# Planck constants

Discussion in 'Physics & Math' started by Mind Over Matter, Jan 13, 2012.

1. ### Mind Over MatterRegistered Senior Member

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Why there are such a things as Planck time and Planck length?
Isn't it obvious that our spacetime ticks at the rate of Planck time?
If it ticks then it must be discrete, don't you agree?

3. ### DeeCeeValued Senior Member

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1,793
"Why there are such a things as Planck time and Planck length?"

Because Mr Planck did the work and developed a mathmatical model that helped predict the behavior of subatomic particles.

"Isn't it obvious that our spacetime ticks at the rate of Planck time?"

Not to me. Pehaps I need educating.

If it ticks then it must be discrete, don't you agree?

No, but you can explain your quantum theory of time if you think it's that important.

Bugger me I'm here again.
Dee Cee

5. ### mathmanValued Senior Member

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Neither of the statements in bold are accurate. The best that can be said of the Planck time unit is that things get very fuzzy at that interval.

7. ### AcitnoidsRegistered Senior Member

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All Planck units represent a limit to the speed of light.
No. The Planck length is similar to the event horizon of a black hole - hence the large value for the Planck mass.
All Planck units arise from the scientific law that says nothing can travel faster than the speed of light. Once you pass this hypothetical boundary the known laws of physics begin to break down. If you're asking; is the speed of light a discrete increment? Then yes, I agree (depending on the frames of refference of course). If you're asking; do the Planck units represent some sort of special fundamental timeframe? Then no, I disagree.

Last edited: Jan 14, 2012
8. ### DeeCeeValued Senior Member

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1,793
Forgive me if I'm wrong but are not the planck constants simply the point at which quantum indeterminacy takes over the math and nothing smaller can be predicted or indeed measured.

DeeCee

9. ### James RJust this guy, you know?Staff Member

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Because if you try to put some fundamental constants of nature into an equation that gives a dimensionally-correct time or length, that's what pops out.

Not to me.

What do you think it means for spacetime to "tick"? Where and how does it tick? And what do you mean by "at the rate of Planck time"?

How do you know it ticks?

10. ### Boris2Valued Senior Member

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i don't think this is true. more true to say that information can't travel faster than light.

11. ### Mind Over MatterRegistered Senior Member

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Here is an excerpt from an article that was published in the Sept, 2011 issue of the "Physics Essays" journal entitled "The Inflaton Space-time Model: Making Sense of the Standard Models of Particle Physics and Cosmology":

If time isn't flowing, then it must be "ticking". The article goes on to argue that the "flow of time", is made real only by our memories; that is to say, that the "flow of time" is part of subjective reality, whereas time is more incremental for objective reality.

I have read both Barbour's and Greene's books and what they contend is that reality increments as a series of NOW's, an idea that is not unlike my contention that at the ground of reality, time is the result of the incrementation of configurations of s-points. Of course some will ask for evidence. However since there is no evidence that at the ground of reality space is either discrete or continuous, perhaps a philosophical construction can be built on the basis of either assumption.

12. ### Crunchy CatF-in' *meow* baby!!!Valued Senior Member

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It's a result of mathematical models that describe reality.

It is neither obvious that spacetime ticks or ticks at the rate of planck time.

No, so little is known about time that if-then agreements aren't valuable beyond some kind of esoteric social bonding.

Last edited: Jan 15, 2012
13. ### wlminexBannedBanned

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1,587
IMO (Only!) . . . . most likely . . . time, in space-time, 'ticks' (vibrates) at the base equivalent frequency (yet undetected) of whatever the energy level is of . . . . dark matter . . . . .(re: SQR) . . . . i.e., EXTREMELY HIGH (est. 10^120 ergs/cc energy density). Energy density of the observable universe is lower (10^60 ergs/cc, or so) . . . it's kind of a harmonics sort of thingy. . . . again, IMO

Last edited: Jan 15, 2012
14. ### James RJust this guy, you know?Staff Member

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36,484
This is the Physics & Math forum, not a place for your personal fantasies.

Please do not post in the science areas of sciforums if you only want to post unfounded opinions you dreamt up.

15. ### wlminexBannedBanned

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1,587
James . . . just responding to MOM's post (above). . . and YOUR post #6!!! . . . BTW: you are welcome to delete/move any of my posts to what location you feel is appropriate . . . BTW: Is it Sciforum "policy" not to allow 'opinions' (e.g., IMO) to be posted? . . . you are certainly not at a loss when it comes to offering YOUR opinions!

Regards . . . wlminex

16. ### Mind Over MatterRegistered Senior Member

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How do we know it ticks? Well, have a look at the attached formula from Wikipedia.

Planck time is defined based on reduced Planck constant, gravitational constant and speed of light in a vacuum. If quantum theory stands then it seems we can not have smaller time then Planck's time.

17. ### AcitnoidsRegistered Senior Member

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I could have been clearer. I should have said; "The five base Planck units arise from ... along with other physical constants." The other constants that go into finding the five base units are the Gravitational constant, the fine structure constant, h-bar, Planck's constant, Boltzmann constant, permittivity of free space, elementary chanrge and the speed of light. The constants listed in bold are used to calculate all the base Planck units. Derived Planck units (momentum, energy, force, power, density, angular frequency, pressure, current, voltage and Impedance) are consequences of the base Planck units (length, mass, time, charge and temperature) and the physical constants that go into them.
Entanglement says otherwise. It's believed that anything with mass/energy cannot travel faster than light.

18. ### Mind Over MatterRegistered Senior Member

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I would appreciate if anyone could comment on my post #13, especially the attached file.

19. ### rpennerFully WiredValued Senior Member

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In no way did these posts argue that time ticks.

$t_{\tiny \textrm{Planck}} \; = \; \sqrt{\frac{\hbar G}{c^{\tiny 5}}} \; \approx \; 5.39106(32) \, \times \, 10^{\tiny -44} \, \textrm{s}$
http://physics.nist.gov/cgi-bin/cuu/Value?plkt
Google's database has slightly different values.
(Half) the uncertainty of Newton's gravitational constant dominates the relative uncertainty of the Planck time.

The Planck time is one of a set of natural units. They are called natural units because when G, c or Planck's reduced constant are expressed in them, numerically these values are 1 and can be omitted from the expression without changing the numerical calculation. The Planck mass is huge compared to the mass of an electron or neutrino.

One of the biggest problems with the naive conception of a universe that has a universal time that ticks discretely is that such a universe violates relativity.

20. ### Mind Over MatterRegistered Senior Member

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However this would not be the case if the tick rates were a function of local variables, with time being discrete but asynchronous.

21. ### Mind Over MatterRegistered Senior Member

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I'm still lost. Do you know how units of measure work?
You shoot general statements that it's hard to respond to.
What relativity you are talking about? Do you know why there are special and general relativity theories?

22. ### rpennerFully WiredValued Senior Member

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No, it is not obvious that spacetime ticks, and therefore it is not obvious that time is discrete.
No, that is not what it implies. Quantum theory says that it is inappropriate to talk about spacetime as a smooth manifold at very small length and time scales where the quantum nature of gravity is important to every particle interaction. Thus even if experiment gives no reason to doubt the standard model of particle physics (which is built on special relativity and quantum physics) we know to describe the universe near or below the scale of the Planck length, or Planck time, we will at least need to combine general relativity and quantum physics. (Or like in string theory, be able to derive quantum gravity from other principles.)
If time ticks, then the universe prefers a certain definition of simultaneous above all others. You seem to think that relativity is preserved if this definition is local, but that only scratches the surface of agreement with quasi-static general relativity. But special relativistic effects indicate that simultaneity is not a local function of position only, but also of state of motion.

Basic Lorentzian time dilation rules out a local master clock, since moving clocks tick slower according to both observers that we consider stationary and those that we consider moving.

If there was some sort of local master clock, ticking away the progress of time, then space and time would be different things entirely and we wouldn't see a universe that respects the relativistic interval : $(ds)^2 = c^2(dt)^2 - (d\vec{x})^2$.
Units of measure in common use are man-made conventions for communication and trade.
The meter is standard world-wide so that French companies can bid on building factories in China with parts from South Africa.
At one time, people had standard measures locked up in a vault, and they would carefully replicate them and transport the replicas so that everyone would know what was "one foot" or "nine stone". But the modern second and meter are based on physical phenomena which can be reproduced time and time again.
Natural units are an extension of this, based largely on $\hbar{/tex] which is used in quantum physics which relates momentum and wavelength, energy and time; [I]c[/I] which relates time and space, mass, momentum and energy; and [I]G[/I] which relates mass to trajectories in time and space. So in any set of units, we can write: [tex]\vec{p} = \hbar \vec{k} \\ \left| \vec{p} \right| = \hbar \left| \vec{k} \right| = \hbar \frac{2 \pi}{\lambda} \\ E = 2 \pi \hbar f = \hbar \omega \\ \vec{v} = c^2 \frac{\vec{p}}{E} \\ E^2 = c^4 m^2 + c^2 \left| \vec{p} \right|^2 \\ \frac{\ddot{\vec{r}}}{\vec{r}} = -G M \left| \vec{r} \right|^{-3} \\ T_{\tiny \textrm{orbit}} = 2 \pi \sqrt{\frac{a^3}{G M}} \\ R_{\tiny \textrm{S}} = 2 G c^{-2} M \\ d\phi_{\tiny \textrm{orbit}} \approx \frac{6 \pi G}{c^2} \frac{M}{a (1 - e^2)}$
In SI units, all these quantities will have specific numerical values. In British Imperial units they will have different numerical values. But in natural units, $\hbar$, c and G all have the value 1.

Thus, when the convention of using natural units are used, the following are just as useful:
$\vec{p} = \vec{k} ; \quad \left| \vec{k} \right| = \frac{2 \pi}{\lambda} ; \quad E = 2 \pi f = \omega ; \quad \vec{v} = \frac{\vec{p}}{E} ; \quad E^2 = m^2 + \left| \vec{p} \right|^2 ; \quad \frac{\ddot{\vec{r}}}{\vec{r}} = - M \left| \vec{r} \right|^{-3} ; \quad T_{\tiny \textrm{orbit}} = 2 \pi \sqrt{\frac{a^3}{M}} ; \quad R_{\tiny \textrm{S}} = 2 M ; \quad d\phi_{\tiny \textrm{orbit}} \approx \frac{6 \pi M}{a (1 - e^2)}$

So natural units help us study physics by concentrating our limited faculties on the math that relates the assumptions of the physical model to detailed predictions that come from it. There is a Planck unit of mass, but it does not quantize masses. (It's about 70 quadrillion times larger than the top quark, the heaviest known fundamental particle or about 1/70th the mass of a mosquito or about the mass of a grain of sand.) The Planck units of time and space are inconveniently small, but that does not imply time and space are divided into chunks of that size. The Planck energy is about human scale, being about the amount of energy your body radiates every 32 weeks. The Planck momentum is on the order of 10 times your momentum while walking. Still other Planck units are very large.

If we want to convert a quantity from meters into feet, we must multiply by a conversion factor that is equal to one and cancel the units.

$381 \, \textrm{meters} = 381 \, \textrm{meters} \times \frac{1 \, \textrm{foot}}{0.3048 \, \textrm{meters}} = \frac{381}{0.3048} \, \textrm{feet} = 1250 \, \textrm{feet}$
In the same way, if I have $E = mc^2$ I can use $E_{\tiny \textrm{Planck}} = \sqrt{\frac{\hbar c^5}{G}} \; \quad m_{\tiny \textrm{Planck}} = \sqrt{\frac{\hbar c}{G}}$ to write $\frac{ E_{\tiny \textrm{Planck}} }{ \sqrt{\frac{\hbar c^5}{G}} } = 1 = \frac{ m_{\tiny \textrm{Planck}} }{ \sqrt{\frac{\hbar c}{G}} }$. Then I convert $E = E \times \frac{ E_{\tiny \textrm{Planck}} }{ \sqrt{\frac{\hbar c^5}{G}} } = E_0 E_{\tiny \textrm{Planck}}$ where $E_0$ is the dimensionless quantity: $\frac{ E }{ \sqrt{\frac{\hbar c^5}{G}} }$ and likewise for m, we go from $E = mc^2$ to $E_0 E_{\tiny \textrm{Planck}} = m_0 m_{\tiny \textrm{Planck}} c^2$. But because of the way the units are constructed, $\frac{m_{\tiny \textrm{Planck}}}{E_{\tiny \textrm{Planck}}}c^2 = \sqrt{c^{-4}} c^2 = 1$ so the dimensionless expression $E_0 = m_0$ is equivalent to the dimensional statement $E = mc^2$ when the definitions of $E_0 = \frac{ E }{ \sqrt{\frac{\hbar c^5}{G}} } ; \quad m_0 = \frac{ m }{ \sqrt{\frac{\hbar c}{G}} }$ are used.

What about units do you think I don't understand?

Yes, and I already explained that common to both special and general relativity is the concept that invariant interval between events is more important than any one observers clock. So a locally ticking universe would break the space-time symmetry of special relativity's Lorentz invariance and general relativity's local Lorentz invariance.

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