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Dimensional Analysis of Energy in higher dimensions
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In 3 spatial dimensions, $$[E] = [ML^2 T^-2]$$
Would it change in higher dimensions? If yes, then what would be the dimensions for 4 spatial dimensions?
units mass-energy dimensional-analysis spacetime-dimensions
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Qmechanic♦
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In 3 spatial dimensions, $$[E] = [ML^2 T^-2]$$
Would it change in higher dimensions? If yes, then what would be the dimensions for 4 spatial dimensions?
units mass-energy dimensional-analysis spacetime-dimensions
edited 1 hour ago
Qmechanic♦
97.8k121651055
asked 2 hours ago
Shubham Gothwal
113
New contributor
Shubham Gothwal 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 |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
In 3 spatial dimensions, $$[E] = [ML^2 T^-2]$$
Would it change in higher dimensions? If yes, then what would be the dimensions for 4 spatial dimensions?
units mass-energy dimensional-analysis spacetime-dimensions
edited 1 hour ago
Qmechanic♦
97.8k121651055
asked 2 hours ago
Shubham Gothwal
113
New contributor
Shubham Gothwal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In 3 spatial dimensions, $$[E] = [ML^2 T^-2]$$
Would it change in higher dimensions? If yes, then what would be the dimensions for 4 spatial dimensions?
units mass-energy dimensional-analysis spacetime-dimensions
units mass-energy dimensional-analysis spacetime-dimensions
edited 1 hour ago
Qmechanic♦
97.8k121651055
asked 2 hours ago
Shubham Gothwal
113
New contributor
Shubham Gothwal 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
Qmechanic♦
97.8k121651055
asked 2 hours ago
Shubham Gothwal
113
New contributor
Shubham Gothwal 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
Qmechanic♦
97.8k121651055
edited 1 hour ago
Qmechanic♦
97.8k121651055
edited 1 hour ago
Qmechanic♦
97.8k121651055
97.8k121651055
asked 2 hours ago
Shubham Gothwal
113
New contributor
Shubham Gothwal 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
Shubham Gothwal
113
asked 2 hours ago
Shubham Gothwal
113
113
New contributor
Shubham Gothwal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Shubham Gothwal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Shubham Gothwal 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 |Â
add a comment |Â
1 Answer
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Suppose for a moment that we're specifically interested in the kinetic energy of a single, non-relativistic particle, so that $E=frac12mvecv^2$. I include the vector notation for the velocity here because it's useful to keep in mind that $vecv^2=vecvcdotvecv$.
If the particle is moving in two dimensions, then $vecv=v_xhatx+v_yhaty$ and $vecvcdotvecv=v_x^2+v_y^2$, which has units of velocity squared. In three dimensions, $vecv = v_xhatx+v_yhaty+v_zhatz$ and $vecvcdotvecv = v_x^2+v_y^2+v_z^2$, which still has the units of velocity squared. In four dimensions, $vecv = v_xhatx+v_yhaty+v_zhatz+v_whatw$, and $vecvcdotvecv=v_x^2+v_y^2+v_z^2+v_w^2$, which, again, still has the units of squared velocity. This is because adding two quantities of the same units together does not change their units. Indeed, this holds for any number of dimensions, and suggests that the units of energy should be unchanged.
We can also see this more generally by the formal definition of work, which is the definition of change in energy:
$$W=int_C vecFcdot dvecs$$
for an object being acted on by a force $vecF$ moving along a path $C$ with arclength parameter $dvecs$. This is the formal, most general definition, and holds no matter how many dimensions of space you have. Once again, you can see that this definition involves a dot product. The dot product takes two vectors in any number of dimensions and outputs a one-dimensional quantity (i.e. a number). The work only cares about one component of the force, namely, the one pointing along the one-dimensional path that the particle is taking. No matter how many spatial dimensions this one-dimensional path is embedded in, the work only cares about things happening along one of those dimensions. As such, the units of work should always be the same, and, since work has the same units as energy (otherwise we wouldn't be able to add them together), the units of energy should also always be the same.
answered 2 hours ago
probably_someone
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If you like the answer, mark it as accepted!
– bRost03
38 mins ago
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Suppose for a moment that we're specifically interested in the kinetic energy of a single, non-relativistic particle, so that $E=frac12mvecv^2$. I include the vector notation for the velocity here because it's useful to keep in mind that $vecv^2=vecvcdotvecv$.
If the particle is moving in two dimensions, then $vecv=v_xhatx+v_yhaty$ and $vecvcdotvecv=v_x^2+v_y^2$, which has units of velocity squared. In three dimensions, $vecv = v_xhatx+v_yhaty+v_zhatz$ and $vecvcdotvecv = v_x^2+v_y^2+v_z^2$, which still has the units of velocity squared. In four dimensions, $vecv = v_xhatx+v_yhaty+v_zhatz+v_whatw$, and $vecvcdotvecv=v_x^2+v_y^2+v_z^2+v_w^2$, which, again, still has the units of squared velocity. This is because adding two quantities of the same units together does not change their units. Indeed, this holds for any number of dimensions, and suggests that the units of energy should be unchanged.
We can also see this more generally by the formal definition of work, which is the definition of change in energy:
$$W=int_C vecFcdot dvecs$$
for an object being acted on by a force $vecF$ moving along a path $C$ with arclength parameter $dvecs$. This is the formal, most general definition, and holds no matter how many dimensions of space you have. Once again, you can see that this definition involves a dot product. The dot product takes two vectors in any number of dimensions and outputs a one-dimensional quantity (i.e. a number). The work only cares about one component of the force, namely, the one pointing along the one-dimensional path that the particle is taking. No matter how many spatial dimensions this one-dimensional path is embedded in, the work only cares about things happening along one of those dimensions. As such, the units of work should always be the same, and, since work has the same units as energy (otherwise we wouldn't be able to add them together), the units of energy should also always be the same.
answered 2 hours ago
probably_someone
13.6k12249
If you like the answer, mark it as accepted!
– bRost03
38 mins ago
add a comment |Â
up vote
4
down vote
Suppose for a moment that we're specifically interested in the kinetic energy of a single, non-relativistic particle, so that $E=frac12mvecv^2$. I include the vector notation for the velocity here because it's useful to keep in mind that $vecv^2=vecvcdotvecv$.
If the particle is moving in two dimensions, then $vecv=v_xhatx+v_yhaty$ and $vecvcdotvecv=v_x^2+v_y^2$, which has units of velocity squared. In three dimensions, $vecv = v_xhatx+v_yhaty+v_zhatz$ and $vecvcdotvecv = v_x^2+v_y^2+v_z^2$, which still has the units of velocity squared. In four dimensions, $vecv = v_xhatx+v_yhaty+v_zhatz+v_whatw$, and $vecvcdotvecv=v_x^2+v_y^2+v_z^2+v_w^2$, which, again, still has the units of squared velocity. This is because adding two quantities of the same units together does not change their units. Indeed, this holds for any number of dimensions, and suggests that the units of energy should be unchanged.
We can also see this more generally by the formal definition of work, which is the definition of change in energy:
$$W=int_C vecFcdot dvecs$$
for an object being acted on by a force $vecF$ moving along a path $C$ with arclength parameter $dvecs$. This is the formal, most general definition, and holds no matter how many dimensions of space you have. Once again, you can see that this definition involves a dot product. The dot product takes two vectors in any number of dimensions and outputs a one-dimensional quantity (i.e. a number). The work only cares about one component of the force, namely, the one pointing along the one-dimensional path that the particle is taking. No matter how many spatial dimensions this one-dimensional path is embedded in, the work only cares about things happening along one of those dimensions. As such, the units of work should always be the same, and, since work has the same units as energy (otherwise we wouldn't be able to add them together), the units of energy should also always be the same.
answered 2 hours ago
probably_someone
13.6k12249
If you like the answer, mark it as accepted!
– bRost03
38 mins ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Suppose for a moment that we're specifically interested in the kinetic energy of a single, non-relativistic particle, so that $E=frac12mvecv^2$. I include the vector notation for the velocity here because it's useful to keep in mind that $vecv^2=vecvcdotvecv$.
If the particle is moving in two dimensions, then $vecv=v_xhatx+v_yhaty$ and $vecvcdotvecv=v_x^2+v_y^2$, which has units of velocity squared. In three dimensions, $vecv = v_xhatx+v_yhaty+v_zhatz$ and $vecvcdotvecv = v_x^2+v_y^2+v_z^2$, which still has the units of velocity squared. In four dimensions, $vecv = v_xhatx+v_yhaty+v_zhatz+v_whatw$, and $vecvcdotvecv=v_x^2+v_y^2+v_z^2+v_w^2$, which, again, still has the units of squared velocity. This is because adding two quantities of the same units together does not change their units. Indeed, this holds for any number of dimensions, and suggests that the units of energy should be unchanged.
We can also see this more generally by the formal definition of work, which is the definition of change in energy:
$$W=int_C vecFcdot dvecs$$
for an object being acted on by a force $vecF$ moving along a path $C$ with arclength parameter $dvecs$. This is the formal, most general definition, and holds no matter how many dimensions of space you have. Once again, you can see that this definition involves a dot product. The dot product takes two vectors in any number of dimensions and outputs a one-dimensional quantity (i.e. a number). The work only cares about one component of the force, namely, the one pointing along the one-dimensional path that the particle is taking. No matter how many spatial dimensions this one-dimensional path is embedded in, the work only cares about things happening along one of those dimensions. As such, the units of work should always be the same, and, since work has the same units as energy (otherwise we wouldn't be able to add them together), the units of energy should also always be the same.
answered 2 hours ago
probably_someone
13.6k12249
Suppose for a moment that we're specifically interested in the kinetic energy of a single, non-relativistic particle, so that $E=frac12mvecv^2$. I include the vector notation for the velocity here because it's useful to keep in mind that $vecv^2=vecvcdotvecv$.
If the particle is moving in two dimensions, then $vecv=v_xhatx+v_yhaty$ and $vecvcdotvecv=v_x^2+v_y^2$, which has units of velocity squared. In three dimensions, $vecv = v_xhatx+v_yhaty+v_zhatz$ and $vecvcdotvecv = v_x^2+v_y^2+v_z^2$, which still has the units of velocity squared. In four dimensions, $vecv = v_xhatx+v_yhaty+v_zhatz+v_whatw$, and $vecvcdotvecv=v_x^2+v_y^2+v_z^2+v_w^2$, which, again, still has the units of squared velocity. This is because adding two quantities of the same units together does not change their units. Indeed, this holds for any number of dimensions, and suggests that the units of energy should be unchanged.
We can also see this more generally by the formal definition of work, which is the definition of change in energy:
$$W=int_C vecFcdot dvecs$$
for an object being acted on by a force $vecF$ moving along a path $C$ with arclength parameter $dvecs$. This is the formal, most general definition, and holds no matter how many dimensions of space you have. Once again, you can see that this definition involves a dot product. The dot product takes two vectors in any number of dimensions and outputs a one-dimensional quantity (i.e. a number). The work only cares about one component of the force, namely, the one pointing along the one-dimensional path that the particle is taking. No matter how many spatial dimensions this one-dimensional path is embedded in, the work only cares about things happening along one of those dimensions. As such, the units of work should always be the same, and, since work has the same units as energy (otherwise we wouldn't be able to add them together), the units of energy should also always be the same.
answered 2 hours ago
probably_someone
13.6k12249
answered 2 hours ago
probably_someone
13.6k12249
answered 2 hours ago
probably_someone
13.6k12249
answered 2 hours ago
probably_someone
13.6k12249
13.6k12249
If you like the answer, mark it as accepted!
– bRost03
38 mins ago
add a comment |Â
If you like the answer, mark it as accepted!
– bRost03
38 mins ago
If you like the answer, mark it as accepted!
– bRost03
38 mins ago
If you like the answer, mark it as accepted!
– bRost03
38 mins ago
add a comment |Â
Shubham Gothwal is a new contributor. Be nice, and check out our Code of Conduct.
Shubham Gothwal is a new contributor. Be nice, and check out our Code of Conduct.
Shubham Gothwal is a new contributor. Be nice, and check out our Code of Conduct.
Shubham Gothwal is a new contributor. Be nice, and check out our Code of Conduct.
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