What is the Minkowski Metric and How Does it Relate to Special Relativity?

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The Minkowski metric is fundamental to understanding special relativity, as it defines the geometric structure of spacetime. It features a signature of (-1, 1, 1, 1), indicating the distinction between time and spatial dimensions. The dot product of vectors in Minkowski space incorporates this signature, where the negative term corresponds to the time coordinate. Beginners can grasp relativity concepts without delving deeply into tensor calculus, focusing instead on the physics and operational interpretations of measurements. Resources like Schutz's books and Thorne's course are recommended for foundational learning in theoretical physics.
  • #31
Say i amend the list to this:

1) General Relativity from A to B by Geroch
2) Spacetime Physics by Wheeler
3) The Geometry of Physics: An Introduction by Frankel
4) A First Course in General Relativity by Schutz

Would i be better served by replacing 'The Geometry of Physics' with 'Flat and Curved Space-time'? I don't want to have more than 4 books just now... But because i learn quickly i would like to have two complex book in there, which is why i included the one by Frankel and Schutz... progressing will keep me motivated.
 
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  • #32
If we want to get into what's boring and what's not, Frankel's book is a little plodding, to be honest. Very pedagogically sound, but a little longwinded. It's a good book for remedial reading on particular topics of differential geometry, particularly when reading advanced GR books.

I only know Flat and Curved Spacetime by reputation.

I can heartily recommend Burke's Spacetime, Geometry, Cosmology, which is similar in intent, but again it's out of print, so you'll have to find it at the library. Burke is fun to read and imparts a ton of useful knowledge.

If you're dead set on buying a 4th book, consider Exploring Black Holes, which would make a nice sequel to Spacetime Physics.
 
  • #33
Daverz said:
Sorry to muddy the waters, but Hartle is also worth considering for a first text.
See the "Resources" section of this article on teaching relativity by Robert Wald.

It's too bad Wald (whose article was published in the American Journal of Physics: http://link.aip.org/link/?AJPIAS/74/471/1 ) didn't make it to this recent conference:
AAPT Topical Workshop: "Teaching General Relativity to Undergraduates"
http://www.aapt-doorway.org/TGRU/

check out the articles, talks, and posters
 
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  • #34
aeroboyo said:
So i guess i'd like to know why the y and z coordinates don't appear to vary but only the x and t coordinate of the event.

Landau and Lifschitz (The Classical Theory of Fields), say that "In this case clearly only the coordinate x and time t are subject to change."* What's yer problem :rolleyes:

The following argument is adapted from Rindler, Relativity: Special, General, and Cosmological.

Suppose we have the standard setup of frames S and S', with S' moving in the +x direction with velocity v in the S frame.
Since xz and x'z' planes are always lined up, y=0 must imply y'=0. Since the relationship between y and y' is linear (there's another argument for that), we must have y = By' (so that y=0 when y'=0).

Now consider the following transformation of the coordiates ("xz reversal"):

<br /> x \leftrightarrow -x&#039;, y \leftrightarrow y&#039;, z \leftrightarrow -z&#039;<br />

What this does is reverse the role of S and S'. After the xz reversal, we have S moving in the +x' direction with velocity v in the S' frame. The same argument as above still applies, just with the role of y and y' reversed. So y' = By. Then B=\pm 1, and since y \rightarrow y&#039; continuously as v \rightarrow 0 we must have B=1. The same argument holds for z with an xy reversal.

* To be fair to L&L, the full argument is that every rotation in 4-space can be resolved into six rotations, in the planes xy, xz, yz, xt, yt, and zt, and that S and S' in the standard setup are related by a rotation in the xt plain, which obviously doesn't affect y and z.
 
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  • #35
The Classical theory of Fields by Landau is about special/general relativity right? What do you think of their series in theoretical physics?
 
  • #36
Books, books

Hi, again, aeroboyo,

aeroboyo said:
Say i amend the list to this:

1) General Relativity from A to B by Geroch
2) Spacetime Physics by Wheeler
3) The Geometry of Physics: An Introduction by Frankel
4) A First Course in General Relativity by Schutz

Would i be better served by replacing 'The Geometry of Physics' with 'Flat and Curved Space-time'? I don't want to have more than 4 books just now... But because i learn quickly i would like to have two complex book in there, which is why i included the one by Frankel and Schutz... progressing will keep me motivated.

Clearly you will making your contribution to the economy this shopping season :-/

If you do purchase Spacetime Physics make sure you find a used copy of the FIRST edition (no longer in print) because the second edition drops a key topic ("rapidity", the hyperbolic analog of angle) which you will need to make the table I outlined.

Rob recommended a book which on second thought I also think would be much more appropriate for you right now than Frankel's Geometry of Physics, the book on mathematical methods by Boas. This really impressed me when I first saw it some years ago as covering a very good selection of topics in a well balanced way, plus it has some good problems. Could be a great way to teach yourself a whole lotta math in a hurry.

aeroboyo said:
The Classical theory of Fields by Landau is about special/general relativity right? What do you think of their series in theoretical physics?

This series is sans doubt one of the great classics of the literature, although in places it might be a bit dated. However, I imagine that volumes 1 and 2 at least will never go out of style! Volume 2 (The Classical Theory of Fields) is remarkable for presenting both Maxwell's theory of EM and gtr in one volume. However, this is a graduate text--- on second thought, I'd tend to caution against biting off more than you can chew. Especially since money is tight, you might want to try one of the two popular books, say the one by Geroch, plus Boas, Mathematical Methods.

Chris Hillman
 
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  • #37
hey Criss Hillman,

i have ordered the 1966 first edition of Spacetime physics from amazon, given that a few in this thread have said it's the best version. I'm curious about the significance of this table you've mentioned a few times now? I get the impression that it's a 'path' towards learning this stuff. I have already ordered the titles i listed, but i guess i could cancel the Geometry of Physics one in favour of Boas or Geroch's book. But what really is the difference? Don't all three cover the same things? I think the appeal of the Frankel one to me is that i read it expresses physical laws in terms of geometry (which apparently makes them more lucid and revealing). I'm intrigued by that.
 
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  • #38
aeroboyo said:
The Classical theory of Fields by Landau is about special/general relativity right? What do you think of their series in theoretical physics?

Mechanics was really useful in grad school. I have the next 3 volumes, but haven't used them much.

Do you have the Feynman Lectures yet? Maybe you could ask for the Definitive Edition set for Christmas. o:)
 
  • #39
Feynman Lectures?

Are they a graduate kind of text? Because i think i should work through the books I've already ordered first... as someone pointed out, i'd best not bite off more than i can chew.

Daverz, I am wondering what is the 'geometric' approach to physics? The reviews of Spacetime Physics metioned that it's a 'geometric approach', and I've also seen that description used in other contexts too... what advantages is there to it? cheers.
 
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  • #40
The Feynman Lectures were Richard Feynman's Freshman and Sophomore physics lectures at CalTech.

By geometric, they mean that they use spacetime diagrams and the invariant interval a lot instead of relying solely on algebraic methods.
 
  • #41
I'm just going to begin my self study by reading a book on linear alegbra, that should be a good first step. I actually didn't even know what linear algebra was until a few minutes ago. I assumed it was just like high school algebra and not something that was important! I'm using the online 'Linear Algebra' text by Jim Hefferon and it seems like a good one.

PS does 'Geometry of Physics' by Frankel cover linear alegbra or does it assume prior knowledge of it? I'm only asking because I've noticed that Boar's maths methods text does cover linear algebra (i haven't found a table of contents for Frankels text anywhere which is why I'm asking).
 
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  • #43
whoa, i can now see why the consensus was that Frankels text might not be suitable for a novice like myself... thanks Daverz you've been a great help.

After comparing the table of contents of Boas maths methods text and Frankels text, i can see that Frankels is a much more complex, higher level book. Boas on the other hand covers more generic things like linear alegbra and a lot of other topics. But does Boas text also cover the necessary mutli-variable calculus to start tackling Frankels text?

If it does, perhaps Boas text would give me a decent grounding in maths so i could confidently move onto something more difficult like Frankels text...
 
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  • #44
aeroboyo said:
what is a metric?
aeroboyo, you asked this question in your original post. On the one hand, I don't see that it has been answered, and yet on the other hand, I see that you have progressed in your reading and may have already settled this question in your mind. I attempt to answer it for anyone reading this thread who wants to know. I invite real physicists to correct any errors I make.

As the name metric suggests, it is a way of measuring things. In this case, it is the interval that is being measured. In Euclidean space, we are familiar with the expression:

dL = \sqrt{dx^2 + dy^2 + dz^2}

valid for Euclidean spaces and it might seem natural to extend this to a Euclidean 3+1 space as:

dL = \sqrt{dt^2 + dx^2 + dy^2 + dz^2}

where t is the fourth dimension. However, through no fault of any physicist, this expression has no practical use, except perhaps to mathematicians. The world is not Euclidean. This expression:

dL = \sqrt{-dt^2 + dx^2 + dy^2 + dz^2}

on the other hand is useful in that it is invariant under Lorentz transformations. This is called the Minkowski metric.

This metric depends upon space being flat, which in turn depends upon there being no matter in it. In a space that is curved by the presence of matter, the metric is different, as you will see as you read further. Think of the way we measure large distances on the surface of the Earth. We use great circles rather than straight lines. In this sense, we use a non-Euclidean metric for 3 dimensional space. In GR, you will require a similar generalization of the Minkowski metric.
 
  • #45
I understand the Minkowski metric to be the space-time interval (distance) between two events in flat 4D space. Invarient under Lortenz transformation means that the distance between two events is the same no matter which frame of reference those events are observed from. So it is invarient under Lorentz transforms, but is it invarient under other transformations? Like the Poincare group? If i understand it right, a lorentz transformation is only a rotation in the xt plane, but there can also be rotations in the other plans and translations... I'm learning about matrix notation in linear algebra just now, it's starting to make sense.

Also, just because adding a negative to the time dimension makes the metric invarient under a Lorentz transformation, how can we be sure that that is representive of reality? I read something about how the Maxwell equations were initially found not to be invarient under galilean transformations, and so people thought the Maxwell equations were wrong. They tinkered with Maxwells equations until they were invarient under galilean transformations, but that introduced fields that there's no evidence of in reality.

PS this will sound ignorant, but why the 'd' infront of each term?
 
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  • #46
aeroboyo said:
I understand the Minkowski metric to be the space-time interval (distance) between two events in flat 4D space. Invarient under Lortenz transformation means that the distance between two events is the same no matter which frame of reference those events are observed from. So it is invarient under Lorentz transforms, but is it invarient under other transformations? Like the Poincare group? If i understand it right, a lorentz transformation is only a rotation in the xt plane, but there can also be rotations in the other plans and translations... I'm learning about matrix notation in linear algebra just now, it's starting to make sense.

The metric at a point is actually a tensor that maps two input vectors at that point to an output scalar number... just like the dot-product. Here is some suggestive notation: for 4-vectors \vec V and \vec W
\vec V \cdot \vec W = V^a g_{ab} W^b
(As written, this is not quite the same thing as matrix multiplication... a little more manipulation is needed.)

A rotation in the xt plane is called a Lorentz boost. A general Lorentz transformation is composed of boosts and "ordinary" rotations [and inversions].

aeroboyo said:
Also, just because adding a negative to the time dimension makes the metric invarient under a Lorentz transformation, how can we be sure that that is representive of reality? I read something about how the Maxwell equations were initially found to be invarient under galilean transformations, and so people thought the Maxwell equations were wrong. They tinkered with Maxwells equations until they were invarient under galilean transformations, but that introduced fields that there's no evidence of in reality.

One way to answer your question about "reality" is to first emphasize the operational meaning of things in relativity with Radar measurements (as described in Geroch, Bondi, Ellis-Williams). Then, you'll have a better "physical" feeling as to how the signature (+---) relates to the principle of relativity and the speed-of-light principle.

One really has to spell out in detail what one means by "Maxwell's Equations" to properly identify what set of transformations they are invariant under. In one formulation, one can write them down in terms of differential forms without a metric. So, one really needs to explicitly specify the field equations, constitutive relations, and possibly transformation laws for the fields.

aeroboyo said:
PS this will sound ignorant, but why the 'd' infront of each term?

The d describes an infinitesimal displacement... just like in Euclidean geometry.
 
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  • #47
aeroboyo said:
I understand the Minkowski metric to be the space-time interval (distance) between two events in flat 4D space. Invarient under Lortenz transformation means that the distance between two events is the same no matter which frame of reference those events are observed from. So it is invarient under Lorentz transforms, but is it invarient under other transformations? Like the Poincare group? If i understand it right, a lorentz transformation is only a rotation in the xt plane, but there can also be rotations in the other plans and translations...

PS this will sound ignorant, but why the 'd' infront of each term?
Yes for the Poincare group, no for any transformation outside of this group. Indeed, I believe the definition of the Poincare group is "those transformations that leave the interval invariant".

The Poincare group includes translations along the three axes and rotations about the three axes and, by virtue of being a group, any combination of these. As stated above, the interval is invariant under all members of this group.

The d is for 'differential'. In the examples I gave, they could all be replaced by \Delta and would correctly give distances. However, in curved spaces, they change from point to point and the metric has to be integrated along the path in order to get the distance. Therefore, you can think of them as deltas for now and don't worry about them being 'd's until you get to curved space.

Where this explanation overlaps that of robphy, you would do well to give his priority. As I had said, I am not a physicist.
 
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  • #48
aeroboyo said:
But does Boas text also cover the necessary mutli-variable calculus to start tackling Frankels text?

Looks like Boas's chapters 3-6 would give you what you need.
 
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  • #49
mommy is buying me Boas for Xmas :) Only chapters 3 to 6? I guess if that's the case, Frankel doesn't require prior knowledge of infinite series, complex numbers etc
 
  • #50
jimmysnyder said:
The d is for 'differential'. In the examples I gave, they could all be replaced by \Delta and would correctly give distances. However, in curved spaces, they change from point to point and the metric has to be integrated along the path in order to get the distance.
Sorry, this is wrong.
It is the metric that changes from point to point. It is because the entries in the metric tensor are constants that you can replace d with delta. And it is because they are not constants in curved space that you can't.
 
  • #51
If it changes from point to point, i can see why the 'd' is required, and why 'd' is unecessary when the metric is the same for all points.

I'm currently learning about linear algebra from an excellent online pdf, and I'm at the point where it's explaining that the solution set of a linear system with n unkowns is a linear surface in R^n, the linear surface having k-dimensions, with k being the number of free variables when the linear system is expressed in echelon form. Well i do understand that, but i don't really understand what a 3-dimensional surface in 4D space would look like, for example.

(i'd like to thank every1 for explaining questions that i have while learning this stuff, it's helping me hugely)
 
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  • #52
aeroboyo said:
If it changes from point to point, i can see why the 'd' is required, and why 'd' is unecessary when the metric is the same for all points.

I'm currently learning about linear algebra from an excellent online pdf, and I'm at the point where it's explaining that the solution set of a linear system with n unkowns is a linear surface in R^n, the linear surface having k-dimensions, with k being the number of free variables when the linear system is expressed in echelon form. Well i do understand that, but i don't really understand what a 3-dimensional surface in 4D space would look like, for example.

(i'd like to thank every1 for explaining questions that i have while learning this stuff, it's helping me hugely)

I deeply recommend that you lose the "look-like" meme when dealing with more than 3 dimensions and learn to satisfy yourself with "Well, it's analogous to a surface in three space". This is apparently harder for some people to do than for others, but you'll just be spinnng your wheels until you try.
 
  • #53
Ok. A system of linear equations with 3 unkowns, and 2 free variables would be a 2 dimensional linear surface in 3D space.

That's easy to picture: it's just a plane, like a piece of paper. I guess your right in that once one starts dealing with geometry in 4D and up then one has to be satisfied with not being able to picture it.
 
  • #54
aeroboyo said:
mommy is buying me Boas for Xmas :) Only chapters 3 to 6? I guess if that's the case, Frankel doesn't require prior knowledge of infinite series, complex numbers etc

You should know complex numbers. You can probably leave the functions of a complex variable chapter for later. You'll need the infinite series chapter to understand many solutions to ODEs.
 
  • #55
my first equation post... anyway this is what i was talking about in the previous post... Bellow would be the easy to picture, plane in 3 dimensions.
 
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  • #56
ok so take a quick look at this solution set:

{ \left( {\begin{array}{*{20}c}<br /> 1 \\<br /> 0 \\<br /> 5 \\<br /> \end{array}} \right) + t\left( {\begin{array}{*{20}c}<br /> 1 \\<br /> 1 \\<br /> { - 8} \\<br /> \end{array}} \right) + s\left( {\begin{array}{*{20}c}<br /> { - 3} \\<br /> 4 \\<br /> { - 4.5} \\<br /> \end{array}} \right)|t,s \in \Re \}

The first column vector is one that is its canonical position right... so it goes from the origon to the point (1,0,5). Then the second column vector states that from this point you draw a vector with direction (1,1,-8) at the point (1,0,5) and the magnitude of this vector is the free variable t so it can be any length. Is that correct way to interpret the solution set? I'm trying to write this out because today is the 1st day of learning about the gemoetry of vectors so this helps me learn :)
 
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  • #57
aeroboyo,

...just to keep the focus of a given thread, it might be better to discuss that problem in a new thread in General Math, Linear and Abstract Algebra, or [even if it's not official homework] in the Homework section.
 
  • #58
Ok. The reason i started taking about geometry of vectors here is that i have an inkling that it is relevant to minkowski space... for example, four-velocity... which I'm assuming would be a line (1 dimensional linear surface) in 4 dimension space... something like this:

{ \left( {\begin{array}{*{20}c}<br /> 1 \\<br /> 0 \\<br /> 5 \\<br /> 2 \\<br /> \end{array}} \right) + t\left( {\begin{array}{*{20}c}<br /> 1 \\<br /> 1 \\<br /> { - 8} \\<br /> 3 \\<br /> \end{array}} \right)|t \in \Re \}

could be a four-velocity vector i think... I'm just trying to apply what I'm learning from linear algebra to special relativity as i go along ;)
 
  • #59
That might be a representation in one particular coordinate system, yes.
 
  • #60
I hadn't thought about that, i guess that wouldn't be invariant when transformed into different coordinate systems... just the fact that the first 'canonical' vector in that set defines a point relative to the origon in that particular coordinate system, would mean that in a different coordinate system it would have to be different. I guess that's where expressing things as invariant tensors comes in, which is something i probably won't grasp until I've worked through boas.

I'm assuming that also, that represents the average four-velocity, because t isn't taken as a limmit. I'm guessing that instantaneous four-velocity might be expressed like this:
{ \left( {\begin{array}{*{20}c}<br /> 1 \\<br /> 0 \\<br /> 5 \\<br /> 2 \\<br /> \end{array}} \right) + dt\left( {\begin{array}{*{20}c}<br /> 1 \\<br /> 1 \\<br /> { - 8} \\<br /> 3 \\<br /> \end{array}} \right)|t \in \Re \}
So now the free variable is infintismally small, and therefore the velocity vector would have to be as well. I haven't read this anywhere I'm just guessing.
 
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