# Length contraction and time dilation: a few examples

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 4, Problems 4.4-4.5.

Here are a few examples of length contraction and time dilation.

Example 1 A rod is moving at a speed ${\beta}$ relative to an observer ${S}$ such that ${S}$ measures the length of the rod to be half its rest length. The speed can be found from

 $\displaystyle L$ $\displaystyle =$ $\displaystyle \frac{L_{0}}{\gamma}=\frac{L_{0}}{2}\ \ \ \ \ (1)$ $\displaystyle \gamma$ $\displaystyle =$ $\displaystyle 2=\frac{1}{\sqrt{1-\beta^{2}}}\ \ \ \ \ (2)$ $\displaystyle \beta$ $\displaystyle =$ $\displaystyle \sqrt{1-1/\gamma^{2}}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\sqrt{3}}{2}=0.866 \ \ \ \ \ (4)$

Example 2 A train travelling at speed ${\beta=0.8}$ passes an observer ${S}$ standing on a platform whose rest length is ${P_{0}=60\mbox{ m}}$. In frame ${S}$, the ends of the train are observed to be exactly at the ends of the platform at the same time ${t=0}$. In other words, ${S}$ measures the length of the train to be ${60\mbox{ m}}$.

${S}$ measures the time taken for the train to pass him as

$\displaystyle t=\frac{60}{0.8}=75\mbox{ m} \ \ \ \ \ (5)$

[All times are divided by ${c}$ but we’re using units such that ${c=1}$ so that time is measured in metres.]

${S}$ measures the length of the train to be contracted from its rest length by the factor ${\gamma}$, so an observer ${S'}$ on the train measures the train’s length as

 $\displaystyle L_{0}$ $\displaystyle =$ $\displaystyle 60\gamma\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{60}{\sqrt{1-\left(0.8\right)^{2}}}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{5}{3}60=100\mbox{ m} \ \ \ \ \ (8)$

The rest length of the platform is ${P_{0}=60\mbox{ m}}$ so to ${S'}$ the platform appears contracted to a length

$\displaystyle P=\frac{P_{0}}{\gamma}=\frac{3}{5}60=36\mbox{ m} \ \ \ \ \ (9)$

To find how long ${S'}$ thinks it takes for the train to pass ${S}$, we need to realize that ${S'}$ must use two clocks in his frame to do this measurement (one at each end of the train), as opposed to the single clock that ${S}$ uses to measure how long it takes the train to pass him. Therefore, it is the ${S}$ frame measurement that is slow, and the time measured by ${S'}$ is

 $\displaystyle \Delta t'$ $\displaystyle =$ $\displaystyle \gamma\Delta t\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{5}{3}75=125\mbox{ m} \ \ \ \ \ (11)$

Although ${S}$ says that the two ends of the train are at the two ends of the platform at the same time ${t=0}$, ${S'}$ will not observe the two ends of the train to be at the ends of the platform at the same time. Suppose we define the origins of the two frames to coincide when the rear of the train is at the rear end of the platform, so that

$\displaystyle \left(t_{r},x_{r}\right)=\left(t_{r}',x_{r}'\right)=\left(0,0\right) \ \ \ \ \ (12)$

In frame ${S}$, the front end of the train is at the front of the platform at ${t_{f}=0}$, so

$\displaystyle \left(t_{f},x_{f}\right)=\left(0,60\right) \ \ \ \ \ (13)$

We can use a Lorentz transformation to find ${t_{f}'}$:

 $\displaystyle t_{f}'$ $\displaystyle =$ $\displaystyle \gamma\left(t_{f}-\beta x_{f}\right)\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -60\gamma\beta\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -80\mbox{ m} \ \ \ \ \ (16)$

So ${S'}$ thinks the front end passed the front of the platform ${80\mbox{ m}}$ before the back end passes the back of the platform.

To check this is consistent with the results above, ${S'}$ thinks the platform is 36 m long so it will take the back end of the train a time of ${36/0.8=45\mbox{ m}}$ to travel the length of the platform, so that it passes the front end at ${t'=+45\mbox{ m}}$. The total time taken for the train to pass the front end of the platform is thus ${t=80+45=125\mbox{ m}}$, in agreement with the time taken to pass ${S}$ calculated above.

# The light clock and time dilation

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 4, Problem 4.3.

One of the original thought experiments that led to the equation of time dilation is the light clock. In a railway carriage a light source is placed on one side of the carriage and a mirror on the wall directly opposite the light. The line between the light and mirror is perpendicular to the direction of the train’s motion, so that the light beam travels across the car rather than along it. To an observer in the train (frame ${S'}$), the time taken for the light to make a round trip to the mirror and back is

$\displaystyle \Delta t'=\frac{2d}{c} \ \ \ \ \ (1)$

where ${d}$ is the width of the carriage.

The train now moves at speed ${u}$ relative to an observer standing beside the track (in frame ${S}$). This observer sees the light follow a diagonal path across the carriage and back, but the speed of the light is, of course, still ${c}$. How long ${\Delta t}$ does ${S}$ say that the light takes to make the round trip?

Since lengths perpendicular to the motion are unaffected, ${S}$ says that the width of the carriage is still ${d}$. In time ${\Delta t}$ the train travels a distance ${u\Delta t}$ in frame ${S}$ and since the light takes the same time on each leg of its journey, it reaches the opposite wall in time ${\Delta t/2}$ during which time the train has travelled a distance ${u\Delta t/2}$. The path of the light beam is therefore the hypotenuse of a right-angled triangle with sides of lengths ${d}$ across the train and ${u\Delta t/2}$ along it. The component of the light’s velocity perpendicular to the direction of motion is therefore

$\displaystyle u_{\perp}=c\frac{d}{\sqrt{d^{2}+\left(u\Delta t/2\right)^{2}}} \ \ \ \ \ (2)$

so the time taken to cross the train once is

 $\displaystyle \frac{\Delta t}{2}$ $\displaystyle =$ $\displaystyle \frac{d}{u_{\perp}}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{c}\sqrt{d^{2}+\left(u\Delta t/2\right)^{2}} \ \ \ \ \ (4)$

Solving for ${\Delta t}$ we get

$\displaystyle \Delta t=\frac{2d}{c}\frac{1}{\sqrt{1-u^{2}/c^{2}}}=\gamma\Delta t' \ \ \ \ \ (5)$

That is, the time measured by ${S}$ is longer than that measured by ${S'}$ by the factor ${\gamma}$, so that ${S}$ thinks that the clock in frame ${S'}$ runs slow. This is the time dilation effect.

# Lorentz transformations and causality

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 4, Problem 4.2.

To determine whether two observers can disagree about the temporal order of two events, we can calculate the invariant interval between the events. Using four-vector notation:

$\displaystyle \Delta s^{2}\equiv\left(\Delta x\right)_{i}\left(\Delta x\right)^{i} \ \ \ \ \ (1)$

This gives three possible types of pairs of events:

1. Timelike: If ${\Delta s^{2}<0}$, then it is possible to find a frame in which the two events occur at the same spatial point, but at different times, since it is the time component ${-\left(\Delta x^{0}\right)^{2}}$ which is negative.
2. Lightlike: If ${\Delta s^{2}=0}$ then ${c^{2}\left(\Delta t\right)^{2}=\Delta x^{2}}$ (if the motion is along the ${x}$ axis; the argument is similar for arbitrary directions), so the events can be connected by a light signal.
3. Spacelike: If ${\Delta s^{2}>0}$, then it is possible to find a frame in which the two events occur at the same time but at different places. Different observers may disagree about which event occurs first.

However, causality is also preserved directly from the Lorentz transformations. Using relativistic units where ${c=1}$, they are, in 2 dimensions:

 $\displaystyle t'$ $\displaystyle =$ $\displaystyle \gamma\left(t-\beta x\right)\ \ \ \ \ (2)$ $\displaystyle x'$ $\displaystyle =$ $\displaystyle \gamma\left(x-\beta t\right) \ \ \ \ \ (3)$

Suppose we observe two events, 1 and 2, in frame ${S}$, and that

$\displaystyle \Delta x\equiv x_{2}-x_{1}=\alpha\left(t_{2}-t_{1}\right)\equiv\alpha\Delta t \ \ \ \ \ (4)$

where ${\alpha}$ is a positive constant (that is, we’re assuming the observer ${S}$ sees event 2 to the right of event 1, and time ${t_{2}}$ is after ${t_{1}}$). If ${\alpha=1}$, then ${\Delta x=\Delta t}$ and the two events could be connected by a light signal, so event 2 could be caused by event 1. If ${\alpha<1}$, then ${\Delta x<\Delta t}$ so it is possible to travel from ${x_{1}}$ to ${x_{2}}$ at less than the speed of light, which means that event 2 could also be caused by event 1. If ${\alpha>1}$ then it is impossible to travel from event 1 to event 2 at less than the speed of light, so the two events cannot be causally connected.

From the Lorentz transformations, we get

 $\displaystyle \Delta t'$ $\displaystyle =$ $\displaystyle \gamma\left(\Delta t-\beta\Delta x\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \gamma\Delta t\left(1-\alpha\beta\right) \ \ \ \ \ (6)$

Since ${\beta\le1}$, if ${\alpha\le1}$ then ${\Delta t'}$ has the same sign as ${\Delta t}$, so both observers will always agree about the order in which the events occur. In other words, if event 1 can cause event 2, then event 1 must always precede event 2 in all reference frames.

However, if ${\alpha>1}$, then if ${\frac{1}{\alpha}<\beta\le1}$, the sign of ${\Delta t'}$ is opposite to the sign of ${\Delta t}$ so that the two observers will disagree about the order of the events. Since the two events are not causally connected in this case, there is no need for the observers to agree on their order.

# Lorentz transformations: derivation from symmetry

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 4, Problem 4.1.

Although we’ve already looked at special relativity several times in this blog, it’s worth working through Chapter 4 in Carroll & Ostlie since they offer a few different ways of looking at some of our previous results.

We can start with the Lorentz transformations. The derivation we studied most recently is that from Griffiths’s book on electromagnetism, in which he first derives the time dilation and length contraction effects and then uses these to derive the Lorentz transformations. Carroll & Ostlie take a somewhat simpler and more elegant approach, but there are still a few points that could be filled in.

The arguments rely on using various symmetries, and also the postulate of the constancy of the speed of light to finish things off.

First, we can use translational invariance to show that the Lorentz transformations must be linear. We’ve already shown that this is the case using a rather involved argument, but in fact there is a simple criterion that can be applied. We start by using the line-painting thought experiment to show that coordinates perpendicular to the direction of relative motion are unaffected, so if we use our usual two coordinate systems ${S}$ and ${S'}$, with ${S}$ at rest relative to the observer, ${S'}$ moving with speed ${u}$ in the ${+x}$ direction and the two frames aligned so that all three of their coordinate axis pairs (${x}$ and ${x'}$, ${y}$ and ${y'}$ and ${z}$ and ${z'}$) are parallel with the origins coinciding at ${t=t'=0}$, then

 $\displaystyle y'$ $\displaystyle =$ $\displaystyle y\ \ \ \ \ (1)$ $\displaystyle z'$ $\displaystyle =$ $\displaystyle z \ \ \ \ \ (2)$

For the remaining two coordinates, the most general linear transformation is

 $\displaystyle x'$ $\displaystyle =$ $\displaystyle a_{11}x+a_{12}y+a_{13}z+a_{14}t\ \ \ \ \ (3)$ $\displaystyle t'$ $\displaystyle =$ $\displaystyle a_{41}x+a_{42}y+a_{43}z+a_{44}t \ \ \ \ \ (4)$

Why linear? Well, suppose we consider the length of a rod in the two frames. The rod is at rest in ${S}$ with one endpoint at ${x_{1}=0}$ and the other at ${x_{2}=L}$. We know that ${S}$ and ${S'}$ disagree about the length of the rod, but one thing we are sure of is that each observer will obtain only one result for the length. In ${S}$ the length is ${L}$ and in ${S'}$ the length is ${L'}$. But suppose we changed the origin in ${S}$ and ${S'}$ by shifting it along the ${x}$ axis by a distance of 1. Then in ${S}$, where the rod is at rest, the coordinates of its endpoints are now ${x_{1}=-1}$ and ${x_{2}=L-1}$ so that the length is still given by ${x_{2}-x_{1}=L}$. Now whatever the Lorentz transformation is, it has to give ${L'}$ for the length as measured by ${S'}$. In the original frames (before we shifted the origins) the length in ${S'}$ (taking the rod to lie on the ${x}$ axis so that ${y=z=0}$)

 $\displaystyle L'$ $\displaystyle =$ $\displaystyle x_{2}'-x_{1}'\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a_{11}\left(x_{2}-x_{1}\right)+a_{14}\left(t_{2}-t_{1}\right)\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a_{11}\left(x_{2}-x_{1}\right)\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a_{11}\left(L-0\right)\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a_{11}L \ \ \ \ \ (9)$

where the third line is true because the two events defining the measurement of the length of the rod occur at the same time in ${S}$ so ${t_{1}=t_{2}}$. Now if we use the shifted origins

 $\displaystyle L'$ $\displaystyle =$ $\displaystyle a_{11}\left(x_{2}-x_{1}\right)+a_{14}\left(t_{2}-t_{1}\right)\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a_{11}\left(L-1-\left(-1\right)\right)\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a_{11}L \ \ \ \ \ (12)$

Thus shifting the origin leaves the length ${L'}$ the same. However, if we use a non-linear transformation ${f\left(x\right)}$ instead of the ${a_{11}x}$ term, then with the original origins

$\displaystyle L'=f\left(L\right)-f\left(0\right) \ \ \ \ \ (13)$

and with the shifted origins

$\displaystyle L'=f\left(L-1\right)-f\left(-1\right) \ \ \ \ \ (14)$

and in general these two values won’t be equal. Even if we do find some non-linear function that gives the same values for these particular choices for ${x_{1}}$ and ${x_{2}}$, what we really need is a transformation that gives the same values for ${L'}$ for all lengths, at any location on the ${x}$ axis. The only transformation that does that is linear.

So much for translational symmetry. Next, we can apply rotational symmetry. If we rotate both coordinates systems by ${180^{\circ}}$ about the ${x}$ axis so that ${y}$ goes to ${-y}$ and ${z}$ to ${-z}$, all we’ve done is change the coordinate system used to describe the problem; we haven’t actually changed any of the events that occur. Thus, the equations 3 and 4 must give the same results with ${y\rightarrow-y}$ and ${z\rightarrow-z}$. By choosing an event with ${y\ne0}$ and ${z=0}$ we have

 $\displaystyle a_{11}x+a_{12}y+a_{14}t$ $\displaystyle =$ $\displaystyle a_{11}x-a_{12}y+a_{14}t\ \ \ \ \ (15)$ $\displaystyle a_{12}y$ $\displaystyle =$ $\displaystyle -a_{12}y \ \ \ \ \ (16)$

so we conclude that ${a_{12}=0}$. Choosing ${y=0}$ and ${z\ne0}$ gives us ${a_{13}=0}$. Thus 3 becomes

$\displaystyle x'=a_{11}x+a_{14}t \ \ \ \ \ (17)$

A similar argument applied to 4 gives ${a_{42}=a_{43}=0}$ so

$\displaystyle t'=a_{41}x+a_{44}t \ \ \ \ \ (18)$

The origin of ${S'}$ is moving to the right with speed ${u}$ in ${S}$, so at time ${t}$ its ${x}$ coordinate is ${x=ut}$, but ${x'=0}$ always since the origin of ${S'}$ is at rest in ${S'}$. Therefore 17 becomes for this origin

 $\displaystyle 0$ $\displaystyle =$ $\displaystyle a_{11}ut+a_{14}t\ \ \ \ \ (19)$ $\displaystyle a_{14}$ $\displaystyle =$ $\displaystyle -a_{11}u \ \ \ \ \ (20)$

The transformations up to this point are

 $\displaystyle x'$ $\displaystyle =$ $\displaystyle a_{11}\left(x-ut\right)\ \ \ \ \ (21)$ $\displaystyle y'$ $\displaystyle =$ $\displaystyle y\ \ \ \ \ (22)$ $\displaystyle z'$ $\displaystyle =$ $\displaystyle z\ \ \ \ \ (23)$ $\displaystyle t'$ $\displaystyle =$ $\displaystyle a_{41}x+a_{44}t \ \ \ \ \ (24)$

To get the final three constants, we need to invoke the constancy of the speed of light ${c}$. Suppose that at ${t=t'=0}$ a pulse of light is generated at the common origins of ${S}$ and ${S'}$. Because ${c}$ is the same in both frames, both observers will see a spherical shell of light expand from their respective origins. The equations of this shell in the two frames are of the same form:

 $\displaystyle x^{2}+y^{2}+z^{2}$ $\displaystyle =$ $\displaystyle \left(ct\right)^{2}\ \ \ \ \ (25)$ $\displaystyle x'^{2}+y'^{2}+z'^{2}$ $\displaystyle =$ $\displaystyle \left(ct'\right)^{2} \ \ \ \ \ (26)$

The last equation gives

 $\displaystyle a_{11}^{2}\left(x-ut\right)^{2}+y^{2}+z^{2}$ $\displaystyle =$ $\displaystyle c^{2}\left(a_{41}x+a_{44}t\right)^{2}\ \ \ \ \ (27)$ $\displaystyle x^{2}\left(a_{11}^{2}-a_{41}^{2}c^{2}\right)+y^{2}+z^{2}$ $\displaystyle =$ $\displaystyle t^{2}\left(c^{2}a_{44}^{2}-u^{2}a_{11}^{2}\right)+2xt\left(c^{2}a_{41}a_{44}+ua_{11}^{2}\right) \ \ \ \ \ (28)$

Comparing this with 25 we get

 $\displaystyle a_{11}^{2}-a_{41}^{2}c^{2}$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (29)$ $\displaystyle c^{2}a_{44}^{2}-u^{2}a_{11}^{2}$ $\displaystyle =$ $\displaystyle c^{2}\ \ \ \ \ (30)$ $\displaystyle c^{2}a_{41}a_{44}+ua_{11}^{2}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (31)$

Multiply the first equation by ${u^{2}}$ and add to the second, and multiply the first equation by ${u}$ and subtract the third:

 $\displaystyle c^{2}a_{44}^{2}-a_{41}^{2}u^{2}c^{2}$ $\displaystyle =$ $\displaystyle c^{2}+u^{2}\ \ \ \ \ (32)$ $\displaystyle -a_{41}^{2}c^{2}u-c^{2}a_{41}a_{44}$ $\displaystyle =$ $\displaystyle u \ \ \ \ \ (33)$

Multiply the second equation by ${-u}$ and add to the first:

 $\displaystyle c^{2}a_{44}^{2}+uc^{2}a_{44}a_{41}-c^{2}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (34)$ $\displaystyle a_{44}^{2}+ua_{44}a_{41}-1$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (35)$

Using 31:

 $\displaystyle a_{44}^{2}-\frac{u^{2}}{c^{2}}a_{44}^{2}$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (36)$ $\displaystyle a_{44}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{1-u^{2}/c^{2}}} \ \ \ \ \ (37)$

From 30:

 $\displaystyle a_{11}^{2}$ $\displaystyle =$ $\displaystyle \frac{c^{2}}{u^{2}}\left(a_{44}^{2}-1\right)\ \ \ \ \ (38)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{c^{2}}{u^{2}}\left(\frac{u^{2}/c^{2}}{1-u^{2}/c^{2}}\right)\ \ \ \ \ (39)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{1-u^{2}/c^{2}}\ \ \ \ \ (40)$ $\displaystyle a_{11}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{1-u^{2}/c^{2}}} \ \ \ \ \ (41)$

Finally, from 29

 $\displaystyle a_{41}^{2}$ $\displaystyle =$ $\displaystyle \frac{1}{c^{2}}\left(a_{11}^{2}-1\right)\ \ \ \ \ (42)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{c^{2}}\left(\frac{u^{2}/c^{2}}{1-u^{2}/c^{2}}\right)\ \ \ \ \ (43)$ $\displaystyle a_{41}$ $\displaystyle =$ $\displaystyle \frac{u}{c^{2}\sqrt{1-u^{2}/c^{2}}} \ \ \ \ \ (44)$

Putting it all together gives the familiar Lorentz transformations:

 $\displaystyle x'$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{1-u^{2}/c^{2}}}\left(x-ut\right)\ \ \ \ \ (45)$ $\displaystyle y'$ $\displaystyle =$ $\displaystyle y\ \ \ \ \ (46)$ $\displaystyle z'$ $\displaystyle =$ $\displaystyle z\ \ \ \ \ (47)$ $\displaystyle t'$ $\displaystyle =$ $\displaystyle \frac{t-ux/c^{2}}{\sqrt{1-u^{2}/c^{2}}} \ \ \ \ \ (48)$

# Shaula (Lambda Scorpii)

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 3, Problem 3.19.

As another example of calculating the colour indices of a star using the blackbody radiation rate, we’ll look at the star Shaula (${\lambda}$ Scorpii), which has a surface temperature of around 22000 K. We can use the formulas:

 $\displaystyle U-B$ $\displaystyle =$ $\displaystyle -2.5\log\frac{\lambda_{B}^{5}\left(e^{hc/\lambda_{B}k_{B}T}-1\right)\Delta\lambda_{U}}{\lambda_{U}^{5}\left(e^{hc/\lambda_{U}k_{B}T}-1\right)\Delta\lambda_{B}}+C_{U-B}\ \ \ \ \ (1)$ $\displaystyle B-V$ $\displaystyle =$ $\displaystyle -2.5\log\frac{\lambda_{V}^{5}\left(e^{hc/\lambda_{V}k_{B}T}-1\right)\Delta\lambda_{B}}{\lambda_{B}^{5}\left(e^{hc/\lambda_{B}k_{B}T}-1\right)\Delta\lambda_{V}}+C_{B-V} \ \ \ \ \ (2)$

where

 $\displaystyle C_{U-B}$ $\displaystyle =$ $\displaystyle -0.87\ \ \ \ \ (3)$ $\displaystyle C_{B-V}$ $\displaystyle =$ $\displaystyle +0.65 \ \ \ \ \ (4)$

Plugging in the numbers, we get

 $\displaystyle U-B$ $\displaystyle =$ $\displaystyle -1.076\ \ \ \ \ (5)$ $\displaystyle B-V$ $\displaystyle =$ $\displaystyle -0.227 \ \ \ \ \ (6)$

The measured values given by Carroll & Ostlie are ${U-B=-0.90}$ and ${B-V=-0.23}$ so the blackbody values are quite close to those measured.

The apparent visual magnitude of Shaula is ${V=1.62}$ and its parallax as measured by Hipparcos is ${0.00464^{\prime\prime}}$. The distance of Shaula from Earth is therefore

$\displaystyle d=\frac{1}{p}=215.517\mbox{ pc} \ \ \ \ \ (7)$

The absolute visual magnitude is therefore

 $\displaystyle M_{V}$ $\displaystyle =$ $\displaystyle V+5-5\log d\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -5.047 \ \ \ \ \ (9)$

If it were 10 parsecs from Earth, it would be the second brightest object (after the moon) in the night sky, outshining even Venus.

# Colour indices in terms of blackbody radiation rate

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 3, Problem 3.18.

We can write the apparent magnitude ${m_{\lambda}}$ of a star for some wavelength band in terms of the flux ${F_{\lambda}}$ in that band received on Earth from the star as

 $\displaystyle m_{\lambda}$ $\displaystyle =$ $\displaystyle -2.5\log\int F_{\lambda}S_{\lambda}d\lambda+C_{\lambda} \ \ \ \ \ (1)$

The constant${C_{\lambda}}$ depends on the wavelength interval over which the integral is done, and on the sensitivity ${S_{\lambda}}$ of the detector, so it will be different for different filters. If we measure the apparent magnitudes in the three standard bands ${U}$, ${B}$ and ${V}$ and treat the star as a blackbody, we can work out the three constants ${C_{U}}$, ${C_{B}}$ and ${C_{V}}$. Since colour indices are commonly used to classify stars, we can work out similar equations for these indices.

 $\displaystyle U-B$ $\displaystyle =$ $\displaystyle -2.5\log\int F_{U}S_{U}d\lambda+C_{U}+2.5\log\int F_{B}S_{B}d\lambda-C_{B}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -2.5\log\frac{\int F_{U}S_{U}d\lambda}{\int F_{B}S_{B}d\lambda}+C_{U-B} \ \ \ \ \ (3)$

where ${C_{U-B}\equiv C_{U}-C_{B}}$. Since the argument of the log is now dimensionless, the constant ${C_{U-B}}$ is independent of the units used to measure flux. In their example 3.6.2, Carroll & Ostlie use a star with surface temperature of ${T=42000\mbox{ K}}$ and measured colour indices of ${U-B=-1.19}$ and ${B-V=-0.33}$. The standard filters are

• ${U}$: ${365\pm34\mbox{ nm}}$;
• ${B}$: ${440\pm49\mbox{ nm}}$;
• ${V}$: ${550\pm44.5\mbox{ nm}.}$

If we make the assumptions that ${S=1}$ for each filter within these bands and ${S=0}$ outside these bands, and that the flux doesn’t change much over the bandwidth in each filter, we can approximate the relation above by

$\displaystyle U-B\approx-2.5\log\frac{F_{U}\Delta\lambda_{U}}{F_{B}\Delta\lambda_{B}}+C_{U-B} \ \ \ \ \ (4)$

with a similar relation for ${B-V}$:

$\displaystyle B-V\approx-2.5\log\frac{F_{B}\Delta\lambda_{B}}{F_{V}\Delta\lambda_{V}}+C_{B-V} \ \ \ \ \ (5)$

As we’re dealing with ratios of flux for the same star, we can express these equations in terms of the blackbody radiation rate:

 $\displaystyle U-B$ $\displaystyle =$ $\displaystyle -2.5\log\frac{B_{U}\Delta\lambda_{U}}{B_{B}\Delta\lambda_{B}}+C_{U-B}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -2.5\log\frac{\lambda_{B}^{5}\left(e^{hc/\lambda_{B}k_{B}T}-1\right)\Delta\lambda_{U}}{\lambda_{U}^{5}\left(e^{hc/\lambda_{U}k_{B}T}-1\right)\Delta\lambda_{B}}+C_{U-B}\ \ \ \ \ (7)$ $\displaystyle B-V$ $\displaystyle =$ $\displaystyle -2.5\log\frac{\lambda_{V}^{5}\left(e^{hc/\lambda_{V}k_{B}T}-1\right)\Delta\lambda_{B}}{\lambda_{B}^{5}\left(e^{hc/\lambda_{B}k_{B}T}-1\right)\Delta\lambda_{V}}+C_{B-V} \ \ \ \ \ (8)$

Plugging in the values for the star mentioned above, we get

 $\displaystyle C_{U-B}$ $\displaystyle =$ $\displaystyle -0.87\ \ \ \ \ (9)$ $\displaystyle C_{B-V}$ $\displaystyle =$ $\displaystyle +0.65 \ \ \ \ \ (10)$

Given these values, we can now estimate the colour indices for any star if we know its temperature. For the Sun, ${T=5777\mbox{ K}}$ and plugging this into 7 and 8, we get

 $\displaystyle U-B$ $\displaystyle =$ $\displaystyle -0.222\ \ \ \ \ (11)$ $\displaystyle B-V$ $\displaystyle =$ $\displaystyle +0.571 \ \ \ \ \ (12)$

The measured values given by Carroll & Ostlie are ${U-B=+0.195}$ and ${B-V=+0.65}$ so the agreement isn’t great, especially for ${U-B}$.

# Bolometric magnitude from flux

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 3, Problem 3.17.

The apparent magnitude of a star at a particular wavelength can be written in terms of the flux observed at that wavelength on Earth by

 $\displaystyle m$ $\displaystyle =$ $\displaystyle -2.5\log\int F_{\lambda}S_{\lambda}d\lambda+C\ \ \ \ \ (1)$ $\displaystyle C$ $\displaystyle \equiv$ $\displaystyle M_{Sun}+2.5\log\int F_{\lambda,10,Sun}S_{\lambda}d\lambda \ \ \ \ \ (2)$

Here ${S_{\lambda}}$ is the sensitivity function and indicates what fraction of the actual flux a particular telescope receives at a given wavelength. This formula is actually not well-formed since whenever we use a transcendental function such as the logarithm, its argument should be dimensionless. It is true that if we combine the two terms into a single logarithm, we get

$\displaystyle m=-2.5\log\frac{\int F_{\lambda}S_{\lambda}d\lambda}{\int F_{\lambda,10,Sun}S_{\lambda}d\lambda}+M_{Sun} \ \ \ \ \ (3)$

giving a dimensionless argument for the log term. However, in the original form, the constant ${C}$ depends on the units used for the flux.

It seems to be standard to use ${\mbox{watts m}^{-2}}$ for total flux, so the units of ${F_{\lambda}}$ are ${\mbox{watts m}^{-2}\mbox{ nm}^{-1}}$ if the wavelength ${\lambda}$ is given in nanometres.

For a bolometric magnitude, we set ${S_{\lambda}=1}$ for all ${\lambda}$. The bolometric flux for the Sun at the distance of Earth is

$\displaystyle \int_{0}^{\infty}F_{\lambda}d\lambda=1365\mbox{ W m}^{-2} \ \ \ \ \ (4)$

Taking the apparent bolometric magnitude of the Sun as ${m_{Sun}=-26.83}$ we get

$\displaystyle C_{bol}=-18.992 \ \ \ \ \ (5)$

As a consistency check, we can plug in the values for Dschubba (Delta Sco), whose flux at Earth is ${F=6.44\times10^{-8}\mbox{ W m}^{-2}}$:

$\displaystyle m=-2.5\log\left(6.44\times10^{-8}\right)-18.992=-1.01 \ \ \ \ \ (6)$

which just gives us the bolometric magnitude we had before.

# Vega as a blackbody

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 3, Problem 3.16.

The blackbody radiation rate in a wavelength interval ${d\lambda}$ is

$\displaystyle B_{\lambda}\left(T\right)d\lambda=\frac{2hc^{2}}{\lambda^{5}\left(e^{hc/\lambda k_{B}T}-1\right)}d\lambda \ \ \ \ \ (1)$

The luminosity ${L_{\lambda}}$ within the same wavelength interval is the radiation rate times the surface area of the star:

$\displaystyle L_{\lambda}d\lambda=4\pi R^{2}B_{\lambda}=\frac{8\pi hc^{2}R^{2}}{\lambda^{5}\left(e^{hc/\lambda k_{B}T}-1\right)}d\lambda \ \ \ \ \ (2)$

The flux ${F_{\lambda}}$ received by an observer on Earth at a distance ${r}$ from the star is then

$\displaystyle F_{\lambda}d\lambda=\frac{L_{\lambda}}{4\pi r^{2}}=\frac{2hc^{2}}{\lambda^{5}\left(e^{hc/\lambda k_{B}T}-1\right)}\frac{R^{2}}{r^{2}}d\lambda \ \ \ \ \ (3)$

The apparent magnitude ${m}$ in some wavelength interval (for example, one of the ${U}$, ${B}$ and ${V}$ magnitudes) is

$\displaystyle m=M_{Sun}-2.5\log\frac{\int F_{\lambda}S_{\lambda}d\lambda}{\int F_{\lambda,10,Sun}S_{\lambda}d\lambda} \ \ \ \ \ (4)$

where ${M_{Sun}}$ is the absolute magnitude of the Sun over the same wavelength interval and ${F_{\lambda,10,Sun}}$ is the flux ${F_{\lambda}}$ for the Sun at a distance of 10 pc. The function ${S_{\lambda}}$is the sensitivity function and indicates what fraction of the star’s light is received at a given wavelength ${\lambda}$. Thus ${0\le S_{\lambda}\le1}$ for all ${\lambda}$. Since the values for the Sun in 4 are same for every star, we can write it as

 $\displaystyle m$ $\displaystyle =$ $\displaystyle -2.5\log\int F_{\lambda}S_{\lambda}d\lambda+C\ \ \ \ \ (5)$ $\displaystyle C$ $\displaystyle \equiv$ $\displaystyle M_{Sun}+2.5\log\int F_{\lambda,10,Sun}S_{\lambda}d\lambda \ \ \ \ \ (6)$

${C}$ depends on the wavelength interval over which the integral is done, and on the sensitivity ${S_{\lambda}}$ of the detector, so it will be different for different filters and telescopes. In practice, ${C_{U}}$, ${C_{B}}$ and ${C_{V}}$ are all chosen so that ${U}$, ${B}$ and ${V}$ are all zero for the star Vega. Unfortunately, this doesn’t mean that Vega is actually the same brightness when viewed in these three regions of the spectrum. We can find the wavelength band in which Vega actually appears brightest by calculating the colour indices using the above formulas.

 $\displaystyle U-B$ $\displaystyle =$ $\displaystyle M_{Sun}-2.5\log\left[\frac{\int F_{\lambda}S_{\lambda}d\lambda}{\int F_{\lambda,10,Sun}S_{\lambda}d\lambda}\right]_{U}-M_{Sun}+2.5\log\left[\frac{\int F_{\lambda}S_{\lambda}d\lambda}{\int F_{\lambda,10,Sun}S_{\lambda}d\lambda}\right]_{B}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2.5\log\frac{\left[\int F_{\lambda}S_{\lambda}d\lambda\right]_{B}}{\left[\int F_{\lambda}S_{\lambda}d\lambda\right]_{U}}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2.5\log\frac{\left[\int B_{\lambda}S_{\lambda}d\lambda\right]_{B}}{\left[\int B_{\lambda}S_{\lambda}d\lambda\right]_{U}} \ \ \ \ \ (9)$

where in each case the integrals in square brackets are evaluated over the wavelength range corresponding to the subscript ${U}$ or ${B}$. If the wavelength filters are narrow enough and we take the sensitivity function ${S_{\lambda}}$ to be 1 inside the filter’s range and 0 outside, we can approximate the integrals by calculating ${B_{\lambda}}$ at the midpoint of the wavelength range and just multiplying by the filter’s bandwidth ${\Delta\lambda}$. That is, we get

 $\displaystyle U-B$ $\displaystyle \approx$ $\displaystyle 2.5\log\frac{B_{\lambda_{B}}\Delta\lambda_{B}}{B_{\lambda_{U}}\Delta\lambda_{U}}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2.5\log\frac{\lambda_{U}^{5}\left(e^{hc/\lambda_{U}k_{B}T}-1\right)\Delta\lambda_{B}}{\lambda_{B}^{5}\left(e^{hc/\lambda_{B}k_{B}T}-1\right)\Delta\lambda_{U}} \ \ \ \ \ (11)$

There’s a similar relation for ${B-V}$:

$\displaystyle B-V\approx2.5\log\frac{\lambda_{B}^{5}\left(e^{hc/\lambda_{B}k_{B}T}-1\right)\Delta\lambda_{V}}{\lambda_{V}^{5}\left(e^{hc/\lambda_{V}k_{B}T}-1\right)\Delta\lambda_{B}} \ \ \ \ \ (12)$

The standard filters are

• ${U}$: ${365\pm34\mbox{ nm}}$;
• ${B}$: ${440\pm49\mbox{ nm}}$;
• ${V}$: ${550\pm44.5\mbox{ nm}.}$

Using a temperature of ${T=9600\mbox{ K}}$ for Vega, we get

 $\displaystyle U-B$ $\displaystyle =$ $\displaystyle +0.161\ \ \ \ \ (13)$ $\displaystyle B-V$ $\displaystyle =$ $\displaystyle -0.539 \ \ \ \ \ (14)$

Thus a blackbody with ${T=9600\mbox{ K}}$ as an approximation to Vega would appear brightest in the blue region, since ${U>B}$ and ${B so ${B}$ is the smallest (brightest) magnitude.

# Welcome to Physics Pages

This blog consists of my notes and solutions to problems in various areas of mainstream physics. An index to the topics covered is contained in the links in the sidebar on the right, or in the menu at the top of the page.

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# Average of product of two waves

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 9, Post 11.

A common calculation that is required when analyzing any system that varies with a sinusoidal period is a time average over one cycle. For example, a monochromatic plane wave with amplitude ${A}$, direction ${\mathbf{k}}$, frequency ${\omega}$ and phase ${\delta}$ can be written as

 $\displaystyle f$ $\displaystyle =$ $\displaystyle A\cos\left(\mathbf{k}\cdot\mathbf{r}-\omega t+\delta\right)=\Re\tilde{A}e^{i\left(\mathbf{k}\cdot\mathbf{r}-\omega t\right)}\ \ \ \ \ (1)$ $\displaystyle \tilde{A}$ $\displaystyle =$ $\displaystyle Ae^{i\delta} \ \ \ \ \ (2)$

Now suppose we have two waves with the same direction and frequency, but different amplitudes and phases. Then

 $\displaystyle f$ $\displaystyle =$ $\displaystyle A\cos\left(\mathbf{k}\cdot\mathbf{r}-\omega t+\delta_{a}\right)\ \ \ \ \ (3)$ $\displaystyle g$ $\displaystyle =$ $\displaystyle B\cos\left(\mathbf{k}\cdot\mathbf{r}-\omega t+\delta_{b}\right) \ \ \ \ \ (4)$

The average of the product of these waves over a single cycle is then

$\displaystyle \left\langle fg\right\rangle =\frac{\omega AB}{2\pi}\int_{0}^{2\pi/\omega}\cos\left(\mathbf{k}\cdot\mathbf{r}-\omega t+\delta_{a}\right)\cos\left(\mathbf{k}\cdot\mathbf{r}-\omega t+\delta_{b}\right)dt \ \ \ \ \ (5)$

We can transform this integral by defining

 $\displaystyle \theta$ $\displaystyle \equiv$ $\displaystyle \mathbf{k}\cdot\mathbf{r}-\omega t\ \ \ \ \ (6)$ $\displaystyle d\theta$ $\displaystyle =$ $\displaystyle -\omega dt\ \ \ \ \ (7)$ $\displaystyle \left\langle fg\right\rangle$ $\displaystyle =$ $\displaystyle \frac{AB}{2\pi}\int_{0}^{2\pi}\cos\left(\theta+\delta_{a}\right)\cos\left(\theta+\delta_{b}\right)d\theta \ \ \ \ \ (8)$

We’ve used the limits of 0 and ${2\pi}$ since any interval of ${2\pi}$ covers one complete cycle of ${\theta}$.

The two cosines have the same period and differ only in their phase, so we will get the same result from the integral if we replace them by

 $\displaystyle \cos\left(\theta+\delta_{a}\right)\cos\left(\theta+\delta_{b}\right)$ $\displaystyle \rightarrow$ $\displaystyle \cos\theta\cos\left(\theta+\delta_{a}-\delta_{b}\right)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \cos^{2}\theta\cos\left(\delta_{a}-\delta_{b}\right)-\cos\theta\sin\theta\sin\left(\delta_{a}-\delta_{b}\right) \ \ \ \ \ (10)$

We now have

 $\displaystyle \left\langle fg\right\rangle$ $\displaystyle =$ $\displaystyle \frac{AB}{2\pi}\cos\left(\delta_{a}-\delta_{b}\right)\int_{0}^{2\pi}\cos^{2}\theta d\theta-\frac{AB}{2\pi}\sin\left(\delta_{a}-\delta_{b}\right)\int_{0}^{2\pi}\cos\theta\sin\theta d\theta\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}AB\cos\left(\delta_{a}-\delta_{b}\right)-0\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}AB\cos\left(\delta_{a}-\delta_{b}\right)\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\Re\left(fg^*\right)=\frac{1}{2}\Re\left(f^*g\right) \ \ \ \ \ (14)$

Thus we can get the answer using complex notation without doing any integrals.

This applies to vector products as well, since the components of vector products are just products of scalar functions. For example, the time average of the Poynting vector becomes, when the electric and magnetic fields are written in complex notation:

$\displaystyle \left\langle \mathbf{S}\right\rangle =\frac{1}{2\mu_{0}}\Re\left(\tilde{\mathbf{E}}\times\tilde{\mathbf{B}}^*\right) \ \ \ \ \ (15)$

The electromagnetic energy density in the fields has a time average of

 $\displaystyle \left\langle u_{em}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{4}\Re\left(\epsilon_{0}\tilde{\mathbf{E}}\cdot\tilde{\mathbf{E}}^*+\frac{1}{\mu_{0}}\tilde{\mathbf{B}}\cdot\tilde{\mathbf{B}}^*\right)\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{4}\left(\epsilon_{0}\tilde{\mathbf{E}}\cdot\tilde{\mathbf{E}}^*+\frac{1}{\mu_{0}}\tilde{\mathbf{B}}\cdot\tilde{\mathbf{B}}^*\right)\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{4\mu_{0}}\left(\frac{1}{c^{2}}\tilde{\mathbf{E}}\cdot\tilde{\mathbf{E}}^*+\tilde{\mathbf{B}}\cdot\tilde{\mathbf{B}}^*\right) \ \ \ \ \ (18)$

We dropped the ${\Re}$ in line 2 since the quantity in parentheses is automatically real anyway, and in the last line we used

$\displaystyle \mu_{0}\epsilon_{0}=\frac{1}{c^{2}} \ \ \ \ \ (19)$