Dtyjlui

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.

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 $3times10^8 text{m}^4$.

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.

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.

$V=int_{V} dx_1dx_2dx_3dx_4$, isn't it?

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.