Is the acceleration of two objects with different temperature same with the same force?
up vote
1
down vote
favorite
$F=ma$ is not the case at relativistic speeds. But what is the total energy including mass which resists against force?
general-relativity acceleration
New contributor
Andy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
1
down vote
favorite
$F=ma$ is not the case at relativistic speeds. But what is the total energy including mass which resists against force?
general-relativity acceleration
New contributor
Andy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$F=ma$ is not the case at relativistic speeds. But what is the total energy including mass which resists against force?
general-relativity acceleration
New contributor
Andy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$F=ma$ is not the case at relativistic speeds. But what is the total energy including mass which resists against force?
general-relativity acceleration
general-relativity acceleration
New contributor
Andy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Andy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
Buzz
3,01621323
3,01621323
New contributor
Andy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Andy
62
62
New contributor
Andy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Andy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Andy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
At relativistic speeds it's easier to work with the more fundamental version of Newton's second law, namely $$textbf{F}=frac{d(gamma m textbf{u})}{dt}.$$
$m$ is the body's Lorentz-invariant mass or 'rest mass'. This includes (multiplied by $frac{1}{c^2})$ the body's internal energy' and therefore raising the temperature of a body increases its mass. $textbf{u}$ is the body's velocity and $gamma=(1-frac{u^2}{c^2})^{-frac{1}{2}}.$ [$gamma m textbf{u}$ is the body's momentum.]
The equation has been tested experimentally by observing charged particle trajectories in electric and magnetic fields, and comparing with the predictions of the equation
$$q(textbf{E}+textbf{v} times textbf{B})=frac{d(gamma m textbf{u})}{dt}.$$
You'll recognise the left hand side as the Lorentz force.
A body of larger mass but the same charge will, for example, go in a wider circle in a given magnetic field, just as in Newtonian Physics, but it would be very difficult to detect this happening by raising the temperature of a macroscopic charged body, as the effect on its mass would be so small.
You can derive relativistic forms of $textbf{F} =mtextbf{a}$ from $textbf{F}=frac{d(gamma m textbf{u})}{dt},$ but they are rather messy, involving different values for 'mass' (that is the coefficient of $a$) according to whether the force is parallel to the direction of motion or transverse to it.
add a comment |
up vote
0
down vote
You are right to be suspicious, and indeed (relativistically), the answer is no. The higher-temperature object will experience less acceleration (even - and this is important, in the "relativistically correct" sense of "proper acceleration"), and can thus be considered to have more mass.
The reason for this is that the mass of a system - here a collection of vibrating atoms or molecules - is not necessarily equal to the sum of its parts. It is always at least this much, but typically is more. That added mass is due to various forms of energy possessed by the system, thanks to $E = mc^2$. And one of these is thermal energy (kinetic energy of random vibrations of atoms relative to each other.). It is important to point out, though, that this is a distinct phenomenon from what is called "relativistic mass", which is the notion that the energy of translation of the system as a whole, or perhaps, of its center of mass, should be counted as a form of mass, which makes mass frame dependent. This increased mass is not frame dependent, since no reference frame shift will change the patterns of relative motions of system components with respect to each other, and thus it is a real increase in mass. It is also not best thought of as the result of relativistic mass increases of the individual particles because if we treat any particle individually we have the same concern, rather it's better thought of as inhering within the system as a whole, as an additional term in the total accounting required to compute its mass.
In practice, of course, the contributions are very small, and typically too small to measure. For example, if I have a 1 kg jug of water at room temperature, say 295 K, versus one with the water boiling hot at 373 K, then the difference in mass $Delta m$ can be found by
$$Delta m = frac{m_0 c_mathrm{th} Delta T}{c^2}$$
where $m_0$ is the mass at the "baseline" temperature (here 295 K), thus here $m_0 = 1 mathrm{kg}$, $Delta T$ is the temperature difference, here $373 - 295 = 78 mathrm{K}$, and $c_mathrm{th}$ (subscripted to disambiguate against the Einstein constant, $c$) is the specific heat capacity for water, $4184 frac{mathrm{J}}{mathrm{kg cdot K}}$. With these parameters the $Delta m$ is about $3.6 times 10^{-12} mathrm{kg}$, or $3.6 mathrm{ng}$. This is about the same mass change that would result as from dropping a single human cell (~3 ng) into the original water. I do not believe we have instruments sensitive enough to measure such tiny mass changes on the order of one part in $10^{12}$ as of this writing, but I could be wrong.
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
At relativistic speeds it's easier to work with the more fundamental version of Newton's second law, namely $$textbf{F}=frac{d(gamma m textbf{u})}{dt}.$$
$m$ is the body's Lorentz-invariant mass or 'rest mass'. This includes (multiplied by $frac{1}{c^2})$ the body's internal energy' and therefore raising the temperature of a body increases its mass. $textbf{u}$ is the body's velocity and $gamma=(1-frac{u^2}{c^2})^{-frac{1}{2}}.$ [$gamma m textbf{u}$ is the body's momentum.]
The equation has been tested experimentally by observing charged particle trajectories in electric and magnetic fields, and comparing with the predictions of the equation
$$q(textbf{E}+textbf{v} times textbf{B})=frac{d(gamma m textbf{u})}{dt}.$$
You'll recognise the left hand side as the Lorentz force.
A body of larger mass but the same charge will, for example, go in a wider circle in a given magnetic field, just as in Newtonian Physics, but it would be very difficult to detect this happening by raising the temperature of a macroscopic charged body, as the effect on its mass would be so small.
You can derive relativistic forms of $textbf{F} =mtextbf{a}$ from $textbf{F}=frac{d(gamma m textbf{u})}{dt},$ but they are rather messy, involving different values for 'mass' (that is the coefficient of $a$) according to whether the force is parallel to the direction of motion or transverse to it.
add a comment |
up vote
3
down vote
At relativistic speeds it's easier to work with the more fundamental version of Newton's second law, namely $$textbf{F}=frac{d(gamma m textbf{u})}{dt}.$$
$m$ is the body's Lorentz-invariant mass or 'rest mass'. This includes (multiplied by $frac{1}{c^2})$ the body's internal energy' and therefore raising the temperature of a body increases its mass. $textbf{u}$ is the body's velocity and $gamma=(1-frac{u^2}{c^2})^{-frac{1}{2}}.$ [$gamma m textbf{u}$ is the body's momentum.]
The equation has been tested experimentally by observing charged particle trajectories in electric and magnetic fields, and comparing with the predictions of the equation
$$q(textbf{E}+textbf{v} times textbf{B})=frac{d(gamma m textbf{u})}{dt}.$$
You'll recognise the left hand side as the Lorentz force.
A body of larger mass but the same charge will, for example, go in a wider circle in a given magnetic field, just as in Newtonian Physics, but it would be very difficult to detect this happening by raising the temperature of a macroscopic charged body, as the effect on its mass would be so small.
You can derive relativistic forms of $textbf{F} =mtextbf{a}$ from $textbf{F}=frac{d(gamma m textbf{u})}{dt},$ but they are rather messy, involving different values for 'mass' (that is the coefficient of $a$) according to whether the force is parallel to the direction of motion or transverse to it.
add a comment |
up vote
3
down vote
up vote
3
down vote
At relativistic speeds it's easier to work with the more fundamental version of Newton's second law, namely $$textbf{F}=frac{d(gamma m textbf{u})}{dt}.$$
$m$ is the body's Lorentz-invariant mass or 'rest mass'. This includes (multiplied by $frac{1}{c^2})$ the body's internal energy' and therefore raising the temperature of a body increases its mass. $textbf{u}$ is the body's velocity and $gamma=(1-frac{u^2}{c^2})^{-frac{1}{2}}.$ [$gamma m textbf{u}$ is the body's momentum.]
The equation has been tested experimentally by observing charged particle trajectories in electric and magnetic fields, and comparing with the predictions of the equation
$$q(textbf{E}+textbf{v} times textbf{B})=frac{d(gamma m textbf{u})}{dt}.$$
You'll recognise the left hand side as the Lorentz force.
A body of larger mass but the same charge will, for example, go in a wider circle in a given magnetic field, just as in Newtonian Physics, but it would be very difficult to detect this happening by raising the temperature of a macroscopic charged body, as the effect on its mass would be so small.
You can derive relativistic forms of $textbf{F} =mtextbf{a}$ from $textbf{F}=frac{d(gamma m textbf{u})}{dt},$ but they are rather messy, involving different values for 'mass' (that is the coefficient of $a$) according to whether the force is parallel to the direction of motion or transverse to it.
At relativistic speeds it's easier to work with the more fundamental version of Newton's second law, namely $$textbf{F}=frac{d(gamma m textbf{u})}{dt}.$$
$m$ is the body's Lorentz-invariant mass or 'rest mass'. This includes (multiplied by $frac{1}{c^2})$ the body's internal energy' and therefore raising the temperature of a body increases its mass. $textbf{u}$ is the body's velocity and $gamma=(1-frac{u^2}{c^2})^{-frac{1}{2}}.$ [$gamma m textbf{u}$ is the body's momentum.]
The equation has been tested experimentally by observing charged particle trajectories in electric and magnetic fields, and comparing with the predictions of the equation
$$q(textbf{E}+textbf{v} times textbf{B})=frac{d(gamma m textbf{u})}{dt}.$$
You'll recognise the left hand side as the Lorentz force.
A body of larger mass but the same charge will, for example, go in a wider circle in a given magnetic field, just as in Newtonian Physics, but it would be very difficult to detect this happening by raising the temperature of a macroscopic charged body, as the effect on its mass would be so small.
You can derive relativistic forms of $textbf{F} =mtextbf{a}$ from $textbf{F}=frac{d(gamma m textbf{u})}{dt},$ but they are rather messy, involving different values for 'mass' (that is the coefficient of $a$) according to whether the force is parallel to the direction of motion or transverse to it.
edited 41 mins ago
answered 1 hour ago
Philip Wood
7,4223616
7,4223616
add a comment |
add a comment |
up vote
0
down vote
You are right to be suspicious, and indeed (relativistically), the answer is no. The higher-temperature object will experience less acceleration (even - and this is important, in the "relativistically correct" sense of "proper acceleration"), and can thus be considered to have more mass.
The reason for this is that the mass of a system - here a collection of vibrating atoms or molecules - is not necessarily equal to the sum of its parts. It is always at least this much, but typically is more. That added mass is due to various forms of energy possessed by the system, thanks to $E = mc^2$. And one of these is thermal energy (kinetic energy of random vibrations of atoms relative to each other.). It is important to point out, though, that this is a distinct phenomenon from what is called "relativistic mass", which is the notion that the energy of translation of the system as a whole, or perhaps, of its center of mass, should be counted as a form of mass, which makes mass frame dependent. This increased mass is not frame dependent, since no reference frame shift will change the patterns of relative motions of system components with respect to each other, and thus it is a real increase in mass. It is also not best thought of as the result of relativistic mass increases of the individual particles because if we treat any particle individually we have the same concern, rather it's better thought of as inhering within the system as a whole, as an additional term in the total accounting required to compute its mass.
In practice, of course, the contributions are very small, and typically too small to measure. For example, if I have a 1 kg jug of water at room temperature, say 295 K, versus one with the water boiling hot at 373 K, then the difference in mass $Delta m$ can be found by
$$Delta m = frac{m_0 c_mathrm{th} Delta T}{c^2}$$
where $m_0$ is the mass at the "baseline" temperature (here 295 K), thus here $m_0 = 1 mathrm{kg}$, $Delta T$ is the temperature difference, here $373 - 295 = 78 mathrm{K}$, and $c_mathrm{th}$ (subscripted to disambiguate against the Einstein constant, $c$) is the specific heat capacity for water, $4184 frac{mathrm{J}}{mathrm{kg cdot K}}$. With these parameters the $Delta m$ is about $3.6 times 10^{-12} mathrm{kg}$, or $3.6 mathrm{ng}$. This is about the same mass change that would result as from dropping a single human cell (~3 ng) into the original water. I do not believe we have instruments sensitive enough to measure such tiny mass changes on the order of one part in $10^{12}$ as of this writing, but I could be wrong.
add a comment |
up vote
0
down vote
You are right to be suspicious, and indeed (relativistically), the answer is no. The higher-temperature object will experience less acceleration (even - and this is important, in the "relativistically correct" sense of "proper acceleration"), and can thus be considered to have more mass.
The reason for this is that the mass of a system - here a collection of vibrating atoms or molecules - is not necessarily equal to the sum of its parts. It is always at least this much, but typically is more. That added mass is due to various forms of energy possessed by the system, thanks to $E = mc^2$. And one of these is thermal energy (kinetic energy of random vibrations of atoms relative to each other.). It is important to point out, though, that this is a distinct phenomenon from what is called "relativistic mass", which is the notion that the energy of translation of the system as a whole, or perhaps, of its center of mass, should be counted as a form of mass, which makes mass frame dependent. This increased mass is not frame dependent, since no reference frame shift will change the patterns of relative motions of system components with respect to each other, and thus it is a real increase in mass. It is also not best thought of as the result of relativistic mass increases of the individual particles because if we treat any particle individually we have the same concern, rather it's better thought of as inhering within the system as a whole, as an additional term in the total accounting required to compute its mass.
In practice, of course, the contributions are very small, and typically too small to measure. For example, if I have a 1 kg jug of water at room temperature, say 295 K, versus one with the water boiling hot at 373 K, then the difference in mass $Delta m$ can be found by
$$Delta m = frac{m_0 c_mathrm{th} Delta T}{c^2}$$
where $m_0$ is the mass at the "baseline" temperature (here 295 K), thus here $m_0 = 1 mathrm{kg}$, $Delta T$ is the temperature difference, here $373 - 295 = 78 mathrm{K}$, and $c_mathrm{th}$ (subscripted to disambiguate against the Einstein constant, $c$) is the specific heat capacity for water, $4184 frac{mathrm{J}}{mathrm{kg cdot K}}$. With these parameters the $Delta m$ is about $3.6 times 10^{-12} mathrm{kg}$, or $3.6 mathrm{ng}$. This is about the same mass change that would result as from dropping a single human cell (~3 ng) into the original water. I do not believe we have instruments sensitive enough to measure such tiny mass changes on the order of one part in $10^{12}$ as of this writing, but I could be wrong.
add a comment |
up vote
0
down vote
up vote
0
down vote
You are right to be suspicious, and indeed (relativistically), the answer is no. The higher-temperature object will experience less acceleration (even - and this is important, in the "relativistically correct" sense of "proper acceleration"), and can thus be considered to have more mass.
The reason for this is that the mass of a system - here a collection of vibrating atoms or molecules - is not necessarily equal to the sum of its parts. It is always at least this much, but typically is more. That added mass is due to various forms of energy possessed by the system, thanks to $E = mc^2$. And one of these is thermal energy (kinetic energy of random vibrations of atoms relative to each other.). It is important to point out, though, that this is a distinct phenomenon from what is called "relativistic mass", which is the notion that the energy of translation of the system as a whole, or perhaps, of its center of mass, should be counted as a form of mass, which makes mass frame dependent. This increased mass is not frame dependent, since no reference frame shift will change the patterns of relative motions of system components with respect to each other, and thus it is a real increase in mass. It is also not best thought of as the result of relativistic mass increases of the individual particles because if we treat any particle individually we have the same concern, rather it's better thought of as inhering within the system as a whole, as an additional term in the total accounting required to compute its mass.
In practice, of course, the contributions are very small, and typically too small to measure. For example, if I have a 1 kg jug of water at room temperature, say 295 K, versus one with the water boiling hot at 373 K, then the difference in mass $Delta m$ can be found by
$$Delta m = frac{m_0 c_mathrm{th} Delta T}{c^2}$$
where $m_0$ is the mass at the "baseline" temperature (here 295 K), thus here $m_0 = 1 mathrm{kg}$, $Delta T$ is the temperature difference, here $373 - 295 = 78 mathrm{K}$, and $c_mathrm{th}$ (subscripted to disambiguate against the Einstein constant, $c$) is the specific heat capacity for water, $4184 frac{mathrm{J}}{mathrm{kg cdot K}}$. With these parameters the $Delta m$ is about $3.6 times 10^{-12} mathrm{kg}$, or $3.6 mathrm{ng}$. This is about the same mass change that would result as from dropping a single human cell (~3 ng) into the original water. I do not believe we have instruments sensitive enough to measure such tiny mass changes on the order of one part in $10^{12}$ as of this writing, but I could be wrong.
You are right to be suspicious, and indeed (relativistically), the answer is no. The higher-temperature object will experience less acceleration (even - and this is important, in the "relativistically correct" sense of "proper acceleration"), and can thus be considered to have more mass.
The reason for this is that the mass of a system - here a collection of vibrating atoms or molecules - is not necessarily equal to the sum of its parts. It is always at least this much, but typically is more. That added mass is due to various forms of energy possessed by the system, thanks to $E = mc^2$. And one of these is thermal energy (kinetic energy of random vibrations of atoms relative to each other.). It is important to point out, though, that this is a distinct phenomenon from what is called "relativistic mass", which is the notion that the energy of translation of the system as a whole, or perhaps, of its center of mass, should be counted as a form of mass, which makes mass frame dependent. This increased mass is not frame dependent, since no reference frame shift will change the patterns of relative motions of system components with respect to each other, and thus it is a real increase in mass. It is also not best thought of as the result of relativistic mass increases of the individual particles because if we treat any particle individually we have the same concern, rather it's better thought of as inhering within the system as a whole, as an additional term in the total accounting required to compute its mass.
In practice, of course, the contributions are very small, and typically too small to measure. For example, if I have a 1 kg jug of water at room temperature, say 295 K, versus one with the water boiling hot at 373 K, then the difference in mass $Delta m$ can be found by
$$Delta m = frac{m_0 c_mathrm{th} Delta T}{c^2}$$
where $m_0$ is the mass at the "baseline" temperature (here 295 K), thus here $m_0 = 1 mathrm{kg}$, $Delta T$ is the temperature difference, here $373 - 295 = 78 mathrm{K}$, and $c_mathrm{th}$ (subscripted to disambiguate against the Einstein constant, $c$) is the specific heat capacity for water, $4184 frac{mathrm{J}}{mathrm{kg cdot K}}$. With these parameters the $Delta m$ is about $3.6 times 10^{-12} mathrm{kg}$, or $3.6 mathrm{ng}$. This is about the same mass change that would result as from dropping a single human cell (~3 ng) into the original water. I do not believe we have instruments sensitive enough to measure such tiny mass changes on the order of one part in $10^{12}$ as of this writing, but I could be wrong.
answered 4 mins ago
The_Sympathizer
3,512923
3,512923
add a comment |
add a comment |
Andy is a new contributor. Be nice, and check out our Code of Conduct.
Andy is a new contributor. Be nice, and check out our Code of Conduct.
Andy is a new contributor. Be nice, and check out our Code of Conduct.
Andy is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f446060%2fis-the-acceleration-of-two-objects-with-different-temperature-same-with-the-same%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown