What is the Lorentz-invariant metric for space-time in special relativity?

In summary: The answer is no.does not satisfy the requirement for a metric m that m(t1,r1, t2,r2) = 0 if and only if (t1,r1) = (t2,r2). Isn't this only a requirement for a Riemannian metric? Spacetime instead is described by a pseudo-Riemannian metric.Thank you both for your responses, and also for the welcome from bcrowell :)
  • #1
zlasner
6
0
Hi,

I have been wondering if there is a Lorentz-invariant quantity that satisfies the definition of a metric for space-time.

The space-time interval s2 = t2 - r2 [where r is the vector (x,y,z)] does not satisfy the requirement for a metric m that m(t1,r1, t2,r2) = 0 if and only if (t1,r1) = (t2,r2). For instance, ANY two points on the path of a beam of light have a space-time interval of 0. Also, the space-time interval can be either positive or negative, which violates one of the conditions of a mathematical metric.

From what I've been able to dig up, there's some way to map 4-dimensional space-time coordinates to 5-dimensional hyperbolic space and then there is a Lorentz-invariant metric of hyperbolic space, but I couldn't figure out or find what that mapping is (and this might not even be correct in the first place). If anyone could provide a formula for a Lorentz-invariant metric of space-time or point me in that direction, I'd be much obliged!
 
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  • #2
Welcome to PF!

zlasner said:
I have been wondering if there is a Lorentz-invariant quantity that satisfies the definition of a metric for space-time.
I don't think you want it to be Lorentz-invariant. If it were Lorentz-invariant it would have to be a scalar. The metric is a rank-2 tensor.

zlasner said:
The space-time interval s2 = t2 - r2 [where r is the vector (x,y,z)] does not satisfy the requirement for a metric m that m(t1,r1, t2,r2) = 0 if and only if (t1,r1) = (t2,r2). For instance, ANY two points on the path of a beam of light have a space-time interval of 0. Also, the space-time interval can be either positive or negative, which violates one of the conditions of a mathematical metric.

From what I've been able to dig up, there's some way to map 4-dimensional space-time coordinates to 5-dimensional hyperbolic space and then there is a Lorentz-invariant metric of hyperbolic space, but I couldn't figure out or find what that mapping is (and this might not even be correct in the first place). If anyone could provide a formula for a Lorentz-invariant metric of space-time or point me in that direction, I'd be much obliged!

I think the answer is no.
 
  • #3
zlasner said:
does not satisfy the requirement for a metric m that m(t1,r1, t2,r2) = 0 if and only if (t1,r1) = (t2,r2).
Isn't this only a requirement for a Riemannian metric? Spacetime instead is described by a pseudo-Riemannian metric.
 
  • #4
Thank you both for your responses, and also for the welcome from bcrowell :)

bcrowell, I might have been a imprecise out of ignorance, but when I say "metric" I really do mean a scalar. From http://en.wikipedia.org/wiki/Metric_tensor" [Broken]:
a metric tensor is a type of function defined on a manifold (such as a surface in space) which takes as input a pair of tangent vectors v and w and produces a real number (scalar) g(v,w) in a way that generalizes many of the familiar properties of the dot product of vectors in Euclidean space

So what I mean by "metric" in those terms is the scalar g(v,w), not the tensor that produces it.

Jesse's post was quite to the point, and the comparison of the Riemannian metric and pseudo-Riemannian metric helped to clarify the issue for me. So in light of that, my question might have been more precisely framed as follows: is there any Riemannian metric that can be applied to space-time? (I gather the answer is no.)

My concern about this is mainly physical: if we take the space-time interval to represent something like the "distance" between two events in space-time (an event being defined by its time and position coordinates), and that "distance" can be 0 between apparently different events, doesn't this imply that those events are the same event in some deeper sense? And if we choose not to use the space-time interval as defined above for our "distance" between events in space-time, what should replace it?

I am quite ready to believe that there is no such distance function for space-time (as you've already indicated), but that doesn't make the physical implications any less troubling to me.

Any thoughts would again be much appreciated.
 
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  • #5
zlasner said:
bcrowell, I might have been a imprecise out of ignorance, but when I say "metric" I really do mean a scalar. From http://en.wikipedia.org/wiki/Metric_tensor" [Broken]:

So what I mean by "metric" in those terms is the scalar g(v,w), not the tensor that produces it.
OK, I just don't think that's what people normally mean by "the metric."

zlasner said:
My concern about this is mainly physical: if we take the space-time interval to represent something like the "distance" between two events in space-time (an event being defined by its time and position coordinates), and that "distance" can be 0 between apparently different events, doesn't this imply that those events are the same event in some deeper sense?
No.

zlasner said:
And if we choose not to use the space-time interval as defined above for our "distance" between events in space-time, what should replace it?
There's no good candidate to replace it.
 
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  • #6
zlasner said:
Thank you both for your responses, and also for the welcome from bcrowell :)

bcrowell, I might have been a imprecise out of ignorance, but when I say "metric" I really do mean a scalar. From http://en.wikipedia.org/wiki/Metric_tensor" [Broken]:

So what I mean by "metric" in those terms is the scalar g(v,w), not the tensor that produces it.

Jesse's post was quite to the point, and the comparison of the Riemannian metric and pseudo-Riemannian metric helped to clarify the issue for me. So in light of that, my question might have been more precisely framed as follows: is there any Riemannian metric that can be applied to space-time? (I gather the answer is no.)

My concern about this is mainly physical: if we take the space-time interval to represent something like the "distance" between two events in space-time (an event being defined by its time and position coordinates), and that "distance" can be 0 between apparently different events, doesn't this imply that those events are the same event in some deeper sense? And if we choose not to use the space-time interval as defined above for our "distance" between events in space-time, what should replace it?

I am quite ready to believe that there is no such distance function for space-time (as you've already indicated), but that doesn't make the physical implications any less troubling to me.

Any thoughts would again be much appreciated.

For those with a primarily physics background, this definition of metric is a little strange. However, reading through the whole wikipedia entry, I see it really is equivalent to the more traditional formulation. Please read the rest of this article. Toward the end, metric signature and limitations on 'arclength' are discussed quite well.
 
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  • #7
zlasner said:
From what I've been able to dig up, there's some way to map 4-dimensional space-time coordinates to 5-dimensional hyperbolic space and then there is a Lorentz-invariant metric of hyperbolic space, but I couldn't figure out or find what that mapping is (and this might not even be correct in the first place). If anyone could provide a formula for a Lorentz-invariant metric of space-time or point me in that direction, I'd be much obliged!

These links might be helpful:

http://en.wikipedia.org/wiki/N-sphere

http://iopscience.iop.org/1742-6596/229/1/012038/pdf/1742-6596_229_1_012038.pdf

The usual 4 dimensional Minkowski metric is [itex]ds^2 = x0^2 - x1^2 - x2^2 - x3^2 [/itex]

where x0 is the time dimension and x1,x2 and x3 are the spatial dimensions. If we propose an additional spatial dimension that is on an equal footing with the other 3 spatial dimensions, then would we get away with simply using:

[itex]ds^2 = x0^2 - x1^2 - x2^2 - x3^2 - x4^2 [/itex]

?

or if the additional dimension is an additional temporal dimension would this work:

[itex]ds^2 = x0^2 - x1^2 - x2^2 - x3^2 + x4^2 [/itex]

??

These additional dimensions are not normally detectable, so if they exist, they are probably not on an equal footing with the regular 4 dimensions we are used to and would presumably have to be scaled down in some way.
 
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  • #8
zlasner said:
My concern about this is mainly physical: if we take the space-time interval to represent something like the "distance" between two events in space-time (an event being defined by its time and position coordinates), and that "distance" can be 0 between apparently different events, doesn't this imply that those events are the same event in some deeper sense? And if we choose not to use the space-time interval as defined above for our "distance" between events in space-time, what should replace it?

This is why it is said that relativity splits things into present, past, and elsewhere. Along a timelike curve, the integrated metric is defined to be the time of an observer whose worldline is the curve. This is why it is said that time does not pass for photons (all frequencies of light travel at the same speed, and are "frozen" relative to each other). So when it was found that neutrinos "oscillate", this was due to neutrinos of different frequencies traveling at different speeds, which meant that at least one sort of neutrino had mass and did not travel at the speed of light.
 
  • #9
Thanks, everyone. Your responses and links helped to settle it in my mind in different ways (at least settle it as much as might be done without a bit of time for some of the information to sink in). Although if anyone wants to add something else, of course feel free to do so!
 

1. What is the metric for space-time in special relativity?

The metric for space-time in special relativity is known as the Minkowski metric, named after the German mathematician Hermann Minkowski. It is a four-dimensional metric that combines space and time into a single entity and is used to calculate distances between events in space-time.

2. How is the metric for space-time in SR different from the metric in classical physics?

The metric for space-time in special relativity is different from the metric in classical physics in that it takes into account the effects of time dilation and length contraction. In classical physics, the metric is simply the Pythagorean theorem, but in special relativity, it is a more complex formula that takes into account the relativistic effects of moving at high speeds.

3. What is the significance of the metric in SR?

The metric for space-time in special relativity is significant because it is the mathematical framework that allows us to understand and describe the effects of relativity. It is the basis for many of the key principles of special relativity, such as time dilation, length contraction, and the invariance of the speed of light.

4. Can the metric for space-time in SR be applied to all situations?

The metric for space-time in special relativity is applicable to any situation where the laws of physics are consistent with special relativity. This includes any situation where objects are moving at high speeds or in the presence of strong gravitational fields. However, it may not be applicable in situations involving quantum mechanics or general relativity.

5. How is the metric for space-time in SR used in practical applications?

The metric for space-time in special relativity is used in many practical applications, such as GPS systems, particle accelerators, and satellite communications. It allows for accurate calculations of time and distance in situations where the effects of special relativity must be taken into account. It is also used in theoretical physics to understand the behavior of particles and objects at high speeds.

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