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There is no controversy or ambiguity. It is possible to define mass in two different ways, but: (1) the choice of definition doesn't change anything about predictions of the results of experiment, and (2) the definition has been standardized for about 50 years. All relativists today use invariant mass. If you encounter a treatment of relativity that ...


9

It is possible for particles to get masses at loop level while they are absent at tree level as long as there is no (non-anomalous) symmetry that forbids it. However, in most models if there is a particle doesn't have a bare mass then its due to a symmetry which then protects it from getting masses at loop level. This is often a subtle topic due to chiral ...


7

There's no controversy about whether mass increases or not, there's controversy about what you call mass. One possible definition is that you consider some object's rest frame, and call the $\tfrac{F}{a}$ you measure there (for small accelerations) the mass. This notion of mass can't change with speed because, by definition, it's always measured in a frame ...


6

To properly understand what is going on you need to understand general relativity. Massless particles, like photons, travel on null geodesics and mass bends spacetime so the null geodesics are not straight lines. The problem is that neither you nor your teacher understand general relativity so this isn't a very convincing argument. But here is an argument to ...


5

Yes, massive particles such as W-bosons, Z-bosons, quarks, and leptons couple to the Higgs field via the cubic (Yukawa) interaction, so they may also exchange the virtual Higgs. Yes, because the virtual particle is massive, one gets the Yukawa potential that includes the exponential dumping with distance. This "Higgs force" is much less fundamental and ...


5

In addition to JeffDror answer : In the case of scalar fields, this is definitely the case. In order to get a massless field, you need to fine-tune one parameter of the Lagrangian. In the case of the a $\varphi^4$ theory defined by the parameters ($m_\Lambda$, $g_\Lambda$, $\Lambda$), i.e. the bare mass, interaction and the UV cut-off, one need to fine-tune ...


3

Some people say that 'mass increases with speed'. Some people say that the mass of an object is independent of its speed. I understand how some things in physics are a matter of ... definitions. But, I can't get my head around how both be 'true' is any sense of the word. Either mass increases or it doesn't, right? Can't we just measure it... ...


3

I can think of two simple ways. Lever The first would be to eschew the scale altogether and build a rudimentary scale with a stick and a pivot, and balance out your nephew's weight with a bunch of water in some buckets, or 2L bottles. Knowing the density of water, you could equate the weight of your nephew to a volume of water. If you are interested in ...


3

The mass of a small fluctuation is usually defined as $$ \pm m^2= \frac{d^2V}{d\phi^2}\biggr|_\text{VEV}$$ The sign depends on your conventions. This makes sense in analogy with the canonical free field potential $$V_\text{free}=\pm \frac{1}{2}m^2\phi^2$$ for which the above formula is clearly right. More generally, we can expect any (reasonably smooth, ...


2

Briefly, the formula $E=mc^2$ applies only particles at rest in an inertial frame of reference. Since there is no rest frame for a photon, no inertial reference frame in which a photon is at rest, one cannot apply the formula $E=mc^2$ to a photon. In more detail, the four-momentum of a particle has components $(\frac{E}{c}, \vec p)$ The four-momentum for ...


2

As has been shown in this paper, pointed out by Matthew in the comments, the expression found is indeed correct and can be understood from a holographic point of view. I now reproduce the argument from relevant section (number 5) of the paper: It seems unusual from the gravitational point of view to have a mass for a solution that is a natural vacuum, ...


2

The question is what do we need the matter content of the universe for. As I understand it, in the usual case we want to find the conserved quantity associated with a certain conserved current gained by the projection of the energy-momentum tensor into a Killing vector, as for example in the paper by Abott and Deser. The requirement of asymptotical ...


2

A negative binding energy would make the vaccum unstable. For example suppose a virtual electron and positron pop out of the vacuum. This costs energy to create the particles, but if their binding energy could be greater in magnitude than their rest masses then they could bind to form an energy state lower than the vacuum from which they were created. The ...


1

This is exactly the source of energy exploited by the fission of high-mass nuclei (i.e. nuclear power). You start with a heavy nucleus (say $^{235}\mathrm{U}$) and add a neutron. What you get out is two lighter nuclei (often, but not always, krypton-92 and barrium-141) and several neutrons: $$ ^{235}\mathrm{U} + n \to ^{92}\mathrm{Kr} ^{141}\mathrm{Ba} + ...


1

This is admittedly a late answer. Hopefully it will clear up some of the confusion. If you ignore aerodynamic drag, ignore that the coefficient of rolling friction varies with load, and ignore a number of other factors such as friction between the axle and the bearings that support it, then yes, stopping time / stopping distance is independent of vehicle ...


1

There is geometrical significance. You are sooo close. You are in Euclidean space, but you should be in hyperbolic space. As @fqq points out, you have stumbled upon rapidity, a parameter in hyperbolic geometry that is the analog of angle in Euclidean geometry. In Euclidean geometry an angle (in radians) is a parameter that measures the Euclidean length ...


1

A partial answer: "I do not see a coupling between the MSSM fields and the messenger fields" There are gauge interactions between observable and messenger fields. Let us discuss with fig $(1)$ page $11$ of this reference. You have not direct interactions between observable sector gauginos ($\lambda$) and s-fermions ($\tilde f$), and the superfield ...


1

when low mass object hits high mass object it is reflected gaining opposite velocity almost the same as initial velocity. If I jump onto the wall why my body is not reflected? I know that collision is not fully elastic but it should be at least similar. Human body is not elastic: it cannot be deformed/ compressed in any way and then return to ...


1

The rocket is in free fall along with the book. The nearest gravitating bodies are very far away, so whatever meager acceleration they cause will be almost exactly the same on the rocket and on the book. Suppose you instead went on a very close flyby of a neutron star. Now the book will fall rapidly, and away from the rocket's center of mass. The only way ...


1

The correct general equation is $E^2=m^2c^4+p^2c^2$, since $m=0$, $E=pc$. A photon's momentum is $p=\hbar k$, where k is the wavenumber ($k=\frac{2\pi}{\lambda}$). This would be consistent with $E=h\nu$ where $\nu$ is the frequency.


1

What you read is correct. I am not sure if those were the exact words of your teacher but according to the general theory of relativity, sun doesn't "attract" the photon (or any other body for that matter). In fact gravity is not even a real force. Let me briefly state what the theory of relativity has to say about gravity without going into the complicated ...


1

I think it is a question of the reference frame. You pick a reference frame tied to your object (the rest frame) then the mass is always the same in that frame and it is the rest mass of the object, m_0. If you pick another reference frame in which your object can move, its mass will be different and will definitely depend on its velocity. Its expression ...


1

You could probably make this precise to about 1e-7 to 1e-8, or so, maybe better, if you are willing to invest more money than is needed for a tabletop experiment, i.e. with an experiment on a satellite. A very good absolute gravimeter, like this one http://www.microglacoste.com/fg5xspecs.php achieves almost ppb precision, which is mostly limited by ...


1

Within special relativity definitions mass is the positive root of the square root of the dot product in four vector space. With this definition there is no way a mass can be negative by construction. Two particles at rest will have zero momenta and their masses will add linearly for the minimum invariant mass of their system. Once they get momentum the ...


1

The problem is that the main reason for the consumption of power is the inefficiency of human muscles. If I run a marathon on level ground then I shouldn't have used any energy at all because my potential energy hasn't changed (there will be some loss of energy to air resistance, but at the speed I run that's negligable). However experience suggests that ...



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