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Old Oct 28th 2017, 04:03 PM   #41
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Another example of a dual use of a term is the term scalar. High school science texts use the term to mean "a number" wherein tensor analysis and relativity the same term refers only to numbers which remain unchanged by a change on frame of reference or change in coordinates.

This caused me a world if headache a while back with an internet nutcase who refused to acknowledge this fact and was completely ignorant of the facts of how its used in tensor analysis and relativity. And that was even after I gave him many independent sources from various textbooks. Yet the nutcase kept on flaming. He even followed me to the relativity newsgroup and tried to convince everyone else that what he thought it meant was the truth and I was/am a crackpot. There's no shortage of those kinds of people in forums across the internet.
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Old Oct 28th 2017, 04:52 PM   #42
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Is energy not then a frame dependent scalar?
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Old Oct 28th 2017, 05:30 PM   #43
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Originally Posted by studiot View Post
Is energy not then a frame dependent scalar?
The terms "frame dependant" and "scalar" contradict each other. By definition of the term scalar a quantity such as energy which is independent of any frame of reference or coordinate system. Something like proper mass is a scalar though.

Let me elaborate that whether something is a scalar or not depends on the class of coordinate systems under consideration. For example; kinetic energy is an invariant under rotation of coordinate axes which makes energy a Cartesian scalar. Energy does change under a Lorentz transformation which means that its not a Lorentz scalar.
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Old Oct 29th 2017, 04:30 PM   #44
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Originally Posted by Pmb View Post
The terms "frame dependant" and "scalar" contradict each other. By definition of the term scalar a quantity such as energy which is independent of any frame of reference or coordinate system. Something like proper mass is a scalar though.

Let me elaborate that whether something is a scalar or not depends on the class of coordinate systems under consideration. For example; kinetic energy is an invariant under rotation of coordinate axes which makes energy a Cartesian scalar. Energy does change under a Lorentz transformation which means that its not a Lorentz scalar.
I think you will find that just as Physicists use a much more restriced definition of vectors than Mathematicians, so too the Maths crowd has a wider definition of scalars than just frame invariants.

In many textbooks you will find somewhere towards the beginning some kind of 'discalimer' which runs along the lines of

For the purposes of this book we will take, scalars/vectors/tensors .../widgets to mean ......

Tensors of rank 0 are scalars and I think it is better to use this term for this purpose.

But invariants are not the most common purpose of scalars.

I like this approach (due to NASA)

Multiplying a vector by a scalar has the effect of changing only the magnitude, but leaving the direction unchanged. The ouptut is another vector.

Multiplying a vector by a vector (cross product) also outputs a vector but with new magnitude and the direction changed to a perpendicular.

To change the direction (and magnitude) to any desired value you need a tensor multiplying a vector.

But there is more. A scalar is a number?

This is also officially a scalar


$\displaystyle \left[ {\begin{array}{*{20}{c}}5 & 0 & 0 \\0 & 5 & 0 \\0 & 0 & 5 \\\end{array}} \right]$

and what about this equation

speed = frequency x wavelength

Are the quantities scalars or what?

In the words of that immortal sage

It's Complicated
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Old Oct 30th 2017, 05:02 PM   #45
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Originally Posted by studiot View Post
I think you will find that just as Physicists use a much more restriced definition of vectors than Mathematicians, so too the Maths crowd has a wider definition of scalars than just frame invariants.
I'm well aware of that since math was one of my majors in college. Note that physicists have a solid understanding of math and some physics textbook authors don't use the term "scalar" to mean invariant.

The irritating part of that idiot was that the subject being discussed was relativity and I was trying to explain how relativists use the term. The flamers problem is arrogance.
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Old Oct 31st 2017, 04:10 PM   #46
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There are other terms which people disagree about. One is the term energy and another the term weight. More often than not people attempt to define energy as the ability to do work. Others, such as myself, hold that its not a term that as of yet has a solid definition. For example, Feynman wrote
It is important to realize that in physics today, we have no knowledge of what energy is. We do not have a picture that energy comes in little blobs of a definite amount. It is not that way. However, there are formulas for calculating some numerical quantity, and we add it all together it gives “28” - always the same number. It is an abstract thing in that it does not tell us the mechanism or the reasons for the various formulas.
There was a paper which explains why "ability to do work" wasn't a good definition of energy. For instance, the defining property of energy is that its a constant (aka integral of motion). "Ability to do work" doesn't even address that aspect. Regarding weight, many textbooks define it as the gravitational force on a body. The American Journal of Physics published two articles on this point. The one by A.P. French explains that using it that way would imply that any body in freefall, such as the astronaut's on board the ISS, aren't weightless. There's an argument to be had where weightless is a frame dependent notion but that's something that belongs in GR rather than Newtonian physics. The other article I mentioned defines weight by the force required to hold a body at rest in a gravitational field, which takes into account the other inertial forces acting on it such as the centrifugal force. There's also some debate about what force is. Newton defined it as [b]F[p/b] = dp/dt. Most sources ignore that and define in terms of the way Euler defined it, i.e. F = ma. And no. These aren't the same things in all cases. It royally fails in relativity whereas [b]F[p/b] = dp/dt never fails. I guess teachers prefer F = ma because they use that in high school course where only algebra is required. Then again, who can say what goes through everyone's minds, huh?
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