46
$\begingroup$

The title says it all.

If I was on a bus at 60 km/h, and I started walking on the bus at a steady pace of 5 km/h, then I'd technically be moving at 65 km/h, right?

So my son posed me an interesting question today: since light travels as fast as anything can go, what if I shined light when moving in a car?

How should I answer his question?

$\endgroup$
5
  • 1
    $\begingroup$ If you like this question you may also enjoy reading this Phys.SE post. $\endgroup$
    – Qmechanic
    Oct 2, 2013 at 12:18
  • 20
    $\begingroup$ This is the amazing thing about relatvivity that the velocity of light does not add the same way as small velocities do! Your son is on the path which led Einstein to develop relativity theory. $\endgroup$
    – Slaviks
    Oct 2, 2013 at 13:10
  • 1
    $\begingroup$ This should answer your question in less than five minutes: youtube.com/watch?v=vVKFBaaL4uM $\endgroup$ Oct 2, 2013 at 17:47
  • 1
    $\begingroup$ This is exactly the question Einstein asked, which led him to discover special relativity. $\endgroup$ Oct 2, 2013 at 18:26
  • 5
    $\begingroup$ Nobody mentioned Michelson-Morley experiment yet - amazing, as that is the empirical basis of everything below, and is simple enough for a child to understand the results (if not the techniques) of: en.wikipedia.org/wiki/Michelson%E2%80%93Morley_experiment $\endgroup$ Oct 2, 2013 at 22:11

8 Answers 8

59
$\begingroup$

If I was on a bus at 60 km/h, and I started walking on the bus at a steady pace of 5 km/h, then I'd technically be moving at 65 km/h, right?

Not exactly right. You would be correct if the Galilean transformation correctly described the relationship between moving frames of reference but, it doesn't.

Instead, the empirical evidence is that the Lorentz transformation must be used and, by that transformation, your speed with respect to the ground would be slightly less than 65 km/h. According to the Lorentz velocity addition formula, your speed with respect to the ground is given by:

$$\dfrac{60 + 5}{1 + \dfrac{60 \cdot 5}{c^2}} = \dfrac{65}{1 + 3.333 \cdot 10^{-15}} \text{km}/\text h \approx 64.9999999999998\ \text{km}/\text h$$

Sure, that's only very slightly less than 65 km/h but this is important to your main question because, when we calculate the speed of the light relative to the ground we get:

$$\dfrac{60 + c}{1 + \dfrac{60 \cdot c}{c^2}} = c$$

The speed of light, relative to the ground remains c!

$\endgroup$
12
  • 10
    $\begingroup$ True, but usually we think of the fact that the speed of light is constant as being a law of physics, and the Lorentz transformation is just a consequence of that law. $\endgroup$ Oct 2, 2013 at 22:06
  • 14
    $\begingroup$ A correct answers, but hardly comprehensible to a 6 or 8 year old. $\endgroup$ Oct 2, 2013 at 22:19
  • 4
    $\begingroup$ @PieterGeerkens No, but the idea of Alfred's answer is some thing children of that age love to hear - I think it would be a grand and appropriate thing to tell them that "it's actually a teeny bit less that 65kmh, and the faster you go, the more the rule that speeds add together in the way that children understand is broken". $\endgroup$ Oct 2, 2013 at 22:57
  • 3
    $\begingroup$ @WetSavannaAnimalakaRodVance: You honestly believe that bright children like being talked down to like that? Shame on you! $\endgroup$ Oct 3, 2013 at 2:46
  • 4
    $\begingroup$ @PieterGeerkens In some cases yes: children are all very different. Education research shows that this age is the very age where a system's abstract properties are only just beginning to take root in their minds - see here - so algebraic explanations befuddle many children. It's important to know your student well and watch their reaction to you carefully - clearly geek-girl or boy is not going to forgive you for a too simplistic answer so you need to be ready to shift gears in a split second. But, given what we know about childrens ... $\endgroup$ Oct 3, 2013 at 4:18
30
$\begingroup$

You should tell your son that this very question was asked by, explored by, and eventually answered by the some of the brightest physicists of the 19th century. Eventually two scientists named Michelson and Morley came up with an experiment to measure this effect, and were amazed to discover that it didn't exist! Rather:

Light travelled at exactly the same speed in all directions, regardless of any velocity of its emitter.

This result astounded the physicists of the world, and led to the development of the Special Theory of Relativity by Einstein.

$\endgroup$
4
  • 4
    $\begingroup$ +1 Great answer. I love how it touches on history of science, the empirical method, and the story of this discovery. Sure to lead to OP's son wanting to find out more. $\endgroup$
    – Ergwun
    Oct 3, 2013 at 2:34
  • $\begingroup$ They were actually checking the velocity of it's emitter respect to the ether, which they wanted to prove. Only to find that there was no ether. $\endgroup$ Oct 3, 2013 at 10:40
  • $\begingroup$ @FrankPresenciaFandos: Yes, by measuring the increased speed of light in the direction in which the Earth was moving relative to the aether. $\endgroup$ Oct 3, 2013 at 11:10
  • 2
    $\begingroup$ "this very question was asked by, explored by, and eventually answered by the some of the brightest physicists of the 19th century" In other words, in asking this very question, OP's son was in some very good company! $\endgroup$
    – user
    Jun 30, 2016 at 15:22
5
$\begingroup$

You can start in answering his question by explaining the Doppler shift for acoustical waves.

The Doppler effect (or Doppler shift), named after the Austrian physicist Christian Doppler, who proposed it in 1842 in Prague, is the change in frequency of a wave (or other periodic event) for an observer moving relative to its source. It is commonly heard when a vehicle sounding a siren or horn approaches, passes, and recedes from an observer. The received frequency is higher (compared to the emitted frequency) during the approach, it is identical at the instant of passing by, and it is lower during the recession.

The relative changes in frequency can be explained as follows. When the source of the waves is moving toward the observer, each successive wave crest is emitted from a position closer to the observer than the previous wave. Therefore each wave takes slightly less time to reach the observer than the previous wave. Therefore the time between the arrival of successive wave crests at the observer is reduced, causing an increase in the frequency. While they are travelling, the distance between successive wave fronts is reduced; so the waves "bunch together". Conversely, if the source of waves is moving away from the observer, each wave is emitted from a position farther from the observer than the previous wave, so the arrival time between successive waves is increased, reducing the frequency. The distance between successive wave fronts is increased, so the waves "spread out".

Your son's expectation works on this intuitive background.

But light waves, in contrast to sound waves which need air to reach our ears, do not need a medium to reach our eyes. This is evident in that the light from stars reaches us through the vacuum of space where there is no medium. People used to hypothesize a medium for light, aether but experiments proved, as the other answers state correctly, that the velocity of light was constant, c, no matter what the motion of the emitter or absorber. Thus no, there will be no change in the velocity measured of the emitted light whether we are sitting on the ground, forward or backward or sideways, or in the car itself.

There is an effect though. Light that has been emitted by a source moving towards us does not change its velocity but it does change its frequency to a higher value; if it is receding, to a lower value. As the energy of the photons is given by E=h*nu it means that it gains an extra energy or loses some due to the relative motions of observer and emitter.

This has been very useful for astrophysics. For example that is how we know the relative motions of stars with respect to us. Light comes from spectra of atoms and we know them here in the lab. They are distinctive and identify whether we see light from iron or oxygen or hydrogen in a gas state. The change in frequency of the spectral lines will tell us of the motion of the star relative to us. There exist many applications of this method.

$\endgroup$
1
  • $\begingroup$ "... if it is receding, to a lower value. As the energy of the photons is given by E=h*nu it means that it gains an extra energy or loses ... " What is your opinion on Doppler effect in respect of "gravitational redshift". I doubt if a "violet shift" your energy rule seems to refer to depends of direction of emission. In other words, in my opinion any light pistol moved close to c emits higher (frequency) in any direction. Very helpful answer. $\endgroup$ Nov 7, 2022 at 19:18
5
$\begingroup$

As a matter of technicality, I believe that shining from a moving vehicle on earth will in fact be faster* than the light shining from a stationary vehicle on earth, however both forms of light would be moving slower than the speed of light (c), which is referencing the speed of light in a vacuum.

This is because the medium that the light is moving through (air) slows down the light by about 88km/s (according to Wikipedia).

That said, light in a vacuum emitted from a moving object should travel at the same speed as light in a vacuum emitted from a stationary object for the reasons outlined by all the other answers to this question.


* so long as the light is propagating in air that is traveling with the vehicle, such as the air between the headlight bulb and the headlight casing

$\endgroup$
7
  • 6
    $\begingroup$ The first paragraph is wrong. Even in media where the speed of light is less than $c$, the speed of propagation is still independent of the speed of the source. $\endgroup$
    – user10851
    Oct 2, 2013 at 23:22
  • $\begingroup$ @ChrisWhite, I'd be more than happy to update my answer if you'd elaborate on why that's the case. My knowledge of physics is superficial at best, so I'd much rather have more good information than lead people astray if I'm wrong. $\endgroup$
    – zzzzBov
    Oct 3, 2013 at 2:53
  • 9
    $\begingroup$ Waves - whether they are sound or light or water - are local phenomena, which means how the wave decides to move is determined by its immediate surroundings, not by whatever the source may have been doing far away. The speed of the source affects the frequency of the wave, but two waves of the same frequency in the same medium will move at the same speed, which for light in air is, as you said, a bit slower than $c$. $\endgroup$
    – user10851
    Oct 3, 2013 at 3:04
  • $\begingroup$ @ChrisWhite, then in that case: if a two light waves were traveling in two separate media of the same substance in the same direction, and those media were moving in relation to each other in that direction, wouldn't the light waves be traveling at different speeds with regards to each other? $\endgroup$
    – zzzzBov
    Oct 3, 2013 at 4:34
  • 5
    $\begingroup$ If the air was moving with the car, the speed of light could be faster that the speed of light through still air. But if it is only the car that is moving and not the air, then both stationary and moving flashlights will put out light of the same speed, ie c minus 88 km/s. $\endgroup$ Oct 3, 2013 at 5:14
4
$\begingroup$

Second postulate (invariance of c) of the special theory of relativity goes like this:

As measured in any inertial frame of reference, light is always propagated in empty space with a definite velocity c that is independent of the state of motion of the emitting body.

Or, that objects travelling at speed c in one reference frame will necessarily travel at speed c in all reference frames. This postulate is a subset of the postulates that underlie Maxwell's equations in the interpretation given to them in the context of special relativity. So basically, there exists an absolute constant 0 < c < (infinity) with the above property. So you can shine light while travelling at the speed of light and it will still go at c, not more, not less. Ref: wiki.

$\endgroup$
3
$\begingroup$

One essential postulate of special relativity is that light moves at the same velocity in all reference frames. Somebody standing next to the moving bus will observe the light travelling just as fast somebody who is on the bus sees it. It might not be intuitive, but it is consistent both with experiment and the mathematical framework of the theory.

$\endgroup$
1
  • 1
    $\begingroup$ Should read "... in all inertial reference frames.". $\endgroup$ Jun 30, 2016 at 15:38
-1
$\begingroup$

Your son is correct, from the perspective of an outside observer who is not on the bus. The same is correct with the light. While 65km/h is negligible in relation to the speed of light, he is right. If I am traveling at 90% c and I shine a light ahead of me, the light will leave me at the speed of light. If you, standing still, observed me, the light would still leave my flashlight at c. Weird right, something has to give to balance this out, and this is time. Which means in my example, someone is time traveling. Remember that everything you see with your eyes, you are seeing it in the past, not now at the exact moment. Einstein realized this when he would ride the train and look at the clock on a building.

$\endgroup$
1
  • 1
    $\begingroup$ This question has been answered countless times in PSE. A little search in PSE will not harm. $\endgroup$
    – Markoul11
    Jan 2, 2022 at 16:11
-1
$\begingroup$

Ok, so that's a classic question that depicts how wrong we think of reality, or maybe how much diffrent reality is from how we think of it.


Every object moves in spacetime at one and only one speed. The speed of C=1ls/s (aka: one lightsecond per second, also known as "the speed of light" but in reality its the speed of everything that moves in spacetime).


The only difference between light and say a car... Is that light moves at the speed of C solely in space. Where a car moves at the speed of C in both space and time combined.

That combination of speeds (your speed in time and in space) is called your speed in spacetime and its related to the Lorentz factor γ.

$$γ\ =\ \frac{1}{\sqrt{1-\frac{u^{2}}{c^{2}}}}$$

where:

  1. v is the relative velocity between inertial reference frames
  2. c is the speed of light in a vacuum
  3. β is the ratio of v to c
  4. t is coordinate time
  5. τ is the proper time for an observer (measuring time intervals in the observer's own frame)

Now... Lets take a look at a photon and a car.

A car is moving at 60km/h (or 16.6 m/s) in space and using the Lorentz factor we can see that the car is ALSO moving at (299,247,994.113 m/s) in time. If we combine 16.6 and 299,247,994.113 for spacetime, we get 299,792,458 m/s which is exactly, the speed of C (or the "speed of light") in spacetime. So your car is moving at the speed of light in spacetime.

Now a photon is even easier!

A photon is moving at 299.792.458 m/s in space. And using the Lorentz factor we can find that the photon is moving at exactly 0 m/s in time!

So again, if we combine 299.792.458 and 0 for spacetime we will get 299.792.458 m/s which is the speed that the photon is traveling in spacetime.

Both the photon and the car are traveling at the same exact speed, in spacetime! The speed of C. The car is mostly moving in time, the photon is ONLY moving in space, that's the only difference between them.


So, your question was "If I am travelling on a car at around 60 km/h, and I shine a light, does that mean that the light is travelling faster than the speed of light?"

Lets rephrase that question to make it more accurate:

"If I am traveling on a car at around 60km/h IN SPACE (and 299,247,994.113 m/s in time), and I shine a light, does that mean that the light is traveling faster than the speed of light in space?"

NO

Your car will be moving at the speed of light but mostly moving in time and a little bit (16.6 m/s) in space. The photons from your flashlight will be moving at exactly the speed of light in spacetime but ONLY in space and not in time. Both you and the photons will continue to move in spacetime at exactly the speed of light, your separate trajectories though won't be the same. You will continue to travel mostly in time and the photons will travel only in space and time for them won't be passing.

That's the real reality. The reality that we observe is false, because we don't see (or can't see) the real picture. We can't see that the spacetime around us is 4 dimensional and hyperbolic. Also we forget that things also move in time and not just in space. That's why we see all those effects of relativity (like time dilation and lenth contraction). If we could see spacetime from a bird's eye view, we would see that the only thing that you can really do is change your "trajectory" in spacetime (performing a hyperbolic rotation). You can't really accelerate in spacetime because the more you speed up in space the more you slow down in time, the net result will continue to be the speed of C everytime!

$\endgroup$
12
  • $\begingroup$ This seems to be largely just nonsense, but even if it made sense, it wouldn't be helpful. Insofar as it says anything at all, it seems to be saying is that in relativity, you can always parameterize a worldline in such a way that all of its tangent vectors have length 1. That's true. It's also true that in Newtonian physics, you can always parameterize a worldline in such a way that all of the tangent vectors have length 1. So if this observation is supposed to somehow illuminate the difference between Newton and Einstein, it fails utterly. $\endgroup$
    – WillO
    Jul 28, 2022 at 14:24
  • $\begingroup$ I honestly don't understand why this is nonsense... What part specifically as an example is nonesense? No this explanation doesn't suggest that you can parameterize a worldline in such a way that all of its tangent vectors have length 1. If my explanation wasn't clear you can also take a look at this video for more clarity youtu.be/au0QJYISe4c?t=290 $\endgroup$
    – Nuke
    Jul 28, 2022 at 23:56
  • $\begingroup$ For starters, what does "move in spacetime" mean? Motion generally refers to a change in location as a function of time. In other words, motion takes place in space, not in spacetime. An object is represented by a curve in spacetime. Curves are mathematical objects which do not change any more than the number 3 changes. The only remotely plausible definition of velocity in this context is the tangent vector to that curve, which depends entirely on how you choose to parameterize that curve. You are choosing a parameterization (so as to make all the tangent vectors CONTINUED $\endgroup$
    – WillO
    Jul 29, 2022 at 0:19
  • $\begingroup$ CONTINUED have length 1, so your conclusion that everything has a velocity of magnitude 1 holds true only because you chose your definition that way. You could do exactly the same thing in classical mechanics by choosing your parameterization correctly. The point is that a) "speed of an object in spacetime" is intrinsically meaningless, b) you've chosen to assign it a completely arbitrary definition; c) you've concluded that we can learn something from the definition you chose --- but all we actually learn is that you happened to make this particular choice. $\endgroup$
    – WillO
    Jul 29, 2022 at 0:21
  • $\begingroup$ Moreover --- the whole bit about "slowing down in time" is of course ridiculous. You proceed through time at a rate of 1 second per second by your own clock, always. If you are moving relative to me, your clock will tick slower in my frame than it does in yours. That's a fact about frames, not clocks. Your clock is not going to tick any differently just because I choose to move relative to you, or you relative to me. $\endgroup$
    – WillO
    Jul 29, 2022 at 0:22

Not the answer you're looking for? Browse other questions tagged or ask your own question.