J.C.MacSwell

Just a quick question: I recall my physics teacher in high school saying as you approach the speed of light, it takes more energy to accelerate at a constant rate, because you get heavier, and you'd eventually get so heavy you'd collapse into a black hole. Is that true/false/unrelated to issues of frame or oversimplified? If it's too off topic for here I apologize.

The bold is false. You are in other reference frames relative to which you are close to the speed of light, yet you are in no danger of collapsing into a black hole in those frames any more than you are in your rest frame.

I had to think about it for a spell. All the mass on the stick is equivalent to shrinking the rings to collapse onto the stick, Therefore the vector linking a point on the stick to m is stronger as the mass is the same around the ring, but the distance is greater for each dM on the stick. Priojecting the force from the shell effectively results in losing some of the force due to off axis cancelation of vector forces perpendicular to the axis.

 

Somewhat repetitive:

Thje mass on the stick is same as the total mass of a ring circling the stick, but the distance to a point on the stick is less than the distance from m to the ring. Thnerefore by distance effects alone the stick should provide a greater force.

 

When considering that the force from the ring was decreased by the cosine projection, no such loss is seen in the stick arrangement. - all forces are preserved.

 

For these reasons I pick the stick to contribute more force than the shell.

That is correct. So you see that, yet you are continuously treating the sphere as if the mass was all concentrated along the same axis when you do your calculations. This causes your error.

Does he arrive shorter?

LOL. If you are having us on that is brilliant.

Answer:

If he arrives at speed with is body aligned in the direction of travel (effectively a fly by)...yes.

If he arrives by coming to rest...no.

Geiestke.

Imagine a sphere lying with it's COM directly along the x axis from a test mass

Now replace the sphere with a magic stick.

1. The magic stick has the same COM as the sphere. It's COM is positioned in the same place as the COM of the sphere was.

2. The axis of the stick lies along the x axis

3. The magic stick has the same length as the sphere's diameter

4. The magic stick has the same mass as the sphere

5. The magic stick has the same mass distribution as the sphere in the x direction but it is magically concentrated along the x axis

Now the question is: Is the force of gravity between the test mass and the magic sphere greater, lesser, or the same as it was between the test mass and the sphere? Do you see the difference?

Counting them is not enough. They contribute different amounts to the total, and not simply because their projected distance along the axis is different.

 

Your differential mass dm is located r from your point, P. There are two contributions from each dm: one along the vector r, and one perpendicular to it. If we look at the "twin" of this point, projected through the center, the angle to it is smaller, thus, the perpendicular contribution to the force is smaller — it makes a larger radial contribution. Simply pairing them doesn't work and when you do the math you find that the method you are using is flawed.

 

Doing the on-axis calculation doesn't show this effect. You need to look off-axis to see it.

I think that is his answer right there if he thinks about it carefully, even without doing the math he should see where he is erring.

In the above the closer "twin" contributes more to the resultant force though, correct? The net force along the axis is greater for the closer twin.

The only reason I ask is that I'm not sure Geistke feels we see this.

Im thick. He has but one body and it is really compressed in a time frame. How do we know what effect this might or might not have?

From the laws of physics proven by experiment.

27 for me. What do I win?

 

What I do is put the sphere at 10 say.

Then arbitrarily I set 9 as the COM of the closest half shell and 11 as the COM of the farthest half shell.

Or if I waant to segment the shel in quarters m1 at 8.5, m2 at 9.5 m3 at 10.5 and m4 at 11.5. This moves the cg found using two 1/2 masses toward m. The more segments the more accuracy and the cg creeping toward m slows with successive doubling of the segments.

 

Wings are spicy meatballs of fire when ready, set, go west, young man about town council.

LOL. You can say that again!

(just not in this thread )

J.C.McSwell

 

I don't know if you picked up another 'asymmetry' in the mass distribution but each dM in the shell development has an intrinsic error due to the closest/farthest problem we have been discussing. I eliminiated the gross complexities by assuming each dM was a differential sphere with a unit vector r defined along the x axis and where the resiultant plane bisects the differential sphere and where the plane is always perpendicular to x, duh, that's what thr r does right?. There is of course, the gross closest/farthest conditions we have been discussing.

Not sure if If I follow what you are saying exactly, but by continuously using the COMs along the x axis in your calculations you end up with higher forces, because you are generally picking a point closer to the test mass than the average distance from the test mass of all the points it represents.

If, say, your segments were an infinitesimally thin set of discs perpendicular to the x axis, and you used the COM of each disc, each point chosen to represent the disc would be the closest point on each disc to the test mass and lead to an overestimation of the force, since you continuously underestimate the effective distance.

For some shapes this effective distance is further than the COM, for others it is closer, but for the sphere it is exactly the right distance.

This is the 50th anniversary of when Chubby Checker signed with Cameo-Parkway Records. He didn't actually write The Twist, but his version was accompanied by the stylistic dance that revolutionized the modern world of music. Geeks everywhere no longer had to worry about stepping on girl's toes since Checker made it possible to dance apart. Of course, we missed out on a lot of holding a girl close when Chubby came between us.

I have some chubby checkers. They stay on the board better than my thin checkers, especially outside on a windy day.

Geistke. The half spheres cannot be modeled as mass points at their respective centers of mass for this purpose (the inverse square law). You agree with that, correct? Yet you are using that assumption to prove that it cannot be true for the sphere itself.

The half spheres, not being spheres or spherical shells, cannot be modeled this way.

Spheres and spherical shells can be modeled this way.

Look in particular at the closer half sphere, the half that is responsible for the majority share of the force. If it had to be replaced by a point mass (of equal mass) that would bring about an equivalent share of the force, this point mass would have to be positioned further from the test mass than the center of mass of the half sphere is positioned. I don't think you realize this point. It is not so much that your logic is wrong, as it is the assumptions you are making.

This Forum is missing an off topic thread so ill be the first to start it!

Here you can submit stupid pics and talk about anything you desires (within good reason, keep it clean folks!)

How about a bucket of dirt. Really dirty dirt. I don't want to talk about it.

And I have no pictures of it.

Absorbing more CO2 pretty much just means a higher growth rate, and we already have plenty of plants with rapid growth, such as water hyacinth and kudzu vines.

 

For an animal, chloroplasts are useless, and especially for an endotherm. The rate at which animals use energy vastly surpasses the rate photosynthesis can produce it.

So my mum's threat some 40 years ago about growing potatoes in my ears - nothing to worry about?

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?