A A better way of talking about time?

[Dale]

”Energy is S” fixes the coordinate system for which you want the time component of the 4-momentum

Sure, it fixes the coordinate system, but fixing the coordinate system doesn’t make it independent of the coordinate system.

Similarly, the tension in a spring depends on the length. You can fix the length, but that doesn’t make the tension independent of the length.

When you fix the coordinate system, a quantity that is coordinate dependent becomes well-defined, not coordinate-independent. I think you are confounding well-defined with coordinate-independent.

the value of the "energy in coordinate system S" depends on what you pick for coordinate system S.
…
"energy in coordinate system S" is frame independent as well.

Even after your explanation I don’t think this is good. It is certainly confusing, and unnecessarily so.

I am not completely sure which quantity you are referring to by “energy in coordinate system S”. I think it is $e_t \cdot p$, which depends on $t$.

It is well defined and, as you say, can be calculated in any other frame. But none of that removes the fact that the coordinate $t$ is literally right there in the formula and changes when you change coordinates.

 

[Orodruin]

Sure, it fixes the coordinate system, but fixing the coordinate system doesn’t make it independent of the coordinate system.

I never said ”energy” was coordinate system independent. I said ”energy in coordinate system S” is coordinate independent. The latter fixes the system in which energy is to be found. You can find that energy from computations made ising any other coordinate system and it will still be the same energy.

Similarly, the tension in a spring depends on the length. You can fix the length, but that doesn’t make the tension independent of the length.

Again, I never claimed so.

When you fix the coordinate system, a quantity that is coordinate dependent becomes well-defined, not coordinate-independent. I think you are confounding well-defined with coordinate-independent.

I disagree. If you have a set of basis vectors, the coefficients of any other vector in terms of that basis are scalar quantities. They are literally found by the inmer product of the (dual) basis and the vector. It does not matter whether I compute that inner product using coordinates that are based on the basis or a different coordinate system. However, the coefficients are quite obviously going to depend on the basis you use for expressing the vector.

I am not completely sure which quantity you are referring to by “energy in coordinate system S”. I think it is et⋅p, which depends on t.

Indeed, and t is still a scalar function in any other coordinate system. It just does not have the expression $t = t’$.

It is well defined and, as you say, can be calculated in any other frame. But none of that removes the fact that the coordinate t is literally right there in the formula and changes when you change coordinates.

Yes, t is there, but it is not equal to t’. It is $t = \gamma(t’ + vx’)$. This is just the regular inverse Lorentz transformation.

Obviously $e_t \cdot p$ need not be equal to $e_t’ \cdot p$.

Let us take it back to Euclidean geometry in 2D. Given an orthonormal basis $\{\vec e_i\}$ you can express any vector $\vec v$ as $\vec v = v^1\vec e_1 + v^2 \vec e_2$. The quantities $v^i$ do not depend on whatever coordinates you pick. You can choose to use the Cartesian coordinates based on $\{\vec e_i\}$ or you can use Cartesian coordinates based* on $\{\vec e_i’\}$ (or any other curvilinear or oblique coordinates), the $v^i$ will still be given by $\vec e_i \cdot \vec v$.

* By this we mean that the coordinates of a point $p$ are the $x^i$ in $\vec x = x^i \vec e_i$ with $\vec x$ being the position vector relative to some chosen origin.

 

[Yuras]

You're quibbling over words.

You are missing the whole point. "Mass" and "mass of" are the same physical quantity.

 

[Dale]

I disagree.

Fine, I think we will leave it at this disagreement then. I think you are using “coordinate independent” to mean “well defined”.

I have to say though that I am quite offended at your repeated insinuation that my statements are disingenuous. While I think your wording is confusing and almost guaranteed to produce misunderstandings in students, I wouldn’t attack your character for using it, because I recognize your intentions are good. You should have given me the same courtesy and not implied dishonesty on my part.

 

[Orodruin]

I still disagree with the claim that I am confounding well-defined with invariant. As I have shown, the coefficients of a given basis are expressable in invariant form and it is only when you change the basis - ie, change the coordinate system defining the basis rather that the coordinates you use for calculation (if you even use coordinates) - that the quantity changes.

I am quite offended at your repeated insinuation that my statements are disingenuous

I am sorry to have caused offense. It was not my intention. We will leave it at disagreement.

 

[PeterDonis]

You are missing the whole point. "Mass" and "mass of" are the same physical quantity.

No, they're not. "Mass" is just an unattached word; it has no meaning by itself. "Mass of" has an obvious blank space after it where a specific object whose mass we are interested in gets specified; then it has meaning. You do not appear to grasp this crucial distinction.

 

[Nugatory]

This thread has drifted in a generally unproductive direction so is closed.