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Mass-energy in relativity
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If an objects mass when measured in its frame of reference (even if that frame is moving) is invariant then why does it require more and more energy to accelerate it. I can understand this when measured by an observer in a reference frame in relative motion to it, they would measure the mass as increased but to the moving object, there would be no change. I'm running myself in circles a little bit here but maybe someone can help?
Thank you
David
special-relativity energy mass relativity
asked 3 hours ago
greenplasticdave
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If an objects mass when measured in its frame of reference (even if that frame is moving) is invariant then why does it require more and more energy to accelerate it. I can understand this when measured by an observer in a reference frame in relative motion to it, they would measure the mass as increased but to the moving object, there would be no change. I'm running myself in circles a little bit here but maybe someone can help?
Thank you
David
special-relativity energy mass relativity
asked 3 hours ago
greenplasticdave
332
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If an objects mass when measured in its frame of reference (even if that frame is moving) is invariant then why does it require more and more energy to accelerate it. I can understand this when measured by an observer in a reference frame in relative motion to it, they would measure the mass as increased but to the moving object, there would be no change. I'm running myself in circles a little bit here but maybe someone can help?
Thank you
David
special-relativity energy mass relativity
asked 3 hours ago
greenplasticdave
332
If an objects mass when measured in its frame of reference (even if that frame is moving) is invariant then why does it require more and more energy to accelerate it. I can understand this when measured by an observer in a reference frame in relative motion to it, they would measure the mass as increased but to the moving object, there would be no change. I'm running myself in circles a little bit here but maybe someone can help?
Thank you
David
special-relativity energy mass relativity
special-relativity energy mass relativity
asked 3 hours ago
greenplasticdave
332
asked 3 hours ago
greenplasticdave
332
asked 3 hours ago
greenplasticdave
332
asked 3 hours ago
greenplasticdave
332
asked 3 hours ago
greenplasticdave
332
332
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add a comment |Â
2 Answers
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One needs to carefully distinguish between coordinate acceleration (which is frame dependent) and proper acceleration (which is frame invariant).
Coordinate acceleration is the 2nd derivative with respect to time of the object's position in some (inertial) coordinate system.
Proper acceleration is essentially the acceleration measured by an ideal accelerometer attached to the object - all observers agree on what the accelerometer reads at some event.
In principle, an object can have constant proper acceleration of say, $1,g$. However, it is not possible in principle for an object to have constant coordinate acceleration since that implies the object eventually reaches and then exceeds $c$.
Thus, an object with constant proper acceleration is observed to have arbitrarily small coordinate acceleration as the object's speed gets arbitrarily close to $c$.
As an aside, the notion of a relativistic mass that is speed dependent is generally considered outdated. It's certainly not necessary.
answered 2 hours ago
Alfred Centauri
46.4k344139
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(This answer elaborates on the footnote in Alfred Centauri's answer.)
This is partly a matter of definition. Sometimes the same word "mass" is used with a different meaning, and I think this difference in language may be contributing to the confusion. There are two different quantities associated with the word "mass":
A quantity that physicists usually call "mass", denoted $m$, that is an intrinsic property of the object and does not depend on how it is moving. (I haven't actually conducted a survey of professional physicists, but in my experience, this is how they usually use the word.)
A synonym for the object's energy $E$, but expressed in mass-like units as $E/c^2$. This is sometimes called the object's "relativistic mass", denoted $m_R$, and it does depend on how the object is moving (because the object's energy does).
When the object is not moving, these two different quantities are equal to each other: $m_R=m$. When the "relativistic mass" language is used, the quantity $m$ (which physicists usually just call "mass") is called "rest mass."
Here's the same answer again, with equations to help clarify things:
The energy $E$, momentum $p$, speed $v$, and mass $m$ of an object are related to each other according to these equations:
$$
E^2 - (pc)^2 = (mc^2)^2
hskip2cm
v = fracpc^2E
$$
where $c$ is the speed of light. The $m$ in the first equation is what physicists usually mean when they use the word "mass". It is an intrinsic property of the object and does not depend on the object's speed. The object's energy $E$ and momentum $p$ do depend on the speed, and they do so in such a way that the combination $E^2-(pc)^2$ does not depend on the speed. That's why this particular combination is interesting, and that's why the $m$ on the right-hand side of the equation deserves a special name: mass.
To relate this to the "relativistic mass" $m_R$ (which, again, is not used by the majority of physicists in my experience), re-arrange the second equation shown above to get
$$
p = fracEc^2v.
$$
If we use $m_R$ as an abbreviation for $E/c^2$, then this becomes
$$
p = m_R v,
$$
which looks superficially like the more familiar low-speed approximation $p=mv$. This resemblance might be comforting to some, but it is also misleading, because the energy $E$ (and therefore $m_R$) is a function of $v$. The momentum $p$ is not really proportional to the velocity $v$ (except approximately when $vll c$).
answered 2 hours ago
Dan Yand
5889
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
One needs to carefully distinguish between coordinate acceleration (which is frame dependent) and proper acceleration (which is frame invariant).
Coordinate acceleration is the 2nd derivative with respect to time of the object's position in some (inertial) coordinate system.
Proper acceleration is essentially the acceleration measured by an ideal accelerometer attached to the object - all observers agree on what the accelerometer reads at some event.
In principle, an object can have constant proper acceleration of say, $1,g$. However, it is not possible in principle for an object to have constant coordinate acceleration since that implies the object eventually reaches and then exceeds $c$.
Thus, an object with constant proper acceleration is observed to have arbitrarily small coordinate acceleration as the object's speed gets arbitrarily close to $c$.
As an aside, the notion of a relativistic mass that is speed dependent is generally considered outdated. It's certainly not necessary.
answered 2 hours ago
Alfred Centauri
46.4k344139
add a comment |Â
up vote
3
down vote
One needs to carefully distinguish between coordinate acceleration (which is frame dependent) and proper acceleration (which is frame invariant).
Coordinate acceleration is the 2nd derivative with respect to time of the object's position in some (inertial) coordinate system.
Proper acceleration is essentially the acceleration measured by an ideal accelerometer attached to the object - all observers agree on what the accelerometer reads at some event.
In principle, an object can have constant proper acceleration of say, $1,g$. However, it is not possible in principle for an object to have constant coordinate acceleration since that implies the object eventually reaches and then exceeds $c$.
Thus, an object with constant proper acceleration is observed to have arbitrarily small coordinate acceleration as the object's speed gets arbitrarily close to $c$.
As an aside, the notion of a relativistic mass that is speed dependent is generally considered outdated. It's certainly not necessary.
answered 2 hours ago
Alfred Centauri
46.4k344139
add a comment |Â
up vote
3
down vote
up vote
3
down vote
One needs to carefully distinguish between coordinate acceleration (which is frame dependent) and proper acceleration (which is frame invariant).
Coordinate acceleration is the 2nd derivative with respect to time of the object's position in some (inertial) coordinate system.
Proper acceleration is essentially the acceleration measured by an ideal accelerometer attached to the object - all observers agree on what the accelerometer reads at some event.
In principle, an object can have constant proper acceleration of say, $1,g$. However, it is not possible in principle for an object to have constant coordinate acceleration since that implies the object eventually reaches and then exceeds $c$.
Thus, an object with constant proper acceleration is observed to have arbitrarily small coordinate acceleration as the object's speed gets arbitrarily close to $c$.
As an aside, the notion of a relativistic mass that is speed dependent is generally considered outdated. It's certainly not necessary.
answered 2 hours ago
Alfred Centauri
46.4k344139
One needs to carefully distinguish between coordinate acceleration (which is frame dependent) and proper acceleration (which is frame invariant).
Coordinate acceleration is the 2nd derivative with respect to time of the object's position in some (inertial) coordinate system.
Proper acceleration is essentially the acceleration measured by an ideal accelerometer attached to the object - all observers agree on what the accelerometer reads at some event.
In principle, an object can have constant proper acceleration of say, $1,g$. However, it is not possible in principle for an object to have constant coordinate acceleration since that implies the object eventually reaches and then exceeds $c$.
Thus, an object with constant proper acceleration is observed to have arbitrarily small coordinate acceleration as the object's speed gets arbitrarily close to $c$.
As an aside, the notion of a relativistic mass that is speed dependent is generally considered outdated. It's certainly not necessary.
answered 2 hours ago
Alfred Centauri
46.4k344139
answered 2 hours ago
Alfred Centauri
46.4k344139
answered 2 hours ago
Alfred Centauri
46.4k344139
answered 2 hours ago
Alfred Centauri
46.4k344139
46.4k344139
add a comment |Â
add a comment |Â
up vote
1
down vote
(This answer elaborates on the footnote in Alfred Centauri's answer.)
This is partly a matter of definition. Sometimes the same word "mass" is used with a different meaning, and I think this difference in language may be contributing to the confusion. There are two different quantities associated with the word "mass":
A quantity that physicists usually call "mass", denoted $m$, that is an intrinsic property of the object and does not depend on how it is moving. (I haven't actually conducted a survey of professional physicists, but in my experience, this is how they usually use the word.)
A synonym for the object's energy $E$, but expressed in mass-like units as $E/c^2$. This is sometimes called the object's "relativistic mass", denoted $m_R$, and it does depend on how the object is moving (because the object's energy does).
When the object is not moving, these two different quantities are equal to each other: $m_R=m$. When the "relativistic mass" language is used, the quantity $m$ (which physicists usually just call "mass") is called "rest mass."
Here's the same answer again, with equations to help clarify things:
The energy $E$, momentum $p$, speed $v$, and mass $m$ of an object are related to each other according to these equations:
$$
E^2 - (pc)^2 = (mc^2)^2
hskip2cm
v = fracpc^2E
$$
where $c$ is the speed of light. The $m$ in the first equation is what physicists usually mean when they use the word "mass". It is an intrinsic property of the object and does not depend on the object's speed. The object's energy $E$ and momentum $p$ do depend on the speed, and they do so in such a way that the combination $E^2-(pc)^2$ does not depend on the speed. That's why this particular combination is interesting, and that's why the $m$ on the right-hand side of the equation deserves a special name: mass.
To relate this to the "relativistic mass" $m_R$ (which, again, is not used by the majority of physicists in my experience), re-arrange the second equation shown above to get
$$
p = fracEc^2v.
$$
If we use $m_R$ as an abbreviation for $E/c^2$, then this becomes
$$
p = m_R v,
$$
which looks superficially like the more familiar low-speed approximation $p=mv$. This resemblance might be comforting to some, but it is also misleading, because the energy $E$ (and therefore $m_R$) is a function of $v$. The momentum $p$ is not really proportional to the velocity $v$ (except approximately when $vll c$).
answered 2 hours ago
Dan Yand
5889
add a comment |Â
up vote
1
down vote
(This answer elaborates on the footnote in Alfred Centauri's answer.)
This is partly a matter of definition. Sometimes the same word "mass" is used with a different meaning, and I think this difference in language may be contributing to the confusion. There are two different quantities associated with the word "mass":
A quantity that physicists usually call "mass", denoted $m$, that is an intrinsic property of the object and does not depend on how it is moving. (I haven't actually conducted a survey of professional physicists, but in my experience, this is how they usually use the word.)
A synonym for the object's energy $E$, but expressed in mass-like units as $E/c^2$. This is sometimes called the object's "relativistic mass", denoted $m_R$, and it does depend on how the object is moving (because the object's energy does).
When the object is not moving, these two different quantities are equal to each other: $m_R=m$. When the "relativistic mass" language is used, the quantity $m$ (which physicists usually just call "mass") is called "rest mass."
Here's the same answer again, with equations to help clarify things:
The energy $E$, momentum $p$, speed $v$, and mass $m$ of an object are related to each other according to these equations:
$$
E^2 - (pc)^2 = (mc^2)^2
hskip2cm
v = fracpc^2E
$$
where $c$ is the speed of light. The $m$ in the first equation is what physicists usually mean when they use the word "mass". It is an intrinsic property of the object and does not depend on the object's speed. The object's energy $E$ and momentum $p$ do depend on the speed, and they do so in such a way that the combination $E^2-(pc)^2$ does not depend on the speed. That's why this particular combination is interesting, and that's why the $m$ on the right-hand side of the equation deserves a special name: mass.
To relate this to the "relativistic mass" $m_R$ (which, again, is not used by the majority of physicists in my experience), re-arrange the second equation shown above to get
$$
p = fracEc^2v.
$$
If we use $m_R$ as an abbreviation for $E/c^2$, then this becomes
$$
p = m_R v,
$$
which looks superficially like the more familiar low-speed approximation $p=mv$. This resemblance might be comforting to some, but it is also misleading, because the energy $E$ (and therefore $m_R$) is a function of $v$. The momentum $p$ is not really proportional to the velocity $v$ (except approximately when $vll c$).
answered 2 hours ago
Dan Yand
5889
add a comment |Â
up vote
1
down vote
up vote
1
down vote
(This answer elaborates on the footnote in Alfred Centauri's answer.)
This is partly a matter of definition. Sometimes the same word "mass" is used with a different meaning, and I think this difference in language may be contributing to the confusion. There are two different quantities associated with the word "mass":
A quantity that physicists usually call "mass", denoted $m$, that is an intrinsic property of the object and does not depend on how it is moving. (I haven't actually conducted a survey of professional physicists, but in my experience, this is how they usually use the word.)
A synonym for the object's energy $E$, but expressed in mass-like units as $E/c^2$. This is sometimes called the object's "relativistic mass", denoted $m_R$, and it does depend on how the object is moving (because the object's energy does).
When the object is not moving, these two different quantities are equal to each other: $m_R=m$. When the "relativistic mass" language is used, the quantity $m$ (which physicists usually just call "mass") is called "rest mass."
Here's the same answer again, with equations to help clarify things:
The energy $E$, momentum $p$, speed $v$, and mass $m$ of an object are related to each other according to these equations:
$$
E^2 - (pc)^2 = (mc^2)^2
hskip2cm
v = fracpc^2E
$$
where $c$ is the speed of light. The $m$ in the first equation is what physicists usually mean when they use the word "mass". It is an intrinsic property of the object and does not depend on the object's speed. The object's energy $E$ and momentum $p$ do depend on the speed, and they do so in such a way that the combination $E^2-(pc)^2$ does not depend on the speed. That's why this particular combination is interesting, and that's why the $m$ on the right-hand side of the equation deserves a special name: mass.
To relate this to the "relativistic mass" $m_R$ (which, again, is not used by the majority of physicists in my experience), re-arrange the second equation shown above to get
$$
p = fracEc^2v.
$$
If we use $m_R$ as an abbreviation for $E/c^2$, then this becomes
$$
p = m_R v,
$$
which looks superficially like the more familiar low-speed approximation $p=mv$. This resemblance might be comforting to some, but it is also misleading, because the energy $E$ (and therefore $m_R$) is a function of $v$. The momentum $p$ is not really proportional to the velocity $v$ (except approximately when $vll c$).
answered 2 hours ago
Dan Yand
5889
(This answer elaborates on the footnote in Alfred Centauri's answer.)
This is partly a matter of definition. Sometimes the same word "mass" is used with a different meaning, and I think this difference in language may be contributing to the confusion. There are two different quantities associated with the word "mass":
A quantity that physicists usually call "mass", denoted $m$, that is an intrinsic property of the object and does not depend on how it is moving. (I haven't actually conducted a survey of professional physicists, but in my experience, this is how they usually use the word.)
A synonym for the object's energy $E$, but expressed in mass-like units as $E/c^2$. This is sometimes called the object's "relativistic mass", denoted $m_R$, and it does depend on how the object is moving (because the object's energy does).
When the object is not moving, these two different quantities are equal to each other: $m_R=m$. When the "relativistic mass" language is used, the quantity $m$ (which physicists usually just call "mass") is called "rest mass."
Here's the same answer again, with equations to help clarify things:
The energy $E$, momentum $p$, speed $v$, and mass $m$ of an object are related to each other according to these equations:
$$
E^2 - (pc)^2 = (mc^2)^2
hskip2cm
v = fracpc^2E
$$
where $c$ is the speed of light. The $m$ in the first equation is what physicists usually mean when they use the word "mass". It is an intrinsic property of the object and does not depend on the object's speed. The object's energy $E$ and momentum $p$ do depend on the speed, and they do so in such a way that the combination $E^2-(pc)^2$ does not depend on the speed. That's why this particular combination is interesting, and that's why the $m$ on the right-hand side of the equation deserves a special name: mass.
To relate this to the "relativistic mass" $m_R$ (which, again, is not used by the majority of physicists in my experience), re-arrange the second equation shown above to get
$$
p = fracEc^2v.
$$
If we use $m_R$ as an abbreviation for $E/c^2$, then this becomes
$$
p = m_R v,
$$
which looks superficially like the more familiar low-speed approximation $p=mv$. This resemblance might be comforting to some, but it is also misleading, because the energy $E$ (and therefore $m_R$) is a function of $v$. The momentum $p$ is not really proportional to the velocity $v$ (except approximately when $vll c$).
answered 2 hours ago
Dan Yand
5889
edited 1 hour ago
answered 2 hours ago
Dan Yand
5889
answered 2 hours ago
Dan Yand
5889
answered 2 hours ago
Dan Yand
5889
5889
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