Would the units of energy be the same if there were a fourth spatial dimension?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
13
down vote

favorite
1












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?










share|cite|improve this question









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 the 4-dimensional case, would the fourth dimension be pretty much the same thing as the first three spatial dimensions?
    – Nat
    7 mins ago














up vote
13
down vote

favorite
1












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?










share|cite|improve this question









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 the 4-dimensional case, would the fourth dimension be pretty much the same thing as the first three spatial dimensions?
    – Nat
    7 mins ago












up vote
13
down vote

favorite
1









up vote
13
down vote

favorite
1






1





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?










share|cite|improve this question









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






share|cite|improve this question









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.











share|cite|improve this question









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.









share|cite|improve this question




share|cite|improve this question








edited 5 mins ago









Nat

3,38931731




3,38931731






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 11 hours ago









Shubham Gothwal

694




694




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.











  • In the 4-dimensional case, would the fourth dimension be pretty much the same thing as the first three spatial dimensions?
    – Nat
    7 mins ago
















  • In the 4-dimensional case, would the fourth dimension be pretty much the same thing as the first three spatial dimensions?
    – Nat
    7 mins ago















In the 4-dimensional case, would the fourth dimension be pretty much the same thing as the first three spatial dimensions?
– Nat
7 mins ago




In the 4-dimensional case, would the fourth dimension be pretty much the same thing as the first three spatial dimensions?
– Nat
7 mins ago










1 Answer
1






active

oldest

votes

















up vote
22
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.






share|cite|improve this answer
















  • 1




    If you like the answer, mark it as accepted!
    – bRost03
    9 hours ago










Your Answer




StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "151"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: false,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);






Shubham Gothwal is a new contributor. Be nice, and check out our Code of Conduct.









 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f433863%2fwould-the-units-of-energy-be-the-same-if-there-were-a-fourth-spatial-dimension%23new-answer', 'question_page');

);

Post as a guest






























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
22
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.






share|cite|improve this answer
















  • 1




    If you like the answer, mark it as accepted!
    – bRost03
    9 hours ago














up vote
22
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.






share|cite|improve this answer
















  • 1




    If you like the answer, mark it as accepted!
    – bRost03
    9 hours ago












up vote
22
down vote










up vote
22
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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 11 hours ago









probably_someone

13.7k12250




13.7k12250







  • 1




    If you like the answer, mark it as accepted!
    – bRost03
    9 hours ago












  • 1




    If you like the answer, mark it as accepted!
    – bRost03
    9 hours ago







1




1




If you like the answer, mark it as accepted!
– bRost03
9 hours ago




If you like the answer, mark it as accepted!
– bRost03
9 hours ago










Shubham Gothwal is a new contributor. Be nice, and check out our Code of Conduct.









 

draft saved


draft discarded


















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.













 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f433863%2fwould-the-units-of-energy-be-the-same-if-there-were-a-fourth-spatial-dimension%23new-answer', 'question_page');

);

Post as a guest













































































Popular posts from this blog

How to check contact read email or not when send email to Individual?

Displaying single band from multi-band raster using QGIS

How many registers does an x86_64 CPU actually have?