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I know there is 4 dimensional objects due to communicating with my friend group. I never found how would you find the volume because it was off the science channel.

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    $\begingroup$ A bunch of comments removed. One should have been posted as an answer. Another was a suggestion for clarification which I'm assuming has been addressed because the question now has an accepted answer. The rest were replies to these, and the replies were kind of going off the rails. People, if somebody's rude to you in the comments, there's no internet law that says you have to win the argument. If you see something particularly unpleasant, flag for moderator attention; then disengage. It'll be okay, I promise. $\endgroup$
    – rob
    Jan 25 '19 at 2:20
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    $\begingroup$ It's not appropriate to use the comments for debates. Have a look a the tour and the help center to see how our reopen process works. It is rare for the diamond moderators to overturn the community's decision on a question. My personal opinion is that the off-topic close vote was a good decision, but I'm glad you got an answer anyway. $\endgroup$
    – rob
    Jan 25 '19 at 2:36
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    $\begingroup$ Luna, I voted to close because the question showed no effort of prior research, and I don't see the question being useful to future readers. I didn't vote because someone else told me to, and the vote was not personal. $\endgroup$ Jan 25 '19 at 3:15
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    $\begingroup$ I don't understand how adding "due to deep research" changes much (v3). Furthermore, I think that it doesn't matter if the objects physically exist or not, since you still have mathematical analogues of volume for higher dimensions, the way you have area for two dimensions. It's more of a math question. You can easily extend the principle of physics.stackexchange.com/a/456457 to an indefinitely large number of dimensions. The only 'problem' would be that those results probably won't be physically meaningful, but that's a completely different subject. $\endgroup$
    – user191954
    Jan 25 '19 at 4:10
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I know there is 4 dimensional objects [...]

I don't know if I would say it this way. There are 4-dimensional regions of spacetime. For example, a book that exists over some amount of time will sweep out a 4-dimensional region of spacetime, but the book is not a 4-dimensional object.

In relativity we don't have a preferred set of coordinates, and therefore it's not trivial to define volume the way we ordinarily would. In a 4-dimensional spacetime, we can define 1-volume (length), 2-volume, 3-volume, and 4-volume. We would like such a definition to have the following properties:

  1. The definition is coordinate-independent (either implicitly or manifestly).

  2. Any two m-volumes can be compared in terms of their ratio.

  3. For any m nonzero vectors, the m-volume of the parallelepiped they span is nonzero if and only if the vectors are linearly independent.

It turns out that the only m-volume that allows a definition simultaneously satisfying all these properties is the 4-volume, which is the one you asked about. It is not self-evident that there is such a definition or what it should be. For a parallelepiped with edges given by vectors $a$, $b$, $c$, and $d$, the 4-volume is defined by

$V=\epsilon_{ijkl}a^ib^jc^kd^l,$

where $\epsilon$ is the Levi-Civita tensor. If the Levi-Civita tensor is defined in one set of coordinates, then its value in other sets of coordinates can be determined by the usual transformation law for a 4-tensor. It does turn out that if you define the Levi-Civita tensor with elements $\pm1$ in one set of Minkowski coordinates, it also has that form after you boost to a different set of Minkowski coordinates. However, this is by no means an obvious property. In general, we have $\epsilon_{ijkl}=\sqrt{|\operatorname{det}g|}$, where $g$ is the metric.

As an example suggested by PM2Ring in a comment, suppose we have a $1\ \text{m}^3$ box, and it exists for 1 s. Since we're working in SI units, the metric is $c^2dt^2-dx^2-...$,. The determinant of the metric is is $-c^2$, and the volume is $3\times10^8\ \text{m}^4$.

There are several other answers that have been posted to the effect that this is somehow trivial, and all we have to do is multiply length x width x height x time. To see that this is not quite right, consider what happens when you try to define 1-volume that way. The 1-volume is simply the length, and we do not have a coordinate-independent way to define the length of a stick. If the stick is a rigid object sweeping out a 2-dimensional ribbon through spacetime, then you can define an instantaneous rest frame and a proper length. It is simply not true that you can naturally define such a thing for any one-dimensional segment. For example, the metric doesn't give us any nontrivial way to measure the length of a lightlike line segment.

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  • $\begingroup$ Can you give a simple numerical example? Eg, in a flat region of spacetime, does a 1m cube over a time interval of 1 second have a 4-volume of -c m^4? $\endgroup$
    – PM 2Ring
    Jan 24 '19 at 21:27
  • $\begingroup$ @PM2Ring: In your example, to get units of m^4, you would have to be using a metric $c^2dt^2-dx^2-...$,. Then the determinant of the metric is is $-c^2$, and the answer is not 1 m^4, it's $3\times10^8\ \text{m}^4$. That's actually a nice example. I've added it to the answer. $\endgroup$
    – user4552
    Jan 24 '19 at 22:07
  • $\begingroup$ Thanks, Ben. I wrote -c m^4 as shorthand for -299792458 $m^4$. I wasn't sure how to handle the metric signature, although I admit I was uncomfortable with negative volume. ;) $\endgroup$
    – PM 2Ring
    Jan 24 '19 at 22:14
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    $\begingroup$ Oh, I see what you meant. The volume is a signed quantity, but the sign is determined by the handedness of the basis vectors you use to span it, just as it would be if you treat area as a signed quantity in the Euclidean plane. The sign isn't determined by the signature of the metric. $\endgroup$
    – user4552
    Jan 24 '19 at 22:24
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    $\begingroup$ "In general, we have $\epsilon_{ijkl} =\sqrt{|\det g|}$" seems to be missing a bit. I think you meant to say "In general, the volume form is $E_{ijkl}=\sqrt{|\det g|} \epsilon_{ijkl}$ with $g$ the metric and $\epsilon$ as above." $\endgroup$
    – tobi_s
    Jan 25 '19 at 1:25
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Why could we not? It is not that hard to generlize the concept of the Riemann-Integral to $\mathbb{R}^n$, and given a measurable subset, we can integrate the characterstic function which gives us the measure of our $n$-dimensional object.

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  • $\begingroup$ ok that makes sense would a black hole be an example of a 4D object? $\endgroup$ Jan 24 '19 at 20:32
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    $\begingroup$ no... we live in a 3d universe. $\endgroup$
    – cxx
    Jan 24 '19 at 20:34
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    $\begingroup$ space-time. Time is a dimension that is separate from the 3 space ones. It does not contribute to volume. $\endgroup$
    – cxx
    Jan 24 '19 at 20:55
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    $\begingroup$ Why a downvote? $\endgroup$
    – James
    Jan 24 '19 at 21:07
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    $\begingroup$ I've removed a string of comments which are about a user's personal theory. Please limit comments to suggestions for improvements to a post. $\endgroup$
    – rob
    Jan 25 '19 at 2:31
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It is an abstraction and we don't think of it as "volume", but people in warehousing do this all the time: if you rent warehouse space, the amount to pay is some rate times unit volume times unit time, that is some "space-time (4D) volume". Of course time is not measured in the same units as space, so in terms of physics this 4D volume is in units of m³·s (or rather days for warehousing!)

The 4D volume calculation is classically done by multiplying the 3D volume of each item by the time it is stored in the warehouse and sum, but you could well imagine to integrate in another order.

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  • $\begingroup$ This answer is not really right, for the same reasons as the answers by James and two other people. $\endgroup$
    – user4552
    Jan 24 '19 at 21:19
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    $\begingroup$ @BenCrowell Well, I understand that asking this question in a physics forum leads one to think of spacetime in relativity, however the OP does not specifically refer to it. $\endgroup$
    – Joce
    Jan 24 '19 at 21:30
  • $\begingroup$ It's somewhat unclear, since the OP originally didn't put in any tags and Qmechanic added the relativity-related tags. But if she's not actually asking a relativity question, then this isn't even the appropriate site for the question. $\endgroup$
    – user4552
    Jan 24 '19 at 22:11
  • $\begingroup$ it is related to space because 4D is space-time and there are 2 parts to space-time @BenCrowell $\endgroup$ Jan 25 '19 at 0:56
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$V=\int_{V} dx_1dx_2dx_3dx_4$, isn't it?

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  • $\begingroup$ I have no clue which is why I am asking I think you are be right $\endgroup$ Jan 24 '19 at 20:47
  • $\begingroup$ can you please give me an example like a 4 D cube like object is $\endgroup$ Jan 24 '19 at 20:50
  • $\begingroup$ 4-dimentional space is an abstraction, an extension of our 3D space. The volume of a 4D cube is $L^4$. $\endgroup$ Jan 24 '19 at 20:55
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    $\begingroup$ This requires more justification, for the same reasons as James's answer. $\endgroup$
    – user4552
    Jan 24 '19 at 21:08
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    $\begingroup$ @Luna 4D objects can be handled by math. It is not so hard as you would think. The problem is that we don't know, what you know from it already, and you need to learn a lot. Ask for more specific 4D objects, for example, "How to calculate the volume of a 4-sphere" on the Math SE. Then learn some analysis. $\endgroup$
    – peterh
    Jan 28 '19 at 23:16

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