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

[Meatbot]

If m=0, then e=0(c^2), so e=0

 

[f95toli]

E=m2^2 is not the complete equation.
There is also a momentum term and since light carries momentum there is no contradiction.
The complete equation is
E=sqrt(mc^2+(pc)^2)

sqrt=square root

In most cases the momentum term can be neglected, but obviously not for light.

 

[Dale]

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.

 

[dst]

Because mass is defined after not before.

e = hf for a photon. And you can get a virtual mass for it through e = mc^2, useful or not. It comes from the relativistic equation of mass/kinetic energy.

e^2 - p^2 = m^2c^4

But since RHS is 0, then e^2 = p^2.Edit: I seem to post very slow? 2 replies...

 

[Meatbot]

And you can get a virtual mass for it through e = mc^2, useful or not.

What is the point of saying a photon has zero mass if that's only true when it's not moving (which can't happen (can it? (I love nested parentheses.)))? Why do we not say that the photon has the "virtual" mass in actuality?

 

[dst]

What is the point of saying a photon has zero mass if that's only true when it's not moving (which can't happen (can it? (I love nested parentheses.)))? Why do we not say that the photon has the "virtual" mass in actuality?

Because of pair production, it can spontaneously become real mass, i.e. electron/positron pair. Only when the photon is colliding with something to transfer momentum, however. So a photon with a certain energy can create 'real mass', in a way.

 

[Meatbot]

Because of pair production, it can spontaneously become real mass, i.e. electron/positron pair. Only when the photon is colliding with something to transfer momentum, however. So a photon with a certain energy can create 'real mass', in a way.

What's the difference between virtual mass and real mass?

 

[rbj]

there is also different qualifications of the word "mass". photons, if they really do travel at the speed of c (relative to any observer), cannot have rest mass a.k.a. invariant mass (what i'll denote as m₀). but if you think of momentum as simply

p = m v

then the mass of a body that is moving past an observer (at a relative speed of v) appears to that observer to be heavier than the rest mass:

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

but that is for a regular old object or particle with rest mass m₀. note that the relativistic mass approaches infinity as v approaches c. if you use this (relativistic) m in E = m c^2, then the E is the total energy of the particle (in the reference frame of the observer that the particle is whizzing past) and, for photons, is the same energy as

E = h \nu

so, if we're talking relativistic mass, photons have a mass of

m = \frac{E}{c^2} = \frac{h \nu}{c^2}but if you turn the previous equation around a little:

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

you will see that for any finite m, the rest mass, m₀, must go to zero if the speed of the photon, v, is c. photons are called "massless", because their rest mass is zero. but their relativistic mass is not zero and is, indeed, proportional to their energy which is proportional to the frequency of radiation.

 

[ZapperZ]

If m=0, then e=0(c^2), so e=0

Please read our FAQ in this forum.

Zz.

 

[Meatbot]

photons are called "massless", because their rest mass is zero. but their relativistic mass is not zero and is, indeed, proportional to their energy which is proportional to the frequency of radiation.

Ok, I understand that now. But why do we even bother to use the concept of rest mass when everything is in motion relative to something else? Nothing is ever at rest. Or is it used in calculating the apparent mass to different observers?

That also brings to mind the perspective of the photon. From its perspective, can't it "see" itself as being at rest and everything else is moving at c? Then, if it's massless at rest AND not moving, then it doesn't have mass OR momentum and thus has no energy at all. If you have no mass and no momentum and no energy how can you even be said to exist at all?

 

[rbj]

Ok, I understand that now. But why do we even bother to use the concept of rest mass when everything is in motion relative to something else? Nothing is ever at rest.

an inertial object is at rest in its own frame of reference.

Or is it used in calculating the apparent mass to different observers?

yes. but i have to be careful here.

it is true that the terminology differentiating rest mass (the apparent mass in an objects own frame of reference) from the apparent inertial mass (a.k.a. "relativistic mass") of an object whizzing past an observer at 0.99c, has been "falling out of favor".

That also brings to mind the perspective of the photon. From its perspective, can't it "see" itself as being at rest and everything else is moving at c?

contemplating the "perspective" or frame of reference of a photon gets one in trouble. we really can't deal with a reference frame (ostensibly with an observer) that whizzes past another at a speed of c. the physics doesn't work out.

Then, if it's massless at rest AND not moving, then it doesn't have mass OR momentum and thus has no energy at all. If you have no mass and no momentum and no energy how can you even be said to exist at all?

we get into arguments here about this at PF. I'm a relatively old electrical engineer who had a college level modern physics course in the 70s that had no problem with terminology of relativistic mass and of the masses of photons. back then we said:

total energy of a body:

E = m c^2

where

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

where m₀ is the "rest mass" or what they now call "invariant mass" of the body. nowadays they just like to use the symbol "m" for that, but, to keep our terminology straight for this discussion, let's not do that. now the total energy of a body was the energy that it had at rest (used to be called the "rest energy") plus the additional energy it gets from being in motion (a.k.a. "Kinetic Energy"):

E = m c^2 = E_0 + KE

and since the kinetic energy of a body (from the perspective of some observer) is zero if it's not moving (relative to that observer) and the apparent inertial mass of the body not moving is the same of the rest mass (m = m₀), then the rest energy of the body is

E_0 = m_0 c^2

and then, in general, the kinetic energy of the body (the total energy less the rest energy) is

KE = E - E_0 = m c^2 - m_0 c^2 = m_0 c^2 \left(1 - \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \right)

now we all agree with the last two equations since they are expressed with no reference to this "relativistic mass" that has fallen out of favor, and if you allow the velocity v to be much much smaller than the speed of light c, then you will get the more familiar classical expression for kinetic energy:

KE = m_0 c^2 \left(1 - \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \right) \approx \frac{1}{2} m_0 v^2

another difference in nomenclature or expression i have with the current trend is that while we all agree that this expression relating (rest) mass, momentum, and total energy is:

E^2 = E_0^2 + (p c)^2 = m_0^2 c^4 + p^2 c^2

i would disagree that this comes as a first principle and that the expression for relativistic momentum, p, is derived from it. i would say that the apparent momentum of a body is derived first from the kinematics and the knowledge of what relativistic time-dilation does, we get

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

and if we continue to interpret momentum as the product of the inertial mass, m, and velocity, v,

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

then we get the above expression for the inertial mass or "relativistic mass":

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

but this mass is not only a property of the body or particle itself, and i think that is why its use is commonly discouraged. but it is, in my opinion, a useful quantity and concept to keep. this relativistic mass, m makes the famous Einstein equation work, not only for rest energy, but for the total energy of a body or particle:

E = m c^2

that can validly represent the total energy of a body or particle if the relativistic mass is used for m. but if m means only the "rest mass" (what I've been calling "m₀") then the above equation means only the "rest energy" (what I've been calling "E₀"), devoid of kinetic energy, not the total energy of the body.

now, about photons, i have always qualified the term "massless" in regard to photons. photons have mass, but they do not have rest mass. the photon total energy is:

E = h \nu

and if that is equated to the relativistic total energy

E = h \nu = m c^2

then it is clear that the photon has a relativistic mass of

m = \frac{E}{c^2} = \frac{h \nu}{c^2}

which is neither zero nor infinite. people here don't like me saying this, but what we do agree on is that, assuming the speed of the photon is c (for any observer), the momentum of the photon is

p = \frac{h \nu}{c}.

but i would say it is because

p = m v = m c = \frac{h \nu}{c^2} c = \frac{h \nu}{c}.

the other guys (that discourage talk of "relativistic mass") would say it's because

E^2 = m_0^2 c^4 + p^2 c^2

and because photons are "massless" (as a first principle), then m₀ = 0 and

E^2 = (h \nu)^2 = (0)^2 c^4 + p^2 c^2

and solving that you get

p = \frac{h \nu}{c}.

I would not say that photons are massless as a first principle. i would turn the relativistic mass equation around and express the rest mass in terms of the relativistic mass (and v and c):

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

but if the apparent inertial mass or relativistic mass m is finite (as it is for a photon), then the rest mass must go to zero if v = c. so photons, although the have inertial mass (which i think is a necessary component to have a non-zero momentum), do not have any rest mass. nor do they have rest energy, all of their energy is "kinetic" (that might not be the best word for it, so i am bracing for pervect or someone to slap me down for saying it).

this expression of the concepts makes sense to me, and i believe is pedagogically more accessable than how most of the real physicists here like to express it (without the use of relativistic mass, so then given the only mass they recognize, rest mass or "invariant mass", then photons are massless). but, again, i am not a physicist. i am an electrical engineer who also does math for a living. i get into other pedagogical disputes here regarding the dirac delta "function", but that's a different topic.

 

[Dale]

But why do we even bother to use the concept of rest mass

Have you learned about the spacetime interval? In special relativity certain concepts that we once thought were independent and absolute, like time, were found to be frame variant. And yet, we were able to find a spacetime "distance" that is not frame variant and that all observers can agree on. That is the length of the spacetime interval.

Similarly for energy and momentum. They are combined in the same way that time and space are combined, and the length of the energy-momentum 4-vector is the invariant mass. It is something that all observers can calculate and agree on and it is the same as the rest mass. For light the length of that energy-momentum 4-vector is 0 so it has 0 invariant or rest mass. It is the only mass that all observers will agree on.

 

[Meatbot]

it is true that the terminology differentiating rest mass (the apparent mass in an objects own frame of reference) from the apparent inertial mass (a.k.a. "relativistic mass") of an object whizzing past an observer at 0.99c, has been "falling out of favor".

Why has it been falling out of favor? I was under the impression that this mass gain was real and not virtual. If its real then you have to deal with it as if it were. Is it just because it's harder to do the math if you have to keep in mind that the mass is different for different observers?

contemplating the "perspective" or frame of reference of a photon gets one in trouble. we really can't deal with a reference frame (ostensibly with an observer) that whizzes past another at a speed of c. the physics doesn't work out.

Apparently, as I was shown yesterday. You get division by zero in the Lorentz equation. But my question is why can't we deal with it sonehow? Is our equation wrong? Is a photon doing something qualitatively different than other moving objects so it requires a different equation altogether? Perhaps its irrelevant and a photon does not really move at c or perhaps there are no photons. What new physics is required to resolve this?

 

[lightarrow]

Why has it been falling out of favor? I was under the impression that this mass gain was real and not virtual. If its real then you have to deal with it as if it were. Is it just because it's harder to do the math if you have to keep in mind that the mass is different for different observers?

Also, but even because you would have a different mass in the motion direction and in the perpendicular direction; also for other reasons. Rest mass is the only meaningful mass in SR.

Apparently, as I was shown yesterday. You get division by zero in the Lorentz equation. But my question is why can't we deal with it sonehow? Is our equation wrong? Is a photon doing something qualitatively different than other moving objects so it requires a different equation altogether? Perhaps its irrelevant and a photon does not really move at c or perhaps there are no photons. What new physics is required to resolve this?

If you think that rest mass is meaningless for photons, think about the fact that "two" photons not traveling in the same direction, have a total rest mass =/= 0:

E^2 = (cp)^2 + (mc^2)^2

If the two photons don't travel in the same direction, you can find a ref frame in which they travel in opposite directions, so that total momentum p = 0, so E^2 = (mc^2)^2 --> m = E/c^2 =/= 0.

Edit: you can compute mass in a different ref frame because mass is invariant = doesn't change with ref frame.

 

[lightarrow]

That also brings to mind the perspective of the photon. From its perspective, can't it "see" itself as being at rest and everything else is moving at c? Then, if it's massless at rest AND not moving, then it doesn't have mass OR momentum and thus has no energy at all. If you have no mass and no momentum and no energy how can you even be said to exist at all?

The problem is that "photon's perspective" doesn't exist because to assign a physical meaning to that concept means to find a meaningful ref frame co-moving with a photon and that frame cannot be find in physics. So you can't do that reasoning.

 

[twinsen]

meaningful ref frame co-moving with a photon and that frame cannot be find in physics.

What exactly does this mean? that we can't travel as fast as light so there is no way we can create a refference frame?

Im also interested in how the energy of a photon is derived as E=h\nu
Has this come from experimentation?

Alex

 

[CJames]

rbj, I'm a fan of using relativistic mass as well, even though it can be confusing for some, because it's so much easier to work with mathematically (for me anyway) than relativistic kinetic energy

 

[chooserslain]

What exactly does this mean? that we can't travel as fast as light so there is no way we can create a refference frame?

Im also interested in how the energy of a photon is derived as E=h\nu
Has this come from experimentation?

Alex

No, we can travel with the speed of light. But even when we travel with the speed of light in the same direction with the photon, the photon, with respect to us, still travel with the speed=c.

There is no reference frame that light, in that frame, travels with speed different from c.
That because the space-time is distorted. As we try to travel faster, in our reference frame, the more space is stretched and time is condensed. So, the light, due to the fact that distance is stretched and time is condensed, has it velocity= distance/time still remaining c.

You can use the Lorentz fomula of velocity addition to verify:

A body travels with speed vx in a rest reference frame RF1. The speed of it when we observe it while we are resting in a reference RF2 which is traveling with speed v0 with respect to the RF1 is vx':

vx'=(vx-v0)/(1-vx.v0/c^2)

Let's consider the body as a photon, so vx=c. Let's try to travel with the speed of light, so our speed v0->c, then the speed of the photon in our reference frame, RF2, is
vx'=lim (as vx->c) (vx-c)/(1-vx.c/c^2)=c

In general, no matter what we are travelling, following the photon, with the speed of c/100, c/10, c/2 or 0,9999c, or even c, the space will stretched, time will condensed, so as that the photon still travel with its speed=c with respect to us. That the weird characteristic of photon.

 

[jtbell]

No, we can travel with the speed of light.

Eh? How do we accomplish that?

 

[jostpuur]

No, we can travel with the speed of light. But even when we travel with the speed of light in the same direction with the photon, the photon, with respect to us, still travel with the speed=c.

This is incorrect use of the velocity addition formula.

 

[lightarrow]

What exactly does this mean? that we can't travel as fast as light so there is no way we can create a refference frame?

AFAIK, SR and Lorentz transforms cannot be applied when v = c; also note that a "co-moving frame of reference" is not simply a "co-moving object" but a system of coordinates and clocks put in every point of the space which is co-moving with you.

Im also interested in how the energy of a photon is derived as E=h\nu
Has this come from experimentation?

It's an experimental result and cannot be derived in any way; it's part of the "essence" of quantum mechanics.
If you can derive it I propose you for the Nobel prize

 

[rbj]

rbj, I'm a fan of using relativistic mass as well,

nice to know one is not alone.

even though it can be confusing for some

dunno if it's pedagogically more confusing than the 4-vector explanation ("like uhh, what's a 4-vector?" or "uhh, man, uh, what's a Minkowski space-time?") that somehow leads directly to

E^2 = (m_0 c^2)^2 + p^2 c^2

 

[Dale]

dunno if it's pedagogically more confusing than the 4-vector explanation ("like uhh, what's a 4-vector?" or "uhh, man, uh, what's a Minkowski space-time?")

Personally I disagree most strongly. I did not understand SR at all until I was introduced to 4-vectors and Minkowski spacetime diagrams. I think the framework they provide is incredibly clear and, for me, it got rid of years of "pedagogical confusion".

This is probably why we disagree about the value of the relativistic mass concept.

 

[rbj]

This is probably why we disagree about the value of the relativistic mass concept.

i think you're right. for me, my first (and only formal introduction) to special relatitivity was after the first two semesters of physics (classical mechanics and E&M). we had the M-M experiment, we had the postulates, we had the light clock, we had length contraction, we had Lorentz transformation and velocity addition, then we had a little thought experiment with two identical balls, each with one of two observers whizzing past each other at some known speed v and a little collision they make at a direction that is perpendicular to the direction of v. the reason why the mass (of the other guy's ball) had to increase by a factor of \gamma was because the vertical velocity of the ball appeared to be slower by the same factor because of time dilation. from that thought experiment was where relativistic momentum and, if you divide by v, relativistic mass was derived. then from that we got to E=mc² and that E was not rest energy (nor was the m rest mass).

 

[lightarrow]

i think you're right. for me, my first (and only formal introduction) to special relatitivity was after the first two semesters of physics (classical mechanics and E&M). we had the M-M experiment, we had the postulates, we had the light clock, we had length contraction, we had Lorentz transformation and velocity addition, then we had a little thought experiment with two identical balls, each with one of two observers whizzing past each other at some known speed v and a little collision they make at a direction that is perpendicular to the direction of v. the reason why the mass (of the other guy's ball) had to increase by a factor of \gamma was because the vertical velocity of the ball appeared to be slower by the same factor because of time dilation. from that thought experiment was where relativistic momentum and, if you divide by v, relativistic mass was derived. then from that we got to E=mc² and that E was not rest energy (nor was the m rest mass).

...but, as you know, you can't write that formula for massless objects like photons, so it's not as general as:
E^2 = (cp)^2 + (mc^2)^2.
Your relativistic mass can't be anything else than "energy" called with another name so it's useless to call it "mass".

 

[Dale]

I have a much easier time accepting momenum and energy as non-linear functions of velocity than accepting mass as a function of velocity.

 

[thestien]

sorry to barge in here but in space wouldn't mass be redundent because every thing would weigh 0? and is space friction free? which would then mean it would require little to no energy to acclereate to the speed of light?or have i missed somthing very fundermental?

 

[Doc Al]

sorry to barge in here but in space wouldn't mass be redundent because every thing would weigh 0? and is space friction free? which would then mean it would require little to no energy to acclereate to the speed of light?or have i missed somthing very fundermental?

You've missed the basic fact that it takes energy to accelerate an object. An object in space might be "weightless", but it still has mass.

 

[thestien]

define mass? mass on Earth is weight maybe you use the word mass in a different context to what i was thinking?
would it take more energy to move say a person in space than a house in space?
i know that as u drive a car on Earth the faster u go the more force you build up trying to slow you down so the more enegry you need to keep accelerating.
so would that same principle work in space i.e. is there a resistance in space? and i don't mean space near a star i mean the in between parts? oh to make things clear I am not saying u are wrong just trying to expand my understanding :)

 

[Doc Al]

define mass? mass on Earth is weight maybe you use the word mass in a different context to what i was thinking?

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

would it take more energy to move say a person in space than a house in space?

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

i know that as u drive a car on Earth the faster u go the more force you build up trying to slow you down so the more enegry you need to keep accelerating.

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

so would that same principle work in space i.e. is there a resistance in space? and i don't mean space near a star i mean the in between parts?

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?

 

[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