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14

The equation you are mentioning is the gravitation force derived by Newton. This force doesn't apply to particles such as photons for two reasons: Photons are too small, and you can't use Newtonian physics to describe their properties. Photons travel too fast (their velocity is the speed of light) and at such a velocity Newtonian mechanics cannot be ...


8

$-i ħ \nabla$ is the momentum operator. You have to apply it to a wave function to get the actual momentum. Consider the plane wave solution to the Schrödinger equation: $\Psi = e^{i \mathbf{k} \cdot \mathbf{r} - \omega t}$. Applying the momentum operator gives $-i ħ \mathbf{k} \Psi$. You can see the eigenvalue has units of momentum. (If you can't see ...


6

That's because the relation $p=\gamma mv$ doesn't hold universally. As you just showed yourself, using this relation for a photon would lead to a contradiction because the energy of a photon isn't zero. A heuristic way of seeing why this relation won't hold for a photon is by recognizing that $$p=\gamma mv =m\frac{d x}{d\tau}$$ but a photon doesn't ...


5

Buckminsterfullerene (C-60) is about as close as you could get to spherical with 60 atoms. Earth's radius is 6,371,000 meters and Buckminsterfullerene's radius is 3.5 nanomoeters (nm). A carbon atom's radius is about 0.07nm (which I would approximate as the degree of deviation from roundness of Buckminsterfullerene). Therefore, Buckminsterfullerene ...


5

An obvious difference between the two ways of thinking about it you mention is that in the case of the Higgs mechanism, there is an observable particle excitation of the field associated with it, which was found recently. Furthermore it should be noted that the Higgs mechanism only concerns the mass generation of some elementary particles. The mass of ...


5

Your logic is ultimately wrong because that equation doesn't reveal the true nature of gravity. According to general relativity, objects themselves bend space-time. Imagine space like a rubber sheet. If you stretch it and place a mass in the middle and roll a ping-pong ball past the mass, it will curve towards the mass. Similarly when space-time is being ...


5

You can understand the expression by attempting the limit for $v\rightarrow c$ and $m\rightarrow0$. Notice that $\gamma\rightarrow\infty$ when $v\rightarrow c$. Therefore $m v \gamma$ is an undetermination of the form $0\cdot\infty$. From your expressions, you cannot say that $E=0$. The limit does not exist, and this implies that this expression is not valid ...


4

There is no way to be 100% sure, but we can put upper limits on the mass. Massless particles don't have a rest frame, so it doesn't make sense to talk about time dilation in the photon's frame. A massive photon would have a rest frame, so you could eventually catch up to it and move alongside it. List of experimental limits on photon mass more ...


3

Ah, all that talk about curved space-time. Well, there is a simpler argument. The fundamental axiom of general theory of relativity, "principle of equivalence", says: The effect of a homogeneous gravitational field is equivalent to that of a reference frame in uniform acceleration in the direction opposite to that of the gravitational field. All that ...


3

Your understanding is spot on, as is PhotonicBoom's Answer. Something that might give you a bit more insight along the lines that you are thinking is if I answer your question backwards: the property we call "mass" (or "rest mass") is acquired by a particle with a rest mass of nought when that particle is confined in some way. If you look at my thought ...


3

Model the child as a rod of mass $m$ and length $l$ standing on the ground vertically, with center of mass at height $l/2$, with feet glued to the ground but the rest of the body able to rotate. When upright, the potential energy is $mgl/2$. When lying on the ground, the potential is 0. So when falling, the child hits the ground with energy $mgl/2$. ...


3

by putting $p=\gamma mv$ and then get a value for $m$ (which will be 0 for a photon) and therefore rendering the equation to $E=0$ First, let's write this out in full (in 1D) $$p = \frac{m v}{\sqrt{1 - \frac{v^2}{c^2}}} $$ Then, solve for $m$ $$m = p\frac{\sqrt{1 - \frac{v^2}{c^2}}}{v}$$ Now, holding $p$ constant, see that the limit of $m$ as $v ...


2

I) On one hand $-i\hbar{\bf\nabla}$ is the Schrödinger (position) representation of the the canonical/conjugate momentum operator $\hat{\bf p}$ in order to satisfy the CCR $$\tag{1} [x^i, p_j]~=~\hbar {\bf 1}~\delta^i_j. $$ II) On the other hand, $m\hat{\bf v}$ is the kinetic/mechanical momentum operator. (Let us for simplicity imagine a non-relativistic ...


2

It appears that the statement is half-right.... As answered by @DumpsterDoofus, the factor for a scaled object falling a scaled height, is the fourth power. However, if we scale an object that is traveling at a velocity of $x \frac{\text{body-lengths}}{\text{second}}$, then a fifth power for the kinetic energy is correct. The size and thus mass in the ...


2

I think you are missing that the source is actually referring to he observable universe explicitly - just somewhat indirect: It is not obvious as it is talking about "visible/observable universe", but about "total mass of the visible matter" and "mass density of visible matter". As far as I can see, any references to mass etc in the universe are covered by ...


2

Any body as you say with rest mass cannot fully reach the speed of light, as you would need to supply an infinite amount of energy to accelerate it to that exact speed. We do know thought that all massless particles do travel at the speed of light. Are they pure energy? They are, but then, everything is as we know from Einstein's relation $E = mc^2$. I ...


2

In this context, the right-handed neutrino is a singlet under the Standard Model gauge groups. Only the right-handed neutrino is allowed a Majorana mass. The left-handed term is not gauge invariant. If the SM and right-handed neutrino fields were embedded into a Grand Unified gauge group, the right-hand term would break that symmetry. It is expected ...


2

Newton's formula is an approximation of how "gravity really works". We actually still don't know how gravity really works, but we have vastly refined our understanding of it thanks to Einstein's general theory of relativity. Gravity is simply a measure of curvature of a 4-dimensional manifold we earthlings call space-time. Local concentrations of mass or ...


2

In Newtons theory of gravity photons are not affected by gravity (created by masses). So your conclusion is correct. But in General Relativity the curves of free objects like test particles or photons (geodesics) are determined by the space-time geometry. The geometry is described by the metric which is given by the energy and mass distribution of the ...


2

Critical mass is actually more about 'the right number of nuclei in a specified space'. As we are talking about solid matter this equivalently translates to a given number of atoms (or molecules depending on your matter). And this furthermore translates to our everyday mass. But it's 'not about mass', it is just a practical way to specify the quantity. So ...


2

The energy being used is lost internally, in your arm. Your arm is not a solid, it has joints around which it is free to move. The way to hold them in place is to have your muscles act against gravity. Now muscles are not "solids" either: they are made with filaments which can slide relative to one another, these filaments are connected by molecules ...


1

The bomb doesn't "care" what its mass might look like to an observer in another frame. If you calculate critical mass you don't worry how big it might seem to observers located in billions of other possible frames of reference. Local frame of reference is the only one valid for making calculations concerning the occurence of local phenomena.


1

I'll provide an answer in non-relativistic quantum mechanics. The short answer is that momentum and mass commute, so a particle can have a well-defined momentum and mass simultaneously. But really, mass isn't considered an operator in quantum mechanics; it's a parameter, a number. So for some system, it is presumed that the mass is known always. There's no ...


1

http://www.quantumdiaries.org/2014/04/04/moriond-2014-new-results-new-explorations-but-no-new-physics/ Top quark mass has been revised upward.


1

you are correct in your thinking, because you are using newton's laws. However, newton's laws were not completely accurate. For circumstances like ones with photons and extreme gravity like black holes, you must use Einstein's general theory of relativity. These laws say that all particles follow the shortest path along spacetime, including photons. But ...


1

Unfortunately I can't add comments yet. How much knowledge theoretical physics do you know, before I make too many assumptions (or lack thereof!)? I don't think that this question can have an acceptable answer. This will depend greatly on what model for the Dark Energy (DE) you are using. There are dozens, and new ones continuily being created, and we do ...


1

The harsh answer is, "by solving Einstein's field equations". That is an extremely more difficult problem than using Newton's law -- the equations take up at least a page when you write them out. However for the case of light near a black hole, you can treat the black hole as unaffected by the light, and instead solve the null geodesic equation. This is ...


1

In special relativity the energy is related to mass and momentum by $E^2 = (pc)^2 + (mc^2)^2$, where $p$ is the momentum. $m$ here is the rest mass of the particle, so for the photons case there is only energy from the momentum. The $E = mc^2$ you are likely familiar with ignores the momentum term, and hence only involves the rest mass. Photons are the ...


1

The easiest way to determine the uniformity of your slabs of wood is to take an x-ray. Seriously. This image is of an x-ray of a piece of wood where the dark regions are cavities created by termites eating away at it (the termites are the white things inside the cavities). If you don't want to take an x-ray (or can't afford one), then just assume that ...


1

What you're asking is essentially whether anything can rotate faster than the speed of light. Just like how it would take infinite energy to accelerate an object to the speed of light in a straight line, it would also take an infinite amount of energy to rotationally accelerate an object to the speed of light. In any practical sense, this tower would be ...



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