Caitlyn Cole

2022-04-26

Why the E of the time component 4-momentum is the total energy and not another?

The time component of the 4-momentum is $E/c$, and I saw that it is the "total energy" and from here we can derive the formula ${E}^{2}=(pc{)}^{2}+{m}^{2}{c}^{4}$

$E$ is the total energy? Can't it be some multiple of it or just some other energy?

The time component of the 4-momentum is $E/c$, and I saw that it is the "total energy" and from here we can derive the formula ${E}^{2}=(pc{)}^{2}+{m}^{2}{c}^{4}$

$E$ is the total energy? Can't it be some multiple of it or just some other energy?

nrgiizr0ib6

Beginner2022-04-28Added 1 answers

No, it can't be any other energy. But it is interesting to derive the equation. The fact is that the equation for the relation of energy and momentum comes from special relativity, and only after that can we write the relationship as a 4 vector, with m being the rest mass, independent of coordinate frame (i.e., in whatever frame you are you still use the rest mass. I'll not use the subscript zero for rest mass). I'll use p for momentum, you can think of it as a 3 vector, where ${p}^{2}$ denotes it's norm squared.

So the fact is the derivation of the equation comes first, then we realize it's a 4 vector. Not the other way around.

The term relativistic mass comes considering the dynamics of a particle where energy imparted on a particle originally at rest is the work done, with force the rate of change of momentum. Then it comes out that

$KE=(\gamma -1)m{c}^{2}$

or total energy

$E=m{c}^{2}$

Then with 3-momentum the relativistic mass times velocity

$cp=c\gamma mv$

one squares and substracts both equations to get

${E}^{2}-(pc{)}^{2}={\gamma}^{2}(m{c}^{2}{)}^{2}-(c\gamma {)}^{2}{p}^{2}=$

${\gamma}^{2}{m}^{2}{c}^{4}[1-{v}^{2}/{c}^{2}]=$

$=(m{c}^{2}{)}^{2}{\gamma}^{2}[1-{v}^{2}/{c}^{2}]$

But since ${\gamma}^{2}=1/(1-{v}^{2}/{c}^{2})$ , the entity in the [] parenthesis and the $\gamma $ term together is just 1.

So we get

${E}^{2}-(pc{)}^{2}=(m{c}^{2}{)}^{2}$

Which is the equation for the norm of the 4 vector squared. So you see that the mass term is an invariant, E is the 0 component, and pc the other 3 spatial components of the energy momentum 4 vector.

The equations were derived first in special relativity, then we saw that it was easy to arrange it as a 4 vector.

Alternatively, you can see that one can go the other way by requiring the invariance, and then figuring the momentum and energy both needed the relativistic mass. The relativistic mass is not used much nowadays, but once you get the first term for kinetic energy, and take the total energy to be kinetic plus rest mass energy, it all falls out whether you use the term or not.

Notice also that the equation has the right Newtonian limit (you can see it in the wiki article), and also reproduces the equation for the energy of a zero mass particle, m=0, where E = pc, consistent with the formulas for energy and momentum in terms of frequency and wavelength.

Hope this helps. Just look online for the derivation of the kinetic energy formula.

Best to understand this first, then the 4 vector is just some notation.

So the fact is the derivation of the equation comes first, then we realize it's a 4 vector. Not the other way around.

The term relativistic mass comes considering the dynamics of a particle where energy imparted on a particle originally at rest is the work done, with force the rate of change of momentum. Then it comes out that

$KE=(\gamma -1)m{c}^{2}$

or total energy

$E=m{c}^{2}$

Then with 3-momentum the relativistic mass times velocity

$cp=c\gamma mv$

one squares and substracts both equations to get

${E}^{2}-(pc{)}^{2}={\gamma}^{2}(m{c}^{2}{)}^{2}-(c\gamma {)}^{2}{p}^{2}=$

${\gamma}^{2}{m}^{2}{c}^{4}[1-{v}^{2}/{c}^{2}]=$

$=(m{c}^{2}{)}^{2}{\gamma}^{2}[1-{v}^{2}/{c}^{2}]$

But since ${\gamma}^{2}=1/(1-{v}^{2}/{c}^{2})$ , the entity in the [] parenthesis and the $\gamma $ term together is just 1.

So we get

${E}^{2}-(pc{)}^{2}=(m{c}^{2}{)}^{2}$

Which is the equation for the norm of the 4 vector squared. So you see that the mass term is an invariant, E is the 0 component, and pc the other 3 spatial components of the energy momentum 4 vector.

The equations were derived first in special relativity, then we saw that it was easy to arrange it as a 4 vector.

Alternatively, you can see that one can go the other way by requiring the invariance, and then figuring the momentum and energy both needed the relativistic mass. The relativistic mass is not used much nowadays, but once you get the first term for kinetic energy, and take the total energy to be kinetic plus rest mass energy, it all falls out whether you use the term or not.

Notice also that the equation has the right Newtonian limit (you can see it in the wiki article), and also reproduces the equation for the energy of a zero mass particle, m=0, where E = pc, consistent with the formulas for energy and momentum in terms of frequency and wavelength.

Hope this helps. Just look online for the derivation of the kinetic energy formula.

Best to understand this first, then the 4 vector is just some notation.

albusgks

Beginner2022-04-29Added 2 answers

Consider a particle of charge $q$ in an electric field $\mathbf{e}$. Then from $p=\gamma m\mathbf{v}$ and $E=\gamma m$ (I am using units where $c=1$), if we assume

$\frac{d\mathbf{p}}{dt}=q\mathbf{e}=\mathbf{F},$

we then have

$\frac{dE}{dt}=\mathbf{F}.\mathbf{v}$

In English, the variation of $E$ is the work of the force exerted on the particle, which naturally lead to calling $E$ the energy of the particle. This only defines it up to an additive constant though. But if we require that $(E,\mathbf{p})$ is a 4-vector, ${E}^{2}-{\mathbf{p}}^{2}$ shall have the same value in all inertial frames, which does not leave any freedom.

But more importantly than those theoretical consideration, there are the experimental validations. Consider a particle whose speed is known by time-of-flight measurement, which hits a target in which it stops. With calorimetry, one can figure out how much energy the target gained, energy which must be the kinetic energy of the particle. This measurement matches the relativistic formula ${E}_{c}=(\gamma -1)m$. This has been the bread and butter of particle physics experiment for nearly 100 years. The earliest precision measurement is probably that of Bertozzi [1].

[1] W. Bertozzi. Speed and Kinetic Energy of Relativistic Electrons. American Journal of Physics, 32:551–555, July 1964.

$\frac{d\mathbf{p}}{dt}=q\mathbf{e}=\mathbf{F},$

we then have

$\frac{dE}{dt}=\mathbf{F}.\mathbf{v}$

In English, the variation of $E$ is the work of the force exerted on the particle, which naturally lead to calling $E$ the energy of the particle. This only defines it up to an additive constant though. But if we require that $(E,\mathbf{p})$ is a 4-vector, ${E}^{2}-{\mathbf{p}}^{2}$ shall have the same value in all inertial frames, which does not leave any freedom.

But more importantly than those theoretical consideration, there are the experimental validations. Consider a particle whose speed is known by time-of-flight measurement, which hits a target in which it stops. With calorimetry, one can figure out how much energy the target gained, energy which must be the kinetic energy of the particle. This measurement matches the relativistic formula ${E}_{c}=(\gamma -1)m$. This has been the bread and butter of particle physics experiment for nearly 100 years. The earliest precision measurement is probably that of Bertozzi [1].

[1] W. Bertozzi. Speed and Kinetic Energy of Relativistic Electrons. American Journal of Physics, 32:551–555, July 1964.

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