4
$\begingroup$

First I will have to explain my question. Look at the image below. This shows doppler shift when an object is moving horizontally to the direction of the wave. Keep the word 'horizontally' in mind. Now this happens because:

I will quote Jim from his answer for Redshifting of light from a moving light source

We all know that light is a wave, when you turn on your headlights and drive in reverse, the light is doppler shifted because of the motion of source. When not moving, each cycle of the light wave is emitted from the same position; it has a specific set of wavelengths. The distance between one crest of a wave and the next crest is equal to the speed of light, c, times the period of the light (which is determined by the oscillations in your headlights and won't change when you are in motion). When you drive backwards, the distance between one crest and the next becomes the period times c plus the period times your backwards velocity (approximately); the second crest is not emitted at the same location as the first, so it extends the wavelength. From your perspective, the emitted wave would not be red-shifted at all, but from a stationary observer's perspective it is.

enter image description here

So now my question is, imagine a car which has a torch attached to one of its windows. The torch is switched on and the car begins to move. When the car moves, its movement is in the opposite axis from the propagation of the wave. So each crest will be released from a different location while the first crest is already on its way in a straight line. I will try to represent this graphically.

enter image description here

The representation is very estimate. It just shows how would the light bend as each crest is released from a different location. Please explain this to me. Will the light actually bend? Why or Why not?

Edit

What I have concluded from the answers is that first a photon is emitted and then it continues as a wave and is in no way attached to other photons. Is this right? If I got this then I got the answer for my question.

Improve this question
$\endgroup$
  • 1
    $\begingroup$ The other guys have what I think are good answers and I cant do better, but my understanding is that this is similar to turning on a hose and watching the flow of water come out. Lets pretend there is no gravity for visualization purposes. Water will spray out in a nice straight line and continue on forever. Now if you slowly move your arm and change the direction of the hose, the new water "sections" will come out staying perpendicular to the hose nozzle. But the water that left the hose when it was first turned on is already far away on its journey and nothing will stop that... $\endgroup$ – Steve Hatcher Jun 18 '14 at 13:56
  • $\begingroup$ You are confusing direction of light with frequency of light. $\endgroup$ – Val Jun 19 '14 at 19:40
  • $\begingroup$ thinking about it in terms of photons is not the answer. You will get the correct result using Maxwell's equations (which satisfy relativity) which treats light as a classical (in the sense of non-quantum) electromagnetic wave. $\endgroup$ – user16007 Jun 20 '14 at 0:22
  • $\begingroup$ No Val, I am just building upon the concept. Read Steve Hathcer's comment. $\endgroup$ – rahulgarg12342 Jun 20 '14 at 2:57
  • $\begingroup$ @SteveHatcher That is a fine analogy but it has an error. Water molecules are not connected in the sense the troughs and crests of a light wave are connected. So water coming out of a hose is completely a different thing as compared to a light ray being emitted. $\endgroup$ – rahulgarg12342 Jul 11 '14 at 19:58
1
$\begingroup$

The light is not bending, although the paths you draw appear as if they do. If you treat the light as a photon (because of wave-particle duality), you will see that the newly emitted photons are simply emitted at an offset from each previous one. Thus the wave isn't bending, it's simply being generated by a source at a different location. Evaluated in extreme conditions (i.e. when v_car -> c), relativistic effects may begin to be relevant, but those seem to be beyond the scope of the question.

Improve this answer
$\endgroup$
  • $\begingroup$ You see when light is propagating, you consider it as a wave not as a particle. When it hits something, it becomes a photon. So when it is generated, as you can see from the quote, it is generated as a sequence of crests and troughs. $\endgroup$ – rahulgarg12342 Jun 14 '14 at 6:27
  • $\begingroup$ @rahulgarg12342 light can be seen as waves or particles. In this case, wave-like nature is more convenient to think about Doppler effect, but particle-like is more convenient for this trajectories. If you want to ease your mind, just follow the crests, they will behave like particles. $\endgroup$ – Davidmh Jun 14 '14 at 13:56
  • $\begingroup$ I know but I don't just want to interpret it to understand it. Lets think differently and practically rather than just solve a problem under just for convenience. This is similar to doppler shift and the source in both cases is the same. So we cannot say that when you switch on your headlights, then light is emitted as a wave and when you place a torch on a window, it will always be emitted as a particle. $\endgroup$ – rahulgarg12342 Jun 14 '14 at 15:24
  • 1
    $\begingroup$ Light waves are always an emergent phenomenon from the zillions of photons that make it up. One uses the framework that is more convenient for solving a problem. No need to point to your left ear with your right hand behind your back. $\endgroup$ – anna v Jun 18 '14 at 3:36
  • $\begingroup$ @annav I completely agree with your point but then lets take my problem in two ways. One in which light is emitted as zillions of photons and one in which it is emitted as a wave. I am asking the latter problem. $\endgroup$ – rahulgarg12342 Jun 18 '14 at 5:11
1
$\begingroup$

The light will not bend. The blue line is not the trajectory of the light ray (whether you view it as a wave or a photon does not matter), it represents the distribution of electric and magnetic fields at a certain point in time. View it as a snapshot of the wave, if you like. What you have drawn is a simplified version of this diagram:

enter image description here

The arrow labelled "Direction" does not represent the wave's path as a function of time, but the wave's spatial extent at a certain point in time. Even if this spatial distribution is "bent", that does not mean that light follows a curved trajectory.

Improve this answer
$\endgroup$
  • $\begingroup$ Why? If the spatial distribution is bent, why will it not follow a straight line? I know our brain will think it is following a straight line but I am talking about the actual thing rather than what are brain will interpret. A snapshot will obviously be straight but I am talking about the entire light ray. Please help! $\endgroup$ – rahulgarg12342 Jun 18 '14 at 4:59
  • $\begingroup$ ^ I meant If the spatial distribution is bent, why will it follow a straight line? I know it will not be the light ray that will bend but the moving source that will change its spatial distribution. But isn't it almost the same thing. This is what sort of happens in a gravitational redshift. The gravity causes a change in the spatial distribution but we call it the bending of light. $\endgroup$ – rahulgarg12342 Jun 18 '14 at 5:12
  • $\begingroup$ @rahulgarg12342: The whole spatial distribution is bent and as a whole, and if you consider this to be the ray of light, then yes, the ray of light is bent. But if you look at an individual point within that spatial distribution, e.g. a crest of the wave that is emitted at a certain moment and follow it, it will move on a straight line. $\endgroup$ – Frederic Brünner Jun 18 '14 at 6:05
  • $\begingroup$ Yes I understand that a specific crest will not bend. That answered my question. I was talking about the light ray as a whole. Well that leads to another point, can this also lead to time dilation? Since the light ray is bent and covers a longer distance so it would take a little more time? $\endgroup$ – rahulgarg12342 Jun 18 '14 at 7:07
  • $\begingroup$ @rahulgarg12342: With respect to an observer at rest with respect to the source, the frequency at which crests arrive will change with time. This is also known as the Doppler-effect. $\endgroup$ – Frederic Brünner Jun 18 '14 at 8:22
0
$\begingroup$

The excentrical Doppler effect is not an easy problem. It has afaik no analytical solution (if you employ special relativity). In the classical limit it might be easier.

Since you want the light to be treated as wave, you also need to throw away notion of "bending". A wave spreads out in every direction from every point. A light ray is just superposition of point waves into a wave front. "Bending" of light makes more sense when you talk about photons.

Perhaps this will become clearer if you follow try to calculate the classical!Doppler effect:

Let's look at the crests the torch produces. If the car moves at constant velocity, the origin of crests will be equally spaces on the car's path (your x-axis). Let's say a crest is produced on every meter, with the car going one meter per second. Then the time a crest needs to arrive at the observer is $\Delta t_x=\frac{1}{c}\sqrt{e^2+x^2}$, where $c$ is the speed of light, $e$ is the excentricity of the observer (how far he is from the car's path) and $x$ is the distance the car has travelled.

Since the crests are not sent all at the same time, we have to add that time to the time of sending: $t^{\text{sent}}_x=\frac{x}{v}$ where $v$ is $1\frac{m}{s}$. The time of arrival thus is $t^\text{arrival}_x = \frac{x}{v}+\frac{1}{c}\sqrt{e^2+x^2}$. The frequency of sending light is 1Hz (one crest per second). The frequency the observer will perceive is the inverse of the difference between two crests: $$ f_x = \frac{1}{t^{\text{arrival}}_{x+1}-t^{\text{arrival}}_x} $$ Which is a rather complicated expression. :( But fortunately the computer can visualize it:

This is what the frequency will look like

Improve this answer
$\endgroup$
0
$\begingroup$

Your drawing has some mistakes: the wave is not spatial in that sense, the peaks and valleys are a representation of the strength of the electric (and or magnetic) field at that position. So, your light ray should be unidimensional. For instance, the actual wave for the car at rest should be a line, the light will not move back and forth as it travels up. Plus in addition, in you drawing you are implicitly assuming a classical addition of velocities by assuming that the wave behaves non-relativistically. It is an axiom of special relativity that any observer will perceive the light as moving stight and at the same speeed regardeless of the motion of the source of the light. All weird effect of relativity start at this assumption. So, both observers will see a ONE-dimensional light ray moving along a straight path: the locations of the peaks within that line will have nothing to do with the perceived path.

Improve this answer
$\endgroup$
  • $\begingroup$ I am not even talking about the perceived path! Please read my question again. $\endgroup$ – rahulgarg12342 Jun 18 '14 at 19:01
  • $\begingroup$ it will not bend because it always move in a straight line. Your graph is misleading you because it is wrong:a ray of light is not physicaly a 2D wave, but a 1D one. You are confusing the intensity of the electromagnetic field (represented as a 2D wave), as if it were representing the position of the light itself. So the curved line you get in your graph cannot be interpreted in spatial terms. Your graph is conceptually wrong, or at least your interpretation of what it means. $\endgroup$ – user16007 Jun 18 '14 at 21:25
-1
+50
$\begingroup$

rahulgarg12342,

Looks like either your description or your drawing is misleading. You say that the torch is fixed to the car and sends light beam perpendicular to the direction of the movement, and yet on the drawing the torch seems to be hand-held and directed at particular point on the ground, and so the angle keeps changing all the time.

Anyway, whatever you had in mind, your puzzle can be solved by using the Principle of Relativity and assuming it is the ground moving, and not the car. In such case the light beam is always a straight line. And if the torch is just fixed, as you said, the light is sweeping the surroundings all the time.

(In case the torch was actually meant to be hand-held and changing the angle. Remember one thing: when we are watching a light beam in the dark we are just seeing a multiple of photons - each being also a wave itself - and not a single wave that goes all the way from the torch to the destination point. Also when it looks as if we are seeing the photons as they are traveling, all we can actually see are the photons that never made it to the destination point. What we can see are only photons reflected back toward our eye by particles (of air), and therefore those photons never reached "the end", unless they got reflected off the end-point, but still, we cannot see their route. We cannot see see a photon "on its way" from afar, because in order to see a photon it must come directly to our eye.)

If you clarify your question a little bit, I will be able to elaborate more.

Improve this answer
$\endgroup$
  • $\begingroup$ You did not interpret my question in the right way. The first ray is emitted in the first frame and goes as a straight line. When more crests are produced and emitted, the car is constantly changing its position. Therefore each crest is emitted from a different position. But since the original crest was already emitted in a straight line, the other crest have to bend in order to follow the first path. It is not a handheld torch. $\endgroup$ – rahulgarg12342 Jun 18 '14 at 7:11
  • 1
    $\begingroup$ Light is not a continuous "wavy string". It is a set of photons, and each of them behaves like a wave itself. If a photon is emitted, it is no longer "attached" to the source and to other photons. When a car is moving, the torch is sending multiple photons parallel to each other. They are not creating one wave with many crests. $\endgroup$ – bright magus Jun 18 '14 at 7:33
  • $\begingroup$ Well, that is just one aspect of it. Remember we have to consider wave particle duality. You are only considering it as a photon. In a wave it is very well attached to the other crests. Look to the above answer for clarification. $\endgroup$ – rahulgarg12342 Jun 18 '14 at 7:55
  • $\begingroup$ If you emit just one photon it is still a particle and a wave. You only need one photon to have a wave. $\endgroup$ – bright magus Jun 18 '14 at 8:19
  • $\begingroup$ So first a photon is emitted and then it propagates as a wave? $\endgroup$ – rahulgarg12342 Jun 19 '14 at 5:08
-1
$\begingroup$

Other answrs havent covered effect of Coriolis force on the emitted waves. Light bends here because of Coriolis force on the waves because of the rotation of the earth, just like you twist and fall from your bicycle when it is moving. In book situations our teachers always give us simplified situatons, but life is considerably complex.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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