# Do all objects really fall at the same velocity to the ground in vacuum?

Discussion in 'Physics & Math' started by pluto2, Sep 23, 2012.

1. ### hansdaValued Senior Member

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If particle Neutrino velocity is depended upon momentum , then its speed at the source and at the detector must be different.

If momentum of particle Neutrino is varying, whether it is increasing or decreasing?

Considering conservation of momentum, how the additional momentum of particle Neutrino can be accounted for?

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3. ### rpennerFully WiredStaff Member

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Poor, mathless boy, blind to all but pictures, has no physics in his soul.

//Edit.. Or as haiku:

Poor, mathless, lost kid.
Blind to all but cartoon sketches.
No physics in soul.

Why? What do you know about neutrinos that would require this to be true?

Last edited: Oct 1, 2012

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5. ### hansdaValued Senior Member

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You only said that speed of particle Neutrino is variable(not constant) and depended upon its momentum.

[note: Mass of particle Neutrino should change because of Lorentz transformation of mass at a relativistic speed.]

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7. ### rpennerFully WiredStaff Member

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This is true of electrons, Congressmen, and planets. Speed and momentum are related when $m$, the invariant mass of the particle, is non-zero.
That is not invariant mass which you are speaking of. That is a quantity sometimes called "relativistic mass" $\frac{m}{\sqrt{1 - \left( \frac{\vec{v}}{c} \right)^2}} = \frac{\left| \vec{p} \right| }{\left| \vec{v} \right| } = \frac{ \sqrt{(mc^2)^2 + (\vec{p} c)^2} }{ c^2 }$ which doesn't behave like mass because the relationship between force and acceleration is not a simple proportionality as in the Newtonian approximation $\vec{F} \propto \vec{a}$. Specifically for perpendicular forces, $\vec{F}_{\tiny \perp} = m \left( 1 - \left( \frac{\vec{v}}{c} \right)^2 \right)^{ \tiny - \frac{1}{2}} \vec{a}_{\tiny \perp}$ while for forces parallel to the direction of movement, $\vec{F}_{\tiny \parallel} = m \left( 1 - \left( \frac{\vec{v}}{c} \right)^2 \right)^{ \tiny - \frac{3}{2}} \vec{a}_{\tiny \parallel}$.

8. ### FarsightValued Senior Member

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3,475
I'm not one for souls, rpenner, and as you know I'm not mathless. I focus on the terms, the meaning, the understanding. Best if you avoid siding with Guest, who contributes only spoiler posts, it will do you no good. By the way, you didn't respond to my post about energy/mass/momentum/pair production/QED. Anybody with physics in his soul would have been genuinely interested in that, but you blanked it. A couple of reams of impress-the-kiddies latex and a snide quip is no substitute for physics, rpenner. Poor, meaningless lost kid, blind to all but math, no physics in soul.

Re your above post to Hansda, you need to understand rest mass and that symmetry between momentum and inertia. Rest mass $m$ is not some magic invariant cosmic treacle thing that remains fixed in $\vec{v} = \frac{\vec{p} c^2}{\sqrt{(m c^2)^2 + (\vec{p} c)^2}}$.
Pay attention to this post, and pay attention to what Einstein said: the mass of a body is a measure of its energy-content. He talked of kinetic energy, which relates to motion. Move past that body and its relativistic mass is your measure of the energy. It increases with your relative motion. Come on, think man. Compton scattering is where a photon moves an electron, and the photon is diminished as a result. You could repeat until you had no photon left. And yet the electron (and positron) is made from a photon via pair production. Make the leap: the electron is effectively made of motion. Its rest mass is nothing more than a measure of the internal kinetic energy which is not moving in aggregate with respect to you. The photon it was made from has no rest mass because it is kinetic energy moving at c with respect to you. There's a sliding scale in between, rpenner, and it cuts both ways: more energy, more mass, less speed, more mass. None of that energy-momentum appears as mass when it goes at c, some of it appears as effective mass when it goes at less than c, and when it goes at zero, it is indeed effective, and all of it does. You know this from trapping a photon in a mirror-box, and opening the box to simulate a radiating body losing mass. And you know that we have never seen a neutrino at rest. So please try to address the issue of neutrino oscillation and give Hansda something more than mathematical incantations thrown out like Latin.

9. ### rpennerFully WiredStaff Member

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I was taught it was rude to stare and I have no obligation to respond to you. But now that you insist on posting, I call upon the moderators to enforce their proclamations against tripe.

10. ### Guest254Valued Senior Member

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1,056
Wrong illustration? I have written down two vector fields and I have also computed the curl (or rot) of both these vector fields in their respective domains. Mathematica has kindly produced the streamlines for those vector fields for me. Judging by your comments, I can only assume you think Mathematica has made a mistake. What do you think the vector fields look like?

11. ### AlphaNumericFully ionizedModerator

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6,697
For all intents and purposes, you are. You don't even understand the mathematics behind the picture you've quoted and the picture Guest made. That is stuff you learn in the first year of any physics or maths degree and is ESSENTIAL to understand electromagnetism and Maxwell's equations, something you claim to be a world expert in. You are deluding yourself if you think you have any kind of maths capability in regards to this stuff, as illustrated by you failing to grasp what Guest posted.

And what has this arm waving gotten you? 6 wasted years, rejections from all journals, a self published book no one cares about and work which is not even not even wrong. You cannot provide any working model of anything in the real world, using your work. Given you complain about such things in regards to string theory you cannot ignore such a complaint about your own work.

You have no idea what physics is.

Ah, the "I don't understand the mathematics so it must be nonsense" approach. The problem is you don't understand maths expected of teenagers. As I've told you many times, you are for all intents and purposes innumerate.

12. ### FarsightValued Senior Member

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3,475
It isn't tripe, rpenner. There's a clear issue with pair production occurs because pair production has occurred. If you'd prefer to duck my post concerning thats interesting physics matter and respond instead with abusive comments whilst crying for a moderator, that's up to you.

13. ### hansdaValued Senior Member

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2,182
So, that is not true for particle Neutrino. IE, You mean to say speed of particle Neutrino is constant and it does not depend(vary) on its momentum. Whereas speed of electron is depended(vary) upon its momentum.

At a relativistic speed, which mass should be used for calculating kinetic energy and momentum of a particle? Invariant mass(non-zero) or "relativistic mass"?

14. ### rpennerFully WiredStaff Member

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Incorrect, like electrons, Congressmen and planets, neutrinos also have a non-zero invariant mass. It happens to be very small compared to an electron.

Since this contradicts what I wrote, I don't see why you came to that conclusion. Not only does the speed of the neutrino depend on its momentum, I told you what that relationship was: $\vec{v} = \frac{\vec{p} c^2}{\sqrt{(m c^2)^2 + (\vec{p} c)^2}}$
$\vec{v}$ is velocity.
$c$ is the speed of light.
$m$ is the invariant mass.
$\vec{p}$ is the momentum.

If $m = 0$ then the above reduces to saying the speed is constant at the speed of light; but for neutrinos, electrons, Congressmen and planets, we have evidence that $m \gt 0$.

Invariant mass, always. But you have to use the correct formulas.
$E^2 = (mc^2)^2 + (\vec{p} c)^2$ and $E \vec{v} = c^2 \vec{p}$ work for free particles of all masses and for all values of momentum.

$E = \sqrt{(mc^2)^2 + (\vec{p} c)^2} = c^2 \frac{ \left| \vec{p} \right| }{\left| \vec{v} \right| }$ are solutions for E that work for all values.
$E = \frac{m c^2}{ \sqrt{ 1 - \left( \frac{ \vec{v} }{c} \right)^2 }}$ works only when $m > 0$
$E \approx mc^2 + \frac{m}{2} \vec{v}^2 + \frac{3 m}{8 c^2} \vec{v}^4$ is an approximation that works only when $m > 0$ and $\left| \vec{v} \right| \lt \lt c$.
$E \approx mc^2 + \frac{1}{2 m} \vec{p}^2 - \frac{1}{8 m^3 c^2}\vec{p}^4$ is an approximation that works only when $m > 0$ and $\left| \vec{p} \right| \lt \lt mc$.

$\left| \vec{p} \right| = \sqrt{\frac{E^2}{c^{2}} - m^2 c^2 }$ and $\vec{p} = \frac{E \vec{v}}{c^2}$ are solutions for $\vec{p}$ (or its magnitude) that work for all values.
$\vec{p} = \frac{m \vec{v}}{ \sqrt{ 1 - \left( \frac{ \vec{v} }{c} \right)^2 }}$ works only when $m > 0$.
$\vec{p} \approx \left( 1 + \frac{\vec{v}^2}{2 c^2} + \frac{3 \vec{v}^4}{8 c^4} \right) m \vec{v}$ is an approximation that works only when $m > 0$ and $\left| \vec{v} \right| \lt \lt c$.
$\left| \vec{p} \right| \approx \frac{E}{c} - \frac{m^2 c^3}{2 E} - \frac{m^4 c^7}{8 E^3}$ is an approximation that works only when $m c^2 \lt \lt E$.

Last edited: Oct 3, 2012
15. ### FarsightValued Senior Member

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3,475
It depends on the particle. If the particle concerned is a photon, you don't use the invariant mass because it's zero. If the particle concerned is an electron, you do. The electron invariant mass is clear cut, you can measure it fairly easily. See wiki for that. You can't do the same for a neutrino. You can't effectively hold a neutrino in your hand and weigh it. You don't know what the invariant mass is, so you can't use it to calculate the energy-momentum of a neutrino. Neutrinos were proposed by Pauli in 1930 to account for missing energy-momentum in beta decay. Not missing mass. We measure the energy-momentum of a neutrino, here's a paper on it. We don't calculate it using the invariant mass. The non-zero neutrino mass is inferred from neutrino oscillation, see this paper on arXiv. Here's an excerpt from it:

"Just as light passing through matter slows down, which is equivalent to the photon gaining a small effective mass, so neutrinos passing through matter also result in the neutrinos slowing down and gaining a small effective mass. The effective neutrino mass is largest when the matter density is highest, which in the case of solar neutrinos is in the core of the Sun. In particular electron neutrinos generated in the core of the Sun will be subject to such matter effects. It turns out that neutrino oscillations, which would be present in the vacuum due to neutrino mass and mixing, will exhibit strong resonant effects in the presence of matter as the effective mass of the neutrinos varies along the path length of the neutrinos"

16. ### rpennerFully WiredStaff Member

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4,833
And if kinetic energy $KE = E - mc^2$, then we have
$KE = \sqrt{ (mc^2)^2 + (\vec{p} c)^2 } - mc^2 = c^2 \left( \frac{ \left| \vec{p} \right| }{\left| \vec{v} \right| } - m \right)$
$KE = m c^2 \left(\frac{1}{ \sqrt{ 1 - \left( \frac{ \vec{v} }{c} \right)^2 }} - 1 \right)$ when $m > 0$
$KE \approx \frac{1}{2} m \vec{v}^2 + \frac{3 m}{8 c^2} \vec{v}^4$ is an approximation that works only when $m > 0$ and $\left| \vec{v} \right| \lt \lt c$.
$KE \approx \frac{1}{2 m} \vec{p}^2 - \frac{1}{8 m^3 c^2}\vec{p}^4$ is an approximation that works only when $m > 0$ and $\left| \vec{p} \right| \lt \lt mc$.

Also, at the high speed limit we have
$E \approx \left| \vec{p} \right| c + \frac{m^2c^3}{2 \left| \vec{p} \right|} - \frac{m^4 c^5}{8 \left| \vec{p} \right|^3}$ when $mc \lt \lt \left| \vec{p} \right|$
and
$E \approx \left| \vec{p} \right|c + \left| \vec{p} \right| ( c - \left| \vec{v} \right| ) + \frac{\left| \vec{p} \right|}{c} \left( c - \left| \vec{v} \right| \right)^2$ when $c - \left| \vec{v} \right| \lt \lt c$

Also
$\left| \vec{p} \right| \approx \sqrt{ 2 m ( E - mc^2) } \left( 1 + \frac{ E - mc^2 }{4 mc^2} - \frac{ ( E - mc^2)^2 }{32 (mc^2)^2 } \right)$ when $E - mc^2 \lt \lt mc^2$
or
$\left| \vec{p} \right| \approx \sqrt{ 2 m ( KE ) } \left( 1 + \frac{ KE}{4 mc^2} - \frac{ ( KE )^2 }{32 (mc^2)^2 } \right)$ when $KE \lt \lt mc^2$
And in terms of kinetic energy:
$\left| \vec{p} \right| = \sqrt{ \left( 2 m + \frac{KE}{c^2} \right) \, KE }$ and $\vec{p} = \left( m + \frac{KE}{c^2} \right) \vec{v}$

17. ### Guest254Valued Senior Member

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1,056
Wrong illustration? I have written down two vector fields and I have also computed the curl (or rot) of both these vector fields in their respective domains. Mathematica has kindly produced the streamlines for those vector fields for me. Judging by your comments, I can only assume you think Mathematica has made a mistake. What do you think the vector fields look like?

18. ### hansdaValued Senior Member

Messages:
2,182
Do you have any reference for all these equations?

You mentioned about kinetic energy K E. What about potential energy P E?

For particle Neutrino, what are the equations for its Kinetic Energy(KE) and Potential Energy(PE)?

If particle Neutrino is affected by gravity, it should have potential energy.

19. ### EmilValued Senior Member

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a[sub]1[/sub]=Gm[sub]2[/sub]/r[sup]2[/sup] ; a[sub]2[/sub]=Gm[sub]1[/sub]/r[sup]2[/sup]
The distance D between m[sub]1[/sub] and m[sub]2[/sub] after time t
D=r-a[sub]1[/sub]t[sup]2[/sup]/2-a[sub]2[/sub]t[sup]2[/sup]/2=r-t[sup]2[/sup]/2(Gm[sub]2[/sub]/r[sup]2[/sup]+Gm[sub]1[/sub]/r[sup]2[/sup])=r-t[sup]2[/sup]G/2r[sup]2[/sup](m[sub]1[/sub]+m[sub]2[/sub])

D=r-t[sup]2[/sup]G/2r[sup]2[/sup](m[sub]1[/sub]+m[sub]2[/sub])

20. ### rpennerFully WiredStaff Member

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L.B. Okun "The Concept of Mass" Physics Today 42(6) 31-36 (1989)
And hundreds of scientific articles and papers and textbooks since 1905.
And the rules of vector algebra and differential calculus which have been largely in place since the 17th century.
A particle in a potential is not a free particle. Potentials are just one type of dynamical system, so you would want to study Hamiltonian or Lagrangian physics and general differential equations to get a full set of solutions.
Neutrinos are well approximated as free particles under all laboratory conditions.
No one has been able to make a source of neutrinos with known energies and momenta, or prepared a beam of neutrinos with velocities less than 10,000 times the escape velocity of Earth, so the effect of gravity on the trajectory of neutrinos is negligible. To first approximation, the momentum in the vertical direction changes to satisfy $0 = \frac{g E}{c^2} \Delta h + \Delta E$
or $1 - \frac{g \Delta h}{c^2} = \frac{ \sqrt{(mc^2)^2 + c^2 p_x^2 + c^2 p_y^2 + c^2 (p_z + \Delta p_z)^2} }{ \sqrt{(mc^2)^2 + c^2 p_x^2 + c^2 p_y^2 + c^2 p_z^2}} \approx 1 + \frac{c^2 p_z}{E^2} \Delta p_z$
or $\Delta p_z \approx - \frac{g E^2}{c^4 p_z} \Delta h$
or in the relativistic limit of a beam of neutrinos headed up or down: $\frac{\Delta p_z}{p_x} \approx - \frac{g}{c^2} \Delta h \approx - \frac{\Delta h}{9.16475 \times 10^{15} \; \textrm{meters}}$.
So a neutrino beam climbing 1 km loses about 0.00000000001% of its momentum.

21. ### Mars RoverBannedBanned

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If a neutrino was climbing radially away a black hole gravity well from just outside the event horizon, would this affect the neutrino mass and oscillation? And if a neutrino is emitted at its customary aproximate lights peed from far away source towards a black hole event horizon, can the neutrino accelerate further due to gravity acceleration as neutrino closes towards the event horizon? Does its mass or oscillation get affected such that it cannot exceed light speed despite black hole gravity accelerations on top of initial speed?

22. ### rpennerFully WiredStaff Member

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Discussion of neutrino oscillation would require a discussion of quantum physics, and involving strong gravity would require a discussion of quantum field theory in curved space time. That discussion would not be in the least illuminating to the present audience who think that algebraic rearrangement of terms leads to new physics formulas.

In an otherwise empty universe, it is conceivable that a neutrino could loose so much energy that it would fall back into a black hole some time after being emitted, but without a precise knowledge of the neutrino mass, there is no reliable way to predict the order of magnitudes of such a relationship. There is an extremely small set of parameters which would allow a neutrino emitted from conventional radioactive processes to orbit a black hole, but a neutrino popping into existence just outside the event horizon is either going to escape or fall back in.

No neutrino aimed at a black hole is going to move at or faster than light speed. Similarly, invariant mass remains invariant mass, and oscillations happen at the same period as measured in proper time because the mass eigenstates are not the same as flavor eigenstates.

23. ### Mars RoverBannedBanned

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A photon and a neutrino (or anti-neutrino?) is produced together from same muon decay a million miles from a supermassive black hole and aimed radially at event horizon. It is agreed that the photon is already @ light speed, it cannot speed up (cannot increase energy by "falling" faster), so increases energy by becoming higher frequency or shorter wavelength or whatever inherent change is according to theory. Meanwhile neutrino (or anti) which left source at same instant has non-zero rest mass (whatever that is, even if not determined by measurement). If a neutrino (or anti) has rest mass, gravity acceleration while "falling" can affect its speed and so its momentum increases from that of its initial momentum from rest mass and emission speed?

What is it that stops neutrino reaching light speed from its initial near light speed while "falling" due to gravity acceleration time between its source and event horizon?