typo on your part considering you know the ideal gas law). This is counterintuitive to people who note that liquid water is heavier than air. However, the molecular weight of water is 18 g/mol, compared to the heavier 28 g/mol for N₂ and 32 g/mol for O₂. Since the density of gases is related to the molecular weight of the gas, this makes moist air less dense than dry air.

Moist air is heavier though, when it has liquid water suspended in the air. I think a lot of people consider that moist air.

Nope.

Nice Link, but can you direct us to where it makes the time dilation distinction between expansion recession and recession due to movement otherwise?

If we are both on the CMB rest frame but very far apart and I send a probe that accelerates toward you, closes the gap, decelerates less than it accelerated originally to come to a halt in your lap, is it then in the same time frame it started? I don't know, and saw nothing in that link that would indicate either way when I scanned it.

Edit: I hadn't read NTWK's latest post. I will re-check and try and see if it makes sense to me.

Re-edit: Checked it rather quickly but not sure that's the key. Isn't the expansion of the Lorentz interval not equal in all directions if you are not at rest wrt the CMB? How can it be invariant for GR if that is the case? (I can see how it is invariant for SR)

An average hippo is 1500 kg. Hippo's sort of float, therefore their density is similar to water. Therefore a hippo is 1.5 m3 on average.

 

1 mole of hippo's is therefore 1.5 * 6.022*10^23 = 9.033*10^23 m3 in volume.

The volume of the earth is 4/3*pi*r^3 = 4/3*pi*(6.378*10^6)^3 = 1.086*10^21 m3 (the volume of the hippos far exceeds our tiny earth).

 

We then must add these two volumes, calculate the total radius, and subtract the earth's radius:

 

total volume = 9.044*10^23

total radius = 60*10^6 m

radius earth = 6.378*10^6

 

So the layer of hippos is about 54*10^6 m, or 54000 km thick.

 

So, not miles deep... more like the Earth becomes the size of Jupiter

 

Sorry for adding a completely pointless post - I just enjoy silly calculations such as these. Time to get serious again.

How about a mole of moles?The furry kind that can't see very well. That must be closer to the moon's size.

Actually they should weigh the same in a similar gravitational field. Their mass should also be same. Unlike weight, mass is the amount of material an object contains and does not change under a different gravitational field.

 

 

 

If the boat is floating it's weight is equal to the weigth of the volume of water displaced.

Fair enough, you have used pretty much the same definition of weight in each case.

However, if you used a scale, 10 kg of lead would weigh more than 10 kg of feathers due to the greater buoyant force of the atmosphere on the feathers. That is more or less the definition some others have used in this thread. They have allowed buoyancy to be factored into the terms heavier or lighter. Still valid, but they are weighing things differently.

OTOH, mass is always defined the same way. So on the moon, with no buoyant forces from the atmosphere 10 kg of lead or feathers weigh the same, even on a scale.

The difference might be subtle, but become quite significant with respect to something as "light as air".

Does that mean he could age and die in his frame while staying young when viewed from outside frames?

Not if while and when are defined in the same frame.

I still don't get it. Why is mass always compared to volume? Mass is just the quantity of matter, regardless of the volume. So shouldn't two objects of two different volumes still have the same weight regardless of their volume if their mass is the same or is mass always a function of volume?

 

That is what I have been taught but it could be wrong.

Which has more weight, 10 kg of lead or 10 kg of feathers? What about if you weighed them on the moon?

Which has more mass, 10 lb of lead or 10 lb of feathers?

A lot depends on how you define your terms. Does a boat weigh nothing when floating in water - or just enough to displace an equal weight of water?

So if there was no atmosphere like on the Moon then denser air will not be lighter than less dense air?

Assuming the same mass of air they would weigh the same on the moon.

Assuming the same volume the denser air would weigh more due to the additional mass in the same volume.

No atmosphere means no buoyant forces to complicate matters.

Most would say less dense air is lighter generally, but it depends on what you are comparing and what you mean by lighter.

.

 

Who made that claim? If anyone here made it, its wrong.

Split a sphere in half. In fact, split a sphere any way you want into two separate parts. Now compute the gravitational force induced by the two parts of the sphere at some test point inside the sphere. No matter where you place the test point and no matter how you split the sphere into two parts, the forces from these two parts on the test point will be equal in magnitude and opposite in sign.

 

Just to be clear, any time I referred to this ,it was with regard to a mass outside the sphere.

I know I mentioned it with regard to a mass inside a hoop.

I think it was obvious to Newton, no calculation required, that the closer half sphere contributed more to the gravitational force on an external mass, than the further half.

Every point on the closer half has a symmetrically opposite point. Each of those on the closer half contributes more gravitational force than the corresponding point on the further half.

Not sure if Newton would have needed to give it even that much thought, to draw the above conclusion.

Was it not obvious to you, or did you have to think about it? (I assume you didn't have to calculate it).

Edit: Adding after seeing your last post:

Geistke, there exists a point closer than the COM where half of the force contribution lies further away and half lies closer.

However, for the purposes of modeling the gravitational force using the inverse square law the COM is exactly where to place a point mass for a sphere, not closer.

J.C.,

 

I never said that the single hoop can be modeled as a single point mass. The force can nonetheless be calculated. Calculating the force for a point along the hoop's axis is not that hard. Knowing how to do that is central to understanding the shell theorem.

 

I am asking geistkei, and not you, to provide this calculation.

I thought your post #11 was in part replying to my post #10.

Sorry about that.

Length contraction without a corrosponding contraction of the other dimensions will lead to distorton which should be appearent without regard to size. I take it that the view is that this would not cause damage because it is not viewed in the frame of the spaceman. This takes me back to where I was at post #50.

The distortion is the natural shape of the object viewed from a chosen frame at significant relative velocity. Any lack of distortion, would be cause for alarm. That could lead to the spaceman's death.

Beware the spaceman speeding by at relativistic speeds yet apparently undistorted.

He is considerably "stressed".

The reason I asked the question is because knowing the answer to that question is central to knowing how to calculate the total force exerted by a spherical shell. You have to integrate over a surface. In short, you have to do a double integral. By far the easiest way to do this double integral is to slice the spherical shell into infinitesimally small circular hoops.

When you integrate for a hoop though, say for a test mass inside the hoop, you would get a resultant force towards the nearest point on the hoop, where inside a spherical shell you would have no net force at all.

A single hoop could be modeled as a point mass at the COM only as an approximation at a distance.

It is the geometry of a sphere that conspires with the inverse square law to get the shell theorem. It doesn't work with any other shapes or groups of masses unless you carefully "tune" them.

It is amazing that with everything else he did, Newton practically invented calculus to do this.

But that is why we have a master clock. So we do not have to wait for the time of transmission.

If you take the master clock with you on a trip it seizes to be a master clock...unless it worked on something outside of known physics.

Just because we see the lightning before we hear the thunder does not mean they were not simultaneous or synchronized.

So you do the math based on distance, the speed of light and speed of sound and happily are confident that they were simultaneous.

Do the same based on the distance and speed of light between the twins...and they each still think the other has slurred speech.

The gravitational net force towards a hoop does not act as if the mass was concentrated at the COM. In this particular case what Geistke describes makes sense (it would act as if the mass was closer to the object than the COM when outside the hoop but in the plane of the hoop) and I think he is confused with regard to the sphere being similar to that or to a group of masses.

Edit: I thought Geistke's point was that the closer half of the sphere contributed to the gravitational force more than the further half, which is true and of course obvious to Newton. The resultant force acting as if at the COM is pretty much unique to the sphere, or shell of the sphere. It is exact and not a close approximation.

How can all that be true? It sounds like a contradiction. I gave the two twins a master clock slaved to Earth's time frame and can not see how they would differ. And I thought I had those d@mned twins figured out.

Let's say they stopped accelerating in a symmetrical predetermined way and started signaling each other as planned.

They would both agree that the master clock had slowed the same amount relative to their own.

But even after accounting for the time for transmission of the signals they would each think the other twin's clock had slowed, by more in fact than the master clock had slowed.

I asked this in a thread in the relativity forum and it got buried and never got answered.

 

Does the separation of matter caused by the expansion of space experience the same time dilation and length contraction as a similar distance change and speed from a rocket ride?

I think this is a great question.

I am thinking that if two galaxies are at rest with the CMB but are separating at relativistic speeds due to a increase in distance from expansion, they will share the same time frame.

I don't know. I think it would depend on how you define "time frame".

At rest with the CMB, or on the Big Bang Track is reference frame I like to use, but can it be considered an inertial frame? A continuum of inertial frames? Can a standard SR inertial frame extend cosmic distances and still give valid predictions?

I am thinking that if the two twins blasted off from Earth experiencing equal acceleration but in opposite directions, They also would experience the same time frame. Despite there increasing separation speed.

Earth would judge them to be in the same time frame. They each would each consider the Earth's time frame to have slowed the same amount, relative to their own, but they would not agree at all that they were in the same time frame.

At the same time (so to speak), wouldn't each be at rest with respect to some distant point on the Big Bang Track, at rest in an inertial frame that included a Big Bang Track "rest point"?

Anyone have the answer and where am I going wrong?

I don't, I have more questions than answers, but I think when you extrapolate local physics to cosmic distances and times, you must lose some degree of certainty.

Someone with a better understanding of Astronomy, Cosmology and General Relativity like Martin may be able to give you a more definitive answer.

I have a question about the forces and aerodynamics at play during a paraglider's flight.

 

As I understand it normally flight can be thought of simply in terms of Thrust, Lift, Gravity and Drag.

 

Imagining a paraglider in flight above a ridge, with no forward movement, remaining at a constant location relative to a spot on the ground.

 

How does thrust apply in this situation?

 

Is thrust even present? and if not what balances the forces of drag?

 

How do the stated forces balance with respect to my stated normal understanding of aerodynamics?

 

 

Thanks.

What Mokele said.

In particular what you are describing sounds like a paraglider heading (and falling relative to the flow) upwind in a rising air current.

Gravity has a slight forward component relative to the flow.

Lift, a force perpendicular to the flow, has a slight forward component relative to the ground.

The drag, a force parallel to the flow, has a slight upward component.

It can all balance out so that the paraglider is stationary.

This note does not rewrite gravity physics, nor does it attempt to -- it is a correction to Newton's Shell theorem wherein it is claimed that the force equation indicates the location of the point where all mass of the shell is concentrated;, that the total force on a test mass m from the mass on the surface of a shell located a distance d from m that the total force can be considered concentrated at a point located at the center of the sphere.

 

This note concerns the location of the center of the force which cannot possibly be located at the sphere (center of mass) geometric center. The results of integrating the force term to get s F = GmM/d^2, where d is the distance between m and the sphere center, but the expression is only one of force and one cannot assume automatically that the center of force is located a distance d from m. I make the point the sphere half closest to m contributes a larger share of the total force than does the shell half farther from m.

 

To avoid confusipon, I use the term 'mass-force-center' (not seen in the literature) to mean that point from which the test mass m considers the location of the source of the total [concentrated] mass bearing on it from the mass M of the shell.

 

Newton's Shell Theorem –Bad mathematics. Bad physics,

 

This part shows how to determine the point where m sees the force of two equal masses in a line added to provide the total force of the two masses and from this determine the mass-force-enter.

Take three mass point objects m1 = m2 = m3 = 1 unit mass, G=1 unit gravitation constant, and using init distances the force of attraction between m1 and m3 separated by 10 unit distance is calculated using the universal law of gravity expression, F = Gm1m2/r^2 (minus sign omitted). F12 = (1)(1)(1)/10^2. Similarly, the force generated between m1 and m3 separated by 12 unit distance F13 = 1/144. When the masses are arranged along a common axis the total force of m2 and m3 on m1 is F123 = 1/100 + 1/144 = (1)(1)(1 + 1)/r^2 . Rearranging the terms, 2/r^2 = 244/(100)(144), or, r^2 = (2)(14400)/244 or R^2 = 118.03. The result r = 10.86 and, 10 < r < 11, where 11 is the location of the m1m2 system center of mass (COM).

 

The combined forces' COM acting on m1 is located a distance 11 from m1. However, the center of mass-force (CMF) is located at 10.86, which is off set from the COM in the direction of m1. The total forces of the mass of this spherical shell is calculated manually (see above) using mirrored image pairs of masses on the shell where one member of the each pair is in an opposite hemispheres, one closest to m1, one farthest from m1. Clearly when all forces are calculated each and all CMFs of each calculation are located in the nearest sphere segment to m1, contrary to Newton's Shell theorem that without any physical basis, claimed that m1 may consider the mass of the sphere concentrated at the COM of the sphere.

 

The links below are consistent examples of the rote acceptance of a developed shell theorem, referenced as an unchallenged law of physics and communicated as scientific gospel, chiseled in stone, as it were, leaving the authors and subsequent decuples immune from any heretical criticism.

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

http://www.physclips.unsw.edu.au/jw/NewtonShell.pdf

http://www.absoluteastronomy.com/topics/Shell_theorem

From inspection of the sphere and m1 externally located at some point r from the sphere center it is obvious that using the concept of "inverse distance squared" (and the universal law of gravity) as a starting point, the mass of M in the hemisphere closest to m1 will contribute a greater share of the total force on m1 than the mass in the farthest hemisphere, hence the CMF for the entire mass M is located on the m-M axis off set from the COM in the direction of m.

 

Where, and how, does the Newton Shell theorem (NST) place the CMF at the sphere COM? It doesn't! The NST model begins with a ring of differential mass, dM, oriented perpendicular to and centered on r. Then, summing all forces for each dM on each ring and integrated over the surface of the sphere producing the calculated total force acting on m, which says nothing, absolutely nothing, regarding the location of the CMF. The Wikipedia model referenced above states after dF is integrated to F = GmM/r^2 that,

"The shell really does act as though all the mass is concentrated at the center!"

 

The big problem here is that the developed algorithm made no inclusion for determining the location of the CMF. The intuitive assumption that the CMF is located at the COM of the sphere was arbitrarily (instinctively) made (or so I surmise) from the 'r^2' term in the expression, that clearly is an expression for determining the total gravitational force of mass M on m, only.

 

Another flaw is seen in the expression when obtaining the net vector force in the m-M direction derived by taking the cosine projection of force in the "r" direction only, and from this, supposedly, the inference is that 'the CMF followed the projection of the force onto the m-M axis' – the projection of force vectors is mathematically proper (forces perpendicular to the m-M axis 'cancel'', or so we are told), but to include the scalar quantity location of the CMF is just plain "bad mathematics"; not properly placing the CMF in the nearest hemisphere of M is just plain, "bad physics".

 

Caveat emptor – beware of standing on the shoulders of giants who have been dead for 300 years.

If you know the total net force and direction you have the resultant force. It effectively acts through the center of mass (assuming a homogenous sphere), not from that point. If you want to say the resultant acts through your CMF described that is fine, it is in the same direction, but it is not acting from that point either, the resultant simply is in that direction. The COM is simply an easier point to describe, can be used as the masses change in position, and is the point from which the inverse square is measured.

Why complicate something that is being simplified? - the summing of gravitational forces into a resultant force that consistently acts through the COM.

Do you think Newton didn't realize what you just described?

I still come back to the point that the space man has but one body and if that contracts it could kill him. You seem to be saying that he in effect has two bodies and that since only the one in the other reference plane contracts he wont notice it.

You are in an infinite number of reference frames at near lightspeed.

How ya feelin'? No need to reply if you're dead.

But with relativistic velocities, it does shrink, in the applicable reference frames. It's not just an illusion, as with a mirror. A twenty foot pole moving fast enough will fit inside a ten foot room, in the reference frame where the room isn't moving. Of course, in the reference frame with where the pole isn't moving, the pole is still 20 feet long, and it's the room which has shrunk. Believe it or not, this actually doesn't cause a paradox:

 

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

 

 

 

Interesting. I don't know, but I'm thinking it can't work the same way as motion, if only because at some point they'd all have infinite momentum, and then beyond...

 

Oh, and it's not caused by motion relative to the CMB. "At rest with respect to the CMB" is convenient but not priveleged. It's caused by motion relative to whatever reference frame it's considered in.

The CMB frame is only an inertial frame locally, or more of a continuum of frames. There's lot's of fun to be had with this. If you go strictly by SR anything moving faster than light should be going backwards in time, so in our time frame some of the oldest things we are viewing should be much younger by now!