I Light Clock Relativity: Why Divide T_moving/T_stationary?

[rgtr]

TL;DR Summary

Why do I divide T_moving by T_stationary for light clocks and not minus T_moving - T_stationary?

In special relativity I can get $\gamma$ , $\frac {T_B}{T_A}=\gamma$ Why do I not go ${T_B} - {T_A} = \gamma$ ?

$T_B = \frac {2H} {c^2 - v^2}$ . $T_B$ is the moving light clock.

$T_A = \frac {2H} {c^2}$ . $T_A$ is the stationary light clockI assume LaTeX doesn't work in a messages. The previous sentence was a mistype.I am using this book https://scholar.harvard.edu/files/david-morin/files/relativity_chap_1.pdf page 18 EQ (1.11)

Thanks

 

[Dale]

Summary: Why do I divide T_moving by T_stationary for light clocks and not minus T_moving - T_stationary?

In special relativity I can get γ , TBTA=γ Why do I not go TB−TA=γ ?

It is because $T_B/T_A$ is a function of $v$ only, but $T_B-T_A$ is not

 

[vela]

In special relativity I can get $\gamma$ , $\frac {T_B}{T_A}=\gamma$ Why do I not go ${T_B} - {T_A} = \gamma$ ?

You could, of course, try to define $\gamma$ as the difference, but defined that way, it wouldn't be terribly useful. On the other hand, the quantity $\gamma$ defined the usual way appears all over the place in special relativity, so that definition is very useful.

 

[FactChecker]

It is because $T_B/T_A$ is a function of $v$ only, but $T_B-T_A$ is not

That is the key. In particular, it is very convenient that the times have a constant ratio, whether you are talking about 1 second, one week, or 1000 years. We could try to do it another way, but it would be an unnecessary mess.

 

[Nugatory]

$\gamma$ is defined to be the ratio between the two times because that quantity is more generally useful. Suppose the light clock is sized such that at a given speed $T_A$ is .6 seconds and $T_B$ is 1.2 seconds; the ratio is $\frac{1}{2}$ telling us that one is ticking at half the speed of the other. Now consider what would happen if we were to scale the light clock up by a factor of two: $T_A$ and $T_B$ would both change but one clock is still ticking at half the speed of the other. The ratio $T_A/T_B$ doesn’t change, telling us that that the relative clock rates are unchanged, but the difference $T_A-T_B$ does change, obscuring the underlying fact of the clock rates.

 

[Ibix]

I assume LaTeX doesn't work in a messages.

It doesn't render in new thread preview. It also doesn't render in post preview unless there's already LaTeX on the page, although there you can trick it by previewing and refreshing the page (copy your text to clipboard first, for safety).

In special relativity I can get $\gamma$ , $\frac {T_B}{T_A}=\gamma$ Why do I not go ${T_B} - {T_A} = \gamma$ ?

In this case, if you used the difference between times you'd have to recalculate your version of $\gamma$ for every possible duration. The ratio is the same whether $T_A$ is one second or 70 years.

Edit: I type too slow, it seems.

 

[robphy]

If you visualize these as lengths of sides of a triangle in a Minkowski spacetime diagram, then the ratio of sides is a natural construction.

Note that $\gamma$(hyperbolic cosine of the rapidity-angle) is dimensionless. So it can’t be the difference of two elapsed times.

 

[Herman Trivilino]

Why do I not go ${T_B} - {T_A} = \gamma$ ?

For one thing, that definition doesn't give you a dimensionless quantity. As others have pointed out, it just wouldn't be a very useful definition.

 

[rgtr]

It is because $T_B/T_A$ is a function of $v$ only, but $T_B-T_A$ is not

Why does it need to be a function of v? And what do you mean by a function of v?

 

[Ibix]

Why does it need to be a function of v? And what do you mean by a function of v?

It's not that it's a function of $v$, it's that it's a function of $v$ only. $T_A-T_B$ depends on the value of $T_B$ while $T_A/T_B$ doesn’t, so the latter is a more generally useful quantity. It is the ratio of any pair of time intervals measured in those frames, depending only on the relative velocity. The difference depends on the length of the time interval as well.

 

[Dale]

Why does it need to be a function of v? And what do you mean by a function of v?

It is a function of $v$ ONLY. In other words, we can say that $\gamma$ is a result of $v$. In contrast, the difference is a function of both $v$ and $T_B$. In other words we cannot say that it is a result of $v$, but we have to say it is a result of $v$ and $T_B$. When we are trying to describe the result of moving fast then things that are purely due to $v$ are simpler and preferred. Other things are not just the result of moving fast, but the result of moving fast and something else.

For example, the relativity of simultaneity is a function of $v$ and $\Delta x$. It is therefore more complicated and students find it harder to learn and understand.

 

[PeroK]

Why does it need to be a function of v? And what do you mean by a function of v?

The ratio (gamma factor) is constant, as long as $v$ remains constant. The difference is not constant. And, as mentioned above, this makes it less useful.

 

[robphy]

Summary: Why do I divide T_moving by T_stationary for light clocks and not minus T_moving - T_stationary?

In special relativity I can get $\gamma$ , $\frac {T_B}{T_A}=\gamma$ Why do I not go ${T_B} - {T_A} = \gamma$ ?

$T_B = \frac {2H} {c^2 - v^2}$ . $T_B$ is the moving light clock.

$T_A = \frac {2H} {c^2}$ . $T_A$ is the stationary light clock

[snip]

I am using this book https://scholar.harvard.edu/files/david-morin/files/relativity_chap_1.pdf page 18 EQ (1.11)

Thanks

Why does it need to be a function of v? And what do you mean by a function of v?

One can try to define whatever one wants along as it's consistent.
One then hopes the definition is useful.\begin{align*}
\frac{T_B}{T_A} \stackrel{(1.11)}{=}\frac{c}{\sqrt{c^2-v^2}}=\frac{1}{\sqrt{1-v^2/c^2}}\stackrel{(1.12)}{\equiv}\gamma
\end{align*}

The $\gamma$ factor here is ubiquitous in special relativity

So, $\gamma$ is already defined and is (according to Morin) ubiquitous [and presumably, useful]
... and is called the $\gamma$-factor (not $\gamma$-difference).
It is dimensionless since it is the ratio of two times.
$\gamma\equiv\frac{1}{\sqrt{1-v^2/c^2}}$ is a "function of $v$ alone" (since $c$ is a constant) and "not a function of $H$".
That is, the $\gamma$-factor depends on the relative-velocity $v$ and not on the size of transverse light-clock. Doubling $H$ will double $T_B$ and double $T_A$, but not their ratio $\gamma$.

---

You can try to define the difference of times, but you can't use $\gamma$ since it's already taken.
Let's call it $\Delta T \equiv T_B-T_A$, the time-difference
\begin{align*}
{T_B}-{T_A} = \frac{2H}{\sqrt{c^2-v^2}} - \frac{2H}{c} =\frac{2H}{c}\left( \frac{1}{\sqrt{1-v^2/c^2}}-1\right) \equiv \Delta T
\end{align*}
$\Delta T$ is a function of $v$ and $H$.
It has dimensions of a time since it is the difference of two times.
Doubling $H$ will double $T_B$ and double $T_A$, and thus double the difference $\Delta T$.
It is consistent and has its uses in various situations, but (it turns out) is not as useful as $\gamma$ in studying various aspects of relativity. (Later, you see that it's needed in studying the Michelson-Morley experiment. It could be helpful in the study of the Doppler factor. The energy analogue of the difference is helpful for relativistic kinetic energy.)

---

It turns out a lot special relativity reduces to studying
triangles on a position-vs-time graph (called a spacetime diagram)
and one often deals with the analogue of a right-triangle in the Euclidean plane.

In Euclidean circular trigonometry, the ratio of sides is more useful that the difference of sides.
The ratio of sides tells us (via various definitions and formulae) something about the relative-slope
and the angle between the sides. Doubling the size of the triangle doesn't change the slope or the angle.
In relativity (using hyperbolic trigonometry), the relative-velocity is the analogue of the slope.

 

[malawi_glenn]

Note that you can still calculate $T_B - T_A = (\gamma - 1)T_A = (1- 1/\gamma )T_B$ so, as you see, the difference $T_B-T_A$ depends on $T_A$ or $T_B$.

 

[rgtr]

One can try to define whatever one wants along as it's consistent.
One then hopes the definition is useful.\begin{align*}
\frac{T_B}{T_A} \stackrel{(1.11)}{=}\frac{c}{\sqrt{c^2-v^2}}=\frac{1}{\sqrt{1-v^2/c^2}}\stackrel{(1.12)}{\equiv}\gamma
\end{align*}So, $\gamma$ is already defined and is (according to Morin) ubiquitous [and presumably, useful]
... and is called the $\gamma$-factor (not $\gamma$-difference).
It is dimensionless since it is the ratio of two times.
$\gamma\equiv\frac{1}{\sqrt{1-v^2/c^2}}$ is a "function of $v$ alone" (since $c$ is a constant) and "not a function of $H$".
That is, the $\gamma$-factor depends on the relative-velocity $v$ and not on the size of transverse light-clock. Doubling $H$ will double $T_B$ and double $T_A$, but not their ratio $\gamma$.

---

You can try to define the difference of times, but you can't use $\gamma$ since it's already taken.
Let's call it $\Delta T \equiv T_B-T_A$, the time-difference
\begin{align*}
{T_B}-{T_A} = \frac{2H}{\sqrt{c^2-v^2}} - \frac{2H}{c} =\frac{2H}{c}\left( \frac{1}{\sqrt{1-v^2/c^2}}-1\right) \equiv \Delta T
\end{align*}
$\Delta T$ is a function of $v$ and $H$.
It has dimensions of a time since it is the difference of two times.
Doubling $H$ will double $T_B$ and double $T_A$, and thus double the difference $\Delta T$.
It is consistent and has its uses in various situations, but (it turns out) is not as useful as $\gamma$ in studying various aspects of relativity. (Later, you see that it's needed in studying the Michelson-Morley experiment. It could be helpful in the study of the Doppler factor. The energy analogue of the difference is helpful for relativistic kinetic energy.)

---

It turns out a lot special relativity reduces to studying
triangles on a position-vs-time graph (called a spacetime diagram)
and one often deals with the analogue of a right-triangle in the Euclidean plane.

In Euclidean circular trigonometry, the ratio of sides is more useful that the difference of sides.
The ratio of sides tells us (via various definitions and formulae) something about the relative-slope
and the angle between the sides. Doubling the size of the triangle doesn't change the slope or the angle.
In relativity (using hyperbolic trigonometry), the relative-velocity is the analogue of the slope.

Thanks for the excellent explanation it was very clear now.
Question for anyone.

In this picture it shows a light clock. Let's use the moving light clock example.
Am I essentially calculating the b component of moving clock.
Assume the moving frame is the B frame.
Assume the stationary frame is the A frame
https://simple.wikipedia.org/wiki/Light_clock

Or essentially the b component of the picture below. Am I essentially calculating Light time for the vertical component?
Does this have any more profound implications?
Did I make any mistakes in my thinking ?

triangle image

 

[robphy]

Thanks for the excellent explanation it was very clear now.

I'm glad it was useful.

Question for anyone.
...

I think it would be better if you started a new thread with your new question.
Your new question could link back to this first question
(so, one could trace back your sequence of questions).