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Online TimFox

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Re: Physics Question - ma = mg
« Reply #75 on: June 25, 2021, 10:43:41 pm »
I agree about the confusion in English-language contexts between pound as unit of force or weight, and as unit of mass.  The usual statement on a US food package label is “net wt 1 lb 2 oz” (on a Cheerios box), which in strict physics language would be “mass that weighs 1 lb 2 oz at mean sea level”.  In metric lands, however, I have seen pressure gauges calibrated in “kgf/cm2” instead of N/m2 or Pa or dyn/cm2.
The infamous NASA mission failure a while back involved changing vendors on small rockets and misreading the force units.
By the way, my mnemonic for conversion between N and lb is to remember that 1 kg weighs 9.8 N and 2.2 lb (approximately).
 

Offline RJSV

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Re: Physics Question - ma = mg
« Reply #76 on: June 25, 2021, 11:17:44 pm »
Ok, Tim I think you've supplied some KEY info, to clarify portions of this thread.
   So, if I read right, a definite precise definition and use of 'MOTION' was given by Newton. Fantastic, even tho proves me wrong, somewhat.
   So then, as MOTION was given the precise definition, as what we today might call 'momentum'.
Then, will the units be kg-meters/ seconds (squared) ?
  Plus, maybe the English system direct equiv would be
   Slug-feet / second (squared) ?
Although I realize, maybe the English system avoids that form, and using 'pounds' as an equation substitute is ok as long as a standard is consistent, and even though the language structure isn't strictly correct?
   ?? Did I get that right ?

   As to relativistic effects, after 2 years college courses on CLASSIC Physics, I enrolled in a nuclear physics intro where the instructor described the so- called 'NEUTRINO PARADOX'.
   In this effect, Neutrinos streaming out from the sun would last too long: They were measuring too many reaching sea level, before the very short decay time kicked in. (They decay to mu-mesons).
   The answer came in time dialation where time ticks slower, in the neutrino moving at >90 % light speed.
But, again, I'm saying some problems in this thread are due to injecting relativistic effects too early, into a sort-of 'word salad'  including getting mixed up because MASS does increase (or equiv perceived mass) at relativistic speeds (90% light speed). So, increase in momentum, and now I realize no worries about calling it ' motion of a body', that increase gets all smushed into word salad that reverts to including relativistic mass increase (due to high speed).
WHEW, Thanks Tim, for clarifying your contribution!

But still,.  Nominal Animal for President
 

Online TimFox

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Re: Physics Question - ma = mg
« Reply #77 on: June 25, 2021, 11:20:02 pm »
A couple of side comments about gravitational acceleration g and weight.
1.  Scales.  It was mentioned above that "scales" measure force or weight, not mass.  This is certainly true for spring scales (or modern units with strain gauges) that measure force as an extension of a compliant spring.  However, "balances" that compare the device under test to a standard mass/weight, using leverage as appropriate in a steelyard balance, compare the masses of the DUT and the standard, since the two will balance regardless of the local value of g, but the weight of the standard will change when you take the experiment to the Moon.
2.  Measurement of g.  It is not necessary to know the mass of the object in an experiment to measure the uniform gravitational acceleration in a limited region.  Galileo worked out the period of a simple pendulum around 1602, where the period is independent of the bob mass and the amplitude of the swing (so long as the amplitude is small enough to use the approximation sin(x) = x, the error is small), but depends on the length of the string.  The period T = 2pi x sqrt (L/g) in that limit of small amplitude.  A more precise experimental method is the Kater Pendulum, where the length in the experiment can be better defined.  See  https://en.wikipedia.org/wiki/Kater%27s_pendulum
 

Online TimFox

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Re: Physics Question - ma = mg
« Reply #78 on: June 25, 2021, 11:23:53 pm »
Ok, Tim I think you've supplied some KEY info, to clarify portions of this thread.
   So, if I read right, a definite precise definition and use of 'MOTION' was given by Newton. Fantastic, even tho proves me wrong, somewhat.
   So then, as MOTION was given the precise definition, as what we today might call 'momentum'.
Then, will the units be kg-meters/ seconds (squared) ?
  Plus, maybe the English system direct equiv would be
   Slug-feet / second (squared) ?
Although I realize, maybe the English system avoids that form, and using 'pounds' as an equation substitute is ok as long as a standard is consistent, and even though the language structure isn't strictly correct?
   ?? Did I get that right ?

   As to relativistic effects, after 2 years college courses on CLASSIC Physics, I enrolled in a nuclear physics intro where the instructor described the so- called 'NEUTRINO PARADOX'.
   In this effect, Neutrinos streaming out from the sun would last too long: They were measuring too many reaching sea level, before the very short decay time kicked in. (They decay to mu-mesons).
   The answer came in time dialation where time ticks slower, in the neutrino moving at >90 % light speed.
But, again, I'm saying some problems in this thread are due to injecting relativistic effects too early, into a sort-of 'word salad'  including getting mixed up because MASS does increase (or equiv perceived mass) at relativistic speeds (90% light speed). So, increase in momentum, and now I realize no worries about calling it ' motion of a body', that increase gets all smushed into word salad that reverts to including relativistic mass increase (due to high speed).
WHEW, Thanks Tim, for clarifying your contribution!

But still,.  Nominal Animal for President

In Principia, Newton defined his terms before going into his laws.  In modern nomenclature, momentum (vector) is the product of mass (scalar) and velocity (vector),
P = m v, and has units of kg m/s, not s2.
Force is the product of mass and acceleration, and therefore has the units kg m/s2.  Kinetic energy E = (1/2) m v2, where all terms are scalars, has the units of kg m2/s2.
When you get relativistic, look up the meaning of "rest mass", which is invariant.
« Last Edit: June 25, 2021, 11:28:34 pm by TimFox »
 
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Offline bostonmanTopic starter

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Re: Physics Question - ma = mg
« Reply #79 on: June 27, 2021, 02:38:27 pm »
Quote
It does make sort-of intuitive sense – at least, if you've ever compared the effort needed to do say pull-ups slowly versus fast.  I can do a dozen fast pullups (not the swingy ones – those are cheating via angular momentum –, nor using your legs to kick in the air to get some extra help; just fast but proper ones), but ask me to stay stationary midway, and I'll drop in a few seconds, much faster than it would take to do those pullups.  I think.  Need to verify at the gym (on separate days).

As someone pointed out, this thread has some confusion over people messing terms and units. Hopefully I'm not one of those people as I'm just trying to grasp (what I feel are) basic concepts that confuse me.

I try avoiding the human factor in all this because of the complexity, however, gym topics do help because it's where (at least) I do most "lifting".

Quote
This longish thread has shown how much meat there is in even simple problems.  And how important it is to frame the question properly.  Things which seem obvious to those with long exposure can be either quite confusing or devilishly simple depending on presentation and/or point of view.

I think the "longish thread" is due to me asking the same question, but doing different activities that involve a different part of physics each time. As an example, I started off simply asking why ma=mg based off TBBT. I researched it before asking the question and kept seeing discussions that they are equal because of Earth. This didn't make sense to me because of what I've already stated and now I believe I understand why they are "equal". Then I deviated asking about lifting a box off the floor, and that got moved towards "Work".

As for speed of lifting something, this actually also came from TBBT. Sheldon and the group were moving a time machine up the stairs and (I believe) Howard said why don't we push faster. Sheldon replied stating Work done doesn't matter how fast you push the object.

That got me thinking because if I pick up a box 1m really fast, it feels like I've done far more work than if I lifted it slowly. Now if I lifted the same box 1m really fast ten times, I'd be out of breath whereas ten times slowly and I'd have far more energy at the end.

From what I understand due to this thread, one thing to consider is that I've gave the box kinetic energy picking it up faster because now it needs to transfer that kinetic energy when it stops at 1m than it would be if I picked it up slower. If I understand correctly, it would attribute to me being "out of breath" because I've transferred my kinetic energy to the box and then used energy to stop it at 1m.

It does seem acceleration should factor into some equation because if Ke = 1/2mv^2, then without acceleration, how did the object achieve velocity?

As for doing pullups, I think when discussing the "human factor", too many factors come into play as for getting out of breath and fatigue muscles. If I'm correct, say I use a wench and a motor to lift a box (i.e. no human interaction at all). Excluding friction on the pulleys, wire resistance, etc.. I imagine the Work done by the motor to lift the box would be equal to the Work done by a human to lift the same box the same height.

In other words, the calculation is identical, and I can safely say I burned X calories lifting the box ten times based off converting Joules (i.e. Work) into calories.

Getting back to lifting a box very fast ten times, at a distance of 1m, it's the same Work, however, now I've wasted energy because the box needs to stop at 1m thus the energy being thrown out the window; therefore I'd get tired sooner. On the other hand, if I lifted the box very slow, my arms would begin getting tired.
 

Offline Nominal Animal

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Re: Physics Question - ma = mg
« Reply #80 on: June 27, 2021, 05:50:05 pm »
NOMINAL  ANIMAL for PRESIDENT
I was advised by the reptilians to tell you I politely decline, or else.
 

Offline Nominal Animal

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Re: Physics Question - ma = mg
« Reply #81 on: June 27, 2021, 06:26:14 pm »
As for speed of lifting something, this actually also came from TBBT. Sheldon and the group were moving a time machine up the stairs and (I believe) Howard said why don't we push faster. Sheldon replied stating Work done doesn't matter how fast you push the object.
Sheldon is correct that the time they take does not affect how much energy they need to transfer to the object to push it upwards in Earth's gravity field.

Sheldon is wrong in that the chemical energy required by human bodies to perform such energy transfers, definitely does depend on the rate at which the humans do it.
I don't know what that rate is at all, but provided an example model earlier.  It can be derived from basic principles, how muscle cells contract when given chemical energy, and so on.

So, it is correct – and the core of the joke is – if that one uses "work" strictly according to the physics definition, and completely ignores the human aspect of it.
Which is kinda at the root of all engineer/physicist jokes, I guess.

(I once had to walk out on a Physics 101 lecture, when the professor went "and as you can see, physics isn't as dry as one might think it to be, since 'bar' can be considered both an unit of pressure, as well as the place to relax after a hard days work", and the students dutifully emitted a respectful "laugh".  I love me dad jokes, and the layered nature of that joke (think about it), and loved the prof, but just couldn't handle the herd mentality.  I have hard time with canned laugh tracks, too.)

That got me thinking because if I pick up a box 1m really fast, it feels like I've done far more work than if I lifted it slowly. Now if I lifted the same box 1m really fast ten times, I'd be out of breath whereas ten times slowly and I'd have far more energy at the end.
I'm the opposite: I have ample power, but not much stamina.  (I think those are the terms, not sure.)

Anyway, it is a perfect example of what it means to be efficient at something.

If the object is at rest both in the beginning and in the end, it really does not care how long the lifting takes.  Its own total energy is changed by the exact same amount anyway (the physical work, i.e. m g h, given mass m, acceleration due to gravity g, and the vertical elevation h).

From the human or device causing the energy to be transferred to the object, "work" gives the minimum.  The less efficient we are at it, the more energy we'll waste.
Not using tools to optimize the energy expenditure is like asking friends to help move the object, but tell them to keep one hand behind their back.  Inefficient.

Giving the object more kinetic energy than necessary, or converting it repeatedly into waste heat (say, you lift the object one stair step at a time, lifting the object higher than the next step, then dropping it down), are just ways of doing it inefficiently.

Excluding friction on the pulleys, wire resistance, etc.. I imagine the Work done by the motor to lift the box would be equal to the Work done by a human to lift the same box the same height.
Yes.  It is easier, though, to think of Work from the perspective of the box: when being lifted upwards, measuring from standstill to standstill so ignoring kinetic energy transfers, the height determines exactly how much total energy the box gains.

From the perspective of humans doing the sweaty work, we can either be smart and make sure we spend the minimal energy – and we can do that without any external supplies of energy by using mechanical devices like pulleys –, or we can power through, ignoring the "waste" in chemical energy expenditure.

It is unfortunate that we don't have a separate word to describe the energy difference, since it is confusing.  (Then again, you wouldn't believe some of the physics misconceptions I've observed in graphic artists who have been taught with "interesting" definitions of "gravity", "volume", "depth", et cetera; perfectly valid and very useful in their own domain (in visual/graphic arts), but not at all valid in others, like physics.  Same problem, resulting in utterly hilarious misconceptions.)

In other words, the calculation is identical, and I can safely say I burned X calories lifting the box ten times based off converting Joules (i.e. Work) into calories.
Yes; exactly: because you transferred the energy ten times to the box, did not try to recover any of that energy, you can be sure that by doing it, your body burned at least the amount of chemical energy corresponding to the Work while doing it.
« Last Edit: June 27, 2021, 06:27:46 pm by Nominal Animal »
 

Offline RJSV

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Re: Physics Question - ma = mg
« Reply #82 on: June 27, 2021, 08:43:47 pm »
====.   Since ACCELERATION term is in an equation, therefore there must be movement present... ====

Erroneous because:. Newton started by saying "Let's examine a body in motion ( if I understood the help from Timfox).

   This thread, in 1740, could still encompass all the non-relativistic aspects of our discussion / friendly argument.
Perhaps one my faults is to focus too much on some blatant mis-statements:. But I needed clear examples to quote / make my point.
   Here is my main concern:
   There are equations involving an acceleration constant, where movement changes are described.
BUT, there are other equations where (an acceleration constant) is simply an expression of gravity-related physics, that is the strength of gravity (planet bulk) affects the outcome of a calculation.  You cannot expect an explicit 'change in movement' every time.
   Perhaps it was stupid of me to expect every post to be perfect, (and putting focus on the occasional mis-statements).  The example that bugged me was the post stating that 'MV' increased, therefore MASS increases with velocity increase. Uh, no, it is velocity that got bigger, to cause the parameter 'MV' to increase.
AND injecting relativistic effects into this discussion is just extra confusion, especially when the terms are similar, and often identical. (That being the use of the term 'mass increase'.

   So let me restate that:
    I read that 'MV' increased with speed, therefore the 'mass increased'.
    So sloppy, obviously SLOPPY MATH.
   The I read that at relativistic speeds the apparent mass DOES increase, probably having practical implications.
   BUT, and this is the even more crucial part, one reader takes that and runs back to (my) first complaint, that it's 'V' that caused 'MV' to increase, in a classical physics introspective. (He) takes the relativistic mass increase (valid), and runs back to the classical mode equation, implying that from a practical, everyday (non-relativistic) frame of view, the mass in 'MV' is wildly increased.... O.K. I added 'wildly' etc. to exaggerate my point.
  =========================================
   I recall someone I knew years back who used the term 'Lowest common denominator':. A phrase involving personal interactions among military folks.
   What he meant (I think) is that there is often a choice:
BUT: Do you force highest standards on everyone, or do you 'cave' to the ignorant, misbehaved (and maybe they are incapable of complying anyway.) ?
(Thanks Phil M.)
===========================================
Nom Animal:
   As for presidential bids, I've seen some great minds come out of FINLAND, there. You maybe too smart for that job.
Keep up the good work (helpfully debate).
P.S. I've seen other venues / political meetings where speakers actually had to PAY money, for their time at the microphone. Dave J. perhaps you could 'monitize' a longer thread, charging by the word...
(Just kidding)
 

Offline Nominal Animal

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Re: Physics Question - ma = mg
« Reply #83 on: June 27, 2021, 11:47:04 pm »
The example that bugged me was the post stating that 'MV' increased, therefore MASS increases with velocity increase. Uh, no, it is velocity that got bigger, to cause the parameter 'MV' to increase.
AND injecting relativistic effects into this discussion is just extra confusion, especially when the terms are similar, and often identical. (That being the use of the term 'mass increase'.
It is complicated, because whenever we talk about "bending spacetime", we must talk in relativistic terms; but at the same time, Newtonian physics and relativistic physics do and must agree at the simpler limit (zero velocity, uniform gravitational field, and so on), because system behaviour at simpler limit is measurable and physically verifiable.

No matter how fine the theory, it is only useful if it predicts behaviour physically observed.  That's why many think string theory is "just" math and not physics: it cannot really be used to model something physically measurable, so that its predictions could be compared against physical measurements.  That makes it "not physics" in the sense that physics describes how the universe behaves; maths folks (including string theorists) can work out the details and the tools, and philosophy folks can deal with the "why" part.

For example:

Mass does not bend space-time.  Energy does.

In relativistic physics, "mass" is rarely used.  Instead a quantity called "invariant mass" or "rest mass" is used; this is the mass of the object at rest, as if observed in a frame where the object is not moving.  The invariant mass is invariant, because it is the same quantity no matter what observational frame it is used in.

"Relativistic mass" is the term that best corresponds to what we call "mass" in non-relativistic, or Newtonian, physics.  It includes the space-time bending effects.

Because special relativity says that energy ≡ mass, (mass-energy equivalence, E=mc²), the above can sound like quibbling: I should not say mass does not bend space-time and energy does, because the two are equivalent.
But the above is not quibbling: it is worded so because 'mass' as a term is ambiguous.  In making that statement, I was thinking of the wrong kind of mass, you see.  See?

Simply put, we should avoid claiming mass bends space-time, because term 'mass' is ambiguous; we don't know whether it is used to refer to relativistic mass (which would be correct and match the classical physics definition of mass, but odd because physicists use 'mass' to refer to invariant or rest mass instead), or to invariant or rest mass (which usually 'mass' alone refers to when used in physics).  One interpretation is right but odd, the other is wrong but the more common use for the term.

A case in point: photons.

All electromagnetic radiation from gamma rays to infrared light consists of photons.  Photons are massless bosons: their invariant mass or rest mass is exactly zero (none of that "so tiny we can think of it as zero" stuff; plain and clear zero here), and any number of them can occupy the same state and space (until the energy density is high enough to cause certain interesting stuff to happen).  Bosons are the ones who like company, and fermions –– for example electrons being fermions –– are the ones who occupy their own quantum state, and exclude others from it.

In special relativity, the energy \$E\$ of an elementary particle is defined as
$$E = \sqrt{m_0^2 c^4 + p^2 c^2}$$
where \$m_0\$ is the invariant or rest mass, \$c\$ is the speed of light in vacuum, and \$p\$ is the linear momentum of the particle.  For particles with nonzero invariant mass, the linear momentum \$p\$ is
$$p = \gamma m_0 v = \frac{m_0 v c}{\sqrt{v^2 - c^2}}$$
where \$v\$ is the velocity of the particle, and \$\gamma\$ is the velocity-dependent Lorenz factor; I expanded that to get the nice simple expression on the right side, but note that usually it is kept contracted to \$\gamma\$ for brevity, so the right side expression may look unfamiliar to many.

Note that for small enough velocities, \$p = m_0 v\$.  The correction factor, Lorenz factor \$\gamma = 1 / \sqrt{1 - v^2 / c^2}\$, tells you the error you have for any given velocity if you use Newtonian physics instead of special relativity; see how only the \$p\$ term in total energy has anything to do with particle velocity?
The speed of sound in air is about 330 m/s, so about 0.0000011 of the speed of light.  At that speed, \$\gamma = 1.000000000000605\$, or less than one part in 1,000,000,000,000 over one.  Double-precision floating point numbers – those used for the vast majority of numerical computation on this planet outside financial stuff which uses decimal formats – have barely enough precision to describe that!

Also see how momentum is related to total energy (for elementary particles).  For collections of particles, we'd need to add their internal energy (stored in their structure, their interactions; including angular momentum).  In classical physics, we say kinetic energy \$K = m v^2 / 2\$, but it is important to understand how well that matches the relativistic energy.  If we use \$E(p) = \sqrt{p^2 c^2 + m_0^2 c^4}\$, then \$K = E(p) - E(0)\$.  Expanding the expression for \$K\$ using a Taylor series for small \$v\$ (around \$v = 0\$), we get \$K = m v^2 / 2 + 3 m v^4 / (8 c^2) + \dots\$.  Meaning, even the kinetic energy agrees *exactly* at zero velocity, and for larger velocities, just has relatively small additional terms.  Classical and relativistic physics agree at small velocities even here, as expected.

Photons have zero invariant or rest mass, and for them, linear momentum is
$$p = \frac{h}{\lambda}$$
where \$\lambda\$ is the wavelength of that photon.  Each photon has a single wavelength that only changes when it interacts with other stuff, transferring energy.  It does have other properties like polarization, but those do not affect or contribute to the total energy.  Thus, their total energy is
$$E = \frac{h c}{\lambda}$$

A mind-bending detail is that because the particle velocity depends on the observational frame used for the measurement, so does the total energy of the particle.

To un-bend ones mind, think of the Doppler effect: the wavelength of the light we observe does depend on our own velocity with respect to the light.  We still see it arriving at the speed of light in vacuum, but its energy is shifted.  Because photons, having no rest mass, zip everywhere at the speed of light.  (The velocity only drops when they interact with other stuff; exactly how that happens is quantum mechanics.  That is why we have a constant for it in vacuum, but it drops when zipping through matter.)

In sufficiently small regions, acceleration is indistinguishable from gravity.  Because of this, and the fact that Earths gravity well bends space around it, if we were to observe a single photon at a single moment of time (as measured by the position of that photon zipping along at the speed of light) from the surface of the Earth, and from an orbital space station, they would see the photon having different wavelengths.  Those two are different frames of reference, so there is nothing wrong in their observations differing (by exactly the relative difference of those two reference frames).  That is kinda-sorta one of the basic ideas of relativity.

Also, we have already experimentally proven that even photons themselves do bend space-time.  That tells us invariant mass or rest mass does not bend spacetime, because photons have none (again, not just "close enough to zero for all intents and purposes", but a clear and pure plain real zero), but we have observed them bending space-time.  And relativistic mass is even in principle indistinguishable from the energy, because of Einstein's mass-energy equivalence.  Thus, to avoid misconceptions, it is – I claim! – a good idea to think of *energy* as bending space-time, and relativistic mass being equivalent to energy per Einstein.  As a bonus, we don't have any issues with photons having zero invariant mass, because they are always zipping along at the speed of light (well, duh), and thus always have energy and thus relativistic mass; the zero invariant mass is just not practically relevant other than when working out the details using math.  And as a cherry on top, we easily slide into the habit of assuming 'mass' means 'invariant mass' (or rest mass), and instead of talking about 'relativistic mass' can talk about energy; energy being equivalent to mass, but as a term, not at risk of being confused with something else.

Compare to the thought pretzels one has to twist oneself into, if one insists mass is what bends space-time.  What 'mass', anyway?  No, I think that although you can argue that, that argument is not useful; you need to qualify it with 'invariant' or something else to be precise and not be accidentally misunderstood.
It's the same as when graphics artists talk about 'volume' or 'gravity' as an expression of an emotion evokable by visuals.  If everyone agrees on their definition, and nobody accidentally tries to infer anything about related stuff outside that very small domain, there is no problem.  So there is no technical problem in using those terms.  But, hilarity ensues when a graphics artist, genuinely puzzled, asks a physicist why they don't just use cut pastel tones to reduce gravity.

It is also okay to now realize that the Higgs boson so much talked about a few years ago, has really nothing to do with the bending of space-time, and everything to do with invariant or rest mass.  (Specifically, how "gauge bosons", those that carry the four fundamental forces we know of, have an invariant or rest mass of around 80 GeV/c², while it would be much easier to describe them if they had zero invariant or rest mass.)

See?  As in so many other things, if you just understand the terms correctly in their proper context, things just start falling into place.

They matter much more than just whether or not they are technically correct: they are crucial in building correct understanding in the first place, much beyond technical correctness or minute detail.
« Last Edit: June 27, 2021, 11:53:54 pm by Nominal Animal »
 

Offline T3sl4co1l

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Re: Physics Question - ma = mg
« Reply #84 on: June 28, 2021, 12:46:13 am »
As a visceral example: shine enough light into a mirror box, and it becomes heavier.

Hmm, I forget if this can be understood in more classical terms -- would have to be something about Doppler effects on the radiation pressure, thus manifesting a difference as apparent mass? -- but all that General Relativity requires is that some energy is contained within an object (including rest-mass energy), and that's its effect on spacetime.

The idea of a kugelblitz, is to shine enough light, in from all sides, to a single common point, transmitting enough simultaneous energy to create a black hole.  The equations don't care what form of energy enters, only that enough does to create a singularity in spacetime.  What's inside that singularity?  Mass, energy?  Does it transform?  Moot question -- there's nothing at all at the singularity, it's a, well, singular point. :)  (I mean, as far as we know -- there has to be something quantum mechanical about that.  But it also can't matter because it's beyond the event horizon.)

Tim
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Offline RJSV

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Re: Physics Question - ma = mg
« Reply #85 on: June 28, 2021, 02:08:11 am »
 But STILL:
   I'm saying to the OP 'Boston' question that
                     mg in f=mg
  is NOT movement, the 'g' is called 'acceleration constant' for our case on Earth sized object.
That's a static force equation, a FIELD having potential
to accelerate IF that's a free body to be moved, a la 'Newton'.
   
   The other way; ' f=ma', is for the case when you are pushing something: Your force, again, on a body free to move can cause an accelerated movement. And I believe the physics folks had stated, that the nature of the motion, such as including E= 1/2 mv(squared), cannot be distinguished, as either case has same equations and outcomes.
   That's why I would tend to suggest that 'g' be called an acceleration constant, for specifying a static situation, like weighing your gold bar.
   AND, 'a' is a more general term, signifying how (your) hand pushing a free body out in space in any single isolated case, is going to, YES, actually start moving that sucker, accordingly. You can change that by pushing harder or less hard. And modulate the resulting acceleration alongside.
   ? See any mention, of photons, above?
   ? Rest Mass ?
   ? Year 2021 ?
No, I'm presenting 'Year 1740' arguments, and that approach allows for all the advanced physics that we have today, as well.
 

Offline bostonmanTopic starter

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Re: Physics Question - ma = mg
« Reply #86 on: June 28, 2021, 02:20:18 am »
Quote
That's why I would tend to suggest that 'g' be called an acceleration constant, for specifying a static situation, like weighing your gold bar.
   AND, 'a' is a more general term, signifying how (your) hand pushing a free body out in space in any single isolated case, is going to, YES, actually start moving that sucker, accordingly. You can change that by pushing harder or less hard. And modulate the resulting acceleration alongside.

This is what I initially questioned. I hate to return to the same question, and, in a way, I'm asking rhetorically.

ma can't equal mg all the time. If F=ma, and, to compliment your hand pushing a free body example, the a can be anything. Only in the case of objects (or free bodies) where gravity is acting on it can ma = mg.

Little 'g' should just be called a constant, however, it can't be a constant since I can be standing on a mountain and gravity is less (by a small amount) than someone else standing at sea level.

From what I understood before initiating this thread is that them being equal has little meaning and isn't anything to "brag" about because we live on Earth. Anywhere in the universe ma can equal mg.

Now as for the in depth, informative, and long replies, what I find is that very smart people dislike providing simple answers because much more exists. If I'm pushing a box, I may just consider the pushing factor, whereas, friction should also be included.

Obviously my physics knowledge is low, so I look for the general answers based on simple problems. If I jump right into free bodies, friction, the actual gravity at my height, etc... that will just confuse things causing less focus on the actual question.

I tend to provide long, in depth answers too when my friends ask me simple questions. It's never easy for me to say the sky is blue without explaining that it's a shade of blue because.... blah blah
 

Offline RJSV

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Re: Physics Question - ma = mg
« Reply #87 on: June 28, 2021, 02:34:26 am »
Original Post person: BostonMan:
   That last looked good, but I want to ask, was the original 'g' a variable, such as in saying "The 'g' in the Mars example will be smaller.
OR, is 'g' just a solid constant, for Earth sized cases ?
   Gives even more context.

    Going forward from here, I got no problem, accumulating ever more science 'depth', such as the photon related stuff in posts above. Like the guy said, any innovation in theories must take account what we already know from classical-only.
 

Offline bostonmanTopic starter

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Re: Physics Question - ma = mg
« Reply #88 on: June 28, 2021, 03:01:14 am »
Quote
  That last looked good, but I want to ask, was the original 'g' a variable, such as in saying "The 'g' in the Mars example will be smaller.
OR, is 'g' just a solid constant, for Earth sized cases ?

I assume you meant 'that last post looked good'.

Little 'g' being a constant in the sense that it's understood we are talking about gravity on Earth. Now 9.8m/s^2 is a general rule for physics 101, but, if I dug a five-mile deep hole, gravity would be less because it becomes negligible (???) at the center of Earth. Also, if I stood on a five-mile high mountain, it would be less because I'm further from the surface of Earth.

This means that to argue gravity on Earth is 9.8m/s^2 would be wrong because it can change depending on your height on Earth. Unfortunately, little 'g' can also mean gravity on Mars (or anywhere in the universe) because if I want to know my Weight on mars, I need to use W=mg. To call little 'g' a constant would be wrong, but, I hate to critique TBBT, but I think saying ma = mg is taken out of context.

It's like saying your age is equal to my age, but if your birthday is in March and mine is December, we are the same age for a short period of time. So ma = mg only in cases involving the discussion of Weight on Earth and a free falling object on Earth. Otherwise, a free body may have different acceleration, or, maybe the free body takes place on Mars and now ma = mg thus throwing out 9.8 as little 'g'.

From what I learned about Mass, it's not always the same throughout the universe, however, physics 101 tells me it is. From what I learned from this thread is: mass can change due to small tiny factors; none of which I was aware of.

If anything, I knew gravity is different at different heights and 9.8 is the general number to use for general calculations, but I didn't know mass changed, but, in any case, the fact this was mentioned reminds me to watch how I speak (and type).

My pet peeve is people who argue with statements like 'it's a fact', or 'best ever'. Nothing is a fact unless it can be proven and best ever is just silly.

It's a fact I'm sitting at the keyboard typing, it's not a fact that anyone reading this can argue it's a fact I typed this. I could have had someone type it for me, used speech to text, etc...

Anyway, my point is that whether this thread went much, much deeper, or I (and others) were completely wrong to state things in such a fashion, just reminded me to speak correctly, and, really, it helped me understand how much more exists to Weight, Mass, etc... Now at least I can say 9.8 is good, but know if I were to become a rocket scientists, more exists than just 9.8.



 

Online TimFox

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Re: Physics Question - ma = mg
« Reply #89 on: June 28, 2021, 03:26:00 am »
g is a real acceleration.  Since the gravitational force on a dropped mass is proportional to that mass, the acceleration of a falling body will be independent of its mass. 
Simple algebra: the acceleration of mass m1  is  a = F/m1 = (1/m1) x G m1m2/R2 = G m2/R2, which we define as g.
Here, G is Newton's universal gravitational constant (not g), m1 is the mass of the dropped object, m2 is the other mass (Earth or moon or whatever), and R is the distance between the two objects, specifically between their centers of gravity.  In our case, that is the radius of the Earth or moon.
In a given region where the gravity is uniform, such as a cubic meter at the earth's surface, any object whatsoever, of whatever mass (neglecting air resistance) when dropped freely will be accelerated by an acceleration equal to g.
In another region, such as a cubic meter at the moon's surface, the value of g is different than the value on earth, due to the different mass and radius of the large attractor (moon vs. earth).
I gave the equation for the period of a pendulum, which is independent of the mass, and has g explicitly in the formula:  the period of a given pendulum will therefore be different on the moon or on earth.
There are further details, such as the fact that during our drop experiment the motion of the Earth can be neglected, since it is much heavier.
 

Offline RJSV

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Re: Physics Question - ma = mg
« Reply #90 on: June 28, 2021, 05:28:50 am »
Yup..., BostonMan;
   I think we are in pretty good shape, today. I had wondered if 'g' expressed more explicitly would maybe be; g(subscript Mars) for Martian surface, g(subscript Moon) for moon surface measurements, etc. Thanks for clarification.
   This is all voluntary, remember the eevblog hosts might say "Participate, if you please, but please keep it positive."

   And timfox providing some 'sane' relief.
   And as to veryveryveryvery small extra elements to any parameters under discussion: They are just that, and mostly 'negligable' in everyday live. Now, with modern growth, we know about, and we somewhat know which effects take effect strongly, and when such effects (while still there) can be ignored.
That guidance would maybe fill out the scope, of OP's general question.
    Does ma = mg ? Yes, under some certain circumstances, that being adjusting your applied force, and then additional math, if relativistic conditions dictate, (strong, or very weak).
   It's a volunteer task, blogging, and so makes any frustrations a whole order of magnitude more tolerable!!
 

Offline Nominal Animal

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Re: Physics Question - ma = mg
« Reply #91 on: June 28, 2021, 08:01:44 am »
the 'g' is called 'acceleration constant' for our case on Earth sized object.
Sure; I called it acceleration due to gravity, which I intended to be understood in exact same sense as 'acceleration constant'.

Perhaps this helps:

What is the speed of light, if you are in a dark chamber without a single photon? Does the speed of light still exist?  Yes, of course.

Does the speed of light in vacuum, c, exist outside photons?  Yes, of course.  We now define c as exactly 299,792,458 m/s, using that to derive the exact length of one meter.  (One second is defined as being equal to the time duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the fundamental unperturbed ground-state of the caesium-133 atom.)

Can acceleration exist with zero velocity?  Yes, of course.  Jump up into the air.  At the peak of your trajectory, your velocity is exactly zero, but have about 9.80m/s² of downward acceleration (and that stays pretty much constant from the moment your legs are in fully extended position and no longer touch the ground, until they touch the ground again; as air resistance affects the acceleration a bit).

In calculating forces using formulae like F = m a, one must remember that this does not necessarily involve any movement, because only the sum total of such forces acting on an object are really physically measurable.  Indeed, the box sitting still on the floor is subject to force F1 = m g due to gravity, and to force F2 due to the surface it is sitting on resisting deformation; and from the box being still, we know these forces must cancel out, and are therefore equal but opposite.

If we pick a subset of forces, we can sometimes get very unrealistic ideas about the situation.  For example, one could estimate the gravitational force exerted on the box by the northern hemisphere of the universe, and then claim that the result somehow must be balanced by the static force due to the ground not deforming under the box.  Uh, no, if you pick a silly set of forces, the model itself is then silly, and does not describe reality.

Another way to define force is via an energy potential and the vector gradient operator:
$$\vec{F} = -\nabla V$$
The scalar field \$V\$ is not any kind of energy emanation or all-pervading field that binds us together; it is simply a measure of energy.  On isosurfaces where \$V\$ is constant, you do not need to transfer any energy.  (If we have a frictionless surface here on Earth, in a vacuum chamber so there is no air resistance, an object on that surface would keep their velocity, until they bump into a wall or something.)

For gravitational potential energy between two point-like masses, it is \$V = -G M m / r\$, where \$G\$ is the gravitational constant, \$M\$ and \$m\$ are their masses, and \$r\$ is their distance.  Along the line between those point-like masses, gradient is just partial derivative with respect to \$r\$, and \$F = \partial (G M m / r) / \partial r = - G M m / r^2\$.

Now, if we combine a few of those constants, mainly via \$g = G M / r^2\$, we end up with \$F = m g\$.

What is this magic constant \$g\$?  Why, nothing but the gravitational constant [m³ kg⁻¹ s⁻2] times mass [kg] divided by the squared mean radius of Earth [ m⁻² ].  Dimensional analysis shows that the resulting units are [ m s⁻2 ] = m/s².  Acceleration.  What about the numerical value?  6.674×10⁻¹¹×5.972×10²⁴/(6.371×106)² ≃ 9.82.

Ah-ha!  If we treat gravity as just one form of potential energy, described as \$V = -G M m / r\$ between point-like particles having masses \$M\$ and \$m\$ separated by distance \$r\$, we end up with each of them seeing a force \$F = m g\$ towards their shared center of mass.

So, in this case, \$g\$ has nothing to do with acceleration per se; it is just some constant we derived from Earth's mass and radius, and the gravitational constant \$G = 6.674\cdot10^{-11} \, m^3 \, kg^{-1} \, s^{-2}\$.

The reason it turns out to 9.82 m/s² and not 9.80 m/s², is that because of inertia, and Earth spinning around on its axis, objects near the equator have an inertial force (or fictitious force, as it is caused by inertia and not by any potential energy) away from the axis of rotation of about 0.03 m/s² or so.  This force (again, which is due to inertia, so not caused by any potential energy) is \$F = m \omega^2 r\$, where \$m\$ is the mass of the object rotating, \$r\$ is the radius of rotation, and \$\omega\$ is the angular velocity.  (You can convert \$x\$ RPM to radians per second using \$\omega = 2 \pi x [\text{rad}/\text{s}] / 3600 [\text{RPM}].  \$\omega = 2\pi [\text{rad}/\text{s}]\$ is one turn per second.)

Because the gravity field of the Earth is not consistent with sea level, it is just much simpler to fuzz it all and include that inertial effect in the mess for simplicity (because the other stuff like gravitational "anomalies" – just a word meaning the effects due to Earth being lumpy and fluid inside, not smooth and evenly distributed – mean this approximation is already off by so much that including the inertial effects won't make it any worse).

But, as shown above, just the fact that we have a term that both has the units of acceleration, and in some cases can be physically measured (by observing a weight falling in vacuum), does not mean that actual motion has to be involved.  Just like speed of light in vacuum is a constant, that has uses related to photons and stuff, but is not dependent on there being something zipping along at that velocity.  Or any velocity: remember \$E = m c^2\$.

(I cannot help myself: "But what about electrons orbiting atomic nuclei? Don't they experience centripetal forces?" No, because electrons are delocalized in that fuzzy quantum mechanical way, and not sharp points in orbits around the nuclei like you often see pictured.  The shape, or probability density function, varies depending on the orbital.  Each electron thus delocalized does have a property we call angular momentum, which is deeply involved in how those orbitals get their shapes; and although this angular momentum is split into different parts (spin angular momentum which kinda works out as if the particle was spinning on its axis, but in key aspects isn't; and orbital angular momentum which kinda works as if the delocalized electron had the properties of rotating around the nucleus but doesn't) some confuse the real-world-sounding terms (which even sometimes behave like the properties with those same names in classical, non-quantum mechanics), and end up being the graphics artists I mentioned, asking the physicist why they don't just use pastel tones to reduce gravity.)

It pains me to see intelligent people struggle to comprehend ideas, only because the terms we use are so damn misleading/unintuitive/confusing.

Make no mistake, they are that to me too.  I have a hard time associating any specific "law" or property or constant with just the name of the person who discovered it, because although of historical interest, those names do not help at all in grasping the concepts they represent; you just have to learn them by rote.  And I don't.  At least "invariant mass"/"rest mass" versus "relativistic mass" is descriptive; if they were called "Lorenzian mass" and "Einsteinian mass", I think I'd have to quit physics altogether.
 

Offline RJSV

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Re: Physics Question - ma = mg
« Reply #92 on: June 28, 2021, 08:30:15 am »
That last post a good one. A Nominal animal PHYSICIST beats an EE Everytime.
   Especially the parts explaining about moving or non-moving, in a world having forces and acceleration; framing for the reader the distinctions.
   AND the side-bits on electrons operating in non-intuitive ways, I mean, I had to turn off TV set so I could focus on reading clearly. Nice, and way over my head, but a learning experience.
 

Offline bostonmanTopic starter

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Re: Physics Question - ma = mg
« Reply #93 on: June 28, 2021, 01:15:21 pm »
Without getting too deep with math, or an explanation, how do we know an object has X kg of mass, and gravity (rounding off) is 9.8m/s^2 based on the beginning of time?

We know a 100kg mass is 100kg, but that's based off 9.8m/s^2, but gravity is derived from a Force, and the Force is derived from a known mass and gravity.

If we went back hundreds of years (or a twenty-six-hundred year journey as with TBBT - and I'm not trying to be funny), scientists needed a known gravity to calculate mass; or a known mass to calculate gravity.

I'm sure the math has been worked out several times over hundreds of years, but it seems because each factor is derived from the other, any one can be wrong throwing off the other.

Same with calculating the mass of planets. I believe it's done by looking at how gravity pulls on surrounding stars (?) and thus mass is calculated, however, we needed to start off with the mass of one planet which would be Earth (I should say I assume we started with Earth). If we got the mass of Earth wrong based off using a mass we thought was X kg, then everything derived since is wrong.
 

Online TimFox

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Re: Physics Question - ma = mg
« Reply #94 on: June 28, 2021, 01:51:10 pm »
The mass of an object is completely independent of the local acceleration of gravity g.
The weight of an object is directly proportional to the local acceleration of gravity g.
It is an interesting philosophical question, which requires serious astrophysical investigation, if the universal gravitational constant G has changed over time.
 

Offline CatalinaWOW

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Re: Physics Question - ma = mg
« Reply #95 on: June 28, 2021, 01:55:38 pm »
Without getting too deep with math, or an explanation, how do we know an object has X kg of mass, and gravity (rounding off) is 9.8m/s^2 based on the beginning of time?

We know a 100kg mass is 100kg, but that's based off 9.8m/s^2, but gravity is derived from a Force, and the Force is derived from a known mass and gravity.

If we went back hundreds of years (or a twenty-six-hundred year journey as with TBBT - and I'm not trying to be funny), scientists needed a known gravity to calculate mass; or a known mass to calculate gravity.

I'm sure the math has been worked out several times over hundreds of years, but it seems because each factor is derived from the other, any one can be wrong throwing off the other.

Same with calculating the mass of planets. I believe it's done by looking at how gravity pulls on surrounding stars (?) and thus mass is calculated, however, we needed to start off with the mass of one planet which would be Earth (I should say I assume we started with Earth). If we got the mass of Earth wrong based off using a mass we thought was X kg, then everything derived since is wrong.

While you are sort of conceptually correct, there are literally hundreds of different observations that have to be reconciled and they all converge to the answer we use today.  Including even direct measurements of the gravitational attraction between spheres in a laboratory.  Some truly elegant experimental work to sort out all of the forces and get the precision required.  The problem you are describing is one of the things physicists since Kepler have worked on, even in the last few decades.  Polishing the apple so to speak and adding decimal places to the g constant.
 

Offline bostonmanTopic starter

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Re: Physics Question - ma = mg
« Reply #96 on: June 28, 2021, 02:04:27 pm »
I assumed (to use your term) we've been polishing the apple for years.

At my level, and the level of many on here, I imagine it doesn't make much difference if we use 9.81 or 9.80999, or if a 100kg mass is really 100.00001kg.

It still amazes me that hundreds/thousands of years ago, all these "constants" were discovered without computers, calculators, and just a pencil and paper. If I'm not wrong, many constants were quite close to what we have now.
 

Online TimFox

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Re: Physics Question - ma = mg
« Reply #97 on: June 28, 2021, 02:07:05 pm »
With respect to the mass of planets, Cavendish is credited with "weighing the earth" in 1797 to 1798.
The modern interpretation of his work was to calculate the gravitational constant G by directly measuring the force between two objects in the laboratory, using a torsion balance.  See https://en.wikipedia.org/wiki/Cavendish_experiment
Once G is known, the mass of the earth follows from inserting g (measured at the surface of the earth) and R (radius of the earth, originally calculated by the ancients) into the gravitational force equation.
Orbital mechanics is another interesting field.  When calculating other masses in the Solar System, one needs the dimensions of the orbits.  Early astronomers were able to calculate orbital dimensions in terms of the "astronomical unit" (a.u.), the mean radius of the Earth's orbit, but a precise value for the a.u. waited until radar ranging of the actual distance between the Earth and Venus around 1964.   Since the Sun is continuously losing mass, the a.u. changes slowly with time.
 

Offline David Hess

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Re: Physics Question - ma = mg
« Reply #98 on: June 28, 2021, 02:18:52 pm »
Little 'g' being a constant in the sense that it's understood we are talking about gravity on Earth. Now 9.8m/s^2 is a general rule for physics 101, but, if I dug a five-mile deep hole, gravity would be less because it becomes negligible (???) at the center of Earth. Also, if I stood on a five-mile high mountain, it would be less because I'm further from the surface of Earth.

This means that to argue gravity on Earth is 9.8m/s^2 would be wrong because it can change depending on your height on Earth. Unfortunately, little 'g' can also mean gravity on Mars (or anywhere in the universe) because if I want to know my Weight on mars, I need to use W=mg. To call little 'g' a constant would be wrong, but, I hate to critique TBBT, but I think saying ma = mg is taken out of context.

Commercial load cells are good enough that the variation of surface gravity on Earth is a measurable error term in calibration.
 

Online TimFox

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Re: Physics Question - ma = mg
« Reply #99 on: June 28, 2021, 03:30:39 pm »
Also, inertial navigation systems for guided missiles need to correct for local variations in the gravitational field due to geological effects.
Notes: 
1.  Please stop calling g a physical constant, in the sense that G (universal gravitational constant) is a constant.  Within a limited region where the change in distance to the center of gravity is negligible, g has a constant value (uniform gravitational field), but it changes by a large amount when you go to the moon, and by a small amount when you ascent to the top of Everest.
2.  You do not need to know the mass to determine g, use a pendulum.
 


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