# Special Relativity Problem -- Speeding through a red light to make it look green

By now i must be driving you guys crazy with my relativity questions , but this is my last one! i swear! =]

## Homework Statement

A noted physicist was stopped by police for travelling through a red traffic light. The physicist argued that to because he was travelling fast, the colour of the light had changed from the red wavelength to that of a green wavelength

The judge said that he would let him off the charge of running the red light, but would fine him for speeding at the rate of one cent for every 100km/hr he was travelling over 100km/hr

Was he fined? and how much was his fine?

Assume
c=3x10^8ms-1
frequency of red light=4.5x10^14 Hz
frequency of green light= 6.0 x 10^14 Hz

## Homework Equations

Would this problem also be related to the Dopper Shifty as the change of colour of the traffic lights? And therefore would the doppler shift be appropriate to use in this case ? But i have come across a relativistic doppler formula; f/f0 = √[(1 + v/c)/(1 - v/c)] ?

there is a non relativistic doppler formula ?
hence ;
∆f = f0*(v/c)

v = ∆f/f0 * c

where ;
∆f = 1.5*10^14
f0 = 4.5*10^14
v = 0.33 c

How can i go about this problem and any other ideas of what is involved in solving the problem ?

According to our modern day understanding of physics, light is neither a wave nor a particle. Sometimes, light behaves as a wave, sometimes as a particle, but not both at the same time. Light, as a wave that exhibits wave motion, is a periodic motion characterized, among other things, by a wavelength ##λ~##, a period ##T_0~##, and a wave frequency ##ν~##.
##\dots~λ~\Rightarrow~\begin{cases}\begin{align} & \text {the distance traveled by the wave along the } \nonumber \\
& \text {direction of propagation in one period or one cycle} \nonumber \\
& \text {of the motion before the next cycle is repeated}~\dots\nonumber \end{align}\end{cases}##
##\dots~T_0~\Rightarrow~\begin{cases}\begin{align} & \text {the time to complete one cycle and travel } \nonumber \\ & \text { the distance of one wavelength }~ λ~\dots\nonumber \end{align}\end{cases}##
##\dots~ν~\Rightarrow~\begin{cases}\begin{align} & \text { tells the number of waves passing } \nonumber \\ & \text {through a given point in a given time }~\dots\nonumber \end{align}\end{cases}##
The larger the period ##T_0## (the longer it takes for one complete cycle to pass through), the longer the wavelength ##λ~##, and the smaller the frequency ##ν## (the fewer the number of full cycles that pass through). Being more familiar with length in the Special Theory of Relativity, we consider the wavelength of light in this problem instead of the frequency.
##\dots~## for red light ##~\Rightarrow~##average ##~λ_\rm R## = 685##×10^{-9}## m = 685 nm##~\dots~##
##\dots~## for green light ##~\Rightarrow~##average ##~λ_\rm G## = 533##×10^{-9}## m = 533 nm##~\dots~##
Red light becoming green with ##~λ_\rm R~\gt~λ_\rm G~\Rightarrow~##length contraction##~\dots~##
##\Rightarrow~λ_\rm G## = ##λ_\rm R[1 - (v/c)^2]^{1/2}\Rightarrow~1 - (v/c)^2~##= ##(λ_\rm G/λ_\rm R)^2~##= 0.605443##~\dots~##
##\Rightarrow~(v/c)~##=##~(0.394557)^{1/2}~##= 0.628##~\Rightarrow~v~##= ##(0.628)c~\Rightarrow~v~##=##~1.884 ×10^8~\rm m ⋅ \rm s^{-1}~##...
##\Rightarrow~v~##=##~6.7824 ×10^8~\rm km ⋅ \rm hr^{-1}~\Rightarrow~v/(100~\rm km ⋅ \rm hr^{-1})~##=##~6.7824 ×10^6~##gives the number of times in excess of the 100##~\rm km ⋅ \rm hr^{-1}~##speed limit if the red traffic light appears green to the driver.
Being charged of a fine at the rate of one cent for every 100##~\rm km ⋅ \rm hr^{-1}~##he was traveling over 100##~\rm km ⋅ \rm hr^{-1}~##, the driver must pay a total fine of almost $68,000. berkeman Mentor the driver must pay a total fine of almost$68,000.
@kumusta -- Because this thread is over 10 years old, it is okay for you to post a solution. Just keep in mind that for current schoolwork/homework threads, solutions cannot be provided by us. The student must do the bulk of the work. Thanks.

Everything clear. No solutions for current schoolwork/homework threads.

• berkeman
Thanks too.

kuruman
Homework Helper
Gold Member
For the record, the number provided by @kumusta does not use the relativistic Doppler shift equation. The answer I got with ##\beta = v/c=\frac{7}{25}## from using the equation ##\dfrac{f_{\text{source}}}{f_{\text{receiver}}}=\sqrt{\dfrac{1+\beta}{1-\beta}}## is $30,200 plus change. • • hutchphd, PeroK and berkeman My solution took into account the relativistic effect of length contraction only, if it could be called as such. I failed to consider the fact that time is not absolute but is generally different for different observers. So obviously, the solution that I gave cannot be the correct one. Many thanks to kuruman for pointing it out. • berkeman and kuruman While looking at kuruman's post #6, I noticed that since ##(1 - β) < (1 + β)~##, then he must have used the frequency of green light for ##f_\rm {source}~##and the frequency of red light for ##f_\rm {receiver}~##: ... the answer I got with ##\beta = v/c=\frac{7}{25}## from using the equation ##\dfrac{f_{\text{source}}}{f_{\text{receiver}}}=\sqrt{\dfrac{1+\beta}{1-\beta}}## is$30,200 plus change.
I think it should be the other way around since the source, which is the traffic light, must have shown red and not green. I suspect now that there's also something wrong with kuruman's computations.
Anyway, the answer ##~v = (0.628)c~## that I got must be rather high. Knowing the fact that frequency and wavelength are related by the equation ##~v = fλ = λ/T~##, taking relativistic time dilation into account too, with larger value of ##~T~##, aside from length contraction, would clearly give a smaller value for ##~v~## than the one which treats time as an absolute quantity.

• PeroK
kuruman
Homework Helper
Gold Member
The ratio of the frequencies is ##r=\dfrac{6.0}{4.5}=\dfrac{4}{3}.##

The solution of ##\dfrac{4}{3}=\sqrt{\dfrac{1+\beta}{1-\beta}}## is ##\beta=+\dfrac{7}{25}.##

The solution of ##\dfrac{3}{4}=\sqrt{\dfrac{1+\beta}{1-\beta}}## is ##\beta=-\dfrac{7}{25}.##

Since the fine is for speeding and not velocitying, the sign does not matter and the defendant has to pay the same fine in either case.

berkeman
Mentor
velocitying
TIL a new word at PF... • • jbriggs444 and kuruman
kuruman
Homework Helper
Gold Member
TIL a new word at PF... velocity [ vuh-los-i-tee ]
noun, plural ve·loc·i·ties
rapidity of motion or operation; swiftness;
a high wind velocity

verb (used without object), ve·loc·i·tyed, ve·loc·i·ty·ing
to move in a fixed direction
we velocitied from the Northside to the Southside in record time

• berkeman
berkeman
Mentor
You using the Urban Dictionary again? This is what I get with Google...  kuruman
Homework Helper
Gold Member
You using the Urban Dictionary again? This is what I get with Google... View attachment 289177
Truth be told not even the Urban Dictionary has the word as a verb, I looked. I made it up initially for post #9 and then came up with a dictionary-style definition after seeing your post #10. Hey, that's how the language evolves, no? If enough people use a word in a new context, it will find its way into the real dictionaries.

• berkeman
For the record, it appears that velocitying is the name of a YouTube channel.
View attachment 289187
It's been in YouTube since February 2007? But post #9 was made only yesterday, Thu16Sept2021
Truth be told not even the Urban Dictionary has the word as a verb, I looked. I made it up initially for post #9 and then came up with a dictionary-style definition after seeing post #10 ...

#### Attachments

jbriggs444
Homework Helper
You cannot copyright this word. At least in the U.S. The doctrine of merger applies. In the U.S. you are allowed to copyright the expression of an idea (i.e. the book "Huckleberry Finn") but not the idea of a child floating down a river on a raft with a slave. If there are only a limited number of ways to express an idea, the expression merges with the idea and the expression becomes uncopyrightable.

Trademark, however, follows a different standard. Which explains how the phrase "Super Bowl" can be trademarked, even though it would not be copyrightable.

• hutchphd
It's okay if that is so. But I think no one has the right to claim just a couple of days ago that a new word originated from him when that particular word already existed more than 10 years ago.

kuruman
Homework Helper
Gold Member
It's okay if that is so. But I think no one has the right to claim just a couple of days ago that a new word originated from him when that particular word already existed more than 10 years ago.
I assume you are referring to my posts #11 and #13.

Firstly, I never claimed that I originated a new word. The word "velocity" has existed as a noun for centuries (velocitas in Latin). I acknowledged so explicitly in #11 by showing the existing noun and, beneath it, the verb "to velocity" and then again in #13 where I stated "If enough people use a word in a new context, ##\dots~##"

Secondly, what I made up from the word "velocity" is its use as a verb. To that end, I provided a dictionary-like definition complete with past tense to show that the verb is regular. I included a sentence to illustrate its use in a sentence as an intransitive verb without an object.

Thirdly, the use of "velocitying" in YouTube is irrelevant and only superficially akin to the verb defined in #11, but it is not a verb or part thereof. It is the name of a playlist, therefore a noun, and conveys no action. An example of its use in a sentence might be, "Velocitying is a YouTube playlist" in which, as a noun, it serves as the subject of the copulative verb "is".

Fourthly, I believe that the original question posed by the OP has been fully, albeit belatedly, answered therefore I will henceforth cease and desist posting more silliness on this revenant thread.

• jbriggs444