How many ways there are to conclude that Eo=mc^2 ?

[sami_m]

One way to conclude that E₀=mc² is to look at the work done by constant force:

W = integral[ F dx ] = mc²(1/√1-(v/c)²)-1)

and from that you get that E_kin = E_tot - E₀ , where E₀ = mc²

How many other ways there are to conclude that E₀ = mc² ?

 

[Fredrik]

You can simply define the energy of a particle as the 0 component of the four-momentum p. If p is defined as the four-vector that, in the particle's rest frame, is equal to the mass m times the four-velocity v (which in the rest frame has componennts (1,0,0,0)), then (if we use a -+++ metric and units such that c=1) we have $p^2=m^2v^2=m^2(-1)=-m^2$ and $p^2=-E^2+\vec p^2$. So $-E^2+\vec p^2=-m^2$. This implies $E^2=m^2+\vec p^2$. If we restore factors of c, this turns into $E^2=m^2c^4+\vec p^2c^2$. If $E_0$ denotes the energy when velocity (and momentum) is zero, this implies that $E_0=mc^2$.

If four-momentum is defined as the four-vector p formed by the generators of translations in the 0,1,2,3 directions, then we can define the mass as the non-negative m such that $-m^2=p^2$. Then we can proceed as above.

 

[bcrowell]

One way to conclude that E₀=mc² is to look at the work done by constant force:

W = integral[ F dx ] = mc²(1/√1-(v/c)²)-1)

and from that you get that E_kin = E_tot - E₀ , where E₀ = mc²

How many other ways there are to conclude that E₀ = mc² ?

Welcome to PF! By the way, the method you're using to write up your math is going to be very slow and painstaking for you, will not produce very readable results, and will not work for more complex math. A better way to do it is using LaTeX: https://www.physicsforums.com/help/latexhelp/

I don't think your reasoning in #1 is correct. Your integral is a definite integral, so any constant term cancels out. There is also a more subtle issue, which is that one needs to provide some justification for the assumption that $W=\int F dx$ holds without modification in relativity. In Einstein's first 1905 paper on relativity, he does this in section 10, using an intricate argument that starts from the fact that for a charged particle that is instantaneously at rest, $ma=qE$.

Einstein's original derivation of $E=mc^2$ is not in his first paper on SR. It's the topic of his second paper, and it uses an argument involving an object that emits light simultaneously in two opposing directions. For a critique of this argument, see Ohanian, "Einstein's E = mc^2 mistakes," http://arxiv.org/abs/0805.1400 . My SR book has a presentation of Einstein's argument in section 4.2, along with follow-up discussions in section 4.4 and ch. 9 that take care of the issues that Ohanian complains about.

The most common approach today is to search for a four-vector that is conserved in collisions. An example along these lines by @DrGreg is here: https://www.physicsforums.com/showthread.php?p=2416765 .

This may also be of interest: Sonego and Pin, Deriving relativistic momentum and energy, http://arxiv.org/abs/physics/0402024

 

[Fredrik]

I don't think your reasoning in #1 is correct. Your integral is a definite integral, so any constant term cancels out.

We have $W=\int F\mathrm dx=\int \dot p\dot x\mathrm dt$, where the integration on the right is over a time interval [a,b] such that the velocity is zero at time a. The integrand $\dot p\dot x$ can be rewritten as $\frac{d}{dt}(m\gamma)$ (in units such that c=1). So we end up with
$$W=\int_a^b \frac{d}{dt}(m\gamma)\mathrm dt =\gamma_bm-\gamma_am.$$ If we drop the b from the notation and restore factors of c, we get
$$W=\gamma mc^2-mc^2=E-E_0.$$

 

[bcrowell]

We have $W=\int F\mathrm dx=\int \dot p\dot x\mathrm dt$, where the integration on the right is over a time interval [a,b] such that the velocity is zero at time a. The integrand $\dot p\dot x$ can be rewritten as $\frac{d}{dt}(m\gamma)$ (in units such that c=1). So we end up with
$$W=\int_a^b \frac{d}{dt}(m\gamma)\mathrm dt =\gamma_bm-\gamma_am.$$ If we drop the b from the notation and restore factors of c, we get
$$W=\gamma mc^2-mc^2=E-E_0.$$

But that doesn't imply that $E_0$ has any special interpretation. If this argument were correct, then there would have been no reason for Einstein to publish his second 1905 paper on SR, with a separate argument. He would simply have noted it as a consequence of his argument about mechanical work in his first paper.

 

[Fredrik]

But that doesn't imply that $E_0$ has any special interpretation.

I think I see what you meant when you mentioned a constant before. If we define the total energy E by $E=\gamma m$ (times $c^2$), then $E_0$ is the total energy of an object at rest. But we could add a constant to that definition. We would have $E=\gamma m+A$, but we would still have $W=E-E_0$. So what is the correct value of A?

I don't think this is really an issue. Why not just say that we choose A=0 because this is the choice that gives us the nicest interpretation of E?

The approach to SR that I like best goes like this:

1. Define mathematical terms. (Minkowski spacetime, world line, proper time, momentum, work, etc.)
2. Choose a set of correspondence rules. (Assumptions that specify how to interpret the mathematics as predictions about results of experiments).
3. Use 1-2 to make predictions about results of experiments.
4. Do experiments to determine the accuracy of those experiments.

In this approach, the definitions in step 1 don't need to be justified. Definitions never do. We may be able to show e.g. that two concepts defined in step 1 are related in a way that's very similar to how they're related in pre-relativistic classical mechanics. If such a demonstration makes it easier for us to develop our intuition about SR, then I would consider it a worthwhile thing to do. But I wouldn't consider it essential. The only thing that can really justify what we did in step 1 is step 4.

I would say that the calculation of the work required to accelerate a mass m from speed 0 to speed v is such a demonstration. It tells us that if we define work, force, momentum and energy in certain ways (ways that are similar to how these terms are defined in pre-relativistic classical mechanics), the relationship between work and kinetic energy is the same as in pre-relativistic classical mechanics.

 

[DrStupid]

One way to conclude that E₀=mc² is to look at the work done by constant force:

W = integral[ F dx ] = mc²(1/√1-(v/c)²)-1)

and from that you get that E_kin = E_tot - E₀ , where E₀ = mc²

I don't get it. Can you explain that in detail?

 

[Fredrik]

I don't get it. Can you explain that in detail?

It's explained in some detail (for the more general case of a force that isn't necessarily constant) in post #4. What I left out is the proof of the claim $\dot p \dot x=\frac{d}{dt}(\gamma m)$. It goes like this (still using units such that c=1):
\begin{align*}
& \gamma=\frac{1}{\sqrt{1-\dot x^2}}=(1-\dot x^2)^{-\frac 1 2}\\
& \dot\gamma =-\frac 1 2(1-\dot x^2)^{-\frac 3 2}(-2 \dot x\ddot x) = \gamma ^3\dot x\ddot x\\
& 1+\gamma^2\dot x^2=1+\frac{\dot x^2}{1-\dot x^2}=\frac{1-\dot x^2+\dot x^2}{1-\dot x^2}=\gamma ^2\\
& \dot p=\frac{d}{dt}(\gamma m \dot x)=\dot\gamma m\dot x+\gamma m \ddot x =\gamma ^3\dot x\ddot x m\dot x+\gamma m \ddot x =\gamma m\ddot x(\gamma^2\dot x^2+1) =\gamma ^3 m\ddot x\\
& \dot p \dot x =\gamma^3 m\ddot x \dot x = m\dot\gamma =\frac{d}{dt}(\gamma m).
\end{align*}

 

[DrStupid]

It's explained in some detail (for the more general case of a force that isn't necessarily constant) in post #4.

How does #4 explains Eo = m·c²?

Edit: See bcrowell's post for details (and no, #6 doesn't explain it either).

 

[Fredrik]

How does #4 explains Eo = m·c²?

Edit: See bcrowell's post for details (and no, #6 doesn't explain it either).

I explained how to do the calculation described in the text you quoted, and I had previously explained what that calculation accomplishes. If that's not what you wanted explained, then you shouldn't have asked for an explanation of "that". You should have told us what you wanted explained.

It is impossible to explain why $mc^2$ is the energy of a particle at rest, without a definition of "energy" (and "at rest"). What definition would you like to use?

 

[DrStupid]

If that's not what you wanted explained, then you shouldn't have asked for an explanation of "that". You should have told us what you wanted explained

Just read the title of this thread.

It is impossible to explain why $mc^2$ is the energy of a particle at rest, without a definition of "energy" (and "at rest"). What definition would you like to use?

The same as Einstein used for his derivation of Eo=mc².

 

[Fredrik]

Just read the title of this thread.

If you quote a post and then ask for an explanation of "that", then you're asking for an explanation of what was said in that quote, not for an explanation of what was said in the thread title. That's how quotes work. The whole point of them is to show the reader what you're responding to.

The same as Einstein used for his derivation of Eo=mc².

If you want to discuss that definition here, you're going to have to post it here along with reference (preferably a link to a page where you can download the article). You may have to provide other definitions as well, but this is the minimum requirement to get a discussion started.

And @bcrowell, if you have something to say, you should say it. Why are you "liking" these posts?

 

[DrStupid]

If you quote a post and then ask for an explanation of "that"

And I'm still waiting for the explanation how

W = integral[ F dx ] = mc²(1/√1-(v/c)²)-1)

results in

E_kin = E_tot - E₀ , where E₀ = mc²

PS: You do not need to answer unless you are going to provide this explanation. I will ignore everything else.

 

[stevendaryl]

I'm not sure if it's more or less intuitive than the mathematical proof that assumes energy and momentum are part of a Lorentz 4-vector, but you can give a hand-wavy argument for the relativistic form of momentum and energy by considering collisions, conservation in multiple frames, and the nonrelativistic limit (where the forms for momentum and kinetic energy should reduce to the Newtonian expressions).

You can first establish that spatial momentum is given by \vec{p} = m \gamma \vec{v} (where \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}) It is basically the only choice that reduces to Newtonian momentum in the nonrelativistic limit and which, if it is conserved in one frame, then it is conserved in every frame.

Then, you can try the guess that the energy of a free particle has the form E(v) = g(v) m for some unknown function g(v), which depends on the magnitude of velocity (but not the direction). Now, consider a completely inelastic collision in which two wads of "gum" of mass m that collide and form a larger wad of mass M.

Let F be the center-of-mass frame. In this frame, the initial velocity of one wad is +v (let it be in the x-direction), and the initial velocity of the other wad is -v. Afterward, the larger mass M that results is stationary. Conservation of energy gives:

2 g(v) m = M g(0)

Let F' be a frame that is moving at speed -u in the x-direction, relative to F. Assume that u \ll v. In this new frame, one wad is traveling at speed v' \gt v, and has momentum p' = + m \gamma' v'. The other wad is traveling at speed v'' \lt v, and has momentum p'' = - m \gamma'' v'', where \gamma' = \frac{1}{\sqrt{1 - \frac{v'^2}{c^2}}} and \gamma'' = \frac{1}{\sqrt{1 - \frac{v''^2}{c^2}}}

Now, using the relativistic velocity addition formula and some slightly tedious algebra, you can show, to first order in u, that:
p' \approx \gamma m v + \gamma m u
p'' \approx - (\gamma m v - \gamma m u)

So the total momentum in this frame, before the collision is:
p_{total} = 2 \gamma m u

After the collision, the big wad of mass M is just moving at speed u, which we assumed was very small, so we can use a nonrelativistic approximation: p = M u.

So conservation of momentum in this frame gives:
2 \gamma m u = M u \Rightarrow M = 2 \gamma m

Now, plugging this back into our equation for conservation of energy in the original frame gives:

2 m g(v) = M g(0) \Rightarrow 2 m g(v) = 2 \gamma m g(0)

So we conclude that g(v) = \gamma g(0)

At this point, we compute the form of kinetic energy for low velocity:

K(v) = E(v) - E(0) = m g(v) - mg(0) = m \gamma g(0) - m g(0)

We can use a Taylor series to write \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \approx 1 + \frac{1}{2} \frac{v^2}{c^2} + terms of order v^4/c^4. So to lowest order,

K(v) \approx m g(0) \frac{1}{2} \frac{v^2}{c^2}

Comparing with the Newtonian limit, K(v) = \frac{1}{2} m v^2, tells us that g(0) = c^2

So that tells us the complete form for relativistic energy is E(v) = m g(v) = m \gamma g(0) = m \gamma c^2

Note that the assumption that energy depends only on mass and velocity is kind of bold. In classical physics, that's only true if you ignore "internal energy" such as heat.

 

[Fredrik]

And I'm still waiting for the explanation how
results in
PS: You do not need to answer unless you are going to provide this explanation. I will ignore everything else.

If you're not going to post the definitions you want to use (this is the absolute minimum that would be required from any poster), you're not going to get a better reply than you already got from me.

Don't put something that someone else said in a quote with my name on it.

 

[sami_m]

Let F be the center-of-mass frame. In this frame, the initial velocity of one wad is +v (let it be in the x-direction), and the initial velocity of the other wad is -v. Afterward, the larger mass M that results is stationary. Conservation of energy gives:

2 g(v) m = M g(0)

Wait a moment, what happens if i insist that g(0) = 0 ?
Then this equation does not make any sense, unless g(v) = 0 which is trivial solution.
Should there be then some additional energy after the collision in the forms of photons, some other particles, potential energy etc such that:
E_{something else} = 2 g(v) m
and total momentum:
p_{something else} = 0
to make the larger mass stationary? or is it just impossible possibility that g(0) = 0 ?

I don't think your reasoning in #1 is correct. Your integral is a definite integral, so any constant term cancels out. There is also a more subtle issue, which is that one needs to provide some justification for the assumption that W=∫Fdx" style="position: relative;" tabindex="0" id="MathJax-Element-9-Frame" class="MathJax">W=∫FdxW=∫Fdx holds without modification in relativity. In Einstein's first 1905 paper on relativity, he does this in section 10, using an intricate argument that starts from the fact that for a charged particle that is instantaneously at rest, ma=qE" style="position: relative;" tabindex="0" id="MathJax-Element-10-Frame" class="MathJax">ma=qEma=qE.

Ok thanks:) Yes it is definite integral so there may be a constant term .

There is one other way to calculate the work integral, which is definite integral. But it is just a mathematical trick, nothing else:
W = \int F dx = \int p dv = p_{end}v_{end} - \int v dp = p_{end}v_{end} - (1 - sqrt{ 1 - (v_{end}/c)^{2}})
which suggest that:
g(v) = pv/m + A = \gamma v^{2} + A

 

[stevendaryl]

Let F be the center-of-mass frame. In this frame, the initial velocity of one wad is +v (let it be in the x-direction), and the initial velocity of the other wad is -v. Afterward, the larger mass M that results is stationary. Conservation of energy gives:

2 g(v) m = M g(0)

Wait a moment, what happens if i insist that g(0) = 0 ?

That doesn't sound consistent to me. The left-hand side is nonzero.

The three equations that I have for this collision are:

1. 2 g(v) m = M g(0) (from energy conservation in the COM frame)
   
2. 2 \gamma m = M (from momentum conservation in another frame)
3. m (g(v) - g(0)) = \frac{1}{2} v^2 + higher-order terms (from nonrelativistic limit)

Then this equation does not make any sense, unless g(v) = 0 which is trivial solution.

That's why I conclude g(0) \neq 0.

Should there be then some additional energy after the collision in the forms of photons, some other particles, potential energy etc such that:
E_{something else} = 2 g(v) m
and total momentum:
p_{something else} = 0
to make the larger mass stationary? or is it just impossible possibility that g(0) = 0 ?

As I said in the post, the argument assumes that energy is proportional to mass, and works out the proportionality. That's actually a pretty strong assumption, because in nonrelativistic physics, it's only true for particles with no internal energy.

On the other hand, it's possible it's just a matter of definition. If you want to allow internal energy, then that internal energy will be associated with a "pseudo-mass", that will also have momentum.

 

[stevendaryl]

If you're not going to post the definitions you want to use (this is the absolute minimum that would be required from any poster), you're not going to get a better reply than you already got from me.

I understand that you have to define energy in order to prove anything about it. But you can look at it from the point of view of collisions: If momentum is conserved in every frame, then so is the quantity E_{total} = \sum_j m_j \gamma_j c^2. Whether that's the energy or not is a matter of definition, but it's certainly an interest quantity by the fact that it's conserved.

 

[DrStupid]

Then, you can try the guess that the energy of a free particle has the form E(v) = g(v) m for some unknown function g(v), which depends on the magnitude of velocity (but not the direction).

This is an interesting derivation, but the basic assumption is problematic. In classical mechanics it results in an velocity-independent energy which is obviously wrong. Could you show that E:=g(v)·m is allowed in special relativity if we wouldn't know that the result is correct?

 

[DrStupid]

Ok thanks:) Yes it is definite integral so there may be a constant term .

And the question is, how do you get this term? I still do not understand how your calculation results in Eo=m·c².

 

[Fredrik]

I understand that you have to define energy in order to prove anything about it. But you can look at it from the point of view of collisions: If momentum is conserved in every frame, then so is the quantity E_{total} = \sum_j m_j \gamma_j c^2. Whether that's the energy or not is a matter of definition, but it's certainly an interest quantity by the fact that it's conserved.

Yes, I agree. Another reason to think that this $E_{\text{tot}}$ is an interesting quantity is that the change in it equals the work required to effect that change. This is a reason to think of it as energy.

Just to make sure that there's no misunderstanding: My comment to the other guy doesn't in any way reflect my opinion of your post #14. I have to admit that I still haven't read it. I will take a look later, when I have more time.

 

[Fredrik]

the basic assumption is problematic. In classical mechanics it results in an velocity-independent energy

No, it doesn't. $g(v)=\frac 1 2 v^2$.

Could you show that E:=g(v)·m is allowed in special relativity if we wouldn't know that the result is correct?

Not without a definition of E.

 

[stevendaryl]

This is an interesting derivation, but the basic assumption is problematic. In classical mechanics it results in an velocity-independent energy which is obviously wrong.

No, the guess that E(v) = g(v) m is true classically, as well: In that case, g(v) = \frac{1}{2} v^2

Could you show that E:=g(v)·m is allowed in special relativity if we wouldn't know that the result is correct?

My derivation (I skipped the derivation for momentum, but it's not that different) starts off assuming that in relativistic collisions of point-particles (no interactions other than collisions), that there are two conserved additive quantities: A momentum, which is assumed to be a velocity dependent vector quantity in the same direction as the velocity and proportional to mass, and an energy, which is assumed to be a velocity dependent scalar quantity, also proportional to mass. (Of course vector and scalar here mean 3-vector and 3-scalar, not 4-vector and 4-scalar). Then you demand that the laws governing such collisions be invariant under rotations and translations and Lorentz transformations.

Why make these assumptions? As far as my derivation, it's just a guess, inspired by the corresponding facts about (elastic) collisions in Newtonian physics, where the momentum is given by: \vec{p}(v) = m \vec{v} and the energy is given by: E(v) = \frac{1}{2} m v^2. As I noted in my derivation, in the Newtonian case, the energy E(v) is only conserved for elastic collisions (you can fix this by allowing internal heat energy, but this energy is not assumed to be proportional to mass), so I agree that it's a little unmotivated to assume that the E(v) is conserved, even in inelastic collisions.

 

[DrStupid]

No, the guess that E(v) = g(v) m is true classically, as well: In that case, g(v) = \frac{1}{2} v^2

No, it isn't. In classical mechanics your calculation results in g(v) = g(0). With g(v) = v²/2 you would get E = m·v² = M·0²/2 = 0 in the rest system of the center of mass which makes no sense either. Considering the change of internal energy during the inelastic collision wouldn't solve that problem.

My derivation (I skipped the derivation for momentum, but it's not that different) starts off assuming that in relativistic collisions of point-particles (no interactions other than collisions), that there are two conserved additive quantities: A momentum, which is assumed to be a velocity dependent vector quantity in the same direction as the velocity and proportional to mass, and an energy, which is assumed to be a velocity dependent scalar quantity, also proportional to mass.

In classical mechanics the same assumptions fail in combination with E=g(v)·m. What makes you sure that they work in special relativity?

Don't get me wrong. The result shows that your initial assumption make sense and I can explain why it doesn't work in classical mechanics. But how could we show that without the knowledge of the correct result?

 

[stevendaryl]

No, it isn't. In classical mechanics your calculation results in g(v) = g(0).

Okay, I understand your point, now. Classically, the velocity-dependent energy E=\frac{1}{2} m v^2 is not the total energy and is not conserved in an inelastic collision (sort of by definition of "inelastic").

 

[DrStupid]

Okay, I understand your point, now. Classically, the velocity-dependent energy is not conserved in an inelastic collision (sort of by definition of "inelastic").

The problem is the rest energy. Galilean transformation is the limit of Lorentz transformation for c \to \infty. That implies that the rest energy Eo=m·c² is not limited in classical mechanics with the result that there is no relation between mass and energy as we know it in relativity.

 

[stevendaryl]

The problem is the rest energy. Galilean transformation is the limit of Lorentz transformation for c \to \infty. That implies that the rest energy Eo=m·c² is not limited in classical mechanics with the result that there is no relation between mass and energy as we know it in relativity.

Yes. So it's a little puzzling why SR forces a tighter connection between m and E. Or actually, since "mass" is really nothing more than a parameter in the expressions for energy and momentum, it could be phrased this way: SR forces a tighter connection between E and p than exists classically. Classically, you can change the energy of a moving object (by heating it, for example) without affecting its momentum, but not relativistically.

 

[sami_m]

That doesn't sound consistent to me. The left-hand side is nonzero.

The three equations that I have for this collision are:

1. 2 g(v) m = M g(0) (from energy conservation in the COM frame)
   
2. 2 \gamma m = M (from momentum conservation in another frame)
3. m (g(v) - g(0)) = \frac{1}{2} v^2 + higher-order terms (from nonrelativistic limit)

As I said in the post, the argument assumes that energy is proportional to mass, and works out the proportionality. That's actually a pretty strong assumption, because in nonrelativistic physics, it's only true for particles with no internal energy.

On the other hand, it's possible it's just a matter of definition. If you want to allow internal energy, then that internal energy will be associated with a "pseudo-mass", that will also have momentum.

So this derivation is valid when g(0) is not 0

i mean that if i insist that if g(0) = 0 , i see that there is three possibilities:
A) g(0) just can't be zero
B) Energy equation 1 can't be true because it would give result g(v) = 0 and there is something wrong in this equation.
C) M just can't be stationary and in some cases totally inelastic collision is not possible

Do you think that these B) or C) are not realistic possibilities?

You construct the energy equation just by observing that there is 2 particles before the collision and one particle M after the collision, right?

Now for example if there are additional particles or photons present after the collision, then energy equation construction changes such that:
2g(v) m = M g(0) + \sum_{k} E_k

(-and these additional particles have also some momentum which the sum is zero in COM frame)

right?

now, in this case, if i assume that there are more collision products present, g(0) can be zero.

Generally, there should be additional term in energy equation if g(0) = 0

(OR other option is that in some cases, M can't be stationary and that means that totally inelastic collision is not possible.)

and IF M is stationary, and g(0) = 0, now this new energy term should be E = 2 g(v)m.

 

[sami_m]

And the question is, how do you get this term? I still do not understand how your calculation results in Eo=m·c².

There is derivation in wikipedia, which ends to result W = m\gamma c^2 - mc^2
From that i can see that there must be two energy terms present:
W = [m\gamma c^2 + A(v)] - [mc^2 + A(v)]
By looking this work integral i there is still term A(v), but if it is assumed that A(v) = 0,
then the last term becomes mc^2
But it is not clear to me how this A(v) can be evaluated.

 

[stevendaryl]

So this derivation is valid when g(0) is not 0

Okay, but I would put it the other way around: If the derivation is valid, then g(0) \neq 0

i mean that if i insist that if g(0) = 0

Why would you want to insist that?

i see that there is three possibilities:
A) g(0) just can't be zero
B) Energy equation 1 can't be true because it would give result g(v) = 0 and there is something wrong in this equation.
C) M just can't be stationary and in some cases totally inelastic collision is not possible

Do you think that these B) or C) are not realistic possibilities?

Certainly my equation 1 can be wrong. It's an assumption that the total energy is proportional to the mass. Given that assumption, then you can't have g(0) = 0.

Now for example if there are additional particles or photons present after the collision, then energy equation construction changes such that:
2g(v) m = M g(0) + \sum_{k} E_k

(-and these additional particles have also some momentum which the sum is zero in COM frame)
right?
now, in this case, if i assume that there are more collision products present, g(0) can be zero.

Yes, if you assume that there is no collision that starts with two particles and ends up with one particle, then the argument doesn't hold.

 

[Herman Trivilino]

Yes. So it's a little puzzling why SR forces a tighter connection between m and E. Or actually, since "mass" is really nothing more than a parameter in the expressions for energy and momentum, it could be phrased this way: SR forces a tighter connection between E and p than exists classically. Classically, you can change the energy of a moving object (by heating it, for example) without affecting its momentum, but not relativistically.

Isn't it because the connection between space and time is "tighter" in SR than it is in Newtonian physics?

 

[DrStupid]

but if it is assumed that A(v) = 0

Than you don't conclude but only assume that Eo=mc². You could also assume that Eo is zero or infinit or something between.

 

[Herman Trivilino]

Than you don't conclude but only assume that Eo=mc². You could also assume that Eo is zero or infinit or something between.

The problem with anything other than $mc^2$ is that it doesn't match what's observed in nature.

 

[sami_m]

Okay, but I would put it the other way around: If the derivation is valid, then g(0) \neq 0
Why would you want to insist that?

i think, this to be complete derivation, the case g(0) = 0 must be included there also.

For example if g(v)m = B(v) \frac{1}{2} mv^2 (g(v)m is a product of classical kinetic energy and B(v), that is some function of v)

,then g(0) = 0 but if B(v) \neq 0, then g(v) \neq 0 , when v \neq 0
, and there may be B(v) that is nonzero.

and now the derivation for g(v) does not work. Something may be wrong in the setup or in the equations in this case, instead of that there exist no nonzero B(v).

One possibility that comes to mind is that the observation that there is only one particle M after the collision is wrong observation.

Example: if you look how electron and positron form positronium, the result is not a single positronium, it is i think positronium + 2 photons. (electron and positron both emit photon in the process.)

 

[DrStupid]

The problem with anything other than $mc^2$ is that it doesn't match what's observed in nature.

It would be OK, if sami_m would consider these observations in his derivation even though the result would be an empirical formula. Einstein derived it theoretically and predicted corresponding experimental observations.

 

[sami_m]

It would be OK, if sami_m would consider these observations in his derivation even though the result would be an empirical formula. Einstein derived it theoretically and predicted corresponding experimental observations.

I list here how i can evaluate the term A(v) based on my knowledge:

I think A(v) is nonzero is interesting possibility to think about.

1. Observations tell that the momentum has equation p = \gamma mv and kinetic energy E_{kin} = mc^2(\gamma - 1)

1.b But i think this does not have effect on that there may be also nonzero A(v) -term in the equation
W = E_{kin}(v) = [\gamma mc^2 + A(v)] - [mc^2 + A(v)]

2. In high energy particle collisions, it is possible that A(v) approaches zero when v approaches c (And this may have been verified by the observations), but A(v) may be nonzero in low velocities.

3. both terms in the kinetic energy/work equation should be positive or zero.

from 1,2,3, what i can say about A(v) is that if it is nonzero, then it must be at least that A(c) = 0. And the minimum of A(v) is -mc^2 if both energy terms in the equation are positive or zero. the last term disappears when A has minimum. It is likely that A is more or less well behaving increasing function of v. IF the minimum when v=0, then the last term in work equation totally disappears when v=0.

 

[stevendaryl]

i think, this to be complete derivation, the case g(0) = 0 must be included there also.

I don't think so. The assumption that E = m g(v) implies that g(0) = c^2. The other possibility, which is realized classically, is that E is not a function of m and v, that mass and velocity don't uniquely determine the energy.

 

[DrStupid]

2. In high energy particle collisions, it is possible that A(v) approaches zero when v approaches c (And this may have been verified by the observations), but A(v) may be nonzero in low velocities.

That would be a convincing argument. Can you give me a corresponding example and show that the experimental observations require A(v)=0?

 

[sami_m]

I don't think so. The assumption that E = m g(v) implies that g(0) = c^2. The other possibility, which is realized classically, is that E is not a function of m and v, that mass and velocity don't uniquely determine the energy.

thinking about it, the possibility that if g(0) is allowed to be 0, it may be that mg(v) then is not anymore inner property of the particle -it doesn't describe the inner energy of the particle anymore.

But then mg(v) stil might describe total energy that is associated to kinetic energy and mass energy:

Ekin + Emass = mg(v)

now Emass is not constant anymore, it is also function of velocity. Also it would be zero when v=0. but Emass may approach still mc^2 in higher velocities, but it does not in low velocities.

 

[sami_m]

That would be a convincing argument. Can you give me a corresponding example and show that the experimental observations require A(v)=0?

for seeking if A(c) = 0, or A(v) = 0 i guess it can be done by seeking the maximum of the released total energy per one collision ?

 

[sami_m]

so the relativistic form of the energy would be E= \gamma mc^2 +A(v) = \sqrt{p^2c^2 + m^2c^4 }+A(v) , but nobody talks about possible term A(v).
I am not sure about this , but E may have at least observed to have maximum:
E \to \sqrt{p^2c^2 + m^2c^4} , when v \to c
but there may be room to speculate that A(v) is not zero in low velocities.(?)

 

[Herman Trivilino]

I am not sure about this , but E may have at least observed to have maximum:
E \to \sqrt{p^2c^2 + m^2c^4} , when v \to c

As $v \to c$ the total energy increases beyond all bounds. There is no maximum.

 

[sami_m]

As $v \to c$ the total energy increases beyond all bounds. There is no maximum.

yes,that is true i said wrongly - i mean in high energy collisions it may be possible to see at least:
-whether E_{tot} \to \sqrt { p^2c^2 + m^2c^4} when v \to c ( E_{tot} approach asymptotically this Energy formula) and
-whether E_{mass} = E_{tot} - E_{kin} \to mc^2 when v \to c ( whether E_{mass} has maximum mc^2, when v \to c )

but it may be less clear whether this A(v) \neq 0 when v << c ,what would mean also that E_{mass} = mc^2 + A(v) \neq mc^2 when v &lt;&lt; c

The equation for work or kinetic energy from the work integral was:
W = [ m\gamma c^2 + A(v) ] - [mc^2 + A(v) ]

where this A(v) is usually thought to be zero. But i don't know how to prove it.

 

[Herman Trivilino]

i mean in high energy collisions it may be possible to see at least:
-whether E_{tot} \to \sqrt { p^2c^2 + m^2c^4} when v \to c ( E_{tot} approach asymptotically this Energy formula)

There's no asymptote, either. The total energy of an object increases beyond all bounds as its speed relative to the observer approaches $c$.

The equation for work or kinetic energy from the work integral was:
W = [ m\gamma c^2 + A(v) ] - [mc^2 + A(v) ]

The work-energy theorem is a dynamical relation, valid for objects that cannot posses internal energy, for example, a point particle. If you see it introduced as anything more than that, it presents work as something that cannot be generalized to thermodynamic work. This is why you see introductory physics textbooks changing the way the work-energy theorem is treated in dynamics. They want a definition of work that can be generalized to include not just dynamical considerations, but also thermodynamics.

where this A(v) is usually thought to be zero. But i don't know how to prove it.

I don't think it can be "proven". A proof is something that follows from premises, so in that sense there is no information in the end result of a proof that wasn't already present in the premises. To me, the more important consideration is evidence to support it. Can you think of any experiment that could be done to detect the presence of a nonzero value for $A(v)$?