Lack of Symmetry in X & T: Explained for Beginners

[Chenkel]

Hello everyone,

I'm reading "Special Relativity For the Enthusiastic Beginner" (by David Morin) and so far I really like the book, but I have a little bit of a hiccup in understanding terminology regarding stated symmetries or the lack there of in reference to the Galilean transformation when compared to the Lorentz transformation.

The question I have is super basic, but I think it might be important for me to write down the equations in latex, so I will reply to my thread with a post showing the referenced equations.

Feel free to reply if you have any insights, thank you!

 

[PeroK]

The Galilean Transformation is:$$x' = x - vt, \ t' = t$$Note that the format of these two equations is very different.

The Lorentz Transformation is: $$x' = \gamma(x - vt), \ t' = \gamma(t - \frac v {c^2}x)$$And at the first sight the format of these equations is similar, but not the same. If, however, we take $ct$ to be the time parameter, we have:$$x' = \gamma(x - \frac v c (ct)), \ ct' = \gamma(ct - \frac v c x)$$And we see that these equations have precisely the same format.

This "symmetry" between time and space must be seen in context. The universe has three spatial dimensions, but only one time dimension. But, there is clearly a relationship between time and space i SR that is much deeper than is the case in classical, Newtonian/Galilean physics.

 

[Ibix]

PeroK has beaten me to it, so here's a slightly different presentation of the same thing.

If you pick units where $c=1$ then the Lorentz transforms can be written as a matrix $$\Lambda=\left(\begin{array}{cc}
\gamma&-v\gamma\\
-v\gamma&\gamma
\end{array}\right)$$The Galilean transform is a matrix$$G=\left(\begin{array}{cc}
1&0\\
-v&1
\end{array}\right)$$Just by looking at $G$ can you tell if it was meant to be applied to $(x,t)^T$ or to $(t,x)^T$? What about $\Lambda$?

There are, however, differences between space and time that come from there being three spacelike dimensions. But in a (1+1)d model there's a clear symmetry between $x$ and $t$ which isn't the case in Galilean relativity.

 

[Chenkel]

I appreciated all the replies, I've read them and still have some difficulty understanding, that being said I will attempt to study them more now and in the future until I can (hopefully) gain the insight that they are written with.

I think it's important to note that I feel I am not expert in my understanding of the lack or presence of symmetries, what kind of symmetries does the Lorentz transformations have that Galilean transformations do not in laymen's terms if possible?

The stated equation (eq 1.1) $$\Delta x' = \Delta x' + v \Delta t'$$$$\Delta t = \Delta t'$$The book after mentioning this equation goes on to describe how the laws of physics hold in all non inertial frames, i.e, I believe this to mean that regardless of if the velocity is recorded in one inertial frame or another, the acceleration of the point mass will be the same and so it will have the same EOM regardless of choice of inertial reference frames.

The book ends the page with what the author calls a "remark." The remark confuses me a little in its language and I'm trying to understand what it means as I feel it might be important later on.

The remark is the following:
"Note that the Galilean transformations in 1.1 aren't symmetric in x and t. This isn't necessarily a bad thing, but it turns out that it will in fact be a problem in special relativity, where space and time are treated more on an equal footing. We'll find in Section 2.1 that the Galilean transformations are replaced by the Lorentz transformations, and the latter are in fact symmetric in x and t (up to factors of the speed of light, c)"

I was wondering what it means for a transformation to not be symmetric in two variables (x and t).

I've looked up various kinds of symmetries through online searches but I feel lack the mathematical intuition to know what the author means. When it comes to symmetries in math and physics my knowledge is not robust.

The latter part of the remark says that the Lorentz transformations are symmetric in x and t, and I wonder what that means. I think it's important to mention I do not currently have a solid grasp of Lorentz transformations, but I am developing some intuition regarding the matter, hopefully I will understand Lorentz transformations with a reasonable degree of depth at some point.

Lastly the remark ends saying "x and t (up to factors of the speed of light, c)" I'm not sure what this talk about "factors" means, perhaps the author is referring to the well known property that velocity of anything cannot be faster than the speed of light?

Mostly I feel I understand the information in the book that I've read so far (I'm still very early in the book), but I feel that the questions I have regarding this remark as the author likes to call it are possibly important for understanding the rest of the book.

I hope my questions find you well, please feel free to ask any questions of your own that you might have regarding this post.

Looking forward to your wisdom, thank you.

 

[PeroK]

The stated equation (eq 1.1) $$\Delta x' = \Delta x' + v \Delta t'$$$$\Delta t = \Delta t'$$The book after mentioning this equation goes on to describe how the laws of physics hold in all non inertial frames, i.e, I believe this to mean that regardless of if the velocity is recorded in one inertial frame or another, the acceleration of the point mass will be the same and so it will have the same EOM regardless of choice of inertial reference frames.

Newton's second law involves acceleration, which is the second derivative of position and does not depend on the relative position or relative velocity of the particle. I.e. we have:$$\vec F = m \vec a$$which reduces in one dimensional motion to $$F = ma$$The point to note is that $a$ is independent of the inertial reference frame. If we have:$$x' = x - vt$$then$$a' = \frac{d^2x'}{dt^2} = \frac{d^2x}{dt^2} = a$$This is the basis on which Newton's laws are valid in all inertial reference frames.

The book ends the page with what the author calls a "remark." The remark confuses me a little in its language and I'm trying to understand what it means as I feel it might be important later on.

The remark is the following:
"Note that the Galilean transformations in 1.1 aren't symmetric in x and t. This isn't necessarily a bad thing, but it turns out that it will in fact be a problem in special relativity, where space and time are treated more on an equal footing. We'll find in Section 2.1 that the Galilean transformations are replaced by the Lorentz transformations, and the latter are in fact symmetric in x and t (up to factors of the speed of light, c)"

I was wondering what it means for a transformation to not be symmetric in two variables (x and t).

Symmetry can mean different things in different contexts. In this case, it simply means that things are not the same when it comes to $x$ and $t$. If $t' = t$, then we would need to have $x' = x$ for things to be symmetric.

The latter part of the remark says that the Lorentz transformations are symmetric in x and t, and I wonder what that means. I think it's important to mention I do not currently have a solid grasp of Lorentz transformations, but I am developing some intuition regarding the matter, hopefully I will understand Lorentz transformations with a reasonable degree of depth at some point.

Lastly the remark ends saying "x and t (up to factors of the speed of light, c)" I'm not sure what this talk about "factors" means, perhaps the author is referring to the well known property that velocity of anything cannot be faster than the speed of light?

The factor of $c$ is what we use when we move from $t$ to $ct$.

Mostly I feel I understand the information in the book that I've read so far (I'm still very early in the book), but I feel that the questions I have regarding this remark as the author likes to call it are possibly important for understanding the rest of the book.

They are. It is important to really understand this material.

 

[Chenkel]

If t′=t, then we would need to have x′=x for things to be symmetric.

I'm trying to understanding why this would be the precondition for symmetry of the Galilean transformation.

Perhaps I'm misunderstanding the idea of symmetry. I've heard of a few of different symmetries in math and physics, but I am still a beginner when it comes to recognizing them.

I understand that it is said that the Galilean transformation is not symmetric in x and t, but I'm wondering what that means according to a rigourous definition of symmetry.

You say with any symmetry that context is important, I hope I illustrated the context of the remark by the author (David Morin) effectively, but I feel I am still trying to understand the statement in context myself so I cannot be sure if I have given all sufficient information for the question to be answered.

Hopefully as I study the posts by yourself (Perok) and Ibix I might come around to an understanding.

I'm definitely going to be taking a look at these posts now and later, thanks again for the insights!

 

[PeroK]

I'm trying to understanding why this would be the precondition for symmetry of the Galilean transformation.

Perhaps I'm misunderstanding the idea of symmetry. I've heard of a few of different symmetries in math and physics, but I am still a beginner when it comes to recognizing them.

I understand that it is said that the Galilean transformation is not symmetric in x and t, but I'm wondering what that means according to a rigourous definition of symmetry.

You may be over thinking this. I'm not sure Morin expects you to look for a rigorous definition of symmetry here. I suspect it's supposed to be obvious that in the Galilean transformation $x$ and $t$ are fundamentally different.

In the Lorentz transformation, we have the following:$$x' = f(x,ct), \ ct' = f(ct, x)$$and that's what we mean by symmetry in this case. Here, $f$ is the same function and formally:$$f(X,T) = \gamma(X - \frac v c T)$$where $X$ and $T$ are dummy variables.

Does that help? I would say that at this level that sort of formal definition of symmetry gets in the way and may not be much help. I think that you're supposed to see the symmetry.

PS but, if you don't see it you don't see it.

 

[haushofer]

I was wondering what it means for a transformation to not be symmetric in two variables (x and t).

Just interchange x (x') with ct (ct') in the Lorentz transfo's. You'll then just interchange the eqns for x' and t'. So we say these eqns treat space and time symmetrically.

Do the same for the Galilei transfo with x (x') and t (t'). You get different eqns because t' does not contain x.

 

[Ibix]

Perhaps it helps to say that the (1+1)d Lorentz transforms are symmetric under the interchange of spacelike and timelike coordinates. That is (as has already been said) if you interchange $x$ with $t$ and likewise their primed equivalents, you get back to where you started. I'm just stating explicitly what symmetry is meant.

 

[Chenkel]

In the Lorentz transformation, we have the following:x′=f(x,ct), ct′=f(ct,x)and that's what we mean by symmetry in this case. Here, f is the same function and formally:f(X,T)=γ(X−vcT)where X and T are dummy variables.

I think I might be starting to get it (hopefully).

I think what you are saying is you get a valid expression of a point in space time regardless of the order of the arguments (x and ct) that you pass to the function that performs the Lorentz transformation. I'm not sure of the full implications of this, but I am going to assume this idea is what the author was suggesting.

Conversely if we have a function that represents the Galilean transformation with arguments (x and t) you cannot interchange the arguments to the function and get a consistent point in Galilean space time.

For example (with the Galilean transformation) if you represent it as:$$f(X, T) = X + vT$$ then you cannot interchange X and T to go back and forth between time and position.

In otherwords with the Lorentz transformation you can just interchange the time and position as arguments to the function and you effectively swap the "outputs" between transformed time and transformed position.

Thanks again to everyone for helping in my understanding

 

[vanhees71]

Perhaps it helps to say that the (1+1)d Lorentz transforms are symmetric under the interchange of spacelike and timelike coordinates. That is (as has already been said) if you interchange $x$ with $t$ and likewise their primed equivalents, you get back to where you started. I'm just stating explicitly what symmetry is meant.

Sure, you can exchange $x$ and $t$ and switch from one sign convention of the metric to the other in (1+1) dimensions. Then you flip the "spacelike" and "timelike" vectors, but you are still left with a (1+1)d Minkowski space. In this sense to work too much with (1+1)d simplifications is dangerous, because then students may get the (in my opinion wrong) impression as if space and time were "completely symmetric" in SR, which is not the case. In Nature it's pretty clear that we have a (1+3)d (or equivalently (3+1)d) spacetime, and that thus there are clearly 3 spatial and 1 temporal dimension in the sense of the signature of the fundamental form (aka "metric") of Minkowski spacetime.

 

[Ibix]

In Nature it's pretty clear that we have a (1+3)d (or equivalently (3+1)d) spacetime, and that thus there are clearly 3 spatial and 1 temporal dimension in the sense of the signature of the fundamental form (aka "metric") of Minkowski spacetime.

Yes, there are limits to the symmetry. The Lorentz boost matrices remain symmetric under interchange of the time axis and whatever spatial axis is parallel to the frame velocities, but the metric isn't, as you say.

This does all point at time and space not being quite as separate as they are in Galilean relativity, but I wouldn't spend as much time thinking about Morin's wording as I have writing about it...

 

[vanhees71]

Another point is that for (1+d)D Minkowski space with $d \geq 2$ only the rotation free boost matrices are symmetric. The general Lorentz-transformation matrix is not symmetric.

 

[Chenkel]

The Galilean Transformation is:$$x' = x - vt, \ t' = t$$Note that the format of these two equations is very different.

The Lorentz Transformation is: $$x' = \gamma(x - vt), \ t' = \gamma(t - \frac v {c^2}x)$$And at the first sight the format of these equations is similar, but not the same. If, however, we take $ct$ to be the time parameter, we have:$$x' = \gamma(x - \frac v c (ct)), \ ct' = \gamma(ct - \frac v c x)$$And we see that these equations have precisely the same format.

This "symmetry" between time and space must be seen in context. The universe has three spatial dimensions, but only one time dimension. But, there is clearly a relationship between time and space i SR that is much deeper than is the case in classical, Newtonian/Galilean physics.

I believe I am understanding these equations now that I look at them again, thank you again for making the symmetries of space time in the Lorentz transformations clear.