Do the Energy-Momentum Transformations apply to Photons?

[tade]

I apologize in advance for this dumb question, I think I know the answer but I just want to be sure.A photon has energy E = pc = hf

Do the Energy-Momentum transformations:

apply exactly to photons?

Or must we introduce certain corrective terms? Let's say all this takes place in free space.

 

[Andrew Mason]

I apologize in advance for this dumb question, I think I know the answer but I just want to be sure.A photon has energy E = pc = hf

Do the Energy-Momentum transformations:

apply exactly to photons?

Or must we introduce certain corrective terms? Let's say all this takes place in free space.

The speed of the photon, of course is c in all frames but the wavelength (which determine its p and E) will depend on motion of the observer relative to the source. For that you need to apply the relativistic. doppler factor.

AM

 

[tade]

The speed of the photon, of course is c in all frames but the wavelength (which determine its p and E) will depend on motion of the observer relative to the source. For that you need to apply the relativistic. doppler factor.

AM

Can we also apply the Energy-Momentum transformations without introducing corrective terms?

 

[Ibix]

The energy and momentum transformations come from the requirement that the energy-momentum four-vector has an invariant length in all inertial frames. So yes, they must apply to all things including photons.

Why do you ask?

 

[tade]

The energy and momentum transformations come from the requirement that the energy-momentum four-vector has an invariant length in all inertial frames. So yes, they must apply to all things including photons.

Why do you ask?

I just want to be sure that we can apply them exactly to photons like they are applied to massive particles, without the need for any corrective terms.

 

[PeroK]

I apologize in advance for this dumb question, I think I know the answer but I just want to be sure.A photon has energy E = pc = hf

Do the Energy-Momentum transformations:

apply exactly to photons?

Or must we introduce certain corrective terms? Let's say all this takes place in free space.

Yes, in fact the energy-momentum of a photon is an eigenvector of the Lorentz Transformation; and, the Doppler red and blue-shift factors are the eigenvalues!

 

[tade]

Yes, in fact the energy-momentum of a photon is an eigenvector of the Lorentz Transformation; and, the Doppler red and blue-shift factors are the eigenvalues!

Do these apply exactly to photons without the need for any additional corrective terms?

 

[PeroK]

Do these apply exactly to photons without the need for any additional corrective terms?

I've no idea what corrective terms you may be talking about. The Lorentz Transformation applies to both particles and light rays traveling at $c$. One of the first things you should do with the LT is check that it transforms a light ray traveling at $c$ into a light ray traveling at $c$ in the new frame; likewise, that it transforms the energy-momentum for a photon into the energy-momentum for a photon.

 

[PeterDonis]

Do these apply exactly to photons without the need for any additional corrective terms?

Here's a way to check: in the original frame (which is the primed frame the way you have the equations written), you have $E' = p' c$. Plug that into the two transformation equations and verify that you obtain $E = p c$.

 

[Andrew Mason]

If you are using the Lorentz transformations for the photon, where v=c, γ would be undefined. In any event, the wavelength does not depend on the speed of the photon. It depends on thd speed of the source. So where does the speed of the source appear in the Lorentz transformations?

AM

 

[PeroK]

If you are using the Lorentz transformations for the photon, where v=c, γ would be undefined. In any event, the wavelength does not depend on the speed of the photon. It depends on thd speed of the source. So where does the speed of the source appear in the Lorentz transformations?

AM

In the Lorentz Transformation $v$ and $\gamma$ relate to the relative velocity between the reference frames.

 

[Andrew Mason]

Exactly! v is the relative speed of two different inertial reference frames. But, since a photon does not have or define an inertial reference frame the Lorentz transformations, and the energy-momentum transformations derived from them, do not apply to a photon.

AM

 

[PeroK]

Exactly! v is the relative speed of two different inertial reference frames. But, since a photon does not have or define an inertial reference frame the Lorentz transformations, and the energy-momentum transformations derived from them, do not apply to a photon.

AM

That's an unorthodox point of view!

Neither reference frame needs to coincide with the rest frame of any particle under consideration.

 

[Andrew Mason]

That's an unorthodox point of view!

I am not sure what you mean. How do the transformations apply to a photon? Can you show us?

AM

 

[PeroK]

I am not sure what you mean. How do the transformations apply to a photon? Can you show us?

AM

I'll let Michael Fowler from the University of Virginia do the work. The last section on this page "Photon Energies in Different Frames" shows how to derive the relativistic Doppler shift formula from the E-M transformation for a photon.

http://galileo.phys.virginia.edu/classes/252/energy_p_reln.html

 

[PeterDonis]

If you are using the Lorentz transformations for the photon, where v=c

He's not. He's just showing that the energy and momentum for a photon transform appropriately when you change frames, using an ordinary Lorentz transformation where $v < c$. See below.

the Lorentz transformations, and the energy-momentum transformations derived from them, do not apply to a photon.

Sure they do; they preserve the property $E = pc$ for a photon in every frame. What's more, by plugging in $E = pc$ in the transformation equations, you can see the Doppler factor pop out.

 

[Dale]

Do the Energy-Momentum transformations:

apply exactly to photons?

Yes. Note that $v$ is the velocity between two inertial frames so $v< c$

Can we also apply the Energy-Momentum transformations without introducing corrective terms?

Yes.

I just want to be sure that we can apply them exactly to photons like they are applied to massive particles, without the need for any corrective terms.

Yes.

Do these apply exactly to photons without the need for any additional corrective terms?

Yes.

Note, pulses of light do not have a rest frame, but the energy-momentum of a pulse of light in one inertial frame is related to the energy-momentum in another inertial frame through the transformation equations posted.

 

[Dale]

Exactly! v is the relative speed of two different inertial reference frames. But, since a photon does not have or define an inertial reference frame the Lorentz transformations, and the energy-momentum transformations derived from them, do not apply to a photon.

Consider a person throwing a ball on a train. You can express the momentum of the ball in the frame of the train or in the frame of the embankment. It is moving in both frames, but its energy-momentum is perfectly valid. There is no need to consider the ball's frame.

Similarly with light. If a person is shining a light on a train then you can express the energy-momentum of the light in the train frame or in the embankment frame. The energy-momentum transformation is the same for the ball and the light. The fact that the light does not have a frame where it is at rest does not mean that the energy-momentum transformation equation is invalid or inapplicable between other valid inertial frames.

 

[Andrew Mason]

Consider a person throwing a ball on a train. You can express the momentum of the ball in the frame of the train or in the frame of the embankment. It is moving in both frames, but its energy-momentum is perfectly valid. There is no need to consider the ball's frame.

Similarly with light. If a person is shining a light on a train then you can express the energy-momentum of the light in the train frame or in the embankment frame. The energy-momentum transformation is the same for the ball and the light. The fact that the light does not have a frame where it is at rest does not mean that the energy-momentum transformation equation is invalid or inapplicable between other valid inertial frames.

I must apologize to everyone for the confusion that I may have caused by my last couple of posts. You are right - for some reason I was thinking about a transformation to the rest frame of the moving particle. Of course, and as you point out, this would not have much use since in the rest frame p is always 0 and E is just the rest energy, $m_0c^2$

The Lorentz and Energy-momentum transformations can be used to compare the frequencies/wavelengths of the photon as measured in two different inertial frames moving at speed v relative to each other. While the transformations may be used to calculate the energy and momentum of a particle with mass measured by an observer traveling at speed v relative to the rest frame of the particle, they cannot be used that way with a photon (i.e. by letting v = c), since a photon does not have or define an inertial reference frame.

AM

corrected and added to - see post below.

 

[PeterDonis]

Of course, and as you point out, this would not have much use since in the rest frame E and p are necessarily always 0.

If you mean a photon, there is no such thing as the rest frame of a photon. A photon always moves at $c$; there is no frame in which it is at rest.

If you mean a particle with nonzero rest mass $m$, then in its rest frame, $p = 0$ but $E = m$; the energy is not zero.

 

[Andrew Mason]

If you mean a photon, there is no such thing as the rest frame of a photon. A photon always moves at $c$; there is no frame in which it is at rest.

If you mean a particle with nonzero rest mass $m$, then in its rest frame, $p = 0$ but $E = m$; the energy is not zero.

I think I will quit posting on major holidays...I should have said that $E=m_0(c^2)$ in the rest frame, and as I pointed out in an earlier post, photons do not define an inertial frame.

AM

 

[tade]

Yes. Note that $v$ is the velocity between two inertial frames so $v< c$Yes. Yes.Yes.

Note, pulses of light do not have a rest frame, but the energy-momentum of a pulse of light in one inertial frame is related to the energy-momentum in another inertial frame through the transformation equations posted.

Thank you Dale. You answered all of the one question that I have hahaa

 

[sweet springs]

Two photons created by electron positron annihilation define rest frame. This is an example of a group of photons that defines a rest frame.

 

[jtbell]

Two photons created by electron positron annihilation define rest frame.

This is true not only in e-p annihilation. For any system of two or more photons, with momenta in different directions, there is a frame in which their total momentum is zero. You can consider this to be the "rest frame" of the system.

 

[Dale]

Two photons created by electron positron annihilation define rest frame. This is an example of a group of photons that defines a rest frame.

It is usually called a "center of momentum frame" rather than a rest frame. The photons are not at rest in any frame.

 

[sweet springs]

In that sense "rest frame" should be applicable for a single particle or a rigid body, not for a group of particles or compound materials, e.g. the floating boat on a river flow is not in the "rest frame".

 

[vanhees71]

Yes, in fact the energy-momentum of a photon is an eigenvector of the Lorentz Transformation; and, the Doppler red and blue-shift factors are the eigenvalues!

That's only true if the boost (it's not a general Lorentz transform being discussed here but a boost) is in the direction of the photon's momentum. Indeed, a boost has two space-like and two light-like eigenvectors. For simplicity take a boost in $x$ direction as mentioned in #1. It's representing matrix is given by
$$\Lambda=\begin{pmatrix} \cosh \eta & -\sinh \eta &0 &0 \\ -\sinh \eta & \cosh \eta &0 &0 \\ 0 & 0 & 1 & 0 \\ 0&0&0&1 \end{pmatrix}.$$
Obviously the two space-like vectors $(0,0,1,0)$ and $0,0,1,0$ are eigenvectors with eigenvalue 1 of this matrix.

The other two must be in the $x^0x^1$ plane, because the eigenvalues must be Minkowski-orthogonal to the already found eigenvectors. It's easy to see that these are the two light-like vectors
$$a^{\pm}=\begin{pmatrix}1 \\ \pm 1 \\0 \\0 \end{pmatrix}.$$
That corresponds to photons, one running in the positive the other in the negative $x^1$ direction. The eigenvalues found by
$$\Lambda a^+=(\cosh \eta - \sinh \eta) =\exp (-\eta) a^+, \quad \Lambda a^-=(\cosh \eta + \sinh \eta) a^-=\exp \eta a^-.$$
These are indeed the Doppler factors for the longitudinal Doppler effect, because we have
$$\beta=\frac{v}{c} = \tanh \eta$$
and thus
$$\exp(\mp \eta)=\frac{\cosh \eta \mp \sinh \eta}{\sqrt{\cosh^2 \eta - \sinh^2 \eta}} = \frac{1 \mp \beta}{\sqrt{1-\beta^2}} = \sqrt{\frac{(1 \mp \beta)^2}{(1+\beta)(1-\beta)}} = \sqrt{\frac{1 \mp \beta}{1\pm \beta}}.$$
So if the boost is in direction of the photon (upper sign for $\beta>0$) we have a red shift (energy in the new frame larger than in the original), otherwise a blue shift.

In the general case, i.e., when the boost is not in direction of the photon's motion, you have a Doppler shift and in addition an aberration of the photon's direction!

 

[Dale]

In that sense "rest frame" should be applicable for a single particle or a rigid body

Yes.

e.g. the floating boat on a river flow is not in the "rest frame".

If your boat is moving rigidly then it would have a rest frame. The river flow would clearly not be rigid.

 

[sweet springs]

The theory of relativity excludes rigid bodies. Proton, which is made of quarks and gluons, is not a particle. In these strict sense rest frame could be applicable for a few kind of single elementary particles, e.g. mass of proton could not be observed because proton, as complex of particles, cannot be at rest.

 

[vanhees71]

As I said before, the mass of an object is defined by $m^2 c^2=p_{\mu} p^{\mu}$, where $p^{\mu}$ is the total energy-momentum four vector of the object. A proton has a very well defined mass, because it's a stable hadron (as far as we know).

 

[Dale]

In these strict sense rest frame could be applicable for a few kind of single elementary particles

Which is precisely why the term "center of momentum frame" is so common in the particle physics literature.

This convention is not universally used, but it is more descriptive and is appealing to many scientists.

 

[vanhees71]

Indeed the term "center of momentum frame" is the most precise expression for "rest frame", if not the only correct one at all, of a composite system.

 

[PeterDonis]

The theory of relativity excludes rigid bodies.

More precisely, it excludes bodies in which internal forces are propagated faster than light. But it is perfectly possible to have rigid motions of bodies, under appropriate conditions.