For the record. Nitrocellulose is definitely diamagnetic. I have an estimate of -212.66 ≤ Xd (μemu/mol) ≤ -57.5 based on Pascal Constants.
Can you give a reference to that paper Bobo?
Build a man a fire, warm him for a day,
Set a man on fire, warm him for the rest of his life.
I believe this is the paper I had in mind. It's not quite what we're talking about here, but it shows how these interactions are nontrivial. I think there are others if you search for "electric force between two conductive spheres".
I keep going back and forth in my mind as to whether the diamagnetism would make any difference. On the one hand, I'm pretty sure there's a theorem that states that charge won't redistribute when two neutral conductors are brought close to each other. On the other hand, magnets exist...
I believe electromagnetism should have significant effect on this experiment only if the net charge of both balls is not zero. It should be possible to estimate how much the product q1*q2 should be in order to match the magnitude of the gravitational force.
The reason for that is that the Newtonian attraction falls like 1/r^2, and r is of order meters, and therefore macroscopic. Any magnetic or electrical polarizability repulsion or attraction would fall at least as 1/r^3, and would be insignificant at the scale of the problem, so only Coulomb attraction should play any role, since it also falls like 1/r^2.
Besides, another issue is that you can never isolate the system completely, even in a perfect vacuum the walls of the container would emit black body radiation, and small momentum transfer from the photons could also affect the time. It should be possible to estimate the temperature the system should be in order to measure this effect.
Because the balls are traveling so slowly and they have so little momentum it's entirely possible that they will not collide at all if a repulsive 1/r^3 term is introduced.
However, I don't believe two diamagnets will repel each other because their magnetism is only induced.
If the room is considered to be "very large" and the room and balls are in thermal equilibrium then the affect from the black body radiation of the room can be shown to be isotropic save for where the room is "shadowed" by the other ball. However, in that case the radiation from the other ball is pushing the two apart. But that effect doesn't quite cancel out the affect of the shadow. This effect is because the view factor of two spheres is less than the view factor of a sphere and a coaxial disc (which the shadow can be modeled as). So that results in a net attractive force. (Assuming the surface is diffuse and grey, which it's not of course...)
However, if the surface of the container and the balls were assumed to be at or very close to 0K then there's no problem. Gravity would again easily dominate.
Build a man a fire, warm him for a day,
Set a man on fire, warm him for the rest of his life.
I don't think I was very clear about the photon thing. Consider as a thought experiment. The walls are so hot that their black body radiation is emitting photons of frequency in deep gamma rays. The photons are very energetic and hit the balls. In this limit, the motion is essentially Brownian and gravity doesn't have much effect.
The other limit is the one you mentioned, absolute zero. But the question is, how close? milikelvin, nanokelvin? In physics, we can only say dimensionless quantities are small, not temperature.
One way to estimate is the following. Calculate the energy of the gravitational interaction of the balls. It is E = Gm1m2/r. Substituting, we get something around 5 x 10 -17 J.
Now, you can estimate the amount of electric charge they should have in order to have the same energy. Suppose both balls have a charge q. The energy is E = k0*q2/r. To get the same energy, you need to have q around 7 * 10-13 C. The elementary charge is around 10-19, so it is a significant amount. In practice, you would need to prepare the balls beforehand to have this charge. If you don't, you can neglect this effect.
Now to the temperature, you can estimate it using Boltzmann constant k. We have E=kT, so the temperature where motion should start to become Brownian would be around T = 106 K, which is very large, so room temperature is perfectly OK!
If you know a bit of particle physics you can guess that gravity is the dominant effect without doing any calculation. The thing is, if you use natural units h/2pi = c = 1, there is only one relevant dimension in physics, which we conventionally use as mass.
The "strength" of a force is given by the coupling constant. The nuclear force has a range of a few femtometers, and the weak interaction even smaller than that, so you can ignore them. The electromagnetic coupling is dimensionless (fine-structure constant), so it has the same strength at all scales (actually, quantum mechanics changes that, but you can ignore in this problem). The gravitational coupling (Newton's constant) has dimensions of mass-2, or equivalently length2.
That is the tricky thing about gravity. People say it is weak in comparison to the electric force, but they neglect a key point. Gravity depends on the dimensions of the problem, for very large lengths it will eventually overtake the electric interaction. Now you use common sense, it took many centuries for humanity to describe electricity, and gravity is known since antiquity. Therefore, it should be natural that at a length of order meters, it should be the most relevant force.
Even if we assume that the ping pong balls wouldn't melt, because of the crazy amount of photons interacting at "deep gamma" emission levels of BB radiation I wouldn't expect Brownian behavior to dominate.
We see Brownian motion in collision between objects where the momentum is sufficiently large compared to the mass. And even at an energy of 500MeV, photons only carry about a vanishingly small amount of momentum compared to a ping pong ball's mass, which is about 10^27 MeV/c^2.
Build a man a fire, warm him for a day,
Set a man on fire, warm him for the rest of his life.
Again, you are using the word "small" in a completely unphysical way. Saying that 1 meter is small is meaningless. What you can say is 1 meter is small compared to the length of a road, which is of order km. Then if you have an object of 1 m driving on a road, you can treat it as a particle.
Likewise you cannot say that momentum is small, what you can say is that it is small compared to the momentum impaired by the gravitational force. And if you calculate the energies at 10^6 K you will see they are the same order of magnitude.
Can someone explain the following to me?
The entropy of a black hole, famously, is proportional to its surface area (or, more specifically, the surface area of its event horizon). To motivate the argument for why this is the case, matter is imagined to "fall" into the black hole, though of course, from the perspective of an external observer, time slows down and length contracts until, as time heads off to infinity, the matter is compressed to within the thickness of a Planck-length, just beyond the horizon. This is the same for any matter that has ever fallen into the black hole, and so the whole black hole is compressed into a thin shell of matter at the event horizon. There's nothing inside. Black holes are essentially two-dimensional surfaces rather than 3D solid objects. Hence the entropy scales as the surface area, rather than the volume as for every other object in the universe.
Here's my problem with the above argument: as more matter falls into the black hole, the mass increases. Therefore the radius of the event horizon increases. Therefore matter that was previously compressed against the event horizon is very definitely now inside that event horizon, even if it hasn't moved. Although matter stops at the event horizon, that doesn't prevent the event horizon expanding to swallow the matter anyway. So isn't the conclusion that a black hole is defined by its event horizon fundamentally flawed?
(Incidentally there is a more sophisticated version of this argument which is based on information theory, and has cute finesses like adding matter in the form of photons with a wavelength equal to the Schwarzschild radius, so you can't specify where they entered the black hole. But it still relies on this idea of the black hole existing as a pile-up at the event horizon, which still looks to me like a flawed conclusion.)
To motivate the argument for why this is the case, matter is imagined to "fall" into the black hole, though of course, from the perspective of an external observer, time slows down and length contracts until, as time heads off to infinity, the matter is compressed to within the thickness of a Planck-length, just beyond the horizon.
Note also that any light emanating from that falling matter will red-shift until it becomes completely invisible (ie. it essentially stops emitting light). Thus, AFAIK, it wouldn't be even theoretically possible to observe the matter falling forever from the outside.
Although matter stops at the event horizon
I'm not sure it's correct to say that matter stops at the event horizon. An external observer will never see the matter crossing the event horizon (both because it "slows down" as it approaches it, and because it red-shifts to invisibility, from the perspective of the external observer), but that doesn't mean it just literally stops.
(Note that I'm a complete layman on this subject, so I may well be talking out of my ass.)
Here's my problem with the above argument: as more matter falls into the black hole, the mass increases. Therefore the radius of the event horizon increases. Therefore matter that was previously compressed against the event horizon is very definitely now inside that event horizon, even if it hasn't moved. Although matter stops at the event horizon, that doesn't prevent the event horizon expanding to swallow the matter anyway. So isn't the conclusion that a black hole is defined by its event horizon fundamentally flawed?
I am struggling to understand your argument, but from my perspective it is just a confusion about which theories are applied.
In classical General Relativity, the black hole is defined only by some finite amount of physical quantities, its mass, angular momentum and charge (if you consider electromagnetism). In order to get the entropy, you need to model quantum mechanics. Look at the Bekenstein-Hawking formula, it involves h_bar, so it is an essentially quantum statement.
Of course, we do not have a complete theory of quantum gravity, so any treatment must be approximate, but Hawking's idea is more or less the following. You can calculate the Casimir energy of a Schwarzschild spacetime just like you do it for flat space, once you introduce some mathematical techniques to cancel the divergences, which are worse in this case. Just as the ordinary Casimir effect involves vacuum fluctuations, the fluctuations causing the Hawking effect cause radiation emission near the horizon. It turns out that if you interpret this energy thermodynamically, you get that the entropy is proportional to the area. It is generally well accepted that the quantization of GR must lead to some sort of "heat bath" (see here).
So, yeah, while I am not sure I get your argument, I don't think you can argue against black hole entropy only invoking classical arguments, since it's a quantum phenomenon.
(Note that I'm a complete layman on this subject, so I may well be talking out of my ass.)
No, what you say is correct. But it doesn't help answer the question.
It might indicate that your question is malformed because of a wrong assumption. You seem to assume that matter that's falling into a black hole slows down and never actually crosses the event horizon (assuming the event horizon doesn't move), and just stays stationary there, forming a plank-thick crust.
I doubt that's true from any perspective, not even from an external one. How can you say that there's a crust of matter surrounding the black hole almost touching its event horizon, if you can't even observe it?
(Even if you could observe such a thing, it makes me think that it would lead to a paradoxical observation of this crust of matter being so dense as to be within its own Schwarzschild radius... which is too complicated for me to even try to guess how it would work.)
Perhaps I have used too much jargon and wasn't understood properly. Let me rephrase my argument in more simple terms.
Let us assume that the Bekenstein-Hawking formula is correct, that is:
S = kA/4l^2, with l = sqrt(G*h_bar/c^3).
Notice that there is the reduced planck constant h_bar in this formula. Now look at the Einstein Field Equations, there is no h_bar there!
Now, you wonder. How is it possible that you derive a formula for the entropy containing a physical constant that does not appear in the equations governing the theory? The answer is simple, you don't do that. To derive the entropy formula, you need quantum mechanics, that is where h_bar appears. Pure, classical GR does not contain quantum mechanics, so whatever theory is being used to derive the Hawking formula, it is not pure, classical GR.
But, wait, let us try something different. Quantum mechanics should reproduce classical mechanics at some limit, so the Hawking formula should apply to classical GR at some limit. This limit is just h_bar = 0. Now, if you substitute that in the formula, you get that the entropy is infinite.
So, as long as you use classical GR, you should always arrive at some problems defining the entropy of a black hole, it does not matter if you do a calculation, talk about a redshift, or compression of matter that falls there. Incidentally, that is not in conflict with the Hawking formula, it also predicts that the definition of entropy should break down at the classical level. It is also not in conflict with standard quantum theory, it was actually the breakdown of statistical mechanics for physical systems that led to the development of quantum mechanics in the first place.
Sorry for interfering, but I have two very interesting questions.
1) Consider you have some object standing still.
If you wanna increase its speed to 1 km/h you need to spend X energy.
From now on, it's moving with speed 1 km/h and you need to spend Y energy to increase its speed to 2 km/h.
From formulae of kinetic energy, we know, that it's mv2/2 which means, that Y > X
Why Y > X?
Why increasing of speed of moving object in same direction require more energy than standing still object?
2) Consider this object hits into wall, and all of its energy going into thermal energy.
Considering same mv2/2 all going into thermal energy, it means that thermal energy after hit of object at speed 2 km/h is 4 times more than thermal energy after speed 1 km/h.
Does it mean, that temprature also 4 times more?
Hello r57shell,
1) The fact that kinetic energy grows quadratically with velocity is a consequence of Newton's laws. Sure, you can look at the definition of work in terms of force and distance and derive kinetic energy from there, but that would simply shift the question to why work should be defined like that. It is possible to prove that, if you assume additivity of velocities (that velocities are added when you boost from one frame to another), the only reasonable definition for energy is that it is proportional to velocity squared. But again, it is a consequence of either Galilean invariance or Newton's laws, it is not true in relativity, because additivity of velocities does not hold in that theory.
2) Under some assumptions, yes, the amount of heat is four times greater, and if the temperature of the material grows linearly with the amount of heat, the temperature will rise four times more. What you are describing is something similar to Joule's experiment, which determined that it is possible to convert mechanical work into heat.
Energy is to a physicist like money is to an economist. Energy cannot be observed directly, just as money by itself has no value if no merchant is willing to accept it. What you can observe are conversions between one form of energy to another (just like sales in a market), and physical theories describe quantitatively how this conversion takes place.
I have had some similar questions about kinetic energy before. I think at high school I asked my teacher this question (or something similar):
Consider a mass of 2, such that kinetic energy is simply velocity squared. Accelerating a mass from 0 to 1 takes 1 unit of energy, and accelerating from 1 to 2 takes 3 units of energy. But an acceleration from 1 to 2 is just an acceleration from 0 to 1 in a different reference frame. When you change reference frame, where does the extra kinetic energy come from?
Years later, I think I have an answer (but I'd be happy to hear other's thoughts). It's based on two principles:
1. momentum is conserved
2. the exact distribution of kinetic energy depends on the reference frame, but the amount of work done in an interaction is frame-independent.
Conservation of momentum is important because you can't accelerate an object in a vacuum; to accelerate an object, the momentum has to come from somewhere, and that will have an effect on the energy requirements of the whole interaction.
So, to reframe the original question, we have to add in the extra parts of the system. To keep it simple, we'll have two equal masses that are accelerating away from each other. Then in the two reference frames, we have:
* mass A accelerates from 0 to -1, mass B accelerates from 0 to 1
* mass A decelerates from 1 to 0, mass B accelerates from 1 to 2
In the first reference frame, we start with 0 kinetic energy, and end with 1+1=2 kinetic energy.
In the second frame, we start with 1+1=2 kinetic energy, and end with 0+4=4 kinetic energy.
The total kinetic energy is different, but the change in kinetic energy (aka the work done) is 2 units of energy, independent of reference frame.
The original question undercounted the energy requirements in the first frame, and overcounted the energy requirements in the second frame, because it ignored the extra energy needed for (or provided by) mass A.
So: changing reference frame doesn't magically make energy appear out of nowhere. It might change the distribution of kinetic energy, it might change the total amount of energy, but it does not change the energy required to accelerate an object by a given amount.
The two famous postulates of special relativity are that the laws of physics are the same for all inertial frames of reference, and the speed of light in vacuum is the same for all inertial frames of reference.
Is the same true for non-inertial frames of reference.
Yes. In particular, General Relatively achieves this as a consequence of using a symmetric tensor to describe the gravitational potential. (As this construction allows for general covariance.)
If you're curious to the math behind General Relativity I recommend Leonard Susskind's lectures:
https://www.youtube.com/watch?v=JRZgW1YjCKk&list=PLXLSbKIMm0kh6XsMSCEMnM02kEoW_8x-f
Build a man a fire, warm him for a day,
Set a man on fire, warm him for the rest of his life.
The two famous postulates of special relativity are that the laws of physics are the same for all inertial frames of reference, and the speed of light in vacuum is the same for all inertial frames of reference.
Do we know that a frame of reference is inertial by observing that there's not a single body moving at speeds higher than c? For instance, the Earth is not an inertial system by the simple observation that, relative to it, the stars (except the Sun) travel at a much higher speed.
In essence, a frame of reference is inertial when an accelerometer in that frame of reference would show an acceleration of zero (in all possible directions, including rotational acceleration).
It is a bit trickier than that. In Newtonian mechanics and special relativity, the definition of an inertial frame of reference is, strictly speaking, circular: the laws of physics are defined in terms of inertial frames, and the inertial frames are defined in terms of the laws of physics. Saying things about accelerometers and rotation just displace the problem, but it is still there.
In general relativity, things are slightly better: it states that the laws of physics are the same for all reference frames, and defines a local inertial frame to be a freely falling reference frame.
The two famous postulates of special relativity are that the laws of physics are the same for all inertial frames of reference, and the speed of light in vacuum is the same for all inertial frames of reference.
Is the same true for non-inertial frames of reference.
No, in Rindler coordinates (which represent a uniformly accelerated frame in SR), for example, you can calculate that the speed of light varies according to its distance from you, so it is neither constant or equal to c for this observer.
Amaraticando wrote:
Do we know that a frame of reference is inertial by observing that there's not a single body moving at speeds higher than c?
No, a counterexample is the universe expanding with a constant Hubble parameter (steady state). In this setting, the frame is not inertial, and every motion occurs with velocities below c, since velocities higher than that are not in the observable universe. This is a specific feature of this model, though.
The thing about reference frames is, you have to postulate whether a given frame is inertial or not. Once you do that, you know that all other frames moving with constant velocity relative to it will be inertial, too. Then you can calculate the physics, do an experiment and see if everything works. If it does, your frame is inertial, if it is not, either your theory is wrong, or you picked the wrong frame to be inertial. In any case, it is experiment that determines the truth, like anything in physics.
marzojr wrote:
In general relativity, things are slightly better: it states that the laws of physics are the same for all reference frames, and defines a local inertial frame to be a freely falling reference frame.
I don't agree with this (popular) viewpoint. Every theory can be cast in a way that it's true on an arbitrary frame. For example, pick Newtonian mechanics in an inertial frame and evaluate the path an object will take. Now, go to any other frame and change the theory so that you have forces that make the object follow the same path and, boom, you have exactly the same results without making reference to inertial frames.
What GR says is simply that the equations should be "pretty" (i.e. diffeomorphically invariant), if you drop that requirement you can derive any theory that you want. I think GR is better defined by the fact that, unlike other physical theories, the geometry where interactions take place is also dynamical (space-time can store and release energy, etc.).
It's because the kinetic energy is the integral of force over distance, rather than over time as you seem to be intuitively expecting.
Yes, this makes new flavor of understanding what exactly is conserved in "law of Conservation of Energy"
It's kinda weird that this energy doesn't even depending on time.
No matter what you do, no matter how fast you do, work is same.
Just because of this defenition.
From other hand, what is conserved is exactly THIS value, not other value.
p4wn3r wrote:
1) The fact that kinetic energy grows quadratically with velocity is a consequence of Newton's laws. Sure, you can look at the definition of work in terms of force and distance and derive kinetic energy from there, but that would simply shift the question to why work should be defined like that.
Yes, exactly, shift is to "why work should be defined like that?"
My answer atm is: because what is conserved is exactly this thing.
p4wn3r wrote:
But again, it is a consequence of either Galilean invariance or Newton's laws, it is not true in relativity, because additivity of velocities does not hold in that theory.
I guess Newton's laws is enough. Galilean invariance is about conservations of laws itself. Using galilean transformation, kinetic energy changes. I'm unsure about total energy though.
Anyway, my question was intended to crash misconceptions about what is energy, and energy conservations.
I guess, I succeeded. But I can't grasp "valid" conception yet.
I don't agree with this (popular) viewpoint. Every theory can be cast in a way that it's true on an arbitrary frame. For example, pick Newtonian mechanics in an inertial frame and evaluate the path an object will take. Now, go to any other frame and change the theory so that you have forces that make the object follow the same path and, boom, you have exactly the same results without making reference to inertial frames.
What GR says is simply that the equations should be "pretty" (i.e. diffeomorphically invariant), if you drop that requirement you can derive any theory that you want.
That is exactly backwards: any theory can be written in diffeomorphically invariant form: there is even a formulation of Newtonian mechanics (plus Newtonian gravity) written in such a form, and this one is valid (i.e., had the same form) on all reference frames. As a bonus, gravity is even related to the curvature of space-time in this theory as well (FYI, this version of Newtonian gravity is a lot more complex than I am letting on).
So this is not a requirement of GR in the sense that it does not constrain the theory in any meaningful way.
p4wn3r wrote:
I think GR is better defined by the fact that, unlike other physical theories, the geometry where interactions take place is also dynamical (space-time can store and release energy, etc.).
This is a much better point: but it is also false (as mentioned above).
