# Extra energy of photon in doppler effect

## Main Question or Discussion Point

Assume a object in motion emits a beam of light at an observer.
The observer detects a frequency that is higher than the original frequency.
That is the measured energy of the photon by this observer is higher than it was while emitting.
How to account for this extra energy?
thanks,
kalpak

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Doc Al
Mentor
The source is moving. The energy of a photon is frame dependent.

mfb
Mentor
And the moving sender loses a small amount of kinetic energy by sending a photon in forward direction. This energy is not present in the system of the sender, which can be used as explanation for the different photon energy for different observers.

Thanks for the prompt replies.
If the photon is taking away energy from the sending system, then if the sender measures the frequency of the emitted light, it should also be frequency shift.
How can one verify that?

phyzguy
Energy is frame dependent. Forget photons and think of a basketball. In the frame where it is at rest, the basketball has no kinetic energy. In a frame moving with respect to the basketball, the basketball is moving and has some kinetic energy. Do you then ask where the kinetic energy of the basketball came from?

The comparison between photons and basketball (or any other projectile ) is wrong.
A photon has the same measured velocity in both the sender and observer frames, not so with any other projectile.

If you eye-butt a photon, there's some extra energy in the collision, the energy came from your muscles.

When photon-rocket is decelerating in space, the kinetic energy of the rocket goes into the photons.

PAllen
2019 Award
The comparison between photons and basketball (or any other projectile ) is wrong.
A photon has the same measured velocity in both the sender and observer frames, not so with any other projectile.
The analogy with basketballs is valid. In SR, KE is not just related to velocity as it is for Newtonian mechanics. The frame dependence of basketball energy and momentum is no different in character than for photons. Consider as another comparison, neutrinos. These are taken to have a small mass, but so small that, in practice, they move at a rate indistinguishable from c. Yet, though moving at essentially c for almost all observers, their KE varies with frame just as it does for basketballs (and light).

The frame dependence for all cases is captured using 4-momentum formulated as follows:

(E/c, p)

With E = total energy, and p=3-momentum.
This transforms as a vector under Lorentz transform between frames, with both E and p changing for any Lorentz transform.

The norm of this vector is mc, producing the well known relation: E^2 - p^2 c^2 = m^2 c^4. For a massless particle, this gives p=E/c. For massive particles, you have p =(1/c)√ (E^2 - (mc^2)^2).

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Thanks for the prompt replies.
If the photon is taking away energy from the sending system, then if the sender measures the frequency of the emitted light, it should also be frequency shift.
How can one verify that?

Photon rocket crew always sees the emitted photons to be normal photons. The photons are doing no work when they are pushing a non-moving rocket, that's what it looks like from the rocket.

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

Thanks for the prompt replies.
If the photon is taking away energy from the sending system, then if the sender measures the frequency of the emitted light, it should also be frequency shift.
How can one verify that?
Photon rocket crew always sees the emitted photons to be normal photons. The photons are doing no work when they are pushing a non-moving rocket, that's what it looks like from the rocket.

http://en.wikipedia.org/wiki/Photon_rocket
For the rocket to measure the frequency shift would necessitate receiving it in the rocket frame in which case the very motion that produced the shift is now acting in opposition to cancel out the shift produced in emission.

For the rocket to measure the frequency shift would necessitate receiving it in the rocket frame in which case the very motion that produced the shift is now acting in opposition to cancel out the shift produced in emission.
Yes, that's how it looks from the outside.

The outside observer says that inside a rocket a photon that bounces between two mirrors becomes red shifted when it does work on a mirror by pushing it, and the photon becomes blue shifted when a mirror does work on it by pushing it.

When we replace one mirror by a black wall, the pushing work is halved, compared to mirror. We could say that first a blue shifted photon pushes the wall, becomes a photon whose energy is halfway between a blue sifted photon and a red shifted photon, which is a normal photon, and then the normal photon does heating work on the wall.

The sender is undergoing time dilation due to motion, so the second is "slower" in the sending system.

Therefore, if the sender were to measure own emitted frequency, it would also be blue shifted (more waves fit in the slower second measurement window).

Dale
Mentor
No, your clock is never time dilated in your own frame, only in other frames. That would violate the principle of relativity.