Why Light Experienced a Doppler Shift?

[Dale]

They use time dilation when deriving Doppler shift

They use time dilation on the reciever and on the emitter. Those do have a rest frame. They do not use time dilation on the light since it does not have a rest frame.

Again, I recommend against using either time dilation or length contraction. Just use the Lorentz transform. It will automatically simplify whenever appropriate.

Do you know how do derive it using the Lorentz transform? @PAllen gave a good starting point in post 32

the light frame itself

There is no such thing.

if the light have specific frequency in S frame, then it must have specific wavelength in S frame, because λ×f=cλ×f=c\lambda \times f = c.

Yes

 

[Mohammad Fajar]

No, it isn't. You are still missing a fundamental point: there is no "light frame". Light is not at rest in any frame; it moves at $c$ in all frames. That means there is no "time attached to the light frame" because there is no "light frame". The only relevant frames are the rest frame of the emitter and the rest frame of the receiver.

So if the light have frequency 10 Hz, when 20 crest passed, is that doesn't mean two second passed for that light? And this applied to the wavelength too.

In classical sense the wave velocity mean how many wavelength passed for every second, how many distance covered by all of wavelength when time passed. This mean we add all wave length to get distance covered per second. So yes, like frequency,time attached to light wave too.

If I have light wave with frequency 10 Hz, and I count 1000 wave crest, then it is mean that light wave have an age 100 second, Or I'am wrong?

 

[Dale]

So if the light have frequency 10 Hz, when 20 crest passed, is that doesn't mean two second passed for that light?

Almost correct. If light has frequency 10 Hz in an observer’s frame then when 20 crests have passed that means 2 seconds passed for that observer.

Light does not simply have frequency 10 Hz. It has frequency 10 Hz in a specific reference frame.

So yes, like frequency,time attached to light wave too.

No. The time that can be considered “attached to” something is called proper time. In this context “proper” means “belonging to” as in “property”.

Light does not have proper time. Massive objects, such as emitters and receivers, those have proper time. So proper time is not attached to light, it is attached to massive objects.

 

[Mohammad Fajar]

Light does not have proper time. Massive objects, such as emitters and receivers, those have proper time. So proper time is not attached to light, it is attached to massive objects.

So why frequency attached to the light wave, if total period not?
(I edit my response in post 37)

 

[Dale]

So why frequency attached to the light wave, if total period not?
(I edit my response in post 37)

Frequency is not just attached to a light wave. It is a description of a light wave in a specific reference frame. Simply saying that a given wave has 10 Hz is incomplete. You also need to specify in which reference frame it has 10 Hz.

If I have light wave with frequency 10 Hz, and I count 1000 wave crest, then it is mean that light wave have an age 100 second, Or I'am wrong?

Yes, this is wrong. Again, you must specify the frame in which the light wave has a given frequency. If it has 10 Hz in my frame and I count 1000 wave crests then I have aged 100 s. Light does not have age (proper time)

 

[Dale]

@Mohammad Fajar I probably should have posted this earlier. I think this is the best reference for the Doppler effect. It covers the Doppler effect for both sound and light together

http://www.mathpages.com/rr/s2-04/2-04.htm

 

[vanhees71]

Sorry, is there any way to derive Doppler shift without using time dilation (and why I cannot use time dilation in the same place where I used Length contraction), or to derive that with full Lorentz tranformation?

If time dilation can I applied to every ticks of light, then why doesn't length contraction?
Are frequency of light not related to the time used to generate every peak of wave?
Are wavelength of light not related to the distance/space between every wave peak?

I really don't understand, why we are still debating this really simple calculation. You apply the Lorentz boost to the wave-four-vector, and that's it. I've given the explicit formula in #24.

To understand physics you need a minimum of math. For the kinematical effects of SR it's rather simple linear algebra of a pseudo-Euclidean (Lorentzian) affine manifold. You waste more time in avoiding math than just learning the necessary mathematical tools, and it's even fun after you start to get used to it! For an introduction to SR, see my SR FAQ manuscript here:

https://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

 

[Orodruin]

You apply the Lorentz boost to the wave-four-vector, and that's it.

Or, more elegantly as it completely removes the need to look at components in a particular frame (did I mention I hate looking at component values?), just take its inner product with the observer 4-velocity, which will give you the increase of the phase function per proper time. But I suspect the OP is not advanced enough to treat relativity using 4-vector formalism. Still, you can do some things without it if you do it correctly. Later, as you learn more, you learn that what you did earlier was just a special way of doing it more elegantly.

For the kinematical effects of SR it's rather simple linear algebra of a pseudo-Euclidean (Lorentzian) affine manifold.

For a person with high-school level knowledge, I do not think this is simple. However, you can do SR kinematics without it - even if it is a mess and you do not get the same level of understanding from doing so.

 

[Dale]

just take its inner product with the observer 4-velocity, which will give you the increase of the phase function per proper time.

Woah, I didn’t know that! That is super convenient.

 

[Orodruin]

Woah, I didn’t know that! That is super convenient.

Well, if you think about it, it really is nothing else than computing the time component of the 4-frequency in the observer's rest frame as the 4-velocity components are just the components of the first row of the Lorentz transformation matrix ... But yes, it is very convenient. It seems I picked up at least one or two tricks teaching this stuff for 5 years ...

Edit (warning, A-level statement): Of course, the real object that you want to concern yourself with is the exterior derivative of the phase function. The observed frequency for an observer with 4-velocity $V$ is just $d\phi/ds = d\phi(V)$.

 

[Dale]

Well, if you think about it, it really is nothing else than computing the time component of the 4-frequency in the observer's rest frame

Yes. It is obvious ... in hindsight.

Somehow this is just never something that I bothered to find alternative ways to calculate. I wrote some Mathematica code for calculating Lorentz transforms something like 12 years ago. It is all based on components. Whenever I needed to do a relativistic Doppler I just pull that out. I never even thought about more elegant ways to do it. I typically work with the four momentum out of habit

 

[PAllen]

Yes. It is obvious ... in hindsight.

Somehow this is just never something that I bothered to find alternative ways to calculate. I wrote some Mathematica code for calculating Lorentz transforms something like 12 years ago. It is all based on components. Whenever I needed to do a relativistic Doppler I just pull that out. I never even thought about more elegant ways to do it. I typically work with the four momentum out of habit

Ah, but for light, the 4 momentum and the propagation vector and the 4 frequency are the same vector up to a constant factor.

 

[vanhees71]

For a person with high-school level knowledge, I do not think this is simple. However, you can do SR kinematics without it - even if it is a mess and you do not get the same level of understanding from doing so.

That's the problem rather than the solution! Highschool physics (and paradoxically also high school math) is made more difficult than necessary by an ununderstandable tendency to avoid the necessary math. Physics is complicated enough without such apparent "didactical" ideas; you don't need to make it more complicated by forbidding to use the appropriate language.

In high school we learned some vector algebra of 2D and 3D Euclidean space, and it would not be too difficult to also present Minkowski space (and be it simplified to the (1+1)-dimensional case).

 

[PAllen]

That's the problem rather than the solution! Highschool physics (and paradoxically also high school math) is made more difficult than necessary by an ununderstandable tendency to avoid the necessary math. Physics is complicated enough without such apparent "didactical" ideas; you don't need to make it more complicated by forbidding to use the appropriate language.

In high school we learned some vector algebra of 2D and 3D Euclidean space, and it would not be too difficult to also present Minkowski space (and be it simplified to the (1+1)-dimensional case).

But if we can present an answer using the math someone already knows, that is much more useful to them than using math they have to learn before they can read an explanation. No math is a different issue, and there are many cases where certain math can't be avoided. But there is substantial physics that can be presented with math most current high school students already know. Stating that high school programs ought to be different is not helpful to someone seeking explanation right now.

 

[Orodruin]

I mean, ideally students would already have a year’s worth of differential geometry before thry start learning any physics at all (even classical mechanics), but this is not the world we live in and for the purposes of most students they can get by perfectly fine without it. The problem only arises when one starts asking questions beyond that where the answer is much clearer with more math.

 

[vanhees71]

That's also asked for too much. You don't need very abstract concepts of differential geometry and fibre bundles to do physics. You can do a lot with one-variable high-school calculus and high-school vector algebra. It only has to be taught in a more comprehensible way with a lot of outlook on applications early on, and applications are by far not limited to physics but can be found nearly everywhere.

 

[Mohammad Fajar]

That's also asked for too much. You don't need very abstract concepts of differential geometry and fibre bundles to do physics. You can do a lot with one-variable high-school calculus and high-school vector algebra. It only has to be taught in a more comprehensible way with a lot of outlook on applications early on, and applications are by far not limited to physics but can be found nearly everywhere.

This is how they derive Doppler shift:
If light have period $t_0$ in $S$ frame and $t$ in $O$ frame when both of them at rest, then when $S$ receding this period increasing about $$ T = t + v t /c $$ where $v t$ is the distance traveled by $S$ and become additional distance traveled by the light to reach $O$ according to $O$ frame. So when $t = \frac{t_0 }{\sqrt{1 - \frac{v^2}{c^2}} }$this period become $$T = t + vt / c = t_0 \frac{1 + v/ c}{\sqrt{1 - v^2/c^2}} = t_0 \frac{\sqrt{1 + v/c} \sqrt{1 + v/c}}{ \sqrt{1 + v/c}\sqrt{1 - v/c}} = t_0 \sqrt{ \frac{ 1 + v/c} { 1 - v/c}}$$. And frequency become $$f = f_0 \sqrt{ \frac{ 1 - v/c} { 1 + v/c}} $$
But what if we not substitute $v t$ in place of this distance, but just measure it in distance unit as $x$. Then we get $$ T = t + x / c = t + \frac{x_0}{c} \sqrt{1 - \frac{v^2}{c^2}} = \frac{t_0 }{\sqrt{1 - \frac{v^2}{c^2}} } + \frac{v t_0 }{c} \sqrt{1 - \frac{v^2}{c^2}} $$ and this is valid according to special relativity.
But you see that this last derivation will give different result from the first one. So they are not equivalent. But for the shake of mathematics we choose the first one, maybe it is like how we choose the root of quadratic formula that suitable with the given problem.

 

[Orodruin]

$$ T = t + x / c = t + \frac{x_0}{c} \sqrt{1 - \frac{v^2}{c^2}} = \frac{t_0 }{\sqrt{1 - \frac{v^2}{c^2}} } + \frac{v t_0 }{c} \sqrt{1 - \frac{v^2}{c^2}} $$

Why are you randomly throwing things into formulas without understanding them? What do you imagine $x_0$ is?

 

[Mohammad Fajar]

Why are you randomly throwing things into formulas without understanding them? What do you imagine $x_0$ is?

It is the distance as measure by $S$ (source).

 

[Dale]

It is the distance as measure by $S$ (source).

The light isn’t at rest in the frame of S. So that isn’t right.

 

[PAllen]

Not sure if this will help resolve the stubborn misapplication of length contraction, but here goes.

Previously Dale suggested you simply use the Lorentz transform, and I suggested specifically transforming x=ct and x=ct+L to see what really happens to L in this case (hint: it does not contract).

I propose to do this generally to see how standard length contraction is a very special case.

Consider x = u t and x = u t + L in some inertial frame in standard coordinates, where u may be any propagation speed. In particular, u=0 implying rest in the starting frame gives rise to standard length contraction, while no other case does. We boost these equations to a primed frame moving v in the +x direction, using the Lorentz transform. With a fair bit of algebra, the following results:

x' = u' t' and x' = u' t' + L'

where u' = (u-v)/(1-uv/c²) i.e, velocity addition formula and

L' = L / (γ (1 - uv/c²))

Note that u'/L' is the frequency, in all cases.

We see immediately that for u=0, you have L' = L/γ, simple length contraction.
However, for u=c we get after simple algebra:

L' = L √ ((1+v/c)/(1-v/c)) [of course, u' = u = c for this case, which falls right out of the general formula]

Another interesting case is u=v. This means the primed observer is moving with the wave (not possible for light, but possible for slower waves). Here you get:

L' = Lγ , i.e. the original frame sees contraction by γ compared to the moving frame - because the waves are at rest in the moving frame.

Yet another interesting case is where L' = L. This arises for:

v = 2u / ( 1 + u²/c²) or equivalently u = (1 - 1/γ) c²/v ; note, this equivalence is quite messy to show.

What does this case mean? It is when v is such that the wave traveling at u in the original frame, is traveling at -u in the primed frame. Thus, it is perfectly expected that the wavelength is the same as the original frame. However, due to nonlinear velocity addition, v is not 2u for this case, but the more complex expressions just given. An example of this case with round numbers is u = c/3 and v = .6c. This gives u' = -c/3 and L' = L.

You can see further that for u and v small that L' ≅ L, and speed and frequency approach Newtonian values.

 

[Dale]

I propose to do this generally to see how standard length contraction is a very special case.

Wow! Excellent post @PAllen. This is PF at its best! Knowledgeable, clear, and directly relevant to the OP’s struggle.

@Mohammad Fajar I highly recommend you study this post and ask questions showing some serious thought about it. This is high quality educational material tailored explicitly for you.

 

[robphy]

From #25

I just don't understand what mechanism have a role to make the wavelength changed for O.
[snip]

For classical Doppler Effect, there is no change in wavelength after it emitted by the source, it is always have that wavelength.

Sometimes one has to reason differently, especially when moving symbols around a set of equations isn't working out.
After being convinced of the correctness of alternate approach, one can then go back and figure out how the equations should have led to it.

My reply in #35 suggested that you draw a spacetime diagram.
(Now that I have a little free time)
I will now follow my own suggestion try to draw the situation on a spacetime diagram.
I've drawn the spacetime diagram on rotated graph paper so that one can easily see (and count!) the tickmarks along the various segments.

Alice (at rest) is the periodic source of light signals with period 10.
Bob travels with velocity 3/5.

An observer measures "length" using that observer's spaceline of simultaneity (which is Minkowski-perpendicular to that observer's worldline).

Suppose Alice has a ruler of length 10, interpreted as "where Alice says the wavefront of the previous signal is located when she emits the next signal".
Note that this "x=10 location in Alice's frame" has a worldline parallel to Alice's worldline.

The "length of a ruler" is the spatial separation between two parallel timelike worldlines.
The "wavelength of a light wave" is the spatial separation between two received wavefronts (i.e. two parallel lightlike-lines).
The "period of a light wave" is the temporal separation between two received wavefronts.
The "wave-speed" equals wavelength/waveperiod.

So, Alice says the length of my ruler is $L=10$, $\lambda_{source}=10$, and $T_{source}=10$.

Since Bob is in relative motion, according to special relativity, his sense of simultaneity is different from Alice's.
(In Galilean physics, his sense of simultaneity is the same as Alice's... in that case, Bob will measure the same wavelength and same ruler length as Alice did.)

So, Bob measures Alice's ruler to be $8=\frac{10}{(\frac{5}{4})}$ units long (length contraction) and
the wavelength to be $20=10(2)$ units long (Doppler effect for receding receiver)
and the period to be $20=10(2)$ units elapsed (Doppler effect for receding receiver).
Thus, the wave-speed=$20/20=1.$
[For the approaching case, you can see $5=10/(2)$ units for wavelength and period, and thus wave-speed=$5/5=1.$]

Now, one can try to write down the "equations" and interpret the "variables" suggested by the diagram.
The equations are actually relationships of the sides of various triangles in this diagram.

 

[vanhees71]

Great, but for sure the equations work out in the same way and much simpler than any diagram. It's a simple Lorentz boost of the wave-four-vector of a electromagnetic plane-wave mode, but because this is a B-level thread it's thought to be forbidden to explain it in this way, and maybe your diagram is helpful after all the math-avoiding debates before. SCNR.

 

[robphy]

Many times one needs multiple representations of a situation to more fully understand a situation.

4-vector and “frame of reference Lorentz transformation” equations (when you know what the variables and their relationships mean) tell you some parts of the story. Diagrams (when you know what the dots and lines and their relationships mean) tell you some other possibly overlapping parts of the story. Words, numerical plots, animations, videos, symmetry arguments, explicitly worked out examples, kinesthetic activities, demonstrations, simulations, etc...

Is the goal trying to get the OP to understand the situation by trying multiple methods (when one doesn’t work, try another ... and another...)
or
is the goal to find the most elegant answer with the fewest number of symbols?

 

[Nugatory]

Is the goal trying to get the OP to understand the situation by trying multiple methods (when one doesn’t work, try another ... and another...)
or
is the goal to find the most elegant answer with the fewest number of symbols?

In a B-level thread, it's the former.