What Does the 'E' in E=mc² Really Represent: Total Energy or Mass Energy?

[karanbir]

i'm really a beginner in this topic. So, I'm really confused even about this simple thing

an object has so many types of energies like energy due to mass in it, kinetic energy, pot. energy etc. So does the 'E' in E=mc2 give sum of all these energies or only the energy due to mass?

also when we consider the wave motion of any particle, is the energy of the wave equal to mc2?

 

[Pengwuino]

The energy is strictly the energy due to the mass or the rest energy. In that formulation, there is no kinetic energy or potential energy.

 

[karanbir]

please answer the second part also

 

[karanbir]

if we say E in E=mc2 is energy due to mass only, then where does the kinetic energy or pot. energy that the object had (before its mass was converted to energy) go?

 

[jartsa]

if we say E in E=mc2 is energy due to mass only, then where does the kinetic energy or pot. energy that the object had (before its mass was converted to energy) go?

Like a power plant on a hill, where does the potentential energy of its fuel go?

If it's a water boiling power plant, potential energy of the fuel turns into potential energy of heat energy.

It may be so that massless potential energy of the fuel turned into massless potential energy of the heat.

Or it may be so that potential energy of the fuel had mass, and potential energy of the heat energy has mass.

I leave the question whether potential energy has mass to wiser people.

 

[supernova1387]

i'm really a beginner in this topic. So, I'm really confused even about this simple thing

an object has so many types of energies like energy due to mass in it, kinetic energy, pot. energy etc. So does the 'E' in E=mc2 give sum of all these energies or only the energy due to mass?

also when we consider the wave motion of any particle, is the energy of the wave equal to mc2?

I like to point out that E=mc^2 is a very misleading formula and unfortunately even in many scientific documentaries , they just use it frequently without knowing what they say. The actual formula derived by Einstein was:

E=√[(mc^2)^2+(pc)^2] (1)

where p is the momentum of the particle relative to the base frame. If the 2 bodies have zero velocity with respect to each other, then p=0 and in this SPECIAL case you get E=mc^2. The kinetic energy is thus included in the second term. The potential energy is normally zero in special relativity because it is very small compare with the rest of the components.

Also I should point out that mass-energy equivalence is valid ONLY if the 2 bodies have zero velocity with respect to each other.

For your second question, if by wave you mean electromagnetic waves which are made of photons, then no. Because in case of a photon, m=0 and you get E=pc. But if the particle has some kind of mass, then you need to consider equation (1) in its general form.

For more info have a look at Fundamentals of Physics by David Halliday , special relativity chapter.

 

[karanbir]

thank you very much supernova1387. Your reply was really helpful.

 

[clamtrox]

I leave the question whether potential energy has mass to wiser people.

An uranium nucleus is heavier than the fission products created when it splits. Likewise, carbon and oxygen molecules weigh more than the carbon mono/dioxides produced in coal burning. Same holds true for a power plant on a hill, but then the thought experiments gets a bit involved, because it's the total mass of earth+power plant which is different.

 

[the_emi_guy]

An uranium nucleus is heavier than the fission products created when it splits.

The *rest mass* of a uranium nucleus is heavier than the *rest masses* of the fission products created when it splits.

The total relativistic mass is conserved.

 

[clamtrox]

The *rest mass* of a uranium nucleus is heavier than the *rest masses* of the fission products created when it splits.

The total relativistic mass is conserved.

If you insist calling a particles energy it's mass, then sure :D I usually call mass mass and energy energy, so there's no confusion.

 

[ZapperZ]

The *rest mass* of a uranium nucleus is heavier than the *rest masses* of the fission products created when it splits.

The total relativistic mass is conserved.

What "relativistic mass"? The "missing mass" is in the binding energy!

I definitely agree with clamtrox here. Mixing up those two, especially when someone isn't familiar with the concept, is causing nothing but confusion.

Zz.

 

[harrylin]

[..]The actual formula derived by Einstein was:

E=√[(mc^2)^2+(pc)^2] (1) [..]

Please show me, I can't find it:
http://www.fourmilab.ch/etexts/einstein/E_mc2/www/

However, I do think that your reply was very useful, as you provided a good equation for m= rest mass.

Moreover, indeed the formula as first derived was slightly different (and it was even just put in words):

" If a body gives off the energy L in the form of radiation, its mass diminishes by L/c² "

Δm = L/c² with L=emitted radiation energy

And that formula was derived for an atom at rest ("a stationary body").
The derivation does not explicitly exclude potential energy; however it only discusses kinetic energy, as it was directed at answering a specific question which was correctly answered as cited.

 

[DrStupid]

The total relativistic mass is conserved.

And so is the total rest mass.

 

[harrylin]

And so is the total rest mass.

That's wrong, see #11.

 

[Naty1]

While posts #2 and #6 are correct, there is an additional perspective that I found very helpful to understand.

Karanbir:

..an object has so many types of energies like energy due to mass in it, kinetic energy, pot. energy etc. So does the 'E' in E=mc2 give sum of all these energies or only the energy due to mass?...if we say E in E=mc2 is energy due to mass only, then where does the kinetic energy or pot. energy that the object had

As #6, explains, E = mc² applies when the momentum [p] [lateral movement] is zero. An equivalent way of looking at this is that the first part [mc² is the total energy in the frame of the matter particle while the second part is it's additional energy in the frame of the observer.

However, there are different types of energy in the frame of the particle. So for example, if we are considering a particle with structure, like an atom with orbital electrons, heating the atom increases the energy in the frame of the particle: the electrons move to more energetic orbitals, the structure now exhibits additional 'mass' due to this random kinetic energy. It has more 'rest mass' than a cold atom.

You can think of this as analogous to a coiled spring: In that situation energy is also contained within the structure, but there it is mostly potential energy; a coiled spring has infinitisimally more 'rest mass' than an uncoiled spring! [It also heats a bit when compressed.]

But you cannot 'heat' a fundamental matter particle, say an electron, because it has no 'structure' with which to absorb energy. No degrees of freedom like an atom. In an atom, it is the structure that absorbs heat energy, not the individual electrons themselves...they just bump up or down in orbitals of different energy levels, say as when reacting with chemical reactions. [Neutrons and protons do have structure...quarks.]

The binding energy mentioned in previous posts is also an analogous energy component, but in that case is an energy component of the nucleus rather the the orbital electrons.

 

[TrickyDicky]

I find some ambiguity in the term rest(invariant) mass, because if rest is considered to be relative in SR, how can it be invariant? I mean if rest mass is the same in all reference frames, then all observers are agreeing about the rest frame, aren't they?

Unfortunately mass (like energy) is not a well defined quantity in GR (momentum-energy is), but it is well defined in SR so it should be possible to clarify the above within SR.

 

[harrylin]

I find some ambiguity in the term rest(invariant) mass, because if rest is considered to be relative in SR, how can it be invariant? I mean if rest mass is the same in all reference frames, then all observers are agreeing about the rest frame, aren't they?

Unfortunately mass (like energy) is not a well defined quantity in GR (momentum-energy is), but it is well defined in SR so it should be possible to clarify the above within SR.

Rest mass simply means that the mass is determined with the particle or object "in rest", which distinguishes it from other definitions. Similar expressions are rest length or proper length, and proper time.

 

[jtbell]

For a specific object, the quantity $mc^2 = \sqrt{E^2 - (pc)^2}$ has the same value in all inertial reference frames, even though E and p vary from one frame to another. Therefore we say m is invariant across inertial reference frames (a.k.a. Lorentz invariant). In some particular frame the object has p = 0 (unless it is a photon or other "massless" particle which can never have p = 0). We say it is "at rest" in that frame, and in that frame mc² = E.

 

[TrickyDicky]

For a specific object, the quantity $mc^2 = \sqrt{E^2 - (pc)^2}$ has the same value in all inertial reference frames, even though E and p vary from one frame to another. Therefore we say m is invariant across inertial reference frames (a.k.a. Lorentz invariant). In some particular frame the object has p = 0 (unless it is a photon or other "massless" particle which can never have p = 0). We say it is "at rest" in that frame, and in that frame mc² = E.

But that frame is agreed by all inertial observers, right? I mean you just said it has the same value in all inertial reference frames. So do you mean we say is "at rest" from any frame, but it is not really "at rest"? Is it some idealized rest?

 

[harrylin]

But that frame is agreed by all inertial observers, right? I mean you just said it has the same value in all inertial reference frames. So do you mean we say is "at rest" from any frame, but it is not really "at rest"? Is it some idealized rest?

To use the topic as example: the "rest energy" of a particle is mc².
Now consider that you are discussing a "high energy" electron (as it's accelerated in a particle accelerator), and you want to know how many times more energy it has got. To answer that, one has to know the energy in the lab frame and the energy that it would have in rest - the rest energy. But of course, it's not at rest.

 

[TrickyDicky]

To use the topic as example: the "rest energy" of a particle is mc².
Now consider that you are discussing a "high energy" electron (as it's accelerated in a particle accelerator), and you want to know how many times more energy it has got. To answer that, one has to know the energy in the lab frame and the energy that it would have in rest - the rest energy. But of course, it's not at rest.

I don't mean that an elementary particle with invariant mass has necessarily to be at rest. That is silly. But at least seems to imply that there's a theoretical rest frame state for the particle. I guess you mean that frame doesn't really exist even if we have an invariant rest mass frame that is agreed by any observer (for instance in the case of the electron 0.51 MeV/c^2).

 

[Nugatory]

But that frame is agreed by all inertial observers, right? I mean you just said it has the same value in all inertial reference frames. So do you mean we say is "at rest" from any frame, but it is not really "at rest"? Is it some idealized rest?

The equation that always holds, in all inertial frames, is $mc^2 = \sqrt{E^2 - (pc)^2}$. The values of m and p are different in different frames, but they are always different in ways that make this equality hold in every frame.

There is one frame where p=0, namely the frame of an (imaginary) observer who is sitting on the back of the electron, riding it through space like a cowboy on a horse. We call that frame the "rest frame of the electron" and we call the mass of the electron in that frame the "rest mass". Aside from the fact that it is sometimes easier to do the math in that frame, there's nothing special or unique about it - it's just a frame in which the electron is at rest and everything else is moving at various speeds, as opposed to a frame in which some other object is at rest and everything else including the electron is moving at various speeds.

Often we choose to think about a problem in a whatever frame makes it easiest. For example, if you wanted to answer the question "How much mass does a coil spring gain when it is compressed", you'd certainly want to solve it in a frame in which the momentum p of the spring is zero - its rest frame. On the other hand, it's damn near impossible to make a free electron stand still in a lab, so if we find ourselves sitting in a lab somewhere contemplating an electron, we might find it easiest and most natural to measure its momentum p in our lab frame then proceed with our calculations in that frame.

 

[TrickyDicky]

There is one frame where p=0, namely the frame of an (imaginary) observer who is sitting on the back of the electron, riding it through space like a cowboy on a horse. We call that frame the "rest frame of the electron" and we call the mass of the electron in that frame the "rest mass". Aside from the fact that it is sometimes easier to do the math in that frame, there's nothing special or unique about it - it's just a frame in which the electron is at rest and everything else is moving at various speeds, as opposed to a frame in which some other object is at rest and everything else including the electron is moving at various speeds.

Well, you keep saying that there is really something special about that frame, namely that it is agreed by all observers that it is at rest (not only by the one sitting on its back, every state of motion considers itself to be at rest wrt itself but the rest of observers not necessarily agree), no other frame has that feature in SR. any other state of motion is not agreed by all observers regardless there own state of motion.

 

[Nugatory]

Well, you keep saying that there is really something special about that frame, namely that it is agreed by all observers that it is at rest (not only by the one sitting on its back, every state of motion considers itself to be at rest wrt itself but the rest of observers not necessarily agree), no other frame has that feature in SR. any other state of motion is not agreed by all observers regardless there own state of motion.

No other frame in SR has the property of being the rest frame for that electron... But every frame in SR has the property of being the one and only rest frame, agreed upon by all observers regardless of their own state of motion, for *something*... so that doesn't make this particular electron's rest frame especially "special" unless there is something special about this particular electron.

 

[TrickyDicky]

No other frame in SR has the property of being the rest frame for that electron... But every frame in SR has the property of being the one and only rest frame, agreed upon by all observers regardless of their own state of motion, for *something*... so that doesn't make this particular electron's rest frame especially "special" unless there is something special about this particular electron.

True, I'm not asking about the frame of a certain particle properties.
It is hard to explain, I'm referring to the concept that the quantity is invariant regardless the state of motion.
Take momentum instead of energy, there is no invariant momentum. So it might sound somewhat confusing that there is an energy invariant when in SR what is conserved is the momentum-energy.

 

[harrylin]

True, I'm not asking about the frame of a certain particle properties.
It is hard to explain, I'm referring to the concept that the quantity is invariant regardless the state of motion.
Take momentum instead of energy, there is no invariant momentum. So it might sound somewhat confusing that there is an energy invariant when in SR what is conserved is the momentum-energy.

If we define invariant momentum like invariant mass at v=0 then we get that invariant momentum - the momentum at zero speed - is 0. I suppose that that is why there is no invariant momentum...

 

[TrickyDicky]

If we define invariant momentum like invariant mass at v=0 then we get that invariant momentum - the momentum at zero speed - is 0. I suppose that that is why there is no invariant momentum...

So you are saying v=0 (therefore p=0) can be defined and agreed by all observers?

 

[harrylin]

So you are saying v=0 (therefore p=0) can be defined and agreed by all observers?

The properties of an object as determined with a reference system relative to which it is in rest are agreed by all observers - and yes, to be precise: zero speed relative to a certain object is agreed by all inertial reference systems.

Thus the group of reference systems relative to which the object has zero speed is called its "rest frame", and its energy relative to that "frame" is called its "rest energy" m₀c²

As a matter of fact, an "inertial frame" is an imaginary or real system with rulers and clocks that are in rest relative to each other - according to that system as well as according to systems in uniform motion relative to it. If this was not agreed by all inertial systems, then no consistent transformations would be possible I'm afraid.

 

[DrStupid]

That's wrong

Only if E and p are not conserved or E²/c² = m²·c² + p² is wrong.

 

[harrylin]

Only if E and p are not conserved or E²/c² = m²·c² + p² is wrong.

I wrote "see #11": there is "missing" rest mass
- http://en.wikipedia.org/wiki/Binding_energy

As a matter of fact, rest mass diminishes with L/c² as the first derivation already showed which I cited (post #12, just above your post!).

Nevertheless, while the extended equation includes radiation, it doesn't include potential energy - it's incomplete.

 

[DrStupid]

As a matter of fact, rest mass diminishes with L/c² as the first derivation already showed which I cited (post #12, just above your post!).

As a mater of fact energy diminishes with L. But that does not mean that energy is not conserved. The total energy of an isolated system is constant and so is the total rest mass of the system.

 

[DrGreg]

The total energy of an isolated system is constant and so is the total rest mass of the system.

That depends what you mean by "total rest mass".

• the sum of the individual rest masses
• the invariant or system mass of the whole system

These are not the same. If you mean (b), you are right. If you mean (a), you are wrong.

 

[DrStupid]

That depends what you mean by "total rest mass".

• the sum of the individual rest masses
• the invariant or system mass of the whole system

Of course I mean (b). (a) is just a number without any physical relevance because it can be changed by choosing different theoretical fragmentations of the same system into sub-systems and the properties of a real system must not depend on its theoretical description.

 

[DrGreg]

That depends what you mean by "total rest mass".

• the sum of the individual rest masses
• the invariant or system mass of the whole system

Of course I mean (b). (a) is just a number without any physical relevance because it can be changed by choosing different theoretical fragmentations of the same system into sub-systems and the properties of a real system must not depend on its theoretical description.

Good. The problem is that most people equate the word "total" with "sum" and would go for option (a), which is why you've had people disagreeing with you in this thread.

 

[TrickyDicky]

I find some ambiguity in the term rest(invariant) mass, because if rest is considered to be relative in SR, how can it be invariant? I mean if rest mass is the same in all reference frames, then all observers are agreeing about the rest frame, aren't they?

Unfortunately mass (like energy) is not a well defined quantity in GR (momentum-energy is), but it is well defined in SR so it should be possible to clarify the above within SR.

Ok, so Ithink the easiest way to clarify this is simply to acknowledge that in SR (unlike in reality) there are truly inertial frames, so rest point masses are unambiguously defined and invariant. And as previously commented by other posters there is nothing special with these rest frames within SR as any point mass can be chosen as the rest frame.
I wonder how this translates to more realistic scenarios where there are no truly inertial frames.
More specifically, shouldn't the equation E=mc2 even though derived from SR, really be more of a GR formula? I guess it can't be given that neither mass nor energy are well defined in GR.
But the situation is a bit paradoxical: invariant mass and energy are well defined within SR and therefore E=mc2 considered an SR formula, but at the same time in more physical terms we accept the equivalence of inertial and gravitational mass and therefore strictly speaking mass is outside the range of SR in as much as gravitation cannot be explained within SR.

 

[the_emi_guy]

If you insist calling a particles energy it's mass, then sure :D I usually call mass mass and energy energy, so there's no confusion.

What "relativistic mass"? The "missing mass" is in the binding energy!

I definitely agree with clamtrox here. Mixing up those two, especially when someone isn't familiar with the concept, is causing nothing but confusion.

Zz.

Are you are saying that the rest mass of an atom should exclude the binding energy (since it is energy and not mass)? Should we also exclude the binding energy of the gluon field that holds its constituent quarks together? Or should we arbitrarily consider that gluon field to really be a gluon particle so that we can include its mass (which turns out to be the lion's share of the quark mass)?.

What is rest mass? As Naty1 pointed out, rest mass includes numerous energy components, both kinetic and potential. 1Kg of uranium consists of the rest mass of quarks, qluon field energy, atomic binding energy, thermal kinetic energy etc.

All of these mass-energy components make up its 1Kg rest mass.

Now it seems odd to me that if I now set this complex 1Kg package of mass-energy in motion, that I am obligated to record the increase in energy in Joules, not Kg, because it happens to be motion.

And, yet, if I place this moving 1Kg into a larger, stationary, container, its kinetic energy is now allowed to be converted to Kg and counts toward rest mass of the larger container.

I know that teaching relativistic mass has gone out of favor, and for some valid reasons, but I think that there is an unintended consequence.

Students should not get the impression that mass and energy are fundamentally different things or that e=mc^2 is a rule that tells how much of one can be converted to the other.

Energy can neither be created nor destroyed, and energy, in all of its forms, has mass.
Mass also can neither be created nor destroyed, and in all of its forms, has energy.
Mass and energy are two names for the same thing, and neither one is changed nor transformed into the other.
Rather than mass being changed into energy, the view of special relativity is that rest mass has been changed to a more mobile form of mass, but remains mass. In the transformation process, neither the amount of mass nor the amount of energy changes, since both are properties which are connected to each other via a simple constant.

 

[harrylin]

Good. The problem is that most people equate the word "total" with "sum" and would go for option (a), which is why you've had people disagreeing with you in this thread.

Exactly. Thanks for figuring that one out.

 

[DrStupid]

Now it seems odd to me that if I now set this complex 1Kg package of mass-energy in motion, that I am obligated to record the increase in energy in Joules, not Kg, because it happens to be motion.

It's just a convention. Mass as given by definition 2 and lex 1 to 3 of Newton's Principia includes all kinds of energy if Lorentz transformation applies. But this definition of mass is outdated. Only rest mass is used in relativity today.

And, yet, if I place this moving 1Kg into a larger, stationary, container, its kinetic energy is now allowed to be converted to Kg and counts toward rest mass of the larger container.

No it isn't allowed - not generally. It depends on the momentum of the system. If you place a photon into an empty container the mass of the container would be zero. If you add another photon with the same energy but opposite momentum than the mass of the container is equivalent to the total energy of both photons. The velocity of the container does not matter.

Students should not get the impression that mass and energy are fundamentally different things or that e=mc^2 is a rule that tells how much of one can be converted to the other.

Of course not.

Energy can neither be created nor destroyed

That's true.

and energy, in all of its forms, has mass.

It would be true for the classical definition of mass (aka as relativistic mass) but with the current meaning of mass (rest mass) it is wrong.

Mass also can neither be created nor destroyed

That's true but you have to keep in mind that rest mass is not additive (see discussion above).

and in all of its forms, has energy.

That's true.

Mass and energy are two names for the same thing

See above. It depends on the definition of mass.

Rather than mass being changed into energy, the view of special relativity is that rest mass has been changed to a more mobile form of mass, but remains mass. In the transformation process, neither the amount of mass nor the amount of energy changes, since both are properties which are connected to each other via a simple constant.

It's easier: If a system at rest looses energy than it looses an equivalent amount of mass too and visa versa. Maybe I missed a corresponding original source but the "conversion of mass into energy" seems to be a concept of popular science only.

 

[harrylin]

[..] Mass and energy are two names for the same thing, and neither one is changed nor transformed into the other. [..]

I would say that Einstein's formulation* in his first paper on this is very accurate, and therefore I disagree with the first of your two statements.

*"The mass of a body is a measure of its energy-content"

It's just a convention. [...] If you place a photon into an empty container the mass of the container would be zero. [..]

Yes, it's just a convention and for consistency with length and time I prefer the "outdated" one; that's a matter of taste. But I think that the example doesn't work: a photon is measured as moving at v=c whatever box you choose it to be in, using any standard reference system. The system mass of such a (mass-less) container includes the relativistic mass of the light and is thus not zero (the common example is a box with mirrors; the mass doesn't switch to zero when the photon happens to be in-between the walls). However, this could be another free choice of definition.

 

[DrStupid]

I prefer the "outdated" one

Me too but to avoid confusions I accustomed myself to indicate when I use the word mass in this (currently) unusual sense. If I talk about mass without such an explanation than I mean rest mass (e.g. in my statement about photons in a virtual box).

 

[harrylin]

Me too but to avoid confusions I accustomed myself to indicate when I use the word mass in this (currently) unusual sense. If I talk about mass without such an explanation than I mean rest mass (e.g. in my statement about photons in a virtual box).

Yes, that's how I understood your statement. The relativistic mass of a box in rest equals the rest mass of a box in rest. BTW I found an account of a real box with photon:
http://www.abc.net.au/science/articles/2007/03/15/1872353.htm

 

[DrStupid]

The relativistic mass of a box in rest equals the rest mass of a box in rest.

That depends what you mean by "in rest". If you mean that the walls of the box are stationary than you are wrong. If you mean that the box has no momentum than you are right but a virtual box containing one photon has momentum in every frame of reference.

 

[stevendaryl]

I wonder how this translates to more realistic scenarios where there are no truly inertial frames.
More specifically, shouldn't the equation E=mc2 even though derived from SR, really be more of a GR formula? I guess it can't be given that neither mass nor energy are well defined in GR.

The SR equation E² = p²c² + m²c⁴ has an analogy in GR:

g_αβ P^α P^β = m² c²

where P^α is the 4-momentum, which is defined in GR, as well as SR.

 

[harrylin]

That depends what you mean by "in rest". If you mean that the walls of the box are stationary than you are wrong. If you mean that the box has no momentum than you are right but a virtual box containing one photon has momentum in every frame of reference.

It seems that we are actually disagreeing on this.
As a photon has no rest mass (it's an impossibility!), I assumed that you meant the mass of a system in which the box is in rest - which is also the time-averaged rest mass of the system if the box has perfect mirrors. Compare:
- http://www.weburbia.com/physics/light_mass.html
- http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html (note: I was slightly involved with that)

It's conceptually similar to a box with loss-less bouncing tennis balls inside: the time-averaged "rest mass" of the system is the sum of E/c² of the parts.

 

[DrStupid]

I assumed that you meant the mass of a system in which the box is in rest

See above: What do you mean with "in rest"? I mean "with no momentum".

 

[harrylin]

See above: What do you mean with "in rest"? I mean "with no momentum".

A virtual mass-less container cannot have momentum at any condition, incl. "not in rest"; and a photon cannot have "no momentum" and cannot be "in rest". In such an example "in rest" can only refer to the displacement of the container itself; expressed in Lorentz coordinates its walls have that dx/dt=0.

 

[DrStupid]

A virtual mass-less container cannot have momentum

Do you really want to talk about the mass of a mass-less container? This discussion makes sense for the container including its content only and that's what I am talking about.

 

[harrylin]

Do you really want to talk about the mass of a mass-less container? This discussion makes sense for the container including its content only and that's what I am talking about.

It appears that you did mean a system with a photon in rest which is impossible in SR, as I already mentioned in post #44. As we didn't advance from there and it has little to do with the topic, I'll leave it at that.

 

[DrStupid]

It appears that you did mean a system with a photon

No, I don't. I am talking about a virtual box containing one or more photons. As I told you in #42 a box containing a single photon has momentum in every frame of reference (due to the momentum of the photon). Thus it has no rest frame and therefore no rest mass. But it has relativistic mass (due to the energy of the photon).