B Why Light Experienced a Doppler Shift?

[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.