A A better way of talking about time?

[Sagittarius A-Star]

So adding the adjective "measured" is kinda excessive.

You citations from Misner contain the same formulations, for example:

Another quote from Mister etc (also section 2.8):

... Ee and Ea measured by emitter and absorber ...

 

[Yuras]

You citations from Misner contain the same formulations, for example:

Fare enough. I interpret these adjectives as a way to disambiguate quantities instead of a completely new term, but there could be other interpretations. Though in general I have an impression that authors don't distinguish energy-as-the-timelike-component and energy-as-p-dot-u throughout the book.

 

[Sagittarius A-Star]

Fare enough. I interpret these adjectives as a way to disambiguate quantities instead of a completely new term, but there could be other interpretations. Though in general I have an impression that authors don't distinguish energy-as-the-timelike-component and energy-as-p-dot-u throughout the book.

The first is not invariant as a component of a 4-vector and the second is invariant, because this is the inner product of two 4-vectors.

 

[Yuras]

The first is not invariant as a component of a 4-vector and the second is invariant, because this is the inner product of two 4-vectors.

I though we agreed that these are two different definitions. What is your point here?
Let's say we have a particle with 4-momentum $p$ and two observers $A$ and $B$ with 4-velocities $u_a$ and $u_b$. To calculate the energy as measured by the observer $A$ you can transform $p$ to the comoving coordinates of $A$ and take the timelike component, or you can just calculate $p\cdot u_a$. To calculate the energy as measured by the observer $B$ you can transform $p$ to the comoving coordinates of $B$ and take the timelike components, or you can calculate $p\cdot u_b$. In both cases you'll get exactly the same result, right?
I mean, you can say that these are two different things, and the same result is just a coincidence. If that's what you are saying, then OK, that's just a matter of taste I guess.

 

[Sagittarius A-Star]

I though we agreed that these are two different definitions. What is your point here?
Let's say we have a particle with 4-momentum $p$ and two observers $A$ and $B$ with 4-velocities $u_a$ and $u_b$. To calculate the energy as measured by the observer $A$ you can transform $p$ to the comoving coordinates of $A$ and take the timelike component, or you can just calculate $p\cdot u_a$. To calculate the energy as measured by the observer $B$ you can transform $p$ to the comoving coordinates of $B$ and take the timelike components, or you can calculate $p\cdot u_b$. In both cases you'll get exactly the same result, right?

No. If you calculate the first inner product i.e. in frame $A$ then the components of $p$ are different from if you calculate the second inner product i.e. in frame $B$.

 

[Dale]

See for example Misner, Thorne, Wheeler, Gravitation, section 2.8

That is a good example. Note that they do not call it the “energy” but rather “the energy of a photon with 4-momentum p, as measured by an observer with 4-velocity u”.

If it's a physical quantity, then it has to objectively exist.

Then by that definition any gauge potential is non-physical, that despite the fact that gauge symmetries are the fundamental description of the standard model.

 

[Sagittarius A-Star]

Another quote from Misner etc (also section 2.8):

In this Misner example, emitter and absorber measure different photon energies because of the Doppler effect and $E=hf$.

Calculation of the Doppler effect using the LT:
https://www.physicsforums.com/threa...lation-in-a-cesium-clock.1058283/post-6978366

Edit: see also posting #71.

 

[Orodruin]

Then by that definition any gauge potential is non-physical, that despite the fact that gauge symmetries are the fundamental description of the standard model.

Gauge potential without gauge fixing is unphysical. You cannot measure the gauge potential. What is physical is equivalence classes of gauge potentials related by arbitrary gauge transformations. This is precisely because we require gauge invariance.

 

[Yuras]

No. If you calculate the first inner product i.e. in frame $A$ then the components of $p$ are different from if you calculate the second inner product i.e. in frame $B$.

OK, I'm confused. I'll reread my message tomorrow and check for typos just in case.

That is a good example. Note that they do not call it the “energy” but rather “the energy of a photon with 4-momentum p, as measured by an observer with 4-velocity u”.

As I said above, I think the "as measured by" part is here for disambiguation. But if you insist on having this, then OK, I yield. At this point I'm mostly interested in defending against your point that I invented the whole thing or it's something non-standard. I hope you agree that the source is pretty standard, don't you?

Then by that definition any gauge potential is non-physical, that despite the fact that gauge symmetries are the fundamental description of the standard model.

I don't see a contradiction here. I hope you agree that 4-momentum objectively exists, while it's invariant under Lorentz transformations. The same way gauge potential objectively exists, while it's invariant under guage transformation. Both are interesting because they have interesting symmetries, while the timelike component of 4-momentum doesn't have any.

In this Misner example, emitter and absorber measure different photon energies because of the Doppler effect and $E=hf$.

I don't see how it invalidates the statement "No coordinates are needed to describe the fact that a specific emitter emitting a specific photon attributes to it the energy $E_a$" and the rest.

 

[Yuras]

At this point I'm mostly interesting in defending against your point that I invented the whole thing or it's something non-standard. I hope you agree that the source is pretty standard, don't you?

Oops, it probably sounds a bit passive-aggressive, sorry. @Dale It's possible that you already saw it, so instead of editing, I'll clarify it. I want clarity in this question because otherwise people will continue referring to this conversation claiming that I inverted the whole concept.

 

[Sagittarius A-Star]

I don't see how it invalidates the statement "No coordinates are needed to describe the fact that a specific emitter emitting a specific photon attributes to it the energy $E_a$" and the rest.

It doesn't invalidate that statement, but (together with posting #55) this is an example to invalidate the statement in posting #54:

In both cases you'll get exactly the same result, right?

 

[Yuras]

No. If you calculate the first inner product i.e. in frame then the components of are different from if you calculate the second inner product i.e. in frame .

Ah, I see what's going on here. You are answering a different question.
I'm saying that timelike component of $p$ in the reference frame of $A$ is numerically the same as $p\cdot u_a$. The same way for $B$ case. I'm asking: do you agree with this?

 

[Sagittarius A-Star]

do you agree with this?

Yes.

 

[Yuras]

Yes.

Great! So as far as I'm concerned we established that these two definitions of energy agree numerically, so for me they are just two different definitions of the same thing. Of course you are free to differentiate between them if you want.

 

[Sagittarius A-Star]

Great! So as far as I'm concerned we established that these two definitions of energy agree numerically

No. Which two definitions?

 

[Yuras]

No. Which two definitions?

The energy as a timelike coordinate of the 4-momentum and energy as a scalar product of 4-momentum and the observer's 4-velocity.
In #54 I literally applied both definitions relative to observer $A$ and got the same results. Then I applied both definitions relative to observer $B$ and again got the same results.

 

[Sagittarius A-Star]

The energy as a timelike coordinate of the 4-momentum and energy as a scalar product of 4-momentum and the observer's 4-velocity.
In #54 I literally applied both definitions relative to observer $A$ and got the same results. Then I applied both definitions relative to observer $B$ and again got the same results.

OK.

 

[Orodruin]

There are many different concepts of ”energy”. None of them privileged or unique. The most common one in SR being the time component of the 4-momentum in a Minkowski frame (note: saying ”time-like component” is insufficient without the understanding that we are dealing with a Minkowski frame. A 4-momentum may have anything between 0 and 4 time-like components dependent on the basis used)

Another possibility not yet mentioned here is the inner product of 4-momentum with the time-like Killing field of a stationary spacetime. Again, in Minkowski space there are several such and they all boil down to the same inner products as mentioned already - the energy then depending on the time-like Killing field chosen.

 

[Dale]

KE depends on observer, but it doesn't depend on coordinates.

I disagree. Usually observers and coordinates are considered to be synonyms. But insofar as they are distinct, it is the coordinate system that determines the KE, not the observer.

For example, say I am driving in a car observing someone throw a ball. I am free to use coordinates where I am at rest, but I am also free to use coordinates where the ground is at rest. The KE of the ball is different in those two coordinate systems even though I am the same observer.

The same observer will measure they same value for KE in rectangular, spherical, polar or any other coordinates.

This is not true because “any other coordinates” also includes boosts. As I just described the same observer will obtain different versions of the KE of a ball depending on if they are using coordinates where the car is at rest or where the ground is at rest.

At this point I'm mostly interesting in defending against your point that I invented the whole thing or it's something non-standard. I hope you agree that the source is pretty standard, don't you?

I agree that the source is standard (authoritative even), and I agree that you didn’t invent the concept or the formula.

I only claim that calling that quantity simply “energy” is your personal usage. It does not appear that MTW supports your usage as they carefully qualify the term in their descriptions.

Oops, it probably sounds a bit passive-aggressive, sorry

No worries at all. My skin is substantially thicker than that (metaphorically speaking).

I want clarity in this question because otherwise people will continue referring to this conversation claiming that I inverted the whole concept

I agree that the concept is a standard one: the energy measured by some device. It is just your calling it simply “energy” that is non-standard, and the MTW quotes use wording that I prefer.

I think the "as measured by" part is here for disambiguation.

The disambiguation is important for communication. That is a big part of pedagogy.

Similarly, it is important to disambiguate between “proper time” and “coordinate time”.

 

[PeterDonis]

In this Misner example, emitter and absorber measure different photon energies because of the Doppler effect and $E=hf$.

Not in that example, no. That example is the "centrifuge and photon" example, and the whole point of it is that there is no Doppler shift in that particular scenario--the emitter and absorber both measure the same frequency/energy. The example is meant to illustrate how much easier it is to calculate that answer using the geometric, coordinate-free viewpoint.

 

[Sagittarius A-Star]

Not in that example, no. That example is the "centrifuge and photon" example, and the whole point of it is that there is no Doppler shift in that particular scenario--the emitter and absorber both measure the same frequency/energy. The example is meant to illustrate how much easier it is to calculate that answer using the geometric, coordinate-free viewpoint.

OK. I didn't read section 2.8 in the book. I only looked at the short citation from @Yuras in posting #50 and wrongly assumed, that this would refer to a general scenario.

 

[Yuras]

I disagree. Usually observers and coordinates are considered to be synonyms. But insofar as they are distinct, it is the coordinate system that determines the KE, not the observer.

Then I guess we are using different terminology again.

I agree that the source is standard (authoritative even), and I agree that you didn’t invent the concept or the formula.

Thanks!

It is just your calling it simply “energy” that is non-standard, and the MTW quotes use wording that I prefer.

Fare enough. Though I don't think MTW actually distinguish between "energy" and "energy measured by". E.g. in the other quote (see #50) they say "No coordinates are needed to describe the fact that a specific emitter emitting a specific photon attributes to it the energy $E_a$" without any "measured by".
In fact, both definitions give the same numerical answer, so the difference between them is artificial in my opinion.

The disambiguation is important for communication. That is a big part of pedagogy.

I meant that "measured by" part was there to disambiguate between two different values, $E_a$ and $E_b$, not between different definitions of energy. MTW use "energy" and "energy measured by" interchangeably. Cf. proper time and coordinate time, which actually are different things since they have different values, and MTW carefully distinguish between them.

 

[Dale]

In fact, both definitions give the same numerical answer, so the difference between them is artificial in my opinion.

The two concepts* are distinct as they transform differently. They do not give the same numerical value on transformation.

* The two concepts being $p^t=E/c$ and $E=g_{\mu\nu}p^\mu u^\nu$. They both require further disambiguation as the coordinate system needs to be specified for $p^t$ to be meaningful and $u$ needs to be specified for $g_{\mu\nu}p^\mu u^\nu$ to be meaningful

 

[Yuras]

The two concepts are distinct as they transform differently. They do not give the same numerical value on transformation.

We discussed this with @Sagittarius A-Star above, see #54. The transformation rules are irrelevant here. In one case you apply Lorentz transformations while in other case you change the observer's 4-velocity. The result is the same. At least that's my understanding of these concepts, and I'm pretty sure that's how MTW understand them. (In fact I learned these concepts from MTW)

 

[Dale]

We discussed this

I didn’t discuss it, and I don’t agree. Clearly, $p^t$ changes under coordinate transformation and $g_{\mu\nu}p^\mu u^\nu$ does not.

 

[Yuras]

I didn’t discuss it, and I don’t agree. Clearly, $p^t$ changes under coordinate transformation and $g_{\mu\nu}p^\mu u^\nu$ does not.

I never said they transform the same way. Of course they don't, and it's obvious enough to make it not worth discussing. I already explained what I meant by "they give the same number" and why transformation rules are irrelevant here.

To be clear: I'm not denying your right to distinguish these two concepts. I'm just explaining why I find the difference artificial.

 

[Dale]

I already explained what I meant by "they give the same number" and why transformation rules are irrelevant here

Well, the transformation rules are certainly relevant if the claim is that two concepts are not distinct and yet they transform differently.

I'm not denying your right to distinguish these two concepts. I'm just explaining why I find the difference artificial

I think failing to distinguish them is pedagogically dangerous

 

[Orodruin]

I didn’t discuss it, and I don’t agree. Clearly, $p^t$ changes under coordinate transformation and $g_{\mu\nu}p^\mu u^\nu$ does not.

This is somewhat disingenious. Clearly $u\cdot p$ changes when you change $u$. You cannot refer to energy as defined by the time component of 4-momentum without picking a reference frame just as you cannot refer to energy as defined by $u\cdot p$ without defining $u$. The time component of a vector $w$ is quite literally equal to $e_t\cdot w$ where $e_t$ is the timelike vector of the coordinate basis tetrad. Changing coordinates obviously changes $e_t$, which is why it is coordinate dependent ($p$ being a geometric object and as such coordinate independent).

 

[Dale]

This is somewhat disingenious.

Not at all.

Clearly u⋅p changes when you change u.

Yes, clearly. But that is not a coordinate transformation. Changing $u$ and changing coordinates are different operations.

For example, if I am driving in my car and I have some energy measuring device with me then the outcome of measuring the energy of a ball is $g_{\mu\nu}p^\mu u^\nu$ where $p$ is the four-momentum of the ball and $u$ is the four-velocity of the device.

Without changing anything physically, I can arbitrarily use coordinates where I am at rest or where the ground is at rest. Doing so changes $p^t$ but does not change $g_{\mu\nu}p^\mu u^\nu$.

I cannot change $u$ by a coordinate transformation. I could change $u$ by physically accelerating the measurement device or by redefining $u$ to refer to a different physical device. Both of those are different physics, not just different mathematics.

You cannot refer to energy as defined by the time component of 4-momentum without picking a reference frame just as you cannot refer to energy as defined by u⋅p without defining u.

Agreed. And defining $u$ is a different thing from defining coordinates. Both things ($p^t$ and $g_{\mu\nu}p^\mu u^\nu$) require some specification ($t$ and $u$ respectively), but they are specifying different things and those things can be changed independently.

Changing coordinates obviously changes et, which is why it is coordinate dependent (p being a geometric object and as such coordinate independent).

Yes, changing coordinates does not change the coordinate independent geometric object $p$. Similarly, changing coordinates does not change the coordinate independent geometric object $u$ either.

 

[Orodruin]

Yes, clearly. But that is not a coordinate transformation. Changing $u$ and changing coordinates are different operations.

But it is if you use the timelike tetrad vector $e_t$ instead of $u$. The step to identify that with the 4-velocity of the corresponding coordinate frame is trivial.

For example, if I am driving in my car and I have some energy measuring device with me then the outcome of measuring the energy of a ball is $g_{\mu\nu}p^\mu u^\nu$ where $p$ is the four-momentum of the ball and $u$ is the four-velocity of the device.

And if you put the same measuring device on the ground $u$ is still the four-velocity of the device, but it is now at rest in the ground frame. The key point is in identifying $u$ with the timelike tetrad vector $e_t$ as already pointed out.

Without changing anything physically, I can arbitrarily use coordinates where I am at rest or where the ground is at rest. Doing so changes $p^t$ but does not change $g_{\mu\nu}p^\mu u^\nu$.

Because what that device is measuring is not the energy in the frame where you are computing things. It is measuring the energy in the rest frame of the device. If you put the device at rest in the ground frame, it will measure $p^t$. This is as things should be.

I cannot change $u$ by a coordinate transformation. I could change $u$ by physically accelerating the measurement device or by redefining $u$ to refer to a different physical device. Both of those are different physics, not just different mathematics.

You are missing the point. Both of the definitions require an additional reference to another object, be it a 4-velocity or a reference frame. The two are interchangeable as a reference frame will have a 4-velocity for an object at rest in that frame. If I pick a 4-velocity $u$ that defines (up to rotations, but that is irrelevant here) a frame. Your device will then measure the energy in that frame and nothing else. Picking a different 4-velocity will select a different frame and therefore result in a different $p^t$.

Agreed. Defining $u$ is a different thing from defining coordinates.

It is not though (again, up to rotations). Picking $u$ selects a particular frame (by letting $u$ be the timelike tetrad vector) - the rest frame of an object. Picking a particular frame specifies the 4-velocity of all objects at rest in that frame.

Both things require some specification, but they are specifying different things and those things can be changed independently.

Yes, and changing coordinates does not change the coordinate independent object $u$ either.

Of course changing coordinates does not change the 4-velocity of a different coordinate system. That would be absurd. But that is not what is being argued.

 

[PeroK]

I think failing to distinguish them is pedagogically dangerous

There are worse things than that.

 

[Dale]

But it is if you use the timelike tetrad vector et instead of u. The step to identify that with the 4-velocity of the corresponding coordinate frame is trivial.

Now that seems disingenuous to me, claiming that $u$ is a coordinate dependent object. This is disagreeing with all of MTW's statements above that some expression containing $u$ does not depend on the coordinates.

And if you put the same measuring device on the ground u is still the four-velocity of the device, but it is now at rest in the ground frame. The key point is in identifying u with the timelike tetrad vector et as already pointed out.

Putting a device on the ground is not the same as a coordinate transformation. I already covered this above where I said to you "I cannot change $u$ by a coordinate transformation. I could change $u$ by physically accelerating the measurement device or by redefining $u$ to refer to a different physical device. Both of those are different physics, not just different mathematics."

The two are interchangeable as a reference frame

Here is where we disagree. The two are not interchangeable, in my opinion. I don't see how you can both consider them interchangeable and also consider $g_{\mu\nu}p^\mu u^\nu$ to be coordinate independent. If $u$ changes with coordinates then it is not coordinate independent and it is not a tensor.

Picking u selects a particular frame

I disagree. See above regarding the device in the car which can be analyzed from the ground coordinates or from the car coordinates. Picking $u$ selects the measuring device, not the coordinates.

 

[Orodruin]

Here is where we disagree. The two are not interchangeable, in my opinion. I don't see how you can both consider them interchangeable and also consider $g_{\mu\nu}p^\mu u^\nu$ to be coordinate independent. If $u$ changes with coordinates then it is not coordinate independent and it is not a tensor.

The timelike vector of a tetrad is coordinate independent. But it also singles out the time direction of a particular frame. The frame dependence of energy is explicit as the expression $e_t \cdot p$ obviously depends on the tetrad chosen.

 

[Dale]

The frame dependence of energy is explicit as the expression et⋅p obviously depends on the tetrad chosen.

So do you consider $g_{\mu\nu}p^\mu u^\nu$ to be a coordinate independent or a coordinate dependent quantity?

 

[Orodruin]

So do you consider $g_{\mu\nu}p^\mu u^\nu$ to be a coordinate independent or a coordinate dependent quantity?

It is coordinate independent, but so is ”the energy in the frame $S$”. You can compute the latter in $S$ by taking ttd time component, but you don’t need to. You can just as well compute it in any other frame by the appropriate inner product with the timelike tetrad vector $e_t$ from $S$, which is equivalent to taking the time component in $S$ - of multiplying by the 4-velocity of an observer at rest in $S$.

 

[Dale]

It is coordinate independent, but so is ”the energy in the frame $S$”. You can compute the latter in $S$ by taking ttd time component, but you don’t need to. You can just as well compute it in any other frame by the appropriate inner product with the timelike tetrad vector $e_t$ from $S$, which is equivalent to taking the time component in $S$ - of multiplying by the 4-velocity of an observer at rest in $S$.

Ok, so you are making a distinction between the frame and the coordinates? With frame referring to the tetrad $S$ in the above, I assume.

That is fine. I was talking about coordinates, but I don’t object to talking about tetrads. I will use the word “tetrad” rather than “frame” to avoid confusion since sometimes people use the word “frame” to refer to coordinates.

You can only claim that the <whatever> in tetrad $S$ is independent of the coordinates if $S$ is independent of the coordinates. So the statement of mine that you objected to as “disingenuous” is correct. In fact, you agree that $g_{\mu\nu}p^\mu u^\nu$ is coordinate independent and $p^t$ is not. And the former is independent precisely because $u$, often the timelike vector of a tetrad, is independent of the coordinates.

 

[Orodruin]

So the statement of mine that you objected to as “disingenuous” is correct. In fact, you agree that $g_{\mu\nu}p^\mu u^\nu$ is coordinate independent and $p^t$ is not. And the former is independent precisely because $u$, often the timelike vector of a tetrad, is independent of the coordinates.

No, I still consider it somewhat disingenuous. By frame above I refer to a standard Minkowski coordinate system. Such a coordinate system immediately identifies a tetrad of orthonormal basis vectors that each project out the components in those coordinates of any vector.

I agree that the value of the "energy in coordinate system $S$" depends on what you pick for coordinate system $S$. What I do not agree with is that this if fundamentally different from $u\cdot p$ depending on the observer you pick (and hence its 4-velocity $u$).

Just as the "energy as seen by observer A" is frame independent, "energy in coordinate system $S$" is frame independent as well. Both depend on what you plug in for A and $S$, respectively.

 

[Dale]

By frame above I refer to a standard Minkowski coordinate system.

In that case I think that you are not being consistent.

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.

You say that it depends on the coordinate system, that it is frame independent, and that a frame is a coordinate system. If you have some way to make this not flat out a contradiction, then the best that could be said is that it is confusing.

What I do not agree with is that this if fundamentally different from u⋅p depending on the observer you pick (and hence its 4-velocity u).

I think it is different, and I think my approach is preferable because it doesn’t produce the apparent contradiction that yours does.

For me, this is clear and unambiguous. $p^t$ is coordinate dependent, $g_{\mu\nu}p^\mu u^\nu$ is coordinate independent. A change in coordinate system does not imply a change in $u$, nor does a change in $u$ imply a change in coordinate system. The two are not the same (they are not even isomorphic)

 

[Orodruin]

In that case I think that you are not being consistent.

There is nothing inconsistent in that. The tetrad of coordinate system $S$ of course exists also in coordinate system $S’$. ”Energy is S” fixes the coordinate system for which you want the time component of the 4-momentum, but I can still compute that using a different coordinate system. Not by taking the time component in that coordinate system but by computing $e_t\cdot p$ in that coordinate system where $e_t$ is the timelike tetrad vector of S. It will not take the form (1,0,0,0) obviously, but I can still compute it. It will sometimes be preferable to Lorentz transforming the entire 4-momentum to S.

A change in coordinate system does not imply a change in u, nor does a change in u imply a change in coordinate system. The two are not the same (they are not even isomorphic)

I do not agree with this. It is clear that $e_t\cdot p$ depends on which coordinate system you use to define $e_t$. It must as kt is $p^t$ of that coordinate system! The tetrad uniquely defines the coordinate system (up to translations and rotations, under which the energy does not change).

 

[PeterDonis]

they say "No coordinates are needed to describe the fact that a specific emitter emitting a specific photon attributes to it the energy $E_a$" without any "measured by".

You're quibbling over words. "Attributes to it" means "measured" in this context; the emitter "attributes" a specific energy to the photon that's emitted because the emitter measures it to have that energy on emission.

MTW use "energy" and "energy measured by" interchangeably.

Only in contexts where it's clear who is doing the measuring, so there is no ambiguity. But just using the word "energy" without that context, as you have been claiming can be done, is meaningless.

 

[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.