# Does a blueshift change the energy of a photon?

## Main Question or Discussion Point

The energy of a photon depends on its wavelength, so theoretically when it is blueshifted it should have more energy right?

Then what if a spaceship with a solar panel on the front is traveling towards the sun at relativistic speeds. An incoming photon undergoes a blueshift from the observer on the spacecraft. So does the solar panel read the same energy as if the light wasn't blueshifted? I see two options here: 1. Either the solar panel reads two different numbers depending on the observer. (almost like 2 realities exist) 2. Or it reads the same because the energy of the photon is not actually based on wavelength

To keep it simple let's imagine we're only talking about one photon, because time dilation might affect the power level the solar panel was reading.

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## Answers and Replies

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Ibix
Science Advisor
There isn't an awful lot of difference between time dilation and blue shift (edit: that's an exaggeration, but they are intimately related). But yes, a ship travelling towards the sun will see more energy from a pulse of light than one moving away. Energy is a frame dependent value, both in relativistic and classical mechanics.

The collision is just like any other. In the frame where the rocket is at rest, all of the energy comes from the light and is absorbed by the rocket. In a frame where the rocket is moving, the kinetic energy of the rocket can either decrease or increase, so it can be either a source or sink for energy.

Incidentally, photons do not keep things simple. Best to get into the habit of thinking in terms of pulses of light (perhaps two identical ones absorbed by different ships, or whatever) until you actually mean to start quantum field theory.

Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
It needs to be stressed that energy is not an observer independent quantity. Hence, a light signal in itself does not have a fixed energy. In order to say a light signal has a certain energy you must specify according to which observer (which is often done implicitly). Energy in general is an observer dependent quantity. This is true also in classical mechanics.

Dale
Mentor
when it is blueshifted it should have more energy right?
Yes.

Either the solar panel reads two different numbers depending on the observer. (almost like 2 realities exist)
No, all frames agree on the outcome of any measurement. The result of the measurement depends on the motion of the measurement device, but all frames agree on the number.

it reads the same because the energy of the photon is not actually based on wavelength
No, the energy is based on the frequency. The frequency is a relative quantity, and so is the energy. Energy is conserved, but not invariant.

Dopper effect is another way for explanation.

There isn't an awful lot of difference between time dilation and blue shift (edit: that's an exaggeration, but they are intimately related). But yes, a ship travelling towards the sun will see more energy from a pulse of light than one moving away. Energy is a frame dependent value, both in relativistic and classical mechanics.

The collision is just like any other. In the frame where the rocket is at rest, all of the energy comes from the light and is absorbed by the rocket. In a frame where the rocket is moving, the kinetic energy of the rocket can either decrease or increase, so it can be either a source or sink for energy.

Incidentally, photons do not keep things simple. Best to get into the habit of thinking in terms of pulses of light (perhaps two identical ones absorbed by different ships, or whatever) until you actually mean to start quantum field theory.
So energy of photons are different in different frame, but momentum must constant, by law of conservation of momentum. E=pc, how could that possible?

Ibix
Science Advisor
So energy of photons are different in different frame, but momentum must constant, by law of conservation of momentum. E=pc, how could that possible?
Momentum will be different in different frames too. Momentum will be conserved during the collision, yes, but different frames do not agree on the value. For example, if you drop a 1kg bag of sugar and it hits the ground at 1m/s then in the rest frame of the Earth the total momentum is 1Ns. In the rest frame of the bag of sugar it's the Earth that is moving at 1m/s, giving a total momentum of 6×1024Ns. Both frames agree that the total momentum is unchanged before and after the collision, however.

Edit: the point is that there's a difference between an invariant quantity, which is one that is the same in all frames (e.g. the speed of light), and a conserved quantity, which is one that does not change with time (e.g. total momentum of a closed system).

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all frames agree on the outcome of any measurement
And could you expand a bit on this by saying how that happens mathematically. Would it be the inner product being invariant?

Dale
Mentor
So energy of photons are different in different frame, but momentum must constant, by law of conservation of momentum. E=pc, how could that possible?
You are confusing “conserved” with “invariant”. They are two completely different qualities. “Conserved” means that the value does not change over time. “Invariant” means that different frames agree on the value.

Energy and momentum are conserved, but they are not invariant.

Mass is both conserved and invariant.

Proper acceleration is invariant but not conserved.

Velocity is neither conserved nor invariant.

Ibix
Science Advisor
And could you expand a bit on this by saying how that happens mathematically. Would it be the inner product being invariant?
Anything you can directly measure must be a Lorentz scalar, of which the inner product is an example, yes. But not the only one - for example a clock measures interval along its worldline.

If you could measure something that wasn't a scalar then frames could disagree about the results which would lead to paradoxes.

Dale
Mentor
And could you expand a bit on this by saying how that happens mathematically. Would it be the inner product being invariant?
Yes, the inner product of two vectors gives a scalar, and all scalars are invariant. All measurements produce scalars.

phyzguy
Science Advisor
Yes, the inner product of two vectors gives a scalar, and all scalars are invariant. All measurements produce scalars.
Why do you and Ibix say that measurements only produce scalars? I can measure energy, for example, which is not a scalar. It is frame dependent of course, but I can still measure it.

PAllen
Science Advisor
2019 Award
Why do you and Ibix say that measurements only produce scalars? I can measure energy, for example, which is not a scalar. It is frame dependent of course, but I can still measure it.
But the reason that an energy measurement is invariant is that it is a scalar. It is an inner product of e.g. a particle 4 momentum and an instrument 4 velocity. As a result, all observers agree on the outcome of this specific measurement

the reason that an energy measurement is invariant is that it is a scalar
Now that contradicts Orodruin's and Ibix's first posts where they say energy is not invariant.

PAllen
Science Advisor
2019 Award
Now that contradicts Orodruin's and Ibix's first posts where they say energy is not invariant.
No, it does not. In speaking of measurements being scalar invariants, one notes that the result of a measurement depends on the 4 velocity of the instrument. Two instruments in relative motion will get different results, but each result is invariant and computable as an inner product in any frame.

Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
Now that contradicts Orodruin's and Ibix's first posts where they say energy is not invariant.
It does not. A device set up to measure energy measures energy measures the energy in a particular frame (often implicitly assumed to be the rest frame of the device). This is a measurement and all observers must agree on what this measurement is. The Lorentz invariant way of writing the quantity that is measured is to take the inner product of the 4-momentum of what is being measured and the 4-velocity of the observer relative to whom you are measuring the energy.

phyzguy
Science Advisor
OK, I understand. If you think of a measurement producing a reading on a display, for example, any observer in any state of motion could read it, so they all must agree on what the display says. The energy measured in frame A (inner product of momentum being measured with 4-velocity of frame A) will be different from the energy measured in frame B (inner product of momentum being measured with 4-velocity of frame B), but everyone must agree on the results of these measurements.

Imager
Gold Member
You are confusing “conserved” with “invariant”. They are two completely different qualities. “Conserved” means that the value does not change over time. “Invariant” means that different frames agree on the value.

Energy and momentum are conserved, but they are not invariant.

Mass is both conserved and invariant.

Proper acceleration is invariant but not conserved.

Velocity is neither conserved nor invariant.

Do we give awards for light bulbs going on over people's heads? If so, you just earned at least three of them!

PeterDonis
Mentor
2019 Award
Now that contradicts Orodruin's and Ibix's first posts where they say energy is not invariant.
To amplify a bit on others' responses to this: suppose there are two observers, A and B, who are in relative motion. Each one carries an energy measuring device and each one uses their own device to measure the energy of the same particle.

"Energy measurements are scalars and are therefore invariant" means: A and B will both agree on the reading on A's device, and A and B will both agree on the reading on B's device, even though A and B are using different frames to calculate what that reading will be.

"Energy is not invariant" means: A's device and B's device will show different readings, because they are in relative motion.

Dale
Mentor
Why do you and Ibix say that measurements only produce scalars? I can measure energy, for example, which is not a scalar. It is frame dependent of course, but I can still measure it.
It seems that you already got it, but one more take might help. If observer A, in relative motion to observer B, measures the energy of something, then both A and B will agree on the number that A’s measuring device produces. However, B will disagree that the number produced is, in fact, a correct measurement of the energy. In B’s frame the energy is a different number and A’s device systematically fails to measure it correctly. (And vice versa, of course)

"Energy measurements are scalars and are therefore invariant" means: A and B will both agree on the reading on A's device, and A and B will both agree on the reading on B's device, even though A and B are using different frames to calculate what that reading will be.

"Energy is not invariant" means: A's device and B's device will show different readings, because they are in relative motion.
Hmmm I think I got it. Let's say A reads a different value as that of B, still A will agree on the reading of B, because he knows that upon a coordinate transformation from his coordinates to the B coordinates he will get the same reading as B?

Dale
Mentor
Hmmm I think I got it. Let's say A reads a different value as that of B, still A will agree on the reading of B, because he knows that upon a coordinate transformation from his coordinates to the B coordinates he will get the same reading as B?
Yes, that is a valid coordinate-based way to look at it.

It is also possible to look at it in a coordinate independent manner.

$E_{measured}=g_{\mu\nu}p^{\mu}u^{\nu}$

Where p is the measured object’s four momentum and u is the measuring device’s four velocity. This is a manifestly invariant quantity.

PeterDonis
Mentor
2019 Award
A will agree on the reading of B, because he knows that upon a coordinate transformation from his coordinates to the B coordinates he will get the same reading as B?
No. (Sorry if I seem to be contradicting @Dale, I think it's because he and I are intepreting what you say differently.)

"A coordinate transformation to the B coordinates" does not change A's state of motion. A can use coordinates in which B is at rest, input his own (A's) motion in those coordinates, and still calculate the same answer for his own reading. (So can B, of course.) He can also perfectly well use coordinates in which he (A) is at rest, input B's motion in those coordinates, and calculate the correct answer for B's reading. That's why both readings are invariants: they come out the same no matter what coordinates you use to calculate them.

If A changes his state of motion to be the same as B's, then he will get the same reading as B. But he can do that without changing the coordinates he uses. It's very important not to confuse the choice of coordinates with an actual change in state of motion.