A A better way of talking about time?

[Orodruin]

I was totally sure I'll find it on wikipedia, but it's not there... Had to come up with something on the fly, so don't expect too much.
Basically you calculate hyperbolic angle (aka rapidity) between the initial 4-velocity $u_0$ and the current one $u$. Start with their inner product $cosh(w)=u_0\cdot u$ and differentiate it by the proper time:
$$sinh(w)\dot w=u_0 \cdot \dot u = |\dot u| sinh(w)$$
The last step is analogous to $cos(\frac{\pi}{2}+w)=-sin(w)$ (note that $u$ and $\dot u$ are orthogonal). I don't know hyperbolic trigonometry enough to justify it right now, so I'm hand waving here, but in practice you of course just calculate $u_0\cdot\dot u$ in the reference frame of $u$:
$$u_0\cdot\dot u=
\begin{pmatrix}
cosh(w) &
sinh(w) &
0 &
0
\end{pmatrix}
\cdot
\begin{pmatrix}
0 \\
|\dot u| \\
0 \\
0
\end{pmatrix}
$$
It's a bit of a cheating, but the statement itself is coordinate independent anyway.
One way or another, after cancelling $sinh$ we get the following:
$$\dot w = |\dot u|$$
If the (magnitude of the) 4-acceleration is constant, then we just have uniform hyperbolic rotation, i.e. a hyperbola.
Note that this method handles variable acceleration as long as it stays in the same plane as $u_0$ and $u$. For more general case you have to stop before the last step:
$$sinh(w)\dot w=u_0 \cdot \dot u$$
(I really hope I didn't make any silly mistake, and all the above actually makes sense :) )

This seems unnecessarily complicated. The quick way is to note that the world-line of an observer with constant proper acceleration is necessarily a hyperbola because anything else would not have constant curvature (just as a curve of constant curvature in Euclidean space is a circle). We therefore have (up to spacetime translations)
$$
t^2 - x^2 = - 1/a^2
$$
(the RHS must be negative for the hyperbola to be time-like)
It is simple to verify that $a$ is the proper acceleration:

Parametrize the hyperbola using the hyperbolic one*
$$
t = \sinh(as)/a, \quad x = \cosh(as)/a
$$
This gives $dX/ds = (\cosh(as),\sinh(as))$ which has norm 1 and is therefore the 4-velocity $V$ with $s$ being the proper time. It follows that
$$
A = \frac{dV}{ds} = a (\sinh(as), \cosh(as))
$$
which squares to $\alpha^2 = -A^2 = a^2$.

* Here I obviously chose $as$ as the argument of the hyperbolic functions because I know that makes $s$ the proper time. However, it is also easy to make this a reasonable guess or just use a rapidity $\theta$ as the argument. In the latter case it immediately follows that rapidity is $as$ by rescaling to normalize 4-velocity ($\theta = as$ this is the relativistic generalization of $v = at$ for constant proper acceleration).

The bottom line is you don’t have to delve deep into differentiating gamma factors and solving an ODE in $t$ to solve the case of constant proper acceleration.

Edit: $\LaTeX$ versions of the hyperbolic functions exist: \sinh -> $\sinh$ and \cosh -> $\cosh$ etc.

 

[Yuras]

Here are some non-basic situations where a diagram is quite useful.

These are great diagrams!
But "basic" is a subjective term. Your examples are exercises, very useful for studding, while practical problems are too messy to draw, harder to interpret graphically (e.g. 4D) and easier to handle algebraically. That's my limited experience anyway.

No, I do not refer to the metric tensor alone.

Sorry, probably I'm just too tired right now, but I don't see your point. Let's hope someone will jump in and address it.

 

[Yuras]

This seems unnecessarily complicated.

But it's more general since it handles varying acceleration.

The bottom line is you don’t have to delve deep into differentiating gamma factors and solving an ODE in to solve the case of constant proper acceleration.

That's right. But for some reason it's always presented as a deep delve into differentiating gamma factors.
In general coordinate-free derivation can't beat coondinate-based one since you can always convert the former to the latter. Every non-trivial coordinate-free one contains a hidden coordinate frame.

 

[Orodruin]

But for some reason it's always presented as a deep delve into differentiating gamma factors.

"Always" seems like a too strong statement. It is actually a false statement as the way I just presented is how I do it in my SR course. "Usually" or "often" may be more accurate, and mainly in very introductory courses. At the core it is also just geometry used to infer that the shape must be a hyperbola. It is the direct analogy of a circle in Euclidean space. I typically spend the entire first lecture (45'+45') just introducing Minkowski geometry and drawing parallels to Euclidean space.

I also have a very pragmatic reason for not digging deep into differentiating gamma factors: Apart from the fact that it just loses students' attention - I don't want to do it because it is just messy and you're bound to get something wrong at some point.

 

[Yuras]

"Always" seems like a too strong statement.

I'll try to use hyperbola only in its strict mathematical sense from now on.

the way I just presented is how I do it in my SR course

I'm jealous of your students. When I was studying SR we were still recovering from "SR can't handle accelerating observers" thing.

 

[robphy]

These are great diagrams!
But "basic" is a subjective term. Your examples are exercises, very useful for studding, while practical problems are too messy to draw, harder to interpret graphically (e.g. 4D) and easier to handle algebraically. That's my limited experience anyway.

The examples provide explanations, both physical and geometrical.
Some with nice numbers so that one can focus on the physics, and less on the arithmetic.

By "practical problems", do you mean not-necessarily nice numbers?
Sure... once one understands the physics and the underlying math, then one can use the formulaic approaches.
The problem for the beginner is that, in my opinion,
the heavy-formulaic approaches are too detached from the physics and the geometry.

Try doing a euclidean geometry problem or analyzing a block on an incline
algebraically or formulaically alone (that is, without drawing an associated diagram).

 

[Orodruin]

I'm jealous of your students. When I was studying SR we were still recovering from "SR can't handle accelerating observers" thing.

I mean, this is another of these common misconceptions that you still see parroted in popular media and "modern" physics courses ... Every year there will be at least one or two students that have this preconception that needs to be rooted out. Again an issue of lower level courses being taught by people who never delved any deeper into relativity and parrot what they learned themselves in the same type of course.

Then we also have my favourite misconception: When considering muons produced in the atmosphere (and reaching the Earth due to being time dilated in the Earth frame) answering the question "how far did the muon travel in its rest frame" with a non-zero number. Nowadays I tend to ask exactly this question to students every year as it is great for hammering in the concept of relative motion when you follow up by telling them that the correct answer is zero.

Try [...] analyzing a block on an incline
algebraically or formulaically alone (that is, without drawing an associated diagram).

Been there, done that.

 

[robphy]

Try [...] analyzing a block on an incline
algebraically or formulaically alone (that is, without drawing an associated diagram).

Been there, done that.

For beginners? (the context of my suggestion).
Maybe you have some amazing students.

 

[PeterDonis]

Every non-trivial coordinate-free one contains a hidden coordinate frame.

I don't think this is true. For a fairly simple counterexample, see Misner, Thorne, & Wheeler, Section 2.8. For a more extended counterexample, see much of Wald's GR textbook, where Wald's abstract index notation is defined and used; the whole purpose of that notation is to be able to keep track of things like tensor indices (what MTW calls "slots") without having to choose any coordinates.

 

[Herman Trivilino]

The pedagogical disaster, at least as measured by observed confusion in various PF threads (such as the one from which this was split) is not teaching relativity of simultaneity alongside time dilation and length contraction and showing how the three together create a consistent whole.

IIRC there was an article some years ago in The Physics Teacher emphasizing that the relativity of simultaneity should be taught before time dilation or length contraction, and repeatedly emphasized because of the stubborn notion in the students' minds that there is one true time (in other words, that simultaneity is absolute).

 

[Dale]

I understand why you are saying that, but... I came from the geometric camp. In my world coordinates are just labels, they can't be important.

It's actually a 4-scalar :
$$E=g_{\alpha\beta}p^\alpha u^\beta$$
I mean, it obviously depends on the observer via $u$, but proper time does that as well.

I agree that this quantity is a scalar, but I disagree that it is the energy. Energy is the timelike component of the four momentum: $p^\mu=(E/c,\vec p)$. So your $E$ does not transform correctly, although they do coincide specifically for $u^\beta=(c,0,0,0)$.

 

[PeterDonis]

I disagree that it is the energy.

It's the energy of an object with 4-momentum $p^\alpha$ as measured by an observer with 4-velocity $u^\beta$.

The "timelike component of 4-momentum" is the energy measured by the observer whose 4-velocity is that same 4-momentum divided by the observer's rest mass $m$. In other words, $E = g_{\alpha \beta} p^\alpha p^\beta / m$; i.e., $u^\beta = p^\beta / m$.

 

[PeterDonis]

your $E$ does not transform correctly

It transforms as a scalar; but the physical meaning of this scalar has to be carefully specified; just calling it "energy" is not enough. See my post #42.

 

[Dale]

It's the energy of an object with 4-momentum $p^\alpha$ as measured by an observer with 4-velocity $u^\beta$.

Yes, I agree with that description. Not just "energy". Maybe "measured energy" for short, but certainly not just "energy".

 

[Yuras]

I agree that this quantity is a scalar, but I disagree that it is the energy. Energy is the timelike component of the four momentum: $p^\mu=(E/c,\vec p)$.

It's a question of terminology to some extend, though not all terminology is equally good.
Your definition doesn't immediately work in curvilinear coordinates or in GR. You need to amend the definition with e.g. some kind of normal coordinates. While my definition works as is. Though it's purely technical issue.
More important in my opinion is that your definition denies energy its objective existence. Physics doesn't depend on particular choice of coordinates, moreover it should work even without any coordinates at all. Does energy not exist until I introduce a coordinate system?
The third point is that your definition forces one to go to a specific coordinates to calculate energy (at least to do the "by definition" calculations), while I can do that in any coordinates, including curvilinear, and get exactly the same result.

So your $E$ does not transform correctly, although they do coincide specifically for $u^\beta=(c,0,0,0)$.

I prefer to think about it differently. "My" $E$ is the actual physical definition of energy, and the timelike component of the four momentum coincides with energy in the observer's frame. So it's the timelike component who doesn't transform correctly.

 

[Yuras]

Yes, I agree with that description. Not just "energy". Maybe "measured energy" for short, but certainly not just "energy".

I believe that true physical quantities are things you can directly measure. So adding the adjective "measured" is kinda excessive. Though I understand what you mean.

 

[robphy]

It's the energy of an object with 4-momentum $p^\alpha$ as measured by an observer with 4-velocity $u^\beta$.

The "timelike component of 4-momentum" is the energy measured by the observer whose 4-velocity is that same 4-momentum divided by the observer's rest mass $m$. In other words, $E = g_{\alpha \beta} p^\alpha p^\beta / m$; i.e., $u^\beta = p^\beta / m$.

I think this should read $E = g_{\alpha \beta} p^\alpha q^\beta / m_q$; (the particle-energy measured by a measurer)
where $q^\beta$ is the 4-momentum of the measurer
so that $u^\beta = \left( q^\beta / m_q \right)$ (the 4-velocity of the measurer, like a "unit-vector" tangent to the measurer-worldline)
because $m_p^2=g_{\alpha \beta} p^\alpha p^\beta$
is the essentially the square of the particle's invariant-mass.

Yes, I agree with that description. Not just "energy". Maybe "measured energy" for short, but certainly not just "energy".

I believe that true physical quantities are things you can directly measure. So adding the adjective "measured" is kinda excessive. Though I understand what you mean.

$g_{\alpha \beta} p^\alpha p^\beta / m_p$ can be called "the mass of the particle" (or its-energy equivalent)

$g_{\alpha \beta} p^\alpha q^\beta / m_q$ can be called "the energy of the particle measured-by-$u^\beta$" (something clearly indicating "measured-by-$u^\beta$" is needed)

 

[Dale]

It's a question of terminology to some extend, though not all terminology is equally good.

It is completely a question of terminology. While I agree that not all terminology is equally good, far worse than poor terminology is using a common term in a different way than the community does. The terminology you are using is certainly non-standard (which should be supported with references to the appropriate source), perhaps even personal (which should be avoided here entirely).

your definition denies energy its objective existence. Physics doesn't depend on particular choice of coordinates

Energy does not have an objective existence, and it does depend on the particular choice of coordinates. This should be as obvious as possible since KE is part of energy and KE depends on the choice of coordinates.

"My" E is the actual physical definition of energy

Please provide a reference for this. If you even have such a reference, it is by far not the common usage of the term.

I believe that true physical quantities are things you can directly measure. So adding the adjective "measured" is kinda excessive. Though I understand what you mean.

It is a different quantity so they need different terms. There are plenty of things in physics that are not directly measured.

 

[Yuras]

Please provide a reference for this. If you even have such a reference, it is by far not the common usage of the term.

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

One must have a variety of coordinate-free contacts between theory and experiment in order
to use the geometric viewpoint. One such contact is the equation E = — p • u for the energy
of a photon with 4-momentum p, as measured by an observer with 4-velocity u.

The geometric definition of energy is used multiple times throughout the section (and throughout the whole book IIRC.)

The terminology you are using is certainly non-standard (which should be supported with references to the appropriate source), perhaps even personal (which should be avoided here entirely).

I think you are missing the context here. The whole thread is about how the "standard" way to present SR is, in my opinion, pedagogically questionable. Yes, almost everywhere energy is defined as the timelike component, but the geometric one is non unprecedented as well. I'm arguing that the latter is much better pedagogically. And to be honest, I don't quite understand why you think these are separate things.

Energy does not have an objective existence, and it does depend on the particular choice of coordinates.

I respectfully disagree. If it's a physical quantity, then it has to objectively exist. See section 2.2. of Misner etc for justification.

This should be as obvious as possible since KE is part of energy and KE depends on the choice of coordinates.

KE depends on observer, but it doesn't depend on coordinates. The same observer will measure they same value for KE in rectangular, spherical, polar or any other coordinates. The whole notion of "measurement" doesn't not depend on coordinates in the first place.
It's like saying that proper time depends on coordinates because it's an integral of $dt$ in the normal coordinates.

 

[Yuras]

Another quote from Misner etc (also section 2.8):

Equally coordinate-free are the photon energies $E_e$ and $E_a$ measured by emitter and absorber. No coordinates are needed to describe the fact that a specific emitter emitting a specific photon attributes to it the energy $E_e$; and no coordinates are required in the geometric formula
$$E=-p\cdot u$$

 

[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.