Why is (ct)^2 negative in 4-dimensional space-time?

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In summary, the distance between two points in 2 dimensions, and in 3 dimensions, is sqrt(x^2 + y^2) . But in 4 dimensions, with the 4th being the change in time, why is it that sqrt(x^2 + y^2 + z^2 - (ct)^2) is the correct distance?
  • #1
nhmllr
185
1
I understand that the distance between 2 points in 2 dimensions
sqrt(x^2 + y^2)
and in 3 dimensions it's
sqrt(x^2 + y^2 + z^2)
but in 4 dimensions, with the 4th being the change in time, why is it
sqrt(x^2 + y^2 + z^2 - (ct)^2) ?

Can somebody please explain?
 
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  • #2
Ah, yes! This is perhaps the most important negative sign in all relativity! If it were just normal (ct)^2, then it would be identical to 4-dimensional Euclidean space. I guess it boils down to the fact that time is NOT just another spatial dimension -- there is something unique about it.

If you recall from elementary discussions of the line element, ds^2, what we're really trying to identify is some invariant quantity -- that is, something all inertial observers will agree on. The simple fact is that if you do not use a negative sign, it is not an invariant quantity.

More mathematically speaking, the negative sign comes from the fact that relativity is limited to special sub-class of all riemannian manifolds known as Lorentzian manifolds (those that have - + ++ signature). The language of general relativity (and consequently, special relativity), is formulated in terms of these Lorentzian manifolds which make a distinction between null, timelike, and spacelike events.
 
  • #3
Good question, I wouldn't be able to come up with a derivation for that expression, but it has to do with the fact that you want to get an expression for meter out of the temporal part of the spacetime interval equation. It also accounts for Lorentz transformation too, imagine that x, y, and z are actually vxt, vyt and vzt, now time is ct. If you plug in c for v, you'll get S2= (ct)2-(ct)2 = 0 And I think this is due to length contraction and/or time dilation at relativistic speeds.
 
  • #4
i don't recall the derivation in a detailed manner, but if you consider time as a fourth dimension when trying to derive the lorentz factor for time dilation using a light clock you will get -c^2t^2. I assume it works similarly for the lorentz factor for length contraction, the lorentz factor being gamma.
 
  • #5
Nabeshin said:
Ah, yes! This is perhaps the most important negative sign in all relativity! If it were just normal (ct)^2, then it would be identical to 4-dimensional Euclidean space. I guess it boils down to the fact that time is NOT just another spatial dimension -- there is something unique about it.
Except, it's not that straight forward. If you have two reference frames that are moving relative to each other, time coordinate of one has projection onto spatial coordinates of the other, and in both, you'll have exactly the same negative sign in the metric. So it isn't something unique about a particular direction in space-time. It's something about the structure of space-time that gives you an ability to pick one of infinite number of possible directions to be different than the rest of the directions you are going to pick.

And yeah, this doesn't pack very well in a brain that's used to Eucledian space. I've seen pure mathematicians get downright upset when I suggested that a metric does not have to be positive.
 

1. What does -(ct)^2 represent in physics?

-(ct)^2 is a mathematical term used in the special theory of relativity to represent the time component of spacetime, where c is the speed of light and t is time. It is often referred to as the spacetime interval.

2. Why is -(ct)^2 used in the special theory of relativity?

-(ct)^2 is used in the special theory of relativity to account for the effects of time dilation and length contraction, which are fundamental principles in understanding the behavior of objects moving at high speeds.

3. How is -(ct)^2 related to the famous equation E=mc^2?

E=mc^2 is a special case of -(ct)^2, where the distance component is set to 0. This equation is used to calculate the energy equivalent of a mass, and it is derived from the more general spacetime interval equation -(ct)^2.

4. Can -(ct)^2 be positive?

No, -(ct)^2 is always negative. This is because in the special theory of relativity, time is treated as a 4th dimension and is denoted by a negative sign in the spacetime interval equation, while the other 3 dimensions (x, y, z) are denoted by positive signs.

5. How is -(ct)^2 used in practical applications?

-(ct)^2 is used in various practical applications, such as GPS systems, particle accelerators, and nuclear reactors. It helps in accurately predicting the behavior of moving objects and in understanding the effects of time and space on them.

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