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What do the small terms in the series expansion of relativistic energy mean?
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Through the fabulous Feynman Lectures of Physics and the introduction of relativistic mass, Richard Feynman made a link between the increase in kinetic energy of a heated molecule of gas, and its relativistic increase of mass.
The explanation presented here (http://www.feynmanlectures.caltech.edu/I_15.html section 15.8) takes the expression of the mass as a function of velocity, according to Einstein's relativity:
$$
m = m_0 frac1sqrt1-fracv^2c^2 = m_0 Big(1-fracv^2c^2Big)^-1/2
$$
He then expands it in a power series, which gives the following terms:
$m_0$ which is the mass at rest, $frac12 m_0 v^2 Big(frac1c^2Big)$ which is the apparent increase of mass caused by velocity (kinetic energy divided by $c^2$), and then... negligible terms, which makes sense due to $c^-4$, $c^-6$, etc.
Mathematics being of a great help in such precision physics, I wondered whether the following terms would actually physically represent something, like some sort of other velocity-dependent energy. Actually, due to the small proportion of these terms at low velocities, this would be non-negligible only at very high velocities, like particles very close to the celerity of speed.
My question is about the way to interpret the formula. Should we associate some other energy to $frac38 c^4 m_0 v^4$ and the following terms (but still related to speed...)? Or do we have to consider that the simple formula for kinetic energy $E_k = frac12 m_0 v^2$ was a good approximation with small speeds, but is not totally correct in the relativistic world and shall take also the following terms?
special-relativity mass-energy approximations
edited 20 mins ago
David Z♦
62.4k23135251
asked 6 hours ago
Teuxe
355
add a comment |Â
up vote
2
down vote
favorite
Through the fabulous Feynman Lectures of Physics and the introduction of relativistic mass, Richard Feynman made a link between the increase in kinetic energy of a heated molecule of gas, and its relativistic increase of mass.
The explanation presented here (http://www.feynmanlectures.caltech.edu/I_15.html section 15.8) takes the expression of the mass as a function of velocity, according to Einstein's relativity:
$$
m = m_0 frac1sqrt1-fracv^2c^2 = m_0 Big(1-fracv^2c^2Big)^-1/2
$$
He then expands it in a power series, which gives the following terms:
$m_0$ which is the mass at rest, $frac12 m_0 v^2 Big(frac1c^2Big)$ which is the apparent increase of mass caused by velocity (kinetic energy divided by $c^2$), and then... negligible terms, which makes sense due to $c^-4$, $c^-6$, etc.
Mathematics being of a great help in such precision physics, I wondered whether the following terms would actually physically represent something, like some sort of other velocity-dependent energy. Actually, due to the small proportion of these terms at low velocities, this would be non-negligible only at very high velocities, like particles very close to the celerity of speed.
My question is about the way to interpret the formula. Should we associate some other energy to $frac38 c^4 m_0 v^4$ and the following terms (but still related to speed...)? Or do we have to consider that the simple formula for kinetic energy $E_k = frac12 m_0 v^2$ was a good approximation with small speeds, but is not totally correct in the relativistic world and shall take also the following terms?
special-relativity mass-energy approximations
edited 20 mins ago
David Z♦
62.4k23135251
asked 6 hours ago
Teuxe
355
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Through the fabulous Feynman Lectures of Physics and the introduction of relativistic mass, Richard Feynman made a link between the increase in kinetic energy of a heated molecule of gas, and its relativistic increase of mass.
The explanation presented here (http://www.feynmanlectures.caltech.edu/I_15.html section 15.8) takes the expression of the mass as a function of velocity, according to Einstein's relativity:
$$
m = m_0 frac1sqrt1-fracv^2c^2 = m_0 Big(1-fracv^2c^2Big)^-1/2
$$
He then expands it in a power series, which gives the following terms:
$m_0$ which is the mass at rest, $frac12 m_0 v^2 Big(frac1c^2Big)$ which is the apparent increase of mass caused by velocity (kinetic energy divided by $c^2$), and then... negligible terms, which makes sense due to $c^-4$, $c^-6$, etc.
Mathematics being of a great help in such precision physics, I wondered whether the following terms would actually physically represent something, like some sort of other velocity-dependent energy. Actually, due to the small proportion of these terms at low velocities, this would be non-negligible only at very high velocities, like particles very close to the celerity of speed.
My question is about the way to interpret the formula. Should we associate some other energy to $frac38 c^4 m_0 v^4$ and the following terms (but still related to speed...)? Or do we have to consider that the simple formula for kinetic energy $E_k = frac12 m_0 v^2$ was a good approximation with small speeds, but is not totally correct in the relativistic world and shall take also the following terms?
special-relativity mass-energy approximations
edited 20 mins ago
David Z♦
62.4k23135251
asked 6 hours ago
Teuxe
355
Through the fabulous Feynman Lectures of Physics and the introduction of relativistic mass, Richard Feynman made a link between the increase in kinetic energy of a heated molecule of gas, and its relativistic increase of mass.
The explanation presented here (http://www.feynmanlectures.caltech.edu/I_15.html section 15.8) takes the expression of the mass as a function of velocity, according to Einstein's relativity:
$$
m = m_0 frac1sqrt1-fracv^2c^2 = m_0 Big(1-fracv^2c^2Big)^-1/2
$$
He then expands it in a power series, which gives the following terms:
$m_0$ which is the mass at rest, $frac12 m_0 v^2 Big(frac1c^2Big)$ which is the apparent increase of mass caused by velocity (kinetic energy divided by $c^2$), and then... negligible terms, which makes sense due to $c^-4$, $c^-6$, etc.
Mathematics being of a great help in such precision physics, I wondered whether the following terms would actually physically represent something, like some sort of other velocity-dependent energy. Actually, due to the small proportion of these terms at low velocities, this would be non-negligible only at very high velocities, like particles very close to the celerity of speed.
My question is about the way to interpret the formula. Should we associate some other energy to $frac38 c^4 m_0 v^4$ and the following terms (but still related to speed...)? Or do we have to consider that the simple formula for kinetic energy $E_k = frac12 m_0 v^2$ was a good approximation with small speeds, but is not totally correct in the relativistic world and shall take also the following terms?
special-relativity mass-energy approximations
special-relativity mass-energy approximations
edited 20 mins ago
David Z♦
62.4k23135251
asked 6 hours ago
Teuxe
355
edited 20 mins ago
David Z♦
62.4k23135251
asked 6 hours ago
Teuxe
355
edited 20 mins ago
David Z♦
62.4k23135251
edited 20 mins ago
David Z♦
62.4k23135251
edited 20 mins ago
David Z♦
62.4k23135251
62.4k23135251
asked 6 hours ago
Teuxe
355
asked 6 hours ago
Teuxe
355
asked 6 hours ago
Teuxe
355
355
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
The formula $E_k=frac12m_0v^2$ is an accurate approximation for small values of $v/c$, but it begins to diverge from the correct value as $v$ becomes an appreciable fraction of $c$. The higher order terms you speak of simply measure the extent of the error, which is small when $vll c$ but becomes important as $vrightarrow c$.
Put a different way, the terms you're asking about aren't a different kind of energy - they're just corrections to the kinetic energy which become important at high speeds.
answered 5 hours ago
J. Murray
6,0052519
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The formula $E_k=frac12m_0v^2$ is an accurate approximation for small values of $v/c$, but it begins to diverge from the correct value as $v$ becomes an appreciable fraction of $c$. The higher order terms you speak of simply measure the extent of the error, which is small when $vll c$ but becomes important as $vrightarrow c$.
Put a different way, the terms you're asking about aren't a different kind of energy - they're just corrections to the kinetic energy which become important at high speeds.
answered 5 hours ago
J. Murray
6,0052519
add a comment |Â
up vote
3
down vote
accepted
The formula $E_k=frac12m_0v^2$ is an accurate approximation for small values of $v/c$, but it begins to diverge from the correct value as $v$ becomes an appreciable fraction of $c$. The higher order terms you speak of simply measure the extent of the error, which is small when $vll c$ but becomes important as $vrightarrow c$.
Put a different way, the terms you're asking about aren't a different kind of energy - they're just corrections to the kinetic energy which become important at high speeds.
answered 5 hours ago
J. Murray
6,0052519
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The formula $E_k=frac12m_0v^2$ is an accurate approximation for small values of $v/c$, but it begins to diverge from the correct value as $v$ becomes an appreciable fraction of $c$. The higher order terms you speak of simply measure the extent of the error, which is small when $vll c$ but becomes important as $vrightarrow c$.
Put a different way, the terms you're asking about aren't a different kind of energy - they're just corrections to the kinetic energy which become important at high speeds.
answered 5 hours ago
J. Murray
6,0052519
The formula $E_k=frac12m_0v^2$ is an accurate approximation for small values of $v/c$, but it begins to diverge from the correct value as $v$ becomes an appreciable fraction of $c$. The higher order terms you speak of simply measure the extent of the error, which is small when $vll c$ but becomes important as $vrightarrow c$.
Put a different way, the terms you're asking about aren't a different kind of energy - they're just corrections to the kinetic energy which become important at high speeds.
answered 5 hours ago
J. Murray
6,0052519
edited 33 mins ago
answered 5 hours ago
J. Murray
6,0052519
answered 5 hours ago
J. Murray
6,0052519
answered 5 hours ago
J. Murray
6,0052519
6,0052519
add a comment |Â
add a comment |Â
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