Teaching Relativity in a College Physics course

[nrqed]

I guess it comes down to what one means by "observable".

Suppose the muon is created at event A and dies at event B. A clock carried by the muon is the only inertial clock that experiences both of these events. If I want to measure the time, then I have to say "Well, event A is simultaneous with event A' on my worldline, event B is simultaneous with event B' on my worldline, and my inertial clock (assuming that I am am inertial observer) measures the time difference between A' and B' to be $\Delta t$."

But, since A and B are not on my worldine and thus not experienced by me, I have to use a simultaneity convention to do this.

Hi George,

Ok but to me this sounds more like philosophy than science. A physical observable is something that can be measured and whose result can be used to test predictions from a mathematical formalism (e.g. Special Relativity). Orodruin says that only proper time is a physical observable, that the time measure in any other frame is not. I guess that distances are not physical observables according to that point of view since how can you measure a distance in a frame independent way? One cannot. Ok, so that leaves only one observable: proper time. So one can we test the Lorentz transformations?? I guess that according to that point of view, they are meaningless since they involve things that are not physical observables (like time intervals that are not proper time!). So I guess the equations are meaningless, we cannot even test them in a lab!
This is like saying that the x and y components of a vector (in introductory physics) are meaningless since the depend on the coordinate system used to measure them. It is true that there is a quantity with a deeper meaning here: the magnitude of the vector. But saying that the x and y components are not physical observables and they are meaningless is absurd since they can be measured and used to make predictions using equations from theory, predictions that can then be tested. If that was not true, all of experimental physics would be meaningless.

 

[George Jones]

Coordinates are not meaningless, but they are conventional. Once a convention has been established, it is very meaningful to ask, e.g., how coordinates transform.

 

[Orodruin]

As I said, you lost me there. Sounds like GR. I am restricting my comments to SR and inertial reference frames.

But this limits your scope of SR. The full formalism and coordinate independence does not become clear until you realize you can use any coordinates. The light cone and Rindler coordinates are just two examples. The entire point with my question was that it is not GR for a particular choice of the scale factor, just hyperbolic coordinates on Minkowski space. The definition of simultaneity is also very reasonable (it is the one we like to use in cosmology!) and different from the typical definition by "equal time coordinate" in Minkowski coordinates.

I am really not sure what that means. I am quite certain a student in a one-week course on SR would feel the same.

Again, we have gone beyond that a long time ago. This conversation started with the claim that it is impossible to use the same units for space and time. I would also keep c if teaching at a lower level. I use c = 1 in my course at master level.

Dimensions, on the other hand, are physical.

Yes, but some things can have the same dimension without physically be the same thing.

But, since A and B are not on my worldine and thus not experienced by me, I have to use a simultaneity convention to do this.

This. There is no way of measuring the muon "lifetime" without adhering to a simultaneity convention. A convention is not an observable and what you are actually measuring may also be encoded in a language which does not refer to a particular coordinate system. What you are really saying is "if I a set of Einstein synchronised clocks then the muon life time will be dilated with respect to these". Now to make this physically observable you will need to define a procedure for setting up such a system.

This is very much like the twin paradox: the muon ages less during its journey than the stationary observer on the earth, which is the reference frame in which the muon began and ended its journey.

No, it is not like the twin paradox. The fact that the twins meet up again is crucial for them to compare their clock measurements.

Such an event cannot happen if the muon's clock ran at the same rate as that of an inertial observer on the earth.

But this line of reasoning is only valid in the Earth rest frame and only with your typical definition of simultaneity! SR does not care about your simultaneity definition and neither does the muon. It only cares about the proper time of its world line.

The highly directional light from a synchrotron in the lab frame is omnidirectional in the rest frame of the electron that emits it. So I am not sure why the highly directional result would be the same in any other coordinate system.

You are obtaining this result by projecting what is happening onto an arbitrary three-dimensional subspace (which happens to be your lab frame). Again it is a result of your will to separate time and space when they are really a single entity. The synchrotron radiation world lines are something much more physical and they really can be described by any set of coordinates but will still be the same object. It is only a matter of your chosen coordinates, just like a vector which was originally pointing in the x-direction in one coordinate system may be pointing in the y-direction in another.

You are saying that only invariant quantities are observables but this is plain wrong.

I am sorry, but this statement is simply false. You can only measure invariant things because regardless of what you measure you will be using a measuring device which will give you a number. You would measure your height by placing yourself at rest next to a length scale. Now, regardless of which frame I use to describe this, the same number is going to appear at the top of your head - the measurement is invariant and your coordinates therefore defined by your instrument. A non-invariant measurement would violate the principle of relativity. What is the x-component of a vector may be frame dependent, but how you measure the x-component in a particular frame is not - you place a ruler in the x-direction of that frame. Now this may be in a different direction in a different set of coordinates, but the measurement in that given direction is still the same and therefore invariant.

Coordinates are not meaningless, but they are conventional. Once a convention has been established, it is very meaningful to ask, e.g., how coordinates transform.

Agreed. I am not arguing against the use of coordinates. I am only warning about ascribing a particularly significant meaning to any (arbitrary) set of coordinates. (This is what a lot of students have difficulties grasping in relativity!) The physics will be the same regardless of the coordinates, but coordinates are generally very important for making quantitative predictions.

 

[Mister T]

Good point. But I expect that the electron kinetic energy is small in comparison to the coulomb potential energy of the protons in the nucleus, let alone the nuclear potential energy of the neutrons and protons.

That's hardly a reason for adopting a worldview that ignores it. The kinetic energies of the constituents of a system make a contribution to the system's mass in the same way that their potential energy and masses do. Take the spring energy you mentioned earlier. Suppose you have two particles, each of mass $m$, on opposite ends of a compressed spring. The rest energy of the system is $2mc^2+\frac{1}{2}kx^2$.

Let the spring release so that the potential energy of the spring is converted into kinetic energy of the particles. Now the rest energy of the system is $2 \gamma mc^2$. (Note that $\gamma=(1-v^2/c^2)^{-1/2}$, where $v$ is the speed of each particle in the rest frame, or center-of-momentum frame).

Rest energy is a relativistic invariant, meaning all inertial observers will agree on its value. The same is true of the mass of the system. Before release it was $2m+\frac{1}{2}kx^2/c^2$. After it's $2 \gamma m$. Again, all inertial observers will agree on these values.

I think the central issue here is the modelling process that is physics. Mass, energy, and momentum are all part of the modelling process. They are inventions of the human mind, not something discovered like a geologist might do when finding a buried fossil. Humans define mass, energy, and momentum; so if humans can find ways to measure them all in the same units in ways that are consistent with their definitions, then there are no physical grounds on which to object.

The same is true of distance and time.

Of course we do not need to bring up all of these details when we teach. We bring in the ones we need to in the process of trying to satisfy the learning objectives we've set out for the students. I think it's easier, not harder, for students to understand that it's okay to measure distance and time in the same units.

 

[Andrew Mason]

No, it is not like the twin paradox. The fact that the twins meet up again is crucial for them to compare their clock measurements.
But this line of reasoning is only valid in the Earth rest frame and only with your typical definition of simultaneity! SR does not care about your simultaneity definition and neither does the muon. It only cares about the proper time of its world line.

To be fair, I said it was like the twin paradox. A muon twin that remained at rest in the Earth frame would not survive to compare its clock with its traveling brother. That is the whole point. The twin would have expired. It would be an ex-muon!

AM

 

[Mister T]

To be fair, I said it was like the twin paradox. A muon twin that remained at rest in the Earth frame would not survive to compare its clock with its traveling brother. That is the whole point. The twin would have expired. It would be an ex-muon!

That's not like the twin paradox because it's a frame-dependent observation. What you describe is true in the Earth frame but observers in other frames will draw different conclusions about that same scenario. For example, in the traveling muon's frame it's the traveling muon that will decay first.

To make it like the twin paradox the two muons must start out in the same place at the same time. And then later be at the same place at the same time again. The difference in the proper times each experiences between departure and return will be invariant. All observers will agree on its value.

 

[Andrew Mason]

That's not like the twin paradox because it's a frame-dependent observation. What you describe is true in the Earth frame but observers in other frames will draw different conclusions about that same scenario. For example, in the traveling muon's frame it's the traveling muon that will decay first.

To make it like the twin paradox the two muons must start out in the same place at the same time. And then later be at the same place at the same time again. The difference in the proper times each experiences between departure and return will be invariant. All observers will agree on its value.

?? In the twin paradox the twin's do not have to meet. If the traveling twin arrives back at Earth and discovers that his twin died about a million years earlier has long since decayed it is still the twin paradox. The point is that the traveling twin lives longer than its stationary twin. All inertial observers would agree that the stationary twin died before the traveling twin.

AM

 

[Orodruin]

In the twin paradox the twin's do not have to meet.

Yes they do, it is fundamental for the appearance of the "paradox".

All inertial observers would agree that the stationary twin died before the traveling twin.

This is just plain wrong if you adhere to your statement above. If the twins just go away from each other, who lives longer will be a frame dependent (or more generally, simultaneity convention dependent) statement.

 

[Andrew Mason]

Yes they do, it is fundamental for the appearance of the "paradox".This is just plain wrong if you adhere to your statement above. If the twins just go away from each other, who lives longer will be a frame dependent (or more generally, simultaneity convention dependent) statement.

I disagree. One twin has to start its journey in the reference frame of the twin, then transition to a reference frame moving at a relativistic speed relative to the initial frame, and then transition back to the initial frame. The other twin has to remain in the initial frame. If that occurs, the transitioning twin will have experienced a proper time that is shorter than that of the stationary twin as measured by all observers.

Other observers may disagree on how much the age difference will be but they will all agree that the stationary twin was a dearly departed muon when its robust twin crashed into the muon detector on the Earth surface.

AM

 

[Mister T]

?? In the twin paradox the twin's do not have to meet. If the traveling twin arrives back at Earth and discovers that his twin died about a million years earlier has long since decayed it is still the twin paradox.

The twins have met in the sense that they are now at the same place at the same time. The muons don't have two meetings. A meeting simply means an event where both are in the same place at the same time. In the twin paradox there are two such events.

If you draw a spacetime diagram you can see that the world lines of the two twins intersect at two places whereas the muon's only intersect at most once. One twin switches inertial reference frames, neither muon does. Remedying this very misunderstanding is what led me to develop the lesson about spacetime geometry. It gives the students a way to visualize the effects of time dilation. The usual didactic explanations, however carefully and skillfully made, often have little impact on the students' comprehension of this most basic feature of relativity.

 

[Orodruin]

The other twin has to remain in the initial frame. If that occurs, the transitioning twin will have experienced a proper time that is shorter than that of the stationary twin as measured by all observers

No, this is simply false. In particular, in the inertial frame where the "transitioning" twin was originally at rest, the "transitioning" twin will always be older as the "stationary" twin is always time dilated with respect to this frame but the "transitioning" twin is not until he makes the transition.

 

[Andrew Mason]

No, this is simply false. In particular, in the inertial frame where the "transitioning" twin was originally at rest, the "transitioning" twin will always be older as the "stationary" twin is always time dilated with respect to this frame but the "transitioning" twin is not until he makes the transition.

I don't follow what you are saying. Are you saying that the space-time interval between the deaths of the twins is not time-like?

AM

 

[Orodruin]

Are you saying that the space-time interval between the deaths of the twins is not time-like?

This depends on the exact setup, but in your case (one twin goes away and then stops and goes back to rest in the inertial frame of the staying twin and if the twins live to be the same proper age) yes. In the case when the twins meet up (before either dies), obviously no.

Let us do the actual maths. In the original inertial frame of the traveling twin, the world-line of the staying twin is given by $x = vt$. The world line of the traveling twin is given by $x = 0$ for $t < t_0$ and $x = v(t-t_0)$ for $t > t_0$. Let us say that the twins both live to have a proper age $\tau$.

The death event of the staying twin will be given by $t_1 = \tau \gamma$ due to the time dilation and consequently also by $x_1 = \tau\gamma v$.

The death event of the traveling twin will be given by $t_2 = \gamma(\tau - t_0) + t_0$ and therefore $x_2 = v\gamma(\tau -t_0)$.

Now obviously $t_2 < t_1$ and so the traveling twin dies first in this frame. Looking at the space-time interval between the deaths, you will find that $\Delta x = x_2 - x_1 = -v \gamma t_0$ and $\Delta t = (1-\gamma) t_0$. This implies that
$$
\Delta t^2 - \Delta x^2 = 2(1-\gamma) t_0^2,
$$
which is obviously negative. The space-time interval is therefore space-like.

 

[Andrew Mason]

Ok. I agree that the space-time interval between the muon deaths is not time-like. So I take back my comment that all inertial observers would agree. If the twin muons were created in the LHC and one stayed at the original location while the other made a few circuits at .999c before stopping where it left his stationary twin, all observers would agree that the stationary twin was older. Thank-you for reminding me that one has to do the math - or at least a space-time diagram!

AM

 

[Mister T]

If the twin muons were created in the LHC and one stayed at the original location while the other made a few circuits at .999c before stopping where it left his stationary twin, all observers would agree that the stationary twin was older.

Note that stopping at the end (or starting at the beginning) is not even required. All that's required is that they be co-located at two separate times, and in between experience different amounts of proper time.

 

[Andrew Mason]

Note that stopping at the end (or starting at the beginning) is not even required. All that's required is that they be co-located at two separate times, and in between experience different amounts of proper time.

Yes. But if it kept going, the situation is symmetrical - it would have appeared to both the traveling muon and the stationary one that the other's clock was moving slower. By putting the traveling muon back in the original Earth reference frame that symmetry is broken. In other words, they both agree that, in the Earth frame, the traveling muon's clock ran slower.

AM

 

[Mister T]

Yes. But if it kept going, the situation is symmetrical -

Are you talking about the muon scenario that Orodruin analyzed or the LHC muon scenario you mentioned in your reply?

For the muon in the LHC just look at one lap. In the lab frame at time t = 0 the muon passes by you. In your frame of reference its next passage is at t = 90 μs, but the elapsed time in the muon frame is only 4 μs. Both of those events occur at the same place in each frame of reference, so the elapsed time between the events is a proper time in each frame. Two co-locations separated by different amounts of proper time. You stayed in the same reference frame the whole time, the muon was never in your reference frame at any time.

 

[Orodruin]

In other words, they both agree that, in the Earth frame, the traveling muon's clock ran slower.

For a particular definition of simultaneity, which is no more physical than the xy-plane is in three dimensions. Neither will live to actually see the other's death.