# 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. ### Mars RoverBannedBanned

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rpenner. If a neutrino and photon are emitted from same position a million miles away from supermassive black hole event horizon, photon will maintain light speed and increase frequency as it closes on event horizon due to gravitational potential increasing energy of photon which cannot "fall" any faster than lights peed it had from emission. But the neutrino has non-zero rest mass (whatever it is in fact even if not measured). Therefor the neutrino is affected by gravity as all particles with rest mass are, they "fall" faster and faster the longer the gravitational acceleration acts while it closes on event horizon.

You say that neutrino, which can accelerate by gravity from its initial emission speed, cannot reach light speed. I agree. I agree also that invariant mass (whatever its rest mass is) is invariant mass.

But can you tell me what prevents neutrino being accelerated to higher than emission speed (or even to light speed), since gravitational acceleration is "always on" all the way from a million miles away and until it "falls" down to the event horizon the rest or invariant mass is responding to gravitational gradient potentials and associated accelerations?

3. ### rpennerFully WiredStaff Member

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The classic description of space-time around a black hole of zero angular momentum and Schwarzschild radius of: $r_s = \frac{2 GM}{c^2}$
is the Schwarzschild metric: $c^2 (d \tau)^2 = \left( 1 - \frac{r_s}{r} \right) c^2 (dt)^2 - \left( 1 - \frac{r_s}{r} \right)^{-1} (dr)^2 - r^2 (d\Omega)^2$
Explicitly solving the geodesic equations for particles with zero angular momentum and non-zero rest mass gives timelike curves through space time:
$\frac{d r}{d t} = \pm c \frac{\left( 1 - \frac{r_s}{r} \right) \sqrt{A^2 - \left( 1 - \frac{r_s}{r} \right) }}{A}$
Where A is a positive number, greater than one for particles moving above escape velocity. In the limit as A goes to infinity this expression happens to equal the expression for null radial geodesics, which can be derived in any number of ways from the Schwarzschild metric:
$\frac{d r}{d t} = \pm c \left( 1 - \frac{r_s}{r} \right)$
Thus we get the speed of a radial trajectory relative to the speed of light: $\left| \beta \right| = \sqrt{ 1 - \frac{r - r_s}{r A^2}} \leq 1$ when $r > r_s$

$\frac{\omega_1}{\omega_2} = \sqrt{\frac{r_1}{r_1 - r_s}} \sqrt{\frac{r_2 - r_s}{r_2}}$
Gravity affects particles massive and massless.
Time-like geodesics are everywhere time-like. Above is an explicit calculation that shows these geodesic trajectories nowhere move faster than the speed of light.

Last edited: Oct 6, 2012

5. ### Mars RoverBannedBanned

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rpenner. I know that gravity affects both massive and massless. I was stating that massive particle (neutrino) falls and speeds up as all (rest) massive particles do but which rest massless particle (photon) does not. How do "time-like geodesics" explain what physically prevents (rest massed) neutrino from speeding up like any other rest massed particle would while falling towards but still well outside the event horizon while falling?

7. ### rpennerFully WiredStaff Member

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The rapidity of the neutrino does increase. Who said it didn't?

If A is very large (as it is for neutrinos) then $\frac{\beta(r)}{\beta_{\infty}} \approx 1 + \frac{r_s}{2 A^2 r} > 1$ which while may be small is still an increase.
If you look at rapidity (again for A very large), the increase is again apparent $\tanh^{-1} \beta(r) \; - \; \tanh^{-1} \beta_{\infty} \approx \frac{1}{2} \ln \frac{r}{r -r_s} > 0$
If you look at gamma factors (limiting ourselves to A>1), the increase is again apparent $\frac{\gamma(r)}{\gamma_{\infty}} = \sqrt{\frac{ 1 - \beta_{\infty}^2 }{1 - \beta^2(r)}} = \sqrt{ \frac{r}{r -r_s} } > 1$

8. ### TachBannedBanned

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5,265

With Reality Check constant droning gone we could have an interesting discussion.
I can see $\frac{d r}{d t}=\pm c (1-\frac{r_s}{r}) \sqrt{1-\frac{(\frac{d\tau}{dt})^2}{1-\frac{r_s}{r}}}$
Then, $\frac{d \tau}{dt}=\frac{1-\frac{r_s}{r}}{E}$ where $E=constant$ is the relativistic energy per unit of mass of the test particle. So, your "A" is my "E":

$\frac{d r}{d t}=\pm c (1-\frac{r_s}{r}) \sqrt{1-\frac{1-\frac{r_s}{r}}{E^2}$

In the case of the photon, $E->\infty$ so:
$\frac{d r}{d t}=\pm c (1-\frac{r_s}{r})$ (or, you can deduce the same from the fact that the photon follows null geodesics)

Last edited: Oct 7, 2012
9. ### hansdaValued Senior Member

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2,279
How momentum is being conserved here?

Here loss of momentum can be related with loss of kinetic energy(KE) and gain of potential energy(PE), if we follow conservation of energy in a gravitational field.

What can be the equation with KE and PE for 'conservation of energy' for a particle Neutrino?

10. ### rpennerFully WiredStaff Member

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It is ironic that you who introduced the subject of motion in a potential now want to talk about conservation of momentum when a potential breaks the symmetry that the laws of physics don't respect one's position in the universe that is required to talk about conservation of momentum. If, however, you treat the Earth and the neutrino symmetrically under the same laws of physics, the solution is obvious. The sum of the Earth and the neutrino's momenta is constant.

The neutrino's loss is the Eath's gain.

Easily derived from expressions seen earlier in this thread. And utterly negligible to present-day experimentalists. As your obsessions are not my own, as the answer is of no practical or pedagogical purpose, and as I am on an iPhone, I see no reason to typeset a trivial calculation.

Instead, you should answer the question yourself, being sure to explain you reasoning and showing your work.

11. ### hansdaValued Senior Member

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'Conservation of momentum' and 'conservation of energy' are very fundamental concepts of Physics. Any physical interaction should follow these principles. Considering Einstein's equation E=mc^2, where energy(E) is conserved; it can be shown that here momentum is also conserved. This conserved momentum is mc or E/c(m is mass, E is energy and c is speed of light).

12. ### rpennerFully WiredStaff Member

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There is no reason to quote the message of the person you are replying to unless you are reading and understanding what was written and making detailed, point-by-point comments.
But not completely fundamental as these are theorems (specifically applications of Noether's theorem) based on certain physical assumptions, like invariance of the laws of physics under translation and invariance of the laws of physics under time translations, respectively. These physical assumptions don't apply to all models of all physical systems in which case, you can't count on them to apply. Kepler's laws assume the sun is at the focus of all planetary ellipses and thus don't have conservation of momentum. Newtonian physics in an acceleration field doesn't conserve momentum either. A quantum particle in a box that slowly changes size does not conserve energy. So when you introduced the concept of potential energy based on position (PE = mgh) you violated the assumption required for the laws of conservation of momentum to apply. A full treatment of a neutrino and the Earth in general relativity would show that in this two-body system momentum is conserved although some amount of momentum would also appear as gravitational radiation. But the when you oversimplify the discussion to talk about potential energy, you to violence to the symmetry with respect to position and thus are responsible for the failure of conservation of momentum to hold.
Is this all you know? What about what I just told you about $E^2 = (mc^2)^2 + (\vec{p}c)^2$ and $E \vec{v} = c^2 \vec{p}$ for a free particle (or an isolated system)?
That is not momentum. Momentum is a vector quantity.

13. ### Beer w/StrawTranscendental Ignorance!Valued Senior Member

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MD, do you have dimensional analysis to go with those equations you used?

14. ### hansdaValued Senior Member

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You are right momentum is a vector quantity. What i am telling is that, if an invariant mass 'm' is converted into Light-Energy 'E' through some interaction; its momentum will also be conserved along with its energy 'E'. Magnitude of this conserved momentum will be E/c (c is speed of light).

15. ### rpennerFully WiredStaff Member

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You contradict yourself. If the magnitude of the conserved momentum is E/c then the momentum of the mass is also E/c and therefore you are saying a body at rest has non-zero momentum.

The correct result is if a particle at rest decays into two photons, those photons zip off in opposite directions with equal energies that sum to $mc^2$, therefore each photon has momentum of magnitude $\frac{1}{2} m c$ but the conservation of momentum means these sum to zero because they are vectors of equal magnitude and opposite direction.

16. ### eramSciengineerValued Senior Member

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a body at rest may also recoil.

17. ### hansdaValued Senior Member

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'Rest' is a relative concept. In reality no mass is at rest. Every mass has some motion. When relative velocity between two masses is zero, it is called as rest with relative to the other mass. So, even if a body is at rest(relative), it has motion and its magnitude of momentum is non-zero.

18. ### AlphaNumericFully ionizedModerator

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They are important but not necessarily universally true. A breakdown in Lorentz symmetry would allow for momentum and energy conservation violation, which is something a lot of experimentalists look for, as it would be extremely interesting if it occurs.

Utterly wrong. $E=mc^{2}$ is for an object at rest. An object which is moving with momentum p will instead satisfy $E^{2} = (mc^{2})^{2} + |\mathbf{p}c|^{2}$. This is Lorentz symmetric, as it rearranges (setting c=1 for convenience) to $-m^{2} = -E^{2} + |\mathbf{p}|^{2}$, which is the 4-vector squared norm $p^{\mu}p_{\mu}$ for $p^{\mu} = (E,\mathbf{p})$.

Please learn some physics before trying to tell people how it supposedly works.

19. ### hansdaValued Senior Member

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If an invariant mass 'm' is converted into light energy 'E' through some interaction, it can be said that E is energy content of mass m.

If E is energy content of mass m, it can be said that E/c is the magnitude of momentum content of mass m.

Derivation is as follows:

If mass m is converted into light energy E, it will generate 'E/hf' number of photons whoose total momentum will be (E/hf)(hf/c)= E/c.

[here h is Planck's constant, f is frequency of photon and c is speed of light.]

20. ### rpennerFully WiredStaff Member

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It appears that you don't know how to sum vector quantities, so you don't know what conservation of momentum actually means. Please consider this fair warning before a physics moderator calls you out in it.

21. ### alansteve777Registered Member

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I don't have any idea but here we have to consider to cases they are the body velocity in vacuum and body velocity after entering into gravitational region. It may comes with constant velocity.

22. ### RealityCheckBannedBanned

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800
Hi rpenner.

So the neutrino's rapidity increases. What stops the neutrino from attaining lightspeed on its way to the black hole horizon given a long enough gravity acceleration period from emitter to horizon? If it is emitted at "almost" lightspeed in the direction of a black hole horizon from a distance and then accelerates due to gravity all the way there, then its initial speed increases cumulatively even if it is by small increments? Is there any scenario where the neutrino reaches 'speed singularity' of its own and converts to pure energy (at which point it does not go any faster, just like a photon?)?

23. ### rpennerFully WiredStaff Member

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The question is baseless and ignores the whole of relativity. In GR, the motion of the neutrino is a time-like geodesic which is not only everywhere time-like (i.e. slower than light) but naturally comes with a parameterization in $\lambda$ such that $\frac{d \tau}{d \lambda}$ is constant. There is no need for anything to stop the neutrino from attaining lightspeed if there is nothing postulated to tend to cause it.
"Almost lightspeed" is a statement from the viewpoint of a particular coordinate system and is not respected by the laws of physics. Relative to itself the neutrino is moving at speed zero.
This acceleration you speak of is coordinate-dependent and is not respected by the laws of physics. An object moving freely in curved space-time has no proper acceleration and its motion is that of a geodesic of space-time.
Correct, if only with respect to a particular coordinate system.
No, that is comic-book physics. The LEP accelerated electrons to a speed about 70.3 miles per year ( 12.9 meters per hour, 21.5 cm per minute ) slower than the speed of light, but this was not "almost" light-speed from the electron's point of view.

Put it another way, and treat change of velocity as a hyperbolic rotation. $\Lambda_{\tiny \frac{a}{b}c} = \begin{pmatrix} \frac{b}{\sqrt{b^2 - a^2}} & \frac{a}{\sqrt{b^2 - a^2}} \\ \frac{a}{\sqrt{b^2 - a^2}} & \frac{b}{\sqrt{b^2 - a^2}} \end{pmatrix}$
Then two times this change of velocity is $\left( \Lambda_{\tiny \frac{a}{b}c} \right)^2 =\Lambda_{\tiny \frac{a}{b}c} \; \Lambda_{\tiny \frac{a}{b}c} = \begin{pmatrix} \frac{a^2 + b^2}{b^2 - a^2} & \frac{2 a b}{b^2 - a^2} \\ \frac{2 a b}{b^2 - a^2} & \frac{a^2 + b^2}{b^2 - a^2} \end{pmatrix} = \Lambda_{\tiny \frac{2 a b}{a^2 + b^2}c}$
Or $2 \otimes \frac{a}{b} c = \frac{a}{b} c \oplus \frac{a}{b} c = \frac{2 a b}{a^2 + b^2}c$
Example: $2 \otimes \frac{1}{4} c = \frac{8}{17}c$
Example: $2 \otimes \frac{1}{2} c = \frac{4}{5}c$
Example: $2 \otimes \frac{3}{4} c = \frac{24}{25}c$
Example: $2 \otimes \frac{24}{25} c = \frac{1200}{1201}c$
Example: $2 \otimes \frac{1200}{1201}c = \frac{2882400}{2882401} c$
Example: $2 \otimes \frac{k}{k+1}c = \frac{2 ( k^2 + k)}{2(k^2 +k) + 1} c$
Example: $n \otimes \frac{1}{3}c = \frac{ 2^n - 1 }{ 2^n + 1 } c$
Example: $n \otimes \frac{k}{k+1}c = \frac{ (2k +1)^n - 1 }{ (2k +1)^n + 1 } c \quad \quad \quad \quad ( k \geq 0 )$
So physical processes which change velocity, even when repeated, do not change velocity to c.

// Edit:

Finally, all of this is predicated on the definition of a neutrino as a particle with certain properties including a non-zero rest mass and for a free particle we have the velocity-momentum relationship: $\vec{v} = \frac{c^2 \vec{p}}{\sqrt{c^2 \vec{p}^2 + m^2 c^4}}$ which applies everywhere in flat-space time or any local coordinates about the neutrino in curved space time.

Last edited: Oct 20, 2012