If light has no mass, how can it have energy according to e=mc2?

[thestien]

yes i realize that maybe i have been trival in by explanation of what i mean.
the equation e=mc^2 means that you could never have enough energy to move at the speed of light(beacuse as u add more fuel to get more energy the mass would increase).
ok use this as a comparison on Earth to launch a rocket into space we need a huge amount of fuel which inreases the rockets weight.
on the moon it would take less fuel to launch the rocket as the total weight of the rocket would be reduced.
so if the rocket was already in space i would assume that getting it to the speed that can escape the pull of Earth's gravity would take less fuel?
so if space is friction free then to get a rocket to a constant speed would only take one volume of fuel so say it may take 1 gallon of fuel to get the rocket to 10mph but then the rocket would continue at 10mph forever or until it was affected by some gravity or somthing.
so my question is why in space would it take more than the same gallon of fuel to push a bigger rocket to 10mph?
you may not see where my logic comes from but I am basing this on dropping things in a vacuum which fall at the same rate and space being a vacuum i thought that a rocket or a whole fleet of rockets would move at the same speed(or fall at the same speed) maybe i still haven't explained what i mean exactly but maybe you understand what i mean?

 

[thestien]

ok if i was in space could i push a rocket and make it move the same as if i pushed an apple for instance?

 

[Doc Al]

yes i realize that maybe i have been trival in by explanation of what i mean.
the equation e=mc^2 means that you could never have enough energy to move at the speed of light(beacuse as u add more fuel to get more energy the mass would increase).

That's not quite right. To accelerate anything (with mass) to the speed of light requires an infinite amount of energy.

ok use this as a comparison on Earth to launch a rocket into space we need a huge amount of fuel which inreases the rockets weight.
on the moon it would take less fuel to launch the rocket as the total weight of the rocket would be reduced.

True, but irrelevant.

so if the rocket was already in space i would assume that getting it to the speed that can escape the pull of Earth's gravity would take less fuel?

If it's already "in space", it has already escaped Earth's gravity.

so if space is friction free then to get a rocket to a constant speed would only take one volume of fuel so say it may take 1 gallon of fuel to get the rocket to 10mph but then the rocket would continue at 10mph forever or until it was affected by some gravity or somthing.

OK.

so my question is why in space would it take more than the same gallon of fuel to push a bigger rocket to 10mph?

Because the rocket is bigger! A 10 ton rocket moving at 10mph has 10 times the energy of a 1 ton rocket moving at the same speed.

you may not see where my logic comes from but I am basing this on dropping things in a vacuum which fall at the same rate and space being a vacuum i thought that a rocket or a whole fleet of rockets would move at the same speed(or fall at the same speed) maybe i still haven't explained what i mean exactly but maybe you understand what i mean?

To the extent that I understand your point, I believe you are mistaken.

ok if i was in space could i push a rocket and make it move the same as if i pushed an apple for instance?

You'd have to push the rocket a lot harder to give it the same acceleration as the apple.

 

[lightarrow]

[...]
you may not see where my logic comes from but I am basing this on dropping things in a vacuum which fall at the same rate and space being a vacuum i thought that a rocket or a whole fleet of rockets would move at the same speed(or fall at the same speed) maybe i still haven't explained what i mean exactly but maybe you understand what i mean?

Without air friction, objects falls on Earth with the same acceleration because gravitational force is proportional to their mass; you *don't* apply the same force to objects with different mass, but a *twofold* force to a twofold massive object!

In the case of a rocket in space it means that to give it the same acceleration of a lighter rocket, you will have to give it a greater force, because it has a greater mass!

Always remember F = ma.
In the *special* case of gravitational force, F = km, where k is a constant. So:
km = ma. You semplify:
k = a.
So acceleration is a constant.

 

[Xeinstein]

The e in e=mc^2 is the "rest energy". The total energy of a particle is the rest energy plus the kinetic energy. So your observation about the equation implies that a massless particle does not have any energy at rest (or in fact at any speed < c) or equivalently that all of a photon's energy is kinetic energy.

Did Einstein say the e in e=mc^2 is the "rest energy"?
I don't think so.
It includes all sorts of energy, i.e. energy of heat

 

[lightarrow]

Did Einstein say the e in e=mc^2 is the "rest energy"?
I don't think so.
It includes all sorts of energy, i.e. energy of heat

That equation is valid only if the object with mass m is stationary and I'm sure Einstein specified it in his writings.

 

[rbj]

That equation is valid only if the object with mass m is stationary and I'm sure Einstein specified it in his writings.

it's not how Einstein originally portrayed it but

E = m c^2

is also valid when E is taken to be the object's total energy (rest energy + kinetic energy) and the m is the relativistic mass.

 

[lightarrow]

it's not how Einstein originally portrayed it but

E = m c^2

is also valid when E is taken to be the object's total energy (rest energy + kinetic energy) and the m is the relativistic mass.

I know that you are an advocate of relativistic mass and that you don't want to change idea about it, however Xeinstein have to know which is the most accepted version in physics.

 

[rbj]

I know that you are an advocate of relativistic mass and that you don't want to change idea about it, however Xeinstein have to know which is the most accepted version in physics.

but you said it doesn't work, unless "m" is the rest mass. but it does work if m is the relativistic mass and E is the sum of rest energy (which is invariant) and the kinetic energy (which depends on who the observer is). whether or not "relativistic mass" really exists, one can take the expression for momentum that we all agree on, from

E^2 = E_0^2 + (p c)^2

E_0 = m_0 c^2

or

p = \frac{1}{c} \sqrt{E^2 - E_0^2}

take that value of momentum, divide by the speed of the object (relative to the same observer)

m = \frac{p}{v} = \frac{1}{c v} \sqrt{E^2 - E_0^2}

and you will get a quantity of dimension mass. turns out that when you simplify, you get

m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}

whether you say that such a mass exists physically or not is immaterial, this is a derived mathematical quantity (having a physical dimension of mass). if you plug that derived quantity of mass into

E = m c^2

then you'll have the total energy (rest energy plus kinetic energy) as observed by an observer in a different frame of reference. E=mc² does have validity where the mass is something other than the rest mass.

 

[basePARTICLE]

On earth, weight is proportional to mass but they are not the same thing. Mass doesn't change when you move into space.

Of course. To accelerate something with more mass requires more energy.

Even if there were no resistive forces acting on the car, it takes energy to accelerate it.

Even with no friction or drag forces in space, it still requires energy to accelerate something.
You do realize that when you change something's speed you are giving it energy?

Excuse the philosophical interpretation, but I thought change in this context, could also imply, being slowed down? Take for example a mass, M moving at constant speed, s at time t1, but while preparing for acceleration at time t2, a monkey, from the Maxwellian era, suddenly drops on itz back.

 

[Doc Al]

Excuse the philosophical interpretation, but I thought change in this context, could also imply, being slowed down?

In the context of the original question, the speed was increasing. But it's certainly true that acceleration can be negative as well as positive.

Take for example a mass, M moving at constant speed, s at time t1, but while preparing for acceleration at time t2, a monkey, from the Maxwellian era, suddenly drops on itz back.

 

[lightarrow]

but you said it doesn't work, unless "m" is the rest mass.

Not exactly. I wrote that the equation E = mc^2 is valid only if the body is at rest and not "unless "m" is the rest mass"; it's not the same thing. Infact, *since* m is the rest mass and E^2 = (cp)^2 + (mc^2)^2, you have that equation only when p = 0.

 

[rbj]

Not exactly. I wrote that the equation E = mc^2 is valid only if the body is at rest and not "unless "m" is the rest mass"; it's not the same thing.

curious on how the difference is salient. the "m" you've been referring to is the rest mass, the "m" I've been referring to is the relativistic mass (like the old textbooks, i use "m₀" for rest mass). so, using this corrected semantic, i would again counter and say that the equation E=mc² is valid for bodies not at rest if E is the total energy of the body (the energy of the body as observed by someone possibly in another reference frame) and m is the relativistic mass of the body (the inertial mass of the body as observed by that same observer in the other reference frame). the energy of the body in that body's reference frame is E₀, the rest energy, and the mass of the body in that body's own reference is m₀, the rest mass. they are (and can only be) related in the same way: E₀=m₀c². the difference between the energy a body has when in motion (relative to some reference frame) and the energy it has when not in motion (relative to the same reference frame) is the kinetic energy.

interpreted this way E=mc² is not only valid for bodies at rest. the point stands.

Infact, *since* m is the rest mass and E^2 = (cp)^2 + (mc^2)^2, you have that equation only when p = 0.

no disagreement here when it is clear what is meant by the symbols (and, I'm making a substitute to make sure there is no ambiguity). the "E" you have in the equation E²=(m₀c²)² + (p c)² is the same E i have in E=mc², but not the same E that you have in E=mc² .

 

[Dale]

Did Einstein say the e in e=mc^2 is the "rest energy"?
I don't think so.
It includes all sorts of energy, i.e. energy of heat

To be honest I don't know if the "E=mc^2 is rest energy" idea follows Einstein's approach or if it is a more modern convention. However, thermal energy is part of rest energy. It is energy that exists in the center of momentum frame.

 

[lightarrow]

Not exactly. I wrote that the equation E = mc^2 is valid only if the body is at rest and not "unless "m" is the rest mass"; it's not the same thing.

curious on how the difference is salient. the "m" you've been referring to is the rest mass, the "m" I've been referring to is the relativistic mass (like the old textbooks, i use "m₀" for rest mass). so, using this corrected semantic, i would again counter and say that the equation E=mc² is valid for bodies not at rest if E is the total energy of the body (the energy of the body as observed by someone possibly in another reference frame) and m is the relativistic mass of the body (the inertial mass of the body as observed by that same observer in the other reference frame). the energy of the body in that body's reference frame is E₀, the rest energy, and the mass of the body in that body's own reference is m₀, the rest mass. they are (and can only be) related in the same way: E₀=m₀c². the difference between the energy a body has when in motion (relative to some reference frame) and the energy it has when not in motion (relative to the same reference frame) is the kinetic energy.

interpreted this way E=mc² is not only valid for bodies at rest. the point stands.

rbj, your point is very clear, I've had the same point for a lot of time, then I discovered that the concept of "relativistic mass" is not very useful.
One example. The relativistic force is:

\vec{F} = \gamma m_0 \vec{a} + m_0\gamma^3 \vec{v}(\vec{a} \cdot \vec{v}/c^2)

where F is the force, v velocity, a acceleration, m_0 rest mass and m relativistic mass.
If acceleration is perpendicular to velocity you get:

\vec{F} = \gamma m_0 \vec{a} = m \vec{a}

and that's fine because you have extended Newton's second law to relativistic speeds.
But if acceleration a is parallel to v you have:

\vec{F} = \gamma m_0 \vec{a} + m_0\gamma^3 \vec{v}(|\vec{a}| |\vec{v}|/c^2) =

\vec{F} = m \vec{a} + m\gamma^2 \vec{v}(|\vec{a}| |\vec{v}|/c^2)

which is completely different, so the concept of relativistic mass don't actually extends Newton's second law.

Another example: (rest) mass is invariant and space-time curvature is invariant too, so you can relate the two concepts. You can't do it with relativistic mass.

 

[rbj]

rbj, your point is very clear, I've had the same point for a lot of time, then I discovered that the concept of "relativistic mass" is not very useful.
One example. The relativistic force is:

\vec{F} = \gamma m_0 \vec{a} + \gamma^3 \vec{v}(\vec{a} \cdot \vec{v})

where F is the force, v velocity, a acceleration an m_0 rest mass.
If acceleration is perpendicular to velocity you get:

\vec{F} = \gamma m_0 \vec{a} = m \vec{a}

and that's fine because you have extended Newton's second law to relativistic speeds.
But if acceleration a is parallel to v you have:

\vec{F} = m \vec{a} + \gamma^3 \vec{v}(|\vec{a}| |\vec{v}|)

which is completely different, so the concept of relativistic mass don't actually extends Newton's second law.

since Newton's 2^nd law is not simply

\vec{F} = m \frac{d\vec{v}}{dt}

but is

\vec{F} = \frac{d(m\vec{v})}{dt}

which is\vec{F} = m \frac{d\vec{v}}{dt} + \vec{v} \frac{dm}{dt}does that extend to the relativistic case, be it perpendicular or parallel? i can tell you that the derivation of E=mc² in my old textbook (from the concept of relativistic mass, retaining momentum as p=mv, and defining the inertial mass to be whatever it has to be to multiply v to get the very same momentum that you take as correct) uses that form of Newton's 2^nd law, which is the same as in the classical case. except, i don't think Newton was considering the possibility of the mass changing. actually, the derivation is for relativistic kinetic energy which comes out to be T=mc²-m₀c² and then there is the interpretation of the two terms as total energy less rest energy.

but my question for you, light, is

\vec{F} = \frac{d\vec{p}}{dt}

still valid?

 

[lightarrow]

since Newton's 2^nd law is not simply

\vec{F} = m \frac{d\vec{v}}{dt}

but is

\vec{F} = \frac{d(m\vec{v})}{dt}

which is

\vec{F} = m \frac{d\vec{v}}{dt} + \vec{v} \frac{dm}{dt}


does that extend to the relativistic case, be it perpendicular or parallel?

The equation:

\vec{F} = \frac{d(m\vec{v})}{dt}

Extends to what you want, since it's the definition of force! I actually used that definition to get the equation

\vec{F} = \gamma m_0 \vec{a} + m_0\gamma^3 \vec{v}(\vec{a} \cdot \vec{v}/c^2)

(by the way, I had forgot m_0/c^2 in the last term in my previous post).

i can tell you that the derivation of E=mc² in my old textbook (from the concept of relativistic mass, retaining momentum as p=mv, and defining the inertial mass to be whatever it has to be to multiply v to get the very same momentum that you take as correct) uses that form of Newton's 2^nd law, which is the same as in the classical case. except, i don't think Newton was considering the possibility of the mass changing. actually, the derivation is for relativistic kinetic energy which comes out to be T=mc²-m₀c² and then there is the interpretation of the two terms as total energy less rest energy.

but my question for you, light, is

\vec{F} = \frac{d\vec{p}}{dt}

still valid?

As I wrote, it's just the definition of force.

 

[rbj]

The equation:

\vec{F} = \frac{d(m\vec{v})}{dt}

Extends to what you want, since it's the definition of force! I actually used that definition to get the equation

\vec{F} = \gamma m_0 \vec{a} + m_0\gamma^3 \vec{v}(\vec{a} \cdot \vec{v}/c^2)

(by the way, I had forgot m_0/c^2 in the last term in my previous post). As I wrote, it's just the definition of force.

then i still fail to see, pedagogically, what's so bad about looking at the topic depicted by the question in the subject line, the way that i do (and have been taught 3 decades ago), differentiating between the concepts of rest mass and relativistic mass. still has nice consistent equations.

these guys keep asking these similar questions. they know that E=mc², they know that E = h \nu, and then they ask a very natural question, "how can it be that m=0?" and i think trying to explain 4-vectors and Minkowski space to them and asking them to accept

E^2 = \left(m c^2 \right)^2 + (p c)^2

as a given fundamental relationship to justify why m=0 just does not cut it pedagogically. for Physics majors that take a dedicated class in SR (or maybe SR and GR, i dunno, i never took a class in GR) to approach it pedagogically the way you do, is fine. but for the rest of us, i still cannot see how that explanation is pedagogically superior. when someone asks the question that is the subject line of this thread, i am not sure that we can assume they know about 4-vectors and Minkowski spacetime. or that they are conceptuallizing it. that's why i think that simply differentiating the concepts of rest mass from relativistic mass and showing why the rest mass of a photon must be zero (assuming that photons move at the speed of c). it can be easily stated in a sentence: "Photons have mass, but their rest mass is zero." it's accurate, even if it uses this currently deprecated concept of relativistic mass. it allows us to keep E=mc², p=mv, E=E₀+T, F=dp/dt, (and E = h \nu). we then derive

E^2 = \left(m_0 c^2 \right)^2 + (p c)^2

from the above plus the knowledge of how relativistic mass is related to rest mass (which can be derived from a simple kinematic setup - if time is dilated, the "moving" observer's pool ball appears to be moving slower to the "stationary" observer, so to preserve momentum, the inertial mass must increase by the same factor).

i still do not get why this is considered to be the pedagogically inferior method to answer, to newbies, the basic question: "If light has no mass, how can it have energy according to E=mc²?" to the alternative.

 

[lightarrow]

then i still fail to see, pedagogically, what's so bad about looking at the topic depicted by the question in the subject line, the way that i do (and have been taught 3 decades ago), differentiating between the concepts of rest mass and relativistic mass. still has nice consistent equations.

these guys keep asking these similar questions. they know that E=mc², they know that E = h \nu, and then they ask a very natural question, "how can it be that m=0?" and i think trying to explain 4-vectors and Minkowski space to them and asking them to accept

E^2 = \left(m c^2 \right)^2 + (p c)^2

as a given fundamental relationship to justify why m=0 just does not cut it pedagogically. for Physics majors that take a dedicated class in SR (or maybe SR and GR, i dunno, i never took a class in GR) to approach it pedagogically the way you do, is fine. but for the rest of us, i still cannot see how that explanation is pedagogically superior. when someone asks the question that is the subject line of this thread, i am not sure that we can assume they know about 4-vectors and Minkowski spacetime. or that they are conceptuallizing it. that's why i think that simply differentiating the concepts of rest mass from relativistic mass and showing why the rest mass of a photon must be zero (assuming that photons move at the speed of c). it can be easily stated in a sentence: "Photons have mass, but their rest mass is zero." it's accurate, even if it uses this currently deprecated concept of relativistic mass. it allows us to keep E=mc², p=mv, E=E₀+T, F=dp/dt, (and E = h \nu). we then derive

E^2 = \left(m_0 c^2 \right)^2 + (p c)^2

from the above plus the knowledge of how relativistic mass is related to rest mass (which can be derived from a simple kinematic setup - if time is dilated, the "moving" observer's pool ball appears to be moving slower to the "stationary" observer, so to preserve momentum, the inertial mass must increase by the same factor).

i still do not get why this is considered to be the pedagogically inferior method to answer, to newbies, the basic question: "If light has no mass, how can it have energy according to E=mc²?" to the alternative.

How then would you explain to them that one photon has rest mass = 0 and two of them (not traveling exactly in the same direction) have total rest mass =/= 0?

 

[Antenna Guy]

How then would you explain to them that one photon has rest mass = 0 and two of them (not traveling exactly in the same direction) have total rest mass =/= 0?

Taken as a system, the average velocity (vector) of the two photons you describe is now less than c.

Regards,

Bill

 

[lightarrow]

How then would you explain to them that one photon has rest mass = 0 and two of them (not traveling exactly in the same direction) have total rest mass =/= 0?

Taken as a system, the average velocity (vector) of the two photons you describe is now less than c.

Regards,

Bill

It's not clear to me how you define the average velocity (vector) of the two photons in this case.

 

[Antenna Guy]

It's not clear to me how you define the average velocity (vector) of the two photons in this case.

Project the individual velocity vectors into an average inertial frame.

Regards,

Bill

 

[lightarrow]

Project the individual velocity vectors into an average inertial frame.

Regards,

Bill

Sorry, but do you mean a normal projection along a coordinate axis? That is, if the velocity vector of every photon makes an angle theta with a coordinate axis then you take c*cos(theta)?

 

[Antenna Guy]

Sorry, but do you mean a normal projection along a coordinate axis? That is, if the velocity vector of every photon makes an angle theta with a coordinate axis then you take c*cos(theta)?

Sort of - v*cos(theta) is the same thing as a vector dot product where one vector (i.e the new direction) is of unit length. The "v" in this case is the magnitude (or length) of a velocity vector - not the vector itself.

Additionally, adding two mutually orthogonal unit vectors perpendicular to the new direction vector would define what I called an "inertial frame". Using vector dot products in the same way, these vectors can be used to account for velocity components not in the new direction of propagation - vis. v*sin(theta)*cos(phi) and v*sin(theta)*sin(phi).

If the "average direction" is accurate, it should be consistent with the net kinetic energy of the system.

Regards,

Bill

 

[rohanv2]

Who the hell just calls the Light has no mass!

There are Photons (particles) in light hence being conserved that light has mass. If it has momentum then it is sure that it has mass hence is mass = 0 then

P= m X v
= 0 X V
= 0 kgm/s ?? wth?

Pst.
Einstein proved E=mc² by the help of momentum of light.

After the light that hits the object the electrons are ejected out which and then the object becomes visible but the amount in which it ejects out the electrons are in very less amount but though [ as told by our teachers ], then it definitely needs energy. While if the mass = 0 then
E= Nm
= m X a X s
= 0 X a X s
= 0 Joule??
How can light light participate even in photoelectric effect without any energy to eject out electrons?

I guess this is enough for proving that light has mass... :D

 

[Dale]

Hi rohanv2, welcome to PF

Calm down, this thread has been dead almost two years now. It really is just a question of definition, a pulse of light has "relativistic mass" but no "rest mass". So pick your favorite flavor and relax.

 

[rohanv2]

Hi rohanv2, welcome to PF

Calm down, this thread has been dead almost two years now. It really is just a question of definition, a pulse of light has "relativistic mass" but no "rest mass". So pick your favorite flavor and relax.

Indeed thanks and sorry! didn't realize that it's dead already. Anyway, I realized that light has no mass, it is just a form of a wave (just a sort of flow of energy). How silly of me. :D