Red/blue shift in classic train/platfom thought experiment?

[melchloboo]

To avoid confusion, I'm talking about:
"A flash of light is given off at the center of the traincar just as the two observers pass each other. The observer onboard the train sees the front and back of the traincar at fixed distances from the source of light and as such, according to this observer, the light will reach the front and back of the traincar at the same time."

So assume the train has no windows and the onboard observer doesn't know which is the front or back of the train. Also, the light bulb is directly above the onboard observer's head when it is turned on. The observer has 3 sensors that detect the color (i.e. wavelength of light). He puts one sensor at each end of the inside of the train car. He puts the third sensor on top of his head (although I'm not really sure this is necessary for my question).

My question is, will the sensors all report different wavelengths of light due to redshift? And if so, based on these differences, wouldn't the onboard observer be able to determine not only his direction of travel, but also his velocity, relative to the light source?

 

[ghwellsjr]

No, they will all report the same wavelength.

 

[melchloboo]

No, they will all report the same wavelength.

Ok, if that is the correct answer, can you help me get my head around it?

I am thinking the sensor at the back of the car is moving towards the photons coming from the light, and the sensor at the front of the car is moving away from them. So why wouldn't there be redshift/blueshift?

Is it because the light source was originally in the same inertial frame?

I have the same question for when the experiment is changed where lightning strikes the outside of the train at both ends:

http://upload.wikimedia.org/wikipedia/commons/9/96/Einstein_train_relativity_of_simultaneity.png

Shouldn't the onboard observer see a shifting in that one too?

 

[PAllen]

There is a little clarification needed here. Doppler shift reflects relative motion between emitter and receiver. If the emitter is on the train (thus has the same motion - none relative to center or either end) then there is no Doppler for any train observers. If the emitter is ground based, then one end sees approaching Doppler (blue shift), the other end sees receding Doppler (red shif), the center observer sees transverse Doppler (which exists only special relativistically, not according to Galilean relativity), which is slight red shift.

 

[ghwellsjr]

I considered the possibility that the light source was ground-based but then he said there were no windows in the car so I figured it couldn't be. Maybe he'll explain.

 

[melchloboo]

There is a little clarification needed here. Doppler shift reflects relative motion between emitter and receiver. If the emitter is on the train (thus has the same motion - none relative to center or either end) then there is no Doppler for any train observers. If the emitter is ground based, then one end sees approaching Doppler (blue shift), the other end sees receding Doppler (red shif), the center observer sees transverse Doppler (which exists only special relativistically, not according to Galilean relativity), which is slight red shift.

1. So in the example where the lightning is simultaneous at both ends of the car, if the car is in motion won't the an onboard observer in the middle see a shift? (yes we need windows for this)

2. "Doppler shift reflects relative motion between emitter and receiver." I'm ok with that for certain waves, such as sound, because a single air molecule through which it propagates isn't itself the measurable wave...I can't measure one air molecule and know anything about the sound. I have to wait for more molecules to arrive and measure them, which is basically what the hairs in our ears do.

But with, a single photon, is that the really case? Perhaps I have a fundamental misunderstanding of the wavelength property of light, because I thought wavelength was a property of each and every photon?
http://en.wikipedia.org/wiki/Photon#Physical_properties
In other words, a photon has a wave property itself, and does not depend on the photons around it for that property, unlike the molecules that make up a sound wave.

Unlike sound, I can measure a single photon and know it's "color", if the photon came from a laser I know all I need to know about the wavelength of the photons the laser produces and I don't need to examine any other photons coming from it. And from that wavelength I know E for that photon. Am I totally wrong on that?

And so, I thought redshift is what happens when the energy of the observer and the energy of the photon interact (i.e. the photon reaches the observer so he can measure it) and the observer's momentum interacts with the photon. The photon's velocity cannot vary from c, but its wavelength changes.

 

[ghwellsjr]

It doesn't matter whether the car is in "motion" or the light source is in "motion", it only matters whether there is relative motion between the source of light and the detector of light.

There is a difference between the Doppler for sound and Relativistic Doppler for light. Doppler for sound does depend both on the motion of the source with respect to the air and the motion of the detector with respect to the air. You can measure the motion of the medium for sound by using Doppler but you cannot measure any medium for light.

The same thing applies for photons. It's really very simple. Why are you struggling with this?

 

[PAllen]

1. So in the example where the lightning is simultaneous at both ends of the car, if the car is in motion won't the an onboard observer in the middle see a shift? (yes we need windows for this)

Yes (assuming lightning hits ground spots coincident with train ends, rather than hitting the train ends), and this will tell train they are moving relative to ground. With windows, they don't need lightning to tell them this. They see ground moving. Ground sees train moving. Wow.

2. "Doppler shift reflects relative motion between emitter and receiver." I'm ok with that for certain waves, such as sound, because a single air molecule through which it propagates isn't itself the measurable wave...I can't measure one air molecule and know anything about the sound. I have to wait for more molecules to arrive and measure them, which is basically what the hairs in our ears do.

But with, a single photon, is that the really case? Perhaps I have a fundamental misunderstanding of the wavelength property of light, because I thought wavelength was a property of each and every photon?
http://en.wikipedia.org/wiki/Photon#Physical_properties
In other words, a photon has a wave property itself, and does not depend on the photons around it for that property, unlike the molecules that make up a sound wave.

Unlike sound, I can measure a single photon and know it's "color", if the photon came from a laser I know all I need to know about the wavelength of the photons the laser produces and I don't need to examine any other photons coming from it. And from that wavelength I know E for that photon. Am I totally wrong on that?

And so, I thought redshift is what happens when the energy of the observer and the energy of the photon interact (i.e. the photon reaches the observer so he can measure it) and the observer's momentum interacts with the photon. The photon's velocity cannot vary from c, but its wavelength changes.

You can analyze the Doppler quantum mechanically or classically, with same result. If you are talking about single photons, the thing to note is that kinetic energy is frame dependent (for photons, neutrinos, baseballs, whatever). Relativistic doppler formula tells you how to relate frequency per emitter to frequency per receiver (and KE = E = h *frequency, for photons). Even for one photon, this simply means emitter sees photon of one energy, receiver sees photon of another energy. If a radioactive decay emits a high energy gamma, there exists a frame where this gamma photon is detected as a long wavelength radio photon.

 

[melchloboo]

Ok, maybe I'm starting to get it, maybe not. I think the answer is that the photons emitted inside the train are carrying the same relative kinetic E as the passengers at either end of the train, so there is no shift in wavelength holding c constant.

But consider the platform observer watching a light in the middle of the train through a window. I understand the platform observer must always see the same c as the train approaches, at a right angle to him, and as the train is leaving. In order to leave the system of the moving train, the photons leaving the train to him as the train approaches have to be more energetic then the photons leaving the train as the train travels away from him, and with a constant c that change in energy has to be reflected through the change in observed wavelength.

But as a light inside the train passes by a platform observer at a right angle, the platform observer will see the exact same color of light as any passenger in the train always sees? i.e. if we create a narrow slit at a right angle to the path of the train for the platform observer such that he can only see the light for that instant when it is perfectly at a right angle to him as the train passes, you are saying the light he sees will be indistinguishable from light emitted from a stationary lamp post from the same position as the train, assuming the same color bulbs are used?

But even at a right angle, where the train is neither approaching or retreating form him, don't photons from the train still carry more E (momentum as they were once on the train) than the photons coming from a stationary lamp post? And shouldn't that E show up as a wavelength different from a stationary lamp post?

 

[PAllen]

But as a light inside the train passes by a platform observer at a right angle, the platform observer will see the exact same color of light as any passenger in the train always sees? i.e. if we create a narrow slit at a right angle to the path of the train for the platform observer such that he can only see the light for that instant when it is perfectly at a right angle to him as the train passes, you are saying the light he sees will be indistinguishable from light emitted from a stationary lamp post from the same position as the train, assuming the same color bulbs are used?

But even at a right angle, where the train is neither approaching or retreating form him, don't photons from the train still carry more E (momentum as they were once on the train) than the photons coming from a stationary lamp post? And shouldn't that E show up as a wavelength different from a stationary lamp post?

There is such a thing as transverse Doppler in special relativity. However, transverse doppler is a redshift, not a blueshift as you are proposing. The key fact is the relation of time and energy (timelike component of 4-momentum is energy, for example). The train clocks are slower per the ground observer, so a transverse photon emitted with a given energy on the train, will be received with less energy by a ground detector.

There is a particular angle of emission where the train emission energy will match the ground receiver energy. This will be at some point approaching orthogonal, where you can think of the approaching blueshift balanced by the time dilation redshift.

 

[Petyab]

It's possible for the sensors to detect different light frequencies. If wave packets interfere with each other then any wavelength in the visible light spectrum could be shifted and delay or expedite the time of travel.

 

[yoron]

in a uniform motion you won't see a difference, but in a acceleration you should see light red shift relative the front of the motion (as it moves away) , blue shift toward the rear as the detector there 'rushes to meet' your light, sent from the middle of the car.

But all uniform motions are at rest, and all experiments you do in a uniform motion will come out the same, as far as I know that is :) And it won't really matter what 'velocity' you have relative something else. That a acceleration does it should have to to do with the constant accelerating 'displacements' in 'time/distance' as I think of it, but I'm not sure if that is enough to explain it. Einstein defined it as 'gravity', and it acts that way (inertia). Which makes a planet a really weird idea :)