# Yet another doppler question (regarding conservation, time, and irradiance)

1. Mar 9, 2010

### ssope

I need the following statements logically-'spellchecked'. The internet, nor books will put it this simply, so I typed up what I feel is a summary, but I don't want to be incorrect in my own mind by doing so. So please proof read my logic and correct it for errors:

The doppler effect IS time dilation. However since light always travels at the same speed, the change in "time" is experienced in nature as a change in energy density of the beam of light, and this is because light does not experience time (otherwise it would decay). The only reason that the doppler effect has a special name in the case of light is because that Doppler fellow who discovered the frequency change for light was born back in the 1800's (before einstein's time).

The doppler effect says that by moving a light source towards an observer the observer will receive more energy per unit per time (because the energy is compressed). This apparent energy change does not violate the conservation of energy because the E in "E=hf" can be broken down into the photon's self energy and the photon's kinetic energy, in which case you subtract the appearant gain in energy due to the frequency change from the photon's self energy (meaning the kinetic energy increases at the same rate as the photon's self energy decreases).

{If correct}

How would I calculate the irradiance (change in flow of energy per unit per time). I'm guessing I'll need to know how fast the laser is moving, and what wattage the laser has, but lets say I have such values (if i'm missing any required input data, let me know).

2. Mar 9, 2010

### Staff: Mentor

Almost none of that is correct. The Doppler effect is not the same as time dilation, time dilation is always shifts towards lower frequencies and does not depend on direction, but the Doppler effect can shift higher or lower depending on the direction. Talking about what photons "experience" is mathematically ill-posed (involves division by 0). I don't know what a photon's self energy would be, but energy is conserved because the overall volume of the fields is decreasing.

3. Mar 9, 2010

### ssope

Time dilation also shifts towards higher frequencies. High altitude clocks tick faster. You also did not correct and/or deal with some other statements in my post.

4. Mar 9, 2010

### Staff: Mentor

Yes, gravitational time dilation can be higher or lower, but is not really related to the Doppler effect. In SR, time dilation always shifts to lower frequencies and is sometimes called the transverse Doppler.
Which specific statement do you want me to correct?

5. Mar 9, 2010

### JesseM

ssope, are you aware that the classical Doppler shift is different than the relativistic Doppler shift? See here for the equations. And the difference is precisely because of the time dilation in the relativistic case! I gave a numerical example showing how it works in post #9 of this thread:

6. Mar 10, 2010

### ssope

If you have a fast moving laser, does the laser and the wall its moving at experience time dilation but the light and the wall do not? Would the laser spit out light slower if it was moving fast?

7. Mar 10, 2010

### ssope

You didn't touch this:

The doppler effect says that by moving a light source towards an observer the observer will receive more energy per unit per time (because the energy is compressed).

8. Mar 10, 2010

### JesseM

It's not really meaningful to ask if the light "experiences" time dilation since light doesn't have any internal clock and in SR it doesn't have its own rest frame either. But in the inertial rest frame of any object moving slower than light, the light always moves at exactly c, never slower or faster, and any slower-than-light object that's not at rest in that frame (like a wall viewed in the in the rest frame of a laser emitter moving towards it, or the laser emitter viewed in the wall's rest frame) will experience time dilation.

9. Mar 10, 2010

### ssope

Would the laser spit out less light per unit time if it was moving fast compared to a laser not moving?

The doppler effect says that by moving a light source towards an observer the observer will receive more energy per unit per time (because the energy is compressed), yes? The doppler effect says that by moving a light source towards an observer the observer will receive more energy per unit per time (because the energy is compressed), yes?

10. Mar 10, 2010

### JesseM

Yes, fewer photons emitted per unit time and less energy emitted per unit time as well. Although if it's moving towards you, the blueshift will cause you to receive more photons/energy per unit time than you would from an identical laser emitter at rest relative to you, in spite of the fact that in your frame the laser emitter moving towards you is really emitting fewer photons/less energy per unit time than the one at rest relative to you.

11. Mar 10, 2010

### ssope

Inituitivly, there should be a true energy gain associated with gravitational blueshifting, is this true?

12. Mar 10, 2010

### JesseM

There is a true gain in the amount of energy you receive per unit time, but not in the amount of energy being emitted per unit time in your rest frame. Here's an altered version of the numerical example I gave before:
So here the source is sending out new pulses once every 25 seconds in your frame, but you are receiving new pulses once every 10 seconds. So if each pulse contains a certain amount of energy, the rate at which you are receiving energy is higher than the rate at which it's actually being emitted in your frame.

13. Mar 10, 2010

### Staff: Mentor

Yes, I did.
That was refering to the usual Doppler. For the gravitational Doppler the extra EM energy comes from a decrease in gravitational PE.

14. Mar 10, 2010

### JesseM

What do you mean by "conserved" in this context? If the source was emitting bullets with equal kinetic energy in your frame, the rate at which you'd receive energy from the bullets would be higher than the rate in your frame at which the source was giving up potential energy to impart to the bullets, no? (obviously energy is still conserved in the sense that the total energy of source + you + bullets is constant over time) If so, would it be different for electromagnetic waves?

15. Mar 10, 2010

### Staff: Mentor

I was refering to the fields. E.g. see http://farside.ph.utexas.edu/teaching/em/lectures/node89.html" [Broken]. The energy density in the volume between the emitter and reciever is constant, but the volume is decreasing. Therefore, more power must be leaving the volume than entering.

Last edited by a moderator: May 4, 2017
16. Mar 10, 2010

### JesseM

Got it, so it would be analogous to the bullet situation where the total energy of receiver + emitter + bullets flying in between them is staying constant from one moment to the next, but the receiver is receiving energy from bullets at a greater rate than the emitter is imparting energy to bullets (and since the distance between emitter and receiver is shrinking, the number of bullets in the air at any given moment, and thus the total energy of the bullets at that moment, is decreasing as well).

Last edited by a moderator: May 4, 2017
17. Mar 10, 2010

### Staff: Mentor

Yes, the two descriptions are perfectly compatible.

18. Mar 11, 2010

### ssope

Why doesn't radioactive decay vary as a function of earth-sun distance. Why didn't the cassini spacecraft show such data? They had to correct tom-tom satilites for time dilation because of the low gravity, so why doesn't radioactive decay vary as a function of gravity? http://www.astroengine.com/?p=1382

Last edited: Mar 11, 2010
19. Mar 11, 2010

### Staff: Mentor

All the measurements were done onboard, so you would not expect there to be any relativistic effects.

20. Mar 11, 2010

### ssope

Are you saying because the onboard systems did not COMPARE themselves to earth bound measurements that there is no question to be asked?

21. Mar 11, 2010

### JesseM

What kind of variation would you expect here? Apparent variation due to different velocities relative to Earth (Doppler shift), or some kind of time dilation (gravitational or velocity-based)? Anyway, I would guess the graph of power output vs. time at the bottom of that page is using onboard time rather than Earth time (though the difference would probably be negligible), and certainly the power output was measured on board.

22. Mar 11, 2010

### ssope

No, not due to velocities, the question is regarding due to gravitational changes. The varation I would expect is that plutonium would decay faster and faster the further it got away from the sun (and/or other gravitational fields)

23. Mar 11, 2010

### JesseM

Well, again, you wouldn't expect any changes whatsoever if the time axis is based on the ship's own clock. And if it's based on Earth clocks, I'd imagine any change in the decay rate due to the Earth being closer to the Sun and experiencing more gravitational time dilation would be totally miniscule, too small to make any appreciable difference. The Sun's Schwarzschild radius is about 3 kilometers, so according to the gravitational time dilation equation given here, a clock at radius r will be slowed down by a factor of $$\sqrt{1 - (3\,km)/r}$$ as compared with a clock at infinity. So, the ratio between tick rates of two clocks at r1 and r2 would be $$\frac{\sqrt{1 - (3\,km)/r_1}}{\sqrt{1 - (3\,km)/r_2}}$$. The radius of the Earth's orbit is about 150 million kilometers, and the article says Cassini's decay rate was measured from 0.7 Earth radii to 1.7 Earth radii (about 105 to 255 million km), so I calculate the ratio of clock ticks at the two different distances to be about 0.9999999916 to 1. And even if you consider the ratio for a clock at the Earth's distance from the Sun (ignoring the gravitational time dilation from the Earth itself) and a clock infinitely far from the Sun, the ratio is still only 0.99999999 to 1.

Last edited: Mar 11, 2010
24. Mar 11, 2010

### ssope

I could use your help calculating the time difference for something orbiting very high above the moon in comparison to the surface of the Earth.

25. Mar 11, 2010

### Staff: Mentor

No. I am saying that the experimental answer to the question agrees with the theoretical prediction. From your previous comments it is pretty clear that you don't understand the theory, so you erroneously expected relativistic effects when there would be none.
Yes, and I read the arxiv paper it cited.
You are really pretty pushy.