1
$\begingroup$

I have the following problem:

"A blue light, emitting at a wavelength $\lambda = 400 (nm)$ in its rest frame, is mounted to the side of a train traveling at a constant speed of $v = 0.6c$ in a station’s frame. Calculate the wavelength of the light measured by an observer standing on the station platform at the exact moment at which the train passes. You may neglect the distance between the observer and the train."

Now, I am thinking the following:

Classically any Doppler shift type effect arises because different signals need to travel different distance due to relative motion between the emitter and the observer in the direction of the "signal" (wave). In this case, no Doppler shift should arise in the situation when the emitter travels at right angles to the direction of the signal. Which is to say, in classical physics. However, in the case of the relativistic Doppler shift, there are two factors causing the shift: The classical just described and time-dilation. Hence, in our problem with the train, only the second one will be at work.

Therefore: $ \Delta t' = \gamma \Delta t $, where $ \Delta t = 1/f $ etc. This should yield: $$ \lambda_3 = \frac{\lambda_3'}{\gamma} = 320 (nm).$$ Is this alright? I have never solved a similar problem before and I am therefore uncertain. How do I know for sure if it is $ \Delta t' = \gamma \Delta t $ or $ \Delta t = \gamma \Delta t' $?

$\endgroup$
  • $\begingroup$ Neither the classical Doppler effect nor the relativistic one depends on distance but on the rate of change of it with time, that is on velocities. So, it's not necessary to give permission to neglect the distance. $\endgroup$ – Frobenius Aug 9 '17 at 14:22
  • $\begingroup$ I understand that your troubles are about time dilation. Take a look in my answer therein : How do I know which observer is running the time faster or slower?. May be useful. $\endgroup$ – Frobenius Aug 9 '17 at 14:57
1
$\begingroup$

The Doppler effect can be observed at an arbitrarily close distance

Let’s consider the case, when beam of light moves at normal to paths of source/emitter. This will be the Transverse Doppler Effect, that reflects rate of relatively moving clock.

Amount of redshift/blueshift will depend on angles of emission/reception.

1) Light emitted at points of closest approach. If photon approaches the observer at right angle, the source must emit the photon at oblique angle. Photon will be redshifted. Observer measures dilation of relatively moving clock.

2) Light received at points of closest approach. If photon was emitted at right angle, it will approach observer at oblique angle. The observer will see blueshift of frequency. Observer may explain, that apparent position of source is different from actual due to aberration of light. He will also explain blueshift as dilation of his own clock, since he moves himself in emitters frame. Since his own clock dilates, everything around him appears running faster, so opposite clock appears as ticking faster.

3) Light is neither emitted nor received at points of closest approach, or has been emitted and received at equal angles. In this case there will be no Doppler Shift, i.e. no dilation at all.

Wikipedia: The transverse Doppler effect can be analysed from a reference frame where the source and receiver have equal and opposite velocities. In such a frame the ratio of the Lorentz factors is always 1, and all Doppler shifts appear to be classical in origin. In general, the observed frequency shift is an invariant, but the relative contributions of time dilation and the Doppler effect are frame dependent.

Transverse Doppler effect in Wikipedia: https://en.wikipedia.org/wiki/Relativistic_Doppler_effect

Thus, resolving tasks of this sort it is necessary to consider angles of emission/reception. Angles of emission/reception are tied with relativistic aberration formula.

$$ \cos {\theta_0} = \frac {\cos {\theta_s} - \frac v c} {1- \frac v c \cos \theta_s} $$

Imagine two observers – Tom and Ben that move relatively to each other. Tom has a laser pointer and Ben has a telescope.

If Tom keeps laser pointer at right angle, Ben must turn his telescope at obligue angle $\alpha = \arcsin v/c$. Otherwise rays of light would not be able to pass through the telescope due to aberration of light. Ben will see blueshift (Tom’s clock is ticking faster)

If Ben keeps his telescope at right angle, Tom must turn his laser pointer backward at the same angle $\alpha$. In this case Ben will see redshift of frequency, i.e. Tom’s clock dilates.

Ben and Tom must always respect each other and not to play with reference frames as if they are their own property. They must agree, who moves and who is at rest and always keep in mind, at which angle turn telescope and laser pointer, if they want to see each other.

Please look for Transverse Doppler Effect: http://mathpages.com/home/kmath587/kmath587.htm

https://www.youtube.com/watch?v=hnphFr2Iai4 https://www.youtube.com/watch?v=FQKp3FU8vR8

EDIT: The task says "Calculate the wavelength of the light measured by an observer standing on the station platform at the exact moment at which the train passes".

Apparently they say that source emitted dispersed light and the light was received at points of closest approach, i.e observer at platform sees that radiation blueshifts, i.e. clock at train is ticking faster. Thus, your answer is correct.

But, observer had to look (or turn telescope/spectrometer) at oblique angle in accordance with relativistic aberration angle.

$\endgroup$
  • $\begingroup$ He does not specify this, but it seems to me that he wants to consider the question precisely through this prism. "Hence, in our problem with the train, only the second one will be at work." $\endgroup$ – Albert Aug 9 '17 at 14:44
  • $\begingroup$ I understand this later, so I deleted my comment simultaneously with your response. Thanks. $\endgroup$ – Frobenius Aug 9 '17 at 14:47

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.