Tag Info

New answers tagged

3

The mass of each of the three mass types of neutrinos is unknown (although a lower limit has been placed, as Anna V's answer indicates), chiefly because they exist in three weak classifications as well as the three mass classifications, and the combinations can not be simultaneously isolated and identified with certainty. The combined mass of all three mass ...


4

The particle data group has a thorough article on neutrino oscillation experiments. Oscillations are the way to measure mass differences between neutrinos, and in table 13.7 experimental values are given for delta(m^2) of the three neutrino species. These put a lower limit to the mass values. From the definition it is obvious that the limits are of the ...


0

Nobdody knows. There are many experiments that try to find this out right now. For example the Karlsruhe Tritium Neutrino (KATRIN) experiment. We only know that they must be very light $\approx$ eV.


1

I can see that this question has been downvoted but I think it still deserves a proper answer. First, it is the Chandrasekhar (one word) limit, named after the Indian-American astrophysicist Subrahmanyan Chandrasekhar. Second, the Chandrasekhar limit does not mean that an object cannot be more massive than 1.4 times the mass of the Sun. There are plenty of ...


3

A tensile design for a space elevator cable requires there to be a counterweight some distance beyond the point of geostationary orbit. The counterweight would be hurled with centrifugal force and would keep the entire cable taut. The cable must not only support a payload, but also must be a tether for the counterweight, and provide for centripetal force ...


1

If you work through the numbers, you will find that all of the air on earth has a mass that is less than 1 millionth the mass of the earth. You can't get more than a tiny fraction of that air close to your falling object, so the effects of wind and other disturbances would FAR outweigh any effects due to gravity, because G is sooooooo small.


1

If you consider that gravity is weak compared to the electromagnetic force because $G \approx 6.67 \times 10^{-11} Nm^2 kg^{-2} $ and $k_e \approx 8,987 \times10^9 N m^2 C^{-2} $ it would require very small distances in order for the gravitational force to be effective, but at this distances the electromagnetic force would be several times higher, ...


1

The object of making the silicon sphere was not to define the kilogram by the mass of sphere, but redefine Avogadro's constant in terms of the number of silicon atoms. Then use the fixed Avogadro's constant to fix the kilogram. Avogadro's constant was defined as the number of atoms in 12 grams of carbon atoms. You can read up on the Avogadro project here ...


1

Con: There are 3 naturally occurring isotopes of silicon, so isotopic purification would be required after elemental purification. I'm unsure about how hard it is to maintain pure silicon. Isotopically, beryllium would be better (100% $^9$Be, naturally), but I don't know about its reactivity, either. Carbon has two naturally occurring isotopes, but we know ...


0

The analogy between how a massive object bends a rubber sheet and how sources of mass-energy-momentum bend spacetime is a poor one. It's essentially comparing the mechanical deformation of a two dimensional sheet in space to the curvature of four dimensional (3+1) spacetime, which have a mathematical similarity, but not a physical one. In Newton's theory of ...


0

If you take away mass in Newtonian mechanics, then any force leads to an infinite response. The spring force then promises an infinite negative-feedback response to any deviation from equilibrium, and depending on how you take the limits involved, you either get infinite sinusoidal motion with a period of 0, or perfect rigidity with no motion and a period of ...


1

The general theory of relativity predicts that kinetic energy will contribute to gravitational mass. Here is a paper that explores the gravitational effect of kinetically energetic particles within a system: http://arxiv.org/PS_cache/gr-qc/pdf/9909/9909014v1.pdf. Here is an interesting article by Frank Helle on the production of gravity by relativistic ...


2

there are many experiments that put different contraints on the neutrino masses. Here is a good collection from the particle data group. To summarize: There are lots of experiments that put upper bounds on the neutron mass. The PDG groups estimate is that $\nu_e < 2eV$, $\nu_{\mu} < 0.19eV$, $\nu_{\tau} < 18MeV$. All with a confidence level of ...


0

A Non-Relativistic Interpretation of Relativistic Results Million years ago, when self-studying special relativity, an exercise was created in order to understand non-relativistically the relativistic result that a force could not accelerate a particle to speed greater than $\:c\:$. The exercise, the Figure and the solution are already in LaTeX and are ...


0

Particles that have no mass, as photons, have momentum based on their energy, more especificaly on the relation $$p=\frac{E}{c}$$ Where $p$ is the momentum, $E$ is the energy and $c$ is the speed of light. You can think of it intuitively by noticing Einstein's famous equation, $E=mc^2$, and substituting it on the classical momentum formula $p=mv$, using ...


0

By inertia I assume you mean momentum. The momentum is related to the energy of the object by: $$ E^2 = p^2c^2 + m^2c^4 $$ and to the velocity by: $$ p = \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}} $$ The momentum does indeed tend to infinity as $v \rightarrow c$, but note that it will never reach an infinite value because no massive object can travel at the ...


1

The equation $E = m^2c^2 + p^2 c^2$ is restricted to Special Relativity. However, in classical physics we have $$ \vec{F} = m \vec{a}, $$ and $$\vec{F} = m \vec{g},$$ whence $$ m \vec{a} = m \vec{g}. $$ This can be written as $$ m \big( \vec{a} - \vec{g}\big) = \vec{0}. $$ From a mathematical point of view we have $$ \big( m = 0 \big) \vee \big( \vec{a} - ...


0

For a friction less drop we simply use Newtons law: $$\frac{dv}{dt} = g$$ and solving for $v$ gives $v(t) = gt$. Introducing the friction component into the differential equation, I make the approximation that the drag is laminar: $$ \frac{dv}{dt} = g - \frac{\gamma}{m} v$$ The solution of this is: $v(t) = \frac{mg}{\gamma} - \frac{mg}{\gamma} ...


1

Lets just say were solving for any specific speed. Not the terminal velocity. I'm still not sure what to do. It might help to start with the speed an object attains after falling a short distance, ignoring air resistance and assuming the effect of gravity is constant. If we fall from Mount Everest (about 8000 meters) to sea level, we can treat gravity ...


1

Okay, so here's the basic physics: it comes down to differential equations. Big objects in air tend to see $v^2$ drag. Plugging this into $F = m a$, this means that something which is falling straight downward with speed $v$ will obey the nonlinear differential equation: $$\frac {dv}{dt} = \frac 1\lambda v^2 - g$$for some length $\lambda$. The terminal ...


1

Mass is a Lorentz invariant quantity! The relativistic mass is not the real mass, it is is just called relativistic "mass" for obvious reasons. This term is abandoned by most textbooks, as it often causes this confusion.


-2

It's quite wrong to assert that relativistic mass is obsolete. Gary Oas of Standford university did a study of a sample of the SR and GR textbooks. Of the ones published between 1995 and 2005 there were 60% of them which use relativistic mass. In this video, Alan H. Guth of MIT describes why the mass of a photon is not zero. ...


0

When you breath in helium, you gain mass, volume, and buoyancy, but lose density and weight.


1

The mechanism for "giving mass" to elementary bosons and fermions is different. With bosons, it is related to the gauge symmetry ($SU(3)_c \times SU(2)_L \times U(1)_Y$) which is partially broken (and become $SU(3)_c \times U(1)_{em})$. The unbroken part imposes its associated bosons (gluons and photon) to be massless to respect this symmetry. With ...


-1

I'm wondering if there is any explanation for why bosons(specifically gauge bosons) can be massless (photon and gluon) but we don't see any fundamental massless fermions. It's because a fermion is a "body", and because "the mass of a body is a measure of its energy content". See Einstein's E=mc² paper. He talks about a body and an electron here. "The ...


2

So, unfortunately, if a box contains a certain amount of kinetic/potential energy stored within it, that energy appears as a mass to the outside world. This is actually the basis for using the so-called "mass defect" (change in mass) in radioactive reactions for detecting the energy released: One over-large Uranium nucleus becomes some smaller parts with ...


0

As in Ben Crowell's Answer, the concept of "Relativistic Mass" is not wrong, but it is awkward. There are several things a loose usage of the word "mass" could imply, all different and thus it becomes a strong convention to talk about the meaning of the word "mass" that is Lorentz invariant - namely the rest mass, which is the square Minkowski "norm" of the ...


3

Suppose your object is a sphere with a radius $r$ and mass $m$. The aerodynamic drag on a sphere is given by: $$ F_{drag} = \tfrac{1}{2}C_d \rho \,\pi r^2 \,v^2 \tag{1} $$ where $\rho$ is the density of the air and $C_d$ is the drag coefficient. The drag coefficient varies with speed (the NASA article I linked shows how $C_d$ changes with speed) but over a ...


-1

Look at the paragraph "gravity and the photon" in this link: In the relativistic framework, i.e. large velocities, any energy is also a relativistic mass: For the photon this means the following equation: m is the relativistic mass of a photon with energy h*nu. Gravity attracts relativistic mass, and the photon has one. Read the link further to ...


0

Imagine 3 objects. One is a flat piece of paper. The second is an identical piece of paper rolled into a ball. The third object is the exact same shape and size as the rolled up piece of paper but it's made of iron. If you drop the flat paper and the rolled up paper at the same time, the rolled up paper hits first because it has less air resistance due to ...


1

You misunderstand. All objects have some escape velocity, which is the velocity needed for anything (photons or matter) to escape from that object's gravitational field. And that's not the velocity it needs to maintain under some sort of constant thrust, but the initial velocity it needs to, shall we say, coast away from the object. For a black hole that ...


1

So what I understand from reading the other answers is this: Here on earth: 1 Kilogram of lettuce: Mass-> 1 Kg, Weight-> 1 Kgf In another hypotetical planet where gravity is half of earth: 1 Kilogram of lettuce: Mass-> 1 Kg, Weight-> 0.5 Kgf Since there is no practical easy way to measure mass, in everyday life we use the kilogram as a unit of weight ...


0

The object's center of mass will fall straight downward, accelerated at g. All objects accelerate at the same rate in a vacuum, regardless of their weight, so neither end will accelerate faster than the other. That said, with air resistance, both ends of the object will feel the same drag force, so the end of the object with less mass will feel a greater ...


2

Gravity, like all cause-effect relationships, propagates at a maximum speed of $c$; indeed from the Einstein field equations small amplitude (linear limit) gravitational waves travel at precisely $c$. A good idea for what is going on comes from an approximation of General Relativity called Gravitoelectromagnetism. This makes an approximate analogy between ...


1

Is it possible to decrease the mass of the object? Perhaps surprisingly the answer is yes. All you need to do is it drop it. Then some of the object's mass-energy, which we call potential energy, is converted into kinetic energy, which ends up getting dissipated. You're then left with a mass deficit. The mass of the object is reduced. It is known ...


1

We know how real numbers (in the mathematical sense) behave. There is no a priori reason however to assume that masses of objects behave as real numbers. The proposed experiment can be seen as a check to see if masses actually can be modeled by (a subset of the) real numbers.


2

What is mass of elementary particles? It is the "length" of the four vector (p_x,p_y,-_z,E), for complex systems it is called rest mass. As the length of three dimensional vectors is not additive ( think adding two opposite momenta), rest masses are not additive, vector algebra has to be used. The invariant mass of your two gammas must be larger than the ...


3

"Rest mass" is probably a bit more abstract in modern science than what you seem to be thinking. It is simply this: if you can put yourself into an inertial frame such that a body is at rest relative to you, then that body's rest mass is defined as that body's total energy measured in this particular inertial frame - multiplied by $c^2$, if you want to ...


1

Where does mass come from in pair production? From the energy-momentum of the gamma photons. Think of photon momentum as resistance to change-in-motion for a wave propagating linearly at c. You've maybe heard of electron spin and the Einstein-de Haas effect which "demonstrates that spin angular momentum is indeed of the same nature as the angular ...


-2

This question goes back to the question that Feynman asked his father about why a ball in a wagon rolled to the back, when he pulled on the wagon. His father replied, "That, nobody knows." In other words, you're asking what relativistic momentum is. We don't even know what non-relativistic momentum is. Sure, we know the effects, we measure them, I've ...


2

Your equations (1) and (2) are dimensionally equivalent. Take your equation (1): $$ F = BI\ell $$ The dimensions $I$ are $CT^{-1}$, so the right hand side are $[B]\,CT^{-1}L$ - I've left the dimensions of $B$ unspecified for now and just written them as $[B]$. Now take your equation (2): $$ F = BQv $$ which has the dimensions $[B]\,CLT^{-1}$. If we set ...


2

Since you have not specified the "real world" size of the ship, let's take a 74-gun ship of the line https://en.wikipedia.org/wiki/Seventy-four_(ship) as the desired type, firing a 36-pound cannon. The bore on such a cannon https://en.wikipedia.org/wiki/36-pounder_long_gun was about 175 mm, with a shot weight of about 39 lb. Shrinking this cannon to a bore ...


1

Short: No Because: Assume material SG ~= 7 Assume ~= non-sonic shot. Say 500 m/S based on this superb reference R = 0.05mm = 5E-5m E= 1/2 mV^2 m kg = 4/3.Pi.R^3 x sg x 1000 kg/m^3 = 4/3x 3.14 x (5E-5)^3 x 1000 x 7 = 3.E-9 kg At say V = 500 m/s E = 0.5 x 3.5E-9 x 500^2 ~~= 500 micro Joule. (438 uJ calc) A VERY solid teaspoon weighs 50 gram. ...


1

I think the content of his statement is that "the assignment of mass to an object as defined using the conservation of momentum experiment is a transitive property". Not all relations are transitive properties (for instance, "Wolves eat Deer, Deer eat grass, but wolves don't eat grass"). For instance, if we had defined "the mass of an object" as a weight ...


1

Who can with certainty fathom the mind of a Feynman? :-) However, it sounds like he is saying that there is no reason to decide with certainty that Gold, Copper and Aluminum do not have properties which affect how they behave when accelerated by a force. An example that has some correspondence is carrying out the same experiment in air. Mount 1 kg spheres ...


2

As wikipedia says about the second law, As Newton's second law is only valid for constant-mass systems, mass can be taken outside the differentiation operator by the constant factor rule in differentiation This is expanded in the article about variable mass systems. You derived the formula $$ F = m\dot v + \dot m v. $$ where $\dot v \equiv ...


0

So, $F = \frac{dp}{dt} = \frac{d}{dt}(mv)=m\frac{dv}{dt}+v\frac{dm}{dt} = m\frac{dv}{dt} =\frac{dp}{dv} \times \frac{dv}{dt} = \frac{dp}{dt} = F $ nope, no problem. Note that $\frac{dm}{dt}$ is assumed zero because the situation described by the equation requires that the mass is constant, and hence time invariant. This is of course inapplicable if somehow ...


0

Even without gravity, momentum conservation will still hold. If you elastically scatter an unknown mass $m$ with an initial (known) velocity $v$ against a known mass at rest, for instance we can take the SI standard of $1$ Kg, then from the resulting measured velocities you should be able to find $m$. Thus Yes, there will still be a mass.



Top 50 recent answers are included