Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Driving a car with headlights on at the speed of light: another question

  1. Jan 5, 2012 #1
    So I remember in basic physics the scenario:

    If you're driving at the speed of light, and you turn on your headlights, can you see them?
    I know the standard answer is yes; that it propagates at speed c no matter the reference frame.

    So how about this:

    Suppose I'm driving my car towards a mirror at a speed v (slightly less than c), which is far off at some distance d. Suppose my headlights give off wavelength λ.

    how long does it take me to see the reflection off the mirror?

    how long does it take for my friend to detect the light at the mirror?
    (given he somehow knew precisely when the headlights were turned on)

    What frequency of light do I observe coming from my headlights, and what frequency do I observe returning from the mirror?

    what frequency is measured at the mirror?

    I'm assuming the reflection will be blue-shifted from the perspective of the car.
    Does that mean if I pulse my headlights for a time T, that the reflection will be a shorter pulse of time T' ? (in order to conserve energy?)
  2. jcsd
  3. Jan 5, 2012 #2
    I thought this would be simple, but now I see SR/GR all over it. :-(
  4. Jan 5, 2012 #3
    I'm just curious you know... If I were riding in a spaceship at speed ~c, and I tried to answer this, I'd be like well - either the light will get there in half the time it takes us, or we'll slam into it together and pulverize my buddy trying to measure it. lol
  5. Jan 5, 2012 #4
    The distance from u 2 the mirror gets lorentz contracted according to gamma, which is the lorentz factor l=√l-v^2/c^2.

    D=r x t, so t which is time to see ur reflection is (d-gamma/300k)•2
  6. Jan 5, 2012 #5


    Staff: Mentor

    You can't drive your car at the speed of light to begin with, so the question as stated this way is not well-posed. See the Usenet Physics FAQ here:


    No, the "standard" answer to the question as you stated it is what I said above.

    This part is OK, but there is no "reference frame" that moves at the speed of light.

    Ok, this way of posing the question is well-defined, since v < c.

    According to your clock, or according to a clock that is at rest with respect to the mirror? We can do the calculation in either frame and then just apply the time dilation factor to get the answer in the other; let's try the latter frame first. In that frame, say you are a distance d from the mirror when you turn on the headlights. The beam will take a time d/c to travel to the mirror. When it reaches the mirror, you will be a distance d(1 - v/c) from the mirror (because you traveled a distance dv/c in time d/c). You are still moving to the right at v, and the reflected beam moves to the left at c, so it takes an additional time d(1 - v/c)/(v + c) for the reflected beam to return to you. That gives a total time d/c[1 + (c - v)/(c + v)], which simplifies to 2d/(v + c).

    (A simpler way to get the same answer is to think of the beam as coming from your reflection in the mirror. That reflection is a distance 2d away from you, and the beam moves to the left from that point at c while you move to the right at v, so your "closure rate" is v + c.)

    The above time is for a clock at rest with respect to the mirror; to get the time that elapses on your clock, just divide by the Lorentz gamma factor for v.

    Just to clarify, your friend is co-located with the mirror and at rest relative to it? If so, the time is d/c.

    The only way he can know is to see the beam, unless you and he pre-arranged some scenario such that he could calculate in advance at what time, according to his clock, you would turn on the headlights.

    Assuming you had a measuring device just in front of your car, at rest relative to the car, it would measure the light coming from your headlights to have wavelength λ. (Since you specified wavelength above, it would be better to ask for answers in terms of wavelength; or else specify the headlight frequency instead.)

    Since the mirror is moving towards you, you will see light returning from the mirror as blueshifted (higher frequency, shorter wavelength) compared to λ.

    A measuring device at rest relative to the mirror will see you coming towards it, so it will measure the light from the headlights to be blueshifted by the same factor as you see the light returning from the mirror.

    As observed by you in the car, yes, the outgoing pulse will last for a longer time than the incoming reflected pulse.
  7. Jan 5, 2012 #6
    If you slam into the mirror at just under the speed of light, is it 7 years of bad luck, 7*gamma or 7/gamma?
  8. Jan 7, 2012 #7
    What is the speed of my car as I would measure it from inside? If time is dilated for me, then observed objects would appear to move slower, right? (slower than the speed measured from a stationary frame)

    suppose there were 'mile markers' I could see from the car - how long would I measure the time taken to pass between them? (say markers every c*60 m, as measured from a stationary frame)

    If I then said v= d/t = c*60/t, then what speed would I think I'm travelling at? (just by observing the time(t) it takes me to pass between the markers.)

    it would be less than the speed observed from a stationary frame, right?
  9. Jan 7, 2012 #8


    User Avatar
    Science Advisor
    Gold Member

    No, you would measure the speed of the milestones passing you at the same speed that they measure you passing them. However, you will also measure the milestones to be closer together than one mile, so if you claimed that they were really one mile apart, you would conclude that you were traveling faster than you really were.
  10. Jan 7, 2012 #9
    So we're saying time is dilated, and the length is contracted. And we both measure the same speed.

    if I use v=d/t

    then d/t = d'/t' (' meaning moving frame)

    if d > d' (contraction in ' frame)

    then d/t = d'/t' iff t > t' , right?
  11. Jan 7, 2012 #10


    Staff: Mentor

    Don't forget the relativity of simultaneity. The other guys time is dilated, length is contracted, and his clocks are desynchronized so that you both measure the same speed.
  12. Jan 7, 2012 #11


    User Avatar

    Staff: Mentor

  13. Jan 7, 2012 #12


    User Avatar
    Science Advisor
    Gold Member

    What d and d' are you proposing in both frames? I presume in the milestone frame the distance between pairs of milestones is one mile and the milestone observer has synchronized clocks placed at each milestone and when you pass each one, a record is made of the time and then a calculation of your speed is made by taking the time difference between a pair of milestone markers and dividing that into one mile, correct?

    But how do you measure the speed of the milestone markers as they pass you? I already said that if you use a clock traveling with you and you measure how long it takes to get from one milestone marker to the next and you assume that they are one mile apart, you will calculate that you are traveling faster than you really are. Do you understand why you cannot calculate your speed that way?

    So what distance would you use and how would you measure the time? What if there was just one milestone marker, how would you measure its speed relative to you? Have you thought about this?
  14. Jan 7, 2012 #13
    lets make it simple - Say I'm very good at counting seconds in my head. so no clocks.

    the distance, d, is the distance between markers relative to a stationary observer/frame.

    I know the distance between markers, prior to getting in the car, as measured in the stationary frame. So whenever I pass a marker I start counting, and when I pass another I stop and punch in my calculator d/t.

    What speed would I calculate? the same as a stationary observer?
  15. Jan 7, 2012 #14

    Doc Al

    User Avatar

    Staff: Mentor

    So your head will serve as a clock, just like any other.



    No. The quantity you calculate will equal gamma*v, where v is the car's velocity in the stationary frame. That quantity is sometimes called the 'proper velocity' (see: Proper velocity), but it's not the velocity of anything in the usual sense, since you're mixing quantities measured in different frames.
  16. Jan 7, 2012 #15
    Ok. thanks for clearing it up.

    Quick question though,

    If there's an atomic clock on my car, I can see how the path of the light needs to be longer, like they show in the section called "Simple inference of time dilation due to relative velocity" at http://en.wikipedia.org/wiki/Time_dilation

    But, they don't mention the length contraction. shouldn't it be used to correct D?
  17. Jan 7, 2012 #16


    User Avatar
    Science Advisor
    Gold Member

  18. Jan 7, 2012 #17
    we just talked about how the distances between markers would be contracted, so wouldn't the distance the mirrors move in the time deltaT' be contracted?

    what v are they using? shouldn't it be multiplied by gamma or something?
  19. Jan 8, 2012 #18


    User Avatar
    Science Advisor
    Gold Member

    Mirrors? I thought there was just one. Are you now thinking of each milestone as having a mirror?

    Are you thinking that you have to correct the distance that you measure between milestones by gamma in order to get the correct velocity? If so, don't concern yourself with gamma. Just come up with a specific way to measure the speed of an object approaching you using your own measurement devices. Think about how it can actually be done in real life.
  20. Jan 9, 2012 #19
    I was asking about this diagram in reference to an atomic clock travelling at speeds near c.

    they show the bottom length of the right triangle as 1/2vΔt'

    since we are viewing it from a different frame, shouldn't that length be contracted?

  21. Jan 9, 2012 #20

    Doc Al

    User Avatar

    Staff: Mentor

    You mean a 'light clock', not an atomic clock.

    No. That's the horizontal distance traveled by the clock. It's not the length of anything. In the rest frame of the clock, that distance is zero. Length contraction is irrelevant.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook