I Angular momentum of an atom within a rigid body in motion

james fairclear
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Considering an atom within a rigid body, does the angular momentum of an electron within the atom vary when the body is put in motion?
Considering an atom within a rigid body, does the angular momentum of an electron within the atom vary when the body is put in motion?

My intuition is that, whether considered in a classical sense or quantum sense, the speed of a given electron in its motion within an atom will be constant and invariant (like c) regardless of whether the rigid body is stationary on Earth or moving away from Earth at 0.99c.
 
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james fairclear said:
Summary:: Considering an atom within a rigid body, does the angular momentum of an electron within the atom vary when the body is put in motion?

Considering an atom within a rigid body, does the angular momentum of an electron within the atom vary when the body is put in motion?

My intuition is that, whether considered in a classical sense or quantum sense, the speed of a given electron in its motion within an atom will be constant and invariant (like c) regardless of whether the rigid body is stationary on Earth or moving away from Earth at 0.99c.
The state of the electron in an atom is usually described relative to the nucleus. If you consider a system where the nucleus is not at rest, then the state of the electron must include an element relating to the state of motion of the nucleus.
 
james fairclear said:
the speed of a given electron in its motion within an atom
Is meaningless since the electron in an atom is not in a state that has a definite speed. The electron has orbital angular momentum but you cannot think of it in a classical sense as being due to a well-defined orbital speed about the nucleus.
 
Thank you for your response.

Does the quantity of the angular momentum of an electron within the atom vary when the body is put in motion?
 
james fairclear said:
Does the quantity of the angular momentum of an electron within the atom vary when the body is put in motion?
It depends on what you mean by "the quantity of the angular momentum of an electron within the atom".

In the usual formulation of non-relativistic QM, the operator referred to by "angular momentum" (or more precisely "orbital angular momentum") tells you the electron's orbital angular momentum relative to the nucleus, so it doesn't depend on the state of motion of the nucleus.

I suppose one could construct a different operator that would include the motion of the nucleus and refer the orbital angular momentum to some chosen point in space relative to which the nucleus is moving, but I've never seen that done.
 
PeroK said:
If you consider a system where the nucleus is not at rest, then the state of the electron must include an element relating to the state of motion of the nucleus.
Can you give a reference where this is actually done? I've never seen this done with respect to individual electrons in an atom, only with respect to entire atoms.
 
In non-relativistic physics it's easy to take into account the motion of the nucleus. For the hydrogen atom you can even solve the energy eigenvalue problem analytically. You just work in center-of-math and relative coordinates as in classical mechanics. This can be found in most introductory QM-textbook.
 
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vanhees71 said:
You just work in center-of-mass and relative coordinates as in classical mechanics.
Yes, and this is equivalent to evaluating the orbital angular momentum of an electron in an atom relative to the nucleus of the atom, regardless of the state of motion of the nucleus.
 
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Nearly! The cm-motion is of coarse that of a free particle. The relative motion is equivalent to the motion of a particle of the reduced mass in the Coulomb interaction potential. Rewriting the solution in terms of the original p- and e-coordinates, you'll find that they are in entangled states. Tomorrow I'll look for a nice AJP paper about this.
 
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  • #10
PeterDonis said:
Can you give a reference where this is actually done? I've never seen this done with respect to individual electrons in an atom, only with respect to entire atoms.
I was referring to what @vanhees71 posted above. That the electron wave function must depend on the motion of the atom.

That said, I may have misunderstood what the OP meant by "invariant".
 
  • #11
PeroK said:
the electron wave function must depend on the motion of the atom.
I'm not sure that's what @vanhees71 was saying. Hopefully he'll be able to post a reference to a paper that will have the actual math.
 
  • #12
Let's take the hydrogen atom in its most simple form, i.e., non-relativistic, no spin. Its Hamiltonian is that of a proton and an electron interacting via the Coulomb potential (in the following I leave out the hats for operators; all observables are understood to be represented by self-adjoint operators, and I use Heaviside-Lorentz units for the Coulomb potential):
$$H=\frac{1}{2m_{\text{p}}} \vec{p}_1^2 + \frac{1}{2m_{\text{e}}} \vec{p}_2^2 -\frac{e^2}{4 \pi |\vec{x}_1-\vec{x}_2|}.$$
We are looking for the energy eigenvalues and eigenfunctions. For that it's always good to use all the symmetries to find a complete set of independent compatible observables, i.e., the symmetries of the problem.

Here we can borrow from the classical analogue of the problem. We have a closed system of two particles interacting via a central interaction potential. Thus the full Galileo group is a symmetry. From this it's clear that it is convenient to use the center-of-mass coordinates ##\vec{R}## and relative coordinates ##\vec{r}##
$$\vec{R}=\frac{1}{M} (m_{\text{p}} \vec{x}_1 + m_{\text{e}} \vec{x}_2), \quad \vec{r}=\vec{r}_2-\vec{r}_1$$
with ##M=m_1+m_2##.

Now we need the canonical momenta to these new position vectors. That's determined by finding ##\vec{P}## and ##\vec{p}## such that
$$[R_i,P_j]=\mathrm{i} \hbar \delta_{ij}, \quad [r_i,p_j]=\mathrm{i} \hbar \delta_{ij}, \quad [r_i,P_j]=[R_i,p_j]=0.$$
It's easy to see that from these commutation relations one has uniquely
$$\vec{P}=\vec{p}_1+\vec{p}_2, \quad \vec{p}=\frac{1}{M}(m_1 \vec{p}_2-m_2 \vec{p}_1).$$
In the new variables the Hamiltonian reads
$$H=\frac{1}{2M} \vec{P}^2+ \frac{1}{2 \mu} \vec{p}^2 -\frac{e^2}{4 \pi |\vec{r}|}.$$
Here ##\mu=m_1 m_2/M## is the reduced mass. Obviously a complete compatible set of observables is
$$\vec{P}, \quad H_{\text{rel}}, \quad \vec{L}_{\text{rel}}^2, \quad L_{\text{rel}3}$$
with
$$H_{\text{rel}}=\frac{1}{2 \mu} \vec{p}^2 -\frac{e^2}{4 \pi |\vec{r}|}, \quad \vec{L}_{\text{rel}}=\vec{r} \times \vec{p}.$$
It's clear that the diagonalization of ##H_{\text{rel}}## works as with the approximation, where the proton is just approximated as "infinitely heavy", i.e., sitting at rest in the origin and just providing an external Coulomb potential for the electron. Here ##H_{\text{rel}}## describes the motion of a quasi-particle of mass ##\mu## in such an external Coulomb potential.

The rest is nicely discussed in the following AJP article by Tommasini et al

https://arxiv.org/abs/quant-ph/9709052

It nicely discusses the energy eigenstates as entangled states between electron and proton, while it's a product state of the free center-mass motion and the relative motion.
 
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  • #13
PeroK said:
The state of the electron in an atom is usually described relative to the nucleus. If you consider a system where the nucleus is not at rest, then the state of the electron must include an element relating to the state of motion of the nucleus.
On the basis that the atom (in my understanding) is bound within a rigid body in uniform translatory motion why would you consider the motion of the electron separately to the motion of the nucleus ?

Would you expect the value for angular momentum of the electron to change?
 
  • #14
james fairclear said:
On the basis that the atom (in my understanding) is bound within a rigid body in uniform translatory motion why would you consider the motion of the electron separately to the motion of the nucleus ?

Would you expect the value for angular momentum of the electron to change?
A rigid body in uniform translatory motion is a fundamentally classical concept. Whereas, electrons bound to an atom requires a QM model.

Ultimately, yes, as basic QM is non-relativistic. If you consider an atom moving at relativistic speed then you would need to update your atomic model. For a free electron, for example, you must replace the non-relativistic Schrodinger equation with the Dirac equation.
 
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  • #15
james fairclear said:
On the basis that the atom (in my understanding) is bound within a rigid body in uniform translatory motion why would you consider the motion of the electron separately to the motion of the nucleus ?

Would you expect the value for angular momentum of the electron to change?
If you want to understand the atom as a bound state of an atomic nucleus and electrons, the model of a rigid body is not sufficient but you have to describe the nucleus and the electrons as a dynamical system of interacting charged particles. The atom as a whole is described by its center of mass, which indeed moves uniformly (as long as you can neglect the interactions of the atom with any other particles around it). You can then choose the rest frame of the center of mass and consider the much more complicated dynamical quantum mechanical problem. Here the energy eigenvalue problem is important, leading to a description of the atom in terms of discrete-energy levels (bound states), describing the static configurations of the electrons around the nucleus. One observable conclusion is the prediction of the characteristic spectrum of the light emitted by the atom.
 
  • #16
vanhees71 said:
Let's take the hydrogen atom in its most simple form, i.e., non-relativistic, no spin. Its Hamiltonian is that of a proton and an electron interacting via the Coulomb potential (in the following I leave out the hats for operators; all observables are understood to be represented by self-adjoint operators, and I use Heaviside-Lorentz units for the Coulomb potential):
$$H=\frac{1}{2m_{\text{p}}} \vec{p}_1^2 + \frac{1}{2m_{\text{e}}} \vec{p}_2^2 -\frac{e^2}{4 \pi |\vec{x}_1-\vec{x}_2|}.$$
We are looking for the energy eigenvalues and eigenfunctions. For that it's always good to use all the symmetries to find a complete set of independent compatible observables, i.e., the symmetries of the problem.

Here we can borrow from the classical analogue of the problem. We have a closed system of two particles interacting via a central interaction potential. Thus the full Galileo group is a symmetry. From this it's clear that it is convenient to use the center-of-mass coordinates ##\vec{R}## and relative coordinates ##\vec{r}##
$$\vec{R}=\frac{1}{M} (m_{\text{p}} \vec{x}_1 + m_{\text{e}} \vec{x}_2), \quad \vec{r}=\vec{r}_2-\vec{r}_1$$
with ##M=m_1+m_2##.

Now we need the canonical momenta to these new position vectors. That's determined by finding ##\vec{P}## and ##\vec{p}## such that
$$[R_i,P_j]=\mathrm{i} \hbar \delta_{ij}, \quad [r_i,p_j]=\mathrm{i} \hbar \delta_{ij}, \quad [r_i,P_j]=[R_i,p_j]=0.$$
It's easy to see that from these commutation relations one has uniquely
$$\vec{P}=\vec{p}_1+\vec{p}_2, \quad \vec{p}=\frac{1}{M}(m_1 \vec{p}_2-m_2 \vec{p}_1).$$
In the new variables the Hamiltonian reads
$$H=\frac{1}{2M} \vec{P}^2+ \frac{1}{2 \mu} \vec{p}^2 -\frac{e^2}{4 \pi |\vec{r}|}.$$
Here ##\mu=m_1 m_2/M## is the reduced mass. Obviously a complete compatible set of observables is
$$\vec{P}, \quad H_{\text{rel}}, \quad \vec{L}_{\text{rel}}^2, \quad L_{\text{rel}3}$$
with
$$H_{\text{rel}}=\frac{1}{2 \mu} \vec{p}^2 -\frac{e^2}{4 \pi |\vec{r}|}, \quad \vec{L}_{\text{rel}}=\vec{r} \times \vec{p}.$$
It's clear that the diagonalization of ##H_{\text{rel}}## works as with the approximation, where the proton is just approximated as "infinitely heavy", i.e., sitting at rest in the origin and just providing an external Coulomb potential for the electron. Here ##H_{\text{rel}}## describes the motion of a quasi-particle of mass ##\mu## in such an external Coulomb potential.

The rest is nicely discussed in the following AJP article by Tommasini et al

https://arxiv.org/abs/quant-ph/9709052

It nicely discusses the energy eigenstates as entangled states between electron and proton, while it's a product state of the free center-mass motion and the relative motion.
The article by Tommasini seems to only deal with a stationary system. I am interested in whether the angular momentum of an electron within an individual atom changes when that atom is in motion.
 
  • #17
PeroK said:
A rigid body in uniform translatory motion is a fundamentally classical concept. Whereas, electrons bound to an atom requires a QM model.

Ultimately, yes, as basic QM is non-relativistic. If you consider an atom moving at relativistic speed then you would need to update your atomic model. For a free electron, for example, you must replace the non-relativistic Schrodinger equation with the Dirac equation.

What would be the QM equivalent descriptive term for "A rigid body in uniform translatory motion"?

So in consideration of a rigid body moving at relativistic speed would you expect the angular momentum of an electron within an atom of that body to vary?
 
  • #18
An example for a system which are described approximately as a rigid body in quantum mechanics are the rotational states molecules, but this holds of course only as long as the involved excitation energies are low enough not to excite vibrational modes. For a nice treatment of the quantized spinning top, see Landau and Lifshitz vol. 3.
 
  • #19
james fairclear said:
What would be the QM equivalent descriptive term for "A rigid body in uniform translatory motion"?
Rigid body is problematic (see below). The nearest to uniform translatory motion for a particle would be a minimum uncertainty Gaussian wave-packet, probably.

james fairclear said:
So in consideration of a rigid body moving at relativistic speed would you expect the angular momentum of an electron within an atom of that body to vary?
A rigid body is not compatible with relativity either. There's a thread about this at the moment.

https://www.physicsforums.com/threads/the-paradox-of-relativity-length-contraction.1010138/

What you have now is a mixture of concepts from three different physical theories: Classical Mechanics, Special Relativity and (non-relativistic) QM. That makes your question fairly meaningless.
 
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  • #20
vanhees71 said:
If you want to understand the atom as a bound state of an atomic nucleus and electrons, the model of a rigid body is not sufficient but you have to describe the nucleus and the electrons as a dynamical system of interacting charged particles. The atom as a whole is described by its center of mass, which indeed moves uniformly (as long as you can neglect the interactions of the atom with any other particles around it). You can then choose the rest frame of the center of mass and consider the much more complicated dynamical quantum mechanical problem. Here the energy eigenvalue problem is important, leading to a description of the atom in terms of discrete-energy levels (bound states), describing the static configurations of the electrons around the nucleus. One observable conclusion is the prediction of the characteristic spectrum of the light emitted by the atom.
PeroK said:
A rigid body in uniform translatory motion is a fundamentally classical concept. Whereas, electrons bound to an atom requires a QM model.

Ultimately, yes, as basic QM is non-relativistic. If you consider an atom moving at relativistic speed then you would need to update your atomic model. For a free electron, for example, you must replace the non-relativistic Schrodinger equation with the Dirac equation.
So considering a rigid body in motion at 0.9c can we envisage in a classical sense the electrons of a constituent atom moving at a speed close to c but taking the usual (stationary body) time period per orbit of the nucleus due to the extra distance they are traveling through space?
 
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  • #21
james fairclear said:
can we envisage in a classical sense the electrons of a constituent atom
No. Electrons in atoms are not classical. They do not have classical trajectories.
 
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  • #22
PeterDonis said:
No. Electrons in atoms are not classical. They do not have classical trajectories.
If we consider a rigid body firstly at rest. Within a constituent atom of the body we can consider the probability of an electron cloud and its effect being in a given location within the vicinity of the nucleus.

If we then consider the body in motion at 0.9c. Since a constituent atom of the body will also have a component of motion of 0.9c in the direction of motion of the body and will therefore have further to travel through space than when the body was at rest can we expect a lesser probability of an electron cloud and its effect being in a given location within the vicinity of the nucleus or will it be the same?
 
  • #23
PeroK said:
A rigid body in uniform translatory motion is a fundamentally classical concept. Whereas, electrons bound to an atom requires a QM model.

Ultimately, yes, as basic QM is non-relativistic. If you consider an atom moving at relativistic speed then you would need to update your atomic model. For a free electron, for example, you must replace the non-relativistic Schrodinger equation with the Dirac equation.
Comparing a body in motion with the same body at rest can we expect a lesser probability of an electron cloud and its effect being in a given location within the vicinity of the nucleus or will it be the same?
 
  • #24
james fairclear said:
If we then consider the body in motion at 0.9c.
Motion is relative. The state you called "body at rest" is in motion at 0.9c in some frame. There is no way to detect "motion" in this sense by any change in physical parameters. So it will have no effect on an atom's electron cloud. QM still has to obey the principle of relativity.
 
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  • #25
To be more specific. Consider 2 identical stationary synchronised clocks A and B at a given location on earth. Clock B is set in motion relative to clock A in an Eastwardly direction at 0.9c. There are 2 physical parameters that are now different in clock B compared to Clock A:

1. Its mass has increased
2. Its tick rate has decreased

Considering any constituent atom of clock B can we expect a lesser probability of an associated electron cloud and its effect being in a given location within the vicinity of the nucleus or will it be the same?
 
  • #26
james fairclear said:
1. Its mass has increased

No it did not. Relativistic mass is not used in modern physics.
 
  • #27
james fairclear said:
There are 2 physical parameters that are now different in clock B compared to Clock A:

1. Its mass has increased
2. Its tick rate has decreased
Both these assertions are false. There is no physical difference associated with relative velocity. Clocks A and B remain physically identical. This is essentially the first postulate of Special Relativity: that there is no such thing as a state of absolute motion.
 
  • #28
weirdoguy said:
No it did not. Relativistic mass is not used in modern physics.
My understanding is that particles with mass cannot be accelerated to c due to their mass increasing exponentially. Is there now a different explanation?
 
  • #29
PeroK said:
Both these assertions are false. There is no physical difference associated with relative velocity. Clocks A and B remain physically identical. This is essentially the first postulate of Special Relativity: that there is no such thing as a state of absolute motion.
But surely if the mass doesn't increase then it will be possible to accelerate the clock to a speed > c?

In the Hafele Keating experiment the clock moving eastwards indicated an earlier time than the stationary ground clock, so surely it must have been ticking at a slower rate whilst in motion?
 
  • #30
james fairclear said:
My understanding is that particles with mass cannot be accelerated to c due to their mass increasing exponentially. Is there now a different explanation?
It's the kinetic energy that increases exponentially. Kinetic energy is frame-dependent (as it always has been), and an increase in kinetic energy does not represent a "physical" change.

The idea of ascribing increasing KE to increasing mass is not very useful. Especially as it gives the impression that the particle is physically changing, which it is not doing.
 
  • #31
james fairclear said:
In the Hafele Keating experiment the clock moving eastwards indicated an earlier time than the stationary ground clock, so surely it must have been ticking at a slower rate whilst in motion?
Not necessarily.

There are several factors in the H-K experiment. The ground clock is only stationary relative to the surface of the Earth. There is no sense in which the ground clock is absolutely stationary. In fact, the westbound clock recorded more time than the ground clock.

Ultimately, an attempt to explain the differences in times due to the concept of absolute states of motion does not work. And, contradicts the principle of relativity.

Instead, all clocks tick physically at the same rate, but they take different paths through spacetime. The clock that takes the shortest path through spacetime records the least time (Eastbound); the ground clock takes a longer path; and the westbound clock takes the longest path.
 
  • #32
PS One problem with the idea that "this clock is really moving and is physically ticking slowly" is that we can always change reference frames. Let's take an example.

For simplicity let's assume that the Earth is not spinning on its axis. Clock A remains at rest on the surface of the Earth and clock B moves round the world. Clock B will show less time than clock A when they are back together.

Now, let's take a frame of reference where the Sun is at rest and the Earth is orbitting the Sun. Clock A is moving on this orbit. And, if clock B moves in the opposite direction to the orbit, then initially clock B slows down! In this frame of reference, it's clock A that is initially "ticking slower than clock B". This contradicts the idea that clock B is really, physically ticking slower than clock A. Note that when clock B gets to the far side of the Earth it must move faster than clock A (in the Sun's frame) and when it gets back to clock A the net result is that it shows less time.

This shows that explanations based on the notion of absolute motion cannot be correct. Instead, the two clocks took different paths through spacetime, with clock B taking the shorter path overall. The lengths of the spacetime paths are invariant (the same in every reference frame) and that is a much better explanation. In other words, it's not that clock A or B ticked slower, but that less time passed for one than for the other.
 
  • #34
james fairclear said:
surely if the mass doesn't increase then it will be possible to accelerate the clock to a speed > c?
Your reasoning is backwards. Particles with nonzero rest mass cannot move at ##c## because of the geometry of spacetime. Particles with nonzero rest mass move on timelike worldlines, and particles that move at ##c##, such as photons, move on null worldlines. Those are two fundamentally different kinds of things and you can't change one into the other.

The "mass increase" you mention (to the extent it makes sense at all--note that there are limitations to the concept of "relativistic mass", as explained in the Insights article that has already been linked to) is a side effect of the above, not a cause of it.
 
  • #35
PeroK said:
Not necessarily.

There are several factors in the H-K experiment. The ground clock is only stationary relative to the surface of the Earth. There is no sense in which the ground clock is absolutely stationary. In fact, the westbound clock recorded more time than the ground clock.

Ultimately, an attempt to explain the differences in times due to the concept of absolute states of motion does not work. And, contradicts the principle of relativity.

Instead, all clocks tick physically at the same rate, but they take different paths through spacetime. The clock that takes the shortest path through spacetime records the least time (Eastbound); the ground clock takes a longer path; and the westbound clock takes the longest path.
Regardless of the different paths taken how is it possible for 3 previously synchronised clocks to indicate different times without their respective tick rates having changed at some point? To my knowledge there is no evidence to support your claim that "all clocks tick physically at the same rate"?
 
  • #36
james fairclear said:
To my knowledge there is no evidence to support your claim that "all clocks tick physically at the same rate"?
That's a deficiency in your knowldege!
 
  • #37
Can you point me to a link that discusses evidence for this claim?
 
  • #38
james fairclear said:
Regardless of the different paths taken how is it possible for 3 previously synchronised clocks to indicate different times without their respective tick rates having changed at some point?
The geometry of spacetime.

Special Relativity is theory of space and time; not a theory about the mechanical workings of clocks.
 
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  • #39
james fairclear said:
Can you point me to a link that discusses evidence for this claim?
There is a voluminous literature on experimental tests of Special Relativity. Please familiarize yourself with it. The claim in question is called the "clock hypothesis" and is an integral part of Special Relativity.
 
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  • #40
james fairclear said:
Regardless of the different paths taken how is it possible for 3 previously synchronised clocks to indicate different times without their respective tick rates having changed at some point?
Consider an analogy: you and I both drive from New York to Washington, DC. You take the direct route, straight down the East Coast; I take a roundabout route by way of Pittsburgh, PA. The odometer on my car registers more distance traveled when we meet up again than yours does. Does that mean the "distance rate" of my odometer was different from yours? No; both of our odometers registered one mile per mile. The route I took just had more miles in it than yours: our paths had different lengths.

Similarly, in a "twin paradox" scenario (or a "triplet" scenario such as the H-K experiment), everyone's clock ticks at the same rate: one second per second. But the different paths through spacetime that each observer takes have different numbers of seconds in them: they have different spacetime lengths. That's why the different observers have aged differently when they meet up again: "elapsed age" is a measure of (timelike) distance traveled through spacetime, just as elapsed odometer mileage on a car is a measure of distance traveled through space.
 
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  • #41
james fairclear said:
Can you point me to a link that discusses evidence for this claim?
It's the operational definition of a clock. It must measure time accurately. That's what a clock does.

It's a consequence of the principle of relativity, although actually it's not often highlighted. If a clock was physically altered by motion, then we could tell how fast a clock was "really" moving by examining it physically.

The principle of relativity demands that a clock is unaffected by motion.

Let me look for a good reference when I get the chance.
 
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  • #42
Of course all physics starts with choosing a spacetime model, but of course that spacetime model is only as good as it describes our experience with real-world measurements. Indeed, the great step by Einstein in 1905 was to recognize that the problem with the lack of Galilei invariance of the Maxwell equations is solved when using a different spactime model, i.e., instead of Galilei-Newton spacetime to use the special-relativistic Einstein-Minkowski spacetime model.

Now of course to use this spacetime model one must use it to define the measurement devices to measure times and distances. According to the best of our models concerning matter, relativistic quantum field theory, particularly QED, which describes the electromagnetic field and the interactions of electrically charged particles, the most reliable definitions of a time unit is to use some electromagnetic transition of a specific atomic system, and for practical purposes, i.e., the most reliable practical way to define a unit of time, is to use Cs-133 atoms and a hyperfine transition to define the second as one of the base units of the international system of units, the SI. There are even more precise realizations of the second under investigation, i.e., using transitions in the optical range or even using a nuclear electromagnetic transition to establish a nuclear clock which is even less sensitive to external perturbations than atomic clocks.

Having now a very precise definition of the unit of time, the Einstein-Minkowski spacetime model tells us that you can use the universal limiting speed, which to the best of our knowledge is identical with the speed of light in vacuum, to define a unit of length, and indeed that's how it's done in the SI, the meter is defined via the second by defining a specific value for the speed of light in vacuo, ##c##.

Having such definitions of time and length units, one can make consistency tests. There are very accurate tests of the "clock hypothesis", i.e., if you have something defining a specific time, measurements of this time should follow the predictions of the underlying spacetime model. One example is to use instable particles, which define (in a statistical way) a time by their lifetime. The lifetime is defined as the expectation value of their survival time when kept at rest wrt. an inertial reference frame. Now, the clock hypothesis says that the only changes of time, when the particle is moving wrt. the inertial frame (the "lab frame") should be do to the kinematical effect of time dilation, and that should hold even when the particles are accelerated. This indeed has been tested to be true in many experiments, e.g., in storage rings where radioactive nuclei are moving in a circle or also with elementary particles like muons (in fact with atmospheric muons this was one of the first tests of this kind).

Other tests use the relativistic Doppler effect for light (particularly the transverse Doppler effect).

https://en.wikipedia.org/wiki/Experimental_testing_of_time_dilation

Famously, then also the special-relativistic spacetime model has to be further refined as soon as the gravitational interaction becomes relevant. Then one has to use the general-relativistic spacetime model, and also the predictions of this spacetime model concerning time measurements, particularly the gravitational time dilation, Shapiro delay, etc. have also been tested at high accuracy, particularly also for strong gravitational interactions as with pulsar timing or via gravitational lensing. At very high precision these predictions of general relativity have been confirmed by all observations. The standard review article for this is (open access)

https://doi.org/10.12942/lrr-2014-4
 
  • #43
PeroK said:
The geometry of spacetime.

Special Relativity is theory of space and time; not a theory about the mechanical workings of clocks.
As you say SR is a theory of space and time. The definition of time is "that which is measured by clocks" so the workings of clocks are inextricably bound up with time.

Regardless of whether or not time is a separate entity something must be physically affecting the tick rate of clocks in motion as otherwise their indicated times would not vary.
 
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  • #44
PeroK said:
The principle of relativity demands that a clock is unaffected by motion
If so then the results of the Hafele Keating experiment disprove the principle.
 
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  • #45
james fairclear said:
If so then the results of the Hafele Keating experiment disprove the principle.
I doubt that Hafele and Keating would have agreed with that!
 
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  • #46
james fairclear said:
Regardless of whether or not time is a separate entity something must be physically affecting the tick rate of clocks in motion as otherwise their indicated times would not vary.
That's simply not the case. Motion is relative, not physical. Absolute motion cannot be detected. So, there is nothing physical about relative motion for a clock to be affected by. A good reference for this is Einstein's original 1905 paper:

It is known that Maxwell’s electrodynamics—as usually understood at the
present time—when applied to moving bodies, leads to asymmetries which do
not appear to be inherent in the phenomena ...

Examples of this sort, together with the unsuccessful attempts to discover
any motion of the Earth relatively to the “light medium,” suggest that the
phenomena of electrodynamics as well as of mechanics possesses no properties
corresponding to the idea of absolute rest
.

They suggest rather that the same laws of electrodynamics and optics will be valid for all frames of reference for which the equations of mechanics hold good. We will raise this conjecture (the purport
of which will hereafter be called the “Principle of Relativity”) to the status of a postulate.


In summary, you are missing the first postulate of relativity and thereby misinterpreting all of the theory of relativity.
 
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  • #47
PeroK said:
That's simply not the case. Motion is relative, not physical. Absolute motion cannot be detected. So, there is nothing physical about relative motion for a clock to be affected by. A good reference for this is Einstein's original 1905 paper:

It is known that Maxwell’s electrodynamics—as usually understood at the
present time—when applied to moving bodies, leads to asymmetries which do
not appear to be inherent in the phenomena ...

Examples of this sort, together with the unsuccessful attempts to discover
any motion of the Earth relatively to the “light medium,” suggest that the
phenomena of electrodynamics as well as of mechanics possesses no properties
corresponding to the idea of absolute rest
.

They suggest rather that the same laws of electrodynamics and optics will be valid for all frames of reference for which the equations of mechanics hold good. We will raise this conjecture (the purport
of which will hereafter be called the “Principle of Relativity”) to the status of a postulate.


In summary, you are missing the first postulate of relativity and thereby misinterpreting all of the theory of relativity.

There is nothing stated or implied in the first postulate that motion is not something physical.

What exactly do you mean by your statement "motion is not physical"?

My original question is "Considering an atom within a rigid body, does the angular momentum of an electron within the atom vary when the body is put in motion?"

Some of the responses provided on this forum suggest that the answer is yes which implies that motion will have a physical effect. If there is no physical effect on a body when put into motion then there should be no need to take into account relativistic effects.
 
  • #48
james fairclear said:
There is nothing stated or implied in the first postulate that motion is not something physical.

What exactly do you mean by your statement "motion is not physical"?
That a state of motion cannot be experimentally distinguished from a state of rest. For example, you might claim to be at rest sitting at your computer. But, the Earth is orbitting the Sun at about ##30,000 m/s##. You can, of course, measure that relative motion. The question is: what is physically different about you to the tune of ##30,000m/s##?

Then, the Milky Way is moving towards the Andromeda galaxy at ##300,000 m/s##. How does that physically affect you?

What all physics since Galileo, through Newton and Einstein answers is that these motions are not physically measurable. This is what Einstein was referring to by the unsuccessful attempts to discover
any motion of the Earth relatively to the “light medium,”
.

In other words, there is no experiment that can tell you the Earth's absolute physical state of motion. Only motion relative to other objects. This is the principle of relativity and the first postulate of special relativity.
 
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  • #49
You have just restated the principles of relative motion.

As to whether there any physical changes to an object in motion is a separate issue. Observationally there are physical changes to synchronised clocks when put in motion as they cease to be synchronised with the relatively stationary clock.

A clock ticking at its highest possible tick rate could be considered to be at rest relative to any other clock ticking at a slower rate.
 
  • #50
james fairclear said:
As to whether there any physical changes to an object in motion is a separate issue. Observationally there are physical changes to synchronised clocks when put in motion as they cease to be synchronised with the relatively stationary clock.
The "stationary" clock is in motion relative to the "moving" clock. It, likewise, ought to be physically altered by the relative motion. Neither clock can claim to be "really" at rest or "really" moving. That is what Einstein was saying above.
james fairclear said:
A clock ticking at its highest possible tick rate could be considered to be at rest relative to any other clock ticking at a slower rate.
All clocks tick at their highest tick rate in a reference frame where they are at rest; and all other clocks are ticking relatively slowly as measured by that clock. Every clock can claim to be ticking normally. As indeed it is.
 
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