# Doppler Effect and Conservation of Energy

1. Nov 7, 2014

### bgq

Hi,

Consider a person who holds a light source of frequency f. Consider an observer who sees this light. The energy of each photon as received by the observer is E = hf. The observer is now fixed. Now suppose that the person who holds the light source starts to move towards the observer, then according to Doppler effect, the observer receives the light with a frequency f2 which is different from f. Therefore, according to this observer, the energy of the photon changes.

I tried to look about this issue on the internet, but they say that when moving with respect to the source, energy is not conserved. However, we can make the observer fixed, and make the source to move. In this case for the observer frame of reference the energy of the photon changes.

How to resolve this considering the principle of conservation of energy?

Thanks to any replies.

2. Nov 7, 2014

### Staff: Mentor

Energy is not frame invariant. Simply consider kinetic energy - different values - different frames. And remember from the POR all motion is relative.

Thanks
Bill

3. Nov 7, 2014

### bgq

Even if the frame does not change (the fixed observer), the energy of the photon will change if the source moves. So the energy changes within the same frame of reference

4. Nov 7, 2014

### jartsa

So there's a moving thing that shoots some kind of projectiles onto a fixed target, and we are interested about the total energy before and after the shooting.

Energy and momentum are conserved in these kind of cases.

OP, can you calculate this: A fighter plane flies at velocity 200 m/s, and shoots backwards with a gun that has muzzle velocity 200 m/s. What is the kinetic energy of the bullets?

5. Nov 7, 2014

### Staff: Mentor

If a gun fires a bullet moves then the kinetic energy of the bullet changes - same thing.

The frame could move - or the gun - physically the same - as guaranteed by the POR.

Thanks
Bill

Last edited: Nov 7, 2014
6. Nov 7, 2014

### Staff: Mentor

Even more instructive calculate it from the point of view of a frame attached to the earth and one attached to the plane.

Thanks
Bill

7. Nov 7, 2014

### bgq

OK,
Lets calculate the total energy.
Consider a fixed observer standing beside a long road. Now consider a car with a lamp on its roof that produces light of frequency f moving towards the observer. According to this observer, the total energy of the system (Car, light) is E1=1/2mv2 + nhf' where f' > f. Now, when the car passes the observer (so now moving away from it) the new total energy becomes E2 = 1/2mv2 + nhf'' where f'' < f. Clearly E2 < E1, even when we consider the total energy of the same system calculated with respect to the same frame of reference

8. Nov 7, 2014

### Staff: Mentor

Precisely what don't you get about energy not being frame invariant?

Thanks
Bill

9. Nov 7, 2014

### bgq

I don't have a problem that energy depends on the frame. What confuses me is the example in my previous reply where energy appears not to be conserved with respect to the same frame.

10. Nov 7, 2014

### Staff: Mentor

What happens in your example when the car is coincident with you - is it going towards you or away from you?

Indeed what is forward and away if you are orientated in the direction of the car?

Should the direction you are facing affect the result of your observation?

Thanks
Bill

11. Nov 7, 2014

### bgq

Neither towards nor away. In this case the received frequency would be the same as that sent by the source. Is this true?

12. Nov 7, 2014

### Staff: Mentor

Your analysis is correct. Its just how you are measuring it has changed - exactly as if it was in a different frame. It's kind of misleading to say that energy increases or decreases due to a Doppler shift, because that would imply that there is some physical process changing the energy of the photon. That's really not the case, it's simply that energy is a quantity for which the value you measure depends on how you measure it.

Remember that energy is part of a four vector in relativity, and you can get different values of the components depending on how you measure it. Of course being a 4 vector 'length' is invariant - but not its components.

Thanks
Bill

Last edited: Nov 7, 2014
13. Nov 7, 2014

### vanhees71

I don't know, how the web site's authors come to this conclusion. Energy conservation of course holds true from the very fundamental principles of special relativistic space-time. Energy is defined as the conserved quantity associated with the invariance of space time under time translations.

Of course, energy is not frame independent but together with momentum it forms a four-vector, the energy-momentum vector,
$$(p^{\mu})=\begin{pmatrix}E/c \\ \vec{p} \end{pmatrix}.$$
Of course, energy momentum conservation holds in each frame of reference, but the values for energy and momentum change when transforming from one frame to the other via a Lorentz transformation.

Since electromagnetics a la Maxwell (and the quantum-theoretical extension, QED) is a relativistic theory, also the energy-momentum relation for photons (note that you must not confuse photons with little particles; they are far from that and a pretty abstract concept which has to be carefully study within relativistic quantum field theory) is Lorentz invariant, i.e., in each reference frame you have
$$p_{\mu} p^{\mu}=0 \; \Rightarrow \; E=p c.$$

14. Nov 7, 2014

### Staff: Mentor

I think we've been making this problem harder than it needs to be.

All that's going on here is that viewed from the road frame, the car is emitting more energetic radiation in the forwards direction than the rearwards one. The easiest way to see this is to imagine that you have a large number of detectors [positioned at intervals along the road; when the car is between any pair of detectors, the one behind the car will always record a lower intensity than the one in front.

The apparent paradox in your scenario comes about because we know that radiation is the same in both directions in the frame in which the car is at rest - and we are trying to carry that frame-dependent observation over into the frame in which the car is not at rest.

15. Nov 7, 2014

### Staff: Mentor

I should have moved this thread into the relativity section a while back.... Doing that now.

16. Nov 7, 2014

### Staff: Mentor

But what happens if you are parallel to the direction of motion rather than perpendicular to it?

I think its simply the relation of what you are measuring it with to the light.

I was wrong - the argument is still the same. I was going to delete my reply - but I decided to leave it there to show we all can get confused.

Thanks
Bill

17. Nov 7, 2014

### jartsa

You guys have no faith in conservation laws of physics.

The "missing" energy must be somewhere.

18. Nov 7, 2014

### Staff: Mentor

19. Nov 7, 2014

### ShayanJ

You have a very wrong assumption in your calculations. You're assuming that the frequency is constant at all points of the wave front. But that's true only in the reference frame in which the lamp is at rest. In other reference frames, the lamp is moving and the frequency at each point of the wave front, depends on its orientation w.r.t. direction of motion. So for every point with a greater frequency, there is a point with lower frequency and the overall change is zero. So I think this has nothing to do with relativity.(we can ask the same thing about sound waves, right?)

20. Nov 7, 2014

### jartsa

Emmy Noether: "Energy is conserved if laws of physics are not changing in time"

Physics teachers: "Use the law of conservation of energy to calculate ..."

Me and OP are talking about that conservation of energy.

http://en.wikipedia.org/wiki/Conservation_law

21. Nov 7, 2014

### pervect

Staff Emeritus
Suppose you have a compressed spring, and a couple of ball-shaped projectiles. The conservation of energy means that if you use the spring to launch the projectiles, the energy will be the same before and after the launch.

The conservation of energy does not mean that the total energy you calculate will be independent of your choice of reference frame. The easiest choice of frame is one where the balls and spring are all at rest before launch, but you can also calculate the energy in other frames, where the balls and spring are initially moving. The value of the system energy "before launch" depends on which reference frame you choose. The conservation of energy just says that if you calculate the energy before and after some event (like using the spring to launch the balls) in a given frame, you'll get the same energy "before" and "after".

The short version of this - energy is conserved, but frame-dependent.

Your question seems to conflate (combuse and combine) energy conservation with frame independence. The two concepts are different.

22. Nov 7, 2014

### Staff: Mentor

Of course energy is conserved. There's no violation of energy conservation going on here, and no "missing" energy that has to be accounted for.

Consider a spherical surface surface enclosing the car. The total energy flux through that surface per unit time will be equal to the total energy emitted by the light source inside the car per unit time; this will be true in all frames and is all that is required for energy to be conserved.

Observers in different frames may see a different distribution of the flux across the surface of a sphere, and there is only one frame, the one in which the car is at rest, in which the flux will be uniform across the entire surface of the sphere.

23. Nov 7, 2014

### jartsa

Of course.

Consider a car moving forwards quite fast, about half percent of c, with 100 W lights at both ends. The car battery provides 200 W of power to the lights. Front light illuminates the road with 101 W of light power. Rear light illuminates the road with 99 W of light power.

Then we move the rear light to the front, road is now illuminated by 202 W of light power, while battery is still delivering 200 W of electric power to the lights.

Last edited: Nov 7, 2014
24. Nov 7, 2014

### Staff: Mentor

Please show your calculation. It seems like you are just making stuff up.

Don't forget to account for the energy and momentum of the car.

Last edited: Nov 7, 2014
25. Nov 7, 2014

### Staff: Mentor

The four momentum is conserved, meaning that energy, momentum, and mass (the 'length' you refer to) are all conserved. Of these only the mass is also invariant.