# I On when the Minkowski metric may be used

#### qtm912

Summary
Can the Minkowski metric be used in flat space in a non inertial, say rotating frame, and related questions
I am trying to get a few concepts straight in my mind. There is no homework question here.
1) If we lived in Minkowski space and had to work in a rotating frame of reference would the Minkowski metric still be the one to use? I assume yes as even if the frame is non inertial the geometry of space time is unchanged and the Minkowski metric is therefore the one to use. Just wanted to check if this was the correct way of thinking about it.
2) Is a Minkowski world consistent with the presence of Newtonian gravity (so even before we introduce ideas from GR). I am thinking here that the presence of Newtonian gravity might in fact alter space time geometry, but not really sure.
3) And what if this was a uniform Newtonian gravitational field in a Minkowski world (unless that is a contradiction). Is the metric is affected and do we then have to use the appropriate g-mu-nu.

As is evident quite a lot of confusion in my mind about how the nature of space time/metric, frames of reference and presence (or absence) of gravitational fields interact.
Thanks.

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#### Ibix

1) If we lived in Minkowski space and had to work in a rotating frame of reference would the Minkowski metric still be the one to use?
Yes. It may be more convenient to express it in terms of a rotating coordinate system, but you deciding to rotate does not make spacetime curve.
2) Is a Minkowski world consistent with the presence of Newtonian gravity
No. Newtonian gravity has an infinite propagation speed, which is inconsistent with relativity which does not allow causation to propagate faster than the speed of light.

Gravitational fields are spacetime curvature. The metric is the mathematical structure that encodes the curvature. A frame of reference is just a coordinate system, one chosen so that the time and space axes are perpendicular (which is not true of every coordinate system).

#### Dale

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If we lived in Minkowski space and had to work in a rotating frame of reference would the Minkowski metric still be the one to use?
The metric is a tensor, meaning that it is a geometric quantity which is the same geometric object regardless of the reference frame. However, in different reference frames the expression in the coordinate basis will be different. So the Minkowski metric will be the same in rotating coordinates but the expression $ds^2=-c^2 dt^2+dx^2+dy^2+dz^2$ will only represent the Minkowski metric in inertial coordinates.

• Orodruin

#### Orodruin

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The metric is a tensor, meaning that it is a geometric quantity which is the same geometric object regardless of the reference frame. However, in different reference frames the expression in the coordinate basis will be different. So the Minkowski metric will be the same in rotating coordinates but the expression $ds^2=-c^2 dt^2+dx^2+dy^2+dz^2$ will only represent the Minkowski metric in inertial coordinates.
As an example, in Rindler coordinates, $T$ and $X$ given by $t = X \sinh(T)$ and $x = X \cosh(T)$, we would have
$$ds^2 = - X^2 dT^2 + dX^2.$$
This is still the Minkowski metric, but in a non-inertial coordinate system.

Compare this to polar coordinates on Euclidean space where the Euclidean metric takes the form
$$ds^2 = dr^2 + r^2 d\varphi^2.$$

• Dale

#### kent davidge

This is still the Minkowski metric, but in a non-inertial coordinate system
why you prefer to say non-inertial coordinate system rather than non-inertial frame?

#### Orodruin

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why you prefer to say non-inertial coordinate system rather than non-inertial frame?
Because a frame is not the same thing as a coordinate system and this deals with issues related to coordinates.

• Dale

#### kent davidge

Because a frame is not the same thing as a coordinate system and this deals with issues related to coordinates.
On another post you said non-inertial coordinates defined non-inertial frames.

#### Orodruin

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On another post you said non-inertial coordinates defined non-inertial frames.
Please link the post. It is not very informative to refer to a post without linking to it.

In general, a non-inertial coordinate system naturally gives a non-inertial frame, but the reverse is not true (i.e., there are non-inertial frames that are not given by a corresponding non-inertial coordinate system).

post number 17

#### Orodruin

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post number 17
The Minkowski coordinates is the inertial frame so clearly it cannot be represented in any other way.
Yes, an inertial frame is a frame corresponding to one naturally defined by Minkowski coordinates. This does not mean any frame is defined by coordinates.

#### Dale

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In general, a non-inertial coordinate system naturally gives a non-inertial frame, but the reverse is not true (i.e., there are non-inertial frames that are not given by a corresponding non-inertial coordinate system).
One example of this is a rotating reference frame. It is very easy to make a rotating reference frame, but turning that into a coordinate system is ambiguous and problematic.

• Orodruin

#### Orodruin

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One example of this is a rotating reference frame. It is very easy to make a rotating reference frame, but turning that into a coordinate system is ambiguous and problematic.
It is also very easy to construct a rotating coordinate system, but the corresponding frame will typically not be orthonormal.

• Dale

#### kent davidge

the corresponding frame will typically not be orthonormal
therefore it represents a non-inertial frame?

#### Orodruin

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therefore it represents a non-inertial frame
Being orthonormal has nothing to do with being non-inertial or not. I can construct an orthonormal frame that is not inertial and I can construct a non-orthonormal one that is inertial.

The Minkowski frames are the orthonormal and inertial frames.

#### pervect

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I've recal seeing three different definitions of a frame of reference, though I don't recall exactly where I saw this discussed. From falllible memory, a "frame of reference" is a bit vague, it could refer to refer to a specific coordinate system, a set of vectors that span a tangent space, or a time-like congruence of worldlines. So for tecnical accuracy, one of the later terms would be less ambiguous, but intelligibility to a broad audience might well be sacrificed by this level of precision. Also, I'm fond of Bohr's dictum: "Never express yourself more clearly than you are able to think".

#### kent davidge

a set of vectors that span a tangent space
Can you please explain, how that is used to discern among different frames? I mean, non-inertial and inertial frames.

Since as in post #14, you can have say, a set of orthonormal vectors and still your frame might be inertial or not. It's ambiguos.

#### Orodruin

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Since as in post #14, you can have say, a set of orthonormal vectors and still your frame might be inertial or not. It's ambiguos.
No it is not. As I have already said, being inertial is not directly related to being orthonormal. Even if you have an orthonormal set of basis vectors, this set can change from event to event.

#### qtm912

My main question has been answered, which I believe is that in flat space the Minkowski metric should be used whether operating from an inertial or non inertial frame. The form of the metric may differ based on the most convenient representation (Rindler example was given) of it but the nature of space time , which the metric encodes, is unchanged.
Newtonian gravity cannot be incorporated into a Minkowski world as it is inconsistent in the sense that it does not allow for special relativity.
I hope that's an accurate summary of the previous posts related to the questions but do correct me if anything said in this post is wrong.

• Dale

#### pervect

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Can you please explain, how that is used to discern among different frames? I mean, non-inertial and inertial frames.

Since as in post #14, you can have say, a set of orthonormal vectors and still your frame might be inertial or not. It's ambiguos.
It's just a definition - every point in a manifold (think of a 2-sphere as an example) has a tangent space (think of a tangent plane to the sphere). Note again that every point has a distinct and different tangent space, i.e. there is a distinct tangent plane at every point on the sphere in our example.

Vectors "live" in the tangent space, because vectors commute. If you walk 500 miles north and 500 miles east on the 2-sphere, you don't wind up in the same spot as if you walk 500 miles east then 500 miles north. Walking north and walking east on a sphere can't be represented by vectors, because vectors have to commute. Vectors, which you can envision as little arrows) do exist in the tangent plane (or in the general case, the tangent space), and they commute as required by the definition of what a vector is.

The tangent space can be described abstractly as a vector space. Specifying a frame singles out a specific set of vectors, called basis vectors, in the tangent space of a manifold that span the tangent space at that point. These basis vectors can be used to define tensors at any point on the manifold.

In our 2-sphere example, a frame of reference would be like a compass, that tells you how to go "north" at any point, and how to go "east".

In a curved space-time, using these definitions, it's not that useful to talk about "inertial" vs "non-inertial" frames of reference.

In a flat space-time, we typically take some shortcuts and lump all the tangent spaces together as being "the same space". I'll leave it as a separate question for why we can get away with this in the case of flat space-time, and why we can't get away with it in a general curved space-time.

What distinguishes the inertial frame of reference from the non-inertial one with this definition of a frame of reference and assuming a flat space-time, is the connection between frames, not the frames themselves. This connection is usually specified by the Christoffel symbols.

• kent davidge