Theoretical Question On The Twins Paradox and Heart Rate

[Orangeator]

Okay, so if two identical twins put on a heart rate monitor that after let's say 400 million beats were detected (at 80bpm that's about 10 years) the monitor killed them. Now one gets in a spaceship and the other stays on earth. The one in the spaceship travels at say 95% of the speed of light. Now let's say the twin returns after 8 years in space traveling at that speed. According to time relativity the one who stayed on Earth should be much older now compared to the other twin who just returned... So that heart beat counter should kill the one who stayed on Earth first correct? As he should reach the 400 million beats first from when they both put the monitor on back when the twin left. Meaning that your heart rate slows down when traveling near the speed of light?

 

[phinds]

Okay, so if two identical twins put on a heart rate monitor that after let's say 400 million beats were detected (at 80bpm that's about 10 years) the monitor killed them. Now one gets in a spaceship and the other stays on earth. The one in the spaceship travels at say 95% of the speed of light. Now let's say the twin returns after 8 years in space traveling at that speed. According to time relativity the one who stayed on Earth should be much older now compared to the other twin who just returned

correct

... So that heart beat counter should kill the one who stayed on Earth first correct?

correct

As he should reach the 400 million beats first from when they both put the monitor on back when the twin left.

correct

Meaning that your heart rate slows down when traveling near the speed of light?

Absolutely not. You said yourself that the traveler was only gone 8 years. You are confusing time dilation with differential aging.

 

[Orangeator]

Absolutely not. You said yourself that the traveler was only gone 8 years. You are confusing time dilation with differential aging.

Ah! Yep, that makes sense now. How did I not see this earlier. Whoops.

 

[phinds]

Ah! Yep, that makes sense now. How did I not see this earlier. Whoops.

Well, it's a VERY common misconception. We have debugged it here on PF, at last count, 1,427 times.

 

[sweet springs]

400 million beats are done in 9.5 years if I calculate properly. Say they are twenty yrs old when the count started and the travel started, both will die in age 29.5. Say the travel takes 9 yrs in Earth time, the Earth twin will die in half a year after their meeting again on the Earth. the Rocket twin survives more.

 

[phinds]

400 million beats are done in 9.5 years if I calculate properly. Say they are twenty yrs old when the count started and the travel started, both will die in age 29.5. Say the travel takes 9 yrs in Earth time, the Earth twin will die in half a year after their meeting again on the Earth. the Rocket twin survives more.

You have omitted the differential aging. The ROCKET man will die half a year after he gets back. The Earth man will already be dead or will be about to die. 95% of c doesn't give a huge amount of differential aging but it does give some.

 

[sweet springs]

9 yrs in the Earth time, I wrote. You are right if it takes 9 yrs in Rocket time.

 

[Dale]

So that heart beat counter should kill the one who stayed on Earth first correct?

Yes. Time dilation affects all clocks, both precise ones like atomic clocks and imprecise ones like heartbeats.

 

[Sorcerer]

Well, it's a VERY common misconception. We have debugged it here on PF, at last count, 1,427 times.

There are so many subtleties in special relativity. I still have some trouble distinguishing between the Doppler effect, the twin paradox, differential aging and how all that relates to what you actually see vs time dilation and how differential aging actually works, even though algebraically it’s a piece of cake.

 

[robphy]

There are so many subtleties in special relativity. I still have some trouble distinguishing between the Doppler effect, the twin paradox, differential aging and how all that relates to what you actually see vs time dilation and how differential aging actually works, even though algebraically it’s a piece of cake.

Draw Spacetime diagrams to help provide physical interpretations to the algebraic formulas.

 

[Ibix]

There are so many subtleties in special relativity. I still have some trouble distinguishing between the Doppler effect, the twin paradox, differential aging and how all that relates to what you actually see vs time dilation and how differential aging actually works, even though algebraically it’s a piece of cake.

I second @robphy. If you can sketch a spacetime diagram in your head, learning to "scissor" the axes apart or together is easy. Then you just have to remember that this is a map, and you can work out what you would see by deciding on your worldline and adding light rays from events of interest.

If you want to see what I'm talking about, I wrote a spacetime diagram program with animated frame changes: http://www.ibises.org.uk/Minkowski.html.

 

[Nugatory]

There are so many subtleties in special relativity. I still have some trouble...even though algebraically it’s a piece of cake.

As you pointed out recently in another thread (Is there any way to derive the time dilation formula?) it's easy to get the time dilation formula from the Lorentz transformations; and those transformations are in turn just fairly straightforward algebra starting from the assumption of constant light speed. So yes, the algebra is a piece of cake.

However, you've also hit on the reason why so many pedagogical treatments insist on dragging you through the workings of the light clock, trains and lightning strikes, and the dreary mechanics of synchronizing clocks: the algebra just plain works, but very few students can build a workable physical intuition with just that algebra as a starting point.

 

[Sorcerer]

Draw Spacetime diagrams to help provide physical interpretations to the algebraic formulas.

I second @robphy. If you can sketch a spacetime diagram in your head, learning to "scissor" the axes apart or together is easy. Then you just have to remember that this is a map, and you can work out what you would see by deciding on your worldline and adding light rays from events of interest.

If you want to see what I'm talking about, I wrote a spacetime diagram program with animated frame changes: http://www.ibises.org.uk/Minkowski.html.

As you pointed out recently in another thread (Is there any way to derive the time dilation formula?) it's easy to get the time dilation formula from the Lorentz transformations; and those transformations are in turn just fairly straightforward algebra starting from the assumption of constant light speed. So yes, the algebra is a piece of cake.

However, you've also hit on the reason why so many pedagogical treatments insist on dragging you through the workings of the light clock, trains and lightning strikes, and the dreary mechanics of synchronizing clocks: the algebra just plain works, but very few students can build a workable physical intuition with just that algebra as a starting point.

Incidentally, in another post I remarked about how so many physics instructors are moving towards introducing the basic concepts of special relativity through spacetime diagrams. This seems to be the trend, and while I personally have difficulty with it (I do not have good physics intuition at all), I can see why instructors want to do it. It is far easier to get an intuitive grasp. What does the Lorentz factor really tell you just by looking at it? Mathematically, all it really tells me is that a speed equal to c gives an undefined value, and one greater than c gives an imaginary value. That tells very little at face value about the nature of space and time, at first glance.

So, advice noted and headed. I'll add mastering spacetime diagrams to my to do list. ;)

 

[robphy]

Incidentally, in another post I remarked about how so many physics instructors are moving towards introducing the basic concepts of special relativity through spacetime diagrams. This seems to be the trend, and while I personally have difficulty with it (I do not have good physics intuition at all), I can see why instructors want to do it. It is far easier to get an intuitive grasp. What does the Lorentz factor really tell you just by looking at it? Mathematically, all it really tells me is that a speed equal to c gives an undefined value, and one greater than c gives an imaginary value. That tells very little at face value about the nature of space and time, at first glance.

So, advice noted and headed. I'll add mastering spacetime diagrams to my to do list. ;)

A spacetime diagram will reveal that the Lorentz factor $\gamma=\frac{1}{\sqrt{1-(v/c)^2}}$ is analogous to the "cosine of an angle", which is used to give the time-component of a 4-vector (for example, an observer's measurement of the energy of a particle's 4-momentum). Think... adjacent side to a hypotenuse. This is all a result of the spacetime viewpoint, which was developed by mathematician Minkowski [and initially not appreciated by the physicist Einstein].

[More details: with $v=c\tanh\theta$, it follows that $\gamma=\frac{1}{\sqrt{1-(v/c)^2}}=\cosh\theta$.

 

[SiennaTheGr8]

@Sorcerer, consider that the magnitude of the (normalized) four-velocity is:

$1 = \gamma^2 - (\gamma \beta)^2$,

and that the hyperbolic "equivalent" of the Pythagorean trig identity is:

$1 = \cosh^2{\phi} - \sinh^2{\phi}$.

These are both equations for the unit hyperbola. This is good motivation to introduce the hyperbolic argument $\phi$ into our relativistic toolkit:

$\cosh{\phi} = \gamma$
$\sinh{\phi} = \gamma \beta$
$\tanh{\phi} = \sinh{\phi} / \cosh{\phi} = \beta$.

In general, the quantity $\phi$ can be interpreted as the hyperbolic "angle" between a pair of timelike four-vectors that are either both future-pointing or both past-pointing. When the vectors are future-pointing and correspond to trajectories of things whose relative motion we care about, we call $\phi$ the "rapidity." (This might be splitting hairs, but I think it makes sense to regard the hyperbolic angle as a more general concept, and to reserve the word "rapidity" for cases where we might use words like "speed.")

If the four-vector formalism is the most "natural" way to do SR, then the rapidity is the most "natural" quantity to use to describe relative motion. To take @robphy's point a bit further, we can then regard not just the Lorentz factor but also the celerity $\gamma \beta$ and even the speed $\beta$ as "subordinate" to $\phi$. They are nothing but the three main hyperbolic functions applied to the rapidity.

 

[Sorcerer]

@Sorcerer, consider that the magnitude of the (normalized) four-velocity is:

$1 = \gamma^2 - (\gamma \beta)^2$,

and that the hyperbolic "equivalent" of the Pythagorean trig identity is:

$1 = \cosh^2{\phi} - \sinh^2{\phi}$.

These are both equations for the unit hyperbola. This is good motivation to introduce the hyperbolic argument $\phi$ into our relativistic toolkit:

$\cosh{\phi} = \gamma$
$\sinh{\phi} = \gamma \beta$
$\tanh{\phi} = \sinh{\phi} / \cosh{\phi} = \beta$.

In general, the quantity $\phi$ can be interpreted as the hyperbolic "angle" between a pair of timelike four-vectors that are either both future-pointing or both past-pointing. When the vectors are future-pointing and correspond to trajectories of things whose relative motion we care about, we call $\phi$ the "rapidity." (This might be splitting hairs, but I think it makes sense to regard the hyperbolic angle as a more general concept, and to reserve the word "rapidity" for cases where we might use words like "speed.")

If the four-vector formalism is the most "natural" way to do SR, then the rapidity is the most "natural" quantity to use to describe relative motion. To take @robphy's point a bit further, we can then regard not just the Lorentz factor but also the celerity $\gamma \beta$ and even the speed $\beta$ as "subordinate" to $\phi$. They are nothing but the three main hyperbolic functions applied to the rapidity.

I can see exactly why hyperbolic geometry would be perfect for SR, due to the speed limit acting as an asymptote. Not too clear on rapidity just yet, since I don't really have much intuitive feel for what it physically is, other than some sort of a mapping of velocity v from a -c < v < c interval to -∞ < $\phi$ < ∞.

 

[robphy]

Technically, it’s not hyperbolic geometry (the negatively curved riemannian space that mathematicians study [associated with Gauss and Bolyai and Lobachevsky]). Rather, it’s hyperbolic trigonometry (the flat Spacetime with a certain pseudoRiemannian signature introduced by Minkowski and developed further by AA Robb).

Rapidity (named by Robb) is the analogue of the angle in Euclidean Geometry. It’s the Minkowski-arclength along the unit hyperbola. It’s also the area of the corresponding sector. One of its features is that it is additive (since arclength is additive), while velocities (akin to slopes) are not additive... except in the Galilean case. As noted above, it can be used a basic parameter to describe the relative separation of two future-timelike rays (representing inertial observers meeting at an event), rather than the velocity (slope). In addition to @SiennaTheGr8 ’s list, $e^\phi$ is the Doppler factor
$k=\sqrt{\frac{1+\beta}{1-\beta}}$, an eigenvalue of the Lorentz boost.

 

[Sorcerer]

I guess my problem with it is this: if you asked me what speed meant, I could, say, go look at a ball and tell you that speed is the distance that ball travels during the time my watch measures. But rapidity, as you describe it, is something that's a bit more abstract; more of a useful four-dimensional tool. How can I visualize the angle between two four dimensional world lines? (I mean in a physical sense, without projecting it onto 2D space)

The analogy to Euclidean geometry certainly helps, but I suppose there comes a point in physics where the language of math has to replace basic physical, visual intuition. After all, how can we possibly visualize a 4-sphere, for example?

Of course, my hypocrisy is revealed in how I learned the basics of elementary SR: through the algebraic properties of the Lorentz factor, deriving it using a triangle formed from a light clock. The light clock made intuitive physical sense to me, but it was Euclidean geometry, specifically the properties of a right triangle, that first helped me to derive that factor myself (starting with vt, ct' and ct as lengths of the triangle, if I recall, and using the Pythagorean theorem to end with the Lorentz factor, written as coordinate time divided by proper time).But clearly I have work to do. I am still not comfortable with the hyperbolic trig functions in the first place, and if I were honest with you, even regular trigonometry I try as much as possible to use 1 = cos²α + sin²α. I really hated memorizing all those formulas and more or less forgot all of them except that one, and of course tangent's relation to sin/cos. Well, I still remember how to write them in terms of e and i, but other than that, I'm a novice.

 

[robphy]

I guess my problem with it is this: if you asked me what speed meant, I could, say, go look at a ball and tell you that speed is the distance that ball travels during the time my watch measures. But rapidity, as you describe it, is something that's a bit more abstract; more of a useful four-dimensional tool. How can I visualize the angle between two four dimensional world lines? (I mean in a physical sense, without projecting it onto 2D space)

Given two lines (in whatever dimension) meeting at a point, there is a plane determined by those lines and the angle can be described on that plane. If you have two lines that meet at a point in 3-D space, is it hard to visualize the angle between those lines? Practically speaking, one would work in the plane containing those lines.

Yes, it would seem that if you are only comfortable with measurements along legs of right triangles, then the notion of angle might seem very mathematical. For instance, the angle is $\tan^{-1} \left(\frac{\Delta y}{\Delta x}\right)$... or the limit of the sum of "opposite sides" of a sequence of triangles that approximate a sector of a unit circle. While the angle measure underlies trig, it may be more important to recognize and focus on the various ratios of sides in a right-tria ngle... and not just focus on one particular ratio $\frac{\Delta y}{\Delta x}$. (Imagine trying to do a geometry problem using only the slopes of lines)

 

[Dale]

I could, say, go look at a ball and tell you that speed is the distance that ball travels during the time my watch measures. But rapidity, as you describe it, is something that's a bit more abstract

Not really. It is the distance the ball travels in your frame during the time the ball measures. It is every bit as concrete, it is just a different measurement than speed.

 

[Sorcerer]

Not really. It is the distance the ball travels in your frame during the time the ball measures. It is every bit as concrete, it is just a different measurement than speed.

Okay that is a pretty easy to grasp concept then. There is no limit on this value, correct? I mean if the ball is moving fast enough to me, then the distance it covers (in my frame) can be immense during its own proper time, so this value should be able to approach infinity, right? (This is what I assumed above based on a wiki article)

If so, do values dependent on rapidity transform like how values dependent on speed do in Galilean physics?

 

[Dale]

There is no limit on this value, correct?

Yes, in particular it is not limited to <c.

If so, do values dependent on rapidity transform like how values dependent on speed do in Galilean physics?

Sorry, I don’t think there is an answer to this question, but if there is I certainly don’t know it

 

[robphy]

If so, do values dependent on rapidity transform like how values dependent on speed do in Galilean physics?

There is a Galilean analogue of rapidity ... it’s the Galilean arc length along a Galilean “unit circle” [a horizontal line on a Spacetime diagram, or a vertical line on a position-vs-time graph]. Because of this, “velocity” transforms like rapidity... that is, Galilean velocities (slopes in Galilean geometry) are additive.
It’s this result (among others) that “common sense” has trouble letting go of that makes relativity non-intuitive.

 

[SiennaTheGr8]

I guess my problem with it is this: if you asked me what speed meant, I could, say, go look at a ball and tell you that speed is the distance that ball travels during the time my watch measures. But rapidity, as you describe it, is something that's a bit more abstract; more of a useful four-dimensional tool. How can I visualize the angle between two four dimensional world lines? (I mean in a physical sense, without projecting it onto 2D space)

You might benefit from reading pp. 33–35 of this PDF from David Morin: https://sites.fas.harvard.edu/~phys15a/handouts/ch10.pdf

 

[SiennaTheGr8]

Here's another way to think about it. In the case of rectilinear motion (acceleration parallel to velocity):

$\mathbf{\alpha} = \dfrac{d}{dt} (\gamma v) = \dfrac{d \phi}{d \tau}$

($c=1$). Assume 1+1-D spacetime (or specify a spatial direction / use component notation).

Here $\alpha$ is the traveler's proper acceleration, $t$ is coordinate time for some inertial observer, $\gamma v$ is the traveler's celerity for that same inertial observer, $\tau$ is the traveler's proper time, and $\phi$ is the traveler's rapidity.

Note that rapidity has the lovely property that changes in it are invariant under a Lorentz boost along the axis of the traveler's motion. So although $\phi$ isn't invariant, the quantity $\Delta \phi$ (or its "infinitesimal" version $d \phi$) is invariant in our 1+1-D rectilinear scenario.

There's a sense in which celerity is to coordinate time as rapidity is to proper time. Don't take that too literally.

 

[DrGreg]

I could, say, go look at a ball and tell you that speed is the distance that ball travels during the time my watch measures. But rapidity, as you describe it, is something that's a bit more abstract;

Not really. It is the distance the ball travels in your frame during the time the ball measures. It is every bit as concrete, it is just a different measurement than speed.

Sorry, Dale, but that is the definition of celerity, not rapidity.

$$\begin{align*}
\text{Velocity } \textbf{v} &= \frac{\text{d}\textbf{x}} {\text{d}t} \\
\text{Celerity } \textbf{w} &= \frac{\text{d}\textbf{x}} {\text{d}\tau} = \gamma \textbf{v} \\
\text{Lorentz factor } \gamma &= \frac{\text{d}t} {\text{d}\tau} \\
\text{Rapidity } \phi &= \tanh^{-1} \left( \frac{1}{c} \frac{\text{d}x} {\text{d}t} \right) = \sinh^{-1} \left( \frac{1}{c} \frac{\text{d}x} {\text{d}\tau} \right) = \ln k \\
c^2 \, \text{d}\tau^2 &= c^2 \, \text{d}t^2 - |\text{d}\textbf{x}|^2 \\
\text{Doppler factor } k &= \sqrt \frac{c + v} {c - v} = \text{e}^\phi \\
w/c &= \sinh \phi \\
\gamma &= \cosh \phi \\
v/c &= \tanh \phi \\
\end{align*}$$

 

[Dale]

Sorry, Dale, but that is the definition of celerity, not rapidity.

Oops, sorry about the mistake and thanks for the correction!

 

[Sorcerer]

Sorry, Dale, but that is the definition of celerity, not rapidity.

$$\begin{align*}
\text{Velocity } \textbf{v} &= \frac{\text{d}\textbf{x}} {\text{d}t} \\
\text{Celerity } \textbf{w} &= \frac{\text{d}\textbf{x}} {\text{d}\tau} = \gamma \textbf{v} \\
\text{Lorentz factor } \gamma &= \frac{\text{d}t} {\text{d}\tau} \\
\text{Rapidity } \phi &= \tanh^{-1} \left( \frac{1}{c} \frac{\text{d}x} {\text{d}t} \right) = \sinh^{-1} \left( \frac{1}{c} \frac{\text{d}x} {\text{d}\tau} \right) = \ln k \\
c^2 \, \text{d}\tau^2 &= c^2 \, \text{d}t^2 - |\text{d}\textbf{x}|^2 \\
\text{Doppler factor } k &= \sqrt \frac{c + v} {c - v} = \text{e}^\phi \\
w/c &= \sinh \phi \\
\gamma &= \cosh \phi \\
v/c &= \tanh \phi \\
\end{align*}$$

View attachment 224081

So rapidity is the inverse cosh of the Lorentz factor? What does that mean physically, now?
EDIT- also now I'm curious about cerelity. Because it if's:

$\gamma \textbf{v}$ Well, that's almost like

$\gamma \textbf{u}$

which if you multiply my mass you have relativistic momentum, and can go from there to get everything else.

 

[PeterDonis]

also now I'm curious about cerelity. Because it if's:

$\gamma \textbf{v}$

Well, that's almost like

$\gamma \textbf{u}$

What do you mean by $\gamma \textbf{u}$?

 

[SiennaTheGr8]

Some treatments of SR systematically use different symbols for "relative velocity between inertial frames" and "velocity of some object as measured in an inertial frame." Often one is consistently $\mathbf u$ and the other is consistently $\mathbf v$. Maybe that's what @Sorcerer means?

 

[Sorcerer]

What do you mean by $\gamma \textbf{u}$?

What I mean is, if you start with, say, a coordinate of some moving object, (cdt, dx), ignoring the y and z directions, of course, and then divide by proper time, written as $\frac{dt}{\gamma\ }$ , you get ($\gamma \textbf{c}$, $\gamma\ u_x$), so that second coordinate, the speed times gamma, is what I mean by $\gamma \textbf{u}$ .

So, if you multiply by invariant mass it becomes γmu, then if you take the time derivative, and then integrate over distance (after doing some manipulations with the differentials and integrating from 0 to u), you end up with a result that is identical to the relativistic kinetic energy equation.

So that's what I meant by all that. I don't know if it's mere coincidence or not, but it seems interesting to me.

 

[PeterDonis]

that second coordinate, the speed times gamma, is what I mean by $\gamma \textbf{u}$.

That's the same thing that @DrGreg means by $\gamma \textbf{v}$.

 

[PeterDonis]

then divide by proper time, written as $\frac{dt}{\gamma\ }$

I think you mean $\frac{dt}{d\tau}$. The 4-velocity vector's components are

$$
U^\mu = \left( \frac{dt}{d\tau}, \frac{d \textbf{x}}{d\tau} \right) = \left( \gamma, \gamma \textbf{v} \right)
$$

(in units where $c = 1$). Multiplying by the invariant mass $m$ then gives the 4-momentum $P^\mu$.

 

[Sorcerer]

I think you mean $\frac{dt}{d\tau}$. The 4-velocity vector's components are

$$
U^\mu = \left( \frac{dt}{d\tau}, \frac{d \textbf{x}}{d\tau} \right) = \left( \gamma, \gamma \textbf{v} \right)
$$

(in units where $c = 1$). Multiplying by the invariant mass $m$ then gives the 4-momentum $P^\mu$.

Well, as I understand it, $\frac{dt}{d\tau}$ = γ, hence proper time must be $\frac{dt}{γ}$ . So dividing dx by that will give $\frac{dx}{\frac{dt}{γ}}$ which is $γ\frac{dx}{dt}$, or $γu$ .But yeah the net result is exactly what you put there. I guess it is indeed a 4-velocity, except I'm ignoring y and z.

So why does that look the same as celerity? Is celerity exactly what the spatial components of 4-velocity are?

 

[PeterDonis]

as I understand it, dtdτdtdτ\frac{dt}{d\tau} = γ, hence proper time must be $\frac{dt}{γ}$ .

If you want to treat differentials that way (which would give many rigorous mathematicians apoplexy, although most physicists wouldn't bat an eye), then yes, you can write $d\tau = dt / \gamma$.

So dividing dx by that will give $\frac{dx}{\frac{dt}{γ}}$ which is $\gamma \frac{dx}{dt}$, or $\gamma u$ .

Yes, which is what @DrGreg was calling $\gamma \textbf{v}$, as I said before (and that's how I wrote it in my post).

So why does that look the same as celerity?

Because, as @DrGreg posted, celerity is equal to $\gamma \textbf{v}$.